Page 1 :
The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth, This eBook is for the use of anyone anywhere at no cost and with, almost no restrictions whatsoever. You may copy it, give it away or, re-use it under the terms of the Project Gutenberg License included, with this eBook or online at www.gutenberg.org, , Title: Plane Geometry, Author: George Albert Wentworth, Release Date: July 3, 2010 [EBook #33063], Language: English, Character set encoding: ISO-8859-1, *** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***
Page 2 :
Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy, and the Online Distributed Proofreading Team at, http://www.pgdp.net, , transcriber’s note—Minor typographical corrections and, presentational changes have been made without comment., , This PDF file is optimized for screen viewing, but may easily be, recompiled for printing. Please see the preamble of the LATEX source, file for instructions.
Page 3 :
PLANE GEOMETRY, BY, , G.A. WENTWORTH, Author of a Series of Text-Books in Mathematics, , REVISED EDITION, , GINN & COMPANY, BOSTON · NEW YORK · CHICAGO · LONDON
Page 4 :
Entered, according to Act of Congress, in the year 1888, by, G.A. WENTWORTH, in the Office Of the Librarian of Congress, at Washington, Copyright, 1899, By G.A. WENTWORTH, ALL RIGHTS RESERVED, 67 10, , The Athenæum Press, GINN & COMPANY · PROPRIETORS · BOSTON · U.S.A.
Page 5 :
iii, , PREFACE., Most persons do not possess, and do not easily acquire, the power of abstraction requisite for apprehending geometrical conceptions, and for keeping, in mind the successive steps of a continuous argument. Hence, with a very, large proportion of beginners in Geometry, it depends mainly upon the form in, which the subject is presented whether they pursue the study with indifference,, not to say aversion, or with increasing interest and pleasure., Great care, therefore, has been taken to make the pages attractive. The, figures have been carefully drawn and placed in the middle of the page, so that, they fall directly under the eye in immediate connection with the text; and, in no case is it necessary to turn the page in reading a demonstration. Full,, long-dashed, and short-dashed lines of the figures indicate given, resulting,, and auxiliary lines, respectively. Bold-faced, italic, and roman type has been, skilfully used to distinguish the hypothesis, the conclusion to be proved, and, the proof., As a further concession to the beginner, the reason for each statement in the, early proofs is printed in small italics, immediately following the statement., This prevents the necessity of interrupting the logical train of thought by, turning to a previous section, and compels the learner to become familiar with, a large number of geometrical truths by constantly seeing and repeating them., This help is gradually discarded, and the pupil is left to depend upon the, knowledge already acquired, or to find the reason for a step by turning to the, given reference., It must not be inferred, because this is not a geometry of interrogation, points, that the author has lost sight of the real object of the study. The, training to be obtained from carefully following the logical steps of a complete, proof has been provided for by the Propositions of the Geometry, and the, development of the power to grasp and prove new truths has been provided, for by original exercises. The chief value of any Geometry consists in the, happy combination of these two kinds of training. The exercises have been, arranged according to the test of experience, and are so abundant that it, is not expected that any one class will work them all out. The methods, of attacking and proving original theorems are fully explained in the first, Book, and illustrated by sufficient examples; and the methods of attacking and, solving original problems are explained in the second Book, and illustrated
Page 6 :
iv, , by examples worked out in full. None but the very simplest exercises are, inserted until the student has become familiar with geometrical methods, and, is furnished with elementary but much needed instruction in the art of handling, original propositions; and he is assisted by diagrams and hints as long as these, helps are necessary to develop his mental powers sufficiently to enable him to, carry on the work by himself., The law of converse theorems, the distinction between positive and negative, quantities, and the principles of reciprocity and continuity have been briefly, explained; but the application of these principles is left mainly to the discretion, of teachers., The author desires to express his appreciation of the valuable suggestions, and assistance which he has received from distinguished educators in all parts, of the country. He also desires to acknowledge his obligation to Mr. Charles, Hamilton, the Superintendent of the composition room of the Athenæum, Press, and to Mr. I. F. White, the compositor, for the excellent typography of, the book., Criticisms and corrections will be thankfully received., G. A. WENTWORTH., Exeter, N.H., June, 1899.
Page 7 :
v, , NOTE TO TEACHERS., It is intended to have the first sixteen pages of this book simply read in, the class, with such running comment and discussion as may be useful to help, the beginner catch the spirit of the subject-matter, and not leave him to the, mere letter of dry definitions. In like manner, the definitions at the beginning, of each Book should be read and discussed in the recitation room. There is a, decided advantage in having the definitions for each Book in a single group so, that they can be included in one survey and discussion., For a similar reason the theorems of limits are considered together. The, subject of limits is exceedingly interesting in itself, and it was thought best to, include in the theory of limits in the second Book every principle required for, Plane and Solid Geometry., When the pupil is reading each Book for the first time, it will be well to let, him write his proofs on the blackboard in his own language, care being taken, that his language be the simplest possible, that the arrangement of work be, vertical, and that the figures be accurately constructed., This method will furnish a valuable exercise as a language lesson, will, cultivate the habit of neat and orderly arrangement of work, and will allow a, brief interval for deliberating on each step., After a Book has been read in this way, the pupil should review the Book,, and should be required to draw the figures free-hand. He should state and, prove the propositions orally, using a pointer to indicate on the figure every, line and angle named. He should be encouraged in reviewing each Book, to, do the original exercises; to state the converse propositions, and determine, whether they are true or false; and also to give well-considered answers to, questions which may be asked him on many propositions., The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus, a rectangle with a constant base b, and, a variable altitude x, will afford an obvious illustration of the truth that the, product of a constant and a variable is also a variable; and that the limit of, the product of a constant and a variable is the product of the constant by the, limit of the variable. If x increases and approaches the altitude a as a limit,, the area of the rectangle increases and approaches the area of the rectangle ab, as a limit; if, however, x decreases and approaches zero as a limit, the area of, the rectangle decreases and approaches zero as a limit.
Page 8 :
vi, , An arithmetical illustration of this truth may be given by multiplying the, approximate values of any repetend by a constant. If, for example, we take, 3, ,, the repetend 0.3333 etc., the approximate values of the repetend will be 10, 33, 333, 3333, ,, ,, ,, etc.,, and, these, values, multiplied, by, 60, give, the, series, 18,, 19.8,, 100 1000 10000, 19.98, 19.998, etc., which evidently approaches 20 as a limit; but the product, of 60 into 31 (the limit of the repetend 0.333 etc.) is also 20., Again, if we multiply 60 into the different values of the decreasing series, 1, 1, ,, , 1 , 1 , etc., which approaches zero as a limit, we shall get the, 30 300 3000 30000, 1, 1, decreasing series 2, 51 , 50, , 500, , etc.; and this series evidently approaches zero, as a limit., The Teacher is likewise advised to give frequent written examinations., These should not be too difficult, and sufficient time should be allowed for, accurately constructing the figures, for choosing the best language, and for, determining the best arrangement., The time necessary for the reading of examination books will be diminished, by more than one half, if the use of symbols is allowed., Exeter, N.H., 1899.
Page 9 :
CONTENTS, , vii, , Contents, GEOMETRY., INTRODUCTION. . . . . . . . . . ., GENERAL TERMS. . . . . . . . . ., GENERAL AXIOMS. . . . . . . . ., SYMBOLS AND ABBREVIATIONS., , 1, ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , PLANE GEOMETRY., BOOK I. RECTILINEAR FIGURES., DEFINITIONS. . . . . . . . . . . . . . . . . ., THE STRAIGHT LINE. . . . . . . . . . . . ., THE PLANE ANGLE. . . . . . . . . . . . . ., PERPENDICULAR AND OBLIQUE LINES., PARALLEL LINES. . . . . . . . . . . . . . ., TRIANGLES. . . . . . . . . . . . . . . . . . ., LOCI OF POINTS. . . . . . . . . . . . . . . ., QUADRILATERALS. . . . . . . . . . . . . ., POLYGONS IN GENERAL. . . . . . . . . . ., SYMMETRY. . . . . . . . . . . . . . . . . . ., EXERCISES. . . . . . . . . . . . . . . . . . ., , 1, 3, 6, 6, , 7, ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , ., ., ., ., ., ., ., ., ., ., ., , 7, 7, 8, 10, 17, 26, 33, 48, 51, 61, 65, 72
Page 10 :
CONTENTS, , BOOK II. THE CIRCLE., DEFINITIONS. . . . . . . . . . . . . ., ARCS, CHORDS, AND TANGENTS., MEASUREMENT. . . . . . . . . . . ., THEORY OF LIMITS. . . . . . . . . ., MEASURE OF ANGLES. . . . . . . ., PROBLEMS OF CONSTRUCTION. ., EXERCISES. . . . . . . . . . . . . . ., , viii, , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , BOOK III. PROPORTION. SIMILAR POLYGONS., THEORY OF PROPORTION. . . . . . . . . . . . . ., SIMILAR POLYGONS. . . . . . . . . . . . . . . . . ., EXERCISES. . . . . . . . . . . . . . . . . . . . . . . ., NUMERICAL PROPERTIES OF FIGURES. . . . . ., EXERCISES. . . . . . . . . . . . . . . . . . . . . . . ., PROBLEMS OF CONSTRUCTION. . . . . . . . . . ., EXERCISES. . . . . . . . . . . . . . . . . . . . . . . ., BOOK IV. AREAS OF POLYGONS., COMPARISON OF POLYGONS. . . ., EXERCISES. . . . . . . . . . . . . . ., PROBLEMS OF CONSTRUCTION. ., EXERCISES. . . . . . . . . . . . . . ., , ., ., ., ., , ., ., ., ., , BOOK V. REGULAR POLYGONS AND, PROBLEMS OF CONSTRUCTION. . . ., MAXIMA AND MINIMA. . . . . . . . . ., EXERCISES. . . . . . . . . . . . . . . . ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., ., ., ., , ., ., ., ., , ., ., ., ., ., ., ., , 89, 89, 91, 109, 111, 119, 135, 158, , ., ., ., ., ., ., ., , 168, 168, 183, 195, 197, 207, 210, 216, , ., ., ., ., , 226, 235, 239, 242, 252, , CIRCLES., 258, . . . . . . . . . . . 274, . . . . . . . . . . . 282, . . . . . . . . . . . 289, , TABLE OF FORMULAS., , 302, , INDEX., , 305
Page 11 :
GEOMETRY., INTRODUCTION., 1. If a block of wood or stone is cut in the shape represented in Fig. 1, it, will have six flat faces., Each face of the block is called a surface; and if the faces are made smooth, by polishing, so that, when a straight edge is applied to any one of them, the, straight edge in every part will touch the surface, the faces are called plane, surfaces, or planes., , Fig. 1., , 2. The intersection of any two of these surfaces is called a line., 3. The intersection of any three of these lines is called a point., 4. The block extends in three principal directions:, From left to right, A to B., From front to back, A to C., From top to bottom, A to D., These are called the dimensions of the block, and are named in the order, given, length, breadth (or width), and thickness (height or depth).
Page 12 :
GEOMETRY., , 2, , 5. A solid, in common language, is a limited portion of space filled with, matter ; but in Geometry we have nothing to do with the matter of which a, body is composed; we study simply its shape and size; that is, we regard a, solid as a limited portion of space which may be occupied by a physical body,, or marked out in some other way. Hence,, A geometrical solid is a limited portion of space., 6. The surface of a solid is simply the boundary of the solid, that which, separates it from surrounding space. The surface is no part of a solid and has, no thickness. Hence,, A surface has only two dimensions, length and breadth., 7. A line is simply a boundary of a surface, or the intersection of two surfaces. Since the surfaces have no thickness, a line has no thickness. Moreover,, a line is no part of a surface and has no width. Hence,, A line has only one dimension, length., 8. A point is simply the extremity of a line, or the intersection of two lines., A point, therefore, has no thickness, width, or length; therefore, no magnitude., Hence,, A point has no dimension, but denotes position simply., 9. It must be distinctly understood at the outset that the points, lines,, surfaces, and solids of Geometry are purely ideal, though they are represented, to the eye in a material way. Lines, for example, drawn on paper or on the, blackboard, will have some width and some thickness, and will so far fail of, being true lines; yet, when they are used to help the mind in reasoning, it is, assumed that they represent true lines, without breadth and without thickness., C, , D, , A, B, , Fig. 2., , F, , 10. A point is represented to the eye by a fine dot, and named by a letter,, as A (Fig. 2). A line is named by two letters, placed one at each end, as BF ., A surface is represented and named by the lines which bound it, as BCDF ., A solid is represented by the faces which bound it.
Page 13 :
GENERAL TERMS., , 3, , 11. A point in space may be considered by itself, without reference to a, line., 12. If a point moves in space, its path is a line. This line may be considered, apart from the idea of a surface., 13. If a line moves in space, it generates, in general, a surface. A surface, can then be considered apart from the idea of a solid., 14. If a surface moves in space, it generates, in general, a solid., H, , D, , E, , A, C, , G, F, , B, , Fig. 3., , Thus, let the upright surface ABCD (Fig. 3) move to the right to the, position EF GH, the points A, B, C, and D generating the lines AE, BF ,, CG, and DH, respectively. The lines AB, BC, CD, and DA will generate, the surfaces AF . BG, CH, and DE, respectively. The surface ABCD will, generate the solid AG., 15. Geometry is the science which treats of position, form, and magnitude., 16. A geometrical figure is a combination of points, lines, surfaces, or, solids., 17. Plane Geometry treats of figures all points of which are in the same, plane., Solid Geometry treats of figures all points of which are not in the same, plane., GENERAL TERMS., 18. A proof is a course of reasoning by which the truth or falsity of any, statement is logically established.
Page 14 :
GEOMETRY., , 4, , 19. An axiom is a statement admitted to be true without proof., 20. A theorem is a statement to be proved., 21. A construction is the representation of a required figure by means of, points and lines., 22. A postulate is a construction admitted to be possible., 23. A problem is a construction to be made so that it shall satisfy certain, given conditions., 24. A proposition is an axiom, a theorem, a postulate, or a problem., 25. A corollary is a truth that is easily deduced from known truths., 26. A scholium is a remark upon some particular feature of a proposition., 27. The solution of a problem consists of four parts:, 1. The analysis, or course of thought by which the construction of the, required figure is discovered., 2. The construction of the figure with the aid of ruler and compasses., 3. The proof that the figure satisfies all the conditions., 4. The discussion of the limitations, if any, within which the solution is, possible., 28. A theorem consists of two parts: the hypothesis, or that which is, assumed; and the conclusion, or that which is asserted to follow from the, hypothesis., 29. The contradictory of a theorem is a theorem which must be true, if the given theorem is false, and must be false if the given theorem is true., Thus,, A theorem:, Its contradictory:, , If A is B, then C is D., If A is B, then C is not D.
Page 15 :
GENERAL TERMS., , 5, , 30. The opposite of a theorem is obtained by making both the hypothesis, and the conclusion negative. Thus,, A theorem:, Its opposite:, , If A is B, then C is D., If A is not B, then C is not D., , 31. The converse of a theorem is obtained by interchanging the hypothesis, and conclusion. Thus,, A theorem:, Its converse:, , If A is B, then C is D., If C is D, then A is B., , 32. The converse of a truth is not necessarily true., Thus, Every horse is a quadruped is true, but the converse, Every quadruped is a horse, is not true., 33. If a direct proposition and its opposite are true, the converse proposition, is true; and if a direct proposition and its converse are true, the opposite, proposition is true., Thus, if it were true that, 1. If an animal is a horse, the animal is a quadruped;, 2. If an animal is not a horse, the animal is not a quadruped;, it would follow that, 3. If an animal is a quadruped, the animal is a horse., Moreover, if 1 and 3 were true, then 2 would be true.
Page 16 :
GEOMETRY., , 34., , 6, , GENERAL AXIOMS., , 1. Magnitudes which are equal to the same magnitude, or equal magnitudes, are equal to each other., 2. If equals are added to equals, the sums are equal., 3. If equals are taken from equals, the remainders are equal., 4. If equals are added to unequals, the sums are unequal in the same order;, if unequals are added to unequals in the same order, the sums are unequal in, that order., 5. If equals are taken from unequals, the remainders are unequal in the, same order; if unequals are taken from equals, the remainders are unequal in, the reverse order., 6. The doubles of the same magnitude, or of equal magnitudes are equal;, and the doubles of unequals are unequal., 7. The halves of the same magnitude, or of equal magnitudes are equal;, and the halves of unequals are unequal., 8. The whole is greater than any of its parts., 9. The whole is equal to the sum of all its parts., 35., , SYMBOLS AND ABBREVIATIONS., , is (or are) greater than., Def. . . . definition., is (or are) less than., Ax. . . . axiom., is (or are) equivalent to., Hyp. . . . hypothesis., therefore., Cor. . . . corollary., perpendicular., Scho. . . . scholium., perpendiculars., Ex. . . . exercise., parallel., ks parallels., Adj. . . . adjacent., angle., ∠s angles., Iden. . . . identical., triangle., 4s triangles., Const.. . . construction., / / parallelogram., Sup. . . . supplementary., / /s, parallelograms., Ext. . . . exterior., circle., Int. . . . interior., s circles., rt. right., st. straight., Alt. . . . alternate., q.e.d. stands for quod erat demonstrandum, which was to be proved., >, <, m, ∴, ⊥, ⊥s, k, ∠, 4, , q.e.f. stands for quod erat faciendum, which was to be done., The signs +, −, ×, ÷, =, have the same meaning as in Algebra.
Page 17 :
PLANE GEOMETRY., BOOK I. RECTILINEAR FIGURES., , DEFINITIONS., A, , B, , C, , D, , E, , F, , Fig. 4., , 36. A straight line is a line such that any part of it, however placed on, any other part, will lie wholly in that part if its extremities lie in that part,, as AB., 37. A curved line is a line no part of which is straight, as CD., 38. A broken line is made up of different straight lines, as EF ., Note. A straight line is often called simply a line., , 39. A plane surface, or a plane, is a surface in which, if any two points, are taken, the straight line joining these points lies wholly in the surface., 40. A curved surface is a surface no part of which is plane., 41. A plane figure is a figure all points of which are in the same plane., 42. Plane figures which are bounded by straight lines are called rectilinear, figures; by curved lines, curvilinear figures., 43. Figures that have the same shape are called similar. Figures that have, the same size but not the same shape are called equivalent. Figures that have, the same shape and the same size are called equal or congruent.
Page 18 :
BOOK I. PLANE GEOMETRY., , 8, , THE STRAIGHT LINE., 44. Postulate. A straight line can be drawn from one point to another., 45. Postulate. A straight line can be produced indefinitely., 46. Axiom.∗ Only one straight line can be drawn from one point to another. Hence, two points determine a straight line., 47. Cor. 1. Two straight lines which have two points in common coincide, and form but one line., 48. Cor. 2. Two straight lines can intersect in only one point., For if they had two points common, they would coincide and not intersect., Hence, two intersecting lines determine a point., 49. Axiom. A straight line is the shortest line that can be drawn from one, point to another., 50. Def. The distance between two points is the length of the straight, line that joins them., 51. A straight line determined by two points may be considered as prolonged indefinitely., 52. If only the part of the line between two fixed points is considered, this, part is called a segment of the line., 53. For brevity, we say “the line AB,” to designate a segment of a line, limited by the points A and B., 54. If a line is considered as extending from a fixed point, this point is, called the origin of the line., , The general axioms on page 6 apply to all magnitudes. Special geometrical axioms will, be given when required., ∗
Page 19 :
THE STRAIGHT LINE., A, , 9, B, , C, , Fig. 5., , 55. If any point, C, is taken in a given straight line, AB, the two parts CA, and CB are said to have opposite directions from the point C (Fig. 5)., Every straight line, as AB, may be considered as extending in either of two, opposite directions, namely, from A towards B, which is expressed by AB, and, read segment AB; and from B towards A, which is expressed by BA, and read, segment BA., 56. If the magnitude of a given line is changed, it becomes longer or shorter., Thus (Fig. 5), by prolonging AC to B we add CB to AC, and AB = AC +, CB. By diminishing AB to C, we subtract CB from AB, and AC = AB−CB., If a given line increases so that it is prolonged by its own magnitude several, times in succession, the line is multiplied, and the resulting line is called a, multiple of the given line., A, , B, , C, , D, , E, , Fig. 6., , Thus (Fig. 6), if AB = BC = CD = DE, then AC = 2AB, AD = 3AB,, and AE = 4AB. Hence,, Lines of given length may be added and subtracted; they may also be multiplied by a number.
Page 20 :
BOOK I. PLANE GEOMETRY., , 10, , THE PLANE ANGLE., F, , E, , D, , Fig. 7., , 57. The opening between two straight lines drawn from the same point is, called a plane angle. The two lines, ED and EF , are called the sides, and, E, the point of meeting, is called the vertex of the angle., The size of an angle depends upon the extent of opening of its sides, and, not upon the length of its sides., 58. If there is but one angle at a given vertex, the angle is designated by, a capital letter placed at the vertex, and is read by simply naming the letter., C, , E, , D, , A, , B, , Fig. 8., , F, , A, , d, , c b, , a, , Fig. 9., , B, , If two or more angles have the same vertex, each angle is designated by, three letters, and is read by naming the three letters, the one at the vertex, between the others. Thus, DAC (Fig. 8) is the angle formed by the sides AD, and AC., An angle is often designated by placing a small italic letter between the, sides and near the vertex, as in Fig. 9., 59. Postulate of Superposition. Any figure may be moved from one, place to another without altering its size or shape., 60. The test of equality of two geometrical magnitudes is that they may, be made to coincide throughout their whole extent. Thus,, Two straight lines are equal, if they can be placed one upon the other so, that the points at their extremities coincide., Two angles are equal, if they can be placed one upon the other so that, their vertices coincide and their sides coincide, each with each.
Page 21 :
THE PLANE ANGLE., , 11, , 61. A line or plane that divides a geometric magnitude into two equal parts, is called the bisector of the magnitude., If the angles BAD and CAD (Fig. 8) are equal, AD bisects the angle BAC., 62. Two angles are called adjacent angles when they have the same vertex, and a common side between them; as the angles BOD and AOD (Fig. 10)., D, , D, , A, , O, , Fig. 10., , A, , B, , C, , Fig. 11., , B, , 63. When one straight line meets another straight line and makes the, adjacent angles equal, each of these angles is called a right angle; as angles, DCA and DCB (Fig. 11)., 64. A perpendicular to a straight line is a straight line that makes a right, angle with it., Thus, if the angle DCA (Fig. 11) is a right angle, DC is perpendicular to, AB, and AB is perpendicular to DC., 65. The point (as C, Fig. 11) where a perpendicular meets another line is, called the foot of the perpendicular., 66. When the sides of an angle extend in opposite directions, so as to be, in the same straight line, the angle is called a straight angle., A, , C, , B, , Fig. 12., , Thus, the angle formed at C (Fig. 12) with its sides CA and CB extending, in opposite directions from C is a straight angle., 67. Cor. A right angle is half a straight angle.
Page 22 :
BOOK I. PLANE GEOMETRY., , 12, , D, A, , O, , A, , Fig. 13., , Fig. 14., , 68. An angle less than a right angle is called an acute angle; as, angle A, (Fig. 13)., 69. An angle greater than a right angle and less than a straight angle is, called an obtuse angle; as, angle AOD (Fig. 14)., 70. An angle greater than a straight angle and less than two straight angles, is called a reflex angle; as, angle DOA, indicated by the dotted line (Fig. 14)., 71. Angles that are neither right nor straight angles are called oblique, angles; and intersecting lines that are not perpendicular to each other are, called oblique lines.
Page 23 :
THE PLANE ANGLE., , 13, , EXTENSION OF THE MEANING OF ANGLES., D, , B, C, , A′, , O, , A, , B′, , Fig. 15., , 72. Suppose the straight line OC (Fig. 15) to move in the plane of the paper, from coincidence with OA, about the point O as a pivot, to the position OC;, then the line OC describes or generates the angle AOC, and the magnitude, of the angle AOC depends upon the amount of rotation of the line from the, position OA to the position OC., If the rotating line moves from the position OA to the position OB, perpendicular to OA, it generates the right angle AOB; if it moves to the position, OD, it generates the obtuse angle AOD; if it moves to the position OA0 , it, generates the straight angle AOA0 ; if it moves to the position OB 0 it generates, the reflex angle AOB 0 , indicated by the dotted line; and if it moves to the, position OA again, it generates two straight angles. Hence,, 73. The angular magnitude about a point in a plane is equal to two straight, angles, or four right angles; and the angular magnitude about a point on one, side of a straight line drawn through the point is equal to a straight angle, or, two right angles., 74. The whole angular magnitude about a point in a plane is called a, perigon; and two angles whose sum is a perigon are called conjugate angles., Note. This extension of the meaning of angles is necessary in the applications, of Geometry, as in Trigonometry, Mechanics, etc.
Page 24 :
BOOK I. PLANE GEOMETRY., C, , a, , Fig. 16., , D, , D, , c, b, d, O, , 14, , B, , Fig. 17., , A, , O, , Fig. 18., , B, , 75. When two angles have the same vertex, and the sides of the one are, prolongations of the sides of the other, they are called vertical angles; as,, angles a and b, c and d (Fig. 16)., 76. Two angles are called complementary when their sum is equal to a, right angle; and each is called the complement of the other; as, angles DOB, and DOC (Fig. 17)., 77. Two angles are called supplementary when their sum is equal to a, straight angle; and each is called the supplement of the other; as, angles DOB, and DOA (Fig. 18)., UNIT OF ANGLES., 78. By adopting a suitable unit for measuring angles we are able to express, the magnitudes of angles by numbers., If we suppose OC (Fig. 15) to turn about O from coincidence with OA, until it makes one three hundred sixtieth of a revolution, it generates an angle, at O, which is taken as the unit for measuring angles. This unit is called a, degree., The degree is subdivided into sixty equal parts, called minutes, and the, minute into sixty equal parts, called seconds., Degrees, minutes, and seconds are denoted by symbols. Thus, 5 degrees, 13 minutes 12 seconds is written 5◦ 130 1200 ., A right angle is generated when OC has made one fourth of a revolution, and contains 90◦ ; a straight angle, when OC has made half of a revolution, and contains 180◦ ; and a perigon, when OC has made a complete revolution, and contains 360◦ .
Page 25 :
THE PLANE ANGLE., , 15, , Note. The natural angular unit is one complete revolution. But this unit would, require us to express the values of most angles by fractions. The advantage of using, the degree as the unit consists in its convenient size, and in the fact that 360 is, divisible by so many different integral numbers., H, , C, G, , F, , E, , D, , B, , A, , Fig. 19., , 79. By the method of superposition we are able to compare magnitudes of, the same kind. Suppose we have two angles, ABC and DEF (Fig. 19). Let, the side ED be placed on the side BA, so that the vertex E shall fall on B;, then, if the side EF falls on BC, the angle DEF equals the angle ABC; if, the side EF falls between BC and BA in the position shown by the dotted, line BG, the angle DEF is less than the angle ABC; but if the side EF falls, in the position shown by the dotted line BH, the angle DEF is greater than, the angle ABC.
Page 26 :
BOOK I. PLANE GEOMETRY., C, , F, , 16, , F, D, , H, , P, C, M, , E, , D, , Fig. 20., , B, , A, , B, , Fig. 21., , A, , 80. If we have the angles ABC and DEF (Fig. 20), and place the vertex, E on B and the side ED on BC, so that the angle DEF takes the position, CBH, the angles DEF and ABC will together be equal to the angle ABH., If the vertex E is placed on B, and the side ED on BA, so that the angle, DEF takes the position ABF , the angle F BC will be the difference between, the angles ABC and DEF ., If an angle is increased by its own magnitude two or more times in succession, the angle is multiplied by a number., Thus, if the angles ABM , M BC, CBP , P BD (Fig. 21) are all equal, the, angle ABD is 4 times the angle ABM . Therefore,, Angles may be added and subtracted; they may also be multiplied by a number.
Page 27 :
PERPENDICULAR AND OBLIQUE LINES., , 17, , PERPENDICULAR AND OBLIQUE LINES., Proposition I. Theorem., 81. All straight angles are equal., A, , C, E, , D, , B, F, , Let the angles ACB and DEF be any two straight angles., To prove that ∠ACB = ∠DEF ., Proof. Place the ∠ACB on the ∠DEF , so that the vertex C shall fall on the, vertex E, and the side CB on the side EF ., Then CA will fall on ED,, (because ACB and DEF are straight lines)., ∴ ∠ACB = ∠DEF ., , 82. Cor. 1. All right angles are equal., , § 47, § 60, q.e.d., , Ax. 7, , 83. Cor. 2. At a given point in a given line there can be but one perpendicular to the line., , For, if there could be two ⊥s , we should have rt. ∠s of different magnitudes;, but this is impossible, § 82., 84. Cor. 3. The complements of the same angle or of equal angles are, equal., Ax. 3, 85. Cor. 4. The supplements of the same angle or of equal angles are, equal., Ax. 3
Page 28 :
BOOK I. PLANE GEOMETRY., , 18, , Note. The beginner must not forget that in Plane Geometry all the points of a, figure are in the same plane. Without this restriction in Cor. 2, an indefinite number, of perpendiculars can be erected at a given point in a given line., , Proposition II. Theorem., 86. If two adjacent angles have their exterior sides in a straight line, these, angles are supplementary., D, , A, , O, , B, , Let the exterior sides OA and OB of the adjacent angles AOD and, BOD be in the straight line AB., To prove that ∠s AOD and BOD are supplementary., Proof., AOB is a straight line., But, , ∴ ∠AOB is a st. ∠., , Hyp., § 66, , ∠AOD + ∠BOD = the st. ∠AOB., , Ax. 9, , ∴ the ∠s AOD and BOD are supplementary., , § 77, q.e.d., , 87. Def. Adjacent angles that are supplements of each other are called, supplementary-adjacent angles., Since the angular magnitude about a point is neither increased nor diminished by the number of lines which radiate from the point, it follows that,, 88. Cor. 1. The sum of all the angles about a point in a plane is equal to, a perigon, or two straight angles., 89. Cor. 2. The sum of all the angles about a point in a plane, on the, same side of a straight line passing through the point, is equal to a straight, angle, or two right angles.
Page 29 :
PERPENDICULAR AND OBLIQUE LINES., , 19, , Proposition III. Theorem., 90. Conversely: If two adjacent angles are supplementary, their exterior, sides are in the same straight line., O, , F, A, , C, , B, , Let the adjacent angles OCA and OCB be supplementary., To prove that AC and CB are in the same straight line., Proof. Suppose CF to be in the same line with AC., Then, ∠s OCA and OCF are supplementary,, § 86, (if two adjacent angles have their exterior sides in a straight line, these angles, are supplementary)., But, ∠s OCA and OCB are supplementary., Hyp., ∴ ∠s OCF and OCB have the same supplement., ∴ ∠OCF = ∠OCB., , § 85, , ∴ CB and CF coincide., , § 60, , ∴ AC and CB are in the same straight line., q.e.d., Since Propositions II. and III. are true, their opposites are true. Hence,, § 33, , 91. Cor. 1. If the exterior sides of two adjacent angles are not in a straight, line, these angles are not supplementary., 92. Cor. 2. If two adjacent angles are not supplementary, their exterior, sides are not in the same straight line.
Page 30 :
BOOK I. PLANE GEOMETRY., , 20, , Proposition IV. Theorem., 93. If one straight line intersects another straight line, the vertical angles, are equal., O, , B, , C, A, , P, , Let the lines OP and AB intersect at C., To prove that, ∠OCB = ∠ACP ., Proof., ∠OCA and ∠OCB are supplementary., , § 86, , ∠OCA and ∠ACP are supplementary,, § 86, (if two adjacent angles have their exterior sides in a straight line, these angles, are supplementary)., ∴ ∠s OCB and ACP have the same supplement., In like manner,, , ∴ ∠OCB = ∠ACP ., ∠ACO = ∠P CB., , § 85, q.e.d., , 94. Cor. If one of the four angles formed by the intersection of two straight, lines is a right angle, the other three angles are right angles., Ex. 1. Find the complement and the supplement of an angle of 49◦ ., Ex. 2. Find the number of degrees in an angle if it is double its complement; if it is one fourth of its complement., Ex. 3. Find the number of degrees in an angle if it is double its supplement;, if it is one third of its supplement.
Page 31 :
PERPENDICULAR AND OBLIQUE LINES., , 21, , Proposition V. Theorem, 95. Two straight lines drawn from a point in a perpendicular to a given line,, cutting off on the given line equal segments from the foot of the perpendicular,, are equal and make equal angles with the perpendicular., C, , A, , E, , F, , K, , B, , Let CF be a perpendicular to the line AB, and CE and CK two, straight lines cutting off on AB equal segments F E and F K from F ., To prove that CE = CK; and ∠F CE = ∠F CK., Proof. Fold over CF A, on CF as an axis, until it falls on the plane at the right, of CF ., F A will fall along F B,, (since ∠CF A = ∠CF B, each being a rt. ∠, by hyp.)., Point E will fall on point K,, (since F E = F K, by hyp.)., ∴ CE = CK,, (their extremities being the same points);, , § 60, , and ∠F CE = ∠F CK,, § 60, (since their vertices coincide, and their sides coincide, each with each)., q.e.d., , Ex. 4. Find the number of degrees in the angle included by the hands of, a clock at 1 o’clock. 3 o’clock. 4 o’clock. 6 o’clock.
Page 32 :
BOOK I. PLANE GEOMETRY., , 22, , Proposition VI. Theorem., 96. Only one perpendicular can be drawn to a given line from a given, external point., P, , A, , D, , C, , B, , P′, , Let AB be the given line, P the given external point, P C a perpendicular to AB from P , and P D any other line from P to AB., To prove that, P D is not ⊥ to AB., Proof. Produce P C to P 0 , making CP 0 equal to P C., Draw DP 0 ., By construction, P CP 0 is a straight line., ∴ P DP 0 is not a straight line,, (only one straight line can be drawn from one point to another )., , § 46, , Hence, ∠P DP 0 is not a straight angle., Since P C is ⊥ to DC, and P C = CP 0 ,, AC is ⊥ to P P 0 at its middle point., ∴ ∠P DC = ∠P 0 DC,, § 95, (two straight lines from a point in a ⊥ to a line, cutting off on the line equal, segments from the foot of the ⊥, make equal ∠s with the ⊥), Since ∠P DP 0 is not a straight angle,, ∠P DC, the half of ∠P DP 0 , is not a right angle., ∴ P D is not ⊥ to AB., , q.e.d.
Page 33 :
PERPENDICULAR AND OBLIQUE LINES., , 23, , Proposition VII. Theorem., 97. The perpendicular is the shortest line that can be drawn to a straight, line from an external point., P, , A, , D, , C, , B, , P′, , Let AB be the given straight line, P the given point, P C the perpendicular, and P D any other line drawn from P to AB., To prove that, P C < P D., Proof. Produce P C to P 0 , making CP 0 = P C., Then, , Draw DP 0 ., , P D = DP 0 ,, § 95, (two straight lines drawn from a point in a ⊥ to a line, cutting off on the line, equal segments from the foot of the ⊥, are equal )., and, But, , ∴ P D + DP 0 = 2P D,, P C + CP 0 = 2P C., P C + CP 0 < P D + DP 0 ., , Const., § 49, , ∴ 2P C < 2P D., ∴ P C < P D., , Ax. 7, q.e.d., , 98. Cor. The shortest line that can be drawn from a point to a given line, is perpendicular to the given line.
Page 34 :
BOOK I. PLANE GEOMETRY., , 24, , 99. Def. The distance of a point from a line is the length of the perpendicular from the point to the line., Proposition VIII. Theorem., 100. The sum of two lines drawn from a point to the extremities of a, straight line is greater than the sum of two other lines similarly drawn, but, included by them., C, , E, O, , A, , B, , Let CA and CB be two lines drawn from the point C to the extremities of the straight line AB. Let OA and OB be two lines similarly, drawn, but included by CA and CB., To prove that, CA + CB > OA + OB., Proof. Produce AO to meet the line CB at E., Then, CA + CE > OA + OE,, and, BE + OE > OB,, § 49, (a straight line is the shortest line from one point to another )., Add these inequalities, and we have, CA + CE + BE + OE > OA + OE + OB., , Ax. 4, , Substitute for CE + BE its equal CB, then, CA + CB + OE > OA + OE + OB., Take away OE from each side of the inequality., CA + CB > OA + OB., , Ax. 5, q.e.d.
Page 35 :
PERPENDICULAR AND OBLIQUE LINES., , 25, , Proposition IX. Theorem., 101. Of two straight lines drawn from the same point in a perpendicular, to a given line, cutting off on the line unequal segments from the foot of the, perpendicular, the more remote is the greater., O, , A, , E, , F, , C, , G, , B, , D, , Let OC be perpendicular to AB, OG and OE two straight lines to, AB, and CE greater than CG., To prove that, OE > OG., Proof. Take CF equal to CG, and draw OF ., Then, OF = OG,, § 95, (two straight lines drawn from a point in a ⊥ to a line, cutting off on the line, equal segments from the foot of the ⊥, are equal )., Produce OC to D, making CD = OC., Then, But, , Draw ED and F D., OE = ED, and OF = F D., , § 95, , OE + ED > OF + F D,, , § 100, , ∴ 2OE > 2OF , OE > OF , and OE > OG., , q.e.d., , 102. Cor. Only two equal straight lines can be drawn from a point to a, straight line; and of two unequal lines, the greater cuts off on the line the, greater segment from the foot of the perpendicular.
Page 36 :
BOOK I. PLANE GEOMETRY., , 26, , PARALLEL LINES., 103. Def. Two parallel lines are lines that lie in the same plane and, cannot meet however far they are produced., Proposition X. Theorem., 104. Two straight lines in the same plane perpendicular to the same straight, line are parallel., A, , B, , C, , D, , Let AB and CD be perpendicular to AC., To prove that AB and CD are parallel., Proof. If AB and CD are not parallel, they will meet if sufficiently prolonged;, and we shall have two perpendicular lines from their point of meeting to the same, straight line; but this is impossible,, § 96, (only one perpendicular can be drawn to a given line from a given external, point)., ∴ AB and CD are parallel., , q.e.d., , 105. Axiom. Through a given point only one straight line can be drawn, parallel to a given straight line., 106. Cor. Two straight lines in the same plane parallel to a third straight, line are parallel to each other., For if they could meet, we should have two straight lines from the point of, meeting parallel to a straight line; but this is impossible., § 105
Page 37 :
PARALLEL LINES., , 27, , Proposition XI. Theorem., 107. If a straight line is perpendicular to one of two parallel lines, it is, perpendicular to the other also., H, , A, , O, , B, , C, , F, , M, E, , K, , N, , Let AB and EF be two parallel lines, and let HK be perpendicular, to AB, and cut EF at C., To prove that, HK is ⊥ to EF ., Proof. Suppose M N drawn through C ⊥ to HK., Then, M N is k to AB., § 104, But, EF is k to AB., Hyp., But, , that is,, , ∴ EF coincides with M N ., , § 105, , M N is ⊥ to HK., , Const., , ∴ EF is ⊥ to HK,, HK is ⊥ to EF ., , q.e.d., , 108. Def. A straight line that cuts two or more straight lines is called a, transversal of those lines.
Page 38 :
BOOK I. PLANE GEOMETRY., , 28, , E, , b c, ad, , A, , f g, eh, , C, , B, , D, , F, , 109. If the transversal EF cuts AB and CD, the angles a, d, g, f are, called interior angles; b, c, h, e are called exterior angles., The angles d and f , and a and g, are called alternate-interior angles; the, angles b and h, and c and e, are called alternate-exterior angles., The angles b and f , c and g, e and a, h and d, are called exterior-interior, angles.
Page 39 :
PARALLEL LINES., , 29, , Proposition XII. Theorem., 110. If two parallel lines are cut by a transversal, the alternate-interior, angles are equal., A, , E, , B, , F, , O, , G, , C, , D, , H, , Let EF and GH be two parallel lines cut by the transversal BC., To prove that, ∠EBC = ∠BCH., Proof. Through O, the middle point of BC, suppose AD drawn ⊥ to GH., Then, AD is likewise ⊥ to EF ,, § 107, (a straight line ⊥ to one of two ks is ⊥ to the other ),, that is,, CD and BA are both ⊥ to AD., Apply the figure COD to the figure BOA, so that OD shall fall along OA., Then, OC will fall along OB,, § 93, (since ∠COD = ∠BOA, being vertical ∠s );, and, C will fall on B,, (since OC = OB, by construction)., Then, the ⊥ CD will fall along the ⊥ BA,, § 96, (only one ⊥ can be drawn to a given line from a given external point)., ∴ ∠OCD coincides with ∠OBA, and is equal to it,, § 60, (two angles are equal, if their vertices coincide and their sides coincide, each, with each)., q.e.d.
Page 40 :
BOOK I. PLANE GEOMETRY., , 30, , Proposition XIII. Theorem., 111. Conversely: When two straight lines in the same plane are cut by, a transversal, if the alternate-interior angles are equal, the two straight lines, are parallel., E, , M, H, , A, , B, N, , C, , K, , D, , F, , Let EF cut the straight lines AB and CD in the points H and K,, and let the angles AHK and HKD be equal., To prove that, AB is k to CD., Proof. Suppose M N drawn through H k to CD., Then, ∠M HK = ∠HKD,, § 110, (being alt.-int. ∠s of k lines)., But, ∠AHK = ∠HKD., Hyp., ∴ ∠M HK = ∠AHK., But, , ∴ M N and AB coincide., , Ax. 1, § 60, , M N is k to CD., , Const., , ∴ AB, which coincides with M N , is k to CD., , q.e.d., , Ex. 5. Find the complement and the supplement of an angle that contains, 37◦ 530 4900 ., Ex. 6. If the complement of an angle is one third of its supplement, how, many degrees does the angle contain?
Page 41 :
PARALLEL LINES., , 31, , Proposition XIV. Theorem., 112. If two parallel lines are cut by a transversal, the exterior-interior, angles are equal., E, , H, , A, , C, , K, , B, , D, , F, , Let AB and CD be two parallel lines cut by the transversal EF , in, the points H and K., To prove that, ∠EHB = ∠HKD., Proof., ∠EHB = ∠AHK,, § 93, (being vertical ∠s )., ∠AHK = ∠HKD,, (being alt.-int. ∠s of k lines)., In like manner, , § 110, , ∴ ∠EHB = ∠HKD., , Ax. 1, , ∠EHA = ∠HKC., , q.e.d., , 113. Cor. The alternate-exterior angles EHB and CKF , and also AHE, and DKF , are equal.
Page 42 :
BOOK I. PLANE GEOMETRY., , 32, , Proposition XV. Theorem., 114. Conversely: When two straight lines in a plane are cut by a, transversal, if the exterior-interior angles are equal, these two straight lines, are parallel., (Proof similar to that in § 111.), Proposition XVI. Theorem., 115. If two parallel lines are cut by a transversal, the two interior angles, on the same side of the transversal are supplementary., E, , H, , A, , C, , K, , B, , D, , F, , Let AB and CD be two parallel lines cut by the transversal EF in, the points H and K., To prove that ∠s BHK and HKD are supplementary., Proof., ∠EHB + ∠BHK = a st. ∠,, (being sup.-adj. ∠s )., But, ∠EHB = ∠HKD,, (being ext.-int. ∠s of k lines)., , § 86, , § 112, , ∴ ∠BHK + ∠HKD = a st. ∠., ∴ ∠s BHK and HKD are supplementary., , § 77, q.e.d.
Page 43 :
TRIANGLES., , 33, , Proposition XVII. Theorem., 116. Conversely: When two straight lines in a plane are cut by a, transversal, if two interior angles on the same side of the transversal are supplementary, the two straight lines are parallel., (Proof similar to that in § 111.), TRIANGLES., 117. A triangle is a portion of a plane bounded by three straight lines;, as, ABC (Fig. 1)., A, , B, , D, , Fig. 1., , A, , C, , B, , Fig. 2., , C, , D, , The bounding lines are called the sides of the triangle, and their sum is, called its perimeter; the angles included by the sides are called the angles, of the triangle, and the vertices of these angles, the vertices of the triangle., 118. Adjacent angles of a rectilinear figure are two angles that have one, side of the figure common; as, angles A and B (Fig. 2)., 119. An exterior angle of a triangle is an angle included by one side and, another side produced; as, ACD (Fig. 2). The interior angle ACB is adjacent to the exterior angle; the interior angles, A and B, are called opposite, interior angles.
Page 44 :
BOOK I. PLANE GEOMETRY., , Scalene., , Isosceles., , 34, , Equilateral., , 120. A triangle is called a scalene triangle when no two of its sides are, equal; an isosceles triangle, when two of its sides are equal; an equilateral, triangle, when its three sides are equal., , Right., , Obtuse., , Acute., , Equiangular., , 121. A triangle is called a right triangle, when one of its angles is a, right angle; an obtuse triangle, when one of its angles is an obtuse angle; an, acute triangle, when all three of its angles are acute angles; an equiangular, triangle, when its three angles are equal., 122. In a right triangle, the side opposite the right angle is called the, hypotenuse, and the other two sides the legs., 123. The side on which a triangle is supposed to stand is called the base of, the triangle. In the isosceles triangle, the equal sides are called the legs, and, the other side, the base; in other triangles, any one of the sides may be taken, as the base., 124. The angle opposite the base of a triangle is called the vertical angle,, and its vertex, the vertex of the triangle., 125. The altitude of a triangle is the perpendicular from the vertex to, the base, or to the base produced; as, AD (Fig. 1).
Page 45 :
TRIANGLES., , 35, , 126. The three perpendiculars from the vertices of a triangle to the opposite, sides (produced if necessary) are called the altitudes of the triangle; the three, bisectors of the angles are called the bisectors of the triangle; and the three, lines from the vertices to the middle points of the opposite sides are called the, medians of the triangle., 127. If two triangles have the angles of the one equal, respectively, to the, angles of the other, the equal angles are called homologous angles, and the, sides opposite the equal angles are called homologous sides., 128. Two triangles are equal in all respects if they can be made to coincide, (§ 60). The homologous sides of equal triangles are equal, and the homologous, angles are equal.
Page 46 :
BOOK I. PLANE GEOMETRY., , 36, , Proposition XVIII. Theorem., 129. The sum of the three angles of a triangle is equal to two right angles., B, , A, , E, , C, , F, , Let A, B, and BCA be the angles of the triangle ABC., To prove that ∠A + ∠B + ∠BCA = 2 rt. ∠s ., Proof. Suppose CE drawn k to AS, and prolong AC to F ., Then, ∠ECF + ∠ECB + ∠BCA = 2 rt. ∠s ,, § 89, (the sum of all the ∠s about a point on the same side of a straight line passing, through the point is equal to 2 rt. ∠s )., But, ∠A = ∠ECF ,, § 112, (being ext.-int. ∠s of the k lines AB and CE),, and, ∠B = ∠BCE,, § 110, (being alt.-int. ∠s of the k lines AB and CE)., Put for the ∠s ECF and BCE their equals, the ∠s A and B., Then, ∠A + ∠B + ∠BCA = 2 rt. ∠s ., , q.e.d., , 130. Cor. 1. The sum of two angles of a triangle is less than two right, angles., 131. Cor. 2. If the sum of two angles of a triangle is taken from two right, angles, the remainder is equal to the third angle., 132. Cor. 3. If two triangles have two angles of the one equal to two angles, of the other, the third angles are equal., 133. Cor. 4. If two right triangles have an acute angle of the one equal, to an acute angle of the other, the other acute angles are equal.
Page 47 :
TRIANGLES., , 37, , 134. Cor. 5. In a triangle there can be but one right angle, or one obtuse, angle., 135. Cor. 6. In a right triangle the two acute angles are together equal to, one right angle, or 90◦ ., 136. Cor. 7. In an equiangular triangle, each angle is one third of two, right angles, or 60◦ ., 137. Cor. 8. An exterior angle of a triangle is equal to the sum of the two, opposite interior angles, and therefore greater than either of them., Proposition XIX. Theorem., 138. The sum of two sides of a triangle is greater than the third side, and, their difference is less than the third side., B, , A, , C, , In the triangle ABC, let AC be the longest side., To prove that AB + BC > AC, and AC − BC < AB., Proof., AB + BC > AC,, (a straight line is the shortest line from one point to another )., Then, or, , § 49, , Take away BC from both sides., AB > AC − BC,, , Ax. 5, , AC − BC < AB., , q.e.d.
Page 48 :
BOOK I. PLANE GEOMETRY., , 38, , Proposition XX. Theorem., 139. Two triangles are equal if two angles and the included side of the one, are equal, respectively, to two angles and the included side of the other., C, , A, , F, , B, , D, , E, , In the triangles ABC, DEF , let the angle A be equal to the angle D,, B to E, and the side AB to DE., To prove that, 4ABC = 4DEF ., Proof. Apply the 4ABC to the 4DEF so that AB shall coincide with its, equal, DE., Then, AC will fall along DF , and BC along EF ,, (for ∠A = ∠D, and ∠B = ∠E, by hyp.)., ∴ C will fall on F ,, (two straight lines can intersect in only one point)., ∴ the two 4s coincide, and are equal., , § 48, § 60, q.e.d., , 140. Cor. 1. Two triangles are equal if a side and any two angles of the, one are equal to the homologous side and two angles of the other., § 132, 141. Cor. 2. Two right triangles are equal if the hypotenuse and an acute, angle of the one are equal, respectively, to the hypotenuse and an acute angle, of the other., § 133, 142. Cor. 3. Two right triangles are equal if a leg and an acute angle of, the one are equal, respectively, to a leg and the homologous acute angle of the, other., § 133
Page 49 :
TRIANGLES., , 39, , Proposition XXI. Theorem., 143. Two triangles are equal if two sides and the included angle of the one, are equal, respectively, to two sides and the included angle of the other., C, , A, , F, , B, , D, , E, , In the triangles ABC and DEF , let AB be equal to DE, AC to DF ,, and the angle A to the angle D., To prove that, 4ABC = 4DEF ., Proof. Apply the 4ABC to the 4DEF so that AB shall coincide with its, equal, DE., Then AC will fall along DF ,, (for ∠A = ∠D, by hyp.);, and C will fall on F ,, (for AC = DF , by hyp.)., ∴ CB = F E,, (their extremities being the same points)., ∴ the two 4s coincide, and are equal., , q.e.d., , 144. Cor. Two right triangles are equal if their legs are equal, each to, each., Note. In § 139 we have given two angles and the included side, in § 143 two, sides and the included angle; hence, by interchanging the words sides and angles,, either theorem is changed to the other. This is called the Principle of Duality, or, the Principle of Reciprocity. The reciprocal of a theorem is not always true, just as, the converse of a theorem is not always true.
Page 50 :
BOOK I. PLANE GEOMETRY., , 40, , Proposition XXII. Theorem., 145. In an isosceles triangle the angles opposite the equal sides are equal., A, , B, , D, , C, , Let ABC be an isosceles triangle, having AB and AC equal., To prove that, ∠B = ∠C., Proof. Suppose AD drawn so as to bisect the ∠BAC., In the 4s ADB and ADC,, AB = AC,, , Hyp., , AD = AD,, , Iden., , and ∠BAD = ∠CAD., , Const., , ∴ 4ADB = 4ADC,, § 143, (two 4s are equal if two sides and the included ∠ of the one are equal,, respectively, to two sides and the included ∠ of the other )., ∴ ∠B = ∠C,, (being homologous angles of equal triangles)., , § 128, q.e.d., , 146. Cor. An equilateral triangle is equiangular, and each angle is two, thirds of a right angle., Ex. 7. If the equal sides of an isosceles triangle are produced, the angles, on the other side of the base are equal.
Page 51 :
TRIANGLES., , 41, , Proposition XXIII. Theorem., 147. If two angles of a triangle are equal, the sides opposite the equal angles, are equal, and the triangle is isosceles., A, , B, , D, , C, , In the triangle ABC, let the angle B be equal to the angle C., To prove that, AB = AC., Proof., Suppose AD drawn ⊥ to BC., In the rt. 4s ADB and ADC,, AD = AD,, , Iden., , and ∠B = ∠C., , Hyp., , ∴ rt. 4ADB = rt. 4ADC,, § 142, (having a leg and an acute ∠ of the one equal, respectively, to a leg and the, homologous acute ∠ of the other )., ∴ AB = AC,, (being homologous sides of equal 4s )., , § 128, q.e.d., , 148. Cor. 1. An equiangular triangle is also equilateral., 149. Cor. 2. The perpendicular from the vertex to the base of an isosceles, triangle bisects the base, and bisects the vertical angle of the triangle.
Page 52 :
BOOK I. PLANE GEOMETRY., , 42, , Proposition XXIV. Theorem., 150. Two triangles are equal if the three sides of the one are equal, respectively, to the three sides of the other., B′, , B, , A, , C, , C′, , A′, , B′, , In the triangles ABC and A0 B 0 C 0 , let AB be equal to A0 B 0 , AC to, BC to B 0 C 0 ., To prove that, 4ABC = 4A0 B 0 C 0 ., 0, 0, 0, Proof. Place 4A B C in the position 4AB 0 C having its greatest side 4A0 C 0, in coincidence with its equal 4AC, and its vertex at B 0 , opposite B; and draw BB 0 ., Since, AB = AB 0, Hyp., A0 C 0 ,, , Since, , ∠ABB 0 = ∠AB 0 B, (in an isosceles 4 the ∠s opposite the equal sides are equal )., CB = CB 0 ,, , Hyp., , ∠CBB 0 = ∠CB 0 B., Hence,, , ∴, , ∠ABB 0, , +, , ∠CBB 0, , =, , ∠AB 0 B, , § 145, , § 145, +, , ∠CB 0 B., , Ax. 2, , ∠ABC = ∠AB 0 C., , ∴ 4ABC = 4AB 0 C,, § 143, (two 4s are equal if two sides and the included ∠ of the one are equal,, respectively, to two sides and the included ∠ of the other )., ∴ 4ABC = 4A0 B 0 C 0 ., , q.e.d.
Page 53 :
TRIANGLES., , 43, , Proposition XXV. Theorem., 151. Two right triangles are equal if a leg and the hypotenuse of the one, are equal, respectively, to a leg and the hypotenuse of the other., A′, , A, , C, , B, , C, , B′, , A′, , C′, , B′, , C′, , In the right triangles ABC and A0 B 0 C 0 , let AB be equal to A0 B 0 , and, AC to A0 C 0 ., To prove that, 4ABC = 4A0 B 0 C 0 ., Proof. Apply the 4ABC to the 4A0 B 0 C 0 , so that AB shall coincide with A0 B 0 ,, A falling on A0 , B on B 0 , and C and C 0 on opposite sides of A0 B 0 ., Then, BC will fall along C 0 B 0 produced,, (for ∠ABC = ∠A0 B 0 C 0 , each being a rt. ∠.)., Since, AC = A0 C 0 ,, Hyp., the 4A0 CC 0 is an isosceles triangle., ∴ ∠C = ∠C 0 ,, (∠s opposite the equal sides of an isosceles 4 are equal )., , § 120, § 145, , ∴ 4s ABC and A0 B 0 C 0 are equal,, § 141, (two right 4s are equal if they have the hypotenuse and an acute ∠ of, the one, equal to the hypotenuse and an acute ∠ of the other )., q.e.d., , Ex. 8. How many degrees are there in each of the acute angles of an, isosceles right triangle?
Page 54 :
BOOK I. PLANE GEOMETRY., , 44, , Proposition XXVI. Theorem., 152. If two sides of a triangle are unequal, the angles opposite are unequal,, and the greater angle is opposite the greater side., A, , E, , C, , B, , In the triangle ACB, let AB be greater than AC., To prove that ∠ACB is greater than ∠B., Proof., On AB take AE equal to AC., Draw EC., , But, , and, , then, , ∠AEC = ∠ACE, (being ∠s opposite equal sides)., , § 145, , ∠AEC is greater than ∠B, § 137, (an exterior ∠ of a 4 is greater than either opposite interior ∠),, ∠ACB is greater than ∠ACE., , Ax. 8, , Substitute for ∠ACE its equal ∠AEC,, , ∠ACB is greater than ∠AEC., Since ∠AEC is greater than ∠B, and ∠ACB is greater than ∠AEC,, ∠ACB is greater than ∠B., , q.e.d., , Ex. 9. If any angle of an isosceles triangle is equal to two thirds of a right, angle (60◦ ), what is the value of each of the two remaining angles?, Ex. 10. One angle of a triangle is 34◦ . Find the other angles, if one of, them is twice the other.
Page 55 :
TRIANGLES., , 45, , Proposition XXVII. Theorem., 153. Reciprocally: If two angles of a triangle are unequal, the sides, opposite are unequal, and the greater side is opposite the greater angle., A, , C, , B, , In the triangle ACB, let the angle C be greater than the angle B., To prove that AB > AC., Proof., Now AB = AC, or < AC, or > AC., But AB is not equal to AC;, for then the ∠C would be equal to the ∠B,, (being ∠s opposite equal sides)., , § 145, , And AB is not less than AC;, for then the ∠C would be less than the ∠B., § 152, Both these conclusions are contrary to the hypothesis that the ∠C is greater, than the ∠B., Hence, AB cannot be equal to AC or less than AC., ∴ AB > AC., , q.e.d., , Ex. 11. If the vertical angle of an isosceles triangle is equal to 30◦ , find, the exterior angle included by a side and the base produced., Ex. 12. If the vertical angle of an isosceles triangle is equal to 36◦ , find, the angle included by the bisectors of the base angles.
Page 56 :
BOOK I. PLANE GEOMETRY., , 46, , Proposition XXVIII. Theorem., 154. If two triangles have two sides of the one equal, respectively, to two, sides of the other, but the included angle of the first triangle greater than the, included angle of the second, then the third side of the first is greater than the, third side of the second., B, , B, , B, , A, , A, , E, , F, , C, , A, , C, , E, , In the triangles ABC and ABE, let AB be equal to AB, BC to BE;, but let the angle ABC be greater than the angle ABE., To prove that, AC > AE., Proof. Place the 4s so that AB of the one shall fall on AB of the other, and, BE within the ∠ABC., Suppose BF drawn to bisect the ∠EBC, and draw EF ., The 4s EBF and CBF are equal., For, BF = BF ,, , § 143, , BE = BC,, , Hyp., , and, , Now, , Iden., , ∠EBF = ∠CBF ., , Const., , ∴ EF = F C., , § 128, , AF + F E > AE., , § 138, , ∴ AF + F C > AE., ∴ AC > AE., , q.e.d.
Page 57 :
TRIANGLES., , 47, , Proposition XXIX. Theorem., 155. Conversely: If two sides of a triangle are equal, respectively, to two, sides of another, but the third side of the first triangle is greater than the third, side of the second, then the angle opposite the third side of the first triangle is, greater than the angle opposite the third side of the second., D, , A, , B, , C, , E, , F, , In the triangles ABC and DEF , let AB be equal to DE, AC to DF ;, but let BC be greater than EF., To prove that the ∠A is greater than the ∠D., Proof. Now the ∠A is equal to the ∠D, or less than the ∠D, or greater than, the ∠D., But the ∠A is not equal to the ∠D;, for then the 4ABC would be equal to the 4DEF ,, § 143, (having two sides and the included ∠ of the one equal, respectively, to two sides, and the included ∠ of the other ),, and BC would be equal to EF ., And the ∠A is not less than the ∠D, for then BC would be less than EF . § 154, Both these conclusions are contrary to the hypothesis that BC is greater than, EF ., Since the ∠A is not equal to the ∠D or less than the ∠D,, the ∠A is greater than the ∠D., , q.e.d.
Page 58 :
BOOK I. PLANE GEOMETRY., , 48, , LOCI OF POINTS., 156. If it is required to find a point which shall fulfil a single geometric, condition, the point may have an unlimited number of positions. If, however,, all the points are in the same plane, the required point will be confined to a, particular line, or group of lines., A point in a plane at a given distance from a fixed straight line of indefinite, length in that plane, is evidently in one of two straight lines, so drawn as to, be everywhere at the given distance from the fixed line, one on one side of the, fixed line, and the other on the other side of it., A point in a plane equidistant from two parallel lines in that plane is, evidently in a straight line drawn between the two given parallel lines and, everywhere equidistant from them., 157. All points in a plane that satisfy a single geometrical condition lie, in, general, in a line or group of lines; and this line or group of lines is called the, locus of the points that satisfy the given condition., 158. To prove completely that a certain line is the locus of points that fulfil, a given condition, it is necessary to prove, 1. Any point in the line satisfies the given condition; and any point not in, the line does not satisfy the given condition., Or, to prove, 2. Any point that satisfies the given condition lies in the line; and any, point in the line satisfies the given condition., Note. The word locus (pronounced lo´kus) is a Latin word that signifies place., The plural of locus is loci (pronounced lo´si)., , 159. Def. A line which bisects a given line and is perpendicular to it is, called the perpendicular bisector of the line.
Page 59 :
LOCI OF POINTS., , 49, , Proposition XXX. Theorem., 160. The perpendicular bisector of a given line is the locus of points equidistant from the extremities of the line., R, , C, , D, , O, A, , P, , B, , Let P R be the perpendicular bisector of the line AB, O any point in, P R, and C any point not in P R., Draw OA and OB, CA and CB., To prove OA and OB equal, CA and CB unequal., Proof., 1. 4OP A = 4OP B,, , § 144, , for P A = P B by hypothesis, and OP is common,, (two right 4s are equal if their legs are equal, each to each)., ∴ OA = OB., 2. Since C is not in the ⊥, CA or CB will cut the ⊥., Let CA cut the ⊥ at D, and draw DB., Then, by the first part of the proof DA = DB., But, CB < CD + DB., That is,, , § 128, , § 138, , ∴ CB < CD + DA., , CB < CA., ∴ P R is the locus of points equidistant from A and B., , § 158,1, q.e.d., , 161. Cor. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line.
Page 60 :
BOOK I. PLANE GEOMETRY., , 50, , Proposition XXXI. Theorem., 162. The bisector of a given angle is the locus of points equidistant from, the sides of the angle., P, B, , F, , D, O, , A, , G, , C, , Q, , Let O be any point equidistant from the sides of the angle P AQ., To prove that O is in the bisector of the ∠P AQ., Proof., Draw AO., Suppose OF drawn ⊥ to AP and OG ⊥ to AQ., In the rt. 4s AF O and AGO,, AO = AO,, , Iden., , OF = OG,, , Hyp., , ∴ 4AF O = 4AGO., , § 151, , ∴ ∠F AO = ∠GAO., , § 128, , ∴ O is in the bisector of the ∠P AQ., Let D be any point in the bisector of the angle PAQ., To prove that D is equidistant from AP and AQ., Proof., Suppose DB drawn ⊥ to AP and DC ⊥ to AQ., In the rt. 4s ABD and ACD,, AD = AD,, , Iden., , ∠DAB = ∠DAC,, , Hyp., , ∴ 4ABD = 4ACD., , § 141, , ∴ DB = DC., , § 128, , ∴ D is equidistant from AP and AQ.
Page 61 :
QUADRILATERALS., , 51, , ∴ the bisector of the ∠P AQ is the locus of points that are equidistant from its, sides., § 158, 2, , QUADRILATERALS., 163. A quadrilateral is a portion of a plane bounded by four straight lines., The bounding lines are the sides, the angles formed by these sides are the, angles, and the vertices of these angles are the vertices, of the quadrilateral., 164. A trapezium is a quadrilateral which has no two sides parallel., 165. A trapezoid is a quadrilateral which has two sides, and only two, sides, parallel., 166. A parallelogram is a quadrilateral which has its opposite sides parallel., , Trapezium., , Trapezoid., , Parallelogram., , 167. A rectangle is a parallelogram which has its angles right angles., 168. A square is a rectangle which has its sides equal., 169. A rhomboid is a parallelogram which has its angles oblique angles., 170. A rhombus is a rhomboid which has its sides equal., , Square., , Rectangle., , Rhombus., , Rhomboid., , 171. The side upon which a parallelogram stands, and the opposite side,, are called its lower and upper bases.
Page 62 :
BOOK I. PLANE GEOMETRY., , 52, , 172. Two parallel sides of a trapezoid are called its bases, the other two, sides its legs, and the line joining the middle points of the legs is called the, median of the trapezoid., B, , A, , P, , Q, , C, , E, , 173. A trapezoid is called an isosceles trapezoid if its legs are equal., 174. The altitude of a parallelogram or trapezoid is the perpendicular, distance between its bases, as P Q., 175. A diagonal of a quadrilateral is a straight line joining two opposite, vertices, as AC.
Page 63 :
QUADRILATERALS., , 53, , Proposition XXXII. Theorem., 176. Two angles whose sides are parallel, each to each, are either equal or, supplementary., A, , B, , a, , x, , C, , c, , M, , c, , D, ′, , ′, , a, , N, , H, , Let BA be parallel to HD, and BC be parallel to M N ., To prove ∠s a, a0 and c equal; a and c0 supplementary., Proof., Let HD and BC prolonged intersect at x., Then, ∠a = ∠x, and ∠a0 = ∠x., Also, Now, , Then, , ∴ ∠a =, , ∠a0 ., , Ax. 1, , ∠c = ∠a0 (§ 93). ∴ ∠c = ∠a., ∠a0 and ∠c0 are supplementary., Put ∠a for its equal,, , § 112, , Ax. 1, § 89, , ∠a0 ., , ∠a and ∠c0 are supplementary., , q.e.d., , 177. Cor. The opposite angles of a parallelogram are equal, and the adjacent angles are supplementary.
Page 64 :
BOOK I. PLANE GEOMETRY., , 54, , Proposition XXXIII. Theorem., 178. The opposite sides of a parallelogram are equal., B, , A, , C, , E, , Let the figure ABCE be a parallelogram., To prove BC = AE, and AB = EC., Proof., Draw the diagonal AC., 4ABC = 4CEA., , § 139, , For AC is common,, ∠BAC = ∠ACE, and ∠ACB = ∠CAE,, (being alt-int. ∠s of k lines)., ∴ BC = AE, and AB = CE,, (being homologous sides of equal 4s )., , § 110, § 128, q.e.d., , 179. Cor. 1. A diagonal divides a parallelogram into two equal triangles., 180. Cor. 2. Parallel lines comprehended between parallel lines are equal., A, , B, , D, , C, , 181. Cor. 3. Two parallel lines are everywhere equally distant., For if AB and DC are parallel, ⊥s dropped from any points in AB to DC,, are equal, § 180. Hence, all points in AB are equidistant from DC.
Page 65 :
QUADRILATERALS., , 55, , Proposition XXXIV. Theorem., 182. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram., B, , C, , A, , E, , Let the figure ABCE be a quadrilateral, having BC equal to AE and, AB to EC., To prove that the figure ABCE is a / /., Proof., Draw the diagonal AC., In the 4s ABC and CEA,, , and, , BC = AE,, , Hyp., , AB = CE,, , Hyp., , AC = AC,, , Iden., , ∴ 4ABC = 4CEA,, § 150, (having three sides of the one equal, respectively, to the three sides of the other )., and, , and, , ∴ ∠ACB = ∠CAE,, , § 128, , ∠BAC = ∠ACE,, (being homologous ∠s of equal 4s )., ∴ BC is k to AE,, , AB is k to EC,, § 111, (two lines in the same plane cut by a transversal are parallel, if the alt.-int. ∠s, are equal )., ∴ the figure ACBE is a / /,, (having its opposite sides parallel )., , § 166, q.e.d.
Page 66 :
BOOK I. PLANE GEOMETRY., , 56, , Proposition XXXV. Theorem., 183. If two sides of a quadrilateral are equal and parallel, then the other, two sides are equal and parallel, and the figure is a parallelogram., B, , C, , A, , E, , Let the figure ABCE be a quadrilateral, having the side AE equal, and parallel to BC., To prove that AB is equal and parallel to EC., Proof., Draw AC., The 4s ABC and CEA are equal,, For, , and, , and, , § 143, , (having two sides and the included ∠ of each equal, respectively)., AC is common,, BC = AE, ∠BCA = ∠CAE,, (being alt.-int. ∠s of k lines)., , Hyp., § 110, , ∴ AB = EC,, ∠BAC = ∠ACE,, (being homologous parts of equal 4s )., , § 128, , ∴ AB is k to EC,, (two lines are k, if the alt.-int. ∠s are equal )., , § 111, , ∴ the figure ABCE is a / /,, (the opposite sides being parallel )., , § 166, q.e.d.
Page 67 :
QUADRILATERALS., , 57, , Proposition XXXVI. Theorem., 184. The diagonals of a parallelogram bisect each other., B, , C, , O, , A, , E, , Let the figure ABCE be a parallelogram, and let the diagonals AC, and BE cut each other at O., To prove that AO = OC, and BO = OE., Proof. In the 4s AOE and COB,, AE = BC,, (being opposite sides of a, , / /)., , ∠OAE = ∠OCB,, , § 178, § 110, , and ∠OEA = ∠OBC,, (being alt.-int. ∠s of k lines)., ∴ 4AOE = 4COB,, § 139, (having two ∠s and the included side of the one equal, respectively, to two ∠s, and the included side of the other )., ∴ AO = OC, and BO = OE,, (being homologous sides of equal 4s )., , § 128, q.e.d., , Ex. 13. The median from the vertex to the base of an isosceles triangle, is perpendicular to the base, and bisects the vertical angle., Ex. 14. If two straight lines are cut by a transversal so that the alternateexterior angles are equal, the two straight lines are parallel., Ex. 15. If two parallel lines are cut by a transversal, the two exterior, angles on the same side of the transversal are supplementary.
Page 68 :
BOOK I. PLANE GEOMETRY., , 58, , Ex. 16. If two straight lines are cut by a transversal so as to make the, exterior angles on the same side of the transversal supplementary, the two lines, are parallel., Proposition XXXVII. Theorem., 185. Two parallelograms are equal, if two sides and the included angle of, the one are equal, respectively, to two sides and the included angle of the other., B, , B′, , C, , A, , A′, , D, , C′, , D′, , In the parallelograms ABCD and A0 B 0 C 0 D 0 , let AB be equal to A0 B 0 ,, AD to A0 D 0 , and angle A to A0 ., To prove that the, , / /s, , are equal., , Proof. Place the / / ABCD on the, coincide with its equal, A0 D0 ., , / /, , A0 B 0 C 0 D0 , so that AD will fall on and, , Then AB will fall on A0 B 0 , and B on B 0 ;, (for ∠A = ∠A0 , and AB = A0 B 0 , by hyp.), Now, BC and B 0 C 0 are both k to A0 D0 and drawn through B 0 ., ∴ BC and B 0 C 0 coincide,, § 105, (through a given point only one line can be drawn k to a given line)., Also DC and D0 C 0 are k to A0 B 0 and drawn through D0 ., ∴ DC and D0 C 0 coincide., , § 105, , ∴ C falls on C 0 ,, (two lines can intersect in only one point),, ∴ the two, , / /s, , coincide, and are equal., , § 48, q.e.d., , 186. Cor. Two rectangles having equal bases and altitudes are equal.
Page 69 :
QUADRILATERALS., , 59, , Proposition XXXVIII. Theorem., 187. If three or more parallels intercept equal parts on one transversal, they, intercept equal parts on every transversal., , A, B, C, D, , H, , E, F, , K, M, P, , G, , Let the parallels AH, BK, CM , DP intercept equal parts HK, KM ,, M P on the transversal HP ., To prove that they intercept equal parts AB, BC, CD on the transversal AD., Proof. Suppose AH, BF , and CG drawn k to HP ., ∠s AEB, BF C, etc. = ∠s HKE, KM F , etc., respectively., , § 112, , But ∠s HKE, KM F , etc. are equal., , § 112, , ∴ ∠s AEB, BF C, etc. are equal., , Ax. 1, , Also ∠s BAE, CBF , etc. are equal., , § 112, , Now AE = HK, BF = KM , CG = M P ,, (parallels comprehended between parallels are equal )., ∴ AE = BF = CG., ∴ 4ABE = 4BCF = 4CDG,, (having two ∠s and the included side of each respectively equal )., ∴ AB = BC = CD., , § 180, Ax. 1, § 139, § 128, q.e.d.
Page 70 :
BOOK I. PLANE GEOMETRY., , 60, , A, , E, , D, , B, , C, , F, , 188. Cor. 1. If a line is parallel to the base of a triangle and bisects one, side, it bisects the other side also., Let DE be k to BC and bisect AB. Suppose a line is drawn through A k, to BC. Then this line is k to DE, by § 106. The three parallels by hypothesis, intercept equal parts on the transversal AB, and therefore, by § 187, they, intercept equal parts on the transversal AC; that is, the line DE bisects AC., 189. Cor. 2. The line which joins the middle points of two sides of a, triangle is parallel to the third side, and is equal to half the third side., A line drawn through D, the middle point of AB, k to BC, passes through, E, the middle point of AC, by § 188. Therefore the line joining D and E, coincides with this parallel and is k to BC. Also, since EF drawn k to AB, bisects AC, it bisects BC, by § 188; that is, BF = F C = 21 BC. But BDEF, is a / / by § 166, and therefore DE = BF = 21 BC., D, , E, , A, , C, , F, , G, , B, , 190. Cor. 3. The median of a trapezoid is parallel to the bases, and is, equal to half the sum of the bases., Draw the diagonal DB. In the 4ADB join E, the middle point of AD, to, F , the middle point of DB. Then, by § 189, EF is k to AB and = 21 AB. In, the 4DBC join F to G, the middle point of BC. Then F G is k to DC and, = 12 DC. AB and F G, being k to DC, are k to each other. But only one line, can be drawn through F k to AB (§ 105). Therefore F G is the prolongation, of EF . Hence, EF G is parallel to AB and DC, and equal to 12 (AB + DC).
Page 71 :
POLYGONS IN GENERAL., , 61, , POLYGONS IN GENERAL., 191. A polygon is a portion of a plane bounded by straight lines., The bounding lines are the sides, and their sum, the perimeter of the, polygon. The angles included by the adjacent sides are the angles of the, polygon, and the vertices of these angles are the vertices of the polygon. The, number of sides of a polygon is evidently equal to the number of its angles., 192. A diagonal of a polygon is a line joining the vertices of two angles, not adjacent; as, AC (Fig. 1)., B, , B, , C, , C, E, , A, , D, , A, , D, , D, F, , E, , Fig. 1., , E, , Fig. 2., , Fig. 3., , 193. An equilateral polygon is a polygon which has all its sides equal., 194. An equiangular polygon is a polygon which has all its angles equal., 195. A convex polygon is a polygon of which no side, when produced,, will enter the polygon., 196. A concave polygon is a polygon of which two or more sides, if, produced, will enter the polygon., 197. Each angle of a convex polygon (Fig. 2) is called a salient angle, and, is less than a straight angle., 198. The angle EDF of the concave polygon (Fig. 3) is called a re-entrant, angle, and is greater than a straight angle., When the term polygon is used, a convex polygon is meant.
Page 72 :
BOOK I. PLANE GEOMETRY., , 62, , 199. Two polygons are equal when they can be divided by diagonals into, the same number of triangles, equal each to each, and similarly placed; for, if the polygons are applied to each other, the corresponding triangles will, coincide, and hence the polygons will coincide and be equal., 200. Two polygons are mutually equiangular , if the angles of the one are, equal to the angles of the other, each to each, when taken in the same order., Figs. 1 and 2., 201. The equal angles in mutually equiangular polygons are called homologous angles; and the sides which are included by homologous angles are called, homologous sides., 202. Two polygons are mutually equilateral , if the sides of the one are, equal to the sides of the other, each to each, when taken in the same order., Figs. 1 and 2., , Fig. 4., , Fig. 5., , Fig. 6., , Fig. 7., , 203. Two polygons may be mutually equiangular without being mutually, equilateral; as, Figs. 4 and 5., And, except in the case of triangles, two polygons may be mutually equilateral without being mutually equiangular; as, Figs. 6 and 7., If two polygons are mutually equilateral and mutually equiangular they are, equal, for they can be made to coincide., 204. A polygon of three sides is called a triangle; one of four sides, a, quadrilateral ; one of five sides, a pentagon; one of six sides, a hexagon; one, of seven sides, a heptagon; one of eight sides, an octagon; one of ten sides, a, decagon; one of twelve sides, a dodecagon.
Page 73 :
POLYGONS IN GENERAL., , 63, , Proposition XXXIX. Theorem., 205. The sum of the interior angles of a polygon is equal to two right, angles, taken as many times less two as the figure has sides., E, , D, , C, , F, , A, , B, , Let the figure ABCDEF be a polygon, having n sides., To prove that ∠A + ∠B + ∠C, etc. = (n − 2)2 rt. ∠s ., Proof. From A draw the diagonals AC, AD, and AE., The sum of the ∠s of the 4s is equal to the sum of the ∠s of the polygon., Now, there are (n − 2) 4s ,, and the sum of the ∠s of each 4 = 2 rt. ∠s ., § 129, ∴ the sum of the ∠s of the 4s , that is, the sum of the ∠s of the polygon is equal, to (n − 2)2 rt. 4s ., q.e.d., , 206. Cor. The sum of the angles of a quadrilateral equals 4 right angles;, and if the angles are all equal, each is a right angle. In general, each angle of, 2(n − 2), right angles., an equiangular polygon of n sides is equal to, n, Ex. 17. How many diagonals can be drawn in a polygon of n sides?
Page 74 :
BOOK I. PLANE GEOMETRY., , 64, , Proposition XL. Theorem., 207. The exterior angles of a polygon, made by producing each of its sides, in succession, are together equal to four right angles., dD, c, , e, , C, , E, , A, , a, , B, , b, , Let the figure ABCDE be a polygon, having its sides produced in, succession., To prove the sum of the ext. ∠s = 4 rt. ∠s ., Proof. Denote the int. ∠s of the polygon by A, B, C, D, E, and the corresponding ext. ∠s by a, b, c, d, e., and, , ∠A + ∠a = 2 rt. ∠s ,, , § 89, , ∠B + ∠b = 2 rt. ∠s ,, (being sup.-adj. ∠s )., , In like manner each pair of adj. ∠s = 2 rt. ∠s ., ∴ the sum of the interior and exterior ∠s of a polygon of n sides is equal to 2n, rt. ∠s ., But the sum of the interior ∠s = (n − 2)2 rt. ∠s, = 2n rt. ∠s − 4 rt. ∠s ., ∴ the sum of the exterior ∠s = 4 rt. ∠s ., , § 205, q.e.d., , Ex. 18. How many sides has a polygon if the sum of its interior ∠s is, twice the sum of its exterior ∠s ? ten times the sum of its exterior ∠s ?
Page 75 :
SYMMETRY., , 65, , SYMMETRY., 208. Two points are said to be symmetrical with respect to a third point,, called the centre of symmetry, if this third point bisects the straight line, which joins them., P, , B, , X′, , O, , X, , C, O, , A′, , C′, , B, A, , B′, , X′, , C, , A, , D, B′, , X, , C′, , P′, , Two points are said to be symmetrical with respect to a straight line, called, the axis of symmetry, if this straight line bisects at right angles the straight, line which joins them., Thus, P and P 0 are symmetrical with respect to O as a centre, and XX 0, as an axis, if O bisects the line P P 0 , and if XX 0 bisects P P 0 at right angles., 209. A figure is symmetrical with respect to a point as a centre of symmetry, if the point bisects every straight line drawn through it and terminated, by the boundary of the figure., 210. A figure is symmetrical with respect to a line as an axis of symmetry, if one of the parts of the figure coincides, point for point, with the other part, when it is folded over on that line as an axis.
Page 76 :
BOOK I. PLANE GEOMETRY., , 66, , B, A, C, X′, , X, C′, , A′, B′, , 211. Two figures are said to be symmetrical with respect to an axis if every, point of one has a corresponding symmetrical point in the other., Thus, if every point in the figure A0 B 0 C 0 has a symmetrical point in ABC,, with respect to XX 0 as an axis, the figure A0 B 0 C 0 is symmetrical to ABC with, respect to XX 0 as an axis.
Page 77 :
SYMMETRY., , 67, , Proposition XLI. Theorem., 212. A quadrilateral which has two adjacent sides equal, and the other two, sides equal, is symmetrical with respect to the diagonal joining the vertices of, the angles formed by the equal sides, and the diagonals are perpendicular to, each other., B, , A, , O, , C, , D, , Let ABCD be a quadrilateral, having AB equal to AD, and CB equal, to CD, and having the diagonals AC and BD., To prove that the diagonal AC is an axis of symmetry, and that it is ⊥ to the, diagonal BD., Proof. In the 4s ABC and ADC,, and, , AB = AD, and BC = DC,, , Hyp., , AC = AC., , Iden., , ∴ 4ABC = 4ADC., , § 150, , ∴ ∠BAC = ∠DAC, and ∠BCA = ∠DCA., Hence, if ABC is turned on AC as an axis until it falls on ADC, AB will fall, upon AD, CB on CD, and OB on OD., ∴ the 4ABC will coincide with the 4ADC., ∴ AC is an axis of symmetry (§ 210) and is ⊥ to BD., , § 208, q.e.d.
Page 78 :
BOOK I. PLANE GEOMETRY., , 68, , Proposition XLII. Theorem., 213. If a figure is symmetrical with respect to two axes perpendicular to, each other, it is symmetrical with respect to their intersection as a centre., Y, , A, , X′, , I, , B, , K, H, , C, , M, , N, , D, , X, , O, , L, , E, G, , F, Y′, , Let the figure ABCDEF GH be symmetrical with respect to the two, perpendicular axes XX 0 , Y Y 0 , which intersect at O., To prove that O is the centre of symmetry of the figure., Proof. Let N be any point in the perimeter., Then, , Now, , But, , Suppose N M I drawn ⊥ to Y Y 0 , IKL ⊥ to XX 0 ., N I is k to XX 0 and IL is k to Y Y 0 ., , § 104, , Draw LO, ON , and KM ., KI = KL,, (the figure being symmetrical with respect to XX 0 )., KI = OM ., ∴ KL = OM , and KLOM is a, , § 208, , § 180, / /., , § 183, , ∴ LO is equal and parallel to KM ., , § 183, , In like manner ON is equal and parallel to KM ., ∴ LON is a straight line., § 105, ∴ O bisects LN , any straight line and therefore every straight line drawn through, O and terminated by the perimeter., ∴ O is the centre of symmetry of the figure., , q.e.d.
Page 79 :
SYMMETRY., , 69, , REVIEW QUESTIONS ON BOOK I., 1. What is the subject-matter of Geometry?, 2. What is a geometric magnitude?, 3. What is an axiom? a theorem? a converse theorem? an opposite, theorem? a contradictory theorem?, 4. Define a straight line; a curved line; a broken line; a plane surface; a, curved surface., 5. How many points are necessary to determine a straight line?, 6. How many straight lines are necessary to determine a point?, 7. On what does the magnitude of an angle depend?, 8. Define a straight angle; a right angle; an oblique angle., 9. Define adjacent angles; complementary angles; supplementary angles;, conjugate angles., 10. Define parallel lines and give the axiom of parallels., 11. If two lines in the same plane are parallel and cut by a transversal, what, pairs of angles are equal? what pairs are supplementary?, 12. Define a right triangle; an isosceles triangle; a scalene triangle., 13. To how many right angles is the sum of the angles of a triangle equal?, the sum of the acute angles of a right triangle?, 14. To what angles is the exterior angle of a triangle equal?, 15. What is the test of equality of two geometric magnitudes?, 16. How does a reciprocal theorem differ from a converse theorem?, 17. State the three cases in which two triangles are equal., 18. State the cases in which two right triangles are equal., 19. What is meant by a locus of points?, 20. Where are the points located in a plane that are each equidistant from, two given points? from two intersecting lines?, 21. Define a parallelogram; a trapezoid; an isosceles trapezoid., 22. When is a figure symmetrical with respect to a centre?
Page 80 :
BOOK I. PLANE GEOMETRY., , 70, , 23. When is a figure symmetrical with respect to an axis?, 24. Must a triangle be equiangular if equilateral? must a triangle be equilateral if equiangular?, 25. When are two polygons said to be mutually equiangular?, 26. When are two polygons said to be mutually equilateral?, 27. Can two polygons of more than three sides be mutually equiangular, without being mutually equilateral? mutually equilateral without being, mutually equiangular?, 28. What line do two points each equidistant from the extremities of a given, straight line determine?, METHODS OF PROVING THEOREMS., 214. There are three general methods of proving theorems, the synthetic,, the analytic, and the indirect methods., The synthetic method is the method employed in most of the theorems, already given, and consists in putting together known truths in order to obtain, a new truth., The analytic method is the reverse of the synthetic method. It asserts that, the conclusion is true if another proposition is true, and so on step by step,, until a known truth is reached. Thus, proposition A is true if proposition B, is true, and B is true if C is true; but C is true, hence A and B are true., If a known truth suggests the required proof, it is best to use the synthetic, form at once. If no proof occurs to the mind, it is necessary to use the analytic, method to discover the proof, and then the synthetic proof may be given., The indirect method, or the method of reductio ad absurdum, is illustrated, on page 45. It consists in proving a theorem to be true by proving its contradictory to be false.
Page 81 :
SYMMETRY., , 71, , 215. Generally auxiliary lines are required, as a line connecting two points;, a line parallel to or perpendicular to a given line; a line produced by its own, length; a line making with another line an angle equal to a given angle., Two lines are proved equal by proving them homologous sides of equal, triangles; or legs of an isosceles triangle; or opposite sides of a parallelogram., Two angles are proved equal by proving them alternate-interior angles or exterior-interior angles of parallel lines; or homologous angles of equal, triangles; or base angles of an isosceles triangle; or opposite angles of a parallelogram., Two suggestions are of special importance to the beginner:, 1. Draw as accurate figures as possible., 2. Draw as general figures as possible.
Page 82 :
BOOK I. PLANE GEOMETRY., , 72, , EXERCISES., Prove by the analytic method:, , A, , B, , D, , C, , E, , Ex. 19. A median of a triangle is less than half the sum of the two adjacent, sides., To prove the median AD < 12 (AB + AC)., Now, AD < 12 (AB + AC),, if, 2AD < AB + AC., This suggests producing AD by its own length to E, and joining BE., Then, AE = 2AD,, and, 2AD < AB + AC if AE < AB + AC., But, AE < AB + BE., § 138, And, But, For, , and, , ∴ AE < AB + AC if AC = BE., AC = BE if 4ACD = 4EBD., , § 128, , 4ACD = 4EBD., , § 143, , CD = DB,, , Hyp., , AD = DE,, , Const., , ∠ADC = ∠BDE., ∴ AE < AB + AC., , § 93
Page 83 :
EXERCISES., , 73, , ∴ AD < 12 (AB + AC)., A, , E, , D, , B, , G, , C, , Ex. 20. A straight line which bisects two sides of a triangle is parallel to, the third side., If AD = DB and AE = EC, to prove DE k to BC., Draw CG k to BA, and produce DE to meet it at G., DE is k to BC if BCGD is a, BCGD is a, Now, And, But, For, , / /, , / /., , if CG = BD., , § 166, § 183, , CG = BD if each is equal to AD., , Ax. 1, , BD = AD., , Hyp., , CG = AD if 4CGE = 4ADE., , § 128, , 4CGE = 4ADE., , § 139, , EC = AE., , Hyp., , ∠GEC = ∠AED., , § 93, , ∠ECG = ∠DAE., , § 110, , ∴ DE is k to BC., Prove by the synthetic method:, Ex. 21. The middle point of the hypotenuse of a right triangle is equidistant, from the three vertices., From D, the middle point, draw DE ⊥ to CB., DE is k to AC (why?), and DE bisects CB (why?)., ∴ D is equidistant from B, A, and C. (Why?)
Page 84 :
BOOK I. PLANE GEOMETRY., A, , 74, , A, , 2a, D, , C, , E, , D, , B, , C, , c b, , E, , a, , B, , Ex. 22. If one acute angle of a right triangle is double the other, the, hypotenuse is double the shorter leg., The median CD = BD = AD (Ex. 21)., Then ∠b = ∠a; and ∠c = ∠2a. (Why?), Now a + 2a = 90◦ . (Why?), ∴ ∠a = 30◦ ; ∠2a = 60◦ ; ∠c = 60◦ ., ∴ 4ACD is equilateral (why?), and AD, half of AB = AC. ∴ AB = 2AC.
Page 85 :
EXERCISES., , 75, , Ex. 23. If two triangles have two sides of the one equal, respectively, to, two sides of the other, and the angles opposite two equal sides equal, the angles, opposite the other two equal sides are equal or supplementary, and if equal the, triangles are equal., Let AC = A0 C 0 , BC = B 0 C 0 , and ∠B = ∠B 0 ., Place 4A0 B 0 C 0 on 4ABC so that B 0 C 0 shall coincide with BC, and ∠A0, and ∠A shall be on the same side of BC., C′, , C, , A, , D, , B, , A′, , C′, , B′, , A′, , B′, , Since ∠B 0 = ∠B, B 0 A0 will fall along BA, and A0 will fall at A or at some, other point in BA, as D. If A0 falls at A, the 4s A0 B 0 C 0 and ABC coincide, and are equal., If A0 falls at D, the 4s A0 B 0 C 0 and DBC coincide and are equal., Since CD = C 0 A0 = CA, ∠A = ∠CDA. (Why?), But ∠s CDA and CDB are supplements. (Why?), ∴ ∠s A and CDB are supplements. (Why?), Draw figures and show that the triangles are equal:, 1. If the given angles B and B 0 are both right or both obtuse angles., 2. If the required angles A and A0 are both acute, both right, or both, obtuse., 3. If AC and A0 C 0 are not less than BC and B 0 C 0 , respectively.
Page 86 :
BOOK I. PLANE GEOMETRY., C, , A, D, , D, , E, A, , 76, , OF, , B, , B, , E, F′, O′, E′ D, F, , C, , Ex. 24. The bisectors of the angles of a triangle meet in a point which is, equidistant from the sides of the triangle., Let the bisectors AD and BE intersect at O. Then O being in AD is, equidistant from AC and AB. (Why?) And O being in BE is equidistant, from BC and AB. Hence, O is equidistant from AC and BC, and therefore, in the bisector CF . (Why?), Ex. 25. The perpendicular bisectors of the sides of a triangle meet in a, point which is equidistant from the vertices of the triangle., Let the ⊥ bisectors EE 0 and DD0 intersect at O. Then O being in EE 0 is, equidistant from A and C. (Why?) And O being in DD0 is equidistant from, A and B. Hence, O is equidistant from B and C, and therefore is in the ⊥, bisector F F 0 . (Why?), Ex. 26. The perpendiculars from the vertices of a triangle to the opposite, sides meet in a point., Let the ⊥s be AH, BP , and CK. Through A, B, C suppose B 0 C 0 , A0 C 0 ,, 0 0, A B , drawn k to BC, AC, AB, respectively. Then AH is ⊥ to B 0 C 0 . (Why?), Now ABCB 0 and ACBC 0 are / /s (why?) and AB 0 = BC, and AC 0 = BC., (Why?) That is, A is the middle point of B 0 C 0 . In the same way, B and C are, the middle points of A0 C 0 and A0 B 0 , respectively. Therefore, AH, BP , and, CK are the ⊥ bisectors of the sides of the 4A0 B 0 C 0 . Hence, they meet in a, point. (Why?)
Page 87 :
EXERCISES., A, , C′, , C, , B′, , P, , K, , 77, , O, B, , C, , H, , G, , H, , O, , F, A′, , D, , A, , E, , B, , Ex. 27. The medians of a triangle meet in a point which is two thirds of, the distance from each vertex to the middle of the opposite side., Let the two medians AD and CE meet in O. Take F the middle point of, OA, and G of OC. Join GF , F E, ED, and DG. In 4AOC, GF is k to AC, and equal to 21 AC. (Why?) DE is k to AC and equal to 12 AC. (Why?) Hence,, DGF E is a / /. (Why?) Hence, AF = F O = OD, and CG = GO = OE., (Why?) Hence, any median cuts off on any other median two thirds of the, distance from the vertex to the middle of the opposite side. Therefore, the, median from B will cut off AO, two thirds of AD; that is, will pass through, O., Note. If three or more lines pass through the same point, they are called concurrent lines., , Ex. 28. If an angle is bisected, and if a line is drawn through the vertex, perpendicular to the bisector, this line forms equal angles with the sides of the, given angle., C, , D, , B, , F, , E, , A, , C, , C, , C, , F, , D, , A, , E, , B, , F, , F, D, G, , A, E, , B, , D, G, , A, , B, , E, , Ex. 29. The bisectors of two supplementary adjacent angles are perpendicular to each other., Ex. 30. If the bisectors of two adjacent angles are perpendicular to each, other, the adjacent angles are supplementary., Ex. 31. The bisector of one of two vertical angles bisects the other., Ex. 32. The bisectors of two vertical angles form one line.
Page 88 :
BOOK I. PLANE GEOMETRY., , 78, , Ex. 33. The bisectors of the two pairs of vertical angles formed by two, intersecting lines are perpendicular to each other., Ex. 34. The bisector of the vertical angle of an isosceles triangle bisects, the base, and is perpendicular to the base., 4ADC = 4BDC (§ 143), E, C, , C, , A, , D, , B, , A, , D, , B, , Ex. 35. The perpendicular bisector of the base of an isosceles triangle, passes through the vertex and bisects the angle at the vertex (§ 160)., Ex. 36. If the perpendicular bisector of the base of a triangle passes, through the vertex, the triangle is isosceles., Ex. 37. Any point in the bisector of the vertical angle of an isosceles, triangle is equidistant from the extremities of the base (Ex. 34, § 160)., Ex. 38. If the bisector of an angle of a triangle is perpendicular to the, opposite side, the triangle is isosceles., Ex. 39. If two isosceles triangles are on the same base, a straight line, passing through their vertices is perpendicular to the base, and bisects the, base (§ 161)., Ex. 40. Two isosceles triangles are equal when a side and an angle of the, one are equal, respectively, to the homologous side and angle of the other.
Page 89 :
EXERCISES., , 79, , D, E, , A, , B, , C, , Ex. 41. The bisector of an exterior angle of an isosceles triangle, formed, by producing one of the legs through the vertex, is parallel to the base. Why, does ∠DAC = ∠B + ∠C? Why is ∠DAE = ∠ABC? Why is AE k to BC?, Ex. 42. If the bisector of an exterior angle of a triangle is parallel to one, side, the triangle is isosceles., D, , C, , A, , B, , Ex. 43. If one of the legs of an isosceles triangle is produced through the, vertex by its own length, the line joining the end of the leg produced to the, nearer end of the base is perpendicular to the base., ∠CBA = ∠A, and ∠CBD = ∠D. (Why?), ∴ ∠ABD = ∠A + ∠D., Ex. 44. A line drawn from the vertex of the right angle of a right triangle, to the middle point of the hypotenuse divides the triangle into two isosceles, triangles., Ex. 45. If the equal sides of an isosceles triangle are produced through, the vertex so that the external segments are equal, the extremities of these, segments will be equally distant from the extremities of the base, respectively.
Page 90 :
BOOK I. PLANE GEOMETRY., A, , 80, , D, , H, , K, , P, , B, , C, , Ex. 46. If through any point in the bisector of an angle a line is drawn, parallel to either of the sides of the angle, the triangle thus formed is isosceles., Ex. 47. Through any point C in the line AB an intersecting line is drawn,, and from any two points in this line equidistant from C perpendiculars are, dropped on AB or AB produced. Prove that these perpendiculars are equal., Ex. 48. If the median drawn from the vertex of a triangle to the base is, equal to half the base, the vertical angle is a right angle., C, , F, A, , E, D, , B, , Ex. 49. The lines joining the middle points of the sides of a triangle divide, the triangle into four equal triangles., Ex. 50. The altitudes upon the legs of an isosceles triangle are equal., Rt. 4BEC = rt. 4CDB (§ 141)., Ex. 51. If the altitudes upon two sides of a triangle are equal, the triangle, is isosceles., Rt. 4BEC = rt. 4CDB (§ 151).
Page 91 :
EXERCISES., , A, , 81, A, , A, A, , E, , E, , D, , O, , E, , D, , D, , D, , H, , E, , K, , F, G, , B, , C, , B, , C, , B, , C, , B, , C, , Ex. 52. The medians drawn to the legs of an isosceles triangle are equal., 4BEC = 4CDB (§ 143)., Ex. 53. If the medians to two sides of a triangle are equal, the triangle is, isosceles., BO = CO, and OE = OD (Ex. 27)., ∠BOE = ∠COD., , ∴ 4BOE = 4COD (§ 143)., , Ex. 54. The bisectors of the base angles of an isosceles triangle are equal., 4BEC = 4CDB (§ 139)., Ex. 55. Opposite Theorem. If a triangle is not isosceles, the bisectors, of the base angles are not equal., Let ∠ABC be greater than ∠ACB; then KC > KB. (Why?), Now CD > BE, if KD is greater than or equal to KE., But suppose KD < KE. Lay off KH = KD and KG = KB, join HG,, and draw GF k to BE., 4KDB = 4KHG. (Why?) ∴ ∠KHG = ∠KDB. (Why?), ∴ ∠KEC is greater than ∠KHG. (Why?) ∴ GF > HE. (Why?), ∠GF C is greater than ∠F CG ( 21 ACB). ∴ CG > GF , and > HE., ∴ KC − KG > KE − KH, or KC + KD > KB + KE, or CD > BE., Ex. 56. State the converse theorem of Ex. 54. Is the converse theorem, true?
Page 92 :
BOOK I. PLANE GEOMETRY., , 82, , A, , E, B, , F, D, , C, , Ex. 57. The perpendiculars dropped from the middle point of the base, upon the legs of an isosceles triangle are equal., 4BED = 4CF D (§ 141)., Ex. 58. State and prove the converse., 4BED = 4CF D (§ 151)., Ex. 59. The difference of the distances from any point in the base produced, of an isosceles triangle to the equal sides of the triangle is constant., Rt. 4DGC = rt. 4DF C. (Why?) ∴ DF = DG., ∴ DE − DF = DE − DG = EG, the ⊥ distance between the two ks , BA, and CH., C, , A, , H, , E, G, B, , C, F, , F, , D, , G, , E, A, , P, , D, B, , Ex. 60. The sum of the perpendiculars dropped from any point in the base, of an isosceles triangle to the legs is constant, and equal to the altitude upon, one of the legs., Let P E and P D be the ⊥s and BF the altitude., Draw P G ⊥ to BF ., EP GF is a parallelogram. (Why?) ∴ GF = P E. It remains to prove, GB = P D., The rt. 4P GB = the rt. 4BDP . (Why?)
Page 93 :
EXERCISES., , 83, , Ex. 61. The sum of the perpendiculars dropped from any point within an, equilateral triangle to the three sides is constant, and equal to the altitude., AD is the altitude, P E, P G, and P F the three perpendiculars. Through, P draw HK k to BC, meeting AD at M ., Then, M D = P E. (Why?), P G + P F = AM (Ex. 60)., A, D, G, H, B, , C, , F, P, , K, , M, , ED, , C, , A, , B, , Ex. 62. ABC and ABD are two triangles on the same base AB, and on, the same side of it, the vertex of each triangle being without the other. If AC, equals AD, show that BC cannot equal BD (§ 154)., Ex. 63. The sum of the lines which join a point within a triangle to the, three vertices is less than the perimeter, but greater than half the perimeter., C, , A, F, E, , O, A, , B, , B, , D, , C, , Ex. 64. If from any point in the base of an isosceles triangle parallels, to the legs are drawn, a parallelogram is formed whose perimeter is constant,, being equal to the sum of the legs of the triangle., Ex. 65. The bisector of the vertical angle A of a triangle ABC, and the, bisectors of the exterior angles at the base formed by producing the sides AB, and AC, meet in a point which is equidistant from the base and the sides, produced (§ 162).
Page 94 :
BOOK I. PLANE GEOMETRY., , 84, , A, A, B, , M, , C, D, F, , O, , E, E, , N, , D, C, , B, , Ex. 66. If the bisectors of the base angles of a triangle are drawn, and, through their point of intersection a line is drawn parallel to the base, the, length of this parallel between the sides is equal to the sum of the segments of, the sides between the parallel and the base., ∠EOB = ∠OBC = ∠OBE., , ∴ BE = EO., , Ex. 67. The bisector of the vertical angle of a triangle makes with the, perpendicular from the vertex to the base an angle equal to half the difference, of the base angles., Let ∠B be greater than ∠A., ∠DCE = 90◦ − ∠A − ∠ACD., ∠ACD = 90◦ − 21 ∠A − 12 ∠B., ∴ ∠DCE = 90◦ − ∠A − (90◦ − 21 ∠A − 21 ∠B) = 12 ∠B − 12 ∠A., C, , D, , C, , O, A, , D, , E, , B, , A, , B, , Ex. 68. If the diagonals of a quadrilateral bisect each other, the figure is, a parallelogram., Prove 4AOB = 4COD.
Page 95 :
EXERCISES., , 85, , A, , B, , D, , C, , Ex. 69. The diagonals of a rectangle are equal., Prove 4ABC = 4BAD., Ex. 70., rectangle., , If the diagonals of a parallelogram are equal, the figure is a, , Ex. 71. The diagonals of a rhombus are perpendicular to each other, and, bisect the angles of the rhombus., Ex. 72. The diagonals of a square are perpendicular to each other, and, bisect the angles of the square., C, , D, , E, , N, , F, , M, A, , B, , Ex. 73. Lines from two opposite vertices of a parallelogram to the middle, points of the opposite sides trisect the diagonal., EBF D is a, , / /, , (why?), and DF is k to EB., , AM = M N , and M N = CN (§ 188)., Ex. 74. The lines joining the middle points of the sides of any quadrilateral,, taken in order, enclose a parallelogram., Prove HG and EF k to AC; and F G and EH k to BD (§ 189)., Then HG and EF are each equal to 12 AC.
Page 96 :
BOOK I. PLANE GEOMETRY., , 86, , C, D, , G, , C, , F, , H, A, , G, , F, , D, , E, , B, , G, , C, , B, , O, H, , D, H, , E, A, , F, , A, , E, , B, , G, , D, , H, , C, , F, , A, , E, , B, , Ex. 75. The lines joining the middle points of the sides of a rhombus,, taken in order, enclose a rectangle. (Proof similar to that of Ex. 74.), Ex. 76. The lines joining the middle points of the sides of a rectangle, (not a square), taken in order, enclose a rhombus., Ex. 77. The lines joining the middle points of the sides of a square, taken, in order, enclose a square., Ex. 78. The lines joining the middle points of the sides of an isosceles, trapezoid, taken in order, enclose a rhombus or a square., SHR and QF P drawn ⊥ to AB are parallel. ∴ P QSR is a / /, and by, Const. is a rectangle or a square., ∴ EF GH is a rhombus or a square (Exs. 76, 77)., S D, , G, , H, , C Q, , F, , D, , C, L, E, , A R, , E, , P B, , D, , A, , F, , L, , B, , A, , F, , C, , A, , K, , E, , K, , D, , C, , B, , B, , Ex. 79. The bisectors of the angles of a rhomboid enclose a rectangle., Ex. 80. The bisectors of the angles of a rectangle enclose a square., Ex. 81. If two parallel lines are cut by a transversal, the bisectors of the, interior angles form a rectangle.
Page 97 :
EXERCISES., D, , C, , G, , E, , 87, , H, , F, , A, , B, , Ex. 82. The median of a trapezoid passes through the middle points of, the two diagonals., The median EF is k to AB and bisects AD (§ 190)., ∴ it bisects DB., Likewise EF bisects BC and BD., Ex. 83. The lines joining the middle points of the diagonals of a trapezoid, is equal to half the difference of the bases., 4BF G = 4DF C. (Why?) ∴ EF = 12 AG (§ 180)., CF = F G, DC = BG., D, , C, , E, , A, , ∴ AG = AB − DC. ∴ EF = 12 (AB − DC), , G, , C, , F, , B, , A, , D, , D, , E, , B, , A, , C, , D, , B, , C, , A, , Ex. 84. In an isosceles trapezoid each base makes equal angles with the, legs., Draw CE k to DB. CE = DB. (Why?) ∠A = ∠CEA, ∠B = ∠CEA,, ∠s C and D have equal supplements., Ex. 85. If the angles at the base of a trapezoid are equal, the other angles, are equal, and the trapezoid is isosceles., Ex. 86. In an isosceles trapezoid the opposite angles are supplementary:, ∠C = ∠D (Ex. 84), , B
Page 98 :
BOOK I. PLANE GEOMETRY., , 88, , Ex. 87. The diagonals on an isosceles trapezoidal are equal., Prove 4ACD = 4BDC., Ex. 88. If the diagonals of a trapezoid are equal, the trapezoid is isosceles., Draw CE and DF ⊥ to AB., 4ADF = 4BCE., , (Why?), , ∴ ∠ADF = ∠CBA., 4ABC = 4BAD., , C, , D, , D, , F, , C, , O, E, , A, , E, , F, , B, , A, , B, , Ex. 89. If from the diagonal DB, of a square ABCD, BE is cut off equal, to BC, and EF is drawn perpendicular to BD meeting DC at F , then DE is, equal to EF and also to F C., ∠EDF = 45◦ , and ∠DF E = 45◦ ; and DE = DF . Rt. 4BEF =, rt. 4BCF (§ 151); and EF = F C., Ex. 90. Two angles whose sides are so perpendicular, each to each, are, either equal or supplementary.
Page 99 :
BOOK II. THE CIRCLE., , DEFINITIONS., 216. A circle is a portion of a plane bounded by a curved line, all points of, which are equally distant from a point within called the centre. The bounding, line is called the circumference of the circle., 217. A radius is a straight line from the centre to the circumference;, and a diameter is a straight line through the centre, with its ends in the, circumference., By the definition of a circle, all its radii are equal. All its diameters are, equal, since a diameter is equal to two radii., 218. Postulate. A circumference can be described from any point as a, centre, with any given radius., 219. A secant is a straight line of unlimited length which intersects the, circumference in two points; as, AD (Fig. 1)., A, , D, , B, , C, Fig. 1., , 220. A tangent is a straight line of unlimited length which has one point,, and only one, in common with the circumference; as, BC (Fig. 1). In this, case the circle is said to be tangent to the straight line. The common point is, called the point of contact, or point of tangency., 221. Two circles are tangent to each other, if both are tangent to a straight, line at the same point; and are said to be tangent internally or externally,, according as one circle lies wholly within or without the other.
Page 100 :
BOOK II. PLANE GEOMETRY., , 90, , 222. An arc is any part of the circumference; as, BC (Fig. 3). Half a, circumference is called a semicircumference. Two arcs are called conjugate, arcs, if their sum is a circumference., 223. A chord is a straight line that has its extremities in the circumference;, as, the straight line BC (Fig. 3)., 224. A chord subtends two conjugate arcs. If the arcs are unequal, the, less is called the minor arc, and the greater the major arc. A minor arc is, generally called simply an arc., A, , D, , SEGMENT, , C, SE, T O, N, RA, AD, QU SEMICIRCLE, , N, Fig. 2., , C, , B, , B, , B, , C, , D, , R, TO, , C, , A, D, , A, E, Fig. 3., , E, , H, , F, G, Fig. 4., , 225. A segment of a circle is a portion of the circle bounded by an arc, and its chord (Fig. 2)., 226. A semicircle is a segment equal to half the circle (Fig. 2)., 227. A sector of a circle is a portion of the circle bounded by two radii, and the arc which they intercept. The angle included by the radii is called the, angle of the sector (Fig. 2)., 228. A quadrant is a sector equal to a quarter of the circle (Fig. 2)., 229. An angle is called a central angle, if its vertex is at the centre and, its sides are radii of the circle; as, ∠AOD (Fig. 2)., 230. An angle is called an inscribed angle, if its vertex is in the circumference and its sides are chords; as, ∠ABC (Fig. 3)., An angle is inscribed in a segment, if its vertex is in the arc of the segment, and its sides pass through the extremities of the arc.
Page 101 :
ARCS, CHORDS, AND TANGENTS., , 91, , 231. A polygon is inscribed in a circle, if its sides are chords; and a circle, is circumscribed about a polygon, if all the vertices of the polygon are in the, circumference (Fig. 3)., 232. A circle is inscribed in a polygon, if the sides of the polygon are, tangent to the circle; and a polygon is circumscribed about a circle if its sides, are tangents (Fig. 4)., 233. Two circles are equal, if they have equal radii., For they will coincide, if their centres are made to coincide., Conversely: Two equal circles have equal radii., 234. Two circles are concentric, if they have the same centre., ARCS, CHORDS, AND TANGENTS., Proposition I. Theorem., 235. A straight line cannot meet the circumference of a circle in more than, two points., M, , O, , H, , K, P, , Let HK be any line meeting the circumference HKM in H and K., To prove that HK cannot meet the circumference in any other point., Proof. If possible, let HK meet the circumference in P ., Then the radii OH, OP , and OK are equal., , § 217, , ∴ P does not lie in the straight line HK., , § 102, , ∴ HK meets the circumference in only two points., , q.e.d.
Page 102 :
BOOK II. PLANE GEOMETRY., , 92, , Proposition II. Theorem., 236. In the same circle or in equal circles, equal central angles intercept, equal arcs; and of two unequal central angles the greater intercepts the greater, arc., C, , P, , O, , B′, , B, , A, , O′, , A′, , P′, , In the equal circles whose centres are O and O 0 , let the angles AOB, and A0 O 0 B 0 be equal, and angle AOC be greater than angle A0 O 0 C 0 ., To prove that 1. arc AB = arc A0 B 0 ;, 2. arc AC > arc A0 B 0 ., Proof. 1. Place the A0 B 0 P 0 on the, with its equal, the ∠AOB., , ABP so that the ∠A0 O0 B 0 shall coincide, , Then A0 falls on A, and B 0 on B., , § 233, , ∴ arc A0 B 0 coincides with arc AB., § 216, 2. Since the ∠AOC is greater than the ∠A0 O0 B 0 , it is greater than the ∠AOB,, the equal of the ∠A0 O0 B 0 ., Therefore, OC falls without the ∠AOB., ∴ arc AC > arc AB., , Ax. 8, , ∴ arc AC > arc A0 B 0 , the equal of arc AB., , q.e.d.
Page 103 :
ARCS, CHORDS, AND TANGENTS., , 93, , 237. Conversely: In the same circle or in equal circles, equal arcs subtend equal central angles; and of two unequal arcs the greater subtends the, greater central angle., To prove that, 1. ∠AOB = ∠A0 O0 B 0 ;, 2. ∠AOC is greater than ∠A0 O0 B 0 ., Proof. 1. Place the A B 0 P 0 on the ABP so that O0 A shall fall on its, equal OA, and the arc A0 B 0 on its equal AB., 0, , Then O0 B 0 will coincide with OB., , § 47, , ∴ ∠A0 O0 B 0 = ∠AOB., § 60, 0 0, 0 0, 2. Since arc AC > A B , it is greater than arc AB, the equal of A B , and, OB will fall within the ∠AOC., ∴ ∠AOC is greater than ∠AOB., , Ax. 8, , ∴ ∠AOC is greater than ∠A0 O0 B 0 ., , q.e.d., , 238. Cor. 1. In the same circle or in equal circles, two sectors that have, equal angles are equal; two sectors that have unequal angles are unequal, and, the greater sector has the greater angle., 239. Cor. 2. In the same circle or in equal circles, equal sectors have equal, angles; and of two unequal sectors the greater has the greater angle.
Page 104 :
BOOK II. PLANE GEOMETRY., , 94, , 240. Law of Converse Theorems. It was stated in § 32 that the converse, of a theorem is not necessarily true. If, however, a theorem is in fact a group, of three theorems, and if one of the hypotheses of the group must be true, and, no two of the conclusions can be true at the same time, then the converse of, the theorem is necessarily true., Proposition II. is a group of three theorems. It asserts that the arc AB, is equal to the arc A0 B 0 , if the angle AOB is equal to the angle A0 O0 B 0 ; that, the arc AB is greater than the arc A0 B 0 , if the angle AOB is greater than the, angle A0 O0 B 0 ; that the arc AB is less than the arc A0 B 0 , if the angle AOB is, less than the angle A0 O0 B 0 ., One of these hypotheses must be true; for the angle AOB must be equal, to, greater than, or less than, the angle A0 O0 B 0 ., No two of the conclusions can be true at the same time, for the arc AB, cannot be both equal to and greater than the arc A0 B 0 ; nor can it be both, equal to and less than the arc A0 B 0 ; nor both greater than and less than the, arc A0 B 0 . In such a case, the converse theorem is necessarily true, and no, proof like that given in the text is required to establish it.
Page 105 :
ARCS, CHORDS, AND TANGENTS., , 95, , Proposition III. Theorem., 241. In the same circle or in equal circles, equal arcs are subtended by equal, chords; and of two unequal arcs the greater is subtended by the greater chord., F, , B′, , B, , A, , O, , O′, , A′, , In the equal circles whose centres are O and O 0 , let the arcs AB and, be equal, and the arc AF greater than arc A0 B 0 ., To prove that, 1. chord AB = chord A0 B 0 ;, , A0 B 0, , 2. chord AF > chord A0 B 0 ., Proof., 1., , Draw the radii OA, OB, OF , O0 A0 , O0 B 0 ., , (in equal, , 2., , The 4s AOB and A0 O0 B 0 are equal., , § 143, , For OA = O0 A0 , and OB = O0 B 0 ,, (radii of equal circles),, , § 233, , s, , and ∠AOB = ∠A0 O0 B 0 ,, equal arcs subtend equal central ∠s )., , § 237, , ∴ chord AB = chord A0 B 0 ., , § 128, , In the 4s AOF and A0 O0 B 0 ,, OA = O0 A0 , and OF = O0 B 0 ., , (in equal, , § 233, , But the ∠AOF is greater than the ∠A0 O0 B 0 ,, § 237, s , the greater of two unequal arcs subtends the greater ∠)., ∴ chord AF > chord A0 B 0 ., , § 154, q.e.d.
Page 106 :
BOOK II. PLANE GEOMETRY., , 96, , 242. Cor. In the same circle or in equal circles, the greater of two unequal, major arcs is subtended by the less chord.
Page 107 :
ARCS, CHORDS, AND TANGENTS., , 97, , Proposition IV. Theorem., 243. Conversely: In the same circle or in equal circles, equal chords, subtend equal arcs; and of two unequal chords the greater subtends the greater, arc., F, , B′, , B, , A, , O, , O′, , A′, , In the equal circles whose centres are O and O 0 , let the chords AB, and A0 B 0 be equal, and the chord AF greater than A0 B 0 ., To prove that, , 1. arc AB = arc A0 B 0 ;, 2. arc AF > arc A0 B 0 ., , Proof., , Draw the radii OA, OB, OF , O0 A0 , O0 B 0 ., , 1., , The 4s OAB and O0 A0 B 0 are equal., , § 150, , For OA = O0 A0 , and OB = O0 B 0 ,, , § 233, , and chord AB = chord, , A0 B 0 ., , ∴ ∠AOB = ∠A0 O0 B 0 ., , 2., , (in equal, , s, , ∴ arc AB = arc A0 B 0 ,, equal central ∠s intercept equal arcs)., , Hyp., § 128, § 236, , In the 4s OAF andO0 A0 B 0 ,, , (in equal, , OA = O0 A0 and OF = O0 B 0 ., , § 233, , But chord AF > chord A0 B 0 ., , Hyp., , ∴ the ∠AOF is greater than the ∠A0 O0 B 0 ., , § 155, , s, , ∴ arc AF > arc A0 B 0 ,, the greater central ∠ intercepts the greater arc)., , § 236, q.e.d.
Page 108 :
BOOK II. PLANE GEOMETRY., , 98, , 244. Cor. In the same circle or in equal circles, the greater of two unequal, chords subtends the less major arc., Proposition V. Theorem., 245. A diameter perpendicular to a chord bisects the chord and the arcs, subtended by it., E, , O, , A, , M, , B, , S, , Let ES be a diameter perpendicular to the chord AB at M ., To prove that AM = BM , AS = BS, and AE = BE., Proof. Draw OA and OB from O, the centre of the circle., For, and, , Likewise, , The rt. 4s OAM and OBM are equal., , § 151, , OM = OM ,, , Iden., , OA = OB., , § 217, , ∴ AM = BM , and ∠AOS = ∠BOS., , § 128, , ∠AOE = ∠BOE., , § 85, , ∴ AS = BS, and AE = BE., , § 236, q.e.d., , 246. Cor. 1. A diameter bisects the circumference and the circle., 247. Cor. 2. A diameter which bisects a chord is perpendicular to it., 248. Cor. 3. The perpendicular bisector of a chord passes through the, centre of the circle, and bisects the arcs of the chord.
Page 109 :
ARCS, CHORDS, AND TANGENTS., , 99, , Proposition VI. Theorem., 249. In the same circle or in equal circles, equal chords are equally distant, from the centre. Conversely: Chords equally distant from the centre are, equal., A, , B, , P, , O, , C, , H, , F, , Let AB and CF be equal chords of the circle ABF C., To prove that AB and CF are equidistant from the centre O., Proof. Draw OP ⊥ to AB, OH ⊥ to CF , and join OA and OC., , and, Hence,, , OP bisects AB, and OH bisects CF ., , § 245, , The rt. 4s OP A and OHC are equal., , § 151, , AP = CH,, , Ax. 7, , OA = OC,, , § 217, , OP = OH., , § 128, , ∴ AB and CF are equidistant from O., Conversely:, To prove, , Let OP = OH., , AB = CF ., Proof. The rt. 4s OP A and OHC are equal., For, OA = OC,, and, OP = OH,, Hence,, AP = CH., , § 151, § 217, Hyp., § 128
Page 110 :
BOOK II. PLANE GEOMETRY., , ∴ AB = CF ., , 100, , Ax. 6, q.e.d., , Proposition VII. Theorem., 250. In the same circle or in equal circles, if two chords are unequal, they, are unequally distant from the centre; and the greater chord is at the less, distance., G, , C, , F, H, , A, , O, , E, , D, B, , In the circle whose centre is O, let the chords AB and CD be unequal, and AB the greater; and let OE be perpendicular to AB and OF, perpendicular to CD., To prove that, OE < OF ., Proof. Suppose AG drawn equal to CD, and OH ⊥ to AG., Draw EH., By hypothesis,, , OE bisects AB, and OH bisects AG., , § 245, , AB > CD., ∴ AB > AG, the equal of CD., ∴ AE > AH., , Ax. 7, , ∴ ∠AHE is greater than ∠AEH., § 152, ∴ ∠OHE, the complement of ∠AHE, is less than ∠OEH, the complement of, ∠AEH., Ax. 5, But, , ∴ OE < OH., , § 153, , OH = OF ., , § 249, , ∴ OE < OF ., , q.e.d.
Page 111 :
ARCS, CHORDS, AND TANGENTS., , 101, , Ex. 91. The perpendicular bisectors of the sides of an inscribed polygon, are concurrent (pass through the same point)., Proposition VIII. Theorem., 251. Conversely: In the same circle or in equal circles, if two chords, are unequally distant from the centre, they are unequal; and the chord at the, less distance is the greater., G, , C, , F, H, , A, , O, , E, , D, B, , In the circle whose centre is O, let AB and CD be unequally distant, from O; and let OE, the perpendicular to AB, be less than OF , the, perpendicular to CD., To prove that, AB > CD., Proof. Suppose AG drawn equal to CD, and OH ⊥ to AG., Then, OH = OF, § 249, Hence,, OE < OH., Draw EH., ∠OHE is less than ∠OEH., § 152, ∴ ∠AHE, the complement of ∠OHE, is greater than ∠AEH, the complement, of ∠OEH., Ax. 5, But, , But, , ∴ AE > AH., , § 153, , AE = 21 AB, and AH = 12 AG., , § 245, , ∴ AB > AG., , Ax. 6, , CD = AG., , Const., , ∴ AB > CD., , q.e.d.
Page 112 :
BOOK II. PLANE GEOMETRY., , 102, , 252. Cor. A diameter of a circle is greater than any other chord., Proposition IX. Theorem., 253. A straight line perpendicular to a radius at its extremity is a tangent, to the circle., , O, , C, M, , H, , A, , B, , Let M B be perpendicular to the radius OA at A., To prove that M B is a tangent to the circle., Proof. From O draw any other line to M B, as OH., Then, OH > OA., , § 97, , ∴ the point H is without the circle., § 216, Hence, every point, except A, of the line M B is without the circle, and therefore, M B is a tangent to the circle at A., § 220, q.e.d., , 254. Cor. 1. A tangent to a circle is perpendicular to the radius drawn to, the point of contact., For OA is the shortest line from O to M B, and is therefore ⊥ to M B, (§ 98); that is, M B is ⊥ to OA., 255. Cor. 2. A perpendicular to a tangent at the point of contact passes, through the centre of the circle., For a radius is ⊥ to a tangent at the point of contact, and therefore a ⊥, erected at the point of contact coincides with this radius and passes through, the centre.
Page 113 :
ARCS, CHORDS, AND TANGENTS., , 103, , 256. Cor. 3. A perpendicular from the centre of a circle to a tangent, passes through the point of contact.
Page 114 :
BOOK II. PLANE GEOMETRY., , 104, , Proposition X. Theorem., 257. Parallels intercept equal arcs on a circumference., F, , A, C, , A′, , B, D, , F′, Fig. 1., , E, , M, , F, , C, , D, , A, , B, , B′, , A, , E, , B, , G, , C, Fig. 2., , H, , D, , F, Fig. 3., , Case 1. Let AB (Fig. 1) be a tangent at F parallel to CD, a secant., To prove that, arc CF = arc DF ., Proof., Suppose F F 0 drawn ⊥ to AB., Then F F 0 is a diameter of the circle., And, , FF0, , § 255, , is also ⊥ to CD., , § 107, , ∴ CF = DF , and CF 0 = DF 0 ., Case 2. Let AB and CD (Fig. 2) be parallel secants., To prove that, arc AC = arc BD., Proof., Suppose EF k to CD and tangent to the circle at M ., Then, arc AM = arc BM ,, and, arc CM = arc DM ., , § 245, , Case 1, , ∴ arc AC = arc BD., Case 3. Let AB and CD (Fig. 3) be parallel tangents at E and F ., To prove that, arc EGF = arc EHF ., Proof., Suppose GH drawn k to AB., , Ax. 3
Page 115 :
ARCS, CHORDS, AND TANGENTS., Then, , arc EG = arc EH,, , and, , 105, , Case 1, , arc GF = arc HF ., ∴ arc EGF = arc EHF ., , Ax. 2, q.e.d., , Proposition XI. Theorem., 258. Through three points not in a straight line one circumference, and, only one, can be drawn., , C, O, , A, , B, , Let A, B, C be three points not in a straight line., To prove that one circumference, and only one, can be drawn through A, B,, and C., Proof., Draw AB and BC., At the middle points of AB and BC suppose ⊥s erected., These ⊥s will intersect at some point O, since AB and BC are not in the same, straight line., The point O is in the perpendicular bisector of AB, and is therefore equidistant, from A and B; the point O is also in the perpendicular bisector of BC, and is, therefore equidistant from B and C., § 160, Therefore, O is equidistant from A, B, and C; and a circumference described, from O as a centre, with a radius OA, will pass through the three given points., The centre of a circumference passing through the three points must be in both, perpendiculars, and hence at their intersection. As two straight lines can intersect, in only one point, O is the centre of the only circumference that can pass through, the three given points., q.e.d.
Page 116 :
BOOK II. PLANE GEOMETRY., , 106, , 259. Cor. Two circumferences can intersect in only two points. For, if, two circumferences have three points common, they coincide and form one, circumference., 260. Def. A tangent from an external point to a circle is the part, of the tangent between the external point and the point of contact., Proposition XII. Theorem., 261. The tangents to a circle drawn from an external point are equal, and, make equal angles with the line joining the point to the centre., B, , A, , O, , C, , Let AB and AC be tangents from A to the circle whose centre is O,, and let AO be the line joining A to the centre O., To prove that AB = AC, and ∠BAO = ∠CAO., Proof., Draw OB and OC., AB is ⊥ to OB, and AC ⊥ to OC,, § 254, (a tangent to a circle is ⊥ to the radius drawn to the point of contact)., The rt. 4s OAB and OAC are equal., For OA is common, and the radii OB and OC are equal., ∴ AB = AC, and ∠BAO = ∠CAO., , § 151, § 217, § 128, q.e.d., , 262. Def. The line joining the centres of two circles is called the line of, centres., 263. Def. A tangent to two circles is called a common external tangent, if it does not cut the line of centres, and a common internal tangent if it, cuts the line of centres.
Page 117 :
ARCS, CHORDS, AND TANGENTS., , 107, , Proposition XIII. Theorem., 264. If two circles intersect each other, the line of centres is perpendicular, to their common chord at its middle point., A, C′, , C, B, , Let C and C 0 be the centres of the two circles, AB the common chord,, and CC 0 the line of centres., To prove that CC 0 is ⊥ to AB at its middle point., Proof., Draw CA, CB, C 0 A, and C 0 B., CA = CB, and C 0 A = C 0 B., , § 217, , ∴ C and C 0 are two points, each equidistant from A and B., ∴ CC 0 is the perpendicular bisector of AB., , § 161, q.e.d., , Ex. 92. Describe the relative position of two circles if the line of centres:, 1. is greater than the sum of the radii;, 2. is equal to the sum of the radii;, 3. is less than the sum but greater than the difference of the radii;, 4. is equal to the difference of the radii;, 5. is less than the difference of the radii., Illustrate each case by a figure., Ex. 93. The straight line drawn from the middle point of a chord to the, middle point of its subtended arc is perpendicular to the chord., Ex. 94. The line which passes through the middle points of two parallel, chords passes through the centre of the circle.
Page 118 :
BOOK II. PLANE GEOMETRY., , 108, , Proposition XIV. Theorem., 265. If two circles are tangent to each other, the line of centres passes, through the point of contact., A, , C, , O, , C′, , B, , Let the two circles, whose centres are C and C 0 , be tangent to the, straight line AB at Q, and CC 0 the line of centres., To prove that O is in the straight line CC 0 ., Proof. A ⊥ to AB, drawn through the point O, passes through the centres C, and C 0 ,, § 255, (a ⊥ to a tangent at the point of contact passes through the centre of the circle)., ∴ the line CC 0 , having two points in common with this ⊥ must coincide with, it., § 47, ∴ O is in the straight line CC 0 ., , q.e.d., , Ex. 95. Describe the relative position of two circles if they may have:, 1. two common external and two common internal tangents;, 2. two common external tangents and one common internal tangent;, 3. two common external tangents and no common internal tangent;, 4. one common external and no common internal tangent;, 5. no common tangent., Illustrate each case by a figure., Ex. 96. The line drawn from the centre of a circle to the point of intersection of the two tangents is the perpendicular bisector of the chord joining, the points of contact.
Page 119 :
MEASUREMENT., , 109, , MEASUREMENT., 266. To measure a quantity of any kind is to find the number of times it, contains a known quantity of the same kind, called the unit of measure., The number which shows the number of times a quantity contains the unit, of measure is called the numerical measure of that quantity., 267. No quantity is great or small except by comparison with another, quantity of the same kind. This comparison is made by finding the numerical, measures of the two quantities in terms of a common unit, and then dividing, one of the measures by the other., The quotient is called their ratio. In other words the ratio of two quantities, of the same kind is the ratio of their numerical measures expressed in terms, of a common unit., a, The ratio of a to b is written a : b, or ., b, 268. Two quantities that can be expressed in integers in terms of a common, unit are said to be commensurable, and the exact value of their ratio can be, found. The common unit is called their common measure, and each quantity, is called a multiple of this common measure., Thus, a common measure of 2 21 feet and 3 23 feet is 16 of a foot, which is, contained 15 times in 2 21 feet, and 22 times in 3 23 feet. Hence, 2 12 feet and, 3 23 feet are multiples of 16 of a foot, since 2 12 feet may be obtained by taking, 1, of a foot 15 times, and 3 23 feet by taking 61 of a foot 22 times. The ratio of, 6, 2 12 feet to 3 32 feet is expressed by the fraction 15, ., 22
Page 120 :
BOOK II. PLANE GEOMETRY., , 110, , 269. Two quantities of the same kind that cannot both be expressed in, integers in terms of a common unit, are said to be incommensurable, and the, exact value of their ratio cannot be found. But by taking the unit sufficiently, small, an approximate value can be found that shall differ from the true value, of the ratio by less than any assigned value, however small., a √, Thus, suppose the ratio, = 2., b, √, Now 2 = 1.41421356 · · · , a value greater than 1.414213, but less than, 1.414214., If, then, a millionth part of b is taken as the unit of measure, the value, a, of lies between 1.414213 and 1.414214, and therefore differs from either of, b, these values by less than 0.000001., By carrying the decimal further, an approximate value may be found that, will differ from the true value of the ratio by less than a billionth, a trillionth,, or any other assigned value., m, m+1, a, but <, , then the error in taking either of these, In general, if >, b, n, n, a, 1, values for is less than , the difference between these two fractions. But by, b, n, 1, increasing n indefinitely,, can be decreased indefinitely, and a value of the, n, ratio can be found within any required degree of accuracy., 270. The ratio of two incommensurable quantities is called an incommensurable ratio; and is a fixed value which its successive approximate values, constantly approach.
Page 121 :
THEORY OF LIMITS., , 111, , THE THEORY OF LIMITS., 271. When a quantity is regarded as having a fixed value throughout the, same discussion, it is called a constant; but when it is regarded, under the, conditions imposed upon it, as having different successive values, it is called, a variable., If a variable, by having different successive values, can be made to differ, from a given constant by less than any assigned value, however small, but, cannot be made absolutely equal to the constant, that constant is called the, limit of the variable, and the variable is said to approach the constant as, its limit., A, , M, , M′, , M ′′, , B, , 272. Suppose a point to move from A toward B, under the conditions that, the first second it shall move one half the distance from A to B, that is, to, M ; the next second, one half the remaining distance, that is, to M 0 ; and so, on indefinitely., Then it is evident that the moving point may approach as near to B as we, choose, but will never arrive at B. For, however near it may be to B at any, instant, the next second it will pass over half the distance still remaining; it, must, therefore, approach nearer to B, since half the distance still remaining, is some distance, but will not reach B, since half the distance still remaining, is not the whole distance., Hence, the distance from A to the moving point is an increasing variable,, which indefinitely approaches the constant AB as its limit; and the distance, from the moving point to B is a decreasing variable, which indefinitely approaches the constant zero as its limit.
Page 122 :
BOOK II. PLANE GEOMETRY., , 112, , E, A, , B, , K, , F, , C, , D, H, , 273. Again, suppose a square ABCD inscribed in a circle, and E, F , H, K, the middle points of the arcs subtended by the sides of the square. If we draw, the lines AE, EB, BF , etc., we shall have an inscribed polygon of double the, number of sides of the square., The length of the perimeter of this polygon, represented by the dotted, lines, is greater than that of the square, since two sides replace each side of, the square and form with it a triangle, and two sides of a triangle are together, greater than the third side; but less than the length of the circumference, for, it is made up of straight lines, each one of which is less than the part of the, circumference between its extremities.
Page 123 :
THEORY OF LIMITS., , 113, , By continually doubling the number of sides of each resulting inscribed, figure, the length of the perimeter will increase with the increase of the number, of sides, but will not become equal to the length of the circumference., The difference between the perimeter of the inscribed polygon and the, circumference of the circle can be made less than any assigned value, but, cannot be made equal to zero., The length of the circumference is, therefore, the limit of the length of, the perimeter as the number of sides of the inscribed figure is indefinitely, increased., § 271, 274. Consider the decimal 0.333 · · · which may be written, 3, 3, 3, + 100, + 1000, + ···, 10, The value of each fraction after the first is one tenth of the preceding, fraction, and by continuing the series we shall reach a fraction less than any, assigned value, that is, the values of the successive fractions approach zero as, a limit., The sum of these fractions is less than 13 ; but the more terms we take, the, nearer does the sum approach 31 as a limit., 275. Test for a limit. In order to prove that a variable approaches a, constant as a limit, it is necessary to prove that the difference between the, variable and the constant:, 1. Can be made less than any assigned value, however small., 2. Cannot be made absolutely equal to zero., 276. Theorem. If the limit of a variable x is zero, then the limit of kx,, the product of the variable by any finite constant k, is zero., 1. Let q be any assigned quantity, however small., q, Then is not 0. Hence x, which may differ as little as we please from 0,, k, q, may be taken less than , and then kx will be less than q., k, 2. Since x cannot be 0, kx cannot be 0., Therefore, the limit of kx = 0, , § 275
Page 124 :
BOOK II. PLANE GEOMETRY., , 114, , 277. Cor. If the limit of a variable x is zero, then the limit of the quotient, of the variable by any finite constant k, is also zero., x, 1, For = × x, which by § 276 can be made less than any assigned value,, k, k, however small, but cannot be made equal to zero., 278. Theorem. The limit of the sum of a finite number of variables, x, y, z, · · · is equal to the sum of their respective limits a, b, c, · · · ., Let d, d0 , d00 , · · · denote the differences between x, y, z, · · · and a, b, c, · · · ,, respectively. Then d + d0 + d00 + · · · can be made less than any assigned quantity q., For, if d, d0 , d00 , · · · are n in number and d is the largest,, d + d0 + d00 + · · · < nd., (1), Since d may be diminished at pleasure, we may make d so small that, q, d < ; and therefore nd < q., n, But by (1), d + d0 + d00 + · · · < nd, and therefore < q., Therefore, the difference between (x + y + z + · · · ) and (a + b + c + · · ·) can, be made less than any assigned quantity, but not zero., Therefore, the limit of (x + y + z + · · ·) = a + b + c + · · · ., § 275, 279. Theorem. If the limit of a variable x is not zero, and if k is any, finite constant, the limit of the product kx is equal to the limit of x multiplied, by k., 1. If a denotes the limit of x, then x cannot be equal to a., § 271, Therefore, kx cannot be equal to ka., 2. The limit of (a − x) = 0. Hence, the limit of ka − kx = 0., Therefore, the limit of kx = ka., , § 276, § 275, , 280. Cor. The limit of the quotient of a variable x by any finite constant, k is the limit of x divided by k., x, 1, the limit of x, 1, For = × x, and, = × the limit of x., k, k, k, k
Page 125 :
THEORY OF LIMITS., , 115, , 281. Theorem. The limit of the product of two or more variables is the, product of their respective limits, provided no one of these limits is zero., If x and y are variables, a and b their respective limits, we may put x = a−d,, y = b−d0 ; then d and d0 are variables which can be made less than any assigned, quantity, but not zero., § 275, Now,, xy = (a − d)(b − d0 ), = ab − ad0 − bd + dd0, ∴ ab − xy = ad0 + bd − dd0 ., Since every term on the right contains d or d0 , the whole right member can, be made less than any assigned quantity, but not zero., § 278, Hence, ab − xy can be made less than any assigned quantity, but not zero., Therefore, the limit of xy = ab., , § 275, , Similarly, for three or more variables., 282. Cor. 1. The limit of the nth power of a variable is the nth power of, its limit., For the limit of the product of the variables x, y, z, · · · to n factors is the, product of their respective limits, the constants a, b, c, · · · to n factors (§ 281)., If the n factors xyz · · · are each equal to x, and the n factors abc · · · are each, equal to a, we have xyz · · · = xn , and abc · · · = an ., Therefore, the limit of xn = an ., 283. Cor. 2. The limit of the nth root of a variable is the nth root of its, limit., For if the limit of x = a, we may put this in the following form,, √, √, the limit of n xn = n an ;, √, √, that is, the limit of n xxx · · · to n factors is n aaa · · · to n factors., Now, xxx · · · is a variable since each factor is a variable, and aaa · · · is a, constant since each factor is a constant., If we denote xxx · · · to n factors by the variable y, and aaa · · · to n factors, by the constant b, we have, p, p, the limit of n y = n b.
Page 126 :
BOOK II. PLANE GEOMETRY., , 116, , 284. Theorem. If two variables are constantly equal, and each approaches, a limit, the limits are equal., Let x and y be two variables, a and b their respective limits, d and d0, the respective differences between the variables and their limits. Then, if the, variables are increasing toward their limits, a = x + d, and b = y + d0 ., Since the equation x = y is always true, we have by subtraction, a − b = d − d0 ., Since a and b are constants, a − b is a constant; therefore, d − d0 , which is, equal to a − b, is a constant., But the only constant which is less than any assigned value is 0. Therefore,, d − d0 = 0. Therefore, a − b = 0, and a = b., If the variables x and y are decreasing toward their limits a and b, respectively, then, a = x − d and b = y − d0 ., Therefore, by subtraction, a − b = d0 − d., Therefore, by the same proof as for increasing variables, a = b., 285. Theorem. If two variables have a constant ratio, and each approaches, a limit that is not zero, the limits have the same ratio., Let x and y be two variables, a and b their respective limits., Let, x, = r, a constant; then x = ry., y, Since x and ry are two variables that are always equal,, the limit of x = the limit of ry., , § 284, , the limit of ry = r × limit of y., But the limit of x is a, and the limit of y is b., Therefore,, a, a = rb; that is, = r., b, , § 279, , Now,
Page 127 :
THEORY OF LIMITS., , 117, , Proposition XV. Problem., 286. To find the ratio of two straight lines., , E, , H, B, , A, K, C, , F, , D, , Let AB and CD be two straight lines., To find the ratio of AB and CD., Apply CD to AB as many times as possible., Suppose twice, with a remainder EB., Then apply EB to CD as many times as possible., Suppose three times, with a remainder F D., Then apply F D to EB as many times as possible., Suppose once, with a remainder HB., Then apply HB to F D as many times as possible., Suppose once, with a remainder KD., Then apply KD to HB as many times as possible., Then, , Suppose KD is contained just twice in HB., HB = 2KD;, F D = HB + KD = 3KD;, EB = F D + HB = 5KD;, CD = 3EB + F D = 18KD;, AB = 2CD + EB = 41KD;, AB, 41KD, 41, ∴, =, = ., CD, 18KD, 18, , q.e.f.
Page 128 :
BOOK II. PLANE GEOMETRY., , 118, , Note. By the same process the ratio of two arcs of the same circle or of equal, circles can be found., If the lines or arcs are incommensurable, an approximate value of the ratio can, be found by the same method.
Page 129 :
MEASURE OF ANGLES., , 119, , MEASURE OF ANGLES., Proposition XVI. Theorem., 287. In the same circle or in equal circles, two central angles have the same, ratio as their intercepted arcs., , C′, , A, A′, , C′, , C, , B′, Fig. 1., , m, , B, , A′, , D, , B′, , Fig. 3., , Fig. 2., , C 0,, , In the equal circles whose centres are C and, let ACB and A0 C 0 B 0, be the angles, AB and A0 B 0 the intercepted arcs., To prove that, ∠A0 C 0 B 0, arc A0 B 0, =, ., ∠ACB, arc AB, Case 1. When the arcs are commensurable (Figs. 1 and 2)., Proof. Let the arc m be a common measure of A0 B 0 and AB., Suppose m to be contained 4 times in A0 B 0 ,, and 7 times in AB., , Then, , arc A0 B 0, =, arc AB, At the several points of division on AB and, , 4, ., 7, A0 B 0 draw radii., , These radii will divide ∠ACB into 7 parts, and ∠A0 C 0 B 0 into 4 parts, equal each, to each,, § 237, (in the same, , equal arcs subtend equal central ∠s )., ∠A0 C 0 B 0, 4, ∴, = ., ∠ACB, 7, ∠A0 C 0 B 0, arc A0 B 0, ∴, =, ., Ax. 1, ∠ACB, arc AB, , , or equal, , s,
Page 130 :
BOOK II. PLANE GEOMETRY., , 120, , Case 2. When the arcs are incommensurable (Figs. 2 and 3)., Proof. Divide AB into any number of equal parts, and apply one of these parts, to A0 B 0 as many times as A0 B 0 will contain it., Since AB and A0 B 0 are incommensurable, a certain number of these parts will, extend from A0 to some point, as D, leaving a remainder DB 0 less than one of these, parts. Draw C 0 D., By construction AB and A0 D are commensurable., arc A0 D, ∠A0 C 0 D, =, ., Case 1, ∴, ∠ACB, arc AB, By increasing the number of equal parts into which AB is divided we can diminish at pleasure the length of each part, and therefore make DB 0 less than any, assigned value, however small, since DB 0 is always less than one of the equal parts, into which AB is divided., We cannot make DB 0 equal to zero, since, by hypothesis, AB and A0 B 0 are, incommensurable., § 269, Hence, DB 0 approaches zero as a limit, if the number of parts of AB is indefinitely increased., § 275, And the corresponding angle DC 0 B 0 approaches zero as a limit., Therefore, the arc A0 D approaches the arc A0 B 0 as a limit,, and the ∠A0 C 0 D approaches the ∠A0 C 0 B 0 as a limit., Therefore,, arc A0 D, arc A0 B 0, approaches, as a limit,, arc AB, arc AB, and, ∠A0 C 0 B 0, ∠A0 C 0 D, approaches, as a limit., ∠ACB, ∠ACB, But, ∠A0 C 0 D, arc A0 D, is constantly equal to, ,, ∠ACB, arc AB, as A0 D varies in value and approaches A0 B 0 as a limit., ∠A0 C 0 B 0, arc A0 B 0, ∴, =, ., ∠ACB, arc AB, , § 271, , § 280, § 280, , § 284, q.e.d.
Page 131 :
MEASURE OF ANGLES., , 121, , 288. A circumference is divided into 360 equal parts, called degrees; and, therefore a unit angle at the centre intercepts a unit arc on the circumference., Hence, the numerical measure of a central angle expressed in terms of the, unit angle is equal to the numerical measure of its intercepted arc expressed, in terms of the unit arc. This must be understood to be the meaning when it, is said that, A central angle is measured by its intercepted arc.
Page 132 :
BOOK II. PLANE GEOMETRY., , 122, , Proposition XVII. Theorem., 289. An inscribed angle is measured by half the arc intercepted between its, sides., B, , B, , B, , A, C, , C, , A, P, Fig. 1., , C, , A, , P, , P, F, Fig. 3., , E, Fig. 2., , 1. Let the centre C (Fig. 1) be in one of the sides of the angle., To prove that the ∠B is measured by 12 the arc P A., Proof., Draw CA., , But, , But, , CA = CB., , § 217, , ∴ ∠B = ∠A., , § 145, , ∠P CA = ∠B + ∠A., , § 137, , ∴ ∠P CA = 2∠B., ∠P CA is measured by arc P A,, (a central ∠ is measured by its intercepted arc)., ∴ ∠B is measured by, , 1, 2, , § 288, , arc P A., , 2. Let the centre C (Fig. 2) fall within the angle P BA., To prove that the ∠P BA is measured by 21 the arc P A., Proof., Draw the diameter BCE., Then, ∠EBA is measured by 12 arc AE,, and, ∠EBP is measured by 12 arc EP ., , Case 1
Page 133 :
MEASURE OF ANGLES., , 123, , ∴ ∠EBA + ∠EBP is measured by 12 (arc AE + arc EP ),, , or, , ∠P BA is measured by, , 1, 2, , arc P A., , 3. Let the centre C (Fig. 3) fall without the angle P BA., To prove that the ∠P BA is measured by 12 the arc P A., Proof., Draw the diameter BCF ., Then, ∠F BA is measured by 12 arc F A,, and, ∠F BP is measured by 12 arc F P ., , Case 1, , ∴ ∠F BA − ∠F BP is measured by 12 (arc F A − arc F P ),, , or, , ∠P BA is measured by, A, , A, , 1, 2, , arc P A., , q.e.d., B, C, , C, , B, , A, C, , Fig. 4., , D, B, Fig. 5., , D, , E, , Fig. 6., , 290. Cor. 1. An angle inscribed in a semicircle is a right angle. For it is, measured by half a semicircumference (Fig. 4)., 291. Cor. 2. An angle inscribed in a segment greater than a semicircle is, an acute angle. For it is measured by an arc less than half a semicircumference;, as, ∠CAD (Fig. 5)., 292. Cor. 3. An angle inscribed in a segment less than a semicircle is an, obtuse angle. For it is measured by an arc greater than half a semicircumference; as, ∠CBD (Fig. 5)., 293. Cor. 4. Angles inscribed in the same segment or in equal segments, are equal (Fig. 6).
Page 134 :
BOOK II. PLANE GEOMETRY., , 124, , Proposition XVIII. Theorem., 294. An angle formed by two chords intersecting within the circumference, is measured by half the sum of the intercepted arcs., C, , B, O, A, , D, , E, , Let the angle COD be formed by the chords AC and BD., To prove that the ∠COD is measured by 12 (CD + AB)., Proof., Suppose AE drawn k to BD., Then arc AB = arc DE,, (parallels intercept equal arcs on a circumference)., Also ∠COD = ∠CAE,, (ext.-int. ∠s of ks )., But ∠CAE is measured by 12 (CD + DE),, (an inscribed ∠ is measured by half its intercepted arc)., , § 257, § 112, § 289, , Put ∠COD for its equal, the ∠CAE, and arc AB for its equal, the arc DE; then, ∠COD is measured by 21 (CD + AB)., q.e.d., , Ex. 97. The opposite angles of an inscribed quadrilateral are supplementary., Ex. 98. If through a point within a circle two perpendicular chords are, drawn, the sum of either pair of the opposite arcs which they intercept is equal, to a semicircumference., Ex. 99. The line joining the centre of the square described upon the, hypotenuse of a right triangle to the vertex of the right angle bisects the right, angle.
Page 135 :
MEASURE OF ANGLES., , 125, , Proposition XIX. Theorem., 295. An angle included by a tangent and a chord drawn from the point of, contact is measured by half the intercepted arc., F, , H, , E, M, , A, , O, , Let M AH be the angle included by the tangent M O to the circle at, A and the chord AH., To prove that the ∠M AH is measured by 12 the arc AEH., Proof., Suppose HF drawn k to M O., Then arc AF = arc AEH,, (parallels intercept equal arcs on a circumference)., Also ∠M AH = ∠AHF ,, (alt.-int. ∠s of ks )., ∠AHF is measured by 21 AF ,, (an inscribed ∠ is measured by half its intercepted arc)., , § 257, § 110, § 289, , Put ∠M AH for its equal, the ∠AHF , and arc AEH for its equal, the arc AF ;, then ∠M AH is measured by 12 arc AEH., , Likewise, the ∠OAH, the supplement of the ∠M AH, is measured by half the, arc AF H, the conjugate of the arc AEH., q.e.d., , Ex. 100. Two circles are tangent externally at A, and a common external, tangent touches them at B and C, respectively. Show that angle BAC is a, right angle.
Page 136 :
BOOK II. PLANE GEOMETRY., , 126, , Proposition XX. Theorem., 296. An angle formed by two secants, two tangents, or a tangent and a, secant, drawn to a circle from an external point, is measured by half the difference of the intercepted arcs., O, , O, , C, , O, , B, , B, B, , A, , A, , E, A, , E, P, , P, , M, , E, C, , P, , The proof of this theorem is left as an exercise for the student.
Page 137 :
MEASURE OF ANGLES., , 127, B, , B, , o, C, , O, , D, , A, , C, , A, , /, B′, , 297. Positive and Negative Quantities. In measurements it is convenient to mark the distinction between two quantities that are measured in, opposite directions, by calling one of them positive and the other negative., Thus, if OA is considered positive, then OC may be considered negative,, and if OR is considered positive, then OD may be considered negative., When this distinction is applied to angles, an angle is considered to be, positive, if the rotating line that describes it moves in the opposite direction, to the hands of a clock (counter clockwise), and to be negative, if the rotating, line moves in the same direction as the hands of a clock (clockwise)., Arcs corresponding to positive angles are considered positive, and arcs corresponding to negative angles are considered negative., Thus, the angle ACB described by a line rotating about C from CA to CB, is positive, and the arc AB is positive; the angle ACB 0 described by the line, rotating about C from CA to CB 0 is negative, and the arc AB 0 is negative., 298. The Principle of Continuity. By marking the distinction between, quantities measured in opposite directions, a theorem may often be so stated, as to include two or more particular theorems.
Page 138 :
BOOK II. PLANE GEOMETRY., , 128, , The following theorem furnishes a good illustration:, 299. The angle included between two lines of unlimited length that cut or, touch a circumference is measured by half the sum of the intercepted arcs., Here the word sum means the algebraic sum and includes both the arithmetical sum and the arithmetical difference of two quantities., b, , c, , a, , b, , a, , b a, o, , o, , o, d, , c, , d c, , b, d, , a, o, , c m, , a, d, , c, , ob, , m, , n, d c, , a b, , a, , no, , b, , c, d, , d, m, , 1. If the lines intersect at the centre, the two intercepted arcs are equal,, and half the sum will be one of the arcs (§ 288)., 2. If the lines intersect between the centre and the circumference, the angle, is measured by half the sum of the arcs (§ 294)., 3. If the lines intersect on the circumference, one of the arcs becomes zero, and we have an inscribed angle (§ 289), or an angle formed by a tangent and, a chord (§ 295). In each case the angle is measured by half the sum of the, intercepted arcs., 4. If the lines intersect without the circumference, then the arc ab is negative and the algebraic sum is the arithmetical difference of the included arcs., When the reasoning employed to prove a theorem is continued in the manner just illustrated to include two or more theorems, we are said to reason by, the Principle of Continuity.
Page 139 :
MEASURE OF ANGLES., , 129, , REVIEW QUESTIONS ON BOOK II., 1. What do we call the locus of points in a plane that are equidistant from, a fixed point in the plane?, 2. What does the chord of a segment become when the segment is a semicircle?, 3. What kind of an angle do the radii of a sector include when the sector, is a semicircle?, 4. What is the difference between a chord and a secant?, 5. What part of a tangent is meant by a tangent to a circle from an external, point?, 6. Two chords are equal in equal circles under either of two conditions., What are the two conditions?, 7. Points that lie in a straight line are called collinear ; points that lie in a, circumference are called concyclic. How many collinear points can be, concyclic?, 8. What is meant by the statement that a central angle is measured by, the arc intercepted between its sides?, 9. What is an inscribed angle? What is its measure?, 10. What kind of an angle is the angle inscribed in a semicircle? in a, segment less than a semicircle? in a segment greater than a semicircle?, 11. What is the measure of an angle included by two intersecting chords?, by two secants intersecting without the circle?, 12. What is the measure of an angle included by a tangent and a chord, drawn to the point of contact?, 13. When are two quantities of the same kind incommensurable?, 14. When are two quantities of the same kind commensurable?, 15. Define a variable and the limit of a variable., in., etc., constitute a variable? Is, 16. Does the series 21 in., 34 in., 78 in., 15, 16, the variable increasing or decreasing?, 17. What is the limit of this variable?, 18. What is the test of a limit?
Page 140 :
BOOK II. PLANE GEOMETRY., , 130, , THEOREMS., Ex. 101. An angle formed by a tangent and a chord is equal to the angle, inscribed in the opposite segment., Ex. 102. Two chords drawn perpendicular to a third chord at its extremities are equal., Ex. 103. The sum of two opposite sides of a circumscribed quadrilateral, is equal to the sum of the other two sides., A, , B, , O, C, , D, , O A, , C, , B, D, , M, , Ex. 104. If the sum of two opposite angles of a quadrilateral is equal to, two right angles, a circle may be circumscribed about the quadrilateral., Let ∠A + ∠C = 2 rt. ∠s . Pass a circumference through D, A, and B, and, prove that this circumference passes through C., Ex. 105. The shortest line that can be drawn from a point within a circle, to the circumference is the shorter segment of the diameter through that point., Let A be the given point. Prove AB shorter than any other line AD from, A to the circumference., Ex. 106. The longest line that can be drawn from a point within a circle, to the circumference is the longer segment of the diameter through that point., D, C, P, , A, , O, , B, , Ex. 107. The shortest line that can be drawn from a point without a circle, to the circumference will pass through the centre of the circle if produced., Ex. 108. The longest line that can be drawn from a point without a circle, to the concave arc of the circumference passes through the centre of the circle.
Page 141 :
MEASURE OF ANGLES., , 131, A, , A, , C, , P, , A, O, , Q, , B, D, , A, , H, , D, M C K, B, O, N, E, , L, , G, , B, , F, , H, E, , O, , O, , C, D, , K, , B, , D, , C, , Ex. 109. The shortest chord that can be drawn through a point within a, circle is perpendicular to the diameter at that point., Ex. 110. If two intersecting chords make equal angles with the diameter, drawn through the point of intersection, the two chords are equal., Rt. 4COM = rt. 4CON ., ∴ OM = ON ., Ex. 111. The angles subtended at the centre of a circle by any two opposite, sides of a circumscribed quadrilateral are supplementary., Ex. 112. The radius of a circle inscribed in an equilateral triangle is equal, to one third the altitude of the triangle., 4OEF is equiangular and equilateral; ∠F EA = ∠F AE., ∴ AF = EF ., ∴ AF = F O = OD., Ex. 113. The radius of a circle circumscribed about an equilateral triangle, is equal to two thirds the altitude of the triangle (Ex. 27)., Ex. 114. A parallelogram inscribed in a circle is a rectangle., Ex. 115. A trapezoid inscribed in a circle is an isosceles trapezoid., Ex. 116. All chords of a circle which touch an interior concentric circle, are equal, and are bisected at the point of contact., Ex. 117. If the inscribed and circumscribed circles of a triangle are concentric, the triangle is equilateral (Ex. 116)., Ex. 118. If two circles are tangent to each other the tangents to them, from any point of the common internal tangent are equal.
Page 142 :
BOOK II. PLANE GEOMETRY., , 132, , Ex. 119. If two circles touch each other and a line is drawn through the, point of contact terminated by the circumferences, the tangents at its ends are, parallel., N, A, , D, P, , C, B, , C, , M, , E, , A, , D, , B, , F, , Ex. 120. If two circles touch each other and two lines are drawn through, the point of contact terminated by the circumferences, the chords joining the, ends of these lines are parallel., ∠A = ∠M P C and ∠B = ∠N P D. ∴ ∠A = ∠B., Ex. 121. If two circles intersect and a line is drawn through each point, of intersection terminated by the circumferences, the chords joining the ends, of these lines are parallel., C, , F, , A, O′, , O, C, , B, , D, , O, A, , D, , E, B, , Ex. 122. Through one of the points of intersection of two circles a diameter, of each circle is drawn. Prove that the line joining the ends of the diameters, passes through the other point of intersection., ∠ABC = ∠ABD = 90◦, , § 290, , Ex. 123. If two common external tangents or two common internal tangents are drawn to two circles, the segments intercepted between the points of, contact are equal., Ex. 124. The diameter of the circle inscribed in a right triangle is equal, to the difference between the sum of the legs and the hypotenuse.
Page 143 :
MEASURE OF ANGLES., A, , A, , M, , O, , B, , K, , C, , H, , D, E, , 133, , C, , B, , Ex. 125. If one leg of a right triangle is the diameter of a circle, the tangent, at the point where the circumference cuts the hypotenuse bisects the other leg., ∠BOE = ∠DOE., ∴ OE and AC are k., , ∠BOE = ∠OAD., ∴ BE = EC (§ 188)., , Ex. 126. If, from any point in the circumference of a circle, a chord and, a tangent are drawn, the perpendiculars dropped on them from the middle, point of the subtended arc are equal. ∠BAM = ∠CAM ., Ex. 127. The median of a trapezoid circumscribed about a circle is equal, to one fourth the perimeter of the trapezoid (Ex. 103)., Ex. 128. Two fixed circles touch each other externally and a circle of, variable radius touches both externally. Show that the difference of the distances from the centre of the variable circle to the centres of the fixed circles, is constant., Ex. 129. If two fixed circles intersect, and circles are drawn to touch both,, show that either the sum or the difference of the distances of their centres from, the centres of the fixed circles is constant, according as they touch (i) one, internally and one externally, (ii) both internally or both externally.
Page 144 :
BOOK II. PLANE GEOMETRY., A, , F′, , B, , P, , E′, , A, , E, M, , O, B, D, , F, , 134, , C, , C, P, , Ex. 130. If two straight lines are drawn through any point in a diagonal, of a square parallel to the sides of the square, the points where these lines, meet the sides lie on the circumference of a circle whose centre is the point of, intersection of the diagonals., 4P OE = 4P OF (§ 143). ∴ OE = OF . 4P OE 0 = 4P OF 0 ., Ex. 131. If ABC is an inscribed equilateral triangle and P is any point in, the arc BC, then P A = P B + P C., Take P M = P B. 4ABM = 4CBP (§ 143) and AM = P C., Ex. 132. The tangents drawn through the vertices of an inscribed rectangle, which is not a square, enclose a rhombus.
Page 145 :
PROBLEMS OF CONSTRUCTION., , 135, , E, , G, D, , B, , C, , H, , O, A, , F, B, , E, , H, , M, , I, , C, N, , G, , D, A, , F, , Ex. 133. The bisectors of the angles included by the opposite sides, (produced) of an inscribed quadrilateral intersect at right angles., Arc AF − arc BM = arc DF − arc CM, and arc AH − arc DN = arc BH − arc CN ., ∴ arc F H + arc M N = arc HM + arc F N ., ∴ ∠F IH = ∠HIM ., Discussion. This problem is impossible, if any two sides of the quadrilateral are parallel., PROBLEMS OF CONSTRUCTION., Note. Hitherto we have supposed the figures constructed. We now proceed to, explain the methods of constructing simple problems, and afterwards to apply these, methods to the solution of more difficult problem.
Page 146 :
BOOK II. PLANE GEOMETRY., , 136, , Proposition XXI. Problem., 300. To let fall a perpendicular upon a given line from a given external, point., C, , ... ...., .., .....O, ., ... ..., , A, , H, , M, , K, , B, , Let AB be the given straight line, and C the given external point., To let fall a ⊥ to the line AB from the point C., From C as a centre, with a radius sufficiently great, describe an arc cutting AB, in two points, H and K., From H and K as centres, with equal radii greater than 12 HK,, describe two arcs intersecting at O., Draw CO,, and produce it to meet AB at M ., CM is the ⊥ required., Proof. Since C and O are two points each equidistant from H and K, they, determine a ⊥ to HK at its middle point., § 161, q.e.f., Note. Given lines of the figures are represented by full lines, resulting lines by, long-dashed, and auxiliary lines by short-dashed lines.
Page 147 :
PROBLEMS OF CONSTRUCTION., , 137, , Proposition XXII. Problem., 301. At a given point in a straight line, to erect a perpendicular to that, line., D, , . . . . ......, . ., . . . ..., R, , A, , H, , O, Fig. 1., , C, , B, , C, , A, , B, , E, Fig. 2., , 1. Let O be the given point in AC. Fig. 1., Take OH and OB equal., From H and B as centres, with equal radii greater than OB, describe two arcs, intersecting at R. Join OR., Then the line OR is the ⊥ required., Proof. O and R, two points each equidistant from H and B, determine the, perpendicular bisector of HB., § 161, 2. Let B be the given point. Fig. 2., Take any point C without AB; and from C as a centre, with the distance CB, as a radius, describe an arc intersecting AB at E., Draw EC, and prolong it to meet the arc again at D., Join BD, and BD is the ⊥ required., Proof., The ∠B is a right angle., , § 290, , ∴ BD is ⊥ to AB., q.e.f., Discussion. The point C must be so taken that it will not be in the required, perpendicular.
Page 148 :
BOOK II. PLANE GEOMETRY., , 138, , Proposition XXIII. Problem., 302. To bisect a given straight line., , C, , A, , D, , B, , E, , To bisect the given straight line AB., From A and B as centres, with equal radii greater than, intersecting at C and E., , 1, 2 AB,, , describe arcs, , Join CE., Then CE bisects AB., , § 161, q.e.f.
Page 149 :
PROBLEMS OF CONSTRUCTION., , 139, , Proposition XXIV. Problem., 303. To bisect a given arc., , D, , C, B, , A, , E, , To bisect the given arc AB., Draw the chord AB., From A and B as centres, with equal radii greater than, intersecting at D and E., , 1, 2 AB,, , describe arcs, , Draw DE., Then DE is the ⊥ bisector of the chord AB., , § 161, , ∴ DE bisects the arc ACB., , § 248, q.e.f.
Page 150 :
BOOK II. PLANE GEOMETRY., , 140, , Proposition XXV. Problem., 304. To bisect a given angle., , ... D.., ....., .. ..., ., ... ., C, , A, , B, , E, , Let AEB be the given angle., From E as a centre, with any radius, as EA, describe an arc cutting the sides, of the ∠E at A and B., From A and B as centres, with equal radii greater than half the distance from, A to B, describe two arcs intersecting at D., Draw DE., Then DE bisects the arc AB at C., , § 303, , ∴ DE bisects the angle E., , § 237, q.e.f., , Ex. 134. To construct an angle of 45◦ ; of 135◦ ., Ex. 135. To construct an equilateral triangle, having given one side., Ex. 136. To construct an angle of 60◦ ; of 150◦ ., Ex. 137. To trisect a right angle.
Page 151 :
PROBLEMS OF CONSTRUCTION., , 141, , Proposition XXVI. Problem., 305. At a given point in a given straight line, to construct an angle equal, to a given angle., G, , F, A, , O, , C, E, , H, , M, , At C in the line CM , construct an angle equal to the given angle A., From A as a centre, with any radius, AE, describe an arc cutting the sides of, the ∠A at E and F ., From C as a centre, with a radius equal to AE,, describe an arc HG cutting CM at H., From H as a centre, with a radius equal to the chord EF ,, describe an arc intersecting the arc HG at O., Draw CO, and ∠ HCO is the required angle., , Why?, q.e.f.
Page 152 :
BOOK II. PLANE GEOMETRY., , 142, , Proposition XXVII. Problem., 306. To draw a straight line parallel to a given straight line through a given, external point., E, , C, , H, , A, , D, , F, , B, , Let AB be the given line, and C the given point., Draw ECD, making any convenient ∠EDB., At the point C construct ∠ECF equal to ∠EDB., , § 305, , Then the line HCF is k to AB., , Why?, q.e.f.
Page 153 :
PROBLEMS OF CONSTRUCTION., , 143, , Proposition XXVIII. Problem., 307. To divide a given straight line into a given number of equal parts., A, , B, , C, O, , Let AB be the given straight line., From A draw the line AO, making any convenient angle with AB., Take any convenient length, and apply it to AO as many times as the line AB, is to be divided into parts., From C, the last point thus found on AO, draw CB., Through the points of division on AO draw parallels to the line CB., These lines will divide AB into equal parts., , § 306, § 187, q.e.f., , Ex. 138. To construct an equilateral triangle, having given the perimeter., Ex. 139. To divide a line into four equal parts by two different methods., Ex. 140. Through a given point to draw a line which shall make equal, angles with the two sides of a given angle., Through the given point draw a ⊥ to the bisector of the given ∠., Ex. 141. To draw a line through a given point, so that it shall form with, the sides of a given angle an isosceles triangle (Ex. 140).
Page 154 :
BOOK II. PLANE GEOMETRY., , 144, , Proposition XXIX. Problem., 308. To find the third angle of a triangle when two of the angles are given., R, , A, , B, , E, , c b a, H, , F, , Let A and B be the two given angles., At any point H in any line EF ,, Then, , construct ∠a equal to ∠A, and ∠b equal to ∠B., , § 305, , ∠c is the ∠ required., , Why?, q.e.f.
Page 155 :
PROBLEMS OF CONSTRUCTION., , 145, , Proposition XXX. Problem., 309. To construct a triangle when two sides and the included angle are, given., D, , b, c, , C, , b, , E, , A, , c, , B, , Let b and c be the two sides of the triangle and E the included angle., Take AB equal to the side c., At A, construct ∠BAD equal to the given ∠E., , § 305, , On AD take AC equal to b, and draw CB., Then 4ACB is the 4 required., , q.e.f.
Page 156 :
BOOK II. PLANE GEOMETRY., , 146, , Proposition XXXI. Problem., 310. To construct a triangle when a side and two angles of the triangle are, given., O, H, K, A, , B, , E, , c, , C, , c, Let c be the given side, A and B the given angles., Take EC equal to the side c., At E construct the ∠CEH equal to ∠A., , § 305, , At C construct the ∠ECK equal to ∠B., Produce EH and CK until they intersect at O., Then 4COE is the 4 required., q.e.f., Remark. If one of the given angles is opposite to the given side, find the third, angle by § 308, and proceed as above., Discussion. The problem is impossible when the two given angles are together, equal to or greater than two right angles., , Ex. 142. To construct an equilateral triangle, having given the altitude., To construct an isosceles triangle, having given:, Ex. 143. The base and the altitude., Ex. 144. The altitude and one of the legs., Ex. 145. The angle at the vertex and the altitude.
Page 157 :
PROBLEMS OF CONSTRUCTION., , 147, , Proposition XXXII. Problem., 311. To construct a triangle when two sides and the angle opposite one of, them are given., D, , B, , A, , b, , b, a, , a, , a, C′, , A, , C, , E, , Let a and b be the given sides, and A the angle opposite a., Case 1. If a is less than b., Construct ∠DAE equal to the given ∠A, , § 305, , On AD take AB equal to b., From B as a centre, with a radius equal to a,, describe an arc intersecting the line AE at C and C 0 ., Draw BC and BC 0 ., Then both the 4s ABC and ABC 0 fulfil the conditions, and hence we have two, constructions., This is called the ambiguous case., , D, , D, B, , B, b, , A, , b, , a, H, , E, , A, , a, E, , Discussion. If the side a is equal to the ⊥ BH, the arc described from B will, touch AE, and there will be but one construction, the right 4ABH., If the given side a is less than the ⊥ from B, the arc described from B will not, intersect or touch AE, and hence the problem is impossible., If the ∠A is right or obtuse, the problem is impossible; for the side opposite a, right or obtuse angle is the greatest side., § 153
Page 158 :
BOOK II. PLANE GEOMETRY., Case 2. If a is equal to b., , 148, , D, , B, b, , a, , A, , C, , E, , If the ∠A is acute, and a = b, the arc described from B as a centre, and with a, radius equal to a, will cut the line AE at the points A and C. There is therefore, but one solution: the isosceles 4ABC., Discussion. If the ∠A is right or obtuse, the problem is impossible; for equal, sides of a 4 have equal ∠s opposite them, and a 4 cannot have two right ∠s or two, obtuse ∠s ., B, , a, E, , C′ A, , b, , B, , a, , a, C, , D, , E, , C′, , b, A, , a, , a, C, , D, , E, , C′, , B, b, , a, A C, , D, , Case 3. If a is greater than b., If the given ∠A is acute, the arc described from B will cut the line ED on, opposite sides of A, at C and C 0 . The 4ABC answers the required conditions, but, the 4 ABC 0 does not, for it does not contain the acute ∠A. There is then only one, solution; namely, the 4ABC., If the ∠A is right, the arc described from B cuts the line ED on opposite sides, of A, and we have two equal right 4s which fulfil the required conditions., If the ∠A is obtuse, the arc described from B cuts the line ED on opposite sides, of A, at the points C and C 0 . The 4ABC answers the required conditions, but the, 4ABC 0 does not, for it does not contain the obtuse ∠A. There is then only one, solution; namely, the 4ABC., q.e.f.
Page 159 :
PROBLEMS OF CONSTRUCTION., , 149, , Proposition XXXIII. Problem., 312. To construct a triangle when the three sides of the triangle are given., C, , b, , c, , a, , a, b, , A, , c, , B, , Let the three sides be c, a, and b., Take AB equal to c. From A as a centre, with a radius equal to b, describe an, arc. From B as a centre, with a radius equal to a, describe an arc, intersecting the, other arc at C., Draw CA and CB., 4CAB is the 4 required., q.e.f., Discussion. The problem is impossible when one side is equal to or greater, than the sum of the other two sides.
Page 160 :
BOOK II. PLANE GEOMETRY., , 150, , Proposition XXXIV. Problem., 313. To construct a parallelogram when two sides and the included angle, are given., H, , D, , E, , m, , C, m, o, , A, , B, , o, , Let m and o be the two sides, and C the included angle., Take AB equal to o., At A construct ∠BAD equal to ∠C., § 305, Take AH equal to m. From H as a centre, with a radius equal to o, describe an, arc, and from B as a centre, with a radius equal to m, describe an arc, intersecting, the other arc at E; and draw EH and EB., The quadrilateral ABEH is the, , / /, , required., , § 182, q.e.f.
Page 161 :
PROBLEMS OF CONSTRUCTION., , 151, , Proposition XXXV. Problem., 314. To circumscribe a circle about a given triangle., A, , E, , F, O, G, , B, , D, , C, , Let ABC be the given triangle., Bisect AB and BC., , § 302, , At E and D, the points of bisection, erect ⊥s ., § 301, Since BC is not the prolongation of AB, these ⊥s will intersect at some point, O., From O, with a radius equal to OB, describe a circle., The, , ABC is the, , required., , Proof., The point O is equidistant from A and B,, and also is equidistant from B and C., , § 160, , ∴ the point O is equidistant from A, B, and C,, and a described from O as a centre, with a radius equal to OB, will pass through, the vertices A, B, and C., q.e.f., , The same construction serves to describe a circumference which shall pass, through three points not in the same straight line; also to find the centre of a, given circle or of a given arc., Note. The point O is called the circum-centre of the triangle.
Page 162 :
BOOK II. PLANE GEOMETRY., , 152, , Proposition XXXVI. Problem., 315. To inscribe a circle in a given triangle., K, , B, M, , E, , A, , H, , C, , Let ABC be the given triangle., Bisect the ∠s A and C., , § 304, , From E, the intersection of the bisectors,, draw EH ⊥ to the side AC., From E as centre, with radius EH, describe the, , § 300, KHM ., , The KHM is the required., Proof. Since E is in the bisector of the ∠A, it is equidistant from the sides AB, and AC; and since E is in the bisector of the ∠C, it is equidistant from the sides, AC and BC., § 162, ∴a, described from E as centre, with a radius equal to EH, will touch the, sides of the 4 and be inscribed in it., q.e.f., Note. The point E is called the in-centre of the triangle.
Page 163 :
PROBLEMS OF CONSTRUCTION., , F, P, C, A, , 153, , B, , O, , Q, ED, , 316. The intersections of the bisectors of the exterior angles of a triangle, are the centres of three circles, each of which will touch one side of the triangle,, and the two other sides produced. These three circles are called escribed circles;, and their centres are called the ex-centres of the triangle.
Page 164 :
BOOK II. PLANE GEOMETRY., , 154, , Proposition XXXVII. Problem., 317. Through a given point, to draw a tangent to a given circle., Case 1. When the given point is on the circumference., A, , C, , O, , M, , M, , O, , E, , H, , Let C be the given point on the circumference whose centre is O., From the centre O draw the radius OC., Through C draw AM ⊥ to OC., Then AM is the tangent required., Case 2. When the given point is without the circle., , § 301, § 253, , Let O be the centre of the given circle, E the given point., Draw OE., On OE as a diameter, describe a circumference intersecting the given circumference at the points M and H., Draw OM and EM ., Then EM is the tangent required., Proof., ∠OM E is a right angle., ∴ EM is tangent to the circle at M ., In like manner, we may prove EH tangent to the given ., , § 290, § 253, q.e.f., , Ex. 146. To draw a tangent to a given circle, so that it shall be parallel, to a given straight line.
Page 165 :
PROBLEMS OF CONSTRUCTION., , 155, , Proposition XXXVIII. Problem., 318. Upon a given straight line, to describe a segment of a circle in which, a given angle may be inscribed., K, , O, , M, , A, , F, , B, E, , Let AB be the given line, and M the given angle., Construct the ∠ABE equal to the ∠M ., , § 305, , Bisect the line AB by the ⊥ OF ., , § 302, , From the point B draw BO ⊥ to EB., § 301, From O, the point of intersection of F O and BO, as a centre with a radius equal, to OB, describe a circumference., The segment AKB is the segment required., Proof., The point O is equidistant from A and B., , § 160, , ∴ the circumference will pass through A., But BE is ⊥ to OB., , Const., , ∴ BE is tangent to the ,, (a straight line ⊥ to a radius at its extremity is tangent to the, 1, 2, , ∴ ∠ABE is measured by arc AB,, (being an ∠ formed by a tangent and a chord )., But any ∠ as ∠K inscribed in the segment AKB is measured by, ∴ the ∠M may be inscribed in the segment AKB., , )., , § 253, § 295, , 1, 2, , arc AB.§ 289, q.e.f.
Page 166 :
BOOK II. PLANE GEOMETRY., , 156, , SOLUTION OF PROBLEMS., 319. If a problem is so simple that the solution is obvious from a known, theorem, we have only to make the construction according to the theorem,, and then give a synthetic proof, if a proof is necessary, that the construction, is correct, as in the examples of the fundamental problems already given., 320. But problems are usually of a more difficult type. The application of, known theorems to their solution is not immediate, and often far from obvious., To discover the mode of application is the first and most difficult part of the, solution. The best way to attack such problems is by a method resembling the, analytic proof of a theorem, called the analysis of the problem., 1. Suppose the construction made, and let the figure represent all, parts concerned, both given and required., 2. Study the relations among the parts with the aid of known theorems,, and try to find some relation that will suggest the construction., 3. If this attempt fails, introduce new relations by drawing auxiliary lines,, and study the new relations. If this attempt fails, make a new trial, and so on, till a clue to the right construction is found., 321. A problem is determinate if it has a definite number of solutions,, indeterminate if it has an indefinite number of solutions, and impossible if it, has no solution. A problem is sometimes determinate for certain relative positions or magnitudes of the given parts, and indeterminate for other positions, or magnitudes of the given parts., 322. The discussion of a problem consists in examining the problem with, reference to all possible conditions, and in determining the conditions necessary, for its solution.
Page 167 :
PROBLEMS OF CONSTRUCTION., , 157, , Ex. 147. Problem. To construct a circle that shall pass through a given, point and cut chords of a given length from two parallels., D, , E, , F, , B, , A, , P, , M, , O, N, , G, C, , Analysis. Suppose the problem solved. Let A be the given point, BC, and DE the given parallels, M N the given length, and O the centre of the, required circle., Since the circle cuts equal chords from two parallels its centre must be, equidistant from them. Therefore, one locus for O is F G k to BC and equidistant from BC and DE., Draw the ⊥ bisector of M N , cutting F G in P . P M is the radius of the, circle required. With A as centre and radius P M describe an arc cutting F G, at O. Then O is the centre of the required circle., Discussion. The problem is impossible if the distance from A to F G is, greater than P M .
Page 168 :
BOOK II. PLANE GEOMETRY., , 158, , C, , G, , E, , m, , H, , A, , D B, , n, , F, , Ex. 148. Problem. To construct a triangle, having given the perimeter,, one angle, and the altitude from the vertex of the given angle., Analysis. Suppose the problem solved, and let ABC be the 4 required,, ACB the given ∠, and CD the given altitude., Produce AB both ways, and take AE = AC, and BF = BC, then EF =, the given perimeter. Join CE and CF , forming the isosceles 4s CAE and, CBF ., In the 4ECF , ∠E + ∠F + ∠ECF = 180◦ (why?), but ∠ECF = ∠ECA +, ∠F CB + ∠ACB., Since ∠E = ∠ECA and ∠F = ∠F CB, we have ∠ECF = ∠E + ∠F +, ∠ACB. ∴ 2∠E + 2∠F + ∠ACB = 180◦ ., ∴ ∠E + ∠F + 12 ∠ACB = 90◦ , and ∠E + ∠F = 90◦ − 12 ∠ACB., By substitution, ∠ECF = 90◦ + 12 ∠ACB., ∴ ∠ECF is known., Construction. To find the point C, construct on EF a segment that will, contain the ∠ECF (§ 318), and draw a parallel to EF at the distance CD,, the given altitude., To find the points A and B, draw the ⊥ bisectors of the lines CE and CF ,, and the points A and B will be vertices of the required 4. Why?, PROBLEMS OF CONSTRUCTION., Ex. 149. Find the locus of a point at a given distance from a given, circumference., Find the locus of the centre of a circle:, Ex. 150. Which has a given radius r and passes through a given point P .
Page 169 :
EXERCISES., , 159, , Ex. 151. Which has a given radius r and touches a given line AB., Ex. 152. Which passes through two given points P and Q., Ex. 153. Which touches a given straight line AB at a given point P ., Ex. 154. Which touches each of two given parallels., Ex. 155. Which touches each of two given intersecting lines., Ex. 156. To find in a given line a point X which is equidistant from two, given points., The required point is the intersection of the given line with the perpendicular bisector of the line joining the two given points (§ 160)., Ex. 157. To find a point X equidistant from three given points., X, , d, , X′, P, , R, Q, , Ex. 158. To find a point X equidistant from two given points and at a, given distance from a third given point., Ex. 159. To construct a circle which has a given radius and passes through, two given points., Ex. 160. To find a point X at given distances from two given points., Ex. 161. To construct a circle which has its centre in a given line and, passes through two given points., Ex. 162. To find a point X equidistant from two given points and also, equidistant from two given intersecting lines (§§ 160 and 162)., Ex. 163. To find a point X equidistant from two given points and also, equidistant from two given parallel lines., Ex. 164. To find a point X equidistant from two given intersecting lines, and also equidistant from two given parallels.
Page 170 :
BOOK II. PLANE GEOMETRY., , C, , d, , X, , Z, A Y, , 160, , B, P, E, , T, D, , Ex. 165. To find a point X equidistant from two given intersecting lines, and at a given distance from a given point., Ex. 166. To find a point X which lies in one side of a given triangle and, is equidistant from the other two sides., X, D, , A, Z, , C, B, , T, , A, , Y, E, , D, B, , F, , E, C, , Ex. 167. A straight railway passes two miles from a town. A place is four, miles from the town and one mile from the railway. To find by construction, the places that answer this description., Ex. 168. In a triangle ABC, to draw DE parallel to the base BC, cutting, the sides of the triangle in D and E, so that DE shall equal DB + EC (§ 162)., A, , d, , F, B, , E, D, , C, , Ex. 169. To draw through two sides of a triangle a line parallel to the third, side so that the part intercepted between the sides shall have a given length., Take BD = d., Ex. 170. Prove that the locus of the vertex of a right triangle, having, a given hypotenuse as base, is the circumference described upon the given, hypotenuse as diameter (§ 290)., Ex. 171. Prove that the locus of the vertex of a triangle, having a given, base and a given angle at the vertex, is the arc which forms with the base a, segment capable of containing the given angle (§ 318)., Ex. 172. Find the locus of the middle point of a chord of a given length, that can be drawn in a given circle.
Page 171 :
EXERCISES., , 161, , Ex. 173. Find the locus of the middle point of a chord drawn from a, given point in a given circumference., B, , C, A, E, , A, , B, C, , O, , M, , D, , O, , D, , A, , M, , O, , M, , B, , Ex. 174. Find the locus of the middle point of a straight line drawn from, a given exterior point to a given circumference., Ex. 175. A straight line moves so that it remains parallel to a given line,, and touches at one end a given circumference. Find the locus of the other end., Ex. 176. A straight rod moves so that its ends constantly touch two fixed, rods which are perpendicular to each other. Find the locus of its middle point., Ex. 177. In a given circle let AOB be a diameter, OC any radius, CD, the perpendicular from C to AB. Upon OC take OM equal to CD. Find the, locus of the point M as OC turns about O., E, , C, M, B, , O, , D, , C, , A, A, , B, , O, D, , Ex. 178. To construct an equilateral triangle, having given the radius of, the circumscribed circle., To construct on isosceles triangle, having given:, Ex. 179. The angle at the vertex and the base (§ 160 and § 318)., Ex. 180. The base and the radius of the circumscribed circle., Ex. 181. The base and the radius of the inscribed circle.
Page 172 :
BOOK II. PLANE GEOMETRY., , 162, , C, , E, , A D B, , F, , Ex. 182. The perimeter and the altitude., Let ABC be the 4 required, EF the given perimeter. The altitude CD, passes through the middle of EF , and the 4s AEC, BF C are isosceles., To construct a right triangle, having given:, Ex. 183. The hypotenuse and one leg., Ex. 184. One leg and the altitude upon the hypotenuse., Ex. 185. The median and the altitude drawn from the vertex of the right, angle., Ex. 186. The hypotenuse and the altitude upon the hypotenuse., Ex. 187. The radius of the inscribed circle and one leg., Ex. 188. The radius of the inscribed circle and an acute angle., Ex. 189. An acute angle and the sum of the legs., Ex. 190. An acute angle and the difference of the legs., A, , P, D, , B, , O, , Q, , F, R, , E, , C, , Ex. 191. To construct an equilateral triangle, having given the radius of, the inscribed circle., To construct a triangle, having given:, Ex. 192. The base, the altitude, and an angle at the base.
Page 173 :
EXERCISES., , 163, , Ex. 193. The base, the altitude, and the ∠ at the vertex., Ex. 194. The base, the corresponding median, and the ∠ at the vertex., Ex. 195. The perimeter and the angles., Ex. 196. One side, an adjacent ∠, and the sum of the other sides., To construct a triangle, having given:, Ex. 197. One side, an adjacent ∠, and the difference of the other sides., Ex. 198. The sum of two sides and the angles., Ex. 199. One side, an adjacent ∠, and the radius of the circumscribed, circle., Ex. 200. The angles and the radius of the circumscribed circle., Ex. 201. The angles and the radius of the inscribed circle., Ex. 202. An angle, and the bisector and the altitude drawn from the, vertex of the given angle., Ex. 203. Two sides and the median corresponding to the other side., Ex. 204. The three medians., To construct a square, having given:, Ex. 205. The diagonal., Ex. 206. The sum of the diagonal and one side., Let ABCD be the square required, CA the diagonal. Produce CA, making, AE = AB. 4s ABC and ABE are isosceles and ∠BAC = ∠BCA = 45◦ .
Page 174 :
BOOK II. PLANE GEOMETRY., , 164, , C, F, N, A, D, E, , A, , O, , G, M, , E, , B, , C, B, , D, , Ex. 207. Given two perpendiculars, AB and CD, intersecting in O, and, a straight line intersecting these perpendiculars in E and F ; to construct a, square, one of whose angles shall coincide with one of the right angles at O, and, the vertex of the opposite angle of the square shall lie in EF . (Two solutions.), To construct a rectangle, having given:, Ex. 208. One side and the angle between the diagonals., Ex. 209. The perimeter and the diagonal., Ex. 210. The perimeter and the angle between the diagonals., Ex. 211. The difference of two adjacent sides and the angle between the, diagonals., To construct a rhombus, having given:, Ex. 212. The two diagonals., Ex. 213. One side and the radius of the inscribed circle., Ex. 214. One angle and the radius of the inscribed circle., Ex. 215. One angle and one of the diagonals., To construct a rhomboid, having given:, Ex. 216. One side and the two diagonals., Ex. 217. The diagonals and the angle between them.
Page 175 :
EXERCISES., , 165, , Ex. 218. One side, one angle, and one diagonal., Ex. 219. The base, the altitude, and one angle., To construct an isosceles trapezoid, having given:, Ex. 220. The bases and one angle., Ex. 221. The bases and the altitude., Ex. 222. The bases and the diagonal., D, , F, , C, , O, A, , EG B, , Ex. 223. The bases and the radius of the circumscribed circle., Let ABCD be the isosceles trapezoid required, O the centre of the circumscribed . A diameter ⊥ to AB is ⊥ to CD, and bisects both AB and CD., Draw CG k to F E. Then EG = F C = 21 DC., To construct a trapezoid, having given:, Ex. 224. The four sides., Ex. 225. The two bases and the two diagonals., Ex. 226. The bases, one diagonal, and the ∠ between the diagonals., To construct a circle which has the radius r and which also:, Ex. 227. Touches each of two intersecting lines AB and CD., Ex. 228. Touches a given line AB and a given circle K., Ex. 229. Passes through a given point P and touches a given line AB., Ex. 230. Passes through a given point P and touches a given circle K.
Page 176 :
BOOK II. PLANE GEOMETRY., , 166, , To construct a circle which shall:, Ex. 231. Touch two given parallels and pass through a given point P ., Ex. 232. Touch three given lines two of which are parallel., Ex. 233. Touch a given line AB at P and pass through a given point Q., Ex. 234. Touch a given circle at P and pass through a given point Q., Ex. 235. Touch two given lines and touch one of them at a given point P ., Ex. 236. Touch a given line and touch a given circle at a point P ., Ex. 237. Touch a given line AB at P and also touch a given circle., Ex. 238. To inscribe a circle in a given sector., Ex. 239. To construct within a given circle three equal circles, so that, each shall touch the other two and also the given circle., Ex. 240. To describe circles about the vertices of a given triangle as, centres, so that each shall touch the two others., B, , I, A, , D, E, H, , F, , G, C, , Ex. 241. To bisect the angle formed by two lines, without producing the, lines to their point of intersection., Draw any line EF k to BA. Take EG = EH, and produce GH to meet, BA at I. Draw the ⊥ bisector of GI., C, , Q, , P, , Q, , P, , E, , A, , A, , P, , D, G, , F, , B, , C, D, , E, , B, , A C, , E, , F, , B, , D, , Ex. 242. To draw through a given point P between the sides of an angle, BAC a line terminated by the sides of the angle and bisected at P .
Page 177 :
EXERCISES., , 167, , Ex. 243. Given two points P , Q, and a line AB; to draw lines from P and, Q which shall meet on AB and make equal angles with AB., Make use of the point which forms with P a pair of points symmetrical, with respect to AB., Ex. 244. To find the shortest path from P to Q which shall touch a, line AB., A, , R, E, , H, , O, , O′, , Q, C, , F, , G, , B, P, , D, , S, , Ex. 245. To draw a common tangent to two given circles., Let r and r0 denote the radii of the circles, O and O0 their centres. With, centre O and radius r − r0 describe a . From O0 draw the tangents O0 M ,, O0 N . Produce OM and ON to meet the circumference at A and C. Draw the, radii O0 B and O0 D k, respectively, to OA and OC. Draw AB and CD., To draw the internal tangents use an auxiliary of radius r + r0 .
Page 178 :
BOOK III. PROPORTION. SIMILAR POLYGONS., , THE THEORY OF PROPORTION., 323. A proportion is an expression of equality between two equal ratios;, and is written in one of the following forms:, a : b = c : d;, , a : b :: c : d;, , c, a, = ., b, d, , This proportion is read, “a is to b as c is to d”; or “the ratio of a to b is equal, to the ratio of c to d.”, 324. The terms of a proportion are the four quantities compared; the first, and third terms are called the antecedents, the second and fourth terms, the, consequents; the first and fourth terms, the extremes, the second and third, terms, the means., Thus, in the proportion a : b = c : d; a and c are the antecedents, b and d, the consequents, a and d the extremes, b and c the means., 325. The fourth proportional to three given quantities is the fourth term, of the proportion which has for its first three terms the three given quantities, taken in order., Thus, d is the fourth proportional to a, b, and c in the proportion, a : b = c : d., 326. The quantities a, b, c, d, e, are said to be in continued proportion,, if a : b = b : c = c : d = d : e., If three quantities are in continued proportion, the second is called the, mean proportional between the other two, and the third is called the third, proportional to the other two., Thus, in the proportion a : b = b : c; b is the mean proportional between a, and c; and c is the third proportional to a and b.
Page 179 :
THEORY OF PROPORTION., , 169, , Proposition I. Theorem., 327. In every proportion the product of the extremes is equal to the product, of the means., Let, a : b = c : d., Then, a, c, = ., § 323, b, d, Whence, ad = bc., q.e.d., Proposition II. Theorem., 328. The mean proportional between two quantities is equal to the square, root of their product., Let, a : b = b : c., Then, b2 = ac., § 327, Whence, extracting the square root,, √, q.e.d., b = ac., Proposition III. Theorem., 329. If the product of two quantities is equal to the product of two others,, either two may be made the extremes of the proportion in which the other two, are made the means., Let, ad = bc., To prove that, a:b=c:d, Divide both members of the given equation by bd., Then, a, c, = ., b, d, Or, a : b = c : d., q.e.d.
Page 180 :
BOOK III. PLANE GEOMETRY., , 170, , Proposition IV. Theorem., 330. If four quantities are in proportion, they are in proportion by alternation; that is, the first term is to the third as the second is to the fourth., Let, a : b = c : d., To prove that, a : c = b : d., Now, ad = bc., § 327, Divide each member of the equation by cd., Then, b, a, = ., c, d, Or, a : c = b : d., q.e.d., Proposition V. Theorem., 331. If four quantities are in proportion, they are in proportion by inversion; that is, the second term is to the first as the fourth is to the third., Let, a : b = c : d., To prove that, b : a = d : c., Now, bc = ad., § 327, Divide each member of the equation by ac., Then, d, b, = ., a, c, Or, b : a = d : c., q.e.d.
Page 181 :
THEORY OF PROPORTION., , 171, , Proposition VI. Theorem., 332. If four quantities are in proportion, they are in proportion by composition that is, the sum of the first two terms is to the second term as the, sum of the last two terms is to the fourth term., Let, a : b = c : d., To prove that, a + b : b = c + d : d., Now, c, a, = ., b, d, Then, a, c, + 1 = + 1;, b, d, that is,, c+d, a+b, =, ., b, d, Or, a + b : b = c + d : d., In like manner, a + b : a = c + d : c., q.e.d.
Page 182 :
BOOK III. PLANE GEOMETRY., , 172, , Proposition VII. Theorem., 333. If four quantities are in proportion, they are in proportion by division; that is, the difference of the first two terms is to the second term as the, difference of the last two terms is to the fourth term., Let, a : b = c : d., To prove that, a − b : b = c − d : d., Now, c, a, = ., b, d, Then, a, c, − 1 = − 1;, b, d, that is,, a−b, c−d, =, ., b, d, Or, a − b : b = c − d : d., In like manner, a − b : a = c − d : c., q.e.d.
Page 183 :
THEORY OF PROPORTION., , 173, , Proposition VIII. Theorem., 334. If four quantities are in proportion, they are in proportion by composition and division; that is, the sum of the first two terms is to their, difference as the sum of the last two terms is to their difference., Let, a:b=c:d, Then, c+d, a+b, =, ., § 332, a, c, And, a−b, c−d, =, ., § 333, a, c, Divide,, c+d, a+b, =, ., a−b, c−d, Or, a + b : a − b = c + d : c − d., q.e.d.
Page 184 :
BOOK III. PLANE GEOMETRY., , 174, , Proposition IX. Theorem., 335. In a series of equal ratios, the sum of the antecedents is to the sum, of the consequents as any antecedent is to its consequent., Let, a : b = c : d = e : f = g : h., To prove that a + c + e + g : b + d + f + h = a : b., Let, c, e, g, a, r= = = = ., b, d, f, h, Then, a = br, c = dr, e = f r, g = hr., And, a + c + e + g = (b + d + f + h)r., Divide by (b + d + f + h)., Then, a, a+c+e+g, =r= ., b+d+f +h, b, Or, a + c + e + g : b + d + f + h = a : b., q.e.d.
Page 185 :
THEORY OF PROPORTION., , 175, , Proposition X. Theorem., 336. The products of the corresponding terms of two or more proportions, are in proportion., Let, a : b = c : d, e : f = g : h, k : l = m : n., To prove that, aek : bf l = cgm : dhn., Now, c e, g k, m, a, = , = , = ., b, d f, h l, n, The products of the first members and of the second members of these, equations give, cgm, aek, =, ., bf l, dhn, Or, aek : bf l = cgm : dhn., q.e.d., 337. Cor. If three quantities are in continued proportion, the first is to, the third as the square of the first is to the square of the second., Proposition XI. Theorem., 338. Like powers of the terms of a proportion are in proportion., Let, a : b = c : d., To prove that, an : bn = cn : dn ., Now, a, c, = ., b, d, Raise to the nth power,, an, cn, =, ., bn, dn, Or, an : bn = cn : dn ., , q.e.d.
Page 186 :
BOOK III. PLANE GEOMETRY., , 176, , 339. Def. Equimultiples of two quantities are the products obtained, by multiplying each of them by the same number. Thus, ma and mb are, equimultiples of a and b., Proposition XII. Theorem., 340. Equimultiples of two quantities are in the same ratio as the quantities, themselves., Let a and b be any two quantities., To prove that, ma : mb = a : b., Now, a, a, = ., b, b, Multiply both terms of the first fraction by m., Then, a, ma, = ., mb, b, Or, ma : mb = a : b., q.e.d., 341. Scholium. In the treatment of proportion, it is assumed that the, quantities involved are expressed by their numerical measures. It is evident, that the ratio of two quantities of the same kind may be represented by a fraction, if the two quantities are expressed in integers in terms of a common unit., If there is no unit in terms of which both quantities can be expressed in integers, it is still possible by taking the unit of measure sufficiently small to find, a fraction that will represent the ratio to any required degree of accuracy. § 269, If we speak of the product of two quantities, it must be understood that, we mean simply the product of the numbers which represent them when they, are expressed in terms of a common unit., In order that four quantities, a, b, c, d, may form a proportion, a and b must, be quantities of the same kind; and c and d must be quantities of the same, kind; though c and d need not be of the same kind as a and b. In alternation,, however, the four quantities must be of the same kind.
Page 187 :
THEORY OF PROPORTION., , 177, , Proposition XIII. Theorem., 342. If a line is drawn through two sides of a triangle parallel to the third, side, it divides those sides proportionally., A, , E, , A, , F, , E, , M, B, , C, , F, , K, B, , H, , C, , In the triangle ABC, let EF be drawn parallel to BC., To prove that, EB : AE = F C : AF ., Case 1. When AE and EB (Fig. 1) are commensurable., Proof. Find a common measure of AE and EB, as M B., Let M B be contained m times in EB, and n times in AE., Then, EB : AE = m : n., At the points of division on BE and AE draw lines k to BC. These lines will, divide AC into m + n equal parts, of which F C will contain m, and AF will contain, n., § 187, ∴ F C : AF = m : n., ∴ EB : AE = F C : AF ., Case 2. When AE and EB (Fig. 2) are incommensurable., , Ax. 1, , Proof. Divide AE into any number of equal parts, and apply one of these parts, to EB as many times as EB will contain it., Since AE and EB are incommensurable, a certain number of these parts will, extend from E to some point K, leaving a remainder KB less than one of these, parts. Draw KH k BC., Then, EK : AE = F H : AF, Case 1, By increasing the number of equal parts into which AE is divided, we can make, the length of each part less than any assigned value, however small, but not zero., Hence, KB, which is less than one of these equal parts, has zero for a limit. § 275, And the corresponding segment HC has zero for a limit.
Page 188 :
BOOK III. PLANE GEOMETRY., Therefore, EK approaches EB as a limit,, , 178, § 271, , and F H approaches F C as a limit., EK, EB, ∴ the variable, approaches, as a limit,, AE, AE, FH, FC, and the variable, approaches, as a limit., AF, AF, But, EK, FH, is constantly equal to, AE, AF, EB, FC, ∴, =, ., AE, AF, , § 280, , Case 1, § 284, q.e.d., , 343. Cor. 1. One side of a triangle is to either part cut off by a straight, line parallel to the base as the other side is to the corresponding part., For, AE : EB = AF : F C., By composition,, AE + EB : AE = AF + F C : AF ., § 332, Or, AB : AE = AC : AF ., 344. Cor. 2. If two lines are cut by any number of parallels the corresponding intercepts are proportional., A, , F, H, , B, , C, , L, , G, , M, , N, , K, , D, , Draw AN k to CD. Then, Now, , Or, , AL = CG, LM = GK, M N = KD., , § 180, , AH : AM = AF : AL = F H : LM, = HB : M N ., AF : CG = F H : GK = HB : KD., , § 343
Page 189 :
THEORY OF PROPORTION., , 179, , Proposition XIV. Theorem., 345. If a straight line divides two sides of a triangle proportionally, it is, parallel to the third side., A, , H, , E, , F, , B, , C, , In the triangle ABC, let EF be drawn so that, AC, AB, =, ., AE, AF, To prove that, EF is k to BC., Proof., From E draw EH k to BC., Then, AB : AE = AC : AH,, § 343, (one side of a triangle is to either part cut off by a line parallel to the base as, the other side to the corresponding part)., But, AB : AE = AC : AF ., Hyp., ∴ AC : AF = AC : AH., , Ax. 1, , ∴ AF = AH., But, , ∴ EF and EH coincide., , § 47, , EH is k to BC., , Const., , ∴ EF , which coincides with EH, is k to BC., , q.e.d., , Ex. 246. Find the fourth proportional to 91, 65, and 133., Ex. 247. Find the mean proportional between 39 and 351., Ex. 248. Find the third proportional to 54 and 3.
Page 190 :
BOOK III. PLANE GEOMETRY., , 180, , 346. If a given line AB is divided at M , a point between the extremities, A and B, it is said to be divided internally into the segments M A and M B;, and if it is divided at M 0 , a point in the prolongation of AB, it is said to be, divided externally into the segments M 0 A and M 0 B., M′, , A, , B, , M, , In either case the segments are the distances from the point of division, to the extremities of the line. If the line is divided internally, the sum of the, segments is equal to the line; and if the line is divided externally, the difference, of the segments is equal to the line., Suppose it is required to divide the given line AB internally and externally in the same ratio; as, for example, the ratio of the two numbers, 3 and 5., x ′, M, , A, , M, , B, , y, , We divide AB into 5 + 3, or 8, equal parts, and take 3 parts from A; we, then have the point M , such that, M A : M B = 3 : 5., (1), Secondly, we divide AB into 5 − 3, or 2, equal parts, and lay off on the, prolongation of AB, to the left of A, three of these equal parts; we then have, the point M 0 , such that, M 0 A : M 0 B = 3 : 5., , (2), , Comparing (1) and (2),, M A : M B = M 0 A : M 0 B., 347. Def. If a given straight line is divided internally and externally into, segments having the same ratio, the line is said to be divided harmonically.
Page 191 :
THEORY OF PROPORTION., , 181, , Proposition XV. Theorem., 348. The bisector of an angle of a triangle divides the opposite side into, segments which are proportional to the adjacent sides., E, , C, , A, , M, , B, , Let CM bisect the angle C of the triangle CAB., To prove that, M A : M B = CA : CB., Proof. Draw AE k to M C, meeting BC produced at E., Then, M A : M B = CE : CB,, § 342, (if a line is drawn through two sides of a 4 parallel to the third side, it divides, those sides proportionally)., Also,, ∠ACM = ∠CAE,, § 110, (being alt.-int. ∠s of k lines);, and, ∠BCM = ∠CEA,, § 112, (being ext.-int. ∠s of k lines)., But, ∠ACM = ∠BCM ., Hyp., ∴ ∠CAE = ∠CEA., ∴ CE = CA., Put CA for its equal, CE, in the first proportion., Then, M A : M B = CA : CB., , Ax. 1, § 147, q.e.d., , Ex. 249. In a triangle ABC, AB = 12, AC = 14, BC = 13. Find the, segments of BC made by the bisector of the angle A., Ex. 250. In a triangle CAB, CA = 6, CB = 12, AB = 15. Find the, segments of AB made by the bisector of the angle C.
Page 192 :
BOOK III. PLANE GEOMETRY., , 182, , Proposition XVI. Theorem., 349. The bisector of an exterior angle of a triangle divides the opposite, side externally into segments which are proportional to the adjacent sides., E, , C, , F, M′, , A, , B, , Let CM 0 bisect the exterior angle ACE of the triangle CAB, and, meet BA produced at M 0 ., To prove that, M 0 A : M 0 B = CA : CB., Proof., Draw AF k to M 0 C, meeting BC at F ., Then, M 0 A : M 0 B CF : CB., § 343, Now, ∠M 0 CE = ∠AF C,, § 112, and, ∠M 0 CA = ∠CAF ,, § 110, (being alt.-int. ∠s of k lines)., But, ∠M 0 CE = ∠M 0 CA., Hyp., ∴ ∠AF C = ∠CAF ., ∴ CA = CF ., Put CA for its equal, CF , in the first proportion., Then, M 0 A : M 0 B = CA : CB., , Ax. 1, § 147, q.e.d., , Question. To what case does this theorem not apply? (See Ex. 41,, page 79.), 350. Cor. The bisectors of an interior angle and an exterior angle at, one vertex of a triangle meeting the opposite side divide that side harmonically., § 347
Page 193 :
SIMILAR POLYGONS., , 183, , SIMILAR POLYGONS., 351. Def. Similar polygons are polygons that have their homologous, angles equal, and their homologous sides proportional., B, , B′, , C, , A, , C′, , A′, E, , D, , E′, , D′, , Thus, the polygons ABCDE and A0 B 0 C 0 D0 E 0 are similar, if the ∠s A, B,, C, etc., are equal, respectively, to the ∠s A0 , B 0 , C 0 , etc., and if, AB : A0 B 0 = BC : B 0 C 0 = CD : C 0 D0 , etc., 352. Def. Homologous lines in similar polygons are lines similarly situated., 353. Def. The ratio of any two homologous lines in similar polygons, is, called the ratio of similitude of the polygons., The primary idea of similarity is likeness of form. The two conditions, necessary to similarity are:, 1. For every angle in one of the figures there must be an equal angle in the, other., 2. The homologous sides must be proportional., Thus, Q and Q0 are not similar; the homologous sides are proportional,, but the homologous angles are not equal. Also R and R0 are not similar; the, homologous angles are equal, but the sides are not proportional., Q, , Q′, , R, , R′, , In the case of triangles, either condition involves the other (see § 354 and, § 358).
Page 194 :
BOOK III. PLANE GEOMETRY., , 184, , Proposition XVII. Theorem., 354. Two mutually equiangular triangles are similar., A, , A′, , E, , H, , B, , C, , B′, , C′, , In the triangles ABC and A0 B 0 C 0 , let the angles A, B, C be equal to, the angles A0 , B 0 , C 0 , respectively., To prove that the 4s ABC and A0 B 0 C 0 are similar., Since the 4s are mutually equiangular, we have only to prove that, AB : A0 B 0 = AC : A0 C 0 = BC : B 0 C 0 ., § 351, Proof. Place the 4A0 B 0 C 0 on the 4ABC so that ∠A0 shall coincide with its, equal, the ∠A; and B 0 C 0 take the position EH., Then, ∠AEH = ∠B, Hyp., , That is,, , ∴ EH is k to BC., , § 114, , ∴ AB : AE = AC : AH., , § 343, , AB : A0 B 0 = AC : A0 C 0 ., Similarly, by placing, on 4ABC, so that ∠B 0 shall coincide with its, equal, the ∠B, we may prove that, 4A0 B 0 C 0, , AB : A0 B 0 = BC : B 0 C 0, , q.e.d., , 355. Cor. 1. Two triangles are similar if two angles of the one are equal,, respectively, to two angles of the other., 356. Cor. 2. Two right triangles are similar if an acute angle of the one, is equal to an acute angle of the other.
Page 195 :
SIMILAR POLYGONS., , 185, , Proposition XVIII. Theorem., 357. If two triangles have an angle of the one equal to an angle of the, other, and the including sides proportional, they are similar., A, , A′, , E, B, , H, C, , B′, , C′, , In the triangles ABC and A0 B 0 C 0 , let ∠A = ∠A0 , and let, AB : A0 B0 = AC : A0 C0 ., To prove that the 4s ABC and A0 B 0 C 0 are similar., In this case we prove the 4s similar by proving them mutually equiangular., Proof. Place the 4A0 B 0 C 0 on the 4ABC, so that the ∠A0 shall coincide with, its equal, the ∠A., Then the 4A0 B 0 C 0 will take the position of 4AEH., Now, AB, AC, = 0 0., 0, 0, AB, AC, That is,, AB, AC, =, ., AE, AH, , Hyp., , ∴ EH is k to BC,, § 345, (if a line divides two sides of a 4 proportionally, it is k to the third side)., ∴ ∠AEH = ∠B, and ∠AHE = ∠C., , § 112, , ∴ 4AEH is similar to 4ABC., , § 354, , ∴ 4A0 B 0 C 0 is similar to 4ABC., , q.e.d.
Page 196 :
BOOK III. PLANE GEOMETRY., , 186, , Proposition XIX. Theorem., 358. If two triangles have their sides respectively proportional, they are, similar., A, , A′, , E, B, , H, C, , B′, , C′, , In the triangles ABC and A0 B 0 C 0 , let, AB : A0 B0 = AC : A0 C0 = BC : B0 C0 ., To prove that the 4s ABC and A0 B 0 C 0 are similar., Proof. Upon AB take AE equal to A0 B 0 , and upon AC take AH equal to A0 C 0 ;, and draw EH., Now, AB : A0 B 0 = AC : A0 C 0 ., Hyp., Or, since, AE = A0 B 0 and AH = A0 C 0 ,, AB : AE = AC : AH., , that is,, But, , ∴ 4s ABC and AEH are similar., , § 357, , ∴ AB : AE = BC : EH;, , § 351, , AB : A0 B 0 = BC : EH., AB : A0 B 0 = BC : B 0 C 0 ., , Hyp., , ∴ BC : EH = BC : B 0 C 0 ., , Ax. 1, , ∴ EH =, But, , B0C 0., , Hence, the 4s AEH and A0 B 0 C 0 are equal., , § 150, , 4AEH is similar to 4ABC., ∴ 4A0 B 0 C 0 is similar to 4ABC., , q.e.d.
Page 197 :
SIMILAR POLYGONS., , 187, , Proposition XX. Theorem., 359. Two triangles which have their sides respectively parallel, or respectively perpendicular, are similar., B′, , x, , y, , A′, , D′, H, , C, , E, C′, A, , E′, B, , K, D, , O, , F′, , F, , A0 B 0 C 0, , Let ABC and, have their sides respectively parallel; and DEF, 0, 0, 0, and D E F have their sides respectively perpendicular., To prove that the 4s ABC and A0 B 0 C 0 are similar; and that the 4s DEF and, are similar., , D0 E 0 F 0, , Proof. 1. Prolong BC and AC to B 0 A0 , forming ∠s x and y., Therefore,, In like manner,, , Then ∠B 0 = ∠x (§ 112), and ∠B = ∠x., , § 110, , ∠B 0 = ∠B, , Ax. 1, , ∠A0 = ∠A., , Therefore, 4A0 B 0 C 0 is similar to 4ABC., 2. Prolong DE and F D to meet D0 E 0 at H and D0 F 0 at K., , § 355, , The quadrilateral EHE 0 O has ∠s EHE 0 and E 0 OE right angles, by hypothesis., Therefore,, ∠E 0 and ∠OEH are supplementary., § 206, But, ∠DEF and ∠OEH are supplementary., § 86, In like manner,, , Therefore, ∠DEF = ∠E 0 ., , § 85, , ∠EDF = ∠D0 ., Therefore, 4DEF is similar to 4D0 E 0 F 0 ., , § 355, q.e.d., , 360. Cor. The parallel sides and the perpendicular sides are homologous, sides of the triangles.
Page 198 :
BOOK III. PLANE GEOMETRY., , 188, , Proposition XXI. Theorem., 361. The homologous altitudes of two similar triangles have the same ratio, as any two homologous sides., C′, , C, , A, , O, , B, , A′, , O′, , B′, , In the two similar triangles ABC and A0 B 0 C 0 , let CO and C 0 O 0 be, homologous altitudes., CO, AC, AB, BC, To prove that 0 0 = 0 0 = 0 0 = 0 0 ., CO, AC, AB, BC, Proof. In the rt. 4s COA and C 0 O0 A0 ,, ∠A = ∠A0 ,, (being homologous 4s of the similar 4s ABC and A0 B 0 C 0 )., , § 351, , ∴ 4s COA and C 0 O0 A0 are similar,, § 356, (two rt. 4s having an acute ∠ of the one equal to an acute ∠ of the other are, similar )., CO, AC, ∴ 0 0 = 0 0., § 351, CO, AC, In the similar 4s ABC and A0 B 0 C 0 ,, AC, AB, BC, = 0 0 = 0 0., § 351, A0 C 0, AB, BC, Therefore,, CO, AC, AB, BC, = 0 0 = 0 0 = 0 0., q.e.d., 0, 0, CO, AC, AB, BC, , Ex. 251. The base and altitude of a triangle are 7 feet 6 inches and 5 feet, 6 inches, respectively. If the homologous base of a similar triangle is 5 feet, 6 inches, find its homologous altitude.
Page 199 :
SIMILAR POLYGONS., , 189, , Proposition XXII. Theorem., 362. If two parallels are cut by three or more transversals that pass through, the same point, the corresponding segments are proportional., E′, , D′, , C′, , B′, , A′, , O, A′, A, , B, , B′, , C′, C, , D′, D, , E′, E, , A, , B, , C, , D, , E, , Let the two parallels AE and A0 E 0 be cut by the transversals OA,, OB, OC, OD, OE in A, A0 , B, B 0 , etc., AB, BC, CD, DE, To prove that 0 0 = 0 0 = 0 0 = 0 0 ., AB, BC, CD, DE, Proof. Since A0 E 0 is k to AE, the pairs of 4s OAB and OA0 B 0 , OBC and, OB 0 C 0 , etc., are similar., § 354, AB, OB, BC, OB, ∴ 0 0 =, and 0 0 =, ., § 351, AB, OB 0, BC, OB 0, (homologous sides of similar 4s are proportional )., AB, BC, ∴ 0 0 = 0 0., Ax. 1, AB, BC, In a similar way it may be shown that, CD, CD, DE, BC, = 0 0 and 0 0 = 0 0 ., q.e.d., 0, 0, BC, CD, CD, DE, Note. A condensed form of writing the above is, , , , , , , AB, OB, BC, OC, CD, OD, DE, =, = 0 0 =, = 0 0 =, = 0 0., A0 B 0, OB 0, BC, OC 0, CD, OD0, DE, A parenthesis about a ratio signifies that this ratio is used to prove the equality, of the ratios immediately preceding and following it.
Page 200 :
BOOK III. PLANE GEOMETRY., , 190, , Proposition XXIII. Theorem., 363. Conversely: If three or more non-parallel straight lines intercept, proportional segments upon two parallels, they pass through a common point., O, , D, , B, , A, , F, , C, , E, , Let AB, CD, EF cut the parallels AE and BF so that, AC : BD = CE : DF., To prove that AB, CD, EF prolonged meet in a point., Proof. Prolong AB and CD until they meet in O., Draw OE., Then, But, , Designate by, , F0, , the point where OE cuts BF ., , AC : BD = CE : DF 0 ., , § 362, , AC : BD = CE : DF ., , Hyp., , ∴ CE : DF 0 = CE : DF ., , Ax. 1, , ∴, ∴, , F0, , DF 0, , = DF ., , coincides with F ., , ∴ EF coincides with EF 0 ., , § 47, , ∴ EF prolonged passes through O., ∴ AB, CD, and EF prolonged meet in the point O., , q.e.d.
Page 201 :
SIMILAR POLYGONS., , 191, , Proposition XXIV. Theorem., 364. The perimeters of two similar polygons have the same ratio as any, two homologous sides., E, , E, , A, , D, A, , B, , C, , D, , B, , C, , Let the two similar polygons be ABCDE and A0 B 0 C 0 D 0 E 0 , and let P, and P 0 represent their perimeters., To prove that, P : P 0 = AB : A0 B 0 ., Proof., AB : A0 B 0 = BC : B 0 C 0 = CD : C 0 D0 , etc., § 351, ∴ AB + BC + etc. : A0 B 0 + B 0 C 0 + etc. = AB : A0 B 0 ,, § 335, (in a series of equal ratios the sum of the antecedents is to the sum of the, consequents as any antecedent is to its consequent)., That is,, P : P 0 = AB : A0 B 0 ., q.e.d., , Ex. 252. If the line joining the middle points of the bases of a trapezoid, is produced, and the two legs are also produced, the three lines will meet in, the same point., Ex. 253. AB and AC are chords drawn from any point A in the circumference of a circle, and AD is a diameter. The tangent to the circle at, D intersects AB and AC at E and F , respectively. Show that the triangles, ABC and AEF are similar., Ex. 254. AD and BE are two altitudes of the triangle CAB. Show that, the triangles CED and CAB are similar.
Page 202 :
BOOK III. PLANE GEOMETRY., , 192, , Ex. 255. If two circles are tangent to each other, the chords formed by, a straight line drawn through the point of contact have the same ratio as the, diameters of the circles.
Page 203 :
SIMILAR POLYGONS., , 193, , Proposition XXV. Theorem., 365. If two polygons are similar, they are composed of the same number of, triangles, similar each to each, and similarly placed., E, , E, , A, , D, A, , B, , C, , D, , B, , C, , Let the polygons ABCDE and A0 B 0 C 0 D 0 E 0 be similar., From two homologous vertices, as E and E 0 , draw diagonals EB, EC, and E 0 B 0 ,, , E0C 0., , To prove that the 4s EAB, EBC, ECD are similar, respectively, to the, 4s E 0 A0 B 0 , E 0 B 0 C 0 , E 0 C 0 D0 ., Proof. The 4s EAB and E 0 A0 B 0 are similar., For, ∠A = ∠A0 ,, and, AE : A0 E 0 = AB : A0 B 0 ., Now, ∠ABC = ∠A0 B 0 C 0 ,, and, ∠ABE = ∠A0 B 0 E 0 ., By subtracting,, ∠EBC = ∠E 0 B 0 C 0 ., Now, EB : E 0 B 0 = AB : A0 B 0, and, BC : B 0 C 0 = AB : A0 B 0, ∴ EB : E 0 B 0 = BC : B 0 C 0 ., E0B0C 0, , ∴ 4s EBC and, are similar., In like manner 4s ECD and E 0 C 0 D0 are similar., , § 357, § 351, § 351, § 351, § 351, Ax. 3, § 351, § 351, Ax. 1, § 357, q.e.d.
Page 204 :
BOOK III. PLANE GEOMETRY., , 194, , Proposition XXVI. Theorem., 366. Conversely: If two polygons are composed of the same number of, triangles, similar each to each, and similarly placed, the polygons are similar., E, , E, , A, , D, A, , B, , C, , D, , B, , C, , In the two polygons ABCDE and A0 B 0 C 0 D 0 E 0 , let the triangles AEB,, BEC, CED be similar, respectively, to the triangles A0 E 0 B 0 , B 0 E 0 C 0 ,, C 0 E 0 D 0 ; and similarly placed., To prove that ABCDE is similar to A0 B 0 C 0 D0 E 0 ., Proof., ∠A = ∠A0, Also,, ∠ABE = ∠A0 B 0 E 0 ,, and, ∠EBC = ∠E 0 B 0 C 0 ., By adding,, ∠ABC = ∠A0 B 0 C 0 ., 0, In like manner, ∠BCD = ∠B C 0 D0 , ∠CDE = ∠C 0 D0 E 0 , etc., Hence, the polygons are mutually equiangular., , , , , AB, EB, BC, EC, CD, Also, 0 0 =, = 0 0 =, = 0 0 , etc., 0, 0, 0, 0, AB, EB, BC, EC, CD, Hence, the polygons have their homologous sides proportional., Therefore, the polygons are similar., , § 351, , § 351, Ax. 2, , § 351, § 351, q.e.d.
Page 205 :
EXERCISES., , 195, , THEOREMS., Ex. 256. If two circles are tangent to each other externally, the corresponding segments of two lines drawn through the point of contact and terminated, by the circumferences are proportional., Ex. 257. In a parallelogram ABCD, a line DE is drawn, meeting the, diagonal AC in F , the side BC in G, and the side AB produced in E., 2, Prove that DF = F G × F E., Ex. 258. Two altitudes of a triangle are inversely proportional to the, corresponding bases., Ex. 259. Two circles touch at P . Through P three lines are drawn,, meeting one circle in A, B, C, and the other in A0 , B 0 , C 0 , respectively. Prove, that the triangles ABC, A0 B 0 C 0 are similar., Ex. 260. Two chords AB, CD intersect at M , and A is the middle point, of the arc CD. Prove that the product AB × AM is constant if the chord AB, is made to turn about the fixed point A., Draw the diameter AE, and draw BE., Ex. 261. If two circles touch each other, their common external tangent, is the mean proportional between their diameters., Let AB be the common tangent. Draw the diameters AC, BD. Join the, point of contact P to A, B, C, and D. Show that AP D and BP C are straight, lines ⊥ to each other, and that 4s CAB, ABD are similar., Ex. 262. If two circles are tangent internally, all chords of the greater, circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle., Draw any two of the chords, and join the points where they meet the, circumferences. The 4s thus formed are similar (Ex. 120).
Page 206 :
BOOK III. PLANE GEOMETRY., D, , 196, , C, E, , A, B, , Ex. 263. In an inscribed quadrilateral, the product of the diagonals is, equal to the sum of the products of the opposite sides., Draw DE, making ∠CDE = ∠ADB. The 4s ABD and ECD are similar;, and the 4s BCD and AED are similar., Ex. 264. Two isosceles triangles with equal vertical angles are similar., Ex. 265. The bisector of the vertical angle A of the triangle ABC intersects, the base at D and the circumference of the circumscribed circle at E., Show that AB × AC = AD × AE.
Page 207 :
NUMERICAL PROPERTIES OF FIGURES., , 197, , NUMERICAL PROPERTIES OF FIGURES., Proposition XXVII. Theorem., 367. If in a right triangle a perpendicular is drawn from the vertex of the, right angle to the hypotenuse:, 1. The triangles thus formed are similar to the given triangle, and to each, other., 2. The perpendicular is the mean proportional between the segments of the, hypotenuse., 3. Each leg of the right triangle is the mean proportional between the hypotenuse and its adjacent segment., C, b′ a, , A, , a′, , F, , b, B, , In the right triangle ABC, let CF be drawn from the vertex of the, right angle C, perpendicular to AB., 1. To prove that 4’s BCA, CF A, BF C are similar., Proof. The rt. 4s CF A and BCA are similar,, , § 356, , since the ∠a0 is common., The rt. 4s BF C and BCA are similar,, , § 356, , since the ∠b is common., Since the 4s CF A and BF C are each similar to 4BCA, they are similar to each, other., § 354, 2. To prove that, AF : CF = CF : F B., Proof. In the similar 4s CF A and BF C,, AF : CF = CF : F B., , § 351
Page 208 :
BOOK III. PLANE GEOMETRY., , 198, , 3. To prove that, AB : AC = AC : AF ,, and, AB : BC = BC : BF ., Proof. In the similar 4s BCA and CF A,, AB : AC = AC : AF, In the similar 4s BCA and BF C,, AB : BC = BC : BF ., , § 351, § 351, q.e.d., , 368. Cor. 1. The squares of the two legs of a right triangle are proportional, to the adjacent segments of the hypotenuse., From the proportions in § 367,3,, 2, , Hence,, , 2, , AC = AB × AF , and BC = AB × BF ., AC, , 2, , BC, , 2, , =, , § 327, , AF, AB × AF, =, ., AB × BF, BF, , 369. Cor. 2. The squares of the hypotenuse and either leg are proportional, to the hypotenuse and the adjacent segment., For, 2, AB, AB × AB, AB, =, ., 2 =, AB × AF, AF, AC, C, , A, , D, , B, , 370. Cor. 3. The perpendicular from any point in the circumference to, the diameter of a circle is the mean proportional between the segments of the, diameter., The chord drawn from any point in the circumference to either extremity of, the diameter is the mean proportional between the diameter and the adjacent, segment., For, the ∠ACB is a rt. ∠., § 290
Page 209 :
NUMERICAL PROPERTIES OF FIGURES., , 199, , Proposition XXVIII. Theorem., 371. The sum of the squares of the two legs of a right triangle is equal to, the square of the hypotenuse., C, , A, , F, , B, , Let ABC be a right triangle with its right angle at C., To prove that, 2, 2, 2, AC + CB = AB ., Proof., Draw CF ⊥ to AB., Then, 2, AC = AB × AF ,, and, 2, CB = AB × BF ., By adding,, 2, 2, 2, AC + CB = AB(AF + BF ) = AB, , § 367, q.e.d., , 372. Cor. 1. The square of either leg of a right triangle is equal to the, difference of the square of the hypotenuse and the square of the other leg., D, , C, , C, , D, , A, , B, , A, , P, , R, , B, , 373. Cor. 2. The diagonal and a side of a square are incommensurable., For, 2, 2, 2, 2, AC = AB + BC = 2AB ., √, ∴ AC = AB 2., 374. Def. The projection of any line upon a second line is the segment, of the second line included between the perpendiculars drawn to it from the, extremities of the first line. Thus, P R is the projection of CD upon AB.
Page 210 :
BOOK III. PLANE GEOMETRY., , 200, , Proposition XXIX. Theorem., 375. In any triangle, the square of the side opposite an acute angle is equal, to the sum of the squares of the other two sides diminished by twice the product, of one of those sides by the projection of the other upon that side., A, , A, , B, , C, , D, , D, , B, , C, , Let C be an acute angle of the triangle ABC, and DC the projection, of AC upon BC., 2, , 2, , 2, , To prove that AB = BC + AC − 2BC × DC., Proof. If D falls upon the base (Fig. 1),, DB = BC − DC., If D falls upon the base produced (Fig. 2),, DB = DC − BC., , In either case,, 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , DB = BC + DC − 2BC × DC., 2, Add AD to both sides of this equality, and we have, 2, , AD + DB = BC + AD + DC − 2BC × DC., , But, , AD + DB = AB, , and, 2, , 2, , AD + DC = AC, , Put AB and AC for their equals in the above equality., Then, 2, 2, 2, AB = BC + AC − 2BC × DC., , § 371, , q.e.d.
Page 211 :
NUMERICAL PROPERTIES OF FIGURES., , 201, , Proposition XXX. Theorem., 376. In any obtuse triangle, the square of the side opposite the obtuse angle, is equal to the sum of the squares of the other two sides increased by twice the, product of one of those sides by the projection of the other upon that side., A, , D, , B, , C, , Let C be the obtuse angle of the triangle ABC, and CD be the projection of AC upon BC produced., 2, , 2, , 2, , To prove that AB = BC + AC + 2BC × DC., Proof., DB = BC + DC., Squaring,, 2, 2, 2, DB = BC + DC + 2BC × DC., 2, Add AD to both sides, and we have, 2, , 2, , 2, , 2, , 2, , AD + DB = BC + AD + DC + 2BC × DC., , But, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , AD + DB = AB , and AD + DC = AC ., 2, , Put AB and AC for their equals in the above equality., Then, 2, 2, 2, AB = BC + AC + 2BC × DC., , § 371, , q.e.d., , Note 1. By the Principle of Continuity the last three theorems may be included, in one theorem. Let the student explain., Note 2. The last three theorems enable us to compute the lengths of the altitudes of a triangle if the lengths of the three sides are known.
Page 212 :
BOOK III. PLANE GEOMETRY., , 202, , Proposition XXXI. Theorem., 377. 1. The sum of the squares of two sides of a triangle is equal to twice, the square of half the third side increased by twice the square of the median, upon that side., 2. The difference of the squares of two sides of a triangle is equal to twice, the product of the third side by the projection of the median upon that side., A, , B, , M, , D, , C, , In the triangle ABC, let AM be the median and M D the projection, of AM upon the side BC. Also, let AB be greater than AC., To prove that, 2, 2, 2, 2, 1. AB + AC = 2BM + 2AM ., 2, , 2, , 2. AB − AC = 2BC × M D., Proof. Since AB > AC, the ∠AM B will be obtuse, and the ∠AM C will be, acute., § 155, Then, 2, 2, 2, AB = BM + AM + 2BM × M D,, § 376, and, 2, 2, 2, AC = M C + AM − 2M C × M D., § 375, Add these two equalities, and observe that BM = M C., Then, 2, 2, 2, 2, AB + AC = 2BM + 2AM ., Subtract the second equality from the first., Then, 2, 2, AB − AC = 2BC × M D., q.e.d., Note. This theorem enables us to compute the lengths of the medians of a, triangle if the lengths of the three sides are known.
Page 213 :
NUMERICAL PROPERTIES OF FIGURES., , 203, , Proposition XXXII. Theorem., 378. If two chords intersect in a circle, the product of the segments of one, is equal to the product of the segments of the other., P, , c′, , c, O, , M, , a′, , N, , a, , Q, , Let any two chords M N and P Q intersect at O., To prove that, OM × ON = OQ × OP ., Proof., Draw M P and N Q., ∠a = ∠a0 ,, (each being measured by, And, , 1, 2, , § 289, arc P N )., , ∠c = ∠c0 ,, (each being measured by 12 arc M Q)., , § 289, , ∴ the 4s N OQ and P OM are similar., , § 355, , ∴ OQ : OM = ON : OP ., , § 351, , ∴ OM × ON = OQ × OP ., , § 327, q.e.d., , 379. Scholium. This proportion may be written, OP, OM, 1, OM, =, , or, =, ;, ON, OQ, ON, OQ, OP, that is, the ratio of two corresponding segments is equal to the reciprocal, of the ratio of the other two segments., Hence, these segments are said to be reciprocally proportional.
Page 214 :
BOOK III. PLANE GEOMETRY., , 204, , 380. Def. A secant from a point to a circle is understood to mean, the segment of the secant lying between the point and the second point of, intersection of the secant and circumference., Proposition XXXIII. Theorem., 381. If from a point without a circle a secant and a tangent are drawn,, the tangent is the mean proportional between the whole secant and its external, segment., A, b, B, a, , C, , D, , a′, , Let AD be a tangent and AC a secant drawn from the point A to the, circle BCD., To prove that AC : AD = AD : AB., Proof., Draw DC and DB., The 4s ADC and ABD are similar., For ∠b is common; and ∠a0 = ∠a,, (each being measured by 21 arc BD)., ∴ AC : AD = AD : AB., , § 355, §§ 289, 295, § 351, q.e.d., , 382. Cor. If from a fixed point without a circle a secant is drawn, the, product of the secant and its external segment is constant in whatever direction, the secant is drawn., For, 2, AC × AB = AD ., § 327
Page 215 :
NUMERICAL PROPERTIES OF FIGURES., , 205, , Proposition XXXIV. Theorem., 383. The square of the bisector of an angle of a triangle is equal to the, product of the sides of this angle diminished by the product of the segments, made by the bisector upon the third side of the triangle., N, b′ b, M, , a′, , P, , O, , a, Q, , Let N O bisect the angle M N P of the triangle M N P ., 2, , To prove that N O = N M × N P − OM × OP ., Proof., Circumscribe the M N P about the 4M N P ., Produce N O to meet the circumference in Q, and draw P Q., For, , § 314, , The 4s N QP and N M O are similar., , § 355, , ∠b = ∠b0, , Hyp., , ∠a = ∠a0, , § 289, , N Q : N M = N P : N O., , § 351, , and, Whence, , ∴ NM × NP = NQ × NO, = (N O + OQ)N O, 2, , But, , = N O + N O × OQ., N O × OQ = M O × OP ., , § 378, , 2, , Whence, , ∴ M N × N P = N O + M O × OP ., 2, , N O = N M × N P = M O × OP ., , Ax. 3, q.e.d.
Page 216 :
BOOK III. PLANE GEOMETRY., , 206, , Note. This theorem enables us to compute the lengths of the bisectors of the, angles of a triangle if the lengths of the sides are known., , Proposition XXXV. Theorem., 384. In any triangle the product of two sides is equal to the product of the, diameter of the circumscribed circle by the altitude upon the third side., N, a, , M, , Q, , O, a′, , P, , Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed about the triangle NMQ., Draw the diameter N P , and draw P Q., To prove that N M × N Q = N P × N O., Proof. In the 4s N OM and N QP ,, ∠N OM is a rt. ∠,, , Hyp., , ∠N QP is a rt. ∠,, , § 290, , ∠a0 ,, , and ∠a =, (each being measured by, Whence, , 1, 2, , § 289, arc N Q)., , ∴ 4s N OM and N QP are similar., , § 356, , N M : N P = N O : N Q., , § 351, , ∴ N M × N Q = N P × N O., , § 327, q.e.d., , Note. This theorem enables us to compute the length of the radius of a circle, circumscribed about a triangle, if the lengths of the three sides of the triangle are, known.
Page 217 :
EXERCISES., , 207, , Ex. 266. If OE, OF , OG are the perpendiculars from any point O within, 2, the triangle ABC upon the sides AB, BC, CA, respectively, show that AE +, 2, 2, 2, 2, 2, BF + CG = EB + F C + GA ., THEOREMS., Ex. 267. The sum of the squares of the segments of two perpendicular, chords is equal to the square of the diameter of the circle., If AB, CD are the chords, draw the diameter BE, draw AC, ED, BD., Prove that AC = ED, and apply § 371., Ex. 268. The tangents to two intersecting circles drawn from any point, in their common chord produced, are equal. (§ 381.), Ex. 269. The common chord of two intersecting circles, if produced, will, bisect their common tangents. (§ 381.), A, D, P, C, , E, , O, B, , Ex. 270. If three circles intersect one another, the common chords all pass, through the same point., Let two of the chords AB and CD meet at O. Join the point of intersection, E to O, and suppose that EO produced meets the same two circles at two, different points P and Q. Then prove that OP = OQ (§ 378); hence, that the, points P and Q coincide., Ex. 271. If two circles are tangent to each other, the common internal, tangent bisects the two common external tangents., Ex. 272. If the perpendiculars from the vertices of the triangle ABC upon, the opposite sides intersect at D, show that, 2, , 2, , 2, , 2, , AB − AC = BD − CD .
Page 218 :
BOOK III. PLANE GEOMETRY., , 208, , Ex. 273. In an isosceles triangle, the square of a leg is equal to the square, of any line drawn from the vertex to the base, increased by the product of the, segments of the base., Ex. 274. The squares of two chords drawn from the same point in a, circumference have the same ratio as the projections of the chords on the, diameter drawn from the same point., Ex. 275. The difference of the squares of two sides of a triangle is equal, to the difference of the squares of the segments of the third side, made by the, perpendicular on the third side from the opposite vertex., Ex. 276. E is the middle point of BC, one of the parallel sides of the, trapezoid ABCD; AE and DE produced meet DC and AB produced at F, and G, respectively. Show that F G is parallel to DA., 4s AGD and BGE are similar; and 4s AF D and EF C are similar., Ex. 277. If two tangents are drawn to a circle at the extremities of a, diameter, the portion of a third tangent intercepted between them is divided, at its point of contact into segments whose product is equal to the square of, the radius., Ex. 278. If two exterior angles of a triangle are bisected, the line drawn, from the point of intersection of the bisectors to the opposite angle of the, triangle bisects that angle., Ex. 279. The sum of the squares of the diagonals of a quadrilateral is, equal to twice the sum of the squares of the lines that join the middle points, of the opposite sides., A, , F, D, , F, , B, , C, , H, , E, C, , A, , B, , D, , Ex. 280. The sum of the squares of the four sides of any quadrilateral is, equal to the sum of the squares of the diagonals, increased by four times the, square of the line joining the middle points of the diagonals., Apply § 377 to the 4s ABC and ADC, add the results, and eliminate, 2, 2, BE + DE by applying § 377 to the 4BDE.
Page 219 :
EXERCISES., , 209, , Ex. 281. The square of the bisector of an exterior angle of a triangle is, equal to the product of the external segments determined by the bisector upon, one of the sides, diminished by the product of the other two sides., Let CD bisect the exterior ∠BCH of the 4ABC. 4s ACD and F CB are, similar (§ 355). Apply § 382., Ex. 282. If a point O is joined to the vertices of a triangle ABC; through, any point A0 in OA a line parallel to AB is drawn, meeting OB at B 0 ; through, B 0 a line parallel to BC, meeting OC at C 0 ; and C 0 is joined to A0 ; the triangle, A0 B 0 C 0 is similar to the triangle ABC., Ex. 283. If the line of centres of two circles meets the circumferences at, the consecutive points A, B, C, D, and meets the common external tangent, at P , then P A × P D = P B × P C., Ex. 284. The line of centres of two circles meets the common external tangent at P , and a secant is drawn from P , cutting the circles at the consecutive, points E, F , G, H. Prove that P E × P H = P F × P G., Draw radii to the points of contact, and to E, F , G, H. Let fall ⊥s on P H, from the centres of the s . The various pairs of 4s are similar., Ex. 285. If a line drawn from a vertex of a triangle divides the opposite, side into segments proportional to the adjacent sides, the line bisects the angle, at the vertex.
Page 220 :
BOOK III. PLANE GEOMETRY., , 210, , PROBLEMS OF CONSTRUCTION., Proposition XXXVI. Problem., 385. To divide a given straight line into parts proportional to any number, of given lines., H, , A, , K, , B, , m, C, n, m, n, p, , E, , p, F, , X, , Let AB, m, n, and p be given straight lines., To divide AB into parts proportional to m, n, and p., Draw AX, making any convenient ∠ with AB., On AX take AC equal to m, CE to n, EF to p., Draw BF ., From E and C draw EK and CH k to F B., Through A draw a line k to BF ., K and H are the division points required., Proof., , AH, HK, KB, =, =, ,, § 344, AC, CE, EF, (if two lines are cut by any number of parallels, the corresponding intercepts are, proportional )., Substitute m, n, and p for their equals AC, CE, and EF ., Then, AH, HK, KB, =, =, ., q.e.f., m, n, p, , Ex. 286. Divide a line 12 inches long into three parts proportional to the, numbers 3, 5, 7.
Page 221 :
PROBLEMS OF CONSTRUCTION., , 211, , Proposition XXXVII. Problem., 386. To find the fourth proportional to three given straight lines., A, , B, , m, , n, , C, , x, , p, D, F, m, n, p, , y, , Let the three given lines be m, n, and p., To find the fourth proportional to m, n, and p., Draw Ax and Ay containing any convenient angle., On Ax take AB equal to m, BC to n., On Ay take AD equal to p., Draw BD., From C draw CF k to BD, meeting Ay at F ., DF is the fourth proportional required., Proof., AB : BC = AD : DF ,, § 342, (a line drawn through two sides of a 4 k to the third side divides those sides, proportionally)., Then, , Substitute m, n, and p for their equals AB, BC, and AD., m : n = p : DF, , q.e.f., , Ex. 287. The square of the altitude of an equilateral triangle is equal to, three fourths of the square of one side of the triangle.
Page 222 :
BOOK III. PLANE GEOMETRY., , 212, , Proposition XXXVIII. Problem., 387. To find the third proportional to two given straight lines., A, , m, n, B, , C, , D, , E, , Let m and n be the two given straight lines., To find the third proportional to m and n., Construct any convenient angle A,, and take AB equal to m, AC equal to n., Produce AB to D, making BD equal to AC., Draw BC., Through D draw DE k to BC, meeting AC produced at E., CE is the third proportional required., Proof., AB : BD = AC : CE,, § 342, (a line drawn through two sides of a 4 parallel to the third side divides those, sides proportionally)., Substitute, in the above proportion, AC for its equal BD., Then, AB : AC = AC : CE,, that is,, m : n = n : CE., q.e.f., , ab, a2, , (2) x = ., c, c, Special cases: (1) a = 2, b = 8, c = 4; (2) a = 3, b = 7, c = 11; (3) a = 2,, c = 3; (4) a = 3, c = 5; (5) a = 2c., Ex. 288. Construct x, if (1) x =
Page 223 :
PROBLEMS OF CONSTRUCTION., , 213, , Proposition XXXIX. Problem., 388. To find the mean proportional between two given straight lines., H, , m, n, A, , m, , C, , n, , B, , E, , Let the two given lines be m and n., To find the mean proportional between m and n., On the straight line AE, take AC equal to m, and CB equal to n., On AB as a diameter describe a semicircumference., At C erect the ⊥ CH meeting the circumference at H., CH is the mean proportional between m and n., Proof., AC : CH = CH : CB, § 370, (the ⊥ let fall from a point in a circumference to the diameter of a circle is, the mean proportional between the segments of the diameter )., Substitute for AC and CB their equals m and n., Then, m : CH = CH : n., q.e.f., , 389. Def. A straight line is divided in extreme and mean ratio, when, one of the segments is the mean proportional between the whole line and the, other segment., √, Ex. 289. Construct x, if x = ab., Special cases: (1) a = 2, b = 3; (2) a = 1, b = 6; (3) a = 3, b = 7.
Page 224 :
BOOK III. PLANE GEOMETRY., , 214, , Proposition XL. Problem., 390. To divide a given line in extreme and mean ratio., G, E, , F, C′, , A, , C, , B, , Let AB be the given line., To divide AB in extreme and mean ratio., At B erect a ⊥ BE equal to half of AB., From E as a centre, with a radius equal to EB, describe a, , ., , Draw AE, meeting the circumference in F and G., On AB take AC equal to AF ., On BA produced take AC 0 equal to AG., Then AB is divided internally at C and externally at C 0 in extreme and mean, ratio., AG : AB = AB : AF ., 2, , AB = AF × AG, = AC(AF + AG), = AC(AC + AB), 2, = AC + AB × AC., 2, 2, ∴ AB − AB × AC = AC ., 2, ∴ AB(AB − AC) = AC ., 2, ∴ AB × CB, = AC ., , § 381, , 2, , AB = AG × AF, = C 0 A(AG − AF ), = C 0 A(C 0 A − AB), 2, = C 0 A − AB × C 0 A., 2, 2, ∴ AB + AB × C 0 A = C 0 A ., 2, ∴ AB(AB + C 0 A), = C 0A ., 2, ∴ AB × C 0 B, = C 0A ., q.e.f.
Page 225 :
PROBLEMS OF CONSTRUCTION., , 215, , Proposition XLI. Problem., 391. Upon a given line homologous to a given side of a given polygon, to, construct a polygon similar to the given polygon., E, , E, A, , D, A, , B, , A0 E 0, , Let, ABCDE., , C, , D, , B, , C, , be the given line homologous to AE of the given polygon, , To construct on A0 E 0 a polygon similar to the given polygon., From E draw the diagonals EB and EC., From E 0 draw E 0 B 0 , E 0 C 0 , and E 0 D0 ,, making 4’s A0 E 0 B 0 , B 0 E 0 C 0 , and C 0 E 0 D0 equal, respectively, to, 4s AEB, BEC, and CED., From A0 draw A0 B 0 , making ∠E 0 A0 B 0 equal to ∠EAB,, and meeting E 0 B 0 at B 0 ., From B 0 draw B 0 C 0 , making ∠E 0 B 0 C 0 equal to ∠EBC,, and meeting E 0 C 0 at C 0 ., From C 0 draw C 0 D0 , making ∠E 0 C 0 D0 equal to ∠ECD,, and meeting E 0 D0 at D0 ., Then A0 B 0 C 0 D0 E 0 is the required polygon., Proof., , The 4s ABE, A0 B 0 E 0 , etc., are similar., , § 354, , Therefore, the two polygons are similar., , § 366, q.e.f.
Page 226 :
BOOK III. PLANE GEOMETRY., , 216, , PROBLEMS OF CONSTRUCTION., Ex. 290. To divide one side of a given triangle into segments proportional, to the adjacent sides (§ 348)., E, , A, , C, , A, , A, , B, , D, , H, , P, , C, , F, , O, , C, , D, , P, , B, O, , O, , B, , O, , E, A, , F, , G, , D, C, P, , B, , Ex. 291. To find in one side of a given triangle a point whose distances, from the other sides shall be to each other in the given ratio m : n., Take AG = m ⊥ to AC, GH = n ⊥ to BC. Draw CD k to OG., Ex. 292. Given an obtuse triangle; to draw a line from the vertex of, the obtuse angle to the opposite side which shall be the mean proportional, between the segments of that side., Ex. 293. Through a given point P within a given circle to draw a chord, AB so that the ratio AP : BP shall equal the given ratio m : n., Draw OP C so that OP : P C = n : m. Draw CA equal to the fourth, proportional to n, m, and the radius of the circle., Ex. 294. To draw through a given point P in the arc subtended by a chord, AB a chord which shall be bisected by AB., On radius OP take CD equal to CP . Draw DE k to BA., C, , B, , B, , G, , A, O, , A, , P, D, , C, , P, , C, , D, , A, , E, , D P, O, , D, P, , A, , E, , F, , O′, , B, , Ex. 295. To draw through a given external point P a secant P AB to a, given circle so that the ratio P A : AB shall equal the given ratio m : n., P D : DC = m : n. P D : P A = P A : P C.
Page 227 :
EXERCISES., , 217, , Ex. 296. To draw through a given external point P a secant P AB to a, 2, given circle so that AB = P A × P B., P C : CD = CD : P D. P A = CD., Ex. 297. To find a point P in the arc subtended by a given chord AB so, that the ratio P A : P B shall equal the given ratio m : n., Ex. 298. To draw through one of the points of intersection of two circles, a secant so that the two chords that are formed shall be in the given ratio, m : n., Ex. 299. Having given the greater segment of a line divided in extreme, and mean ratio, to construct the line., Ex. 300. To construct a circle which shall pass through two given points, and touch a given straight line., Ex. 301. To construct a circle which shall pass through a given point and, touch two given straight lines., Ex. 302. To inscribe a square in a semicircle., M, , C, G, , F, , A D H EB, , N, , Ex. 303. To inscribe a square in a given triangle., Let DEF G be the required inscribed square. Draw CM k to AB, meeting, AF produced in M . Draw CH and M N ⊥ to AB, and produce AB to meet, M N at N . The 4s ACM , AGF are similar; also, the 4s AM N , AF E are, similar. By these triangles show that the figure CM N H is a square. By, constructing this square, the point F can be found., Ex. 304. To inscribe in a given triangle a rectangle similar to a given, rectangle., Ex. 305. To inscribe in a circle a triangle similar to a given triangle.
Page 228 :
BOOK III. PLANE GEOMETRY., , 218, , Ex. 306. To inscribe in a given semicircle a rectangle similar to a given, rectangle., Ex. 307. To circumscribe about a circle a triangle similar to a given, triangle., Ex. 308. To construct the expression, x =, , 2ab c, 2abc, ; that is,, × ., de, d, e, , Ex. 309. To construct two straight lines, having given their sum and their, ratio., Ex. 310. To construct two straight lines, having given their difference, and their ratio., Ex. 311. Given two circles, with centres O and O0 , and a point A in their, plane, to draw through the point A a straight line, meeting the circumferences, at B and C, so that AB : AC = m : n.
Page 230 :
BOOK III. PLANE GEOMETRY., , h=, , 2p, s(s − a)(s − b)(s − c)., c, C, , b, , a, m h, , A, , c, , F, , D, , B, , Ex. 313. To compute the medians of a triangle in terms of its sides., By § 377,, c 2, a2 + b2 = 2m2 + 2, ., 2, Whence, 4m2 = 2(a2 + b2 ) − c2 ., 1p 2, ∴m=, 2(a + b2 ) − c2 ., 2, , 220
Page 231 :
EXERCISES., , 221, , C, b, A, , t, , c, , D, , a, B, , E, , Ex. 314. To compute the bisectors of a triangle in terms of the sides., By § 383,, t2 = ab − AD × BD., By §348,, BD, AD + BD, c, AD, =, =, =, ., B, a, a+b, a+b, bc, ac, ∴ AD =, , and BD =, ., a+b, a+b, Whence, abc2, t2 = ab −, (a + b)2, , , c2, = ab 1 −, (a + b)2, ab{(a + b)2 − c2 }, =, (a + b)2, ab(a + b + c)(a + b − c), =, (a + b)2, ab × 2s × 2(s − c), ., =, (a + b2 ), Whence, 2 p, t=, abs(s − c)., a+b
Page 232 :
BOOK III. PLANE GEOMETRY., , 222, , A, c, B, , a, , b, D, , C, , E, , Ex. 315. To compute the radius of the circle circumscribed about a triangle, in terms of the sides of the triangle., By §384,, AC × AB = AE × AD,, or,, bc = 2B × AD., But, 2p, s(s − a)(s − b)(s − c)., Ex. 312, AD =, a, abc, ∴R= p, ., 4 s(s − a)(s − b)(s − c), Ex. 316. If the sides of a triangle are 3, 4, and 5, is the angle opposite 5, right, acute, or obtuse?, Ex. 317. If the sides of a triangle are 7, 9, and 12, is the angle opposite 12, right, acute, or obtuse?, Ex. 318. If the sides of a triangle are 7, 9, and 11, is the angle opposite 11, right, acute, or obtuse?, Ex. 319. The legs of a right triangle are 8 inches and 12 inches; find the, lengths of the projections of these legs upon the hypotenuse, and the distance, of the vertex of the right angle from the hypotenuse., Ex. 320. If the sides of a triangle are 6 inches, 9 inches, and 12 inches,, find the lengths (1) of the altitudes; (2) of the medians; (3) of the bisectors;, (4) of the radius of the circumscribed circle., Ex. 321. A line is drawn parallel to a side AB of a triangle ABC, cutting, AC in D, BC in E. If AD : DC = 2 : 3, and AB = 20 inches, find DE.
Page 233 :
EXERCISES., , 223, , Ex. 322. The sides of a triangle are 9, 12, 15. Find the segments of the, sides made by bisecting the angles., Ex. 323. A tree casts a shadow 90 feet long, when a post 6 feet high casts, a shadow 4 feet long. How high is the tree?, Ex. 324. The lower and upper bases of a trapezoid are a, b, respectively;, and the altitude is h. Find the altitudes of the two triangles formed by producing the legs until they meet., Ex. 325. The sides of a triangle are 6, 7, 8, respectively. In a similar, triangle the side homologous to 8 is 40. Find the other two sides., Ex. 326. The perimeters of two similar polygons are 200 feet and 300 feet., If a side of the first is 24 feet, find the homologous side of the second., Ex. 327. How long a ladder is required to reach a window 24 feet high, if, the lower end of the ladder is 10 feet from the side of the house?, Ex. 328. If the side of an equilateral triangle is a, find the altitude., Ex. 329. If the altitude of an equilateral triangle is h, find the side., Ex. 330. Find the length of the longest chord and of the shortest chord, that can be drawn through a point 6 inches from the centre of a circle whose, radius is 10 inches., Ex. 331. The distance from the centre of a circle to a chord 10 feet long, is 12 feet. Find the distance from the centre to a chord 24 feet long., Ex. 332. The radius of a circle is 5 inches. Through a point 3 inches, from the centre a diameter is drawn, and also a chord perpendicular to the, diameter. Find the length of this chord, and the distance from one end of the, chord to the ends of the diameter., Ex. 333. The radius of a circle is 6 inches. Find the lengths of the tangents, drawn from a point 10 inches from the centre, and also the length of the chord, joining the points of contact., Ex. 334. The sides of a triangle are 407 feet, 368 feet, and 351 feet. Find, the three bisectors and the three altitudes.
Page 234 :
BOOK III. PLANE GEOMETRY., , 224, , Ex. 335. If a chord 8 inches long is 8 inches distant from the centre of, the circle, find the radius, and the chords drawn from the end of the chord to, the ends of the diameter which bisects the chord., Ex. 336. From the end of a tangent 20 inches long a secant is drawn, through the centre of the circle. If the external segment of this secant is, 8 inches, find the radius of the circle., Ex. 337. The radius of a circle is 13 inches. Through a point 5 inches, from the centre any chord is drawn. What is the product of the two segments, of the chord? What is the length of the shortest chord that can be drawn, through the point?, Ex. 338. The radius of a circle is 9 inches and the length of a tangent, 12 inches. Find the length of a line drawn from the extremity of the tangent, to the centre of the circle., Ex. 339. Two circles have radii of 8 inches and 3 inches, respectively,, and the distance between their centres is 15 inches. Find the lengths of their, common tangents., Ex. 340. Find the segments of a line 10 inches long divided in extreme, and mean ratio., Ex. 341. The sides of a triangle are 4, 5, 5. Is the largest angle acute,, right, or obtuse?, Ex. 342. Find the third proportional to two lines whose lengths are 28 feet, and 42 feet., Ex. 343. If the sides of a triangle are a, b, c, respectively, find the lengths, of the three altitudes., Ex. 344. The diameter of a circle is 30 feet and is divided into five equal, parts. Find the lengths of the chords drawn through the points of division, perpendicular to the diameter., Ex. 345. The radius of a circle is 2 inches. From a point 4 inches from, the centre a secant is drawn so that the internal segment is 1 inch. Find the, length of the secant.
Page 235 :
EXERCISES., , 225, , Ex. 346. The sides of a triangular pasture are 1551 yards, 2068 yards,, 2585 yards. Find the median to the longest side., Ex. 347. The diagonal of a rectangle is d, and the perimeter is p. Find, the sides., Ex. 348. The radius of a circle is r. Find the length of a chord whose, distance from the centre is 12 r.
Page 236 :
BOOK IV. AREAS OF POLYGONS., , 392. Def. The unit of surface is a square whose side is a unit of length., 393. Def. The area of a surface is the number of units of surface it, contains., 394. Def. Plane figures that have equal areas but cannot be made to coincide are called equivalent., Note. In propositions relating to areas, the words “rectangle,” “triangle,” etc.,, are often used for “area of rectangle,” “area of triangle,” etc., , Proposition I. Theorem., 395. Two rectangles having equal altitudes are to each other as their bases., D, , A, , O, , C, , D, , B, , A, , F, , O, , E, , Let the rectangles AC and AF have the same altitude AD., To prove that rect. AC : rect. AF = base AB : base AE., Case 1. When AB and AE are commensurable., Proof. Suppose AB and AE have a common measure, as AO, which is contained, m times in AB and n times in AE., Then, AB : AE = m : n.
Page 237 :
AREAS OF POLYGONS., , 227, , Apply AO as a unit of measure to AB and AE, and at the several points of, division erect ⊥s ., The, rect. AC is divided into m rectangles,, and the, rect. AF is divided into n rectangles., § 107, These rectangles are all equal., , Hence,, Therefore,, , § 186, , rect. AC : rect. AF = m : n., , rect. AC : rect. AF = AB : AE., Case 2. When AB and AE are incommensurable., D, , C, , D, , A, , B, , A, , Ax. 1, H F, , K, , E, , Proof. Divide AB into any number of equal parts, and apply one of them to, AE as many times as AE will contain it., Since AB and AE are incommensurable, a certain number of these parts will, extend from A to some point K, leaving a remainder KE less than one of the equal, parts of AB., Draw KH k to EF ., Then AB and AK are commensurable by construction., Therefore,, rect. AH, AK, =, ., Case 1, rect. AC, AB, If the number of equal parts into which AB is divided is indefinitely increased,, the varying values of these ratios will continue equal, and approach for their respective limits the ratios, rect. AF, AE, and, . (See § 287.), rect. AC, AB, rect. AF, AE, ∴, =, ., § 284, rect. AC, AB, q.e.d., , 396. Cor. Two rectangles having equal bases are to each other as their, altitudes.
Page 238 :
BOOK IV. PLANE GEOMETRY., , 228, , Proposition II. Theorem., 397. Two rectangles are to each other as the products of their bases by their, altitudes., , a, , R, , b, , a′, , R′, b′, , a′, , S, b, , R0, , Let R and, be two rectangles, having for their bases b and b0 , and, for their altitudes a and a0 , respectively., To prove that, R, a×b, = 0, ., R0, a × b0, Proof. Construct the rectangle S, with its base equal to that of R, and its, altitude equal to that of R0 ., Then, a, R, = 0,, § 396, S, a, and, S, b, = 0., § 395, 0, R, b, The products of the corresponding members of these equations give, R, a×b, = 0, ., q.e.d., 0, R, a × b0, , Ex. 349. Find the ratio of a rectangular lawn 72 yards by 49 yards to a, grass turf 18 inches by 14 inches., Ex. 350. Find the ratio of a rectangular courtyard 18 12 yards by 15 21 yards, to a flagstone 31 inches by 18 inches., Ex. 351. A square and a rectangle have the same perimeter, 100 yards., The length of the rectangle is 4 times its breadth. Compare their areas., Ex. 352. On a certain map the linear scale is 1 inch to 5 miles. How, many acres are represented on this map by a square the perimeter of which is, 1 inch?
Page 239 :
AREAS OF POLYGONS., , 229, , Proposition III. Theorem., 398. The area of a rectangle is equal to the product of its base by its altitude., , a, , R, 1, b, , U, 1, , Let R be a rectangle, b its base, and a its altitude., To prove that, the area of R = a × b., Proof. Let U be the unit of surface., a×b, R, =, = a × b,, U, 1×1, (two rectangles are to each other as the products of their bases and altitudes)., But, R, = the number of units of surface in R., § 393, U, ∴ the area of R = a × b., , q.e.d., , 399. Scholium. When the base and altitude each contain the linear unit, an integral number of times, this proposition is rendered evident by dividing, the figure into squares, each equal to the unit of surface. Thus, if the base, contains seven linear units, and the altitude four, the figure may be divided, into twenty-eight squares, each equal to the unit of surface.
Page 240 :
BOOK IV. PLANE GEOMETRY., , 230, , Proposition IV. Theorem., 400. The area of a parallelogram is equal to the product of its base by its, altitude., B, , E, , C, , F, , B, , C E, , a, , A, , a, , D, , b, , F, , A, , b, , D, , Let AEF D be a parallelogram, b its base, and a its altitude., To prove that the area of the, , / /AEF D, , = a × b., , Proof. From A draw AB k to DC to meet F E produced., Then the figure ABCD is a rectangle, with the same base and the same altitude, as the / /AEF D., The rt. 4s ABE and DCF are equal., For AB = CD, and AE = DF ., From ABF D take the 4DCF ; the rect. ABCD is left., From ABF D take the 4ABE; the, , / /AEF D, , § 151, § 178, , is left., , ∴ rect. ABCD m / /AEF D, , Ax. 3, , But the area of the rect. ABCD = a × b., , § 398, , ∴ the area of the, , / /AEF D, , = a × b., , Ax. 1, q.e.d., , 401. Cor. 1. Parallelograms having equal bases and equal altitudes are, equivalent., 402. Cor. 2. Parallelograms having equal bases are to each other as their, altitudes; parallelograms having equal altitudes are to each other as their bases;, any two parallelograms are to each other as the products of their bases by their, altitudes.
Page 241 :
AREAS OF POLYGONS., , 231, , Proposition V. Theorem., 403. The area of a triangle is equal to half the product of its base by its, altitude., H, , C, , a, , A, , b, , B, , D, , Let a be the altitude and b the base of the triangle ABC., To prove that the area of the 4ABC = 21 a × b., , Proof. Construct on AB and BC the parallelogram ABCH., Then, 4ABC = 21 / /ABCH., The area of the, , / /ABCH, , = a × b., , Therefore, the area of 4ABC = 12 a × b., , § 179, § 400, Ax. 7, q.e.d., , 404. Cor. 1. Triangles having equal bases and equal altitudes are equivalent., 405. Cor. 2. Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two, triangles are to each other as the products of their bases by their altitudes., 406. Cor. 3. The product of the legs of a right triangle is equal to the, product of the hypotenuse by the altitude from the vertex of the right angle., Ex. 353. The lines which join the middle point of either diagonal of a, quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts.
Page 242 :
BOOK IV. PLANE GEOMETRY., , 232, , Proposition VI. Theorem., 407. The area of a trapezoid is equal to half the sum of its bases multiplied, by the altitude., H, , E, , b′, , C, , O, , P, a, , A, , F, , b, , B, , Let b and b0 be the bases and a the altitude of the trapezoid ABCH., To prove that the area of the ABCH = 21 a(b + b0 )., Proof., Draw the diagonal AC., Then, the area of the 4ABC = 21 a × b,, and, the area of the 4AHC = 21 a × b0 ., ∴ the area of ABCH = 12 a(b + b0 )., , § 403, Ax. 2, q.e.d., , 408. Cor. The area of a trapezoid is equal to the product of the median by, the altitude., § 190
Page 243 :
AREAS OF POLYGONS., , 233, , C, B, , D, , A, , E, , G, , F, , 409. Scholium. The area of an irregular polygon may be found by dividing the polygon into triangles, and by finding the area of each of these triangles, separately. Or, we may draw the longest diagonal, and let fall perpendiculars, upon this diagonal from the other vertices of the polygon., The sum of the areas of the right triangles, rectangles, and trapezoids thus, formed is the area of the polygon.
Page 244 :
BOOK IV. PLANE GEOMETRY., , 234, , Proposition VII. Theorem., 410. The areas of two triangles which have an angle of the one equal to an, angle of the other are to each other as the products of the sides including the, equal angles., A, , D, , E, , B, , C, , Let the triangles ABC and ADE have the common angle A., 4ABC, AB × AC, To prove that, =, ., 4ADE, AD × AE, Proof., Draw BE., Now, AC, 4ABC, =, ,, 4ABE, AE, and, 4ABE, AB, =, ., § 405, 4ADE, AD, The products of the first members and of the second members of these equalities, give, 4ABC, AB × AC, =, ., q.e.d., 4ADE, AD × AE, , Ex. 354. The areas of two triangles which have an angle of the one, supplementary to an angle of the other are to each other as the products of, the sides including the supplementary angles.
Page 245 :
COMPARISON OF POLYGONS., , 235, , COMPARISON OF POLYGONS., Proposition VIII. Theorem., 411. The areas of two similar triangles are to each as the squares of any, two homologous sides., C, , A, , O, , B, C′, , A′, , O′, , B′, , Let the two similar triangles be ACB and A0 C 0 B 0 ., To prove that, 2, 4ACB, AB, =, 2., 4A0 C 0 B 0, A0 B 0, Proof. Draw the altitudes CO and C 0 O0 ., Then, 4ACB, AB × CO, AB, CO, = 0 0, = 0 0 × 0 0,, § 405, 0, 0, 0, 0, 0, 4A C B, AB ×C O, AB, CO, (two 4s are to each other as the products of their bases by their altitudes)., But, AB, CO, = 0 0., § 361, A0 B 0, CO, (the homologous altitudes of two similar 4s have the same ratio as any two, homologous sides)., AB, CO, Substitute, in the above equality, for 0 0 its equal 0 0 ;, CO, AB, then, 2, 4ACB, AB, AB, AB, = 0 0× 0 0 =, q.e.d., 2., 4A0 C 0 B 0, AB, AB, A0 B 0, , Ex. 355. Prove this proposition by § 410.
Page 246 :
BOOK IV. PLANE GEOMETRY., , 236, , Proposition IX. Theorem., 412. The areas of two similar polygons are to each other as the squares of, any two homologous sides., E, , A, , D, , B, , C, E′, , A′, , D′, , B′, , C′, , Let S and S 0 denote the areas of the two similar polygons ABC etc., and A0 B 0 C 0 etc., To prove that, 2, S : S 0 = AB : A0 B 02 ., Proof. By drawing all the diagonals from any homologous vertices E and E 0 ,, the two similar polygons are divided into similar triangles., § 365, !, 2, 2, AB, BE, 4ABE, 4BCE, =, = etc., § 411, ∴, =, =, 0, 0, 0, 4A B E, 4B 0 C 0 E 0, A0 B 02, B 0 E 02, That is,, , 4ABE, 4BCE, 4CDE, =, =, ., 0, 0, 0, 0, 0, 0, 4A B E, 4B C E, 4C 0 D0 E 0, 2, , 4ABE + 4BCE + 4CDE, 4ABE, AB, ∴, =, =, ., 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4A B E + 4B C E + 4C D E, 4A B E, A0 B 02, 2, , ∴ S : S 0 = AB : A0 B 02, , § 335, q.e.d.
Page 247 :
COMPARISON OF POLYGONS., , 237, , 413. Cor. 1. The areas of two similar polygons are to each other as the, squares of any two homologous lines., 414. Cor. 2. The homologous sides of two similar polygons have the same, ratio as the square roots of their areas.
Page 248 :
BOOK IV. PLANE GEOMETRY., , 238, , Proposition X. Theorem., 415. The square on the hypotenuse of a right triangle is equivalent to the, sum of the squares on the two legs., G, , F, H, , A, , K, B, , C, , D, , L E, , Let BE, CH, AF , be squares on the three sides of the right triangle, ABC., To prove that BE m CH + AF ., Proof. Through A draw AL k to CE, and draw AD and CF ., Since ∠s BAC, BAG, and CAH are rt. ∠s , CAG and BAH are straight, lines., § 90, The, 4ABD = 4F BC., § 143, For, BD = BC,, and, , BA = BF ,, ∠ABD = ∠F BC,, (each being the sum of a rt. ∠ and the ∠ABC)., , § 168, Ax. 2, , Now the rectangle BL is double the 4ABD,, (having the same base BD, and the same altitude, the distance between the, ks AL and BD),, and the square AF is double the 4F BC,, (having the same base F B, and the same altitude AB).
Page 249 :
EXERCISES., , 239, , ∴ the rectangle BL is equivalent to the square AF ., , Ax. 6, , In like manner, by drawing AE and BK, it may be proved that the rectangle, CL is equivalent to the square CH., Hence, the square BE, the sum of the rectangles BL and CL, is equivalent to, the sum of the squares CH and AF ., q.e.d., , 416. Cor. The square on either leg of a right triangle is equivalent to the, difference of the square on the hypotenuse and the square on the other leg., THEOREMS., H, A, , D, K, , B, , C, , E, H, , A, , F, G, , L, G, , K, C, , B, , G, , E, , D, F, , H, , E, , A, , C, , D, , F, B, , Ex. 356. The square constructed upon the sum of two straight lines is, equivalent to the sum of the squares constructed upon these two lines, increased by twice the rectangle of these lines:, Let AB and BC be the two straight lines, and AC their sum. Construct, the squares ACGK and ABED upon AC and AB, respectively. Prolong BE, and DE until they meet KG and CG, respectively. Then we have the square, EF GH, with sides each equal to BC. Hence, the square ACGK is the sum, of the squares ABED and EF GH, and the rectangles DEHK and BCF E,, the dimensions of which are equal to AB and BC.
Page 250 :
BOOK IV. PLANE GEOMETRY., , 240, , Ex. 357. The square constructed upon the difference of two straight lines, is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines., Let AB and AC be the two straight lines, and BC their difference. Construct the square ABF G upon AB, the square ACKH upon AC, and the, square BEDC upon BC (as shown in the figure). Prolong ED to meet AG, in L., The dimensions of the rectangles LEF G and HKDL are AB and AC, and, the square BCDE is evidently the difference between the whole figure and the, sum of these rectangles; that is, the square constructed upon BC is equivalent, to the sum of the squares constructed upon AB and AC, diminished by twice, the rectangle of AB and AC., Ex. 358. The difference between the squares constructed upon two straight, lines is equivalent to the rectangle of the sum and difference of these lines., Let ABDE and BCF G be the squares constructed upon the two straight, lines AB and BC. The difference between these squares is the polygon, ACGF DE, which is composed of the rectangles ACHE and GF DH. Prolong, AE and CH to I and K, respectively, making EI and HK each equal to BC,, and draw IK. The rectangles GF DH and EHKI are equal. The difference, between the squares ABDE and BCGF is then equivalent to the rectangle, ACKI, which has for dimensions AI, equal to AB + BC, and EH, equal to, AB − BC., Ex. 359. The area of a rhombus is equal to half the product of its diagonals., Ex. 360. Two isosceles triangles are equivalent if their legs are equal each, to each, and the altitude of one is equal to half the base of the other., Ex. 361. The area of a circumscribed polygon is equal to half the product, of its perimeter by the radius of the inscribed circle., Ex. 362. Two parallelograms are equal if two adjacent sides of the one, are equal, respectively, to two adjacent sides of the other, and the included, angles are supplementary.
Page 251 :
EXERCISES., , 241, , Ex. 363. If ABC is a right triangle, C the vertex of the right angle, BD, 2, 2, 2, 2, a line cutting AC in D, then BD + AC = AB + DC ., Ex. 364. Upon the sides of a right triangle as homologous sides three, similar polygons are constructed. Prove that the polygon upon the hypotenuse, is equivalent to the sum of the polygons upon the legs., Ex. 365. If the middle points of two adjacent sides of a parallelogram are, joined, a triangle is formed which is equivalent to one eighth of the parallelogram., Ex. 366. If any point within a parallelogram is joined to the four vertices,, the sum of either pair of triangles having parallel bases is equivalent to half, the parallelogram., Ex. 367. Every straight line drawn through the intersection of the diagonals of a parallelogram divides the parallelogram into two equal parts., Ex. 368. The line which joins the middle points of the bases of a trapezoid, divides the trapezoid into two equivalent parts., Ex. 369. Every straight line drawn through the middle point of the median, of a trapezoid cutting both bases divides the trapezoid into two equivalent, parts., Ex. 370. If two straight lines are drawn from the middle point of either leg, of a trapezoid to the opposite vertices, the triangle thus formed is equivalent, to half the trapezoid., Ex. 371. The area of a trapezoid is equal to the product of one of the, legs by the distance from this leg to the middle point of the other leg., Ex. 372. The figure whose vertices are the middle points of the sides of, any quadrilateral is equivalent to half the quadrilateral.
Page 252 :
BOOK IV. PLANE GEOMETRY., , 242, , PROBLEMS OF CONSTRUCTION., Proposition XI. Problem., 417. To construct a square equivalent to the sum of two given squares., B, , S, , R′, , R, , A, , Let R and, , R0, , C, , be two given squares., , To construct a square equivalent to R0 + R., Construct the rt. ∠A., Take AC equal to a side of R0 ,, and AB equal to a side of R; and draw BC., Then, Proof., , Construct the square S, having each of its sides equal to BC., S is the square required., 2, , 2, , 2, , § 415, BC m AC + AB ,, (the square on the hypotenuse of a rt. 4 is equivalent to the sum of the, squares on the two legs)., ∴ S m R0 + R., , q.e.f., , Ex. 373. If the perimeter of a rectangle is 72 feet, and the length is equal, to twice the width, find the area., Ex. 374. How many tiles 9 inches long and 4 inches wide will be required, to pave a path 8 feet wide surrounding a rectangular court 120 feet long and, 36 feet wide?, Ex. 375. The bases of a trapezoid are 16 feet and 10 feet; each leg is, equal to 5 feet. Find the area of the trapezoid.
Page 253 :
PROBLEMS OF CONSTRUCTION., , 243, , Proposition XII. Problem., 418. To construct a square equivalent to the difference of two given squares., B, R, , R′, S, A, , C, , X, , Let R be the smaller square and R0 the larger., To construct a square equivalent to R0 − R., Construct the rt. ∠A., Take AB equal to a side of R., From B as a centre, with a radius equal to a side of R0 ,, describe an arc cutting the line AX at C., Then, Proof., , Construct the square S, having each of its sides equal to AC., S is the square required., 2, , 2, , 2, , AC m BC − AB ,, § 416, (the square on either leg of a rt. 4 is equivalent to the difference of the square, on the hypotenuse and the square on the other leg)., ∴ S m R0 − R., , q.e.f., , Ex. 376. Construct a square equivalent to the sum of two squares whose, sides are 3 inches and 4 inches., Ex. 377. Construct a square equivalent to the difference of two squares, whose sides are 2 21 inches and 2 inches., Ex. 378. Find the side of a square equivalent to the sum of two squares, whose sides are 24 feet and 32 feet., Ex. 379. Find the side of a square equivalent to the difference of two, squares whose sides are 24 feet and 40 feet.
Page 254 :
BOOK IV. PLANE GEOMETRY., , 244, , Ex. 380. A rhombus contains 100 square feet, and the length of one, diagonal is 10 feet. Find the length of the other diagonal., Proposition XIII. Problem., 419. To construct a polygon similar to two given similar polygons and, equivalent to their sum., , R, , A, , R′′, , R′, , B, , A′, , A′′, , B′, , O, , B ′′, , P, , H, , Let R and R0 be two similar polygons, and AB and A0 B 0 two homologous sides., To construct a similar polygon equivalent to R + R0 ., Construct the rt. ∠P ., Take P H equal to A0 B 0 , and P O equal to AB., Draw OH, and take A00 B 00 equal to OH., Upon A00 B 00 , homologous to AB, construct R00 similar to R., Then R00 is the polygon required., Proof., , 2, , 2, , 2, , P O + P H = OH ., Put for P O, P H, and OH their equals AB, A0 B 0 , and A00 B 00 ., Then, 2, AB + A0 B 02 = A00 B 002 ., Now, 2, R, AB, A0 B 02, R0, =, =, ,, and, ., R00, R00, A00 B 002, A00 B 002, By addition,, 2, R + R0, AB + A0 B 02, =, = 1., R00, A00 B 002, ∴ R00 m R + R0 ., , § 415, , § 412, , Ax. 2, q.e.f.
Page 255 :
PROBLEMS OF CONSTRUCTION., , 245, , Proposition XIV. Problem., 420. To construct a triangle equivalent to a given polygon., C, , D, , H, , B, , I, , A, , E, , F, , K, , Let ABCDHE be the given polygon., To construct a triangle equivalent to the given polygon., Let D, H, and E be any three consecutive vertices of the polygon. Draw the, diagonal DE., From H draw HF k to DE., Produce AE to meet HF at F , and draw DF ., Again, draw CF , and draw DK k to CF to meet AF produced at K, and draw, CK., In like manner continue to reduce the number of sides of the polygon until we, obtain the 4CIK., Then 4CIK is the triangle required., Proof. The polygon ABCDF has one side less than the polygon, ACBDHE, but the two polygons are equivalent., For the part ACBDE is common,, and the 4DEF m 4DEH,, § 404, (for the base DE is common, and their vertices F and H are in the line F H k, to the base)., In like manner it may be proved that, ABCK m ABCDF , and CIK m ABCK., , q.e.f.
Page 256 :
BOOK IV. PLANE GEOMETRY., , 246, , Proposition XV. Problem., 421. To construct a square equivalent to a given parallelogram., P, , B, , C, R, , a, , A, , b, , M, , D, , N, , O, , X, , Let ABCD be the parallelogram, b its base, and a its altitude., To construct a square equivalent to the, , / /ABCD., , Upon a line M X take M N equal to a, N O equal to b., Upon M O as a diameter, describe a semicircle., At N erect N P ⊥ to M O, meeting the circumference at P ., Then the square R, constructed upon a line equal to N P , is equivalent to the, / /ACBD., Proof., M N : N P = N P : N O,, § 370, (a ⊥ let fall from any point of a circumference to the diameter is the mean, proportional between the segments of the diameter )., 2, , Therefore,, , ∴ N P = M N × N O = a × b., , § 327, , R m / /ABCD., , q.e.f., , 422. Cor. 1. A square may be constructed equivalent to a given triangle, by, taking for its side the mean proportional between the base and half the altitude, of the triangle., 423. Cor. 2. A square may be constructed equivalent to a given polygon,, by first reducing the polygon to an equivalent triangle, and then constructing a, square equivalent to the triangle.
Page 257 :
PROBLEMS OF CONSTRUCTION., , 247, , Proposition XVI. Problem., 424. To construct a parallelogram equivalent to a given square, and having, the sum of its base and altitude equal to a given line., S, , P, , Q, , R, M, , N, , C, , Let R be the given square, and let the sum of the base and altitude, of the required parallelogram be equal to the given line M N ., To construct a, to M N ., , / /, , equivalent to R, with the sum of its base and altitude equal, , Upon M N as a diameter, describe a semicircle., At M erect M P , a ⊥ to M N , equal to a side of the given square R., Draw P Q k to M N , cutting the circumference at S., Draw SC ⊥ to M N ., Any / / having CM for its altitude and CN for its base is equivalent to R., Proof., SC = P M ., §§ 104, 180, 2, , 2, , ∴ SC = P M = R., M C : SC = SC : CN ,, § 370, (a ⊥ let fall from any point of a circumference to the diameter is the mean, proportional between the segments of the diameter )., Then, 2, SC m M C × CN ., § 327, q.e.f., Note. This problem may be stated as follows:, , To construct two straight lines the sum and product of which are known.
Page 258 :
BOOK IV. PLANE GEOMETRY., , 248, , Proposition XVII. Problem., 425. To construct a parallelogram equivalent to a given square, and having, the difference of its base and altitude equal to a given line., S, , C, , M, , R, , N, R′, B, , Let R be the given square, and let the difference of the base and, altitude of the required parallelogram be equal to the given line M N ., To construct a, equal to M N ., , / /, , equivalent to R, with the difference of its base and altitude, , Upon the given line M N as a diameter, describe a circle., From M draw M S, tangent to the, Through the centre of the, , , and equal to a side of the given square R., , draw SB intersecting the circumference at C and, , B., Then any / /, as R0 , having SB for its base and SC for its altitude, is equivalent, to R., Proof., SB : SM = SM : SC,, § 381, (if from a point without a, a secant and a tangent are drawn, the tangent is, the mean proportional between the whole secant and the external segment)., Then, 2, SM m SB × SC,, § 327, and the difference between SB and SC is the diameter of the , that is, M N ., q.e.f., Note. This problem may be stated: To construct two straight lines the difference, and product of which are known.
Page 259 :
PROBLEMS OF CONSTRUCTION., , 249, , Proposition XVIII. Problem., 426. To construct a polygon similar to a given polygon P and equivalent, to a given polygon Q., , A, , P′, , Q, , P, , A′, , B, , B′, , Let P and Q be the two given polygons, and AB a side of P ., To construct a polygon similar to P and equivalent to Q., Find squares equivalent to P and Q,, and let m and n respectively denote their sides., Find A0 B 0 , the fourth proportional to m, n, and AB., Upon, Then, , A0 B 0 ,, , homologous to AB, construct, , similar to P ., , m : n = AB : A0 B 0 ., ∴, , m2, , :, , n2, , 2, , = AB :, , Const., , 2, A0 B 0 ., , § 338, , P m m2 , and Q m n2 ., , Const., 02, , 2, , But, , § 386, , P 0 m Q., , Proof., , But, , P0, , § 423, , ∴ P : Q = m2 : n2 = AB : A0 B ., 2, , 2, , P : P 0 = AB : A0 B 0 ., , § 412, , ∴ P : Q = P : P 0., , Ax. 1, , ∴ P 0 m Q., , q.e.f., , Ex. 381. To construct a square equivalent to the sum of any number of, given squares., Ex. 382. To construct a polygon similar to two given similar polygons, and equivalent to their difference.
Page 260 :
BOOK IV. PLANE GEOMETRY., , 250, , Proposition XIX. Problem., 427. To construct a square which shall have a given ratio to a given square., D, , x, a, , A, R, , b, B, , m, m, , E, , n, , n, , F, , C, , y, , n, , the given ratio., m, To construct a square which shall be to R as n is to m., Let R be the given square, and, , Take AB equal to a side of R, and draw Ay, making any convenient angle with, AB., On Ay take AE equal to m, EF equal to n, and draw EB., Draw F C k to EB meeting AB produced at C., On AC as a diameter, describe a semicircle., At B erect the ⊥ BD, meeting the semicircumference at D., Then BD is a side of the square required., Proof., Now, Therefore,, But, Therefore,, By inversion,, , Denote AB by a, BC by b, and BD by x., a : x = x : b., , § 370, , a2 : x2 = a : b., , § 337, , a : b = m : n., , § 342, , a2 : x2 = m : n., , Ax. 1, , x2 : a2 = n : m., Hence, the square on BD will have the same ratio to R as n has to m., , § 331, q.e.f.
Page 261 :
PROBLEMS OF CONSTRUCTION., , 251, , Proposition XX. Problem., 428. To construct a polygon similar to a given polygon and having a given, ratio to it., , R, , m, , S, , n, A, , A′, , B, , Let R be the given polygon, and, , n, , B′, , the given ratio., m, To construct a polygon similar to R, which shall be to R as n is to m., Construct a line A0 B 0 , such that the square on A0 B 0 shall be to the square on, AB as n is to m., § 427, Upon A0 B 0 , as a side homologous to AB, construct the polygon S similar to, R., § 391, Then S is the polygon required., Proof., But, Therefore,, , 2, , 2, , S : R = A0 B 0 : AB ., 2, , 2, , § 412, , A0 B 0 : AB = n : m., , Const., , S : R = n : m., , Ax. 1, q.e.f., , Ex. 383. To construct a triangle equivalent to a given triangle, and having, one side equal to a given length l., Ex. 384. To transform a triangle into an equivalent right triangle., Ex. 385. To transform a given triangle into an equivalent right triangle,, having one leg equal to a given length., Ex. 386. To transform a given triangle into an equivalent right triangle,, having the hypotenuse equal to a given length.
Page 262 :
BOOK IV. PLANE GEOMETRY., , 252, , PROBLEMS OF CONSTRUCTION., Ex. 387. To transform a triangle ABC into an equivalent triangle, having, a side equal to a given length l, and an angle equal to angle BAC., Upon AB (produced if necessary), take AD equal to l, draw BE k to CD,, meeting AC (produced if necessary) at E., 4BED m 4BEC., Ex. 388. To transform a given triangle into an equivalent isosceles triangle,, having the base equal to a given length., To construct a triangle equivalent to:, Ex. 389. The sum of two given triangles., Ex. 390. The difference of two given triangles., Ex. 391. To transform a given triangle into an equivalent equilateral, triangle., To transform a parallelogram into an equivalent:, Ex. 392. Parallelogram having one side equal to a given length., Ex. 393. Parallelogram having one angle equal to a given angle., Ex. 394. Rectangle having a given altitude., To transform a square into an equivalent:, Ex. 395. Equilateral triangle., Ex. 396. Right triangle having one leg equal to a given length., Ex. 397. Rectangle having one side equal to a given length., To construct a square equivalent to:, Ex. 398. Five eighths of a given square., Ex. 399. Three fifths of a given pentagon.
Page 263 :
EXERCISES., , 253, , Ex. 400. To divide a given triangle into two equivalent parts by a line, through a given point P in one of the sides., Ex. 401. To find a point within a triangle, such that the lines joining this, point to the vertices shall divide the triangle into three equivalent parts., Ex. 402. To divide a given triangle into two equivalent parts by a line, parallel to one of the sides., Ex. 403. To divide a given triangle into two equivalent parts by a line, perpendicular to one of the sides., PROBLEMS OF COMPUTATION., B, B, a, , A, , c, , c, , h, , h b, , B, , a, , D, , C, , a, h, , A, , a/2 D, , C, , A, , b, , D, , C, , E, , Ex. 404. To find the area of an equilateral triangle in terms of its side., Denote the side by a, the altitude by h, and the area by S., Then, a2, 3a2, a2, h2 a2 −, =, =, × 3., § 372, 4, 4, 4, a√, ∴h=, 3., 2, But, a×h, ., § 403, S=, 2, √, √, a a 3, a2 3, ∴S = ×, =, ., 2, 2, 4
Page 264 :
BOOK IV. PLANE GEOMETRY., , 254, , Ex. 405. To find the area of a triangle in terms of its sides., By Ex. 312,, 2p, s(s − a)(s − b)(s − c)., h=, b, Hence,, b 2p, S= ×, s(s − a)(s − b)(s − c), 2 b, p, = s(s − a)(s − b)(s − c)., , § 403, , Ex. 406. To find the area of a triangle in terms of the radius of the, circumscribed circle., If R denotes the radius of the circumscribed circle, and h the altitude of, the triangle, we have, by § 384,, Multiply by a, and we have,, But, , b × c = 2R × h., , a × b × c = 2R × a × h., a × h = 2S., ∴ a × b × c = 4R × S., abc, ., ∴S =, 4R, , Show that the radius of the circumscribed circle is equal to, , § 403, , abc, ., 4S, , Ex. 407. Find the area of a right triangle, if the length of the hypotenuse, is 17 feet and the length of one leg is 8 feet., Ex. 408. Find the ratio of the altitudes of two equivalent triangles, if the, base of one is three times that of the other., Ex. 409. The bases of a trapezoid are 8 feet and 10 feet, and the altitude, is 6 feet. Find the base of the equivalent rectangle that has an equal altitude., Ex. 410. Find the area of a rhombus, if the sum of its diagonals is 12 feet,, and their ratio is 3 : 5., Ex. 411. Find the area of an isosceles right triangle, if the hypotenuse is, 20 feet.
Page 265 :
EXERCISES., , 255, , Ex. 412. In a right triangle the hypotenuse is 13 feet, one leg is 5 feet., Find the area., Ex. 413. Find the area of an isosceles triangle, if base = b, and leg = c., Ex. 414. Find the area of an equilateral triangle, if one side = 8 feet., Ex. 415. Find the area of an equilateral triangle, if the altitude = h., Ex. 416. A house is 40 feet long, 30 feet wide, 25 feet high to the roof,, and 35 feet high to the ridge-pole. Find the number of square feet in its entire, exterior surface., Ex. 417. The sides of a right triangle are as 3 : 4 : 5. The altitude upon, the hypotenuse is 12 feet. Find the area., Ex. 418. Find the area of a right triangle, if one leg = a, and the altitude, upon the hypotenuse = h., Ex. 419. Find the area of a triangle, if the lengths of the sides are 104 feet,, 111 feet, and 175 feet., Ex. 420. The area of a trapezoid is 700 square feet. The bases are 30 feet, and 40 feet, respectively. Find the altitude., Ex. 421. ABCD is a trapezium; AB = 87 feet, BC = 119 feet, CD = 41, feet, DA = 169 feet, AC = 200 feet. Find the area., Ex. 422. What is the area of a quadrilateral circumscribed about a circle, whose radius is 25 feet, if the perimeter of the quadrilateral is 400 feet? What, is the area of a hexagon that has a perimeter of 400 feet and is circumscribed, about the same circle of 25 feet radius (Ex. 361)?, Ex. 423. The base of a triangle is 15 feet, and its altitude is 8 feet. Find, the perimeter of an equivalent rhombus, if the altitude is 6 feet., Ex. 424. Upon the diagonal of a rectangle 24 feet by 10 feet a triangle, equivalent to the rectangle is constructed. What is its altitude?, Ex. 425. Find the side of a square equivalent to a trapezoid whose bases, are 56 feet and 44 feet, and each leg is 10 feet.
Page 266 :
BOOK IV. PLANE GEOMETRY., , 256, , Ex. 426. Through a point P in the side AB of a triangle ABC, a line, is drawn parallel to BC so as to divide the triangle into two equivalent parts., Find the value of AP in terms of AB., Ex. 427. What part of a parallelogram is the triangle cut off by a line, from one vertex to the middle point of one of the opposite sides?, Ex. 428. In two similar polygons, two homologous sides are 15 feet and, 25 feet. The area of the first polygon is 450 square feet. Find the area of the, second polygon., Ex. 429. The base of a triangle is 32 feet, its altitude 20 feet. What is, the area of the triangle cut off by a line parallel to the base at a distance of, 15 feet from the base?, Ex. 430. The sides of two equilateral triangles are 3 feet and 4 feet. Find, the side of an equilateral triangle equivalent to their sum., Ex. 431. If the side of one equilateral triangle is equal to the altitude of, another, what is the ratio of their areas?, Ex. 432. The sides of a triangle are 10 feet, 17 feet, and 21 feet. Find, the areas of the parts into which the triangle is divided by the bisector of the, angle formed by the first two sides., Ex. 433. In a trapezoid, one base is 10 feet, the altitude is 4 feet, the area, is 32 square feet. Find the length of a line drawn between the legs parallel to, the bases and distant 1 foot from the lower base., Ex. 434. The diagonals of a rhombus are 90 yards and 120 yards, respectively. Find the area, the length of one side, and the perpendicular distance, between two parallel sides., Ex. 435. Find the number of square feet of carpet that are required to, cover a triangular floor whose sides are, respectively, 26 feet, 35 feet, and, 51 feet., Ex. 436. If the altitude h of a triangle is increased by a length m, how, much must be taken from the base a that the area may remain the same?
Page 267 :
EXERCISES., , 257, , Ex. 437. Find the area of a right triangle, having given the segments, p, q, into which the hypotenuse is divided by a perpendicular drawn to the, hypotenuse from the vertex of the right angle.
Page 268 :
BOOK V. REGULAR POLYGONS AND CIRCLES., , 429. Def. A regular polygon is a polygon which is both equilateral and, equiangular. The equilateral triangle and the square are examples., Proposition I. Theorem., 430. An equilateral polygon inscribed in a circle is a regular polygon., C, , D, , B, , A, F, , Let ABC etc. be an equilateral polygon inscribed in a circle., To prove that the polygon ABC etc. is a regular polygon., Proof., The arcs AB, BC, CD, etc., are equal., , § 243, , Hence, arcs ABC, BCD, etc., are equal., , Ax. 2, , Therefore, arcs CF A, DF B, etc., are equal., , Ax. 3, , Therefore, ∠s A, B, C, etc., are equal., § 289, Therefore, the polygon ABC etc. is a regular polygon, being equilateral and, equiangular., § 429, q.e.d.
Page 269 :
REGULAR POLYGONS AND CIRCLES., , 259, , Proposition II. Theorem., 431. A circle may be circumscribed about, and a circle may be inscribed in,, any regular polygon., D, , E, , C, , O, , A, , F, , B, , Let ABCDE be a regular polygon., 1. To prove that a circle may be circumscribed about ABCDE., Proof. Let O be the centre of the circle which may be passed through A, B,, and C., § 258, Then, and, By subtraction,, , For, , and, , Draw OA, OB, OC, and OD., ∠ABC = ∠BCD,, , § 429, , ∠OBC = ∠OCB., , § 145, , ∠OBA = ∠OCD., , Ax. 3, , The 4s OBA and OCD are equal., , § 143, , ∠OBA = ∠OCD,, OB = OC,, , § 217, , AB = CD., , § 429, , ∴ OA = OD., ∴ the circle passing through A, B, C, passes through D., , § 128, , In like manner it may be proved that the circle passing through B, C, and D, also passes through E; and so on., Therefore, the circle described from O as a centre, with a radius OA, will be, circumscribed about the polygon., § 231, 2. To prove that a circle may be inscribed in ABCDE.
Page 270 :
BOOK V. PLANE GEOMETRY., , 260, , Proof. Since the sides of the regular polygon are equal chords of the circumscribed circle, they are equally distant from the centre., § 249, Therefore, the circle described from O as a centre, with the distance from O to, a side of the polygon as a radius, will be inscribed in the polygon (§ 232). q.e.d., , 432. Def. The radius of the circumscribed circle, OA, is called the radius, of the polygon., 433. Def. The radius of the inscribed circle, OF , is called the apothem, of the polygon., 434. Def. The common centre, O, of the circumscribed and inscribed, circles is called the centre of the polygon., 435. Def. The angle between radii drawn to the extremities of any side is, called the angle at the centre of the polygon., By joining the centre to the vertices of a regular polygon, the polygon can, be decomposed into as many equal isosceles triangles as it has sides., 436. Cor. 1. The angle at the centre of a regular polygon is equal to four, right angles divided by the number of sides of the polygon. Hence, the angles, at the centre of any regular polygon are all equal., 437. Cor. 2. The radius drawn to any vertex of a regular polygon bisects, the angle at the vertex., 438. Cor. 3. The angle at the centre of a regular polygon and an interior, angle of the polygon are supplementary., For, ∠s F OB and F BO are complementary., § 135, ∴ their doubles AOB and F BC are supplementary., Ax. 6
Page 271 :
REGULAR POLYGONS AND CIRCLES., , 261, , Proposition III. Theorem., 439. If the circumference of a circle is divided into any number of equal, arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regular, circumscribed polygon., J, , D, , H, , C, , E, , K, , G, A, , B, F, , Suppose the circumference divided into equal arcs AB, BC, etc. Let, AB, BC, etc., be the chords, F BG, GCH, etc., the tangents., 1. To prove that ABCDE is a regular polygon., Proof., The sides AB, BC, CD, etc., are equal., Therefore, the polygon is regular., 2. To prove that To prove that F GHIK is a regular polygon., , § 241, § 430, , Proof. The 4s AF B, BGC, CHD, etc., are all equal isosceles triangles., §§ 295,139, ∴ ∠s F , G, H, etc., are equal, and F B, BG, GC, etc., are equal., ∴ F G = GH = HI, etc., , Ax. 6, , ∴ F GHIK is a regular polygon., , § 429, q.e.d., , 440. Cor. 1. Tangents to a circle at the vertices of a regular inscribed, polygon form a regular circumscribed polygon of the same number of sides as, the inscribed polygon.
Page 272 :
BOOK V. PLANE GEOMETRY., D′, Q, E′ E, , D, , E, , P, , A, C, , O, , D, , C′, , A, , M, , B, , B′, , I, , C, , L, , H, O, , F, , N, A′, , K, , B, , K, , R, , 262, , M, D, , G, , C, H, , A, , E, , F, , B, , 441. Cor. 2. Tangents to a circle at the middle points of the arcs subtended, by the sides of a regular inscribed polygon form a circumscribed regular polygon,, whose sides are parallel to the sides of the inscribed polygon and whose vertices, lie on the radii (prolonged ) of the inscribed polygon., For two corresponding sides, AB and A0 B 0 , are perpendicular to OM, (§§ 248, 254), and are parallel (§ 104); and the tangents M B 0 and N B 0 , intersecting at a point equidistant from OM and ON (§ 261), intersect upon the, bisector of the ∠M ON (§ 162); that is, upon the radius OB., 442. Cor. 3. If the vertices of a regular inscribed polygon are joined to, the middle points of the arcs subtended by the sides of the polygon, the joining, lines form a regular inscribed polygon of double the number of sides., 443. Cor. 4. Tangents at the middle points the arcs between adjacent, points of contact of the sides of a regular circumscribed polygon form a regular, circumscribed polygon of double the number of sides., 444. Cor. 5. The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double the number of sides; and the perimeter, of a circumscribed polygon is greater than the perimeter of a circumscribed, polygon of double the number of sides., For two sides of a triangle are together greater than the third side. § 138
Page 273 :
REGULAR POLYGONS AND CIRCLES., , 263, , Proposition IV. Theorem., 445. Two regular polygons of the same number of sides are similar., D′, , D, E, , Q, , A, , C′, , E′, C, , B, , Q′, , A′, , B′, , Let Q and Q0 be two regular polygons, each having n sides., To prove that Q and Q0 are similar., Proof. The sum of the interior ∠s of each polygon is equal to, (n − 2)2 rt. ∠s ,, § 205, (the sum of the interior ∠s of a polygon is equal to 2 rt. ∠s taken as many, times less two as the polygon has sides)., (n − 2)2 rt. ∠s, Each angle of either polygon =, ,, § 206, n, (for the ∠s of a regular polygon are all equal, and hence each ∠ is equal to the, sum of the ∠s divided by their number )., Hence, the two polygons Q and Q0 are mutually equiangular., Since AB = BC, etc., and A0 B 0 = B 0 C 0 , etc.,, , § 429, , AB : A0 B 0 = BC : B 0 C 0 , etc., Hence, the two polygons have their homologous sides proportional., Therefore the two polygons are similar., , § 351, q.e.d., , 446. Cor. The areas of two regular polygons of the same number of sides, are to each other as the squares of any two homologous sides., § 412
Page 274 :
BOOK V. PLANE GEOMETRY., , 264, , Proposition V. Theorem., 447. The perimeters of two regular polygons of the same number of sides, are to each other as the radii of their circumscribed circles, and also as the, radii of their inscribed circles., D′, , D, E, , O, , A, , C′, , E′, C, , B, , M, , O′, , A′, , M′, , B′, , Let P and P 0 denote the perimeters, O and O 0 the centres, of the two, regular polygons., From O, O0 draw OA, O0 A0 , OB, O0 B 0 , and the ⊥s OM , O0 M 0 ., To prove that P : P 0 = OA : O0 A0 = OM : O0 M 0 ., Proof., Since the polygons are similar,, The 4s OAB and, Now, , O 0 A0 B 0, , P : P 0 = AB : A0 B 0 ., are isosceles., , Also,, , § 364, § 431, , ∠O = ∠O0 ,, , and, , § 445, , § 436, , OA : OB = O0 A0 : O0 B 0 ., ∴ the 4s OAB and O0 A0 B 0 are similar., , § 357, , ∴ AB : A0 B 0 = OA : O0 A0 ., , § 351, , AB : A0 B 0 = OM : O0 M 0 ., , § 361, , ∴P :, , P0, , = OA :, , O 0 A0, , = OM :, , O0 M 0 ., , Ax. 1, q.e.d., , 448. Cor. The areas of two regular polygons of the same number of sides, are to each other as the squares of the radii of the circumscribed circles, and, of the inscribed circles., § 413
Page 275 :
REGULAR POLYGONS AND CIRCLES., , 265, , Proposition VI. Theorem., 449. If the number of sides of a regular inscribed polygon is indefinitely, increased, the apothem of the polygon approaches the radius of the circle as its, limit., O, , A, , P, , B, , Let AB be a side and OP the apothem of a regular polygon of n sides, inscribed in the circle whose radius is OA., To prove that OP approaches OA as a limit, when n increases indefinitely., Proof., OP < OA,, § 97, and, OA − OP < AP ., § 138, ∴ OA − OP < AB, which is twice AP ., § 245, Now, if n is taken sufficiently great, AB, and consequently OA − OP , can be, made less than any assigned value, however small, but cannot be made zero., Since OA − OP can be made less than any assigned value by increasing n, but, cannot be made zero, OA is the limit of OP by the test for a limit., § 275, q.e.d., , 450. Cor. If the number of sides of a regular inscribed polygon is indefinitely increased, the square of the apothem approaches the square of the radius, of the circle as a limit., For, 2, 2, 2, OA − OP = AP ., § 372, But by taking n sufficiently great, AB and consequently AP , the half of, AB, can be made less than any assigned value.
Page 276 :
BOOK V. PLANE GEOMETRY., , 266, , 2, , Therefore, AP , the product of AP by AP , can be made less than any, assigned value; for the product of two finite factors approaches zero as a limit,, if either factor approaches zero as a limit (§ 276); and for a still stronger reason,, the product approaches zero as a limit, if each of the factors approaches zero, as a limit., Proposition VII. Theorem., 451. An arc of a circle is less than any line which envelops it and has the, same extremities., D, , C, , F, , E, A, , B, , Let ACB be an arc of a circle, and AB its chord., To prove that the arc ACB is less than any other line which envelops this arc, and terminates at A and B., Proof. Of all the lines that can be drawn, each to include the area ACB between, itself and the chord AB, there must be at least one shortest line; for all the lines, are not equal., Now the enveloping line ADB cannot be the shortest; for drawing ECF tangent, to the arc ACB at C, the line AECF B < AEDF B, since ECF < EDF ., § 49, In like manner it can be shown that no other enveloping line can be the shortest., Therefore, ACB is the shortest., q.e.d., , 452. Cor. 1. The circumference of a circle is less than the perimeter of, any polygon circumscribed about it., 453. Cor. 2. Any convex curve is less than the perimeter of a polygon, circumscribed about it.
Page 277 :
REGULAR POLYGONS AND CIRCLES., , 267, , Proposition VIII. Theorem., 454. The circumference of a circle is the limit which the perimeters of, regular inscribed polygons and of similar circumscribed polygons approach, if, the number of sides of the polygons is indefinitely increased; and the area of a, circle is the limit which the areas of these polygons approach., , O, , A, A′, , D, C, , B, B′, , Let P and P 0 denote the lengths of the perimeters, AB and A0 B 0, two homologous sides, R and R0 the radii, of the polygons, and C the, circumference of the circle., To prove that C is the limit of P and of P 0 , if the number of sides of the polygons, is indefinitely increased., Proof., Since the polygons are similar by hypothesis,, Therefore,, Whence,, Therefore,, , P 0 : P = R0 : R., , § 447, , P 0 − P : P = R0 − R : R., , § 333, , R(P 0 − P ) = P (R0 − R)., , § 327, , P0 − P =, , P, 0, R (R, , − R)., , Now P is always less than C., , § 273, , 0, ∴ P0 − P < C, R (R − R)., But R0 − R, which is less than A0 C (§ 138), can be made less than any assigned, C, quantity by increasing the number of sides of the polygons; and therefore (R0 − R), R, can be made less than any assigned quantity., § 276, , Hence, P 0 − P can be made less than any assigned quantity.
Page 278 :
BOOK V. PLANE GEOMETRY., , 268, , Since P 0 is always greater than C (§ 452), and P is always less than C (§ 273),, the difference between C and either P 0 or P is less than the difference P 0 − P , and, consequently can be made less than any assigned quantity, but cannot be made zero., Therefore, C is the common limit of P 0 and P ., , § 275, , Let K denote the area of the circle, S the area of the inscribed polygon, and S 0 the area of the circumscribed polygon., 2. To prove that K is the limit of S and S 0 ., Proof., S 0 : S = R02 : R2 ., By division,, S 0 − S : S = R02 − R2 : R2 ., Whence, S, S 0 − S = 2 (R02 − R2 )., R, Therefore,, , Now K is always greater than S., , § 448, § 333, , Ax. 8, , K 02, (R − R2 )., R2, But R02 − R2 , which is equal to (R0 + R)(R0 − R), can be made less than any, K, assigned quantity; and therefore 2 (R02 − R2 ) can be made less than any assigned, R, quantity., § 276, S0 − S <, , Hence, S 0 − S can be made less than any assigned quantity., Since S 0 > K always, and S < K always (Ax. 8), the difference between K and, either S 0 or S is less than the difference S 0 − S, and consequently can be made less, than any assigned quantity, but cannot be made zero., Therefore, K is the common limit of S 0 and S., , § 275, q.e.d.
Page 279 :
REGULAR POLYGONS AND CIRCLES., , 269, , Proposition IX. Theorem., 455. Two circumferences have the same ratio as their radii., , Q′, , Q, , Let C and C 0 be the circumferences, R and R0 the radii, of the two, circles Q and Q0 ., To prove that, C : C 0 = R : R0 ., Proof. Inscribe in the s two similar regular polygons, and denote their perimeters by P and P 0 ., Then, P : P 0 = R : R0 ., § 447, Conceive the number of sides of these regular polygons to be indefinitely increased, the polygons continuing similar., Then P and P 0 will have C and C 0 as limits., But P :, , P0, , § 454, , R0 ., , § 447, , ∴ C : C 0 = R : R0 ., , § 285, q.e.d., , will always be equal to R :, , 456. Cor. The ratio of the circumference of a circle to its diameter is, constant., For, C : C 0 = R : R0 ., § 455, By alternation,, , ∴ C : C 0 = 2R : 2R0 ., , § 340, , C : 2R = C 0 : 2R0 ., , § 330, , 457. Def. The constant ratio of the circumference of a circle to its diameter, is represented by the Greek letter π.
Page 280 :
BOOK V. PLANE GEOMETRY., , 458. Cor. π =, , 270, , C, . ∴ C = 2πR., 2R, Proposition X. Theorem., , 459. The area of a regular polygon is equal to half the product of its apothem, by its perimeter., E, , D, , O, , F, , A, , M, , C, , B, , Let P represent the perimeter, R the apothem, and S the area of the, regular polygon ABC etc., To prove that S = 12 R × P ., Proof., Draw the radii OA, OB, OC, etc., The polygon is divided into as many 4s as it has sides., The apothem is the common altitude of these 4s ,, and the area of each 4 = 12 R multiplied by the base., Hence, the area of all the 4s is equal to, bases., , 1, 2R, , § 403, , multiplied by the sum of all the, , But the sum of the areas of all the 4s is equal to the area of the polygon. Ax. 9, And the sum of all the bases of the 4s is equal to the perimeter of the polygon., Ax. 9, ∴ S = 12 R × P ., , q.e.d., , 460. Def. In different circles similar arcs, similar sectors, and similar, segments are such as correspond to equal angles at the centre.
Page 281 :
REGULAR POLYGONS AND CIRCLES., , 271, , Proposition XI. Theorem., 461. The area of a circle is equal to half the product of its radius by its, circumference., E, , A, , D, , O, , B, , M, , C, , Let R represent the radius, C the circumference, and S the area, of, the circle whose centre is O., To prove that, S = 21 R × C., Proof. Circumscribe any regular polygon about the circle, and denote its, perimeter by P , and its area by S 0 ., Then, S 0 = 12 R × P ., § 459, Conceive the number of sides of the polygon to be indefinitely increased., Then P approaches C as its limit,, 1, 2R, , But, , § 454, , × P approaches 12 R × C as its limit,, , § 279, , and S 0 approaches S as its limit., , § 454, , S 0 = 21 R × P , always., , § 459, , ∴ S = 12 R × C., , § 284, q.e.d., , 462. Cor. 1. The area of a sector is equal to half the product of its radius, by its arc., For the sector and its arc are like parts of the circle and its circumference,, respectively.
Page 282 :
BOOK V. PLANE GEOMETRY., , 272, , 463. Cor. 2. The area of a circle is equal to π times the square of its, radius., For the area of the = 21 R × C = 21 R × 2πR = πR2 ., 464. Cor. 3. The areas of two circles are to each other as the squares of, their radii., For, if S and S 0 denote the areas, and R and R0 the radii,, S : S 0 = πR2 : πR02 = R2 : R02 ., 465. Cor. 4. Similar arcs are to each other as their radii; similar sectors, are to each other as the squares of their radii.
Page 283 :
REGULAR POLYGONS AND CIRCLES., , 273, , Proposition XII. Theorem., 466. The areas of two similar segments are to each other as the squares of, their radii., C, , A, , P, , C′, , B, , A′, , B′, , P′, , Let AC and A0 C 0 be the radii of the two similar sectors ACB and, A0 C 0 B 0 , and let ABP and A0 B 0 P 0 be the corresponding segments., To prove that, 2, 2, ABP : A0 B 0 P 0 = AC : A0 C 0 ., Proof., 2, 2, Sector ACB : Sector A0 C 0 B 0 = AC : A0 C 0 ., § 465, The 4s ACB and A0 C 0 B 0 are similar., , § 357, , 02, , 2, , ∴ 4ACB : 4A0 C 0 B 0 = AC : A0 C ., , § 411, , ∴ sector ACB : sector A0 C 0 B 0 = 4ACB : 4A0 C 0 B 0 ., , Ax. 1, , ∴ sector ACB : 4ACB = sector A0 C 0 B 0 : 4A0 C 0 B 0 ., , § 330, , 2, , ∴, That is,, , sector ACB − 4ACB, 4ACB, AC, =, =, 2., 0, 0, 0, 0, 0, 0, 0, 0, 0, sector A C B − 4A C B, 4A C B, A0 C 0, 2, , 2, , ABP : A0 B 0 P 0 = AC : A0 C 0 ., , § 333, q.e.d.
Page 284 :
BOOK V. PLANE GEOMETRY., , 274, , PROBLEMS OF CONSTRUCTION., Proposition XIII. Problem., 467. To inscribe a square in a given circle., B, , C, O, , A, , D, , Let O be the centre of the given circle., To inscribe a square in the given circle., Draw two diameters AC and BD ⊥ to each other., Draw AB, BC, CD, and DA., Then ABCD is the square required., Proof., The ∠s ABC, BCD, etc., are rt. ∠s ,, (each being inscribed in a semicircle),, , § 290, , and the sides AB, BC, etc., are equal,, (in the same, equal arcs are subtended by equal chords)., , § 241, , Hence the quadrilateral ABCD is a square., , § 168, q.e.f., , 468. Cor. By bisecting the arcs AB, BC, etc., a regular polygon of eight, sides may be inscribed in the circle; and, by continuing the process, regular, polygons of sixteen, thirty-two, sixty-four, etc., sides may be inscribed., Ex. 438. The area of a circumscribed square is equal to twice the area of, the inscribed square., Ex. 439. The area of a circular ring is equal to that of a circle whose, diameter is a chord of the outer circle tangent to the inner circle.
Page 285 :
PROBLEMS OF CONSTRUCTION., , 275, , Proposition XIV. Problem., 469. To inscribe a regular hexagon in a given circle., D, , E, , F, , O, , C, , A, , B, , Let O be the centre of the given circle., To inscribe a regular hexagon in the given circle., From O draw any radius, as OC., From C as a centre, with a radius equal to OC,, describe an arc intersecting the circumference at F ., Draw OF and CF ., Then CF is a side of the regular hexagon required., Proof., The 4OF C is equiangular,, (since it is equilateral by construction)., Hence, the ∠F OC is, , 1, 3, , of 2 rt. ∠s , or, , 1, 6, , of 4 rt. ∠s ., , § 146, § 136, , ∴ the arc F C is 61 of the circumference,, and the chord F C is a side of a regular inscribed hexagon., Hence, to inscribe a regular hexagon apply the radius six times as a chord. q.e.f., , 470. Cor. 1. By joining the alternate vertices A, C, D, an equilateral, triangle is inscribed in the circle., 471. Cor. 2. By bisecting the arcs AB, BC, etc., a regular polygon of, twelve sides may be inscribed in the circle; and, by continuing the process,, regular polygons of twenty-four, forty-eight, etc., sides may be inscribed.
Page 286 :
BOOK V. PLANE GEOMETRY., , 276, , Proposition XV. Problem., 472. To inscribe a regular decagon in a given circle., , O, F, , S, C, , B, , Let O be the centre of the given circle., To inscribe a regular decagon in the given circle., Draw any radius OC,, and divide it in extreme and mean ratio, so that OC shall, be to OS as OS is to SC., , § 389, , From C as a centre, with a radius equal to OS,, describe an arc intersecting the circumference at B., Draw BC., Then BC is a side of the regular decagon required., Proof., Now, and, , Moreover,, , Draw BS and BO., OC : OS = OS : SC,, , Const., , BC = OS., , Const., , ∴ OC : BC = BC : SC., ∠OCB = ∠SCB., , Iden., , Hence, the 4s OCB and BCS are similar., , § 357, , But the 4OCB is isosceles., , § 217, , ∴ 4BCS, which is similar to the 4OCB, is isosceles,, and CB = BS = SO., , § 120, , ∴ the 4SOB is isosceles, and the ∠O = ∠SBO., , § 145
Page 287 :
PROBLEMS OF CONSTRUCTION., , Hence,, , But the ext. ∠CSB = ∠O + ∠SBO = 2∠O., , § 137, , ∠SCB = 2∠O,, , and, , and, , 277, , ∠OBC = 2∠O., ∴ the sum of the ∠s of the 4OCB = 5∠O = 2rt. ∠s ,, ∠O =, , 1, 5, , of 2 rt. ∠s , or, , 1, 10, , of 4 rt. ∠s ., , 1, Therefore, the arc BC is 10, of the circumference,, and the chord BC is a side of a regular inscribed decagon., , Therefore, to inscribe a regular decagon, divide the radius internally in extreme, and mean ratio, and apply the greater segment ten times as a chord., q.e.f., , 473. Cor. 1. By joining the alternate vertices of a regular inscribed, decagon, a regular pentagon is inscribed., 474. Cor. 2. By bisecting the arcs BC, CF , etc., a regular polygon of, twenty sides may be inscribed in the circle; and, by continuing the process,, regular polygons of forty, eighty, etc., sides may be inscribed., If R denotes the radius of a regular inscribed polygon, r the apothem, a, one side, A an interior angle, and C the angle at the centre, show that, √, Ex. 440. In a regular inscribed triangle a = R 3, r = 12 R, A = 60◦ ,, C = 120◦ ., √, √, Ex. 441. In an inscribed square a = R 2, r = 12 R 2, A = 90◦ , C = 90◦ ., √, Ex. 442. In a regular inscribed hexagon a = R, r = 21 R 3, A = 120◦ ,, C = 60◦ ., Ex. 443. In a regular inscribed decagon, √, q, √, R( 5 − 1), 1, , r = R 10 + 2 5, A = 144◦ , C = 36◦ ., a=, 2, 4
Page 288 :
BOOK V. PLANE GEOMETRY., , 278, , Proposition XVI. Problem., 475. To inscribe in a given circle a regular pentedecagon, or polygon of, fifteen sides., D, C, , Q, , B, A, , E, , H, , F, , Let Q be the given circle., To inscribe in Q a regular pentedecagon., Draw EH equal to the radius of the circle,, and EF equal to a side of the regular inscribed decagon., Draw F H., Then F H is a side of the regular pentedecagon required., Proof., The arc EH is 61 of the circumference,, and the arc EF is, Hence, the arc F H is, , 1, 6, , −, , 1, 10 ,, , or, , 1, 15 ,, , 1, 10, , of the circumference., , § 472, , § 469, Const., , of the circumference., , And the chord F H is a side of a regular inscribed pentedecagon., By applying F H fifteen times as a chord, we have the polygon required. q.e.f., , 476. Cor. By bisecting the arcs F H, HA, etc., a regular polygon of thirty, sides may be inscribed; and, by continuing the process, regular polygons of, sixty, one hundred twenty, etc., sides may be inscribed.
Page 289 :
PROBLEMS OF CONSTRUCTION., , 279, , Proposition XVII. Problem., 477. To inscribe in a given circle a regular polygon similar to a given, regular polygon., C′, , D′, , B′, , C, , E′ B, , O′, , A′, , F′, , D, , E, , O, , A, , F, , Let ABC etc. be the given regular polygon, and O 0 the centre of the, given circle., To inscribe in the circle a regular polygon similar to ABC etc., From O, the centre of the given polygon,, draw OD and OC., From, , O0 ,, , the centre of the given circle,, , draw O0 C 0 and O0 D0 ,, making the ∠O0 equal to the ∠O., Draw C 0 D0 ., Then C 0 D0 is a side of the regular polygon required., Proof. Each polygon has as many sides as the ∠O, or ∠O0 , is contained times, in 4 rt. ∠s ., Therefore, the polygon C 0 D0 E 0 etc. is similar to the polygon CDE etc.,, , § 445, , (two regular polygons of the same number of sides are similar )., q.e.f., , Ex. 444. The area of an inscribed regular octagon is equal to that of the, rectangle whose sides are equal to the sides of the inscribed and the circumscribed squares.
Page 290 :
BOOK V. PLANE GEOMETRY., , 280, , Proposition XVIII. Problem., 478. Given the side and the radius of a regular inscribed polygon, to find, the side of the regular inscribed polygon of double the number of sides., D, , A, , C, , B, , O, , H, , Let AB be a side of the regular inscribed polygon., To find AD, a side of the regular inscribed polygon of double the number of sides., Denote the radius by R, and AB by a., From D draw DH through the centre O, and draw OA, AH., In the rt. 4OCA,, , DH is ⊥ to AB at its middle point C., 2, , OC = R2 − 14 a2 ., , Therefore,, , OC =, , DC = R −, , q, , R2 − 41 a2 ., , The ∠DAH is a rt. ∠., , In the rt. 4DAH, AD = DH × DC., q, But DH = 2R, and DC = R − R2 − 14 a2 ., r, q, ∴ AD = 2R(R − R2 − 41 a2 ), q, √, = R(2R − 4R2 − a2 )., , 479. Cor. If R = 1, AD =, , p, , 2−, , § 372, , q, R2 − 41 a2 ,, , and, , 2, , § 161, , √, 4 − a2 ., , § 290, § 367, , q.e.f.
Page 291 :
PROBLEMS OF CONSTRUCTION., , 281, , Proposition XIX. Problem., 480. To find the numerical value of the ratio of the circumference of a, circle to its diameter., , Let C be the circumference, when the radius is unity., To find the numerical value of π., By § 458, 2πR = C., , ∴ π = 12 C when R = 1., , Let S6 be the length of a side of a regular polygon of 6 sides, S12 of 12 sides,, and so on., If R = 1, by § 469, S6 = 1 and by § 479 we have, Form of Computation., Length of Side. Length of Perimeter., p, √, 2, S12 = q2 − 4 − 1, 0.51763809, 6.21165708, p, S24 = 2 − 4 − (0.51763809)2, 0.26105238, 6.26525722, q, p, S48 = 2 − 4 − (0.26105238)2, 0.13080626, 6.27870041, q, p, S96 = 2 − 4 − (0.13080626)2, 0.06543817, 6.28206396, q, p, 0.03272346, 6.28290510, S192 = 2 − 4 − (0.06543817)2, q, p, 0.01636228, 6.28311544, S384 = 2 − 4 − (0.03272346)2, q, p, S768 = 2 − 4 − (0.01636228)2, 0.00818121, 6.28316941, , ∴ C = 6.28317 approximately; that is, π = 3.14159 nearly., , 481. Scholium. π is incommensurable. We generally take, π = 3.1416, and, , 1, = 0.31831., π, , q.e.f.
Page 292 :
BOOK V. PLANE GEOMETRY., , 282, , MAXIMA AND MINIMA., 482. Def. Among geometrical magnitudes which satisfy given conditions,, the greatest is called the maximum; and the smallest is called the minimum., Thus, the diameter of a circle is the maximum among all chords; and the, perpendicular is the minimum among all lines drawn to a given line from a, given external point., 483. Def. Isoperimetric polygons are polygons which have equal, perimeters., Proposition XX. Theorem., 484. Of all triangles having two given sides, that in which these sides include a right angle is the maximum., A, , E, , D, , B, , E, , D, , C, , Let the triangles ABC and EBC have the sides AB and BC equal to, EB and BC, respectively; and let the angle ABC be a right angle., To prove that, 4ABC > 4EBC., Proof., From E draw the altitude ED., The 4s ABC and EBC, having the same base, BC, are to each other as their, altitudes AB and ED., § 405, Now, EB > ED., § 97, But, EB = AB., Hyp., ∴ AB > ED., ∴ 4ABC > 4EBC., , § 405, q.e.d.
Page 293 :
MAXIMA AND MINIMA., , 283, , Proposition XXI. Theorem., 485. Of all isoperimetric triangles having the same base the isosceles triangle is the maximum., H, , C, , A, , E, , K, , F, , D, , B, , M, , P, , Let the 4s ACB and ADB have equal perimeters, and let AC and, CB be equal, and AD and DB be unequal., To prove that 4ACB > 4ADB., Proof., Produce AC to H, making CH = AC; and draw HB., Produce HB, take DP equal to DB, and draw AP ., Draw CE and DF ⊥ to AB, and CK and DM k to AB., The ∠ABH is a right ∠, for it may be inscribed in the semicircle whose centre, is C and radius CA., § 290, ADP is not a straight line, for then the ∠s DBA and DAB would be equal,, being complements of the equal ∠s DBM and DP M , respectively; and DA and DB, would be equal (§ 147), which is contrary to the hypothesis. Hence,, AP < AD + DP , ∴< AD + DB, ∴< AC + CB, ∴< AH., ∴ BH > BP ., ∴ CE(=, Therefore,, , 1, 2 BH), , > DF (=, , 4ACB > 4ADB., , § 102, 1, 2 BP )., , Ax. 7, § 405, q.e.d.
Page 294 :
BOOK V. PLANE GEOMETRY., , 284, , Proposition XXII. Theorem., 486. Of all polygons with sides all given but one, the maximum can be, inscribed in a semicircle which has the undetermined side for its diameter., C, , B, D, , M, , A, , E, , N, , Let ABCDE be the maximum of polygons with sides AB, BC, CD,, DE, and the extremities A and E on the straight line M N ., To prove that ABCDE can be inscribed in a semicircle., Proof. From any vertex, as C, draw CA and CE., The 4ACE must be the maximum of all 4s having the sides CA and CE, and, the third side on M N ; otherwise by increasing or diminishing the ∠ACE, keeping, the lengths of the sides CA and CE unchanged, but sliding the extremities A and, E along the line M N , we could increase the 4ACE, while the rest of the polygon, would remain unchanged; and therefore increase the polygon. But this is contrary, to the hypothesis that the polygon is the maximum polygon., Hence, the 4ACE is the maximum of 4s that have the sides CA and CE., Therefore, the ∠ACE is a right angle., , § 484, , Therefore, C lies on the semicircumference., § 290, Hence, every vertex lies on the circumference; that is, the maximum polygon, can be inscribed in a semicircle having the undetermined side for a diameter. q.e.d.
Page 295 :
MAXIMA AND MINIMA., , 285, , Proposition XXIII. Theorem., 487. Of all polygons with given sides, that which can be inscribed in a circle, is the maximum., H, , H′, , D, , D′, , C, C′, , E′, , E, B, , B′, A, , A′, , Let ABCDE be a polygon inscribed in a circle, and A0 B 0 C 0 D 0 E 0 be a, polygon, equilateral with respect to ABCDE, which cannot be inscribed, in a circle., To prove that that ABCDE > A0 B 0 C 0 D0 E 0 ., Proof., Draw the diameter AH, and draw CH and DH., Upon C 0 D0 construct the 4C 0 H 0 D0 = 4CHD, and draw A0 H 0 ., Since, by hypothesis, a, cannot pass through all the vertices of A0 B 0 C 0 D0 E 0 ,, one or both of the parts ABCH, AEDH must be greater than the corresponding, part of A0 B 0 C 0 H 0 D0 E 0 ., § 486, If either of these parts is not greater than its corresponding part, it is equal to, it,, § 486, (for ABCH and AEDH are the maxima of polygons that have sides equal to, AB, BC, CH, and AE, ED, DH, respectively, and the remaining side, undetermined )., ∴ ABCHDE > A0 B 0 C 0 H 0 D0 E 0 ., Then, , Ax. 4, , Take away from the two figures the equal 4s CHD and C 0 H 0 D0 ., ABCDE > A0 B 0 C 0 D0 E 0 ., , Ax. 5, q.e.d.
Page 296 :
BOOK V. PLANE GEOMETRY., , 286, , Proposition XXIV. Theorem., 488. Of isoperimetric polygons of the same number of sides, the maximum, is equilateral., B, , A, , K, C, , D, , Let ABCD etc. be the maximum of isoperimetric polygons of any, given number of sides., To prove that AB, BC, CD, etc., are equal., Proof., Draw AC., The 4ABC must be the maximum of all the 4s which are formed upon AC, with a perimeter equal to that of 4ABC., Otherwise a greater 4AKC could be substituted for 4ABC, without changing, the perimeter of the polygon., But this is inconsistent with the hypothesis that the polygon ABCD etc. is the, maximum polygon., ∴ the 4ABC is isosceles., ∴ AB = BC., In like manner it may be proved that BC = CD, etc., , § 485, q.e.d., , 489. Cor. The maximum of isoperimetric polygons of the same number of, sides is a regular polygon., For the maximum polygon is equilateral (§ 488), and can be inscribed in a, circle (§ 487), and is, therefore, regular., § 430
Page 297 :
MAXIMA AND MINIMA., , 287, , Proposition XXV. Theorem., 490. Of isoperimetric regular polygons, that which has the greatest number, of sides is the maximum., C, , E, , A, , Q′, , Q, , D, , B, , Let Q be a regular polygon of three sides, and Q0 a regular polygon, of four sides, and let the two polygons have equal perimeters., To prove that Q0 is greater than Q., Proof. Draw CD from C to any point in AB., Invert the 4CDA and place it in the position DCE, letting D fall at C, C at, D, and A at E., The polygon DBCE is an irregular polygon of four sides, which by construction, has the same perimeter as Q0 , and the same area as Q., Then the irregular polygon DBCE of four sides is less than the isoperimetric, regular polygon Q0 of four sides., § 489, In like manner it may be shown that Q0 is less than an isoperimetric regular, polygon of five sides, and so on., q.e.d., , Ex. 445. Of all equivalent parallelograms that have equal bases, the, rectangle has the minimum perimeter., Ex. 446. Of all equivalent rectangles, the square has the minimum perimeter., Ex. 447. Of all triangles that have the same base and the same altitude,, the isosceles has the minimum perimeter., Ex. 448. Of all triangles that can be inscribed in a given circle, the, equilateral is the maximum and has the maximum perimeter.
Page 298 :
BOOK V. PLANE GEOMETRY., , 288, , Proposition XXVI. Theorem., 491. Of regular polygons having a given area, that which has the greatest, number of sides has the least perimeter., , Q, , Q′, , Q′′, , Let Q and Q0 be regular polygons having the same area, and let Q0, have the greater number of sides., To prove, the perimeter of Q > the perimeter of Q0 ., 00, Proof. Let Q be a regular polygon having the same perimeter as Q0 , and the, same number of sides as Q., Then, Q0 > Q00, § 490, (of isoperimetric regular polygons, that which has the greatest number of sides is, the maximum)., But, Q m Q0 ., Hyp., ∴ Q > Q00 ., ∴ the perimeter of Q > the perimeter of Q00 ., But the perimeter of Q0 = the perimeter of Q00 ., ∴ the perimeter of Q > the perimeter of Q0 ., , Hyp., q.e.d., , Ex. 449. To inscribe in a semicircle the maximum rectangle., Ex. 450. Of all polygons of a given number of sides which may be inscribed, in a given circle, that which is regular has the maximum area and the maximum, perimeter., Ex. 451. Of all polygons of a given number of sides which may be circumscribed about a given circle, that which is regular has the minimum area, and the minimum perimeter.
Page 299 :
EXERCISES., , 289, , THEOREMS., Ex. 452. Every equilateral polygon circumscribed about a circle is regular, if it has an odd number of sides., Ex. 453. Every equiangular polygon inscribed in a circle is regular if it, has an odd number of sides., Ex. 454. Every equiangular polygon circumscribed about a circle is regular., Ex. 455. The side of a circumscribed equilateral triangle is equal to twice, the side of the similar inscribed triangle., Ex. 456. The apothem of an inscribed regular hexagon is equal to half, the side of the inscribed equilateral triangle., Ex. 457. The area of an inscribed regular hexagon is three fourths of the, area of the circumscribed regular hexagon., Ex. 458. The area of an inscribed regular hexagon is the mean proportional between the areas of the inscribed and the circumscribed equilateral, triangles., Ex. 459. The square of the side of an inscribed equilateral triangle is, equal to three times the square of a side of the inscribed regular hexagon., Ex. 460. The area of an inscribed equilateral triangle is equal to half the, area of the inscribed regular hexagon., Ex. 461. The square of the side of an inscribed equilateral triangle is, equal to the sum of the squares of the sides of the inscribed square and of the, inscribed regular hexagon., Ex. 462. The square of the side of an inscribed regular pentagon is equal, to the sum of the squares of the radius of the circle and the side of the inscribed, regular decagon.
Page 300 :
BOOK V. PLANE GEOMETRY., , 290, , If R denotes the radius of a circle, and a one side of an inscribed regular, polygon, show that:, p, √, Ex. 463. In a regular pentagon, a = 21 R 10 − 2 5., p, √, Ex. 464. In a regular octagon, a = R 2 − 2., p, √, Ex. 465. In a regular dodecagon, a = R 2 − 3., Ex. 466. If two diagonals of a regular pentagon intersect, the longer, segment of each is equal to a side of the pentagon., Ex. 467. The apothem of an inscribed regular pentagon is equal to half, the sum of the radius of the circle and the side of the inscribed regular decagon., Ex. 468. The side of an inscribed regular pentagon is equal to the hypotenuse of the right triangle which has for legs the radius of the circle and, the side of the inscribed regular decagon., Ex. 469. The radius of an inscribed regular polygon is the mean proportional between its apothem and the radius of the similar circumscribed regular, polygon., Ex. 470. If squares are constructed outwardly upon the six sides of a, regular hexagon, the exterior vertices of these squares are the vertices of a, regular dodecagon., Ex. 471. If the alternate vertices of a regular hexagon are joined by, straight lines, show that another regular hexagon is thereby formed. Find the, ratio of the areas of these two hexagons., Ex. 472. If on the legs of a right triangle as diameters semicircles are, described external to the triangle, and from the whole figure a semicircle on, the hypotenuse is subtracted, the remaining figure is equivalent to the given, right triangle., Ex. 473. The star-shaped polygon, formed by producing the sides of a, regular hexagon, is equivalent to twice the given hexagon.
Page 301 :
EXERCISES., , 291, , Ex. 474. The sum of the perpendiculars drawn to the sides of a regular, polygon from any point within the polygon is equal to the apothem multiplied, by the number of sides., Ex. 475. If two chords of a circle are perpendicular to each other, the sum, of the four circles described on the four segments as diameters is equivalent to, the given circle., Ex. 476. If the diameter of a circle is divided into any two segments,, and upon these segments as diameters semicircumferences are described upon, opposite sides of the diameter, these semicircumferences divide the circle into, two parts which have the same ratio as the two segments of the diameter., Ex. 477. The diagonals that join any vertex of a regular polygon to all, the vertices not adjacent divide the angle at that vertex into as many equal, parts less two as the polygon has sides., PROBLEMS OF CONSTRUCTION., Ex. 478. To circumscribe an equilateral triangle about a given circle., Ex. 479. To circumscribe a square about a given circle., Ex. 480. To circumscribe a regular hexagon about a given circle., Ex. 481. To circumscribe a regular octagon about a given circle., Ex. 482. To circumscribe a regular pentagon about a given circle., Ex. 483. To draw through a given point a line so as to divide a given, circumference into two parts having the ratio 3 : 7., Ex. 484. To construct a circumference equal to the sum of two given, circumferences., Ex. 485. To construct a circumference equal to the difference of two given, circumferences., Ex. 486. To construct a circle equivalent to the sum of two given circles.
Page 302 :
BOOK V. PLANE GEOMETRY., , 292, , Ex. 487. To construct a circle equivalent to the difference of two given, circles., Ex. 488. To construct a circle equivalent to three times a given circle., Ex. 489. To construct a circle equivalent to three fourths of a given circle., Ex. 490. To construct a circle whose ratio to a given circle shall be equal, to the given ratio m : n., Ex. 491. To divide a given circle by a concentric circumference into two, equivalent parts., Ex. 492. To divide a given circle by concentric circumferences into five, equivalent parts., Ex. 493. To construct an angle of 18◦ ; of 36◦ ; of 9◦ ., Ex. 494. To construct an angle of12◦ ; of 24◦ ; of 6◦ ., To construct with a side of a given length:, Ex. 495. An equilateral triangle., Ex. 496. A square., Ex. 497. A regular hexagon., Ex. 498. A regular octagon., Ex. 499. A regular pentagon., Ex. 500. A regular decagon., Ex. 501. A regular dodecagon., Ex. 502. A regular pentedecagon., PROBLEMS OF COMPUTATION., Ex. 503. Find the area of a circle whose radius is 12 inches.
Page 303 :
EXERCISES., , 293, , Ex. 504. Find the circumference and the area of a circle whose diameter, is 8 feet., Ex. 505. A regular pentagon is inscribed in a circle whose radius is R. If, the length of a side is a, find the apothem., Ex. 506. A regular polygon is inscribed in a circle, √ whose radius is R. If, 1, the length of a side is a, show that the apothem is 2 R2 − a2 ., Ex. 507. Find the area of a regular decagon inscribed in a circle whose, radius is 16 inches., Ex. 508. Find the side of a regular dodecagon inscribed in a circle whose, radius is 20 inches., Ex. 509. Find the perimeter of a regular pentagon inscribed in a circle, whose radius is 25 feet., Ex. 510. The length of each side of a park in the shape of a regular, decagon is 100 yards. Find the area of the park., Ex. 511. Find the cost, at $2 per yard, of building a wall around a, cemetery in the shape of a regular hexagon, that contains 16, 627.84 square, yards., Ex. 512. The side of an inscribed regular polygon of n sides is 16 feet., Find the side of an inscribed regular polygon of 2n sides., Ex. 513. If the radius of a circle is R, and the side of an inscribed regular, polygon is a, show that the side of the similar circumscribed regular polygon, 2aR, ., is √, 4R2 − a2, Ex. 514. What is the width of the circular ring between two concentric, circumferences whose lengths are 650 feet and 425 feet?, Ex. 515. Find the angle subtended at the centre by an arc 5 feet 10 inches, long, if the radius of the circle is 9 feet 4 inches., Ex. 516. The chord of a segment is 10 feet, and the radius of the circle, is 16 feet. Find the area of the segment.
Page 304 :
BOOK V. PLANE GEOMETRY., , 294, , Ex. 517. Find the area of a sector, if the angle at the centre is 20◦ , and, the radius of the circle is 20 inches., Ex. 518. The chord of half an arc is 12 feet, and the radius of the circle, is 18 feet. Find the height of the segment subtended by the whole arc., Ex. 519. Find the side of a square which is equivalent to a circle whose, diameter is 35 feet., Ex. 520. The diameter of a circle is 15 feet. Find the diameter of a circle, twice as large. Three times as large., Ex. 521. Find the radii of the concentric circumferences that divide a, circle 11 inches in diameter into five equivalent parts., Ex. 522. The perimeter of a regular hexagon is 840 feet, and that of a, regular octagon is the same. By how many square feet is the octagon larger, than the hexagon?, Ex. 523. The diameter of a bicycle wheel is 28 inches. How many revolutions does the wheel make in going 10 miles?, Ex. 524. Find the diameter of a carriage wheel that makes 264 revolutions, in going half a mile., Ex. 525. The sides of three regular octagons are 6 feet, 7 feet, 8 feet,, respectively. Find the side of a regular octagon equivalent to the sum of the, three given octagons., Ex. 526. A circular pond 100 yards in diameter is surrounded by a walk, 10 feet wide. Find the area of the walk., Ex. 527. The span (chord) of a bridge in the form of a circular arc is, 120 feet, and the highest point of the arch is 15 feet above the piers. Find the, radius of the arc., Ex. 528. Three equal circles are described each tangent to the other two., If the common radius is R, find the area contained between the circles.
Page 305 :
EXERCISES., , 295, , Ex. 529. Given p, P , the perimeters of regular polygons of n sides inscribed in and circumscribed about a given circle. Find p0 , P 0 , the perimeters, of regular polygons of 2n sides inscribed in and circumscribed about the given, circle., Ex. 530. Given the radius R, and the apothem r of an inscribed regular, polygon of n sides. Find the radius R0 and the apothem r0 of an isoperimetrical, regular polygon of 2n sides., MISCELLANEOUS EXERCISES., THEOREMS., Ex. 531. If two adjacent angles of a quadrilateral are right angles, the, bisectors of the other two angles are perpendicular., Ex. 532. If two opposite angles of a quadrilateral are right angles, the, bisectors of the other two angles are parallel., Ex. 533. The two lines that join the middle points of the opposite sides, of a quadrilateral bisect each other., Ex. 534. The line that joins the feet of the perpendiculars dropped from, the extremities of the base of an isosceles triangle to the opposite sides is, parallel to the base., Ex. 535. If AD bisects the angle A of a triangle ABC, and BD bisects, the exterior angle CBF , then angle ADB equals one half angle ACB., Ex. 536. The sum of the acute angles at the vertices of a pentagram, (five-pointed star) is equal to two right angles., Ex. 537. The altitudes AD, BE, CF of the triangle ABC bisect the angles, of the triangle DEF ., Circles with AB, BC, AC as diameters will pass through E and D, E and, F , D and F , respectively., Ex. 538. The segments of any straight line intercepted between the circumferences of two concentric circles are equal.
Page 306 :
BOOK V. PLANE GEOMETRY., , 296, , Ex. 539. If a circle is circumscribed about any triangle, the feet of the, perpendiculars dropped from any point in the circumference to the sides of, the triangle lie in one straight line., Ex. 540. Two circles are tangent internally at P , and a chord AB of the, larger circle touches the smaller circle at C. Prove that P C bisects the angle, AP B., Ex. 541. The diagonals of a trapezoid divide each other into segments, which are proportional., Ex. 542. If through a point P in the circumference of a circle two chords, are drawn, the chords and the segments between P and a chord parallel to the, tangent at P are reciprocally proportional., Ex. 543. The perpendiculars from two vertices of a triangle upon the, opposite sides divide each other into segments reciprocally proportional., Ex. 544. The perpendicular from any point of a circumference upon a, chord is the mean proportional between the perpendiculars from the same, point upon the tangents drawn at the extremities of the chord., Ex. 545. In an isosceles right triangle either leg is the mean proportional, between the hypotenuse and the perpendicular upon it from the vertex of the, right angle., Ex. 546. If two circles intersect in the points A and B, and through A any, secant CAD is drawn limited by the circumferences at C and D, the straight, lines BC, BD are to each other as the diameters of the circles., Ex. 547. The area of a triangle is equal to half the product of its perimeter, by the radius of the inscribed circle., Ex. 548. The perimeter of a triangle is to one side as the perpendicular, from the opposite vertex is to the radius of the inscribed circle.
Page 307 :
EXERCISES., , 297, , Ex. 549. If three straight lines AA0 , BB 0 , CC 0 , drawn from the vertices of, a triangle ABC to the opposite sides, pass through a common point O within, the triangle, then, OA0 OB 0 OC 0, +, +, = 1., AA0 BB 0 CC 0, Ex. 550. ABC is a triangle, M the middle point of AB, P any point in, AB between A and M . If M D is drawn parallel to P C, meeting BC at D,, the triangle BP D is equivalent to half the triangle ABC., Ex. 551. Two diagonals of a regular pentagon, not drawn from a common, vertex, divide each other in extreme and mean ratio., Ex. 552. If all the diagonals of a regular pentagon are drawn, another, regular pentagon is thereby formed., Ex. 553. The area of an inscribed regular dodecagon is equal to three, times the square of the radius., Ex. 554. The area of a square inscribed in a semicircle is equal to two, fifths the area of the square inscribed in the circle., Ex. 555. The area of a circle is greater than the area of any polygon of, equal perimeter., Ex. 556. The circumference of a circle is less than the perimeter of any, polygon of equal area., PROBLEMS OF LOCI., Ex. 557. Find the locus of the centre of the circle inscribed in a triangle, that has a given base and a given angle at the vertex., Ex. 558. Find the locus of the intersection of the altitudes of a triangle, that has a given base and a given angle at the vertex., Ex. 559. Find the locus of the extremity of a tangent to a given circle, if, the length of the tangent is equal to a given line.
Page 308 :
BOOK V. PLANE GEOMETRY., , 298, , Ex. 560. Find the locus of a point, tangents drawn from which to a given, circle form a given angle., Ex. 561. Find the locus of the middle point of a line drawn from a given, point to a given straight line., Ex. 562. Find the locus of the vertex of a triangle that has a given base, and a given altitude., Ex. 563. Find the locus of a point the sum of whose distances from two, given parallel lines is equal to a given length., Ex. 564. Find the locus of a point the difference of whose distances from, two given parallel lines is equal to a given length., Ex. 565. Find the locus of a point the sum of whose distances from two, given intersecting lines is equal to a given length., Ex. 566. Find the locus of a point the difference of whose distances from, two given intersecting lines is equal to a given length., Ex. 567. Find the locus of a point whose distances from two given points, are in the given ratio m : n., Ex. 568. Find the locus of a point whose distances from two given parallel, lines are in the given ratio m : n., Ex. 569. Find the locus of a point whose distances from two given intersecting lines are in the given ratio m : n., Ex. 570. Find the locus of a point the sum of the squares of whose, distances from two given points is constant., Ex. 571. Find the locus of a point the difference of the squares of whose, distances from two given points is constant., Ex. 572. Find the locus of the vertex of a triangle that has a given base, and the other two sides in the given ratio m : n.
Page 309 :
EXERCISES., , 299, , PROBLEMS OF CONSTRUCTION., Ex. 573. To divide a given trapezoid into two equivalent parts by a line, parallel to the bases., Ex. 574. To divide a given trapezoid into two equivalent parts by a line, through a given point in one of the bases., Ex. 575. To construct a regular pentagon, given one of the diagonals., Ex. 576. To divide a given straight line into two segments such that their, product shall be the maximum., Ex. 577. To find a point in a semicircumference such that the sum of its, distances from the extremities of the diameter shall be the maximum., Ex. 578. To draw a common secant to two given circles exterior to each, other such that the intercepted chords shall have the given lengths a, b., Ex. 579. To draw through one of the points of intersection of two intersecting circles a common secant which shall have a given length., Ex. 580. To construct an isosceles triangle, given the altitude and one of, the equal base angles., Ex. 581. To construct an equilateral triangle, given the altitude., Ex. 582. To construct a right triangle, given the radius of the inscribed, circle and the difference of the acute angles., Ex. 583. To construct an equilateral triangle so that its vertices shall lie, in three given parallel lines., Ex. 584. To draw a line from a given point to a given straight line which, shall be to the perpendicular from the given point as m : n., Ex. 585. To find a point within a given triangle such that the perpendiculars from the point to the three sides shall be as the numbers m, n, p., Ex. 586. To draw a straight line equidistant from three given points.
Page 310 :
BOOK V. PLANE GEOMETRY., , 300, , Ex. 587. To draw a tangent to a given circle such that the segment, intercepted between the point of contact and a given straight line shall have a, given length., Ex. 588. To inscribe a straight line of a given length between two given, circumferences and parallel to a given straight line., Ex. 589. To draw through a given point a straight line so that its distances, from two other given points shall be in a given ratio., Ex. 590. To construct a square equivalent to the sum of a given triangle, and a given parallelogram., Ex. 591. To construct a rectangle having the difference of its base and, altitude equal to a given line, and its area equivalent to the sum of a given, triangle and a given pentagon., Ex. 592. To construct a pentagon similar to a given pentagon and equivalent to a given trapezoid., Ex. 593. To find a point whose distances from three given straight lines, shall be as the numbers m, n, p., Ex. 594. Given an angle and two points P and P 0 between the sides of, the angle. To find the shortest path from P to P 0 that shall touch both sides, of the angle., Ex. 595. To construct a triangle, given its angles and its area., Ex. 596. To transform a given triangle into a triangle similar to another, given triangle., Ex. 597. Given three points A, B, C. To find a fourth point P such that, the areas of the triangles AP B, AP C, BP C shall be equal., Ex. 598. To construct a triangle, given its base, the ratio of the other, sides, and the angle included by them., Ex. 599. To divide a given circle into n equivalent parts by concentric, circumferences.
Page 311 :
EXERCISES., , 301, , Ex. 600. In a given equilateral triangle to inscribe three equal circles, tangent to each other, each circle tangent to two sides of the triangle., Ex. 601. Given an angle and a point P between the sides of the angle., To draw through P a straight line that shall form with the sides of the angle, a triangle with the perimeter equal to a given length a., Ex. 602. In a given square to inscribe four equal circles, so that each, circle shall be tangent to two of the others and also tangent to two sides of, the square., Ex. 603. In a given square to inscribe four equal circles, so that each, circle shall be tangent to two of the others and also tangent to one side of the, square.
Page 312 :
TABLE OF FORMULAS., , PLANE FIGURES., NOTATION., P, h, b, b0, R, D, C, r, a, b, c, s, p, m, n, , =, =, =, =, =, =, =, =, =, =, =, =, , perimeter., altitude., lower base., upper base., radius of circle., diameter of circle., circumference of circle., apothem of regular polygon., sides of triangle., 1, (a + b + c)., 2, perpendicular of triangle., segments of third side of triangle adjacent to, sides b and a, respectively., S = area., π = 3.1416.
Page 313 :
TABLE OF FORMULAS., , 303, , FORMULAS., Line Values., PAGE, , Right triangle,, , 2, , b, p2, b 2 : a2, b2 : c 2, a2 + b 2, Any triangle,, a2, Altitude of triangle on side a,, , =, =, =, ::, =, =, , 2, , c × m; a = c × n, m×n, m:n, m:c, c2, b2 + c2 ± 2c × m, , 2p, s(s − a)(s − b)(s − c), a, Median of triangle on side a,, p, m = 12 2(b2 + c2 ) − a2, Bisector of triangle on side a,, 2 p, bcs(s − a), t=, b+c, Radius of circumscribed circle,, abc, R= p, 4 s(s − a)(s − b)(s − c), Circumference of circle,, C = 2πR, “, “, C = πD, Areas., Rectangle,, S = b×h, Square,, S = b2, Parallelogram,, S = b×h, 1, b×h, Triangle,, S=p, 2, “, S = s(s − a)(s − b)(s − c), abc, “, S=, 4R, a2 √, Equilateral triangle,, S=, 3, 4, 1, Trapezoid,, S = 2 h(b + b0 ), Regular polygon,, S = 21 r × P, Circle,, S = 21 R × C, “, S = πR2, Sector,, S = 21 R × arc, h=, , 197, 197, 198, 198, 199, 200,201, 219, 220, 221, 222, 270, 270, 229, 229, 230, 231, 254, 254, 253, 232, 270, 272, 272, 271
Page 314 :
PLANE GEOMETRY., , 304
Page 315 :
INDEX., , PAGE, , PAGE, , Abbreviations, Alternation, Altitude of parallelogram, “ of trapezoid, “ of triangle, Analysis, Angle, “ acute, “ at centre of, regular polygon, “ central, “ exterior of triangle, “ inscribed in circle, “ inscribed in segment, “ oblique, “ obtuse, “ reflex, “ right, “ salient, “ straight, “ vertical, Angles, adjacent, “ alternate-exterior, “ alternate-interior, “ complementary, “ conjugate, “ exterior, “ exterior-interior, “ interior, “ supplementary, , 6, 170, 52, 52, 34, 4, 10, 12, 260, 90, 33, 90, 90, 12, 12, 12, 11, 61, 11, 34, 11,33, 28, 28, 14, 13, 28, 28, 28, 14, , “ supplementary-adjacent, “ vertical, Antecedents, Apothem, Arc, Area, Axiom, “ of parallel lines, Axioms of straight lines, “ general, Axis of symmetry, , 18, 14, 168, 260, 90, 226, 4, 26, 8, 6, 65, , Base of isosceles triangle, “ of parallelogram, “ of triangle, Bases of trapezoid, Bisector, , 34, 51, 34, 52, 11, , Centre of circle, “ of regular polygon, “ of symmetry, Chord, Circle, “ circumscribed, “ inscribed, Circles, concentric, “ escribed, Circum-centre of triangle, Circumference, Commensurable, Complement, , 89, 260, 65, 90, 89, 91, 91, 91, 153, 151, 89, 109, 14
Page 316 :
INDEX., PAGE, , PAGE, , Composition, Conclusion, Concurrent lines, Congruent figures, Consequents, Constant, Construction, Continued proportion, Continuity, Principle of, Contradictory of a theorem, Converse of a theorem, Convex curve, Curved surface, , 171, 4, 77, 7, 168, 111, 4, 168, 127, 4, 5,94, 266, 7, , Decagon, Diagonal, Diameter, Dimensions, Distance, Division, Dodecagon, Duality, Principle of, , 62, 52,61, 89, 1, 8,24, 172, 62, 39, , Equal figures, Equimultiples, Equivalent figures, Ex-centres of triangle, Extreme and mean ratio, Extremes, , 7, 176, 7,226, 153, 213, 168, , Figure, curvilinear, “ geometrical, “ plane, “ rectilinear, Foot of perpendicular, , 7, 3, 7, 7, 11, , 306, , Fourth proportional, , 168, , Geometrical solid, Geometry, Geometry, Plane, “ Solid, , 2,2, 3, 3, 3, , Harmonic division, Heptagon, Hexagon, Homologous angles, “ lines, “ sides, Hypotenuse, Hypothesis, In-centre of triangle, Incommensurable ratio, Intersection, Inversion, Isoperimetric figures, Legs of right triangle, “ of trapezoid, Limit, Line, “ curved, “ of centres, “ straight, Lines, oblique, “ parallel, “ perpendicular, Major arc, Maximum, , 180, 62, 62, 35,62, 183, 35,62, 34, 4, 152, 110, 1, 170, 282, 34, 52, 111, 1,2,3, 7, 106, 7, 12, 26, 11, 90, 282
Page 317 :
INDEX., PAGE, , Mean proportional, Means, Median of trapezoid, Minimum, Minor arc, , 168, 168, 52, 282, 90, , Negative quantities, Numerical measure, , 127, 109, , Octagon, Opposite of a theorem, Origin, , 62, 5, 8, , Parallel lines, Parallelogram, Pentagon, Pentagram, Perigon, Perimeter, Perpendicular bisector, Perpendicular lines, Pi (π), Plane, Point, “ of contact, “ of tangency, Polygon, “ angles of, “ circumscribed, “ concave, “ convex, “ equiangular, “ equilateral, “ inscribed, “ regular, , 26, 51, 62, 295, 13, 33,61, 49, 11, 269, 1,7, 1,2, 89, 89, 61, 61, 91, 61, 61, 61, 61, 91, 258, , 307, PAGE, , Polygons mut. equiangular, “ mutually equilateral, Positive quantities, Postulate, Projection, Proof, Proportion, Proposition, Quadrant, Quadrilateral, Radius of regular polygon, Ratio, Ratio of similitude, Reciprocity, Principle of, Rectangle, Rhomboid, Rhombus, Scholium, Secant, Sector, Segment of circle, “ of line, Semicircle, Semicircumference, Sides of an angle, “ of polygon, “ of triangle, Similar arcs, “ figures, “ polygons, “ sectors, “ segments, “ triangles, , 62, 62, 127, 4, 201, 3, 168, 4, 90, 51,62, 260, 109, 183, 39, 51, 51, 51, 4, 89,204, 90, 90, 8, 90, 90, 10, 61, 33, 270, 7, 183, 270, 270, 184
Page 318 :
INDEX., PAGE, , Square, Superposition, Supplement, Surface, Symbols, Symmetry, Tangent, “ common external, “ common internal, Terms of a proportion, Theorem, Third proportional, Transversal, Trapezium, Trapezoid, “ isosceles, Triangle, “ equiangular, “ equilateral, “ isosceles, “ obtuse, “ right, “ scalene, “ altitudes of, “ angles of, “ bisectors of, “ medians of, “ vertices of, Variable, Vertex of angle, “ of triangle, Vertices of polygon, , 51, 10, 14, 1,2,2, 6, 65, 89,106, 106, 106, 168, 4, 168, 27, 51, 51, 52, 33,62, 34, 34, 34, 34, 34, 34, 34, 33, 35, 35, 33, 111, 10, 34, 61, , 308
Page 319 :
LICENSE., , End of Project Gutenberg’s Plane Geometry, by George Albert Wentworth, *** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***, ***** This file should be named 33063-pdf.pdf or 33063-pdf.zip *****, This and all associated files of various formats will be found in:, http://www.gutenberg.org/3/3/0/6/33063/, Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy, and the Online Distributed Proofreading Team at, http://www.pgdp.net, , Updated editions will replace the previous one--the old editions, will be renamed., Creating the works from public domain print editions means that no, one owns a United States copyright in these works, so the Foundation, (and you!) can copy and distribute it in the United States without, permission and without paying copyright royalties. Special rules,, set forth in the General Terms of Use part of this license, apply to, copying and distributing Project Gutenberg-tm electronic works to, protect the PROJECT GUTENBERG-tm concept and trademark. Project, Gutenberg is a registered trademark, and may not be used if you, charge for the eBooks, unless you receive specific permission. If you, do not charge anything for copies of this eBook, complying with the, rules is very easy. You may use this eBook for nearly any purpose, such as creation of derivative works, reports, performances and, research. They may be modified and printed and given away--you may do, practically ANYTHING with public domain eBooks. Redistribution is, subject to the trademark license, especially commercial, redistribution., , *** START: FULL LICENSE ***, THE FULL PROJECT GUTENBERG LICENSE, PLEASE READ THIS BEFORE YOU DISTRIBUTE OR USE THIS WORK, To protect the Project Gutenberg-tm mission of promoting the free, distribution of electronic works, by using or distributing this work, (or any other work associated in any way with the phrase "Project, Gutenberg"), you agree to comply with all the terms of the Full Project, Gutenberg-tm License (available with this file or online at, http://gutenberg.org/license)., , Section 1. General Terms of Use and Redistributing Project Gutenberg-tm, electronic works, , 309
Page 320 :
LICENSE., , 1.A. By reading or using any part of this Project Gutenberg-tm, electronic work, you indicate that you have read, understand, agree to, and accept all the terms of this license and intellectual property, (trademark/copyright) agreement. If you do not agree to abide by all, the terms of this agreement, you must cease using and return or destroy, all copies of Project Gutenberg-tm electronic works in your possession., If you paid a fee for obtaining a copy of or access to a Project, Gutenberg-tm electronic work and you do not agree to be bound by the, terms of this agreement, you may obtain a refund from the person or, entity to whom you paid the fee as set forth in paragraph 1.E.8., 1.B. "Project Gutenberg" is a registered trademark. It may only be, used on or associated in any way with an electronic work by people who, agree to be bound by the terms of this agreement. There are a few, things that you can do with most Project Gutenberg-tm electronic works, even without complying with the full terms of this agreement. See, paragraph 1.C below. There are a lot of things you can do with Project, Gutenberg-tm electronic works if you follow the terms of this agreement, and help preserve free future access to Project Gutenberg-tm electronic, works. See paragraph 1.E below., 1.C. The Project Gutenberg Literary Archive Foundation ("the Foundation", or PGLAF), owns a compilation copyright in the collection of Project, Gutenberg-tm electronic works. Nearly all the individual works in the, collection are in the public domain in the United States. If an, individual work is in the public domain in the United States and you are, located in the United States, we do not claim a right to prevent you from, copying, distributing, performing, displaying or creating derivative, works based on the work as long as all references to Project Gutenberg, are removed. Of course, we hope that you will support the Project, Gutenberg-tm mission of promoting free access to electronic works by, freely sharing Project Gutenberg-tm works in compliance with the terms of, this agreement for keeping the Project Gutenberg-tm name associated with, the work. You can easily comply with the terms of this agreement by, keeping this work in the same format with its attached full Project, Gutenberg-tm License when you share it without charge with others., 1.D. The copyright laws of the place where you are located also govern, what you can do with this work. Copyright laws in most countries are in, a constant state of change. If you are outside the United States, check, the laws of your country in addition to the terms of this agreement, before downloading, copying, displaying, performing, distributing or, creating derivative works based on this work or any other Project, Gutenberg-tm work. The Foundation makes no representations concerning, the copyright status of any work in any country outside the United, States., 1.E., , Unless you have removed all references to Project Gutenberg:, , 310
Page 321 :
LICENSE., , 1.E.1. The following sentence, with active links to, or other immediate, access to, the full Project Gutenberg-tm License must appear prominently, whenever any copy of a Project Gutenberg-tm work (any work on which the, phrase "Project Gutenberg" appears, or with which the phrase "Project, Gutenberg" is associated) is accessed, displayed, performed, viewed,, copied or distributed:, This eBook is for the use of anyone anywhere at no cost and with, almost no restrictions whatsoever. You may copy it, give it away or, re-use it under the terms of the Project Gutenberg License included, with this eBook or online at www.gutenberg.org, 1.E.2. If an individual Project Gutenberg-tm electronic work is derived, from the public domain (does not contain a notice indicating that it is, posted with permission of the copyright holder), the work can be copied, and distributed to anyone in the United States without paying any fees, or charges. If you are redistributing or providing access to a work, with the phrase "Project Gutenberg" associated with or appearing on the, work, you must comply either with the requirements of paragraphs 1.E.1, through 1.E.7 or obtain permission for the use of the work and the, Project Gutenberg-tm trademark as set forth in paragraphs 1.E.8 or, 1.E.9., 1.E.3. If an individual Project Gutenberg-tm electronic work is posted, with the permission of the copyright holder, your use and distribution, must comply with both paragraphs 1.E.1 through 1.E.7 and any additional, terms imposed by the copyright holder. Additional terms will be linked, to the Project Gutenberg-tm License for all works posted with the, permission of the copyright holder found at the beginning of this work., 1.E.4. Do not unlink or detach or remove the full Project Gutenberg-tm, License terms from this work, or any files containing a part of this, work or any other work associated with Project Gutenberg-tm., 1.E.5. Do not copy, display, perform, distribute or redistribute this, electronic work, or any part of this electronic work, without, prominently displaying the sentence set forth in paragraph 1.E.1 with, active links or immediate access to the full terms of the Project, Gutenberg-tm License., 1.E.6. You may convert to and distribute this work in any binary,, compressed, marked up, nonproprietary or proprietary form, including any, word processing or hypertext form. However, if you provide access to or, distribute copies of a Project Gutenberg-tm work in a format other than, "Plain Vanilla ASCII" or other format used in the official version, posted on the official Project Gutenberg-tm web site (www.gutenberg.org),, you must, at no additional cost, fee or expense to the user, provide a, copy, a means of exporting a copy, or a means of obtaining a copy upon, request, of the work in its original "Plain Vanilla ASCII" or other, form. Any alternate format must include the full Project Gutenberg-tm, , 311
Page 322 :
LICENSE., , License as specified in paragraph 1.E.1., 1.E.7. Do not charge a fee for access to, viewing, displaying,, performing, copying or distributing any Project Gutenberg-tm works, unless you comply with paragraph 1.E.8 or 1.E.9., 1.E.8. You may charge a reasonable fee for copies of or providing, access to or distributing Project Gutenberg-tm electronic works provided, that, - You pay a royalty fee of 20% of the gross profits you derive from, the use of Project Gutenberg-tm works calculated using the method, you already use to calculate your applicable taxes. The fee is, owed to the owner of the Project Gutenberg-tm trademark, but he, has agreed to donate royalties under this paragraph to the, Project Gutenberg Literary Archive Foundation. Royalty payments, must be paid within 60 days following each date on which you, prepare (or are legally required to prepare) your periodic tax, returns. Royalty payments should be clearly marked as such and, sent to the Project Gutenberg Literary Archive Foundation at the, address specified in Section 4, "Information about donations to, the Project Gutenberg Literary Archive Foundation.", - You provide a full refund of any money paid by a user who notifies, you in writing (or by e-mail) within 30 days of receipt that s/he, does not agree to the terms of the full Project Gutenberg-tm, License. You must require such a user to return or, destroy all copies of the works possessed in a physical medium, and discontinue all use of and all access to other copies of, Project Gutenberg-tm works., - You provide, in accordance with paragraph 1.F.3, a full refund of any, money paid for a work or a replacement copy, if a defect in the, electronic work is discovered and reported to you within 90 days, of receipt of the work., - You comply with all other terms of this agreement for free, distribution of Project Gutenberg-tm works., 1.E.9. If you wish to charge a fee or distribute a Project Gutenberg-tm, electronic work or group of works on different terms than are set, forth in this agreement, you must obtain permission in writing from, both the Project Gutenberg Literary Archive Foundation and Michael, Hart, the owner of the Project Gutenberg-tm trademark. Contact the, Foundation as set forth in Section 3 below., 1.F., 1.F.1. Project Gutenberg volunteers and employees expend considerable, effort to identify, do copyright research on, transcribe and proofread, , 312
Page 323 :
LICENSE., , public domain works in creating the Project Gutenberg-tm, collection. Despite these efforts, Project Gutenberg-tm electronic, works, and the medium on which they may be stored, may contain, "Defects," such as, but not limited to, incomplete, inaccurate or, corrupt data, transcription errors, a copyright or other intellectual, property infringement, a defective or damaged disk or other medium, a, computer virus, or computer codes that damage or cannot be read by, your equipment., 1.F.2. LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the "Right, of Replacement or Refund" described in paragraph 1.F.3, the Project, Gutenberg Literary Archive Foundation, the owner of the Project, Gutenberg-tm trademark, and any other party distributing a Project, Gutenberg-tm electronic work under this agreement, disclaim all, liability to you for damages, costs and expenses, including legal, fees. YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE, STRICT, LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT THOSE, PROVIDED IN PARAGRAPH F3. YOU AGREE THAT THE FOUNDATION, THE, TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT BE, LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL, PUNITIVE OR, INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH, DAMAGE., 1.F.3. LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a, defect in this electronic work within 90 days of receiving it, you can, receive a refund of the money (if any) you paid for it by sending a, written explanation to the person you received the work from. If you, received the work on a physical medium, you must return the medium with, your written explanation. The person or entity that provided you with, the defective work may elect to provide a replacement copy in lieu of a, refund. If you received the work electronically, the person or entity, providing it to you may choose to give you a second opportunity to, receive the work electronically in lieu of a refund. If the second copy, is also defective, you may demand a refund in writing without further, opportunities to fix the problem., 1.F.4. Except for the limited right of replacement or refund set forth, in paragraph 1.F.3, this work is provided to you ’AS-IS’ WITH NO OTHER, WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO, WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY PURPOSE., 1.F.5. Some states do not allow disclaimers of certain implied, warranties or the exclusion or limitation of certain types of damages., If any disclaimer or limitation set forth in this agreement violates the, law of the state applicable to this agreement, the agreement shall be, interpreted to make the maximum disclaimer or limitation permitted by, the applicable state law. The invalidity or unenforceability of any, provision of this agreement shall not void the remaining provisions., 1.F.6., , INDEMNITY - You agree to indemnify and hold the Foundation, the, , 313
Page 324 : LICENSE., , trademark owner, any agent or employee of the Foundation, anyone, providing copies of Project Gutenberg-tm electronic works in accordance, with this agreement, and any volunteers associated with the production,, promotion and distribution of Project Gutenberg-tm electronic works,, harmless from all liability, costs and expenses, including legal fees,, that arise directly or indirectly from any of the following which you do, or cause to occur: (a) distribution of this or any Project Gutenberg-tm, work, (b) alteration, modification, or additions or deletions to any, Project Gutenberg-tm work, and (c) any Defect you cause., , Section, , 2., , Information about the Mission of Project Gutenberg-tm, , Project Gutenberg-tm is synonymous with the free distribution of, electronic works in formats readable by the widest variety of computers, including obsolete, old, middle-aged and new computers. It exists, because of the efforts of hundreds of volunteers and donations from, people in all walks of life., Volunteers and financial support to provide volunteers with the, assistance they need, are critical to reaching Project Gutenberg-tm’s, goals and ensuring that the Project Gutenberg-tm collection will, remain freely available for generations to come. In 2001, the Project, Gutenberg Literary Archive Foundation was created to provide a secure, and permanent future for Project Gutenberg-tm and future generations., To learn more about the Project Gutenberg Literary Archive Foundation, and how your efforts and donations can help, see Sections 3 and 4, and the Foundation web page at http://www.pglaf.org., , Section 3., Foundation, , Information about the Project Gutenberg Literary Archive, , The Project Gutenberg Literary Archive Foundation is a non profit, 501(c)(3) educational corporation organized under the laws of the, state of Mississippi and granted tax exempt status by the Internal, Revenue Service. The Foundation’s EIN or federal tax identification, number is 64-6221541. Its 501(c)(3) letter is posted at, http://pglaf.org/fundraising. Contributions to the Project Gutenberg, Literary Archive Foundation are tax deductible to the full extent, permitted by U.S. federal laws and your state’s laws., The Foundation’s principal office is located at 4557 Melan Dr. S., Fairbanks, AK, 99712., but its volunteers and employees are scattered, throughout numerous locations. Its business office is located at, 809 North 1500 West, Salt Lake City, UT 84116, (801) 596-1887, email,
[email protected]. Email contact links and up to date contact, information can be found at the Foundation’s web site and official, page at http://pglaf.org, , 314
Page 325 : LICENSE., , For additional contact information:, Dr. Gregory B. Newby, Chief Executive and Director,
[email protected], , Section 4. Information about Donations to the Project Gutenberg, Literary Archive Foundation, Project Gutenberg-tm depends upon and cannot survive without wide, spread public support and donations to carry out its mission of, increasing the number of public domain and licensed works that can be, freely distributed in machine readable form accessible by the widest, array of equipment including outdated equipment. Many small donations, ($1 to $5,000) are particularly important to maintaining tax exempt, status with the IRS., The Foundation is committed to complying with the laws regulating, charities and charitable donations in all 50 states of the United, States. Compliance requirements are not uniform and it takes a, considerable effort, much paperwork and many fees to meet and keep up, with these requirements. We do not solicit donations in locations, where we have not received written confirmation of compliance. To, SEND DONATIONS or determine the status of compliance for any, particular state visit http://pglaf.org, While we cannot and do not solicit contributions from states where we, have not met the solicitation requirements, we know of no prohibition, against accepting unsolicited donations from donors in such states who, approach us with offers to donate., International donations are gratefully accepted, but we cannot make, any statements concerning tax treatment of donations received from, outside the United States. U.S. laws alone swamp our small staff., Please check the Project Gutenberg Web pages for current donation, methods and addresses. Donations are accepted in a number of other, ways including checks, online payments and credit card donations., To donate, please visit: http://pglaf.org/donate, , Section 5., works., , General Information About Project Gutenberg-tm electronic, , Professor Michael S. Hart is the originator of the Project Gutenberg-tm, concept of a library of electronic works that could be freely shared, with anyone. For thirty years, he produced and distributed Project, Gutenberg-tm eBooks with only a loose network of volunteer support., , 315
Page 326 :
LICENSE., , Project Gutenberg-tm eBooks are often created from several printed, editions, all of which are confirmed as Public Domain in the U.S., unless a copyright notice is included. Thus, we do not necessarily, keep eBooks in compliance with any particular paper edition., , Most people start at our Web site which has the main PG search facility:, http://www.gutenberg.org, This Web site includes information about Project Gutenberg-tm,, including how to make donations to the Project Gutenberg Literary, Archive Foundation, how to help produce our new eBooks, and how to, subscribe to our email newsletter to hear about new eBooks., , 316
