PLANE AND SOLID 
 GEOMETRX 
 
 BY 
 
 WEBSTER WELLS, S.B. 
 
 AUTHOR OF A 8EHIES OF TEXTS ON MATHEMATICS 
 
 AND 
 
 WALTER W. HART, A.B. 
 
 ASSISTANT PROFESSOR OF MATHEMATICS, UNIVERSITY OF WMO^WIN 
 COURSE FOR THE TRAINING OF TEACHERS 
 
 o>Hc 
 
 D. C. HEATH & COMPANY, PUBLISHERS 
 BOSTON NEW YORK CHICAGO 
 
aA4 3 
 
 WELLS AND HAKT'S MATHEMATICS 
 
 First Year Algebra 
 
 A one year course 
 
 New High School Algebra 
 
 A three semester course 
 
 Second Course in Algebra 
 
 Appears in a brief and in an enlarged edition. 
 Suitable for a one or a two semester course following 
 a first year course 
 
 Plane Geometry 
 
 Provision is made for a brief or an extended course 
 
 Solid Geometry 
 
 D. C. HEATH & CO., PUBLISHERS 
 
 EDUcAT«ojv< i)epf: 
 
 Copyright, 1915, 1916, 
 By D. C. Heath & Co. 
 
 I a6 
 
PREFACE 
 
 Greek Geometry, the finest product of deductive thinking 
 which high school pupils encounter, has come down to us 
 through twenty-two centuries practically unchanged in essen- 
 tial content or form. It has been presented in texts, each 
 built upon a preceding one and each good in its day, which 
 have sought to present the great science in accord with the 
 ideals of their time. 
 
 This text is a thorough revision of Wells's Essentials of 
 Geometry in accord with current scientific and pedagogical 
 thought. The scientific ideal is represented the world over 
 by Hilbert's Foundations of Geometry. (Translated by Town- 
 send, Open Court Pub. Co., Chicago.) The pedagogical ideals 
 are represented in this country by the Report of the National 
 Committee of Fifteen. (See Mathematics Teacher, Dec, 1912 ; 
 School Science and Mathematics, 1911 ; Proceedings of N. E. A., 
 1911.) These ideals and the personal experience of one of the 
 authors in teaching high school geometry in recent years have 
 been the determining factors in the making of this text. 
 Permit us to direct attention to some of its features. 
 
 In each Book, the fundamentally important theorems are 
 given first. These theorems present a safe and sane minimum 
 course. These are followed in each Book by one or more 
 groups of theorems or applications which are strictly supple- 
 mentary, — material which either has long appeared in geom- 
 etries in some form or has been introduced in recent years to 
 add to the pupils' interest. Teachers will find no difficulty in 
 
 54 X an 8 
 
IV PREFACE 
 
 making selection from this material, and, on the other hand, 
 will not be embarrassed by omitting any of it. (See pp. 172, 
 210, and 245.) 
 
 The introduction presents only the immediately necessary 
 concepts, notation, and terminology. Emphasis is upon the 
 acquisition of these and of skill in the use of tools, and above 
 all upon the acquisition of the important point of view pre- 
 sented in §§ 48-50. 
 
 The fundamental constructions are placed early in Book I 
 so that pupils can be required to construct their figures ; they 
 are not placed earlier because they cannot be proved earlier. 
 
 Authorities and details of demonstrations which pupils can 
 supply are increasingly omitted from the demonstrations, and 
 often only suggestions are given. The resulting proofs are an 
 incentive to real thinking for all the members of the class ; 
 they do not consume time that can be spent more profitably 
 upon exercises and other valuable supplementary material. 
 
 Pupils are encouraged to plan their proofs instead of plung- 
 ing blindly into a demonstration. (See §§69 and 117.) 
 
 Unnecessary corollaries have been omitted, and dignity and 
 importance is given to those which are included in the text. 
 (See §§ 71, 96, 101.) 
 
 The stages of the proof are plainly marked, the steps are 
 numbered, the reasons are given in full, and the proofs are 
 arranged attractively on the page. 
 
 Carefully selected exercises follow most of the propositions. 
 Notice exercises such as Exercises 2, 23, 45, 63, of Book I, de- 
 signed to teach concretely and inductively the theorems which 
 immediately follow. Notice also the illustrative exercises 
 which set a standard for the pupils' solution. (See pp. 31, 32, 
 157.) Enough exercises are provided for a minimum course. 
 Besides these, there are miscellaneous exercises at the close of 
 each Book, depending upon only the theorems of the minimum 
 course. Finally there will be found from time to time a note 
 like that on page 52, referring to supplementary exercises at 
 the end of the text. (See pp. 52, 59, 83.) Suggestions are 
 
PREFACE V 
 
 given with exercises where experience has shown that a ma- 
 jority of a class require such assistance in order to do effective 
 work. (See Book 1, Ex. 128, 131, etc.) 
 
 Simple applied problems (see Book I, Ex. 15, 37, 39, 40, 41, 
 etc.) and artistic designs (see pp. 1, 47, 50, etc.) exhibit to the 
 pupils some of the uses of geometry. Only simple applica- 
 tions are included in the minimum course. Other applications 
 are introduced among the supplementary exercises at the end 
 of the text and among the supplementary topics at the close 
 of certain of the Books. (See pp. 138, 172, 174, 246, etc.) 
 
 A brief history of geometry is included in the introduction, 
 and other historical references are introduced from time to time 
 throughout the text. (See pp. 29, 36, 46, 240, etc.) 
 
 Axioms are defined in the accepted modern form (p. 22). 
 They are introduced only as they become necessary. (See 
 pp. 22, 29, 50, 82.) In the introduction, their meaning is made 
 clear by suitable preliminary exercises. (See Introduction, Ex. 
 22, 23, 24, 37, etc.) The definitions also are modern and con- 
 sistent, 'even though they are in some cases different from those 
 ordinarily given. (See § § 1,2, 4, 5, 47, the note on p. 27, etc.) 
 For example, after defining a circle as a line, which is correct, 
 there is every reason for also defining a polygon as a line, 
 instead of defining it as a portion of a plane. It may seem 
 strange at first also not to find in the first paragraph of the 
 text the attempted distinction between a physical and a geo- 
 metrical solid, — something that is psychologically impossible 
 for beginners, — but the authors believe firmly that there is 
 much to recommend their own informal statements in § 1. 
 
 The incommensurable cases are dismissed with a mere re- 
 mark on pages 113, 149, and 194, and are treated fully only 
 after the theory of limits is given on page 260. 
 
 The mensuration of the circle is treated informally at first 
 on page 238. The treatment involves nevertheless the basic 
 ideas which are developed more fully in the formal treatment 
 of the same topic which appears as one of the supplementary 
 topics of Book V on page 248. This treatment is as elemen- 
 
VI PREFACE 
 
 tary as the difficulty of the subject permits ; to give less would 
 render the treatment either incomprehensible or incomplete. 
 On the other hand, the treatment is as sound as an elementary 
 presentation renders possible ; to give more would certainly 
 render the subject distasteful to an average high school class. 
 
 In the treatment of the mensuration of the cylinder and 
 the cone, the fundamental limits theorems are assumed on 
 the ground that rigorous proofs are beyond the scope of an 
 elementary course. In the enunciation of the area theorems 
 for portions of the surface of a sphere, changes have been 
 made which enable pupils both to learn and to remember the 
 theorems more readily. 
 
 The course in Solid Geometry is practical in the sense that 
 the mensuration theorems for the common solids are given the 
 place of prominence. Por example, in Book IX the mensu- 
 ration of the sphere is treated in the minimum course, — the 
 mathematically interesting theorems about spherical geometry 
 being grouped as a supplementary topic. Besides this empha- 
 sis given to the mensuration theorems, some natural applica- 
 tions of solid geometry are touched upon in the exercises. 
 
CONTENTS 
 
 PAOB 
 
 Introduction 1 
 
 Book I. Rectilinear Figures 29 
 
 Supplementary Theorems 87 
 
 Miscellaneous Exercises 91 
 
 Book II. The Circle 93 
 
 Measurement of Angles and Arcs 1 12 
 
 Loci 124 
 
 Supplementary Topics 128 
 
 Miscellaneous Exercises 139 
 
 Book III. Proportion — Similar Polygons 141 
 
 Supplementary Topics 172 
 
 Miscellaneous Exercises 189 
 
 Book IV, Areas of Polygons 191 
 
 Supplementary Topics 210 
 
 Miscellaneous Exercises 219 
 
 Book V. Regular Polygons 221 
 
 Mensuration of the Circle — Informal Treatment . . . 238 
 
 Supplementary Topics 245 
 
 Supplementary Exercises — Book I 273 
 
 Supplementary Exercises — Book II 283 
 
 Supplementary Exercises — Book III 289 
 
 Supplementary Exercises — Book IV 294 
 
 Supplementary Exercises — Book V 299 
 
 vn 
 
viii CONTENTS 
 
 PAGE 
 
 Book VI. Lines and Planes — Polyedral Angles .... 307 
 
 Supplementary Topics o 339 
 
 Supplementary Theorems . . . . o . . . 343 
 
 Book VII. Polyedra ........ e 346 
 
 Supplementary Topics . . 376 
 
 Miscellaneous Theorems .... .... 380 
 
 Book VIII. The Cylinder and the Cone . . . . , .388 
 
 Measuring the Cylinder 392 
 
 The Cone 396 
 
 Supplementary Topics 405 
 
 Book IX. The Sphere 409 
 
 Spherical Polygons 415 
 
 Volume of a Sphere . 434 
 
 Supplementary Topics 438 
 
 Supplementary Exercises — Books VI to IX 454 
 
 Index 463 
 
PLANE AND SOLID GEOMETRY 
 
SYMBOLS 
 
 = , is equal to ; equals. 
 >, is greater than. 
 <, is less than. 
 II , is parallel to ; parallel. 
 ±, is perpendicular to. 
 J_, perpendicular. 
 ~, is similar to. 
 ^, is congruent to. 
 
 Any symbol representing 
 plural by affixing the letter s 
 
 Z, angle. 
 
 A, triangle. 
 
 O, parallelogram. 
 
 □, rectangle. 
 
 O, circle. 
 
 .'., therefore. 
 
 = , is identically equal to. 
 
 = , approaches as limit. 
 
 a noun is converted into 
 thus A means angles. 
 
 the 
 
 ABBREVIATIONS 
 
 Adj., 
 
 adjacent. 
 
 Ex., 
 
 exercise. 
 
 Alt, 
 
 alternate. 
 
 Ext., 
 
 exterior. 
 
 Ax., 
 
 axiom. 
 
 Hom.^ 
 
 homologous. 
 
 Con., 
 
 conclusion. 
 
 Hyp/, 
 
 hypothesis. 
 
 Cong., 
 
 congruent. 
 
 Int., 
 
 interior. 
 
 Const., 
 
 construction. 
 
 Rect., 
 
 rectangle. 
 
 Cor., 
 
 corollary. 
 
 Rt, 
 
 right. 
 
 Corres. 
 
 , corresponding. 
 
 St., 
 
 straight. 
 
 Def., 
 
 definition. 
 
 Supp. 
 
 , supplementary. 
 
 It is not necessary to learn any of these symbols until they 
 are introduced in the text. 
 
m- i( 
 
 m 
 
 iWg^' 
 
 ifH 
 
 A View in the Congressional Library, Illustrating the Use 
 OF Geometry in Architecture 
 
GEOMETRY 
 
 INTRODUCTION 
 
 In arithmetic and algebra, frequent reference is made to 
 the rectangle, the square, the triangle, and the circle. These 
 are geometrical figures, and in geometry a careful study of 
 them and of many others is made. 
 
 Geometrical figures are used constantly in architecture. 
 
 Plans of a House 
 They often form the basis of artistic designs. 
 
 A Textile Pattern 
 
 An Autistic Tbay 
 
2 PLANE GEOMETRY 
 
 Our playgrounds are often laid out in geometrical forms. 
 
 A Football Field 
 
 Familiarity with such figures and their properties, and 
 ability to construct and measure them, is both interesting and 
 worth while. It is interesting also to know how man has 
 developed his knowledge of such figures and his skill in using 
 them. 
 
 HISTORY OF GEOMETRY 
 
 Geometry as it is now studied has been handed down to us 
 from the Greeks. The word " geometry " is derived from two 
 Greek words meaning the earth and to measure; this fact is 
 evidence that the Greeks believed that geometry was inti- 
 mately associated with or else had been developed out of the 
 practical business of measuring the earth, — surveying. 
 
 The. Greeks receiyed thei? start in geometry from the Egyp- 
 tians, 'tkales Qf'*3^iiilcs''(.feO-550 b.c.) is given special credit 
 for tBans<plaiut4iig»arknowledgdi3f Egyptian geometry to Greece. 
 
 BitJ- IKe Egyptians* ©r?giimte'*geometry ? Whether they did 
 or not, there is evidence that they had some knowledge of prac- 
 tical geometry. Their pyramids and other marvelous struc- 
 tures point to this fact. Also, there is in the British Museum 
 a papyrus written about 1700 b.c. by an Egyptian, commonly 
 
INTRODUCTION 3 
 
 called Ahmes, which contains among other interesting mathe- 
 matical records some formulae for measuring geometrical 
 figures. This papyrus is a copy of another written before the 
 time of Ahmes. Herodotus, a Greek traveler and historian, 
 is said to be responsible for the story that the Egyptians de- 
 veloped these rules of mensuration because of the necessity of 
 frequently surveying the lands which were inundated by the 
 floods of the Nile. The Egyptians must have obtained their 
 formulae by experiment or by observation. Some of the 
 formulae were incorrect and their formula for measuring the 
 area of a circle was less accurate than that developed later by 
 the Greeks. 
 
 The Greeks became interested in geometry for its own sake 
 as well as for its usefulness. In the three hundred years 
 following the time of Thales, geometry grew into a great 
 science in their schools, far exceeding the geometry of the 
 Egyptians in the number and interest of the facts discovered, 
 and in the accuracy and usefulness of the results. Pythag- 
 oras and Plato were the leaders of two groups of students 
 which were responsible for much of the advance made in the 
 subject. 
 
 Hippocrates (about 420 b.c.) made an attempt to prepare a 
 text on geometry, but it remained for Euclid to write what be- 
 came the standard text. Euclid lived between 330 and 275 b.c. 
 He was one of the first and greatest mathematicians who 
 taught at the University of Alexandria. As a teacher he felt 
 the need of a text by which to lead beginners through the 
 known facts of elementary geometry. He therefore gathered 
 together and systematized these facts in a book known as the 
 Elements. Euclid's Elements has stood as the model for all 
 subsequent texts on the subject. 
 
 During the two thousand years since the time of Euclid, 
 geometry has been studied by all civilized peoples and has been 
 enriched from time to time by their mathematicians. This 
 history is so long and the details are so technical that it is 
 unwise to attempt to give more of it at this time. 
 
PLANE GEOMETRY 
 
 4y 
 
 INFORMAL PREPARATORY GEOMETRY 
 
 1. The adjoining figure is a cube. It has six surfaces. 
 Each surface is bounded by four lines, — 
 straight lines. Each straight line is bounded 
 by two points. 
 
 The surfaces of a cube, which are smooth 
 and flat, are called Plane Surfaces; they are 
 such that a straightedge (ruler) will touch 
 the surface at all points of the straightedge, no matter where 
 the plane surface may be tested. 
 
 2. Plane Geometry is the study of figures like the square, 
 the triangle, the circle, etc., — figures which lie in a plane 
 surface. 
 
 A Plane Geometrical Figure is a combination of points and 
 lines which lie in one plane surface. Only such figures are 
 considered in plane geometry. 
 
 Ex. 1. Test the surface of your desk with your ruler to determine 
 whether the surface is a plane or not. (See § 1.) 
 
 Ex. 2. What are some other objects which have plane surfaces ? 
 
 Ex. 3. How do men who are laying a concrete walk make use of this 
 test in order to make the surface of the walk approximately plane ? 
 
 3. Solid Geometry is the study of figures like the cube, the 
 sphere, the cylinder, the pyramid, the cone, etc. 
 
 Pyramid 
 
 Cylinder 
 
 Cone 
 
INTRODUCTION 5 
 
 4. A Point is represented to the eye by a small dot. . A 
 
 A point is named by placing beside it a capital printed letter; as 
 point A. 
 
 A point represents position only. 
 
 5. A Straight Line is represented to the eye by a mark made 
 by drawing a pencil, a pen, or a piece of crayon along the edge 
 of a straightedge, 
 
 A line represents length only. 
 
 A Curved Line is a line no part of which is straight. 
 
 A Broken Line is a line composed of different successive 
 
 straight lines. 
 
 F Q 
 
 Straight Line Curved Line Broken Line 
 
 6. Lines like the adjoining ones are called closed lines. 
 
 It is apparent that a closed line incloses 
 a portion of the plane. 
 
 7. The word " line " will mean a 
 straight line hereafter unless otherwise specified. 
 
 Ex. 4. Place upon paper a single point, (a) Draw through it one 
 straight line. (6) Can you draw through it another straight line ? (c) A 
 third ? {(l) How many straight lines can be drawn through one point ? 
 
 Ex. 5. Place upon paper a point A and a point B. (a) Draw from 
 .4 to ^ a straight line. (6) What happens when you try to draw a sec- 
 ond straight line from A to B? (c) How many different straight lines 
 do you conclude can be drawn between two points ? 
 
 Ex. 6. Can more than one curved line be drawn between two points ? 
 Illustrate. 
 
 Ex. 7. (a) When walking along a straight line, are you moving con- 
 stantly in the same direction or not ? (6) Answer the same question if 
 you are walking along a curved line. 
 
 Ex. 8. Draw a straight line 2 inches long. Extend it one inch in 
 each direction. 
 
6 PLANE GEOMETRY 
 
 8. It will be assumed as apparent from the preceding exer- 
 cises that : 
 
 (a) One and only one straight line can be drawn through 
 
 two points. 
 This fact is also expressed thus : two points determine a 
 
 straight line, 
 (h) A straight line can he extended indefinitely in each 
 
 directioyi. 
 
 9. The straight line determined by points A and B is called 
 
 the line AB. 
 
 A B 
 
 Ex. 9. Select three points which are not in one straight line. Letter 
 them A, J5, and C. Draw the different straight lines determined by them 
 taken two at a time. Name the straight lines that you get. 
 
 Ex. 10. If four towns are situated so that no three can be connected 
 by one straight road, how many roads must be constructed if each town 
 is to be connected with each of the others by a straight road ? Illustrate 
 by a drawing. 
 
 Ex. 11. Draw the straight line determined by two points. Then turn 
 the straightedge over, and again draw a straight line between the two 
 points. If the edge is a true straightedge, the two straight lines will 
 coincide (form one line). Why is this so ? 
 
 Ex. 12. Make a straightedge by folding a piece of paper. Test it by 
 the method suggested in the preceding exercise. 
 
 Ex. 13. In order to walk across a field in a straight line, a boy selects 
 two objects which are in the direction in which he wishes to go, one of 
 them directly between him and the other. As he walks, he constantly 
 keeps the first object between himself and the second. 
 
 (a) Why can he guide himself in this manner ? 
 
 (6) What two points determine the straight line along which he walks ? 
 
 10. Two lines, straight or curved, ^ ^^^ ^^/ 
 
 intersect if they have one or more 
 
 common points. The common points are called Points of 
 
 Intersection. 
 
INTRODUCTION 
 
 11. Two straight lines can intersect at only one point 
 
 If they were to intersect in two points, ^^ ^7) 
 
 there would be two straight lines through 
 
 these two points, and this is impossible ^^^^ 
 (§8). <^^ ^^ 
 
 This fact is also stated thus : two intersecting straight lines 
 determine a point. 
 
 Ex. 14. Draw three straight lines intersecting by pairs which do not 
 all pass through one point. How many points do they determine ? 
 
 Ex. 15. If there are in a county four straight roads, each of which 
 crosses each of the others, and no three of which meet at one point, how 
 many crossings are there ? Illustrate. C 
 
 Ex. 16. How many points are determined by five straight lines in- 
 tersecting by pairs, no three of which pass through a common point ? / O 
 
 Ex. 17. Can you make any definite statement about the number of 
 points of intersection of two curved lines ? 
 
 12. A Line-segment or Segment is the part of 
 
 a straight line between two points of the line; ^ ^ 
 
 as, segment RS. 
 
 13. Two segments are equal if they can be placed so that 
 the ends of the one are exactly upon the ends of the other. 
 
 The tool for testing the equality of two segments is the dividers, 
 rhe dividers are spread until the points are upon A 
 
 and B respectively. If the dividers can then be placed -^ E. 
 
 with their points on C and D respectively without chang- 
 ing the position of the legs of the dividers, then the two 
 
 segments are equal. 
 
 AB is less than (<) CD if AB equals a part of CD. 
 
 Ex. 18. Determine by means of the dividers the 
 relative lengths of AB and BC; of ^5 and CD; of 
 AB and AD. 
 
 Ex. 19. Draw any segment AB. On a line of 
 indefinite length, mark off from a point of that line a segment equal to 
 2 AB ; also one equal to 3 AB. 
 
8 PLANE GEOMETRY 
 
 Ex. 20. Draw segments AB and CD, with AB greater than CD. 
 (a) On a line of indefinite length, mark off a segment equal to 
 AB + CD. (6) Mark off a segment equal to AB - CD. 
 
 Ex. 21. Let AB and CD be two segments. Sup- ^ ^ 
 
 pose that AB is placed upon CD with point A on point ] — 
 
 C. (a) Where will B fall if AB= CD? C ^ 
 
 (6) Where, if ^^ = i CD? 
 
 (c) Where, if AB is- greater than CD ? 
 
 Ex. 22. Suppose that two segments are each equal to a third seg- 
 ment. How do these two segments compare with each other ? 
 
 Ex. 23. Suppose that two segments are each equal to equal seg- 
 ments. How do these segments compare with each other ? 
 
 Ex. 24. Complete the following sentences : 
 
 (a) If equal segments are added to equal segments, the sums are ••• 
 (6) If equal segments are subtracted from equal segments, the re- 
 mainders are • • • 
 
 14. It will be assumed as apparent that: 
 the straight line-segment is the shortest line 
 between two points. \n 
 
 The Distance between two points is the length of the seg- 
 ment of the straight line between the points. 
 
 To obtain a straight line between two points, a carpenter stretches a 
 piece of twine between the two points. In doing so, he assumes that the 
 shortest line between two points is the straight line. 
 
 Ex. 25. Why are streets usually made straight ? 
 
 Ex. 26. Why do people often " cut across " a vacant corner lot ? 
 
 Ex. 27. Place upon paper points A, B, and C so that they do not all 
 lie upon a straight line. Draw segments AB, BC, and AC. By means 
 of your dividers compare the longest segment with the sum of the other 
 two segments. 
 
 15. A point bisects a segment if it divides the segment into 
 two equal segments. The point is called the Mid-point of the 
 segment. 
 
 A n G S 5 
 
 Thus, C bisects AB ii AC = CB. 
 
INTRODUCTION 9 
 
 It will be assumed as apparent that a segment has only one 
 mid-point. 
 
 Ex. 28. Determine with your dividers whether C does actually bisect 
 AB. If it does, what part of AB is AC? Does D bisect AC? Does 
 E bisect CB ? (See Fig. § 15.) 
 
 Ex. 29. Draw a segment of any length and locate freehand the 
 point which you think bisects the segment. Test the two parts to 
 determine whether you actually located the mid-point of the segment. 
 (Continue this exercise until you can approximately bisect a segment in 
 this manner.) 
 
 Ex. 30. What must be true about halves of equal segments ? 
 
 16. A Circle is a closed curved line all points of which 
 are equidistant from a point within 
 called the Center. 
 
 A Radius of a circle is the distance 
 from the center to any point on the 
 circle; as OA. 
 
 A Diameter of a circle is a seg- 
 ment drawn through the center of. 
 the circle with its ends on the circle ; 
 as BD. 
 
 A Chord of a circle is the segment joining any two points 
 of the circle ; as, chord CE. 
 
 A circle can be drawn with any point as center and any 
 given segment as radius. 
 
 17. Two circles having equal radii can be made to coincide 
 and are called equal circles. Hence : 
 
 All radii of the same circle or of equal circles are equal. 
 Ex. 31. Draw a circle of radius 1 inch. 
 
 Ex. 32. Draw two circles having the same center with radii of 1.5 in. 
 and 2 in. respectively. 
 
 Ex. 33. Draw a circle and a straight line which intersects it. How 
 many points of intersection are there ? 
 
 Ex. 34. Draw two circles that intersect. How many points of 
 intersection are there ? 
 
10 PLANE GEOMETRY 
 
 18. Circles form the basis of numerous designs. 
 
 Can you copy any of these designs ? 
 
 19. A Half-line or Ray is the part of a straight line in one 
 
 direction from a given point on the line; as ^ _^ 
 
 p in 
 ray m. ^ 
 
 Ex. 35. How many end-points has a line-segment ? A ray ? A 
 line? 
 
 20. An Angle (Z) is the figure formed by two rays drawn 
 from the same point. ^B 
 
 This definition was introduced by a mathema- 
 tician, Beitrand, in 1778. q^ 1 ^j^ 
 
 The common point is called the Vertex of the angle. 
 
 The two rays are called the Sides of the angle. 
 
 One may imagine a ray starting from the position OA and 
 turning about point O until it occupies the position OB. OA 
 is then called the initial line, and OB the termiyial line. 
 
 Note 1. — The size of an angle may be thought of as depending upon 
 the amount of turning about the vertex which a line must do to pass from 
 the initial line to the terminal line. 
 
 Note 2. — The portion of the plane over which the line would pass 
 is said to be within the angle. (See Note 2, page 28.) 
 
INTRODUCTION 11 
 
 Note 3. — Since the rays extend indefinitely, it is clear that the size 
 of an angle does not depend upon the length of its sides. 
 
 An angle may be named by the letter at its vertex if there 
 is in the figure only one angle having that vertex ; as Z 0. 
 
 An angle may be indicated by a number placed within the 
 angle near its vertex ; as Z 1. 
 
 An angle may be named by reading the letters A^ 0, B-, as 
 Z AOB, where the vertex letter is placed between the other 
 two letters. 
 
 21. Two angles are equal if they 
 can be made to coincide. 
 
 Thus, ii ZE can be placed upon Z B 
 so that point E is on point J5, line ED 
 on line BA, and line EF on line BC\ 
 then Z E equals Z B. 
 
 22. An Z AOB is less than Z AOC if it /^w^^ 
 equals a part of Z AOC. ^^^^^ A 
 
 Ex. 36. Make on thin paper a tracing of Z DEF (§ 21). Place the 
 tracing over Z ABC and thus determine whether the angles are actually 
 equal. 
 
 Ex. 37. "What must be true about two angles each 
 of which is equal to a third angle ? 
 
 Ex. 38. In the adjoining figure, make a tracing of 
 angle 1. Determine which of the other angles are equal 
 to Z 1. 
 
 23. A line bisects an angle if it divides the angle into two 
 equal angles. 
 
 Thus, 00 bisects Z AOB ii Z\ = Z2. 
 
 The line is called the Bisector of the angle. 
 It will be assumed as apparent that an q ^^ 
 angle has only one bisector. 
 
 Ex. 39. Make a tracing of i^ 1 and determine whether Zl is actually 
 equal to Z 2. Is 00 actually the bisector oiZAOB? 
 
12 
 
 PLANE GEOMETRY 
 
 Ex. 40. What part oiZAOBiaZl? (See Fig. § 23.) 
 
 Ex. 41. Draw any angle. Can you fold the paper so that the crease 
 will bisect the angle ? 
 
 Ex. 42. Draw any angle. Draw a line which you think bisects 
 
 the angle. Test the equality of the two parts of the angle by means of 
 
 tracing paper. (Continue this exercise until you can approximately 
 bisect an angle in this manner.) 
 
 Ex. 43. What must be true about halves of equal angles ? 
 
 Ex. 44. Suppose that Z ABC is placed upon Z DEF so that point B 
 is on point E, and line BA is on line ED. 
 
 Where will BC fall if ZABC is equal to /C /F 
 
 ZDEF'f 
 
 Where will BC fall if Z^^C is less than 
 ZDEF7 
 
 Where will BC fall if ZABC is greater than ZDEF7 
 
 Ex. 45. Complete the following sentences : 
 
 (a) If equal A are added to equal zi, the sums are ••• 
 
 (6) If equal A are subtracted from equal A^ the remainders are 
 
 24. Adjacent Angles are two angles that -^ 
 
 liave a common vertex and a common side 
 between tliem. 
 
 Thus, Z 1 and Z 2 are adjacent angles. 
 
 Ex. 46. In the adjoining figure : 
 (a) Ts Z 1 adjacent to Z 2 ? Why ? 
 (&) Is Z 2 adjacent to Z 3 
 
 Why 
 
 25. Two adjacent angles are readily added, j^ 
 
 Thus, Z AOB + Z BOC = Z AOC. 
 Also, ZAOC-ZBOC=ZAOB. 
 
 Ex. 47. In the figure (§ 25), read the angle which represents : 
 (a) ZAOB + ZBOD; (d) ZBOE-ZDOE; 
 
 (6) ZBOC + ZCOD; 
 
 (c) ZBOC+ZCOD + ZDOE] 
 
 (e) ZAOC + ZCOD-ZBOD; 
 (/) ZAOE-ZDOE- ZCOD. 
 
INTRODUCTION 
 
 13 
 
 26. If one straight line meets another straight line so 
 that the adjacent angles formed are equal, 
 each of these angles is a Right Angle ; as, 
 Z 1 and Z 2. 
 
 27. It will be assumed as apparent that ^ ^ ^ 
 
 all right angles are equal. 
 
 28. An angle is measured by finding how many times it 
 contains another angle selected as unit of measure. 
 
 The usual unit of measure is the Degree, which is one ninetieth of a 
 right angle. 
 
 To express fractional parts of the unit, the degree is divided into sixty 
 equal parts, called minutes, and the minute into sixty equal parts, called 
 seconds. 
 
 Degrees, minutes, and seconds are represented by the symbols °, ', 
 and " respectively. 
 
 Thus, 42° 22' 37" denotes an angle of 42 degrees, 22 minutes, and 37 
 seconds. 
 
 Ex. 48. Point out in your classroom some right angles. 
 
 Ex. 49. Fold a piece of paper so as to make a straight line ; then fold 
 it again so as to form two equal adjacent angles. Then open it out. 
 
 What kind of angles are formed by the lines along which the paper 
 was folded ? 
 
 Ex. 50. How many degrees are there in : 
 i rt. Z ? ^ rt. Z ? 1 rt. Z ? | rt. Z ? 
 
 29. Two lines are Perpendicular (±) if 
 they form a right angle. C" 
 
 Thus, 5^ ± CZ) if Z 1 = Z 2. 
 
 When two lines are perpendicular, the ad- 
 jacent angles are equal (§ 26). 
 
 A practical method of drawing a perpendicular to 
 a line at a point in the line, is to use a pattern right 
 angle as in the adjoining figure. Draughtsmen use 
 the right triangle illustrated. A card with two per- 
 pendicular edges may be used as well. 
 
 B 
 
14 
 
 PLANE GEOMETRY 
 
 Ex. 51. Draw a perpendicular to a line at a point in the line as 
 suggested in § 29. Use a card or else the pattern right angle con- 
 structed in Exercise 49. 
 
 Ex. 52. Draw a perpendicular to a line from a point not in the line, 
 using a pattern right angle. 
 
 Ex. 53. Draw freehand a perpendicular to a line at a point in the 
 line. Test the accuracy of your construction by means of your pattern 
 right angle. (Continue this exercise until you can draw a line which is 
 approximately perpendicular to a given line, either at a point in the line, 
 or from a point not in the line.) y^ 
 
 30. An Acute Angle is an angle which is 
 less than a right angle ; as Z CBA. 
 
 An Obtuse Angle is an angle which is greater 
 than a right angle ; as Z FEB. 
 
 Acute and obtuse angles are called collectively 
 Oblique Angles. 
 
 Two intersecting lines which are not perpendicular are said to be 
 oblique to each other. 
 
 Ex. 54. What kind of angle isZl? Z2? Z3? 
 Z4 ? (Test each with your pattern right angle.) 
 
 Ex. 55. What kind of angle do the hands of a 
 clock form at 3 o'clock ? At 1 o'clock ? At 2 
 o'clock ? At 5 o'clock ? 
 
 Ex. 56. How many degrees are there in each 
 of the angles in Exercise 55 ? 
 
 31. A Straight Angle is an angle 
 whose sides lie in a straight line on 
 opposite sides of its vertex. ^ 
 
 Such an angle would result if a line were to start from the position of 
 line OA and revolve about point one half of a complete revolution. 
 
 32. If AB is any straight line, and 
 C01.AB, then Z 1 and Z2 are each 
 right angles by § 29 ; also, Z AOB is a 
 straight angle by § 31. Hence a straight 
 angle equals two right angles. 
 
INTRODUCTION 
 
 15 
 
 33. Since a straight angle is equal to two right angles 
 (§ 32) and since all right angles are equal (§ 27), it is evident 
 that all straight angles are equal. 
 
 Ex. 57. How many degrees are there in a straight angle ? 
 
 Ex. 58. What part of a straight angle is a right angle ? 
 
 Ex. 59. At what hour do the hands of a clock form a straight angle ? 
 
 34. The sum of all the successive adjacent angles around a 
 point on one side of a straight line is one straight angle. 
 
 Thus, Zl + Z2 + Z3+Z4 = ZAOB= 1 stZ. 
 
 Ex.60. If Zl=Z2=Z3=Z4,howmany 
 degrees are there in each angle ? 
 
 .Ex.61. If Z2=3 times Z1,Z3=Z2, and 2? 
 Z4 = 2 times Zl, how many degrees are there 
 in each angle ? (Use algebraic method. ) 
 
 35. TTie sum of all the successive 
 adjacent angles around a point is two jy 
 sfraight angles. 
 
 What kind of angle is Z 1 ? Z2? Hence Zl + Z2 = ? 
 
 The total angular magnitude around a point is called a 
 Perigon. 
 
 Perigon is from Greek words meaning " the angle around." 
 
 Ex. 62. How many right angles are there in a perigon ? How many 
 degrees ? 
 
 Ex. 63. Through what angle does the minute hand of a clock turn in 
 one half hour ? In one hour ? 
 
 Ex. 64. How large would each angle of the adjoin- 
 ing figure be if the angles were equal angles ? 
 
 Ex. 65. How large would each angle of the adjoin- 
 ing figure be if Z 1 were equal to Z 4, if Z 3 = 2 times 
 Z 1, if Z5 = the sum of Z 3 and Z 1, and if Z2 = the 
 sum of Z3 andZ5? 
 
 36. Two angles are Complementary if their 
 sum is equal to a right angle. Each of the 
 angles is called the Complement of the other. 
 
 Thus, the complement of 40^ is 60°. 
 
 c 
 
 ^^ 
 
 
 
 
 
 A 
 
16 
 
 PLANE GEOMETRY 
 
 Ex. 66. What is the complement of 10^ 
 ofO°? of 45°? ofr/? 
 
 ? of 25"? of 50°? of 90°? 
 
 (Uj 
 
 Ex. 67. How large is the angle which equals its complement 
 algebraic method. ) 
 
 Ex. 68. How large is the angle which equals four 
 times its complement ? 
 
 Ex. 69. If OC±OA and OD ± OB, and if Z2 
 = 70°, compare Z 1 and Z 3. 
 
 Ex. 70. Draw any acute angle. Through the vertex draw a line 
 which will form with one side of the angle the complement of the angle. 
 
 37. Complements of the same 
 angle or of equal angles are equal. 
 
 The complement of w° is (90 — m)° 
 and the complement of n° is (90 — n)°. 
 
 If, now, m = n, then 90 — m must equal 90 — n, for when equals are sub- 
 tracted from equals the remainders are equal. 
 
 38. Two angles are Supplementary if their sum is equal to a 
 straight angle. Each of the angles is called the Supplement of 
 the other. 
 
 Thus, an angle of 150° is the supplement of an 
 angle of 30°. 
 
 Ex. 71. What is the supplement of 80° ? 
 of 0° ? of ic° ? of Sx°? 
 
 of 60°? of 100°? of 90°? 
 
 Ex. 72. How large is the angle which equals its supplement ? 
 
 Ex. 73. How large is the angle which is nine times as large as its 
 supplement ? 
 
 Ex. 74. Draw any angle less than a straight angle. Through its 
 vertex draw a straight line which will form with one side of the angle the 
 supplement of the angle. 
 
 39. If two adjacent angles have their exterior sides in a 
 straight line, they are supplementary. /^ 
 
 In the adjoining figure, Z 1 and Z 2 are adjacent ^ y^ 
 
 angles. OG and OA are their exterior sides. If 
 
 OC and OA lie in a straight line, then ZAOG = 1 st. Z. But Z 1 + Z2 
 
 = ZAOC. Hence, Z 1 and Z 2 are supplementary (§ 38). 
 
INTRODUCTION 17 
 
 Two adjacent angles which are also supplementary are 
 called Supplementary- adjacent Angles. 
 
 40. If two adjacent angles are supplementaiyj their extenor 
 sides lie in a straight line. (Fig. of § 39.) 
 
 For, if Zl + Z2 = 1 St. Z, then Z.AOC must be a straight angle and 
 AC must be a straight Hne. 
 
 Ex. 75. Draw two adjacent angles which are not supplementary. 
 
 41. Supplements of the same angle or of equal angles are 
 equal. 
 
 In the adjoining figure, let Z 1 = Z3, and let AC 
 and jB T be straight lines. Then Z 2 is the supple- 
 ment of Z 1 and Z 4 is the supplement of Z3 ; that 
 is, Z 2 = 180 - Z 1 and Z 4 = 180 - Z 3. Clearly, 
 then, Z 2 must equal Z 4, for, when equal angles 
 are subtracted from equal angles, the remainders 
 are etjual. 
 
 Ex. 76. If AB and CD are straight lines, are 
 Z 1 and Z 2 supplementary ? Z 3 and Z 4 ? Why ? '^ XHT 
 
 (See §39.) 
 
 Suppose that Z1=Z4. Must Z3 then equal ^ /3 / 4^ 
 Z2? Why? 
 
 Ex. 77. In the adjoining figure, if Z 1 equals Z2, 
 then Z 3 must equal Z 4. Prove it. 
 
 42. Two angles are called Vertical Angles when the sides of 
 one are prolongations of the sides of the other. 
 
 Thus, Z 1 and Z 2 are vertical angles. (See Fig. Ex. 78.) 
 
 Ex. 78. If Z 1 = 40"^, how many degrees are, 
 there in Z 3 ? 
 
 How many degrees are there in Z 2 + Z 3 ? "' 
 
 How many degrees are there in Z 2 ? 
 How then do Z 2 and Z 1 compare ? 
 
18 
 
 PLANE GEOMETRY 
 
 Ex. 79. Draw two straight lines that intersect. Make a tracing of 
 one of the angles formed and compare it with its vertical angle. What 
 do you conclude must be true about the vertical angles ? 
 
 43. The Protractor is a tool for measuring angles. 
 
 The point on the protractor will be referred to as the center of the 
 protractor. 
 
 44. Problem. Measure a given angle A VB. 
 
 fmun 
 
 Place the protractor- over the given angle so that its center is on the 
 vertex, F, of the angle, and its edge is on VA. Then read from the pro- 
 tractor the number opposite the point where VB crosses the outer edge 
 of the protractor. This gives the number of degrees in the angle AVB. 
 Thus, in the figure, ZAVB - 50°. 
 
INTRODUCTION 
 
 19 
 
 Ex. 80. Make on paper a tracing of the 
 adjoining figure. On your paper, extend rays 
 OA, OB, OC, OD, and OE until they are 
 about 3 in. in length ; then measure : 
 
 (a) ZAOB; {b)/iAOC; (c) ZBOE; 
 (d) Z COD. 
 
 45. Problem. Construct an angle of given size at a point in 
 a given line. 
 
 Construct an angle of 35° at P in AB. 
 
 Place the protractor with its center on P and its edge on PB as in the 
 figure. Then place a point B on the paper opposite the 35" mark on the 
 protractor. Remove the protractor and draw the line PB. Then angle 
 BPB equals 35°. 
 
 Ex. 81. Construct with the protractor an angle of : 
 
 (a) 70°; (h) 40°; (c) 65°; (d) 100°; (e) 143°. 
 Write below each angle whether it is an acute or an obtuse angle. 
 
 46. A Field Protractor. The figure at top of page 20 repre- 
 sents a simple field protractor which can be made by some mem- 
 ber of the class. With it angles can be measured in the field 
 and thus some elementary surveying problems can be solved. 
 
 On a flat board about 20 inches square, draw a circle of diameter 10 
 inches. Divide its circumference into 360 equal parts. 
 
20 
 
 PLANE GEOMETRY 
 
 Make an arm which may swing about the center of the circle as pivot. 
 
 Let the arm have upon it two " sights " directly in line with the center of 
 
 the circle. At the end of the arm and in line with the sights place a pin. 
 The board may be attached to the end of a 
 stake about 4 feet long, or better to a tripod. 
 
 This instrument can be used to measure 
 angles in the open field. 
 
 Thus, to measure an Z. CAB^ place the in- 
 strument over point A. Make the board stand 
 level. (An inexpensive level would be a great 
 help.) Holding the board stationary, sight 
 first at point O, and read the angle on the 
 
 protractor ; then sight at point 5, and note the angle on the protractor. 
 
 The number of degrees through which the arm is turned in passing from 
 
 ^O to AB is the measure of angle GAB. 
 
 47. Triangle (A). Three points which do not lie in the 
 same straight line determine three segments. 
 Thus, A, B, and C determine the segments 
 AB, BO, and AC. The figure formed by 
 these three segments is called the triangle 
 ABC (A ABC). A, B, and C are the 
 vertices of the triangle; AB, BC, and ^lOare . 
 
 the sides of the triangle ; Z A, A B, and Z C are the angles of 
 the triangle. 
 
 The sides and angles of the triangle are called the parts of 
 the triangle. They are six in number. Opposite each side 
 there is an angle, and opposite each angle there is a side. 
 Thus, Z (7 is opposite side AB. 
 
 48. Experimental Geometry. Many facts about geometrical 
 figures can be discovered by careful drawing, measurement, 
 and observation. 
 
 Ex, 82. On a line AB, at a point P, draw a ray PC making Z APC 
 = 80°. Measure Z CPB. What fact studied previously does this exer- 
 cise verify ? 
 
 Ex. 83. Draw two intersecting straight lines. Measure each of a 
 pair of vertical angles. What fact studied previously does this exercise 
 verify ? 
 
INTRODUCTION 21 
 
 Ex. 84. Construct a A ABC, having AB and BC each 3 inches 
 long and Z B = 40°. Measure Z A and Z C. IIow do they compare ? 
 (A triangle having two equal sides is called an isosceles triangle.) 
 
 Ex. 85. Draw any triangle having two equal sides. Measure the 
 angles opposite the equal sides. How do they compare ? What fact is 
 suggested by Exercises 84 and 85 ? 
 
 Ex. 86. Draw any triangle of reasonably large size. Measure each 
 of its angles. Find their sum. Kepeat the exercise for another triangle 
 of somewhat different shape. Compare your results with those of some 
 other pupils. What seems to be the sum of the angles of a triangle ? 
 
 Ex. 87. Draw a A ABC in which AB and BC are each 3 inches and 
 ZB = 50°. Let E be the mid-point of BC, and F the mid-point of AB. 
 Draw AE and CF. Measure them. What seems to be true ? 
 
 Ex. 88. Draw any triangle ABC of reasonably large size. Let E be the 
 mid-point of AB, and F the mid-point of BC. Draw EF. Compare EF 
 and AC by measuring them. What seems to be the relation between them ? 
 
 Ex. 89. Draw a segment AB. At its center, C, draw a line CD 
 perpendicular to AB. From E, any point on CD, draw AE and BE. 
 Compare them by means of your dividers. Take any other point on CD 
 and repeat the exercise. What fact seems to be suggested ? 
 
 Ex. 90. Let AB be any line segment. Draw CA ± AB at A, and 
 DB LAB At B. Make CA - DB. Draw AD and CB. Compare them 
 either by measurement or by means of the dividers. 
 
 49. Objections to studying geometry only hy the experi- 
 mental method may be given. To get satisfactory results, the 
 figures must be drawn and measured with greater accuracy than 
 is usually possible. Conclusions reached from the study of one 
 or two special figures may be incorrect. Frequently one is mis- 
 led by assuming relations ivhich appear to the eye to be correct. 
 
 Ex. 91. In the first figure above, are ^B and CD straight lines ? 
 Ex. 92. In the second figure above, is AB equal to or less than CD ? 
 
22 PLANE GEOMETRY 
 
 50. Demonstrative Geometry. For the reasons given in § 49 
 and for other reasons, it is customary to study geometry by 
 what is known as the demonstrative method. Statements are 
 not accepted until they are proved to he true, except for a few 
 fundamental ones which are assumed as a foundation. 
 
 51. An Axiom is a statement accepted as true without 
 proof. Usually the truth is very evident. The following 
 are important axioms ; others will be introduced as they be- 
 come necessary. 
 
 AXIOMS 
 
 Ax. 1, Quantities which are equal to the same quantity or to 
 
 equal quantities are equal to each other. (See Ex. 22 
 
 and Ex. 23.) 
 Ax. 2. Any quayitity may he suhstituted for its equal in a 
 
 mathematical expression. 
 Ax. 3. If equals he added to equals, the sums are equal. (See 
 
 Ex. 24.) 
 Ax. 4. If equals he suhtracted from equals, the remainders are 
 
 equal. 
 Ax. 5. If equals he multiplied hy equals, the products are 
 
 equal. 
 Ax. 6. If equals he divided hy equals, the quotients are equal. 
 
 (The divisor must not be zero.) 
 Ax. 7. The whole equals the sum of its parts. 
 Ax. 8. TJie whole is greater than any of its parts. 
 Ax. 9. If a and h are any two magnitudes of the same hind, 
 
 then a is less than h, is equal to h, or is greater than h. 
 Ax. 10. Oyily one straight line can he draivn through two 
 
 points. (§ 8.) 
 Ax. 11. The straight line segment is the shortest line that ca7i 
 
 he drawn hetween two points. (§ 14.) 
 Ax. 12. All right angles are equal. (§ 27.) 
 Ax. 13. An angle has only one hisector. (§ 23.) 
 Ax. 14. A segment has only one mid-point. (§ 15.) 
 
INTRODUCTION 23 
 
 52. A Theorem is a statement which requires proof. Every 
 theorem can be expressed by a sentence which has one clause 
 beginning with " if " and a second clause beginning with 
 " then." 
 
 The clause beginning with "if' is called the Hypothesis; 
 it indicates what is known or assumed. 
 
 The clause beginning with " then " is called the Conclusion ; 
 it states what is to be proved. 
 
 Thus: (Hypothesis) If two sides of a triangle are equals 
 (Conclusion) then the angles opposite are equal. 
 
 53. Some theorems have been proved already in an informal 
 manner. 
 
 INFORMALLY PROVED THEOREMS 
 
 1. Two straight lines can intersect at only one point, (§ 11.) 
 
 2. All radii of the same circle or of equal circles are equal. 
 
 (§ 17.) 
 
 3. A straight angle equals two right angles. (§ 32.) 
 
 4. All straight angles are equal. (§ 33.) 
 
 5. The sum of all the successive adjacent angles around a 
 
 point on one side of a straight line is one straight angle. 
 (§ 34.) 
 
 6. Tlie sum of all the successive adjacent angles around a 
 
 point is two straight angles. (§ 'do.) 
 
 7. Complements of the same angle or of equal angles are 
 
 equal. (§ 37.) 
 
 8. If two adjacent angles have their exterior sides in a straight 
 
 line J they are supplementary. (§ 39.) 
 
 9. If two adjacent angles are supplementary, their exterior 
 
 sides lie in a straight line. (§ 40.) 
 10. Supplements of the same angle or of eqvM angles are equal. 
 (§ 41.) 
 
 54. In a formal demonstration or proof, each statement 
 made is proved by quoting a definition, an axiom, the hypothe- 
 sis, or some previously proved theorem. 
 
24 
 
 PLANE GEOMETRY 
 
 ILLUSTRATIVE DEMONSTRATION 
 
 Theorem. If two straight lijies intersect, the vertical 
 angles are equal. 
 
 Hypothesis. St. lines AB and CD intersect at 0, 
 forming vertical Al and 2. 
 
 Conclusion. Z 1 = Z 2. 
 
 Proof. 1. AB is a straight line. 
 
 2. .*. Z 1 is a supplement of Z 3. 
 
 [If two adj. A have their ext. sides in a st. line, 
 they are supplementary.] 
 
 Hyp. 
 §39 
 
 Hyp. 
 §39 
 § 41 
 
 3. CD is a straight line. 
 
 4. .*. Z 2 is a supplement of Z 3. 
 
 5. .-. Z1 = Z2. 
 
 [Supplements of the same Z are equal.] 
 
 Note. — This theorem was apparently assumed by Thales. 
 
 Ex. 93. Prove in the same manner that Z AOD = Z COB. 
 
 Ex. 94. If Z 3 = 130°, how many degrees are there in each of the 
 other angles in the figure above ? 
 
 Ex. 95. Two lines intersect so that one of the angles is a right angle. 
 How large is each of the other angles formed ? 
 
 55. Besides the proofs of certain theorems, the methods of 
 constructing certain figures are studied in geometry. 
 
 A Problem is a construction to be made. 
 
INTRODUCTION 25 
 
 56. A Postulate is a construction assumed possible. 
 
 The following postulates are necessary at the present time. 
 
 POSTULATES 
 
 1. One straight line can be drawn through two points. (§ 8.) 
 
 2. A straiqht line can he extended indefinitely in each direction. 
 
 (§ 8.) 
 
 3. A circle can be drawn with any point as center and any 
 
 given segment as radius. 
 
 57. The word Proposition is commonly used to designate a 
 theorem or a problem discussed in the text. 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 96. Draw a line to represent the path of a baseball when the 
 pitcher throws an "out-curve." 
 
 Ex. 97. A farmer is setting out trees for an orchard. He first places 
 the trees which are at the ends of a row. How may he then locate the 
 other trees of that row so that they will be in a straight line without using 
 a long line between the two end trees ? What two points determine the 
 straight line formed by the trees ? 
 
 Ex. 98. How do plasterers use in their work the characteristic 
 property of a plane mentioned in § 1 ? 200^ 
 
 Ex. 99. A man wishes a scale drawing of 
 his suburban lot so that he may consult a land- 
 scape architect about the proper planting of it. 
 He made the adjoining rough drawing of the lot, 
 and then obtained the measurements indicated. 
 Make a scale drawing of the lot, letting J inch 
 represent 25 feet. „^_ 
 
 Supplementary and Complementary Angles 
 
 Ex. 100. What is the complement of 40° 30' ? 
 
 Ex. 101. What is the supplement of 56^ 30' ? 
 
 Ex. 102. How many degrees are there in an angle if its complement 
 contains 40° ? 
 
26 PLANE GEOMETRY 
 
 Ex. 103. How many degrees are there in an angle whose supple- 
 ment contains 80° ? 
 
 Ex. 104. Find the angle whose supplement is ten times its comple- 
 ment. 
 
 Ex. 105. Two angles are complementary. The greater exceeds the 
 less by 25°. Find the angles. (Use the algebraic method.) 
 
 Ex. 106. Find the angle which exceeds its supplement by 34°. 
 
 Ex. 107. The sum of the supplement and complement of a certain 
 angle is 140°. Find the angle. 
 
 Ex. 108. Find the number of degrees in the angle the sum of whose 
 supplement and complement is 196°. 
 
 Ex. 109. The supplement of a certain angle exceeds three times 
 its complement by 18°. Find the angle. 
 
 Ex. 110. Prove that the supplement of any angle exceeds its com- 
 plement by one right angle. 
 
 Vertical Angles 
 
 Ex. 111. Prove that the straight line which 
 bisects an angle also bisects the vertical angle. 
 Hyp. OE bisects AAOC. j^ Oi^ is a st. Ihie. 
 Con. OF bisects Z. BOD. 
 
 Ex. 112. Prove that the bisectors of two supplementary adjacent 
 angles are perpendicular. .E /D 
 
 Suggestions. — L1 = \LBCD; L2^h^DCA. \ A ^^ 
 
 Zl + Z2 = ? ^ ^ ^' 
 
 A ^ B 
 
 Ex. 113. If the bisectors of two adjacent angles are perpendicular, 
 the angles are supplementary. (See figure of Ex. 112.) 
 
 Ex. 114. If the bisectors of two adjacent angles make an angle of 
 45°, the angles are complementary 
 
 Ex.115. Hyp. CO bisects 2^:^05. DEA.CO. 
 Con. Z3 = Z4. 
 
 Suggestions. — 1. Compare Z 1 and Z. 2. 
 
 2. Compare L 3 and Z 1 ; Z 4 and Z 2. 
 
INTRODUCTION 
 
 27 
 
 Ex. 116. Hyp. A ABC is a rt. Z. 
 
 Z 3 is complementary to Z. 1. 
 Con. Z3 = Z2. 
 
 D *rr" 
 
 Ex. 117.' Two straight lines intersect so that one angle formed is 
 60°. How large is each of the other angles ? 
 
 Ex. 118. Review the following definitions : 
 
 (a) Segment of a straight line. Qi) Equal angles. 
 
 (6) Equal segments. 
 
 (c) Mid-point of a segment. 
 
 (d) Circle. 
 
 (e) Radius and diameter. 
 (/) Ray or half-line. 
 
 (^) Angle ; vertex ; side. 
 
 (i) Bisector of an angle. 
 
 (j) Right angle ; acute ; obtuse ; 
 straight. 
 
 {Ic) Complementary angles ; supple- 
 mentary ; adjacent ; vertical. 
 
 (Z) Perpendicular lines. 
 
 Ex. 119. What is an axiom ? a theorem ? 
 
 Ex. 120. What is the hypothesis of a theorem ? 
 elusion of a theorem ? 
 
 What is the con- 
 
 Ex. 121. Lay off on a field some irregular piece of ground. Obtain 
 such measurements as will enable you to make a scale drawing of the 
 field. (This exercise is similar to Ex. 99.) 
 
 Ex. 122. Another method for ob- 
 taining measurements for a scale 
 drawing for a piece of ground and 
 objects upon it is to locate the instru- 
 ment for measuring angles at a point 
 like in the adjoining figure. Then 
 find the distances of the other points 
 from and their directions from OB 
 or from OA. 
 
 Select an irregular piece of ground and obtain the measurements which 
 will enable you to make a scale drawing of it and locate upon the draw- 
 ing some of the trees or other objects on the lot. 
 
 Supplementary Notes on Definitions 
 
 Note 1. — Toird and straight line are undefined. (See §4 and §5.) 
 A definition describes a term by means of simpler terms. It is evident 
 
28 PLANE GEOMETRY 
 
 then that there must be some terms which cannot be defined, as there are 
 no terms simpler than them by which to define them. 
 
 No definition of point can be given. 
 
 No satisfactory definition of straight line suitable for high school pupils 
 can be given. n 
 
 Hence point and straight line are left undefined. y^ 
 
 Note 2. — Belating to the definition of an angle f /y^ 
 (§20). ^^0-^ A 
 
 Two rays OA and OB actually form two angles ; V^_^ 
 
 namely, Z 1 and Z2 of the adjoining figure. In Z 1, OA is the initial line 
 (§ 20) and OB is the terminal line ; in Z2, OB is the initial line and OA 
 is the terminal line. Usually, one angle is less than and the other is 
 greater than a straight angle. 
 
 Unless something is said to the contrary, Z J. 0J5 refers to the smaller 
 angle formed by the rays OA and OB. 
 
BOOK I 
 
 KECTILINEAR FIGURES 
 
 58. A Rectilinear Figure is a geometrical figure composed 
 of straight lines only. 
 
 59. Congruence. If asked to compare two sheets of paper 
 as to shape and size, it is natural to place one upon the other to 
 determine whether they can be made to coincide (fit together). 
 
 60. Two geometrical figures are congruent (^) if they can be 
 made to coincide. 
 
 61. Ax. 15. Congruence Axiom. — Two figures which are 
 congruent to the same figure are congruerit to each other. 
 
 Historical Note. — The symbol ^ was introduced by a mathemati- 
 ♦cian, Leibnitz, in 1679. 
 
 62. Superposition is the process of placing one geometrical 
 figure upon another for the purpose of comparing them. 
 Literally, superpose means " place above." 
 
 Postulate. — A geometrical figure may be moved about in space 
 icithout changing any of its parts. 
 
 Ex. 1. Notice the panes of glass in the windows of your schoolroom. 
 Do they appear to be congruent ? Do they coincide now ? 
 
 Ex. 2. Draw A ylBC, having ^5 = 4 in., BC = Q in., and Zi? = 50°. 
 
 (a) Measure ^A, ZC^ and AC. 
 
 (h) Cut your triangle from the paper. Compare it by superposition 
 with the triangles made by other members of your class. 
 
 (c) What do you conclude must be true about all triangles made 
 according to the directions ? 
 
 Ex. 3. Are the statements of the hypothesis assumed to be true or 
 must they be proved to be true ? Answer the same question for the 
 conclusion. 
 
 29 
 
30 PLANE GEOMETRY — BOOK I 
 
 Proposition I. Theorem 
 63. If hvo triangles have two sides and the included 
 angle of one equal respectively to two sides and the 
 included angle of the other, the triangles are congruent. 
 
 B B E 
 
 Hypothesis. In A ABC and A DBF: 
 
 AB = DE; AC=DF; ZA = ZD. 
 
 Conclusion. A ABC ^ A DEF. 
 
 Proof. 1. Place A ABC on A DEF with point A on point 
 D, and side AB on side DE. 
 
 2. Point B will fall on point E. 
 
 [Since AB = DE, by hypothesis.] § 13 
 
 3. AC will fall on DF. 
 
 [Since ZA = ZD, by hypothesis.] § 21 
 
 4 Point C will fall on point F. 
 
 [Since AC = DF, by hypothesis.] § 13 
 
 5. .-. BC must coincide with EF. 
 
 [Only one st. line can be drawn through two points.] Ax. 10 ; § 51 
 
 6. .-. A ABC coincides with A DEF and is congruent to it. 
 
 [Two A are congruent if they can be made to coincide.] § 60 
 
 Ex. 4. After you place A ABC so that A falls on D, does AB fall on 
 DE, or must you place AB on DE ? 
 
 Ex. 5. Where would C fall if AC were equal to | DF? Would the 
 triangles be congruent ? 
 
 Ex. 6. Where would AC fall 'd ZA were less than ZD? 
 
 Ex. 7. Make a free-hand drawing to illustrate the result if a A ABC 
 is superposed on a A DEF, when AC=i DF, ZA=ZD, and AB=^ DE. 
 
RECTILINEAR FIGURES 
 
 31 
 
 64. Application of First Theorem. Illustrative Exercise. 
 
 Hypothesis. 
 
 AB^CD. 
 
 i\ 
 
 
 Z 1 = Z 2. 
 
 ^^o>^ 
 
 Conclusion. 
 
 A ABC ^ A BCD. 
 
 ^^O^ 
 
 Proof. 1. 
 
 In A ABC and A BCD: 
 
 J) 
 
 
 AB=CD', 
 
 Hyp. 
 
 
 Z1 = Z2; 
 
 Hyp. 
 
 
 BC = BC. 
 
 See note below. 
 
 2. 
 
 .'^A ABC ^ A BCD, 
 
 
 [If two A have two sides and the included Z of one 
 equal respectively to two sides and the included Z of the 
 other, the A are congruent.] 
 
 Note. — The symbol = is read. " is identically equal to." 
 thority is required, for any magnitude is equal to itself. 
 
 Ex. 8. Hyp. Zl=Z2; AB = EC. 
 
 Con. AABD^ABCD. ' 
 
 No other au- 
 
 ■^' 
 
 Ex. 9. Hyp. 
 
 Con. 
 
 Suggestion. 
 Why? 
 
 AD and BC are st. lines. 
 A0= OD] B0= OC. 
 AABO^ACDO. 
 
 What do you know about Zl and Z2? 
 
 Ex. 10. Hyp. 
 
 Con. 
 
 Ex. 11. Hyp. 
 
 Con. 
 
 XY = XW. 
 XZ bisects Z X. 
 
 AXYZ^AXZW. 
 
 DBC is a St. line. 
 ABA. DC) DB = BC. 
 A ADB ^ A ABC. 
 
 65. Homologous parts of congruent figures are parts which 
 are similarly located ; they are the parts which coincide when 
 
32 
 
 PLANE GEOMETRY — BOOK I 
 
 the figures are made to coincide. It follows that : homologous 
 parts of congruent figures are equal. 
 
 Thus, in § 63, Z O is homologous to Z F^ and side J5(7 is homologous 
 to side EF. It follows that AC = A F and BC = EF. 
 
 Note. — In congruent triangles, homologous sides lie opposite equal 
 angles and homologous angles lie opposite equal sides. 
 
 Q6. Principle I. To prove that two segments are equal or 
 two angles are equal, try to prove them homologous parts of 
 congruent triangles. 
 
 Illustrative Exercise 
 
 Hyp. BG is a st. line. j 
 AB±BC', DC±BG; 
 
 \ 
 
 
 
 AB=DC. 
 
 1 ^\o 
 
 
 rf 
 
 is mid-point of BC. 
 
 
 2 
 
 \-/ 
 
 Con. AO = OD. 
 
 \J 
 
 T) 
 
 Plan. Try to prove AO and OD 
 homologous sides of congruent A. 
 
 
 Proof. 1. Since ABA.BC, Al = 2^ rt. Z. 
 
 2. Since DC 1. BC,Z2 = ^ rt. Z. 
 
 3. .-. Z1=Z2. 
 
 Def. 
 Def. 
 
 [All rt, Zs are equal.] 
 
 4. In A^jBOand A7)C0: 
 
 AB^DC; Hyp: 
 
 Z1 = Z2; Step 3 
 
 BO=OC. Hyp. 
 
 5. .'. AABO^ADCO. 
 
 [If two A have two sides and the included Z of one equal 
 respectively to two sides and the included Z of the other, the 
 A are congruent.] 
 
 6. .'.AO=OD. 
 
 [Homologous sides of cong. A are equal.] 
 
 Note. — AO lies opposite Z 1 and OD lies opposite Z2; hence they 
 are homologous sides of the congruent triangles. 
 
RECTILINEAR FIGURES 
 
 Ex. 12. Hyp. AB±BC; DC±BC; 
 
 O bisects BC; AB = DC. 
 Con. A0= OD. 
 
 Ex. 13. Hyp. NO bisects Z PNM. 
 MN=NP. 
 Con. Zilf = ZP. 
 
 Ex. 14. Hyp. AB = BC= CD 
 ZB = ZD. 
 Con. AC = CE. 
 
 DE. 
 
 Ex. 15. To obtain tlie distance AB. 
 
 (1) Locate point O from which OA and OB may be 
 measured. (2) Extend AO and 50, making OC=AO 
 and 0D = BO. Then DO = ^i?. Prove it. 
 
 Ex. 16. If AB and CD are two diameters of a 
 circle, prove that AD must equal BC. 
 
 Review Exercises 
 
 Ex. 17. If, in the adjoining figure, Z 3 = Z 7, 
 prove Z 2 = Z 7. ~ 
 
 Ex. 18. If Z 4 = Z 5, prove Z 1 = Z 8. ^~ 
 
 Ex. 19. If Z 3 = Z G, prove Z 1 = Z 5. ^' 
 
 Suf^gestioiis. —1. Recall §41. 2. Of what angle is Zl the supplement? 
 3. Of what angle is Z 5 the supplement? 
 
 Ex.20. If Z4 = Z8, proveZ3 = Z6. 
 
 Ex. 21. When are two figures congruent ? 
 
 Ex. 22. What method of proof is employed in proving Proposition I ? 
 
 Ex. 23. Draw a A ^50, having ^B=4 in., ZA = 60°, and Z 5=80°. 
 Cut your triangle from the paper and compare it with the triangles made 
 by other members of your class. What do you conclude must be true 
 about all triangles made according to the directions given ? 
 
34 PLANE GEOMETRY — BOOK I 
 
 Pkoposition II. Theorem 
 
 67. If two triangles have two angles and the included 
 side of one equal respectively to two angles and the irv- 
 eluded side of the other, the triangles are congruent 
 c 
 
 A B D 
 
 Hypothesis. In A ABO and A DEF: 
 
 AB = DE; ZA=ZD; Z.B=ZE. 
 
 Conclusion. A ABC ^ A DEF. 
 
 Proof. 1. Place A ABC on A DEF, with point A on point 
 D and AB on DE. 
 
 2. Point B will fall on point E. 
 
 [Since AB = BE, by hypothesis.] § 13 
 
 3. AC will fall on DF, C falling somewhere on line DF. 
 
 [Since ZA=ZD,\)y hypothesis. ] § 21 
 
 4. BC will fall on EF, falling somewhere on line EF. 
 
 [Since /.B = AE^hy hypothesis.] § 21 
 
 5. .*. point C must fall on point F. 
 
 [Two St. lines can intersect at only one point.] § 11 
 
 6. .-. A ABC coincides with AZ>^Fandis congruent to it. 
 
 [Two A are congruent if they can be made to coincide. ] § 60 
 
 Ex. 24. Where would AC fall if Z A, above, were less than /.D'> 
 
 Ex. 25. Draw freehand the approximate figure which would result 
 from superposing A ABC on A DEF if AB = DE, AA = ^ZD, and 
 ZB = iZE. 
 
 Ex. 26. After proving A ABC congruent to A DEF, what do you 
 know about : (a)ZC? Why ? (6) About AC? (c) About BC? § 65 
 
RECTILINEAR FIGURES 35 
 
 Ex. 27. In Propositions I and II, how many parts (§ 47) of one triangle 
 are given equal to parts of the other triangle ? ^ 
 
 Ex.28. Hyp. Z1=Z2. /^^"^>^ 
 
 Con. A ABC ^ A BCD. ^^^^ 
 
 Ex. 29. Hyp. AE and BD are st. lines. 
 ZB= ZD: C bisects BD. 
 
 t^ 
 
 Con. ZA = Z£. ^^^ 
 
 Suggestion. — Read § 66. ^ 
 
 Ex.30. Hyp. ^C bisects Z a /, 
 
 BG±AD. 
 Con. AB = BD. 
 
 Ex. 31. If the line which joins two opposite vertices of a quadrilateral 
 (four-sided figure) bisects the angles whose vertices it joins, then the 
 other two angles are equal. b^ 
 
 Hyp. Z1 = Z2; Z3 = Z4. 
 Con. ZB = ZD. 
 
 D 
 
 Note. — Supplementary Exercises 1 and 2, p. 273, can be studied now. 
 
 68. A triangle is Scalene when no two of its sides are equal ; 
 Isosceles when two of its sides are equal ; Equilateral when all 
 its sides are equal ; Equiangular when all its angles are equal. 
 
 Scalene Isosceles Equilateral 
 
 A triangle can be made to " stand upon " any one of its sides. 
 Hence any side of a triangle can be considered its Base. When 
 a side has been selected as base, the opposite vertex is called 
 the Vertex of the triangle, and the angle at that vertex is called 
 the Vertical Angle of the triangle. 
 
 In an iso.sceles triangle, the side which is not one of the 
 equal sides is usually taken as the base ; and then the angle 
 formed by the equal sides is the vertical angle of the isosceles 
 triangle. 
 
36 
 
 PLANE GEOMETRY — BOOK I 
 
 Pkoposition III. Theorem 
 
 69- In a7i isosceles triangle, the angles opposite the 
 equal sides are equal. 
 
 Proof. 
 
 2. 
 
 B 
 
 Hypothesis. In A ABC, AC= BC. 
 
 Conclusion. /.A = /.B. 
 
 Plan. Prove ZA and Z B homologous A of cong. A. 
 1. Let CD bisect Z O. 
 
 In A ACDsLTid A BCD: 
 
 AC=BC; Hyp. 
 
 CD = CD. See Note 1, § 64. 
 
 Z 1 = Z 2. Def . 
 
 [Since CD bisects Z O. ] 
 r.AACD^ABCD. 
 
 [If two /^ have two sides and the included Z of one 
 equal respectively to two sides and the included Z of the 
 other, the A are congruent. ] 
 
 .'.ZA = ZB. 
 
 [Homologous A of cong. ii are equal.] 
 
 Note 1. — Principle I (§ 66) is used. To get two triangles, a construc- 
 tion line, CD, was drawn. This is often necessary. 
 
 Note 2. — This theorem is ascribed to Thales, although this may not 
 be his proof. The proof for this theorem which was given by Euclid ap- 
 pears as Ex. 3, p. 273. It can be studied after § 73. 
 
 Ex. 32. Why are ZA and ZB homologous angles of the congruent 
 triangles in the proof of Proposition III ? 
 
 70. Cor. An equilateral triangle is also equiangular. 
 (Read § 71 at this time.) 
 
 3. 
 
 4. 
 
 63 
 
 §65 
 
RECTILINEAR FIGURES 
 
 37 
 
 71. A Corollary is a theorem which is easily deduced from 
 the theorem with which it is given. For each corollary, draw 
 a figure, form the hypothesis and conclusion, and give the 
 proof. 
 
 Ex. 33. Hyp. AB = AC. 
 Zl =Z2. 
 Con. AABX^AAYC. 
 Suggestion. — Does Z.3 = Z'^'! Why? 
 
 Ex. 34. If AB = AC and if D is the mid-point 
 of BC, Eoi AB, and F of AC, prove that ED=FD. 
 
 Suggestions. — Form the hypothesis and conclusion. 
 Read § (56. Does ^A' = CF ? Why ? 
 
 Ex. 35. After proving DE =DF in Ex. 34, draw EF and prove that 
 ZDEF=ZDFE. 
 
 Ex. 36. In the figure drawn for Ex. 35, prove that Z AFE = Z AEF. 
 
 Ex. 37. If AB and CB are two rafters of 
 equal length in a roof, and if DF and EG are 
 supports, perpendicular to the floor AG, at 
 points equally distant from A and C respec- 
 tively, prove that DF must equal EG. (Form 
 the hypothesis and conclusion first. ) 
 
 Ex. 38. In an isosceles triangle ABC, having AB = AC, point D is 
 any point in the base BC. E ia taken on AC and F on AB so that EC 
 = BD and BF = DC. Prove that DE = DF. (Draw the figure as it is 
 described.) 
 
 Ex. 39. To obtain the distance AB. 
 
 (1) LayoffjBCJLto^^. (2) Lay off (7jE:± to ^O. 
 (3) Place a stake at 0, the mid-point of BC. (4) De- 
 termine, by sighting, a point D on CE so that A, O, 
 and D will be in the same straight line. Then CD 
 = AB. Prove it. 
 
 Ex. 40. To obtain the distance AB. 
 
 (1) Let AC he any convenient segment. (2) Lay 
 off AD, making ZS = Zl. (3) Lay off CD, making 
 Z 4 = Z 2. Then AB = AD. Prove it. 
 
38 PLANE GEOMETRY — BOOK I 
 
 Proposition IV. Problem 
 72. Construct a triangle, having given its three sides. 
 
 Given m, n, and p, the three sides of a triangle. 
 Required to construct the triangle. 
 Construction. 1. Draw AB = m. 
 
 2. With A as center, and n as radius, draw an arc. 
 
 3. With B as center, and p as radius, draw a second arc, in- 
 tersecting the first arc at C. 
 
 4. Draw AC and BO. 
 
 Statement. A ABC is the required triangle, as it has the 
 given sides. 
 
 Discussion. If one side is equal to or greater than the sum 
 of the other two sides, the construction is impossible. 
 
 Ex. 41. A piece of ground is triangular in form. Its sides measure 
 100 rd., 150 rd., and 200 rd., respectively. Make a scale drawing of the 
 triangle, letting 1 in. represent 100 rd. 
 
 Ex. 42. Construct an isosceles triangle whose base is 2 in. and whose 
 equal sides are each 3 in. 
 
 Ex. 43. A girl wants an equilateral triangle whose sides are each 3 in. 
 long, to be used as a pattern in making a patch-work pillow-cover. Con- 
 struct the equilateral triangle. 
 
 Ex. 44. Try to construct a triangle whose sides are 1 in., 3 in., and 
 4 in., respectively. 
 
 Ex, 45. Construct a triangle whose sides are 2 in., 3 in., and 4 in., 
 respectively. Cut the triangle from the paper and compare it by super- 
 position with the triangles made by other members of your class. What 
 do you conclude must be true about all triangles made according to the 
 directions given ? 
 
RECTILINEAR FIGURES 39 
 
 Proposition V. Theorem 
 
 73. If two triangles have the three sides of one equal 
 respectively to the three sides of the other, the triangles 
 are congruent. 
 
 Hypothesis. In A ABC and A DEF : 
 
 AB = DE', BO=EF; B,ndAC=DF. 
 
 Conclusion. A ABC ^ A DEF. 
 
 Proof. 1. Place A DEF so that DE will coincide with AB, 
 E falling on B, and so that F falls at G, on the opposite side 
 of AB from C. Draw CG. 
 
 2. In A ACG, Z 1 = Z 2, since AC= AG. 
 
 [In an isosceles A, the A opposite the equal sides are 
 equal.] § 69 
 
 3. In A BCG, Z 3 = Z 4, since BC= BG. 
 
 4. .•.Z1 + Z3=Z2 + Z4. 
 
 [If equals be added to equals, the sums are equal.] Ax. 3 ; § 51 
 
 5. .'.ZC=ZG,ovZC = ZF. Ax.7;§51 
 
 6. In A ABC and A DEF: 
 
 AC= DF, and BC= EF; Hyp. 
 
 ZC = ZF. Step 5 
 
 7. .-. A ABC ^ A DEF. 
 
 [See Note 1 and §63.] 
 
 Note 1. — From now on, an authority which should be familiar to the 
 student will be omitted from demonstrations in the text. The paragraph 
 reference will be given for the present. The student should supply the 
 authority in full, without consulting the authority quoted, if possible ; 
 otherwise he should look up the reference. When writing out a demon- 
 
40 
 
 PLANE GEOMETRY — BOOK I 
 
 stration or giving the demonstration orally, give all authorities in full as 
 has been done in the text heretofore. 
 
 Note 2. — Three sides determine a triangle ; that is, the shape and size 
 of the triangle cannot change unless one or more of the sides 
 is changed. Practical use is made of this fact as illustrated 
 in the figures below. In each case, three lengths deter- 
 mine a triangle which makes some part of the object rigid. 
 
 ...^^^^f^. 
 
 Roof Truss 
 
 P^^ 
 
 -innnnnnnru 
 Gate 
 
 lOE^ 
 
 Ex. 46. If MN= iVPand MO = OP, then NO bisects 
 /.MNP. 
 
 (Form the Hyp., Con., and give the proof. See § 52.) 
 
 Ex. 47. If the opposite sides of a quadrilateral 
 ABGD are equal, then /.A = ZG. pf ^0 
 
 Ex. 48. In quadrilateral ABCD in Ex. 47, can you prove that /.B 
 = ZD? 
 
 Ex. 49. Why is a shelf bracket made in the form of a triangle ? 
 
 Ex. 50. Can you give any other practical uses of Proposition V ? 
 
 Ex. 51. On segment XY construct isosceles A XYZ and a second isos- 
 celes AXYW. Draw ZW. Prove that AXZW^ 
 A YZW. 
 
 Ex. 52. In the adjoining figure, if AB = DC, and b^ 
 AC = BD, then Z A must equal Z D. 
 
 Suggestion. — Prove A ABC ^ A BCD. 
 Note. — Supplementary Exercises 3 to 5, p. 273, can be studied now. 
 
 ,^< 
 
 Review Questions 
 
 Ex. 53. What is a theorem ? an axiom ? 
 
 Ex. 54. State three theorems by which two triangles can be proved 
 congruent. 
 
 Ex. 55. What are homologous parts of congruent triangles ? 
 
 Ex. 56. State Principle I. 
 
 Ex. 57. What does it mean to bisect an angle ? 
 
RECTILINEAR FIGURES 41 
 
 Proposition VI. Problem 
 
 74. Bisect a given angle. 
 
 Given Z^O-B. 
 
 Required to bisect Z AOB. 
 
 Construction. 1. With as center and a convenient radius, 
 draw an arc intersecting AO a,t C and BO at D. 
 
 2. With C and D as centers and with equal radii, draw arcs 
 intersecting at E. 
 
 3. Draw OE. 
 Statement. OE bisects Z AOB. 
 
 Proof. The proof is to be given by the pupil. 
 
 Suggestions. — Draw CE and DE. Recall § 66. 
 
 Note. — For construction problems, the regular form is to give the 
 statement of the problem, the parts which are "given," that which is 
 "required," the "construction," the "statement," and the "proof." 
 Also it is important to " discuss " the solution finally, in order to decide 
 when the solution is possible, etc. In this problem it is evident that 
 the solution is always possible. 
 
 Ex. 58. Draw an obtuse angle. Divide it into four equal parts. 
 
 Ex. 59. Construct the bisectors of the three angles of a large triangle. 
 What seems to happen ? 
 
 Ex. 60. Three pieces of wood are to be joined as in 
 the figure on the right. Construct to scale (letting 1'' 
 =4") an equilateral triangle ; bisect its angles; on each 
 bisector lay off a point 4" from the vertex ; connect these 
 points. 
 
42 PLANE GEOMETRY - BOOK I 
 
 Proposition VII. Problem 
 
 75. At a given point in a line, construct an angle 
 equal to a given angle. 
 
 ^. 
 
 By 
 
 PC o 
 
 Given / P, and point in line OA. 
 
 Required to construct an angle equal to Z P, having as 
 vertex and OA as side. 
 
 Construction. 1. With P as center and a convenient radius, 
 draw an arc intersecting the sides of Z Pat (7 and D, Draw CD. 
 
 2. With as center and PC as radius, draw arc FE. 
 
 3. With F as center and CD as radius, draw an arc inter- 
 secting arc FE at B. 
 
 4. Draw OB. 
 Statement. Z.AOB = ZP. 
 
 Proof. Draw FB. Proof to be given by the pupil. 
 
 Ex. 61. Construct /\ABC with side AB = 4 in., and ZA and ZB 
 equal to the angles given in the adjoin- 
 ing figure. Measure AC. Should the 
 triangles made by different pupils be 
 congruent ? 
 
 Ex 62. Construct A ABC having Z A equal to the Z A given in Ex. 61, 
 AB = 3 in., and AC = 2 in. Measure BC. 
 
 Note. — Supplementary Exercises 6-9, p. 273, can be studied now. 
 
 76. A line perpendicular to a segment at its mid-point is 
 the Perpendicular-bisector of the segment. 
 
 Ex. 63. Draw a line BC, 3 in. in length. With compasses, locate a 
 point A above BC which is 2 in. from B and 2 in. from C. Locate simi- 
 larly a point D below BC which is 3 in. from B and 3 in. from C. Draw 
 AD, cutting BC at E. (a) Compare BE with EC by means of your 
 dividers, (b) Measure Z BE A. (c) What kind of lines are AD and BC? 
 
RECTILINEAR FIGURES 
 
 Proposition VIII. Theorem 
 
 43 
 
 77. If two points are each equidistant from the ends 
 of a segment, they determine the perpendicular-bisector 
 of the segment 
 
 Hypothesis. C and D are equidistant from the ends of seg 
 ment AB. CD intersects AB at E. 
 
 Conclusion. AE = EB; CD± AB. 
 
 Proof. 1. In A ACD and A CBB : 
 
 AC=CB,kndAD = DB', 
 
 CD = CD. 
 .'. A ACD ^ A BCD. 
 .-. Z3 = Z4. 
 AACE^AECB. 
 (Give the full proof.) 
 
 .-. AE = EB. 
 
 Also Z1=Z2. 
 
 .-. CD±AB. 
 
 [If one straight line meets another straight line so that 
 the adj. A formed are equal, the A are rt. A, and the lines 
 are ±.] §§ 20, 29 
 
 Note. — It is often necessary, as in this proof, to prove one pair of tri- 
 angles congruent in order to obtain two equal angles or two equal segments 
 which are required in turn to prove another pair of triangles congruent. 
 
 Ex. 64. If XZ is the perpendicular-bisector of ABj 
 and Y is a point on XZ, prove A AXY^ A BXY. 
 
 First prove /\AXZ^/\BXZ to get AX=XB; 
 then prove A ^Z F ^ A BZ T to get ^ F = YB. 
 
 2. 
 3. 
 4. 
 
 5. 
 
 6. 
 
 7. 
 
 Hyp. 
 
 Why? 
 §65 
 § 63 
 
 Why? 
 Why? 
 
 Note. — Supplementary Exercises 10-11, p. 274, can be studied now. 
 
44 PLANE GEOMETRY — BOOK I 
 
 . Proposition IX. Problem 
 78. Construct the perpendicular-bisector of a given 
 
 segment. 
 
 x1 
 
 V 
 
 a4 
 
 N 
 
 \A 
 
 \ 
 
 ■^B 
 
 I / 
 
 Given line segment AB. 
 
 Required to construct the perpendicular-bisector of AB. 
 
 Construction. 1. With A and B as centers, and with equal 
 radii, draw arcs intersecting at Q and also at Z>. 
 2. Draw CD intersecting AB at B. 
 
 Statement. E bisects AB. 
 
 Proof. 1. AC= BO, and also AD = BD. 
 
 [Radii of equal circles are equal.] § 17 
 
 2. .*. CD is the perpendicular-bisector of AB. 
 
 [If two points are each equidistant from the ends of a 
 segment, they determine the perpendicular-bisector of the 
 segment.] § 77 
 
 Ex. 65. Divide a given segment into four equal parts. 
 Ex. 66. Draw a triangle of large size. Construct the perpendicular- 
 bisectors of the three sides. What happens ? 
 
 79. A Median of a triangle is the line drawn from a vertex 
 to the mid-point of the opposite side. 
 
 Ex. 67. Draw a triangle of large size. Construct the three medians 
 of the A. What happens ? 
 
 Ex. 68. Prove that the median drawn to the base of an isosceles 
 triangle bisects the vertical angle. (Construct the figure.) 
 
 Note. — Supplementary Exercises 12-14, p. 274, can be studied now. 
 
RECTILINEAR FIGURES 45 
 
 Proposition X. Problem 
 
 SO. At a point in a line, construct a perpendicular 
 to the line. 
 
 A-J- 
 
 \ 
 
 Given O, any point in line AB. 
 Required to construct a perpendicular to AB at (7. 
 Construction. 1. With C as center, and any radius, draw- 
 arcs intersecting AB at D and E respectively. 
 
 2. With D and E as centers and a radius greater than one 
 half DE, draw two arcs intersecting at F. 
 
 3. Draw CF. 
 Statement. CF± AB at C. 
 Proof. 1. Draw DF and EF. 
 
 2. C is equidistant from D and E. Why ? 
 
 3. F is equidistant from D and E. Why ? 
 
 4. .-. CF is the perpendicular-bisector of DE. Why ? 
 
 5. .-. CFJ_^J5at a 
 
 [Since AB and i>^ are the same straight line.] 
 
 81. Proposition X proves that one perpendicular can be 
 draivn to a line at a point in the line. 
 
 It can be proved that only one perpendicular can be drawn to 
 a line at a point in the line. For, if CP and DP were both 
 perpendicular to AB at P, then Z 1 and Z 2 ^ ^ 
 
 would both be right angles and hence would 
 be equal. But Z 1 is greater than Z 2, for 
 the whole is greater than any of its parts. -^ 
 
 Ex. 69. Prove CF perpendicular to DE (§80) by proving that 
 Z FCD = Z FCE, and then using § 26. 
 
46 PLANE GEOMETRY — BOOK I 
 
 Proposition XI. Problem 
 
 82. Construct a perpendicular to a line from a point 
 not in the line, 
 
 G 
 
 •^"i^ 
 
 -fi 
 
 Given line AB and point C not in AB. 
 Required to construct a ± to AB from C. 
 Construction. 1. With C as center and a convenient radius, 
 draw an arc intersecting AB at D and E respectively. 
 
 2. With D and E as centers, and equal radii, draw two arcs 
 intersecting at F. 
 
 3. Draw CF. 
 
 Statement. CF±AB. 
 
 Proof. 1. C is equidistant from D and E. Why ? 
 
 2. F is equidistant from D and E. Why ? 
 
 3. .-. CF is the perpendicular-bisector of DE. Why ? 
 
 4. .-. Ci^±^S. 
 
 [Since AB and J>£' are the same straight line.] 
 
 Historical Note. — This construction is attributed to Oenipodes of 
 Chios (466 B.C.) 
 
 83. Proposition XI proves that one perpendicular can be 
 drawn to a line from a point not in the line. 
 
 It will be proved later (§ 88) that only one perpendicular can 
 he drawn to a line from a point not in the line. 
 
 It will also be proved (§ 164) that the perpendicular is the 
 shortest segment from the pfoint to the line. 
 
 Note. — Supplementary Exercises 15-16, p. 274, can be studied now. 
 
RECTILINEAR FIGURES 
 
 47 
 
 84. The Distance from a point to a line is the length of the 
 perpendicular from the point to the line. 
 
 85. An Altitude of a triangle is the per- 
 pendicular drawn from a vertex to the opposite 
 side or the opposite side extended ; as, AD. 
 
 Ex. 70. How many altitudes does a triangle 
 have ? 
 
 Ex. 71. Construct a triangle whose sides are 2 inches, 3 inches, and 
 4 inches, respectively. Construct the three altitudes of the triangle. 
 
 Ex. 72. Construct an angle of 45°. (Use Prop. XI and Prop. VI.) 
 Construct an angle of 135° ; of 22^° ; of 67^°. 
 
 Note. — The second and third designs below are drawn upon the first 
 figure as a background. Can you discover how the first figure is con- 
 structed ? Can you make an original design similar to these designs ? 
 
 86. An Exterior Angle of a triangle 
 is the angle at any vertex formed by a 
 side of the triangle and the adjacent side 
 extended ; as, Z DC A. 
 
 One interior angle is adjacent to the exterior ^ 
 angle and the other two are remote interior angles, 
 are the remote interior angles of exterior A DC A. 
 
 Ex. 73. Draw a large figure like that in § 86. Measure the exterior 
 Z.DCA and each of the remote interior angles. How does the exterior 
 
 angle compare with the remote interior angles ? 
 
 • 
 
 Ex. 74. In the figure for Prop. IX, p. 44, Z. CEB is an exterior 
 angle of what triangles ? 
 
 O D 
 Thus, Z .A and Z.B 
 
48 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition^ XII. Theorem 
 87. A71 exterior angle of a triangle is greater than 
 
 either remote interior angle. 
 
 Hjrpothesis. 
 Conclusion. 
 
 Z BCD is an exterior Z of A ABC. 
 Z BCD >ZB; also Z BCD > Z A. 
 Part I. Proof. 1. Through 0, the mid-point of BC, draw 
 AG. Extend AO to E, making OE equal to AG. Draw CE. 
 
 2. 
 
 .-. AABG^AGCE. 
 [Give the full proof.] 
 
 .-. Z4 = Z1. 
 Z BCD > Zl. 
 [The whole is greater than any of its parts.] 
 .'.ZBCD>Z4.. 
 (Substitute Z4 for its equal, Z 1.) 
 Part IL Proof. 1. Extend BC to F. 
 
 Why? 
 Ax. 8, § 51 
 
 Ax. 2, § 51 
 
 2. Z ACF > ZA. 
 
 [By a proof similar to that for Part I.] 
 
 3. ZBCD = ZACF. Why? 
 
 4. .'. ZBCD>ZA. 
 
 (Substitute Z 5 (7i> for its equal, ZAGF, in step 2.) 
 
 Ex. 75. In the figure for Prop. XII, prove that ZBOE is greater 
 than Z 4. (Use §87.) 
 
 Ex. 76. Prove also that Z AOC > Z 1. 
 Ex. 77. Compare Z ECF with Z 3. 
 
RECTILINEAR FIGURES ' 49 
 
 88. Cor. Tliere can he only one perpendicular from a point 
 to a line. ^ 
 
 If AB ± DE, then AC cannot be ± DE, for 
 Z2 > A\ and hence Z2 is an obtuse angle. 
 
 ^- V B 
 
 Ex. 78. If straight lines be drawn from a point with- 2 
 
 in a triangle to the extremities of any side, the angle in- 
 cluded by them is greater than the angle included by the 
 other two sides. (Prove Z AXC >ZABC.) ^'^ ^C 
 
 Suggestions. — 1. Compare Z AXC with Z A YC. 2. Compare Z A YC with 
 ZABC. 
 
 Review Questions 
 
 Ex. 79. What is the perpendicular-bisector of a segment ? 
 
 Ex. 80. What is a median of a triangle ? How many medians does a 
 triangle have ? 
 
 Ex. 81. What is an altitude of a triangle ? 
 
 Ex. 82. What is the distance from a point to a line ? 
 
 Ex. 83. (a) What is an exterior angle of a triangle ? (b) How 
 many exterior angles can be formed at one vertex of a triangle ? 
 (c) How do they compare ? 
 
 Ex. 84. (a) How many perpendiculars can be drawn to a line from a 
 point not in the line ? (6) How many at a point of the line ? 
 
 Ex. 85. Five fundamental construction problems have now been 
 taught (§§ 74, 75, 78, 80, 82). (a) State each of them. (6) Are you 
 able to make each of these constructions quickly with ruler and compass 
 alone ? 
 
 Note. — Straightedge and compass alone are employed in making the 
 constructions in elementary geometry. This practice was initiated by 
 Plato (429-348 b.c). Naturally some constructions cannot be made 
 with these tools alone. For example, it is impossible to trisect an angle 
 by ruler and compass alone. 
 
50 PLANE GEOMETRY — BOOK I 
 
 PARALLEL LINES 
 
 ill ill iji ifi ifi fli -^^ 
 
 Some Parallel Line Border Designs 
 
 89. Two straight lines are Parallel (II) if they lie in the 
 same plane and do not meet however far they are extended. 
 
 Note. — Two straight lines in the same plane either intersect or are 
 parallel lines. 
 
 90. Ax. 16. Axiom of Parallels. TJirough a ^ -^.^p^ ^ 
 given point there can he only one parallel to a 
 given line. Thus, XFII MN. 
 
 M N 
 
 91. Cor. If two lines are parallel to a . „ 
 
 third line, they are parallel to each other. 
 
 Hyp. ABWGD) EFWCD. ^ ^ 
 
 Con. AB II EF. 
 
 Proof. If AB and EF are not parallel, they must meet at a point. 
 Through this point there would then be two lines parallel to CD. But 
 this is impossible by the Axiom of Parallels. Hence AB must be paral- 
 lel to EF. 
 
 92. If two lines are cut by a third line, AB, called a Trans- 
 versal, the angles are named as follows : 
 
 A 3, 4, 5, and 6 are called Interior Angles. 
 
 A 1, 2, 7, and 8 are called Exterior Angles. 
 
 AS and 6 are called Alternate-interior 
 Angles ; also A 5 and 4. 
 
 Al and 8 are called Alternate-exterior 
 Angles ; also zi 2 and 7. 
 
 A 2 and 6 are called Corresponding Angles ; also /4 1 and 5, 
 A 3 and 7, and A 4 and 8. 
 
PARALLEL LINES 
 
 51 
 
 Proposition XIII. Theorem 
 
 93. If two lines are cut hy a transversal so that a 
 pair of alternate-interior angles are equal, the lines are 
 parallel. 
 
 >o 
 
 Hypothesis. AB and CD are cut by EF-, Z 1 = Z 2. 
 
 Conclusion. AB II CD. 
 
 Proof. 1. Suppose that AB is not parallel to CD, and that 
 it meets CD at point O on the right of EF, forming A MNO. 
 
 2. Then Z 1 is an exterior angle of A MNO. 
 
 3. .-. Z 1 > Z 2 
 
 [An ext. Z of a A is greater than either remote int. Z.] § 87 
 
 4. ButZl = Z2 Hyp. 
 
 5. .*. AB cannot meet CD on the right of EF. 
 
 6. Similarly it can be proved that AB cannot meet CD on 
 the left of EF. 
 
 7. .-. AB II CD 
 
 [Two lines are II if they lie in the same plane and do not 
 meet however far they are extended.] § 89 
 
 Note. — Read the next paragraph at this time. 
 
 94. The method of proof used in §§91 and 93 is called the 
 Indirect Method of Proof. Notice : (a) it starts by assuming 
 the negative of the conclusion ; (b) it follows up the conse- 
 quences of this assumption until a statement is reached which 
 contradicts a known fact ; (c) this contradiction is made the 
 basis for asserting that the desired conclusion is true. 
 
52 
 
 PLANE GEOMETRY — BOOK I 
 
 95. Principle II. To prove two lines parallel, try to prove 
 a pair of alternate-interior angles equal. 
 
 96. Cor. 1. If two lines are cut by a 
 transversal so that a pair of corresponding 
 angles are equal, the lines are parallel. 
 
 Hyp. AB and CD are cut by EF\ Z2= Z6. 
 
 Con. AB li CD. 
 
 Plan. Try to prove Z 3 = Z 6. Then use § 93. 
 
 97. Cor. 2. If two lines are perpendic- 
 ular to a third line, they are parallel. 
 
 Hyp. AB±XY; CD A. XY. 
 
 Con. AB II CD. 
 
 Plan. Try to prove Z 1 = Z2. Then use § 93. 
 
 X— f- 
 
 -M- 
 
 B 
 
 D 
 
 98. Cor. 3. If two lines are cut by a transversal so that a 
 pair of interior angles on the same side of the transversal are 
 supplementary, the lines are parallel. 
 
 Hyp. AB and CD are cut by EF ; Z 4 + Z 6 = 1 st. Z. 
 Con. AB II CD. 
 
 Plan. Try to prove Z 3 = Z 6. 
 Proof. 1. Z3 + Z4 = 1 St. Z. 
 
 2. Z4 + Z6 = 1 St. Z. 
 
 3. .-. Z3 + Z4 = Z4 + Z6. 
 
 4. .-. Z3 = Z6, and hence^^ll CD. 
 
 §39 
 
 A 
 
 Hyp. 
 
 C 
 
 Why? 
 
 
 Why? 
 
 
 Ex. 86. If tw^o lines are cut by a transversal so that a pair 
 of alternate-exterior angles are equal, the lines are parallel. 
 
 Ex. 87. A carpenter wants a board of length AC v^ith 
 parallel ends, AB and CD. He marks AB and CD by means of 
 his square as in the figure. Why must AB and CD be parallel ? 
 Assume that edge ^O is a straightedge. 
 
 TZyp 
 
 A B 
 
 CD 
 
 Ex. 88. If sides BA and CA of any A ABC are extended their own 
 lengths through vertex ^ to Z> and E respectively, then DE is parallel to 
 BC. 
 
 Suggestion. — Apply § 95 and § 66. 
 
 Note. — Supplementary Exercises 17-21, p. 274, can be studied now. 
 
PARALLEL LINES 
 
 53 
 
 Proposition XIV. Problem 
 
 99. Construct a parallel to a line through a point 
 not in the line. e 
 
 Given line AB and (7, any point not in line AB. 
 Required to construct a parallel io AB through C. 
 Construction (a). 1. Draw FE through (7, meeting AB at O. 
 2. At (7, construct Z 2= Z 1. 
 
 Statement. MN II AB. Why ? 
 
 Proof. To be given by the pupil. 
 
 Construction (6). A second construction is based upon § 96. 
 This construction is left as an exercise for the pupil. 
 
 Ex. 89. The figure adjoining shows 
 how a draughtsman draws a line through 
 C parallel to AB. Why is DE parallel to 
 ABf 
 
 Ex. 90. The adjoining figure shows 
 how a draughtsman draws parallel lines 
 by means of his T-square. Why are the 
 lines parallel ? 
 
 Ex. 91. Construct the figure for § 07 ; then construct the bisectors of 
 
 A 1 and 2. Prove that these bisectors are parallel. 
 
 Ex. 92. In the figure for Prop. XII, p. 48, prove that CE \a parallel 
 to AB. 
 
54 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XY. Theorem 
 
 100. If tivo parallels are cut hy a transversal, alter- 
 nate-interior angles are equal. 
 
 Hypothesis. EFcuts lis AB and CD at G and H. 
 
 Conclusion. Z AGH = Z GHD. 
 
 Proof. 1. Suppose that Z AGH is less than Z GHD. 
 
 2. Then draw i^l^ through H, so that Z (7^3f = Z ^(^^. 
 
 3. .'.LMWAB. 
 
 [If two lines are cut by a transversal so that a pair of alt.- 
 int. A are equal, the lines are II.] 
 
 4. But this is impossible, for CD II AB, by hypothesis. 
 
 [Through a given point, there can be only one II to a 
 given line.] 
 
 5. .-. Z AGH cannot be less than Z GHD. 
 
 6. Similarly, Z AGH cannot be greater than Z GHD. 
 
 7. .'. Z AGH = Z GHD. 
 Note. — Review § 94. 
 
 93 
 
 90 
 
 E 
 
 Ex. 93. If EF cuts parallels AB and CD so ^ 
 that ZS = 30°, how many degrees are there in each 
 of the other angles of the figure ? G 
 
 Ex. 94. If EF, joining two parallels, be bisected 
 and GH be drawn through the mid-point and included 
 between the parallels, then GH will also be bisected 
 by the point. 
 
 1/2 
 
 j> 
 
 3/4 
 5/6 
 
 j\ 
 
 7/8 
 
 F 
 
 G 
 
 E 
 
 K 
 
PARALLEL LINES 
 
 55 
 
 101. Cor. 1. If two parallels are cut by a transversal^ cor- 
 responding angles are equal. 
 
 Hyp. AB II CD; Z2 and Z.Q are corresponding 
 angles. 
 
 Con. Z 2 = Z 6. 
 
 [What do you know about Z 3 and Z 6 ?] 
 
 102. Cor. 2. If a line is perpendicular 
 to one of two parallels, it is perpendicular to 
 the other also. 
 
 Hyp. AB II CD ; XY ± AB. 
 Con. XT±CD. 
 
 [What must be proved about Z 2 ?] 
 
 103. Cor. 3. If two parallels are cut by a transversal, intenor 
 angles on the same side of the transversal are supplementary. 
 
 Hyp. AB II CD ; Z 4 and Z 6 are int. A on the same side of the 
 transversal. (Fig. § 101.) 
 
 Con. Z 4 + Z 6 = 1 St. Z. 
 
 Proof. 1. Z 4 + Z3 = 1 St. Z. § 39 
 
 2. Z6 = Z3. Why? 
 
 8. .-. Z4 + Z6 = l St. Z. 
 
 [Substituting for Z 3 its equal, Z 6.] 
 
 Ax. 2, § 51 
 
 \ Isi Ave. 
 
 ' ^%\ 2nd Ave, 
 
 Ex. 95. If N. W. Street crosses 1st Ave. at an 
 angle of 45°, at what angle does it cross the parallel 
 avenues ? 
 
 Ex. 96. In the adjoining figure, if AB II CD^ 
 and EFWGH, prove: (a) Z 1 = Z 13 ; 
 
 (6) Z 3 + Z 16 = 1 St. Z. " 11/12 15/16 
 
 F H 
 
 Ex. 97. If a line be drawn parallel to the base of an isosceles triangle, 
 cutting the two sides of the triangle, it makes equal angles with these 
 sides. 
 
56 
 
 PLANE GEOMETRY — BOOK I 
 
 104. One theorem is called the Converse of another when the 
 hypothesis and conclusion of the one become the conclusion and 
 hypothesis of the other. Thus, Prop. XV is the converse of • 
 Prop. XIII. 
 
 In Proposition XIII 
 
 In 
 
 Proposition XV 
 
 
 Hyp. EFcMtsABsiud 
 
 Hyp. 
 
 FF cuts AB and 
 
 A 
 
 CD. 
 
 
 CD. 
 
 
 Z3 = Z6. 
 
 
 AB 11 CD. 
 
 C 
 
 Con. ABWCD. 
 
 Con. 
 
 ZS=Z6. 
 
 
 The converse of a given theorem is not always true. 
 Thus, all right angles are equal. The converse would be all 
 equal angles are right angles. Evidently this is not true. 
 
 Ex. 98. Of what statement is Cor. 1 (§ 101) the converse ? Cor. 2 
 (§ ]02) ? Cor. 3 (§ 103) ? 
 
 Ex. 99. In the figure for § 101, prove Z 3 = Z 7. 
 
 Ex. 100. In the figure for § 101, prove Z 3 + Z 5 = 1st Z. 
 
 Ex. 101. If two parallels are cut by a transversal, alternate exterior 
 angles are equal. 
 
 Ex. 102. State the converse of Ex. 101. Is it a true statement ? 
 
 Ex. 103. If two parallels are cut by a transversal, bisectors of a pair 
 of corresponding angles are parallel. 
 
 Ex. 104. If two parallels are cut by a transversal, exterior angles on 
 the same side of the transversal are supplementary. 
 
 Suggestion. — The proof is like that for § 103. 
 
 Ex. 105. State the converse of Ex. 104. 
 
 Ex. 106. If two lines are perpendicular to parallel lines, then they 
 are either parallel or coincide. 
 
 Ex. 107. What is the axiom of parallels ? 
 
 Ex. 108. What are parallel lines ? 
 
 Ex. 109. Is the following a correct statement ? When two lines are 
 cut by a transversal, alternate interior angles are equal. 
 
 Note. — The various theorems about angles made by a transversal of 
 two parallels are found in Euclid. They were probably formulated by 
 the Pythagoreans. 
 
PARALLEL LINES 57 
 
 Proposition XVI. Theorem 
 105. If tioo angles have their sides i^espectively par- 
 allel, they are equal, provided both pairs of parallels 
 extend in the same directions from their vertices, or in 
 opposite directions. 
 
 / 
 
 '-7 
 
 E 
 
 Fig. 1 Fig. 2 
 
 I. (Fig. 1.) Hypothesis. A ABC and DEF have AB II DE 
 and BC II EF. 
 
 Conclusion. Z B = /. E. 
 
 [Proof to be given by the pupil.] 
 
 Suggestion. — Extend BC and ED until they intersect at G. Compare Z. B 
 with Z 2 and Z E with Z 2. Then compare Z B with Z £". 
 
 II. (Fig. 2.) Hypothesis. ^ ABC and i>^i^ have ^B II DE 
 and 5C II EF: 
 
 Conclusion. Z B = ZE. 
 
 Note. — The sides extend in the same direction if they are on the same 
 side of a straight line joining their vertices, and in opposite directions if 
 they are on opposite sides of this line. 
 
 Ex. 110. If two angles have their sides respectively 
 parallel, one pair of parallels extending in the same 
 directions but the other pair extending in opposite 
 directions from their vertices, the angles are supple- f-*- 
 mentary. (Prove Z B -\- Z E = I at. Z.) 
 
 Ex. 111. Two streets cross as in the adjoining 
 figure. If the lot lines at corner C make an angle of 
 70°, determine the number of degrees in the angle 
 formed at each of the other corners. 
 
 J^ 
 
58 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XVII. Theorem 
 
 106. The sum of the angles of any triangle is one 
 straight angle. 
 
 Hypothesis. A ABC is any triangle. 
 
 Conclusion. ZA-\-ZB-\-ZO=lst Z. 
 
 Proof. 1. Extend AC to D, and construct CE parallel to AB. 
 
 2. Z 1 + Z 2 + Z 3 = 1 St. Z. 
 
 [The sum of all the successive adj. A around a point 
 on one side of a st. line is one st. Z.] § 34 
 
 3. Z ^ = Z 1. Const. 
 
 4. Z B = Z2, since BC cuts lis AB and CE. Why ? 
 
 5. .'.ZA + ZB + ZC=lst.Z. 
 [Substituting ZAtor Zl.ZB for Z 2, and Z C for Z 3.] Ax. 2, § 51 
 
 Note. — This theorem is attributed to Eudemus, a pupil of Aristotle. 
 
 Ex. 112. If Z ^ = 70°, and Z 5 = 35°, how large is Z C ? 
 
 Ex. 113. How large is each angle of an equiangular triangle ? 
 
 Ex. 114. Prove Prop. XVII by constructing a line through B parallel 
 to^C 
 
 Ex. 115. The rafters of a "saddle roof " make 
 an angle of 40° with a level line. What angle do 
 the rafters form at the ridge ? 
 
 Ex. 116. AB 
 
 so that AD = AB. 
 
 AC in A ABC. BA is extended to D 
 Prove that CD is perpendicular to BC. 
 
 Suggestions. — 1. Z 1 + Z 4 must be proved a right Z. 
 2. Whatpartof Zl + Z2 + Z3 + Z4isZH-Z4? 
 
 Note. — This is a very important exercise. It may be expressed thus: 
 if the median to one side of a triangle is one half of that side, the angle 
 from which it is drawn is a right angle. 
 
PARALLEL LINES 59 
 
 107. A triangle is a Right Triangle when it has one right 
 angle. 
 
 The Hypotenuse of a right triangle is the side opposite the 
 right angle ; the Legs of a right triangle are the two sides 
 of the triangle including the right angle. 
 
 If the legs of a right triangle are equal, the triangle is 
 called an Isosceles Right Triangle. 
 
 Corollaries to Proposition XVII 
 
 108. Cor. 1. A triangle cannot have two right angles or two 
 obtuse angles. 
 
 109. Cor. 2. The acute angles of a right triangle are com- 
 plementary. 
 
 110. Cor. 3. An exterior angle of a triangle /-^^ 
 equals the sum of the two remote interior angles. 14 
 
 Prove Zl = Z3 + ^4. 
 
 [Z2 + Z1 = ? Z2 + Z3 + Z4 = ? Form an equation.] 
 
 111. Cor. 4. If two angles of one tri- 
 angle equal respectively two angles of 
 another triangle, the third angles are 
 equal. 
 
 Hyp. Z 1 = Z 4, and Z 2 = Z 5. 
 
 Con. Z3 = Z6. 
 
 112. Cor. 5. If two triangles have a side, the opposite angle, 
 and another angle of the one equal re- „ 
 spectively to a side, the opposite angle, 
 
 and another angle of the other, the tri- 
 
 angles are congruent. 
 
 Hyp. AB = DE\ Z.C = ^F; ZB = ZE. 
 
 Con. AABC^ADEF. 
 
 Suggestions. — 1. Prove LA = LB. 
 
 2. Then prove /S ABC^/^ DBF by § 63. 
 
 Note. — Supplementary Exercises 22-35, p. 275, can be studied now. 
 
60 
 
 PLANE GEOMETRY — BOOK I 
 
 Propositiot^ XYIII. Theorem 
 
 113. If two angles have their sides respectively per- 
 jpendicular, they are either equal or supplementary. 
 
 v" \ 
 
 
 JB 
 
 G 
 Hypothesis. A ABC and EDF have 
 
 ABJuDE and BC±FG. 
 
 Conclusion. Z7 = Z.1. 
 
 Z 7 + Z 2 = 1 St. Z. 
 
 Proof. 1. Extend DE until it meets BA at B. 
 until it meets EG at S and intersects DB at 0. 
 
 2. In A BBO said A ODS : 
 
 Z6 = Z5; 
 Z 3 = Z 4. 
 
 3. .-.ZT^Zl. 
 
 4. Z 1 -h Z 2 = 1 St. Z. 
 
 5. .-. Z7 + Z2== 1 St. Z. 
 [Substitute Z 7 for its equal Z 1.] 
 
 Extend BO 
 
 Why? 
 
 Why? 
 
 §111 
 
 Why? 
 
 Ax. 2, § 51 
 They 
 
 Note. — The angles are equal if both are acute or both obtuse 
 are supplementary if one is acute and one is obtuse. 
 
 Ex. 117. If two right triangles have the hypotenuse and an acute 
 angle of one equal respectively to the hypotenuse and an acute angle of 
 the other, they are congruent. (§ 112.) 
 
 Ex. 118. If two right triangles have a leg and the opposite acute angle 
 of one equal respectively to a leg and the opposite acute angle of the other, 
 they are congruent. 
 
PARALLEL LINES 
 
 61 
 
 Proposition XIX. Theorem 
 
 . 114. If two right triangles have the hypotenuse and 
 a leg of one equal respectively to the hypotenuse and a 
 leg of the other , the triangles are congruent. 
 
 Hypothesis. In rt. A ABC and DBF-. 
 
 hypotenuse AB = hypotenuse DE ; BC= EF. 
 Conclusion. AABC^ADEF. 
 
 Proof. 1. Place A ABjO beside A DEF so that BC will 
 coincide with its equal EF, B falling on E, and so that A 
 falls at Gj on the opposite side of EF from D. 
 
 2. .-. Zl + Z2 = 1 St. Z. Why? 
 
 3. .-. DFG is a straight line. § 40 
 
 4. .-. figure EDFG is a triangle. 
 
 .6. .'. ^G = ZD, ov Z.A=Z.D. §69 
 
 6. In A ABC and A DEF: 
 
 AB = DE: Z1 = Z2: ZA=ZD. 
 
 A ABC ^ A DEF. 
 
 §112 
 
 115. Cor. If two equal oblique segments are di'awn to a line 
 from a point in a perpendicular to the line : 
 
 (1) they cut off equal distances from the 
 foot of the perpendicular. (Prove AD=DB.) 
 
 (2) they make eqtial angles with the per- 
 pendicular. (Prove Z 1 = Z 2.) 
 
 (3) they make equal angles with the given 
 line. (Prove Z 3 = Z 4.) 
 
 Note. — Supplementary Exercise 36, p. 276, can be studied now. 
 
62 PLANE- GEOMETRY — BOOK I 
 
 SUMMARY 
 
 116. The student will liave constant use for the foregoing 
 theorems, problems, and facts. 
 
 A. Two triangles are congruent if : 
 
 1. Two sides and the included Z. of one are equal respectively, etc. § 63 
 
 2. Two A and the included side of one are equal respectively, etc. § 67 
 
 3. The three sides of one are equal respectively, etc. § 73 
 
 4. A side, the opposite Z, and another Z of one are equal respectively, 
 etc. § 112 
 
 B. Two right triangles are congruent if : 
 
 1. The hypotenuse and a leg of one are equal respectively, etc. § 114 
 
 2. The hypotenuse and an acute A of one are equal respectively, etc. 
 
 Ex. 117 
 
 3. A leg and the opposite acute Z of one are equal respectively, etc. 
 
 Ex. 118 
 
 C. Two lines are parallel if : 
 
 1. Alt. -int. A made by a transversal are equal. § 93 
 
 2. Corresponding A made by a transversal are equal. § 96 
 
 3. Int. A on the same side of the transversal are supp. § 98 
 
 4. They are parallel to, or perpendicular to, the same line. §§ 91, 97 
 
 D. If a transversal cuts two parallels : 
 
 1. Alt. -int. A are equal. § 100 
 
 2. Corresponding A are equal. § 101 
 
 3. Int. A on the same side of the transversal are supp. § 103 
 
 E. To prove two line segments are equal : 
 
 Try to prove them homologous sides of congruent A. § 66 
 
 E. To prove two angles equal, try to prove that they : 
 
 1. Are homologous A of congruent A. § 66 
 
 2. Are supplements or complements of the same or equal A. §§ 37, 41 
 
 3. Are right A or vertical A. ' §§ 27, 54 
 
 4. Are opposite the equal sides of an isosceles A. § 69 
 
 5. Are alt. -int. A or corresponding A made by a transversal of two 
 parallels. §§ 100, 101 
 
 6. Have their sides respectively II or ±, etc. §§ 105, 113 
 G. To prove an angle is a right angle, try to prove : 
 
 1. It is equal to its supplement. § 26 
 
 2. It is the angle formed by two lines which are -L by § 77. 
 
 3. It is equal to an angle known to be a right Z. 
 
SUMMARY 63 
 
 117. Success in demonstrating unproved theorems comes as 
 a result of knowledge of the facts summarized in § 116, system- 
 atic methods of studying a theorem and planning its demon- 
 stration, and experience and perseverance. 
 
 Directions 
 
 1. Read the theorem carefully, making certain that each 
 word is thoroughly understood. 
 
 2. Draw the figure carefully, constructing it when possible. 
 
 Make the figure general. Thus, if the figure is based upon a triangle, 
 do not draw a right triangle or an isosceles triangle unless told to do so. 
 
 3. Decide upon the hypothesis and conclusion. 
 
 (a) Remember that the hypothesis states the facts about the figure 
 which are assumed, and that the conclusion states the facts which are to 
 be proved. 
 
 (6) If the theorem is stated in the " if . . . then . . ." form (§ 52), 
 the hypothesis and conclusion are evident at once. 
 
 (c) If the theorem is not stated in the " if . . . then ..." form, the 
 declarative sentence in its simplest form will give the conclusion, and the 
 subject of the sentence with its modifiers will give the hypothesis. 
 
 4. Decide upon a plan for the demonstration. 
 
 For the present, the suggestions in § 116 will aid the pupil 
 in solving most exercises. 
 
 Ask " what does the conclusion mean ? " or " how can I prove 
 the conclusion ? " The answers will suggest a plan for the 
 proof. 
 
 Thus, suppose that the conclusion is : EA = EB. 
 Question. How can I prove EA = EB ? 
 Answer. By proving them homologous parts of congruent ^. 
 
 This means that two triangles of which EA and EB are sides must be 
 selected. 
 
 Question. How can I prove two triangles congruent? 
 Answer. By one of the methods given in § 116, A and B. 
 
 This leads to the comparison of the sides and angles of the triangles. 
 Question. What do I know about the sides and A of the ^ ? How can 
 
 I prove these two angles equal ? 
 Answer. § 116, E and F suggest possible answers. 
 
64 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition^ XX. Theorem 
 
 118. I. Any point in the p)erpendicular-bisector of a 
 segment is equidistant from the 
 
 ends of the segment. 
 
 Hypothesis. CD 1. AB -, AD = DB ; 
 
 E is any point in CD. 
 
 Conclusion. EA — EB. 
 
 Plan. Try to prove EA and ^5 horn, 
 sides of cong. A. 
 
 [Proof to be given by the pupil.] 
 
 IT. (Converse.) Any pohit equi- 
 distant from the ends of a segment 
 lies in the perpeyidicular-hisector of 
 the segment. 
 
 Hypothesis. AB is a st. line. 
 PA = PB. 
 
 Conclusion. F lies in the perpendicular-bisector of AB. 
 
 Plan. Let C be the mid-point of AB. Try to prove 
 PC A. AB, by proving Z 1 = Z 2. 
 Proof. 1. AAPC^APCB. 
 
 [Give the full proof.] 
 
 2. .-. Z1 = Z2. Why? 
 
 3. .:PC±AB. 
 
 [If one St. line meets another st. line so that the adj. A 
 formed are equal, the A are rt. A and the lines are ±.] 
 
 §§ 26, 29 
 
 119. Cor. Two obliques, drawn to a line from a point in a 
 perpendicular to the line and cutting off equal distances from the 
 foot of the perpendicular, are equal. 
 
 Note. — Supplementary Exercises 37-38, p. 276, can be studied now. 
 
TRIANGLES 
 
 65 
 
 120. 
 
 Proposition XXI. Theorem 
 I. Any point in the bisector of an angle is equi- 
 
 distant from the sides of the 
 angle. 
 
 Hypothesis. BD bisects 
 ZABC; Pisin BD'yPM±AB; 
 PjSr±Ba 
 
 Conclusion. PM = PN. 
 
 Plan. Try to prove PM and P^hom. parts of cong. A. 
 
 Suggestion. — Recall § 112. 
 
 II. (Converse.) A?i2/ point equidistant from the 
 sides of an angle lies in the bi- 
 sector of the angle. 
 
 Hypothesis. P lies within Z ABC\ 
 PM1.AB', PN±BC; PM=PN. 
 Conclusion. P lies in the bisector 
 
 of Z ABC. 
 Plan. Draw PB. Try to prove 
 PB bisects Z ABC^ by proving Z 3 = Z 4. 
 
 121. Cor. Any point not in the bisector of an angle is 
 unequally distant from the sides of the angle. 
 
 Note. — Supplementary Exercise 39, p. 276, can be studied now. • 
 
 ISOSCELES AND EQUILATERAL TRIANGLES 
 
 122. Review the definitions of isosceles and equilateral 
 triangles in § 68; also review Prop. Ill, § 70, and Ex. 113. 
 
 Notice that an equilateral triangle is a special form of 
 isosceles triangle. Hence, for each theorem about an isosceles 
 triangle there is a corresponding theorem about an equilateral 
 triangle, which may be considered a corollary of the former. 
 Thus, § 70 follows at once from Prop. III. 
 
66 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XXII. Theorem 
 
 123. If two angles of a triangle are equal, the sides 
 opposite are equal, and the triangle is isosceles. 
 
 Hypothesis. In A ABC, ZA = ZB. 
 
 Conclusion. AC = EC. 
 
 Plan. Try to prove AG and BC horn, sides of cong. A. 
 Proof. 1. Construct CD bisecting Z ACB, 
 
 [Complete the proof in good form. ] • 
 
 124. Cor. If a triangle is equiangular, it is also equilateral. 
 
 Ex. 119. Prove that the bisector of the vertical angle of an isosceles 
 triangle is perpendicular to and bisects the base. 
 
 Ex. 120. Prove that the altitude to the base of an isosceles triangle is 
 also the median to the base and bisects the vertical angle. 
 
 Ex. 121. Prove that the altitudes drawn to the eqaal sides of an 
 isosceles triangle are equal. 
 
 Ex. 122. Prove that the medians drawn to the equal sides of an 
 isosceles triangle are equal. 
 
 Ex. 123. A boy wishes to make a saw-buck. Assume that B0= OD 
 and that Z EPF = 40°. Determine the angles at B 
 and D so that the pieces AB and CD will stand firmly 
 upon the ground. Determine the angles at C and A so 
 that CE and FA will be parallel to the ground line. 
 
 Ex. 124. Construct an angle of 60°. 
 
 (Construct an equilateral triangle.) 
 Also construct an angle of : 30° ; 120° ; 150°. 
 
 Ex. 125. If one angle of an isosceles triangle is 60°, the triangle is 
 equilateral. 
 
TRIANGLES 
 
 67 
 
 Ex. 126. The bisectors of the equal angles of an 
 isosceles triangle form with the base another isosceles 
 triangle. 
 
 Ex. 127. If the bisector of the exterior angle at one ^^ ^B 
 
 vertex of a triangle is parallel to the side joining the other two vertices, 
 the triangle is isosceles. 
 
 Ex. 128. If one acute angle of a right triangle is 
 30°, the side opposite is one-half the hypotenuse. 
 
 Suggestion. — Extend BC to Z), making CD equal to 
 
 BC. Prove A ABD is equilateral. _ 
 
 B 
 
 Note. — This is a very important exercise. Pupils 
 should endeavor to remember it, as it will be required in the proofs of 
 certain exercises in geometry. 
 
 Ex. 129. Prove that the perpendiculars drawn from the mid-point of 
 the base of an isosceles triangle to the sides of the triangle are equal. 
 
 Ex. 130. If the equal sides of an isosceles triangle 
 be extended beyond the base, the exterior angles so 
 formed are equal. 
 
 Ex. 131. Prove that the bisector of the exterior angle 
 at the vertex of an isosceles triangle is parallel to the 
 base of the triangle. 
 
 Suggestion. — Comi^2ire LBCD with LA-^ LB (§ 110). 
 
 Ex. 132. In the gable in the front of a garage, 
 the two boards whose upper edges are AB and A C 
 are of equal length and meet at a point .<4 on a line 'b 
 AD which is perpendicular to BC. 
 
 If ZACD = 30°, how large are Z ABD, Z CAD, 
 and Z BAD ? 
 
 Ex. 133. If the perpendiculars drawn from the mid-point of one side 
 of a triangle to the other two sides are equ^, the triangle is isosceles. 
 
 Ex. 134. If the altitudes drawn to two sides of a triangle are equal, 
 the triangle is isosceles. 
 
 r- 
 
 
 
 
 / 
 
 E 
 
 \ 
 
 .1 
 
 / 
 
 \. 
 
 Note. — Supplementary Exercises 40-49, p. 276, can be studied now, 
 
68 
 
 PLANE GEOMETRY — BOOK I 
 
 POLYGONS 
 
 125. A Polygon is a dosed (§ 6) broken line ; as ABCDE, 
 Points A, B, C, etc., are the vertices of 
 
 the polygon ; A A, B, C, etc., are the 
 angles ; AB, BC, CD, etc., are the sides ; 
 the sum of the lengths of the sides is the 
 perimeter oi the polygon; a line joining 
 any two non-consecutive vertices is a di- 
 agonal of the polygon ; as ^C. 
 
 A polygon incloses a portion of the plane called the in- 
 terior of the polygon. 
 
 126. A polygon is Convex if no side, 
 when extended, will pass through the inte- 
 rior of the polygon ; as ABCDE of § 125. 
 
 A polygon is Concave if at least two 
 sides, when extended, will pass through the 
 interior of the polygon ; as FGHIK. 
 
 127. Only convex polygons are considered in this text. A 
 convex polygon having n sides has n vertices. 
 
 128. An Equilateral Polygon is one whose sides are all equal. 
 An Equiangular Polygon is one whose angles are all equal. 
 
 129. Two polygons are mutually equilateral if the sides of 
 one are equal respectively to the sides of the other ; and mutu- 
 ally equiangular if the angles of one are equal respectively to 
 the angles of the other. If two polygons are both mutually 
 equiangular and mutually equilateral, they are congruent. 
 
 130. The principal polygons are named as follows : 
 
 No. OF Sides 
 
 Name of the Polygon 
 
 No. OF Sides 
 
 Name of the Polygon 
 
 3 
 4 
 
 6 
 6 
 
 Triangle 
 Quadrilateral 
 Pentagon 
 Hexagon 
 
 7 
 
 8 
 
 10 
 
 n 
 
 Heptagon 
 Octagon 
 Decagon 
 w-gon 
 
QUADRILATERALS 69 
 
 QUADRILATERALS 
 
 131. A Parallelogram (O) is a quadrilateral whose opposite 
 sides are parallel. 
 
 A pair of parallel sides are called bases; / 7 
 
 the perpendicular distance between them is / / 
 
 called the altitude. 
 
 132. Cor. Two consecutive angles of a parallelogram are 
 suiyplcmentary. 
 
 For, in the figure of § 131, since AB cuts the parallels AD and J5(7, 
 ZA+ZB=l8\../.. (§103.) 
 
 Ex. 135. Construct a CJ ABCD, making ^ / g \ 
 
 AD = 2 in., AB = S in., and B = 60°. After you / ~7 
 
 have constructed the figure, compare the opposite /\\ /3\ 
 
 sides by means of your dividers. e""' 
 
 Proposition XXIII. Theorem 
 
 133. A diagonal of a parallelogram divides it into 
 two congruent triangles. 
 
 Hypothesis. ABCD is a parallelogram. AC is a diagonal. 
 
 Conclusion. A ABC ^ A ACD. 
 
 [Proof to be given by the pupil.] 
 
 Suggestions. — 1. Since AD \\ BC, compare Z 1 and Z 2. 
 2. Compare Z3 and Z4. What are the parallels? 
 
 134. Cor. 1. Tlie opposite sides of a parallelogram are equal. 
 
 135. Cor. 2. TJie opposite angles of a 
 parallelogram are equal. ^~7 y ^ 
 
 136. Cor. 3. Segments of parallels in- Z-^ ^^^ 
 
 eluded between parallels are equal. 
 
70 PLANE GEOMETRY — BOOK I 
 
 Proposition XXIV. Theorem 
 
 137. The diagonals of a parallelogram bisect each 
 other. 
 
 Hypothesis. ABCD is a O. 
 
 Diagonals AC and BD intersect at E. 
 Conclusion. AE = EG\ BE = ED. 
 
 [Proof to be given by the pupil.] 
 
 Note. — The point of intersection of the diagonals of a parallelogram 
 is the Center of the parallelogram. 
 
 Ex. 136. If one angle of a parallelogram is 100°, how large is each of 
 the other angles ? 
 
 Ex. 137. If one angle of a parallelogram is a right angle, the others 
 are also, 
 
 Ex. 138. If two adjacent sides of a parallelogram are equal, all its 
 sides are equal. 
 
 Ex. 139. Two parallels are everywhere equidistant. 
 Hypothesis. CD || AB. 
 
 EF and GH are any two Js to CD and AB. 
 Conclusion. EF = GH. A- 
 
 F K " 
 
 Ex. 140. If perpendiculars BE and DF are drawn 
 to the diagonal ^O of a parallelogram ABCD, then BE = DF. 
 (Construct the figure with ruler and compasses.) 
 
 Ex. 141. If a line be drawn through the center of a parallelogram 
 and terminated by two opposite sides of the parallelogram, it is bisected 
 by the center. 
 
 Ex. 142. Construct the parallelogram whose diagonals are 2 in. and 
 3 in. respectively if the included acute angle is 45°. Measure the longer 
 and shorter sides of the parallelogram. 
 
QUADRILATERALS 71 
 
 Proposition XXV. Theorem 
 
 138. If two sides of a quadrilateral are equal and 
 parallel, the figure is a parallelogram. 
 
 rmC 
 
 Hypothesis. AB=CD', AB II CD. 
 
 Conclusion. ABCD is a parallelogram. 
 
 Plan. AD must be proved II to BC. Try to prove Z 1 = Z 2. 
 
 Proof. 1. lu A ABC and A ACD : 
 
 AB = CD and AC = AC\ Why ? 
 
 Z3 = Z4. 
 [Since lis ^B and CD are cut by ^C] Why ? 
 
 2. .-. A ABC ^ A ACD. Why? 
 
 3. .-. Z1 = Z2. Why? 
 
 4. .-. ^Z> II BC Why ? 
 
 5. .-. ABCD is a parallelogram. § 131 
 
 B, 
 
 Ex. 143. The line joining the mid-points of 
 
 two opposite sides of a parallelogram is parallel to \^y^ v /p. 
 
 the other two sides. /^*^ / ^ 
 
 (Prove AEFD \s Hl O and therefore EF || AD) ^ ° 
 
 Ex. 144. li ABCD is a parallelogram, and E and i^are the mid- 
 points of AB and CD respectively, then AECF is also a parallelogram. 
 
 Ex. 145. Prove that two straight lines are parallel if any two points 
 of one are equidistant from the other. C Recall § 84.) 
 
 Ex. 146. If the diagonals of a quadrilateral bisect each other, the 
 figure is a parallelogram, 
 
 Ex. 147. If from any point in the base of an isosceles 
 triangle parallels to the equal sides be drawn, the perimeter 
 of the parallelogram formed is equal to the sum of the 
 equal sides of the triangle. A^ ^ 
 
 Note. — Supplementary Exercises 50-52, p. 277, can be studied now. 
 
72 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XXYI. Theorem 
 
 139. If the opposite sides of a quadrilateral are 
 equal, the figure is a parallelogram. 
 
 2.' 
 
 -^0 
 
 Proof. 
 2. 
 3. 
 4. 
 
 Note. - 
 
 Hypothesis. AB=CD; BC = AD, 
 
 Conclusion. ABCD is a parallelogram. 
 
 Plan. Try to prove AB II CD, and AD II BC. 
 
 1. A ABC ^ A ACD. Give the full proof. 
 
 .-. Z 1 = Z 2, and hence BC II AD. Why ? 
 
 Also Z 3 = Z 4, and hence AB II CD. Why ? 
 
 .-. .4^ CD is aO. Why? 
 
 Another proof may be given, which is based upon § 138. 
 
 Ex. 148. If E, F, G, and H are mid-points of sides AB, BC, CD, 
 and AD respectively of parallelogram ABCD, then EFGH is a parallelo- 
 
 ^^^"^- ( Prove EF = HG and EH= FG.) 
 
 Ex. 149. Construct a parallelogram having sides 2 in. and 3 in. re- 
 spectively, and with included angle of 45°. Measure its longer diagonal. 
 
 140. There is an important applica- ^. __^^ 
 
 tion of parallelograms in science. If an ^^/ x^ 
 
 object is being pulled in the direction AB 
 
 with a force of 50 lb. and in the direc- ^ ^^^ 
 
 tion AC with a force of 100 lb., it will actually move in the 
 
 direction AD and as if pulled by a force which bears to 100 
 
 lb. the same relation that AD bears to AC. 
 
 Thus, AC= 1" and AD = 11"-, since AC represents 100 lb., 
 AD represents 125 lb. 
 
 Ex. 150. A steamer is being propelled east at the rate of 15 mi. an 
 hour ; the wind is driving it north at the rate of 5 mi. an hour. Deter- 
 mine by a construction the direction in which the steamer will travel and 
 its rate. 
 
QUADRILATERALS 73 
 
 SPECIAL PARALLELOGRAMS 
 
 141. A Rectangle is a parallelogram one of whose angles is 
 a right angle. It can be proved and it is impor- 
 tant to remember that all the angles of a rectangle 
 are right angles. * 
 
 Note. — Since a rectangle is a special parallelogram, every theorem 
 true about parallelograms is true about rectangles. Thus, the diagonals 
 of a rectangle bisect each other. On the other hand, theorems true about 
 a rectangle are not necessarily true about a parallelogram, since a rectangle 
 is a special parallelogram. 
 
 Ex. 151. State some other properties of a rectangle which follow at 
 once from properties of a parallelogram. (See §§ 133-137.) 
 
 Ex. 152. Construct a rectangle whose sides are 1.5 in. and 2 in. 
 respectively. Draw and measure its diagonals. 
 
 Ex. 153. Prove that the diagonals of a rectangle are equal. 
 
 Ex. 154. Prove that a quadrilateral whose angles are all right angles 
 is a rectangle. 
 
 Ex. 155. Prove that a parallelogram whose diagonals are equal is a 
 rectangle. B 
 
 Plan. Try to prove one of its ^i is a right angle. 
 
 (Recall § 116, G 1, and also § 103.) 
 
 Ex. 156. When laying out the lines for the foundation of a rec- 
 tangular building, as ABCD, contractors often measure off AD and DC 
 at riglit angles and of the required lengths. Then AB is measured off 
 equal to CD and at right angles to AD. (See figure for Ex. 155.) 
 
 (a) Why should ABCD then be a rectangle ? 
 
 (6) To test whether ABCD is a true rectangle, AG and BD are 
 measured. If they prove to be equal, it is concluded that the figure is a 
 rectangle. Is this a safe test ? Why ? 
 
 Note. — Supplementary Exercises 53-64, p. 278, can be studied now. 
 
 142. A Rhombus is a parallelogram having two adjacent 
 
 sides equal. It can he proved and it is important / 7 
 
 to remember that all the sides of a rhombus are / / 
 
 equal ; also it is usually implied that the angles Z / 
 
 are not right angles. (See § 143.) 
 
 Ex. 157. State properties of a rhombus which are evident at once 
 because the rhombus is a special parallelogram. (See Note, § 141.) 
 
74 
 
 PLANE GEOMETRY — BOOK I 
 
 Ex. 158. Prove that the diagonals of a rhombus are perpendicular 
 to each other. 
 
 Ex. 159. Prove that the diagonals of a rhombus bisect the angles. 
 
 Note. — Supplementary Exercises 55-57, p. 278, can be studied now. 
 
 143. A Square is a parallelogram having two adjacent sides 
 equal and one angle a right angle. It can he proved and it is 
 important to remember that 
 
 All the angles of a square are right angles and all the sides are 
 equal. 
 
 Note. — The square is a special rectangle and also a special rhombus. 
 Hence every theorem true about a rectangle or a rhombus is true about a 
 square. (See Note, § 141.) 
 
 144. Many artistic designs are made on a network of squares 
 as illustrated below. 
 
 
 !"S 
 
 SI 
 
 Iri: 
 
 "SP 
 
 H"ss:s 
 
 
 H«.. 
 
 .HHH. 
 
 
 
 ir-!:- 
 
 in. 
 
 i-f:.! 
 
 l:i 
 
 is!:! 
 
 :g; 
 
 Bs'nil 
 
 t^^• 
 
 m\n 
 
 
 ^■■■■■■■■■■niSaai 
 
 id>« 
 
     
 SS   
 
 
 
 1 
 
 1 
 
 1 
 
 
 «,7,t 
 
 [hhh. 
 
 
 jgS hi S.! ij !i li 
 
 
 loo 
 
 M 
 
 Corner and Border 
 
 Four Border Designs 
 
 Rug Designs 
 
QUADRILATERALS 75 
 
 Ex. 160. Make a list of facts about the square which may be inferred 
 from known facts about the parallelogram, the rectangle, and the 
 rhombus. 
 
 Ex. 161. How large are the angles into which a diagonal of a square 
 divides its angles ? 
 
 Ex. 162. Construct a square whose diagonals shall be 2 in. in length. 
 
 Ex. 163. Prove that the lines drawn from the ends of one side of a 
 square to the mid-points of the two adjacent sides are equal. 
 
 Ex. 164. Prove that if the diagonals of a quadrilateral ate perpen- 
 dicular to and bisect each other, the figure is a rhombus. 
 
 Ex. 165. If E, F, G, and H are points on the sides, AB, BC, CD, 
 and AD respectively of square ABCD, such that AE = BF = CG = DH, 
 prove that EFGH is a square. 
 
 Suggestions. — 1. Try to prove EFGH i& a O, having 
 two adj. sides equal, and having one Z{Z^) a right angle. 
 (§ 143.) 
 
 2. To prove Z4 a right angle: 
 
 (a)Zl+Z2=? (6) DoesZ3 = Z2? 
 
 (c) 2:i + Z3 + Z4= ? (d) .■.Zi= ? 
 
 Note. — Supplementary Exercises 58-60, p. 278, can be studied now. 
 
 TRAPEZOIDS 
 
 145. A Trapezoid is a quadrilateral which has one and only 
 one pair of parallel sides ; AB and CD are ^ 
 
 called the non-parallel sides. / \ 
 
 The parallel sides of a trapezoid are called the / \ 
 Bases. ^^ ^^ 
 
 The perpendicular distance between the bases is called the Altitude. 
 
 The line joining the mid-points of the non-parallel sides is called the 
 Median of the trapezoid. 
 
 146. An Isosceles Trapezoid is a trapezoid the non-parallel 
 sides of which are equal. 
 
 Ex. 166. Construct the trapezoid having lower base of 4 in., one of 
 its non-parallel sides 2 in., the angle between these two sides being 60°, 
 and the upper base being 1.6 in. 
 
 Ex. 167. If the angles at the ends of one base of a trapezoid are 
 equal, the angles at the ends of the other base are also equal. 
 
76 PLANE GEOMETRY — BOOK I 
 
 Ex. 168. If a" trapezoid is isosceles, the lower base angles are equal. 
 (If AB= CD, prove ZA = ZD. Draw BE || CD. g 
 
 Compare Z AEB with Z D and Z A.) A \ 
 
 Ex. 169. If one pair of base angles of a trape- / \ \ 
 
 zoid are equal, the trapezoid is isosceles. ^ E ^ 
 
 Ex. 170. Prove that the diagonals of an isosceles trapezoid are 
 equal. 
 
 Ex. 171. Prove that the opposite angles of an isosceles trapezoid are 
 
 supplementary. 
 
 Note. — Supplementary Exercises 61-63, p. 278, can be studied now. 
 
 Proposition XXYII. Theorem 
 147. If three or r)%ore parallels intercept equal lengths 
 on one transversal, they intercept equal lengths on all 
 transversals. 
 
 Al 
 
 \B 
 
 \L 
 
 )ih\D 
 
 M 
 
 r^l^F 
 
 Gl 
 
 kf\H 
 
 Hypothesis. AB !1 CD || EF II GH. 
 
 AG cuts the lis at A, C, E, and G. 
 BH cnts the lis at' 5, D, F, and H. 
 AC=CE = EG. 
 Conclusion. BD = DF= FH. 
 
 Plan. Try to prove BD, DF, and FH homologous sides of 
 cong. A. 
 
 Proof. 1. Draw BI, DJ, F/f parallel to AG. 
 
 §91 
 
 EG. Why ? 
 
 Hyp. 
 
 Ax. 1, § 51 
 
 [Complete the proof by proving ^ jBZ)/, DJF, and FHK 
 are congruent, and then proving that BD =. DF -— FH.'] 
 
 2 
 
 
 .'.BIWDJWFK 
 
 3. 
 
 Also 
 
 BI=AC, DJ= CE, and FK 
 
 4. 
 
 But 
 
 AC=CE=EG. 
 
 5. 
 
 
 .-. BI=DJ=FK. 
 
QUADRILATERALS 
 
 77 
 
 148. Cor. 1. If a line bisects one side of a triangle, and is 
 parallel to a second side, it bisects the third side also. 
 
 Hyp. I) is on AB of A ABC ; 
 
 AD = DB; DEWBG. 
 Con. AE = EC. 
 
 Proof. 1. Assume X4r II ^C 
 
 
 'K 
 
 
 c 
 
 f::A 
 
 \ 
 
 149. Cor. 2. If a line is parallel to the 
 
 bases of a trapezoid and bisects one of the 
 
 non-parallel sides, it bisects the other also. 
 
 B ' O 
 
 Note. — Supplementary Exercises 64-66, p. 278, can be studied now. 
 
 Proposition XXVIII. Problem 
 
 150. Divide a given segment into any number of 
 equal parts, ^ 
 
 Ct)(< 
 
 ^V< 
 
 I '1 
 
 P\' 
 
 I 
 I 
 
 I 
 
 "--^.v 
 
 IF 
 
 Given segment AB. 
 
 Required to divide AB into five equal parts. 
 Construction. 1. Draw line AC, making a convenient Z 
 with AB 
 
 2. Upon AC, lay off AD=DE = EF= FG = GIL 
 
 3. Draw IIB. 
 
 4. Through D, E, F, G, and H, draw lines parallel to IIB, 
 meeting? .17? at X, Y, Z, and W. 
 
 Statement. AX=XY=^YZ=Z W= WB. 
 Proof. 1. Assume RS through A parallel to HB. 
 
 2. .'. RSW DXWEYWFZWGWWHB. Why? 
 
 3. .-. AX=XY= YZ= ZW= WB. Why? 
 
 Note. — Supplementary Exercises 66-67, p. 279, can be studied now. 
 
78 PLANE GEOMETRY — BOOK I 
 
 Proposition XXIX. Theorem 
 
 151. If a line joins the midpoints of two sides of a 
 triangle, it is parallel to the third side and equal to 
 one half of it. 
 
 Hypothesis. D is the mid-point of AB, and E is the mid- 
 point of AC in A ABC. 
 
 Conclusion. DE W BG; DE = \ BC 
 
 Plan. Extend DE its own length to F. Try to prove 
 FE = BC, and FE II BG. To do this, try to prove FECB 
 is a lJ. 
 
 Proof. 1. Extend DE to F, making DF= DE. Draw BF. 
 
 2. .-.A FBD ^ A DAE. Give the proof. 
 
 3. .-. Z 1 = Z 2 ; and also BF =AE. Why ? 
 
 4. .-. BF II AC, and .-. BF II EC. Why ? 
 
 5. AhoBF=Ea Why? 
 
 6. .-. ^2^£;(7 is a parallelogram. - Why ? 
 
 7. .-. i^^ or DE is parallel to ^C. Why ? 
 
 8. Aho FE = BC, 3ind .: DE = i BC. Why? 
 Note. — This theorem is very important. 
 
 152. The proof of Proposition XXIX illustrates another 
 valuable device for proving theorems. 
 
 Principle III. To prove that one segment is double another, 
 either double the shorter and prove the result equal to the 
 longer, or halve the longer and prove the result equal to the 
 shorter. The first of these plans is followed in the proof of 
 Proposition XXIX; the second plan will be used in Propo- 
 sition XL. 
 
QUADRILATERALS 79 
 
 Ex. 172. The lines joining the mid-points of the sides of a triangle 
 divide it into four congruent triangles. 
 
 Ex. 173. If J57, F, G, and ^are the mid-points of the sides AB, BC, 
 CD, and AD respectively of a quadrilateral ABCD, then EFGH is a 
 parallelogram. (Draw AC and use Proposition XXIX.) This theorem 
 appeared in a book on geometry by Th. Simpson in 1760. a 
 
 Ex. 174. The lines joining the mid-points of the oppo- 
 site sides of a quadrilateral bisect each other. 
 
 Ex. 175. The mid-point of the hypotenuse of a right 
 triangle is equidistant from the vertices of the triangle. 
 
 (Let AE = EB. Vroxe ED ±AB. Then complete 
 the proof. ) B 
 
 Proposition XXX. Theorem 
 
 153. The median of a trapezoid is parallel to the 
 bases and equal to one half their sum, 
 
 A B 
 
 52^_^::_^:^., 
 
 Hypothesis. ABCD is a trapezoid. 
 
 E is the mid-point of AD and F of BC. 
 Conclusion. . EF II AB and DC. 
 
 EF=i(AB-\-DC). 
 Proof. 1. Extend DC to G, making CO = AB. Draw AC, 
 BG, and AG. 
 
 2. .'.ABGCissLCJ. Why? 
 [Since CG = AB, and CG II AB.-\ 
 
 3. .-. AG passes through i^and is bisected by it. § 137 
 
 4. .-. in A ADG, AE = ED and AF = FG. 
 
 5. .-. EF II DG ; and EF = i; DG. Why ? 
 
 6. .'. EF W DC B,nd AB. 
 
 7. Also EF = i (DC 4- AB), since DG= DC-\- AB. 
 
 Note. — Supplementary Exercises 68-74, p. 279, can be studied now. 
 
80 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XXXI. Theorem 
 154. The sum of the interior angles of a i^olygon 
 
 having n sides is (n — 2) straight angles. 
 
 Hypothesis. Assume a polygon of n sides, like ABCD ••• . 
 Conclusion. The sum of its int. A={n — 2) st. A. 
 Proof. 1. Draw diagonals from B to each of the other 
 vertices. 
 
 2. Each side of the polygon, excepting AB and BC, becomes 
 the base of a triangle whose vertex is at B. Hence there are 
 (n — 2) A formed. 
 
 That is, when n is 4, there are 2 A ; 
 when w is 5, there are 3 A ; 
 when n is 6, there are 4 A ; etc. 
 
 3. The sum of the int. A of each A is 1 st. Z. Why ? 
 
 4. . • . the sum of the int. A of the (n — 2) A is {n — 2) st. A. 
 
 5. But the sum of the int. A of the A = the sum of the 
 int. A of the polygon. 
 
 6. .*. the sum of the int. A of the polygon is (n — 2) st. A. 
 
 Ex. 176. Express in straight angles, in right angles, and in degrees 
 the sum of the angles of a polygon having : 
 
 {a) four sides ; (6) five sides ; (c) six sides ; (d) eight sides. 
 
 Ex. 177. How many degrees are there in each angle of an equi- 
 angular : 
 
 {a) quadrilateral? (6) pentagon? (c) hexagon? {d) octagon? 
 
 Ex. 178. If two angles of a quadrilateral are supplementary, then 
 the other two are also. 
 
 Ex. 179. How many sides has a polygon the sum of whose angles is 
 16 right angles ? 7 straiglit angles ? 1620 degrees ? 
 
POLYGONS 81 
 
 Proposition XXXII. Theorem 
 
 155. If the sides of any polygon he extended in order 
 to form an exterior angle at each vertex, the sum of 
 these exterior angles is two straight angles. 
 
 Hypothesis. Assume a polygon of n sides. 
 
 Extend the sides as in the figure. 
 Conclusion. The sum of ext. A like Z 1, Z 2, Z 3, etc. = 
 
 2 St. A. 
 Proof. 1. The sum of the int. Z and the ext. Z at each 
 vertex = 1 st. Z. § 39 
 
 2. .'. the sum of all the int. and ext. A = n st. A. Why? 
 
 3. But the sum of all the int. Z =(?i - 2) st. A. § 154 
 
 4. .*. the sum of all the ext. Z = 2 st. A. 
 
 Note. — Propositions XXXI and XXXII were proved in their general 
 form by Regiomontanus (1436-1476), although the theorems were known 
 to earlier mathematicians and were proved by them for special cases. 
 
 Ex. 180, Prove the theorem of § 154 by drawing 
 lines from any point within the polygon to the vertices. ^,^^'''T^"""~~--^ 
 (Recall § .35.) ^-- i--'^ 
 
 Ex. 181. State and prove the converse of § 135. \ / ' n / 
 
 Suf/f/estion. — Apply § 154 and § 98. \/'' -J 
 
 Ex. 182. How many sides are there in the polygon 
 the sum of whose interior angles exceeds the sum of its exterior angles 
 by 540^^ ? 
 
 Ex. 183. How many sides has a polygon the sum of whose interior 
 angles equals four times the sum of its exterior angles ? 
 
82 PLANE GEOMETRY — BOOK I 
 
 INEQUALITIES 
 
 156. The symbol for " less than " is < ; for ^^ greater than " 
 is >. 
 
 157. Order of Inequalities, a <h and c < d are two in- 
 equalities of the same order, m < w and x ^ y are two in- 
 equalities of opposite orders. 
 
 158. Axioms for combining Inequalities. 
 
 Ax. 17. If equals be added to unequals, the sums are unequal 
 in the same order. 
 
 Thus, if a < &, then a + c <Cb ■{■ c. 
 
 Ax. 18. If equals he subtracted from unequals, the differences 
 are unequal in the same order. 
 
 Thus, if a < 6, then a — c<ih — c. 
 
 Ax. 19. If unequals be added to unequals in the same order, 
 the sums are unequal in the same order. 
 
 Thus, if a < 6, and c < (?, then a + c < 6 -1- d. 
 
 Ax. 20. If unequals be subtracted from equals or from un- 
 equals of opposite order, the differences are unequal and of order 
 opposite to that of the subtrahend. 
 
 Thus, if a > 6, and c < <?, then a — c > 6 — d. 
 Arithmetical Example. — Since 12 > 7 and 3 < 5, then 12 — 3 should 
 be greater than 7 - 5. Is it ? 
 
 Ax. 21. Ifa>b and b > c, then a > c. 
 
 Ex. 184. Given an arithmetical example for each of the axioms. 
 
 159. Fundamental Inequalities for Segments. 
 
 (a) Any side of a triangle is less thari the sum 
 of the other two sides. 
 
 This follows from Ax. 11, §51. 
 Thus BC<AB-\-AC. 
 
INEQUALITIES 83 
 
 (b) Any side of a triangle is greater than the 
 difference between the other two sides. 
 
 Thus, BOAC-AB. 
 
 For, from (a) BC + AB > AC. Subtracting AB 
 from both members of the inequality, BO AC— g 
 AB, by Ax. 18, § 158. 
 
 Note. — Ex. 188, p. 86, and Supplementary Exercises 75-84, p. 280, 
 can be studied now. 
 
 160. Fundamental Inequality for Angles. 
 
 An exterior angle of a triangle is greater than either remote 
 interior angle of the triangle. (§ 87.) 
 
 Proposition XXXIII. Theorem 
 
 161. If hvo sides of a triangle are unequal, the 
 angles opposite are unequal, the angle oppjosite the 
 greater side being the greater. 
 
 Hypothesis. In A ABC, AC > AB. 
 
 Conclusion. ZB > Z.C. 
 
 Proof. 1. Since AC > AB, take AD = AB. Draw BD. 
 
 § 13 
 
 2. .-. Z1=Z2. Why? 
 
 3. Z 2 is an exterior angle of A BDC. Def. 
 
 4. .-. Z2> Z C. Why? 
 
 5. .-. Z1>ZC. Why? 
 
 6. But ZABC>Z1. Ax. 8, §51 
 
 7. .-. Z ABC > ZC Ax. 21, § 158 
 
 Ex. 185. If a triangle is scalene, all its angles are unequal. 
 
84 
 
 PLAJNTE GEOMETRY — BOOK I 
 
 Proposition XXXIV. Theorem 
 
 162, If tivo angles of a triangle are uJiequaL the 
 sides opposite are uneqiml, the side opposite the greater 
 angle being the greater. 
 
 Hypothesis. Jn A ABC, Z.C < Z B. 
 
 Conclusion. AB < AC. 
 
 Proof. 1. Since Z B > Z C, construct BD, making Z1=ZC. 
 
 2. .'.BD = DC. §123 
 
 3. In A ABD, AB < AD -{- BD. § 159, a 
 
 4. .-. AB<AD + DC, or AB < AC. 
 [Substitute DC for its equal, DB.] 
 
 163. Cor. 1. The hypotenuse of a right tri- 
 angle is greater than either leg of the triangle. 
 
 164. Cor. 2. The perpendicular from a point 
 to a line is the shortest segment from the point to 
 the line. 
 
 165. Cor. 3. If two oblique .segments, 
 drawn from a point in a perpendicular 
 to a line, cut off unequal distances from 
 the foot of the perpendicular, the more 
 remote is the greater. 
 
 Hyp. CD±AB; ED> BF. Con. CE > CF. 
 
 Suggestions. — 1. Take DH= BF, and draw CH. 
 
 2. Prove CH= CF. 
 
 3. Prove Z 2 > Z 1, by comparing eacb with Z 3. 
 
 4. Then complete the proof. 
 
 % 
 
 \ 
 ^ ^ 
 
 i 
 
INEQUALITIES 
 
 85 
 
 Ex. 186. If is any point in the base BC oi isosceles triangle ABC^ 
 then AO is less than AC. (Prove Z^OOZ^CO.) 
 
 Ex. 187. Prove that the median to any side of a triangle is greater 
 than the altitude to that side unless the side is the base of an isosceles 
 triangle. 
 
 Proposition XXXV. Theorem 
 
 166. If two triangles have tioo sides of one equal 
 respectively to two sides of the other, hut the included 
 angle of the first greater than the included angle of the 
 second, then the third side of the first is greater than 
 the third side of the second. 
 
 D 
 
 Hypothesis. In A ABC, and A DEF: 
 
 AB=DE', AC=DF', /.BAOZ.D. 
 Conclusion. BC > EF. 
 
 Proof. 1. Place A DEF in the position ABG, side DE 
 coinciding with its equal AB. 
 
 2. DF falls within Z BAC, taking the position AG. Why ? 
 3*. Construct AH bisecting Z GAC, and meeting BC at H. 
 Draw GH. 
 
 A GAH ^ A ACIL Give the full proof. 
 
 .-. GH= CH. 
 
 In A BUG, BH+ GH>BG. 
 
 BH+CH> BG, or BC > BG. 
 
 Note. — If (r falls on BC, then EF is at once less than BC. 
 within A ABC, the proof is similar to that given in the text. 
 
 Why? 
 
 Why? 
 
 Wliy ? 
 
 If G falls 
 
86 
 
 PLANE GEOMETRY — BOOK I 
 
 Proposition XXXYI. Theorem 
 
 167. If two triangles have two sides of one equal 
 respectively to two sides of the other, hut the third side 
 of the first greater than the third side of the second, 
 then the angle opposite the third side of the first is 
 greater than the angle opposite the third side of the 
 second. 
 
 B 
 Hypothesis. In A ABC and A I)EF: 
 
 AB = DE, AC = BF', BOEF. 
 Conclusion. ^A>Z B. 
 
 Proof. 1. Suppose that Z Ais not greater than Z D ; that 
 is, that Z A either equals Z D or is less than Z D. 
 
 2. 
 
 li ZA 
 
 Why? 
 
 Hyp. 
 
 §94 
 
 §166 
 
 Hyp. 
 
 Z B, then A ABC ^ A BEF. 
 [Give the full proof.] 
 
 ThenBC=EF. 
 But BC> EF. 
 .-. Z A cannot be equal to Z B. 
 liZA<ZB, then BC < EF. 
 But BC > EF. 
 .-. Z A cannot be less than Z B. 
 Since Z A cannot be equal to Z B or less than Z B, then 
 Z A must be greater than Z B. 
 
 Ex. 188. If is any point within A ^J5C, then ^0+0(7 <^^+^ (7. 
 Suggestions. — 1. Extend CO until it intersects 
 AB at R. 2. Compare BO with BR + RO. 
 
 3. Add OC to both members of the inequality. 
 
 4. Compare RO+OC with RA-jrAC, and com- ^ — a^ ^ 
 
 plete the proof. 
 
SUPPLEMENTARY THEOREMS 87 
 
 SUPPLEMENTARY THEOREMS 
 
 168. Three or more lines ordinarily do not pass through a 
 common point. Three or more lines which do pass through a 
 common point are called Concurrent Lines. 
 
 Proposition XXXVII. Theorem 
 
 169. The bisectors of the interior angles of a tri- 
 angle meet at a point ivhich is equidistant from the 
 sides of the triangle. 
 
 Hypothesis. In A ABC : 
 
 AD bisects Z^ A, BE bisects Z B] Ci^ bisects Z C. 
 Conclusion. AD, BE, and CF meet at a point which is 
 equidistant from the sides of A ABC. 
 
 Proof. 1. Let AD and BE meet at point P. Note 1 
 
 2. Since P is in AD, it is equidistant from AC and AB. 
 
 § 120, I 
 
 3. Since P is in BE, it is equidistant from AB and BG. 
 
 4. .-. P is equidistant from AC and BC. Ax. 1, § 51 
 
 5. .-. P lies in CF, the bisector of Z C. § 120, II 
 
 6. Hence AD, BE, and CF meet at P, which is equidistant 
 from AB, AC, and BC. 
 
 Note 1. — This fact may be assumed as evident from the figure, or 
 may be proved as follows : 
 - 1. If AD does not intersect BE, then AD II BE. 
 
 2. Then Z DAB + Z EBA = 1 St. Z. (§ 103.) ' 
 
 3. But this is impossible, since Z DAB + Z EBA < 1 st. Z. Why ? 
 
 Note 2. — The point of intersection of the bisectors of the interior 
 angles of a triangle is called the In-center of the triangle. (See § 226.) 
 
88 PLANE GEOMETRY — BOOK I 
 
 Proposition XXXVIII. Theorem 
 
 170. The perpendicular-bisectors of the sides of a tri- 
 angle r)ieet at a point which is equidistant from the 
 vertices of the triangle. 
 
 Hypothesis. In A ABC, FK, DO, and EH are the perpen- 
 dicular-bisectors of AB, BC, and AC, respectively. 
 
 Conclusion. FK, DO, and EH meet at a point which is 
 equidistant from A, B, and C. 
 
 Proof. 1. Let FK and DO meet at point 0. Note 1 
 
 2. Since is in FK, O is equidistant from A and B. Why ? 
 
 3. Since is in DO, is equidistant from B and C Why ? 
 
 4. .•. is equidistant from A and C. Ax. 1, § 51 
 
 5. .-.0 lies in EH, or ^^ passes through 0. § 118 
 
 6. Hence the perpendicular-bisectors are concurrent at a 
 point which is equidistant from A, B, and C 
 
 Note 1. — This fact may be assumed or be proved as follows : 
 
 1. If i^^does not intersect GD, then FKW GD. 
 
 2. .-. AB, which is ± to FK, is also ± to GD. 
 
 3. But BD ± GD. 
 
 4. .-. either AB 11 jBZ>, or AB coincides with BD. 
 
 5. But this is impossible, since AB and BD intersect. 
 
 Note 2. — The point of intersection of the perpendicular-bisectors of the 
 sides of a triangle is called the Circum-center of the triangle, for a circLe 
 can be drawn with il as center which will pass through the vertices of the 
 triangle. 
 
 Ex. 189. Construct a circle which will pass through the vertices of 
 the triangle the sides of which are 3 in., 3 in., and 4 in., respectively. 
 
SUPPLEMENTARY THEOREMS 
 
 89 
 
 Proposition XXXIX. Theorem 
 171. The altitudes of a triangle meet at a point. 
 
 Hypothesis. In A ABC^ AD, BE, and CF are the altitudes 
 from A, B, and C respectively. 
 
 Conclusion. AD, BE, and CF meet at a point. 
 
 Proof. 1. Drawjy/rthrough^llto^C; 7t 6^ through 5 II to 
 AC; and GH through C II AB. These parallels form A HKG. 
 
 2. Since AD 1 BC, then AD ± HK. Why ? 
 
 3. KACB and ABCII are [U. Why ? 
 
 4. KA = ^(7, and ^£^ = -BO. Why ? 
 
 5. .'. /f^ = AH, and ^Z) is the perpendicular-bisector of KH. 
 
 6. Similarly BE and 02^ can be proved to be the perpendicu- 
 lar-bisectors of KG and GH respectively. 
 
 7. .-. AD, BE, and CF in A HKG meet at a point. § 170 
 
 Note. — The point of intersection of the altitudes of a triangle is 
 called the Ortho-center of the triangle. 
 
 Ex. 190. If /is the ortho-center (Note, § 171) and 
 J is the circum-center (Note, § 170) of triangle ABC^ 
 then BI=2JK and AI=2 JL. 
 
 Suggestions. —1. Recall § 152. 
 
 2. Prove 3/iV^II KL, BIW JK, and AIW JL. 
 
 3. Recall §105. 
 
 Ex. 191. Does the ortho-center of a triangle necessarily fall inside 
 the triangle ? 
 
 Note. — Supplementary Exercise 86, p. 281, can be studied now. 
 
90 PLANE GEOMETRY — BOOK I 
 
 Proposition XL. Theorem 
 
 172. The medians of a triangle meet at a point 
 which lies tivo thirds the distance from each vertex to 
 the mid-point of the opposite side, 
 
 C 
 
 Hypothesis. AD, BE, and CF are the medians of A ABC. 
 
 Conclusion. AD, BE, and CF meet at a point which lies 
 two thirds the distance from each vertex to the mid-point of 
 the opposite side. 
 
 Proof. 1. Let AD and BE meet at point 0. Note 1, § 169. 
 
 2. Let Gf and H be the raid-points of AG and BO respec- 
 tively. Draw ED, GH, EG, and DH. 
 
 3. Then, in A AOB, GH^^AB and GH II AB. Why ? 
 
 4. Similarly, ED = \AB and ED II AB, 
 
 5. .'.EDHG is 2i a. Why? 
 
 6. .-. GD and EH bisect each other. Why ? 
 
 7. .-. OD=OG = AG, and EO=OH= HB. 
 
 8. HcQce AD and BE meet at a point which lies two thirds 
 the distance from ^ to D and from B to E. 
 
 9. In like manner, AD and CF meet at a point which lies 
 two thirds the distance from A to D and from C to F. On 
 AD, this is point 0. 
 
 10. Hence the three medians meet at point 0, which is two 
 thirds the distance from each vertex to the mid-point of the 
 opposite side. 
 
 Note. — The point of intersection of the medians of a triangle is called 
 the Center of Gravity of the triangle. 
 
 This theorem was known to Archimedes. 
 
MISCELLANEOUS EXERCISES 91 
 
 Exercises Solved by Indirect Proofs 
 Ex. 192. If two straight lines are cut by a transversal, and a pair of 
 alternate interior angles are unequal, the lines are not parallel. 
 Suggestion. — Review § 94. 
 
 Ex. 193. If two lines are cut by a transversal and the sum of the 
 
 interior angles on the same side of the transversal is not equal to two 
 right angles, the lines are not parallel. 
 
 Ex. 194. If a point is unequally distant from the ends of a segment, 
 it is not in the perpendicular-bisector of the segment. 
 
 Ex. 195. If a point is not equidistant from the sides of an angle, it is 
 not in the bisector of the angle. 
 
 Ex. 196. Prove that the two altitudes of a parallelogram which has 
 two unequal sides are unequal. 
 
 Ex. 197. If two unequal oblique segments be drawn from a point to a 
 straight line, the greater cuts off the greater distance from the foot of the 
 perpendicular from the point to the line. 
 
 Suggestion. — Recall § 165. 
 
 Miscellaneous Exercises 
 
 Ex. 198. If D, E, and JP'are points on the sides AB, 
 BC\ and AC respectively of equilateral triangle ABC^ 
 such that AD = BE = CF, then A DEF is also equilat- 
 eral. '"'"T 
 
 Ex. 199. Prove that the bisectors of a pair of vertical angles form a 
 straight line. 
 
 Ex. 200. If two lines are cut by a transversal so that a pair of exterior 
 angles on the same side of the transversal are supplementary, the lines are 
 parallel. 
 
 Ex. 201. If perpendiculars be drawn to the sides of an acute angle 
 from a point outside of the angle, they form an angle equal to the given 
 angle. 
 
 Ex. 202. If through any point D in one of the equal 
 sides AB of isosceles A ^-BC, DFhe drawn perpendicular 
 to base BC, meeting CA extended at E, then A ADE is 
 isosceles. 
 
 SuqgeHtion. — Compare LE with Z C, and LBDF vj'Mh 
 LB.' B 
 
 Ex. 203. Prove that the altitudes drawn to homologous sides of con- 
 gruent triangles are equal. 
 
92 
 
 PLANE GEOMETRY — BOOK I 
 
 Ex. 204. If D is mid-point of side BG of A ABC, 
 and BE and CF are perpendiculars from B and C to 
 AD, extended if necessary, prove BE = CF. 
 
 Ex. 205. If a line be drawn through the vertex of 
 
 an isosceles triangle parallel to the base, it bisects the 
 exterior angle at the vertex. 
 
 Ex. 206. Prove that the segments bisecting the base angles of an 
 isosceles triangle and terminating in the opposite sides are equal . 
 
 Ex. 207. If a line be drawn through a point in the ' 
 bisector of an angle parallel to one side of the angle, the 
 bisector, the parallel, and the other side of the angle form 
 an isosceles triangle. qi^^ q 
 
 Ex. 208. If the median to the base of a triangle is perpendicular to 
 the base, the triangle is isosceles. 
 
 Ex. 209. Prove that the sum of the perpendiculars 
 drawn from any point in the base of an isosceles tri- 
 angle to the equal sides of the triangle is equal to the 
 altitude drawn to one of the equal sides. 
 
 Prove 0D+ 0F= CE. 
 
 Suggestions. — 1. Draw OG 1 CE. 
 
 2. Compare OD and EG ; also OF and CG. 
 
 Ex. 210. If two parallels are cut by a transversal, 
 the bisectors of the four interior angles form a rec- 
 tangle. 
 
 Suggestions. — 1. EFGH must be proved a O and 
 one Z. must be proved a right angle. 
 
 2. Recall §§ 93, 103, 106. 
 
 Ex. 211. If the mid-point of any side of a square is joined to the two 
 vertices of the opposite side, the lines so drawn are equal. 
 
 Ex. 212. Prove that the lines drawn from the mid-point of the base of 
 an isosceles triangle tcr mid-points of the sides of the triangle form with the 
 half sides a rhombus. 
 
 Ex. 213. If the lower base AD of trapezoid ABCD is double the upper 
 base BG, and the diagonals intersect at E, prove that GE'm | ^^and that 
 BE \^ I ED. (§152.) 
 
 Ex. 214. If D is any point in side ^(7 of A ABC and E, F, G, and H 
 are the mid-points of AD, CD, BG, and AB, respectively, then EFOH 
 is a parallelogram. 
 
 Suggestion. — Draw BD. 
 
 Note. — Supplementary Exercises 86-108, p. 281, can be studied now. 
 
BOOK II 
 
 THE CIRCLE 
 
 173. Review the definitions given in § 16 and § 17, and the 
 Exercises 31--34, Introduction. The symbol for circle is O. 
 The circle whose center is 0, is denoted by O 0. 
 
 174. Since a circle is a closed line (§ 6), it incloses a portion 
 of the plane called its interior. 
 
 Ex. 1. Draw a circle with radius 1 in. Where will a point lie : («) if 
 its distance from the center is f in. ? (6) If its distance from the center 
 is 1.5 in. '? 
 
 Ex. 2. Draw a circle with diameter 5 in. Cut the circle from paper. 
 Prove, by folding it, that any diameter bisects the circle and the surface 
 within the circle. 
 
 Ex. 3. Draw two circles which intersect. From one of the points of 
 intersection draw the radius of each circle. How does the distance be- 
 tween their centers compare with the sum of their radii ? 
 
 Ex. 4. Prove that a diameter of a circle is greater 
 than any other chord of the circle. 
 
 Suggestion. — Compare CD with AO and OB. 
 Also, compare A0-\- OB with AB. 
 
 Ex. 5. If two circles intersect, the distance between 
 their centers is greater than the difference of their radii. 
 
 175. From the exercises following § 17 and § 174, the fol- 
 lowing facts are evident : 
 
 (a) If a straight line cuts a circle, it intersects it in two and 
 only two points. 
 
 (b) If two circles intersect , they have two and only two points 
 of intersection. 
 
 93 
 
94 PLANE GEOMETRY — BOOK II 
 
 (c) A point is within, on, or outside a circle if its distance from 
 the center is less than, equal to, or greater than the radius. 
 
 (d) A diameter of a circle bisects the circle and the surface with- 
 in it; also, if a line bisects a circle j it is a diameter of the circle. 
 
 The theorem (d) was known to Thales. 
 
 176. One half of a circle is called a Semicircle. 
 A quarter of a circle is called a Quadrant. 
 
 Circles having the same center are called Concentric Circles. 
 
 Peoposition I. Problem 
 
 177. Construct a circle ivhich loill pass through three 
 points which are not in a straight line. 
 
 Given points A, B, and G which are not in a straight line. 
 
 Required to construct a circle which will pass through A, B, 
 and a 
 
 Construction. 1. Draw AB and BC. 
 
 2. Construct the JL bisectors oi AB and BC, meeting at 0. 
 
 Statement. A circle drawn with as center and OA as 
 radius will pass through A, B, and C. 
 
 Proof. OA=OB= Oa Why ? 
 
 Note, — Only one circle can be drawn through three points, for the 
 center must lie on each of the perpendicular-bisectors (§ 118, II) and 
 these lines can intersect at only one point. 
 
 Ex. 6. What would happen if the three points in Proposition I were 
 in a straight line ? 
 
THE CIRCLE 
 
 95 
 
 Ex. 7. (a) Construct a circle which will pass through two given 
 points. 
 
 (6) How many circles can be constructed through two 
 given points ? 
 
 (c) Where do the centerp of all these circles lie ? 
 
 Eac. 8. Construct full size the design for a four-inch 
 square tile. Make the decorative arcs f in. wide. 
 
 178. A polygon is said to be inscribed in a 
 circle when its vertices lie on the circle ; as 
 ABCD. The circle is said to be circumscribed 
 about the polygon. 
 
 CHORDS, ARCS, AND CENTRAL ANGLES 
 
 179. Two points on a circle are the ends of 
 two arcs ; a Minor Arc, as AMB, and a Major A 
 Arc, as ANB. 
 
 Unless the contrary is stated, the minor arc will 
 always be understood when an arc is indicated by 
 means of its extremities. Thus, arc AB means minor 
 arc AB. 
 
 An arc AB will be indicated by a small arc drawn over AB ; as AB. 
 
 180. A Central Angle is an angle whose ver- 
 tex is at the center and whose sides are radii 
 of the circle ; as Z ^OC 
 
 Z ^OC is said to intercept AC. 
 
 AC is said to be iiitercepted by Z. AOC, 
 
 " Intercept " is derived from two Latin words meaning 
 
 between 
 
 and "to take," so that it means " to take between." 
 
 Eac. 9. Construct a circle with radius 2 in. (a) Construct two cen- 
 tral angles which are equal, and a third central angle which is greater 
 than each of the equal central angles. (6) Cut from the paper the two 
 equal central angles. Compare their intercepted arcs by superposition. 
 
 (c) Cut from the paper the third central angle; compare its inter- 
 cepted arc with the arc intercepted by one of the other two angles. 
 
 (d) What do you conclude must be true about the arcs intercepted by 
 equal central angles of a circle ? By unequal central angles ? 
 
96 PLANE GEOMETRY — BOOK II 
 
 Proposition II. Theorem 
 
 181. In the same circle or in equal circles^ if central 
 angles are equal, they intercept equal arcs. 
 
 Hypothesis. © = O i? ; Z AOB = Z CRD. 
 
 Conclusion. AB ^ CD. 
 
 Proof. 1. Place O on Q R, with point on point R, 
 and so that Z AOB coincides with its equal, Z CRD. 
 
 2. Then O will coincide with Q) R. § 17 
 
 3. Point A will fall on point C, since OA = RC § 17 
 
 4. Point B will fall on point D, since OB = RD. § 17 
 
 5. .-. AB coincides with CD and hence AB = CD. 
 
 Ex. 10. Divide a circle into four equal arcs. 
 What kind of central angles must be constructed ? 
 
 Ex. 11. Divide a circle into eight equal arcs. 
 Ex. 12. Divide a circle into -six equal arcs. 
 
 How large must the central angles be ? Recall § 70 and Ex. 113, Book I. 
 
 Ex. 13. Using the construction made in Ex. 12, /^TV^ 
 
 draw a six-pointed star. A"/ ' ^/\ 
 
 Ex. 14. Tell how you can divide a circle into five \/\ /\j 
 equal parts by means of your protractor and straight- \/\'7'y 
 edge. 
 
 Ex. 15. By means of compass, ruler, and protractor, L- — / \ — A 
 draw a five-pointed star in a circle with 2 in. radius, to I ^ ^ j 
 be used as a pattern for a star on a sailor collar. \ //\\/ 
 
THE CIRCLE 
 
 97 
 
 Proposition III. Theorem 
 
 182. In the sam,e circle or in equal circles, if arcs are 
 equal, the central angles ivhich intercept them are equaL 
 
 Hypothesis. Q 0= Q B, AB= CD. 
 
 Conclusion. Z AOB = Z CRD. 
 
 Proof. 1. Since O = O i2, and J2 = CD, the O can 
 be made to coincide with the O Rj and AB with CD, A 
 falling on C, B on D, and on R. 
 
 2. .-. AO will fall on CR and BO on DR. Ax. 10, § 15 
 
 3. .-. Z AOB= Z CRD. Why ? 
 
 Ex. 16. If a radius bisects an arc, it is perpendicular 
 to and bisects the chord which subtends the arc. 
 
 Ex. 17. The twelve spokes of a wheel are spaced 
 so that the points at which they are attached to the rim 
 divide the rim into equal arcs. How many degrees are 
 there in the angle formed by two adjacent spokes ? 
 
 183. It may be proved that : in, the same circle or in eqxtal 
 circles, 
 
 (a) The greater of two unequal central angles intercejits thi 
 greater arc; 
 
 (b) The greater of two unequal arcs is intercepted by the 
 greater central angle. 
 
 184. A chord AB is said to subtend arc AB. Arc AB is 
 said to be subtended by chord AB, 
 
 *' Subtend " is derived from Latin words meaning " to stretch under." 
 
98 
 
 PLANE GEOMETRY — BOOK II 
 
 Pkopositiois' ly. Theorem 
 
 185. In the same circle or in equal circles, if chords 
 are equal, they subtend equal arcs. 
 
 Hypothesis. O = Q R) AB= CD. 
 
 Conclusion. AB = CD. 
 
 Plan. 1. Draw AO, OB, RC, RD. 
 
 2. Prove Z = Z i?, and apply § 181. 
 
 [Proof to be given by the pupil.] 
 
 Ex. 18. Construct an equilateral triangle. Circumscribe a circle 
 about the triangle. (§ 177.) Prove that the vertices of the triangle di- 
 vide the circle into three equal arcs. 
 
 Proposition V. Theorem 
 
 186. In the same circle or in equal circles, if arcs 
 are equal, the chords which subtend them are equal. 
 
 Hypothesis. QO = Q R; AB=CD. (Fig. § 185.) 
 
 Conclusion. AB = CD. 
 
 Plan. Try to prove AAOB^A CRD. Compare Z and 
 
 Z R. § 182 
 
 [Proof to be given by the pupil.] 
 
 Q 
 
 Ex. 19. If C is the mid-point of arc AB, prove that 
 AC is greater than one half AB. j^ 
 
 Draw CB. Compare AB with AC + CB. 
 
THE CIRCLE 
 
 99 
 
 Proposition VI. Theorem 
 
 187. In the same circle or in equal circles, if two 
 minor arcs are unequal, then their chords are unequal, 
 the greater arc heing subtended by the greater chord. 
 
 Hypothesis. O = O /2 ; AB> CD. 
 
 Conclusion. AB > CD. 
 
 Proof. 1. Draw radii AO, BO, CR, and DR. 
 
 2. In A^O^and AC/eZ>: 
 AO=CRsind BO = DR', 
 
 but since AB> CD, Z 0> ZR. 
 
 3. .-. AB > CD. 
 
 Hyp. 
 
 § 183, b 
 §166 
 
 Proposition VII. Theorem 
 
 188. In the same circle or in equal circles, if two 
 chords are unequal, then they subtend unequal minor 
 arcs, the greater chord subtending the greater arc. 
 
 Hypothesis. Q = Q R; AB> CD. (Fig. § 187.) 
 
 Conclusion. AB > CD. 
 
 Plan. Prove ZO>ZR(% 167) and apply § .183, a. 
 
 D 
 
 Ex. 20. Prove that the straight line which bisects 
 the arcs subtended by a chord bisects the chord at right 
 angles. 
 
 Suggestion. —CompaiTe AD and BD; also AC and BC. 
 Apply § 77. 
 
100 
 
 PLANE GEOMETRY — BOOK II 
 
 Pkoposition Yin. Theorem 
 
 189. If a diameter is perpendicular to a chord, it 
 bisects the chord and its subtended arcs. 
 
 D 
 
 /i ix 
 
 V / 3 4 X 
 
 ^K — Y~~y 
 
 Hypothesis. In O 0, diameter CD ± AB at E. 
 
 Conclusion. AE = EB; AC=CB', smd AD = DB. 
 
 Proof. 1. Draw AO and OB. 
 
 2. A AEO ^ A OEB. Give the full proof. 
 
 3. .'. AE = EB, and Z.1 = A2. Why? 
 
 4. ,\AC=GB. §181 
 
 5. Also, AD = DB. Why? 
 
 190. Cor. 1. A line through the center of a circle perpen- 
 dicular to a chord bisects the chord. 
 
 191. Cor. 2. The p>erpendicular-bisector of a 
 chord passes through the center of the circle, and 
 bisects the arcs subtended by the chord. A 
 
 Compare AO and OB. Then use § 118, II 
 
 Ex. 21. If a radius of a circle bisects a chord, it is perpendicular to 
 the chord and bisects the subtended arcs. 
 
 Ex. 22. Determine the center and the radius of the circle of which 
 AB is an arc. 
 
 Draw any two chords and 
 erect the ± bisectors. These 
 must pass through the center. 
 (§ 191.) A B 
 
THE giRCLE 101 
 
 Proposition^' 'IX.. Xu^o,rsm, 
 
 192. In the same circle' or in equal circles , if chords 
 are equal, they are equidistant from the center. 
 
 Hypothesis. In OABC: 
 
 AB=GD; OE±AB', OF A. CD. 
 Conclusion. OE = OF. 
 
 Proof. 1. Draw OA and 0(7. 
 
 2. AE = \AB,CF^\CD,zxiiiAB=CD. Why? 
 
 3. .'.AE=CF. Ax. 6, §51 
 
 4. .-. A AEO ^ A CFO. Give full proof. 
 6. .'.OE=OF. Why? 
 
 Ex. 23. If two intersecting chords are equal, the radius drawn 
 through the point of intersection bisects the angle between them. 
 
 Ex. 24. If a straight line bisects a chord and its subtended arc, 
 then it is perpendicular to the chord. (§ 186. ) 
 
 Ex. 25. If a straight line is drawn cutting two concentric circles in 
 the points A, B^ C, and D, respectively, then AB equals CD. 
 
 Ex. 26. Prove that the perpendicular-bisectors of the sides of an 
 inscribed polygon are concurrent. (§ 168.) 
 
 Ex. 27. On equal chords of a circle, points are taken at equal 
 distances from the ends of the chords. Prove that all these points are 
 equidistant from the center of the circle. 
 
102 PLANE GEOMETRY — BOOK II 
 
 Propqsitiok . X. Theorem 
 
 193. In the same circle or in equal circles, if chords 
 are equidistant from the center, they are equal. 
 
 ^^ — "^^^ 
 
 Hypothesis. 
 
 In O^BC: 
 OE^^AB; 0F1.CD) OE=OF. 
 
 
 Conclusion. 
 
 AB = CD. 
 
 
 Proof. 1. 
 
 Draw OA and OC. 
 
 
 2. Then 
 
 A OEA^AOCF 3ind AE = OF. 
 [Give the full proof.] 
 
 
 3. But 
 
 AB = 2AE and CD = 2 CF. 
 
 Why? 
 
 4. 
 
 .-. AB = CD. 
 
 Why? 
 
 tJx. 28. If a straight line be drawn parallel to the line connecting 
 the centers of two equal circles, and intersecting the circles, then the 
 chords formed in the two circles are equal. 
 
 Ex. 29. If two chords, intersecting within the circle, make equal 
 angles with the radius passing through their point of intersection, they 
 are equal. 
 
 Suggestion. — DrsiW Js to the chords from the center; prove the Js are 
 equal ; then use § 193. 
 
 Ex. 30. If two equal chords of a circle intersect, the segments of one 
 equal respectively the segments of the other. 
 
 194. A straight line which intersects a circle in two points 
 is called a Secant of the circle. 
 
 Ex. 31. If ABE and DCE are two secants 
 of a circle which make equal angles with the 
 line connecting E with the center of the circle, 
 then chord AB = chord DC. (Use § 193.) 
 
THE CIRCLE 103 
 
 Proposition XI. Theorem 
 
 195. In the same circle or in equal circles^ the less of 
 tivo unequal chords is at the greater distance from the 
 center of the circle. 
 
 E 
 
 Hypothesis. In O : 
 
 AB<CD; OF±AB', OG ± OD. 
 
 Conclusion. OF > OG. 
 
 Proof. 1. Since AB < CD, AB < CD. § 188 
 
 2. Let CE = AB. Draw CE. 
 
 3. .\CE = AB. Why? 
 
 4. Draw OH A, CE, intersecting CD at K. 
 
 5. .-. 011= OF. § 192 
 
 6. But OH > OK. Why ? 
 
 7. .\OF>OK. Why? 
 
 8. OK>OG. Why? 
 
 9. .\OF>OG. Why? 
 
 Ex. 32. All equilateral tiiani^'le and a square are inscribed in a circle. 
 Prove that the sides of the triangle are nearer the center than the sides of 
 the square. 
 
 Ex. 33. Chord BY is drawn through one extremity of a diameter 
 AB of circle 0. Radius OX is drawn in Z^BF parallel to BY, inter- 
 secting arc ^ F at X Prove arc ^X equals arc XF. 
 
 Suggestion. — Draw radius OY. 
 
 Ex. 34. AB is a diameter of a circle and XF is an intersecting 
 diameter of a smaller concentric circle. Prove AXB Y is a parallelogram. 
 
104 
 
 PLANE GEOMETRY — BOOK II 
 
 Proposition XII. Theorem 
 
 196. In the same circle or in equal circles, if tivo 
 chords are unequally distant from the center, the more 
 remote is the smaller. 
 
 Hypothesis. In O O : 
 
 OELAB', OF±CD, 0E> OF. 
 Conclusion. AB < CD. 
 
 Proof. 1. Suppose that AB is not less than CD ; that is, 
 suppose that AB = CD or AB > CD. 
 
 2. If AB = CD, then OE = OF. 
 
 But OE > OF. 
 .'.AB is not = to CD. 
 
 3. liAB> CD, then OE < OF. 
 But OE > OF. 
 
 .'. AB is not greater than CD. 
 
 4. ..AB<CD. 
 
 Why? 
 Why? 
 
 §195 
 Hyp. 
 
 197. Tangent Line. Assume that the secant AB turns about 
 the point A in the direction indicated by 
 the arrow. The point B moves closer to 
 the point A. When B finally coincides 
 with A, the line assumes the position XY. 
 XYis called a tangent to the circle. 
 
 A tangent to a circle is a straight line 
 which touches the circle at only one point. 
 
 The circle is also said to be tangent to the line. 
 
 The point where the tangent touches the circle is called the 
 Point of Tangency, or Point of Contact. 
 
THE CIRCLE 
 
 105 
 
 Pkoposition XIII. Theorem 
 
 198. A straight line perpendicular to a radius at its 
 outer extremity is a tangent to the circle. 
 
 Hypothesis. 
 
 Conclusion. 
 
 Proof. 1. 
 
 2. 
 
 3. 
 
 4. 
 
 OC is a radius of O ; ^J5 ± OC at G. 
 
 AB is tangent to O 0. 
 
 Let D be any point in AB except O. 
 Draw OD. 
 .'. OD > 00. Why ? 
 
 point D lies outside of the ©. § 175, c 
 
 5. .*. every point in AB except lies outside the circle, 
 and hence AB is tangent to the circle. § 197 
 
 199. Cor. 1. A tangent to a circle is 
 perpendicular to the radius drawn to the 
 point of contact. 
 
 Since all points in AB except C lie outside the 
 circle, OC Is the shortest segment to AB from 0. 
 Hence OCXAB. 
 
 200. Cor. 2. A line perpendicular to a 
 tangent at its point of contact passes through 
 the center of the circle. 
 
 By Cor. 1, the radius 00 is ± to AB. Hence 
 
 OC and CD must coincide. (Why?) .• 
 passes through the center of the circle. 
 
 CD 
 
 201. Cor. 3. A line from the center of the circle perpen- 
 dicular to a tangent passes through the point of contact. 
 
106 
 
 PLANE GEOMETRY — BOOK II 
 
 Proposition XIV. Theorem 
 
 202. The tangents to a circle from an outside point 
 are equal. 
 
 Hypothesis. AB and AC are tangents to O 0. 
 Conclusion. AB = AC. 
 
 [Proof to be given by the pupil. Recall § 114.] 
 
 Note. — The proof of Prop. XIV is attributed to a mathematician Fink, 
 with the date 1583. The theorem does not appear in Euclid at all. It 
 appears first as a definite theorem in writings of Hero, although it was 
 apparently used by Archimedes. 
 
 Ex. 35. Prove that the tangents to a circle at the extremities of a 
 diameter are parallel. ^ 
 
 Ex. 36. If two circles are concentric, any two chords 
 of the greater which are tangents of the smaller are C 
 equal. 
 
 Ex. 37. Prove that all tangents drawn from the larger of two con- 
 centric circles to the smaller are equal. 
 
 Ex. 38. Prove that the line joining the center of a circle to the point 
 of intersection of two tangents : (a) bisects the angle formed by the 
 radii drawn to the points of contact ; (6) bisects the angle formed by the 
 tangents ; (c) bisects and is perpendicular to the chord joining the points 
 of contact. c 
 
 Ex. 39. Prove that the sum of two opposite sides of ^ 
 a circumscribed quadrilateral is equal to the sum of the e 
 other two opposite sides. 
 
THE CIRCLE 
 
 107 
 
 203. A straight line tangent to each of two circles is called 
 a Common Tangent of the circles; 
 as AB. 
 
 If the circles lie on opposite sides 
 of AB, AB is called a common in- 
 ternal tangent. 
 
 If the circles lie on the same side 
 of AB, AB is called a common ex- 
 ternal tangent. 
 
 The length of a common tangent 
 is the length of the segment between 
 the two points of contact. 
 
 Some uses of common tangents 
 are pictured in the figures below. 
 
 Belts abound Pulleys 
 
 Chain around Wheels 
 
 Ex. 40. Prove that the coniraon internal tangents of two circles are 
 equal. 
 
 Ex. 41. Trove that the common external tangents of two circles are 
 equal, when the circles are unequal. 
 
 Note. — The theorem is also true when the circles are equal. This 
 might be solved as an optional exercise. 
 
 204. Two circles are tangent when 
 they are tangent to the same straight 
 line at the same point. 
 
 They are tangent externally if they 
 lie on opposite sides of the common 
 tangent. 
 
 They are tangent internally if they 
 lie on the same side of the tangent. 
 
 Note. — Supplementary Exercises 1-6, p. 
 283, can be studied now. 
 
108 PLANE GEOMETRY — BOOK II 
 
 Proposition XV. Theorem 
 
 205. If two circles are tangent to each other, their 
 line of ceyiters passes through the point of co7itact. 
 
 Hypothesis. (D and 0' are both tangent to AB at A. 
 
 00' is the line of centers. 
 Conclusion. St. line 00' passes through A. 
 Proof. 1. Draw the radii OA and O'A. 
 
 2. Then OA J_ AB and also O'A _L AB. Why ? 
 
 3. .-. OAO' is a st. line. § 40 
 
 4. .-. St. lines 00' and OAO' coincide. Ax. 10, § 51 
 
 5. .-. 00' passes through A. 
 
 Note. — The theorem has been proved for two © tangent externally. 
 As an optional exercise, it is suggested that the theorem be proved when 
 the (D are tangent internally. 
 
 206. Cor. If the distance between the centers of two circles 
 equals the sum of their radii, the circles are tangent externally. 
 
 For then a point A can be taken on 00' so that OA = one radius and 
 then O'A = the other radius. A perpendicular to 00' at J. will then be 
 tangent to each of the circles. Hence the circles are 
 tangent (§ 204). 
 
 Ex. 42. Study the adjoining figure to determine how 
 to construct it. Construct a figure like it, having the radius 
 of the large circle 1 in. and that of the small circles | in. 
 
 Ex. 43. How many common tangents do two circles have 
 (a) Which are tangent internally ? 
 (&) Which are tangent externally ? 
 
THE CIRCLE 
 
 109 
 
 Proposition XVI. Theorem 
 
 207. If tivo circles intersect, the straight line joining 
 their centers bisects their common chord at right angles. 
 
 Hypothesis. (D and 0' intersect at A and B. 
 AB is the common chord and 00' is the line of centers. 
 
 Conclusion. 00' A. AB and 00' bisects AB. 
 
 Suggestion. — Btslw 6 A, OB, O'A, and O'B. (Apply § 77.) 
 
 Ex. 44. If two circles and 0' intersect at points A and B^ and if 
 00' intersects O O at Xand O 0' at F, then Xand Fare each equidis- 
 tant from A and B. 
 
 Ex. 45. If a straight line be drawn through 
 the point of contact of two circles which are 
 tangent externally, terminating in the circles, 
 the radii drawn to its extremities are parallel. 
 
 Note. — The theorem is stated for two circles which are tangent exter- 
 nally. Investigate its truth for two circles which are tangent internally. 
 
 Ex. 46. If two circles are tangent to each other externally at point 
 A, the tangents to them from any point in their common tangent which 
 passes through A are equal. 
 
 Ex. 47. If two circles are tangent to each other 
 externally at point A, the common tangent which passes 
 through A bisects the other two common tangents. 
 
 Ex. 48. AB and AC are the tangents to a circle from point A^ and 
 D is any point in the smaller of the arcs subtended by the chord BC. If 
 a tangent to the circle at D meets AB at F, and AC at F, prove the per- 
 imeter of A AFF = AB + A'^. 
 
no 
 
 PLANE GEOMETRY — BOOK II 
 
 Proposition XYII. Theorem 
 208. Parallel lines intercept equal arcs on a circle. 
 Case I. When one line is a tangent and one a secant: 
 
 E 
 
 Hypothesis. 
 Conclusion. 
 
 AB 
 
 is tangent to O CED at E ; 
 secant CD li AB. 
 
 CE = DE.' 
 
 
 Proof. 1. 
 
 2. 
 
 3. 
 
 4. 
 
 
 Draw diameter EF. 
 ..EF±AB. 
 .'. EF± CD. 
 .: CE = DE. 
 
 Why? 
 Why? 
 
 § 189 
 
 Case II. When both lines are secants . 
 
 Hypothesis. AB and CD are II secants 
 
 of O ABDC. 
 Conclusion. AC = BD. 
 
 Proof. 1. Assume J5;6^i^ tangent to the 
 O at 6r, and parallel to CD. 
 
 E O F 
 
 (Complete the proof. Compare AG and BG ; also CG and D^. ) 
 
 E 
 
 Case III. When both lines are tan- 
 gent: q 
 
 Hypothesis. AEB and CFD are 
 
 1! tangents to the O at JS' and F. 
 
 Conclusion. EGF= EHF\ 
 
 [Proof to be given by the pupil.] C" 
 
THE CIRCLE 
 
 111 
 
 Ex. 49. Prove that the straight line joining the points of contact 
 of two parallel tangents of a circle is a diameter of the circle. 
 Ex. 50. Prove that an inscribed trapezoid must be isosceles. 
 
 Q 
 
 Ex. 51. The adjoining figure gives the method of b^^^^^^^^'^^^^^^d 
 construction of one form of mansard roof. The 
 chords AB, BC, CD, and DE are equal. 
 
 (a) Construct such a figure for a roof whose span AE is 28', using the 
 scale J" = 1'. 
 
 (6) Is the line BD parallel to AE ? Prove it. 
 
 209. Theorems concerning tangents and tangent circles have 
 unusually wide application in design. 
 
 Direction is naturally indicated by a straight line. 
 
 On a circle, the direction is constantly changing. It is con- 
 venient in both pure and applied mathematics 
 to speak of the direction of a curve at a point ; 
 also it is agreed that this direction shall be the 
 same as the direction of the tangent to the 
 curve at the point. It is this fact which is 
 used in a variety of ways. 
 
 If a road turns a corner as pictured, there is an 
 abrupt change of direction. If a street car line runs 
 along the road, such an abrupt change in direction in 
 the tracks is impossible. For that reason, the arrange- 
 ment of tracks indicated in the adjoining 
 
 figure is employed. A car running from C ^_^ 
 
 towards D passes readily from CA to the arc ^_ 
 
 AB, for on both the straight line and the arc 
 the direction at A is the direction of line 
 CA ; similarly at B. 
 
 Ex. 52. Where must the center be lo- 
 cated in order that the circle will be tangent to 
 both CA and BD in the last figure for § 209 ? 
 
 Ex. 53. What kind of circles should 
 circles AB and EF be ? 
 
 Ex. 54. Determine how to construct 
 the figures at the right. Construct such 
 figures in circles with diameter 3 in. 
 
 ~7| B 
 -'O 
 
112 PLANE GEOMETRY — BOOK II 
 
 MEASUREMENT OF ANGLES AND ARCS 
 
 210. To measure a given magnitude, two steps are necessary. 
 
 (a) Select a quantity of the same kind to be used as the unit 
 of measure. 
 
 (b) Determine the number of times the given magnitude con- 
 tains the unit of measure. This number is called the Numerical 
 Measure of the quantity in terms of the unit employed. 
 
 If the quantity contains the unit itself or any part of it an 
 integral number of times, the quantity can be measured exactly. 
 
 If the quantity does not contain the unit of measure an in- 
 tegral number of times, the quantity can be measured only 
 approxi7nateli/. 
 
 Thus, the diagonal of a square whose side is 1 in. is known to be 
 1.414 + in., where the decimal is a "never ending " decimal. 
 
 211. Two magnitudes of the same kind are said to be Com- 
 mensurable when each contains the same unit of measure, called 
 a Common Measure, an integral number of times. 
 
 Thus, two segments whose lengths are 2| in. and 3\ in. respectively 
 are commensurable, for the common measure ^ in, is contained in the first 
 segment 10 times and in the second 13 times. 
 
 Two magnitudes of the same kind are said to be Incommensur- 
 able when no unit of measure can be found which is. contained 
 an integral number of times in each. 
 
 The diagonal and the' side of a square are incommensurable. 
 
 212. The Ratio of two magnitudes of the same kind is the 
 quotient of their numerical measures in terms of a common 
 measure. 
 
 Thus, the segments of lengths 2^ in. and 3| in., in § 211, have the 
 ratio \^ . 
 
 Ex. 55. What is the measure of a yard in terms of the unit : {a) 1 ft. ? 
 (6) 1 in. ? (c) \ in. ? 
 
 Ex. 56. What is the measure of 1 gallon in terms of the unit : (a) 1 
 qt.? (ft) 1 pt.? (c) 1 gill? 
 
 Ex. 57. What is the ratio of 2 yd. to 1^ ft.? 
 
MEASUREMENT OF ANGLES AND ARCS 113 
 
 Proposition XVIII. Theorem 
 
 213. In the same circle or in equal circles, two central 
 angles have the same ratio as their intercepted arcs. 
 
 Case I. When the angles are commensurable : 
 Hypothesis. In O 0, Z AOB and Z BOC are commensur- 
 able. 
 ConclusioD. ^QB^AB, 
 
 ZBOC ^ 
 
 Proof. 1. Z AOB and Z BOC have a common measure. 
 
 §211 
 
 2. Let the common measure be Z AOD, and let it be con- 
 tained in Z.AOB 4 times and in Z BOC 3 times. 
 
 3. ...^^^ = 1 2^2 
 
 ZBOC 3 _ 
 
 4. The radii drawn from in step (2) divide AB into 4 and 
 BC into 3 arcs which are all equal. Why ? 
 
 6. .-.^ = 1 §212 
 
 6. .*. from steps (3) and (5), 
 
 Z BOC BC 
 
 Case II. TF^e^i the angles are incommensurable : 
 
 The theorem is true also in this case. The proof presents 
 
 certain difficulties which it is wise to postpone at this time. 
 
 This proof is taken up in § 423. 
 
114 PLANE GEOMETRY — BOOK II 
 
 Ex. 58. In the adjoining figure, compare AB and 
 BC. Also compare AB and DC; also BC and DC. 
 
 Ex. 59. A right central angle is what part of a 
 straight angle ? What part therefore is its intercepted 
 arc of a semicircle ? 
 
 Ex. 60. A 60° angle is what part of the perigon ? 
 What part therefore is its intercepted arc of the whole 
 circle ? 
 
 Ex. 61. If a circle is divided into 5 equal parts, what part of the 
 perigon is the central angle which intercepts one of the parts? How 
 many degrees are there in the central angle ? 
 
 Ex. 62. If AB^ any chord of circle O, is extended to a point C so that 
 BC equals the radius of the circle, and CO is drawn, cutting the circle at 
 Z and E respectively, then AE = 3 • BZ. 
 
 Suggestions. — 1. Draw OA and OB. 2. FroYe A AOE = 3 jLBOC. Use 
 § 87 and § 69. 
 
 214. Measuring Angles and Arcs. In § 28, the unit for 
 measuring angles is given as 1 degree, J^- of 
 a right angle or g-^-g- of the perigon. This will 
 be called for the present one angular-degree. 
 Let Z AOB represent 1°. Similarly, we shall 
 speak of angular-minutes and angular-seconds. 
 Thus, 60 angular-seconds equal one angular- 
 minute ; and 60 angular-minutes equal one angular-degree. 
 
 Let a circle be drawn around point as center, and the 
 radii which divide the perigon into 360 equal central angles 
 be imagined. These angles are angular-degrees. They will 
 intercept 360 equal arcs on the circle. Let AB represent one 
 of these arcs. It is the unit for measuring arcs on this circle 
 and on any equal circle. It will be called one arc-degree. 
 
 Evidently on a circle with longer radius, the arc corresponding to AB 
 will be longer. 
 
 In similar manner, each arc-degree could be divided into 60 
 equal parts, called arc-minutes, and each arc-minute into 60 
 equal arc-seconds. A central angle of one angular minute 
 intercepts an arc of one arc-minute. 
 
MEASUREMENT OF ANGLES AND ARCS 115 
 
 215. A central angle has the same measure as its inter- 
 cepted arc, ivhen angular-degrees and arc degrees are used as the 
 respect ice units of measure. 
 
 Let Z AOB represent 1 angular-degree and AAOC any- 
 other central angle. 
 
 Then ZAOC ^AC ^ 213 
 
 ZAOB ^ ^ 
 
 Z. AOO AC 
 
 But — is the numerical measure Z AOC, and — — is 
 
 ZAOB ^ ^ 
 
 the numerical measure of AC, by the definition. 
 
 Hence the measure of Z AOC equals the measure of AC. 
 
 Thus, if Z AOC = 57.29 angular-degrees, then AV= 57.29 arc- 
 degrees. 
 
 From now on, it will be understood that angles are measured 
 in terms of angular-degrees, and arcs in terms of arc-degrees. 
 Also, the following statement of the theorem of § 215 will be 
 employed for convenience : 
 
 A central angle is measured by its intercepted arc. 
 
 Ex. 63. What is an arc-degree ? An angular- degree ? 
 
 Ex. 64. Are all angular-degrees of the same size ? 
 Are all arc-degrees of the same size : (a) on the same or on equal 
 circles ? (6) on Unequal circles ? 
 
 Ex. 65. If ABCD is an inscribed square and is the center of the 
 circle, how may degrees are there in AB ? in Z AOB ? 
 
 Ex. 66. A ABC is an equilateral triangle inscribed in a circle with 
 center O ; how many degrees are there in AB ? in ZAOB ? 
 
 216. An angle is said to be an Inscribed Angle when its 
 vertex is on the circle and its sides are chords 
 of the circle ; as Z ADC. 
 
 Z ABC intercepts the AC; ^C is intercepted 
 hyWxQZABC. 
 
 Such an angle is said to be inscribed in a g 
 
 circle or may be said to be inscribed in the arc ABC 
 
116 
 
 PLANE GEOMETRY — BOOK II 
 
 Proposition XTX. Theorem 
 
 217. An inscribed angle is measured hy one half its 
 intercepted arc. 
 
 Case I. Wlieu 07ie side of the angle is a diameter : 
 
 A 
 
 Hypothesis. 
 
 Conclusion. 
 
 Proof. 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 i^l. 
 
 Why? 
 Why? 
 
 §215 
 
 AC is a diameter ; AB is any other chord of 
 
 OO. _ 
 
 Z BAC is measured by i BC. 
 Draw BO. 
 Z1 = ZB+ZA, 
 ZB = Z A. 
 .-. Z 1 = 2 . Z ^, or Z ^ = 
 
 But Z 1 is measured by BC. 
 .-. Z A is measured by ^ BG. 
 
 Case II. When the center of the Q is 
 within the angle : 
 
 Hypothesis. Center lies within 
 
 . scribed Z BAC. 
 Conclusion. Z.BAC is measured 
 
 iBC. 
 Proof. 1. Draw diameter AD. 
 
 2. Then Z 1 is measured by | BD, 
 and Z 2 is measured by i DC Case I 
 
 3. .-. Z 1 + Z 2 is measured by | (BD + DC), Ax. 3, § 51 
 or Z BAC is measured by | BC, 
 
MEASUREMENT OF ANGLES AND ARCS 
 
 117 
 
 Case III. When the center of the O lies 
 outside the angle : 
 
 Hypothesis. Center lies outside in- 
 scribed Z BAC. 
 Conclusion. Z BAC is measured by 
 \BC. 
 
 Suggestions. — 1. Z BAD is measured by what ? 
 2. LCAD'i 
 
 218. Cor. 1. An angle inscribed in a 
 semicircle is a right angle. 
 
 (If BCis a diameter, prove Z A = \ri. A.) 
 
 Inscribed angles which intercept the sayne arc 
 
 219. Cor. 2. 
 
 are equal. 
 
 Ex. 67. Three consecutive sides of an inscribed quadrilateral sub- 
 tend arcs of 82°, 90°, and 60° respectively. Find each angle of the quad- 
 rilateral. 
 
 Ex. 68. Construct a line perpendicular to a 
 given segment at one extremity of the segment. 
 
 Take any point not in segment AB and draw 
 a circle with as center and OB as radius, cut- 
 ting AB at D. Draw DO meeting the circle at 
 E. Then EB ± AB at B. Prove it. 
 
 Ex. 69. If chords AB and CD intersect at E within the circle, prove 
 that A AEC and A BDE are mutually equiangular. 
 
 Ex. 70. If chords AB and CD extended meet outside the circle at 
 point E^ prove A ADE and A BCE are mutually equiangular. 
 
 Ex. 71. Prove that the opposite angles of an in- 
 scribed quadrilateral are supplementary. 
 
 (Z B is measured by what ? Z 2) ? .'. /. B + /.D'}) 
 
 Note. — Supplementary Exercises 9 to 20, p. 284, can be studied now. 
 
118 
 
 PLANE GEOMETRY — BOOK II 
 
 Proposition XX. Theorem 
 
 220. TJie angle formed hy a tangent and a chord 
 draivn to the point of contact is measured hy one half 
 its intercepted arc. 
 
 A ^..^^ E 
 
 Hypothesis 
 
 Conclusion. 
 
 Proof. 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 Why? 
 Why? 
 
 Why? 
 
 AE is tangent to O CBB at B ; BC is a chord. 
 Z ABC is measured by \ BC. 
 Draw diameter BD\ then BT) J_ AE. 
 Z ABD = 90°, and BCD = 180°. 
 .*. Z ABD is measured by \ BCD. 
 Z CBD is measured by ^ CD. 
 
 Z ABD - Z CBD is measured by i ^OT- i CD. 
 
 Ax. 4, § 51 
 
 6. .-. Z J.5(7 is measured by i ^O. 
 
 7. Similarly, Z.EBC is measured by i J3Z>0. 
 
 Ex. 72. If, in the figure for Prop. XX, BC = llO'^, how many- 
 degrees are there in Z ABC and Z ^50 ? 
 
 Ex. 73. If tangents are drawn to a circle at the extremities of a 
 a chord, they make equal angles with the chord. 
 
 Ex. 74. If two tangents drawn from a point to a circle form an 
 angle of 60°, then each of the tangents equals the chord joining the points 
 of contact. (Prove the triangle is equilateral.) 
 
 Ex. 75. If a tangent be drawn to a circle at the extremity of a chord, 
 the line joining the mid-point of the intercepted arc to the point of con- 
 tact bisects the angle formed by the tangent and the chord. 
 
 Ex. 76. Prove that a tangent to a circle at the mid-point of an arc 
 is parallel to the chord of the arc. (§ 93.) 
 
MEASUREMENT OF ANGLES AND ARCS 119 
 
 Proposition XXI. Theorem 
 
 221. TJie angle formed hy two chords intersecting 
 within a circle is measured by one half the sum of the 
 arcs intercepted hy it and its vertical angle. 
 
 Hypothesis. Chords AB and CD intersect at E within 
 
 OO. _ _ 
 
 Conclusion. Z 1 is measured by ^ {AC + DB). 
 
 Proof. 1. Draw CB. 
 
 2. Z1 = Z3 + Z2. Why ? 
 
 3. Z 3 is measured by J AG. Why ? 
 
 4. Z 2 is measured by | DB. Why ? 
 6. ,-. Z 1 is measured by ^ {AC + DB)- Ax. 3, § 51 
 
 Es. 77. li AC= 70° and Z)J5 = 50°; how many degrees are there in 
 AAECf 
 
 Ex. 78. liAC= 74° and A AEC = 50°, how large is ^ ? 
 
 Suggestion. — Let DB = z°. 
 
 Ex. 79. If two chords intersect at right angles within a circle, the 
 sum of one pair of opposite intercepted arcs is equal to a semicircle. 
 
 B 
 
 Ex. 80. If AX = (fr &nd AB = CB, prove 
 A iif iVJ5 is an isosceles triangle. „j 
 
 Note. — Supplementary Exercises 21 to 23, p. 286, can be studied now 
 
120 PLANE GEOMETRY — BOOK II 
 
 Proposition XXII. Theorem 
 
 222. The angle formed hy tivo secants intersecting 
 outside the circle is measured hy one half the differ- 
 ence between its intercepted arcs. 
 
 Hypothesis. Secants AB and CD intersect at E outside 
 Q 0. _ _ 
 
 Conclusion. Z ^ is measured by ^ (AC — DB). 
 
 Proof. 1. /.l = /.A-\-Z.E. Why? 
 
 2. .-. Z.E = Z 1 — Z A By algebra. 
 
 [Complete the proof. Obtain the measures of Z 1 
 and Z.A and then determine the measure of Z -E".] 
 
 223. Cor. 1. The angle formed hy a A 
 secant and a tangent is measured by one 
 half the differeyice between its intercepted 
 arcs. 
 
 Prove Z ^ is measured by 
 i (BC - DB). 
 
 224. Cor. 2. The angle formed by 
 two tangents is measured by one half 
 the difference between its intercepted 
 arcs. 
 
 Prove Z ^ is measured by ^ 
 
 l(BFD 
 
 BCD). 
 
MEASUREMENT OF ANGLES AND ARCS 121 
 
 Ex. 81. If in § 222 J^ = 100° and BD = 40°, how large is ZE? 
 
 Ex. 82. K in § 222 AC is a quadrant, and ZE is 40°, how large is BD ? 
 Suggestion. — Let BD = z°. 
 
 Ex. 83. If AC in the figure of Prop. XXII is 120°, and ZA = 15°, 
 how large is ZE? 
 
 Ex. 84. If in the figure for ^ 22S Z E = 50° and BD = 70°, how 
 
 how large is BFC ? 
 
 Ex. 85. If in the figure for § 224, BED = | of the circle, how Urge 
 isZ^? 
 
 Ex. 86. If in § 223 BEG = 100°, and CD = 200°, how many degrees 
 are there in angle E ? 
 
 Ex. 87. If ^B is the common chord of two inter- 
 secting circles, and AC and AD are diameters drawn 
 from A, prove that line CD passes through B. 
 
 Suggestioji. — Draw CB and BD, and try to prove 
 CBD a straight line. 
 
 Ex. 88. A square ABCD is inscribed in a circle. A tangent is 
 drawn to the circle at point A. How large is the angle formed by the 
 tangent and side AB? 
 
 Ex. 89. The line joining the mid-points of the arcs subtended by the 
 sides AB and ^C of inscribed A ABC cuts AB at F and AC at G. 
 VToyeAF = AG. 
 
 Ex. 90. If AB and AC are two chords of a circle making equal 
 angles with the tangent at A^ prove AB = AC. 
 
 Ex. 91. If ABCD is an inscribed quadrilateral, and AD and BC ex- 
 tended meet at P, the tangent XY at P to the circle circumscribed about 
 the A ABP is parallel to CD. 
 
 Suggestions. — 1. XY\\ CD if ZDCP = ? 
 
 2. Compare each of these angles with LB AD. Recall Ex. 73. 
 
 Note. — Supplementary Exercises 24 to 30, p. 286, can be studied now. 
 
 225. A circle is said to be inscribed in a 
 polygon when it is tangent to each side of 
 the polygon. 
 
 The polygon is said to be circumscribed 
 about the circle ; as EFGH, 
 
122 
 
 PLANE GEOMETRY — BOOK II 
 
 Pkopositioi^ XXIII. Problem 
 226. Inscribe a circle in a given triangle. 
 
 A 
 
 Given A ABC. 
 
 Required to inscribe a O in A ABG. 
 
 Construction. 1. Construct the bisectors BE and AD oi Z.B 
 and Z A respectively, meeting at point 0. 
 
 2. Construct OMl^AG. 
 
 3. With as center and OiJf as radius, draw a O. 
 Statement. This circle will be tangent to AB, BC, and AC. 
 Proof. 1. is equidistant from the sides of the triangle. 
 
 §169 
 
 2. .'. Js from to the sides are all equal to OM. 
 
 3. .-. AB, BG, and AG are tangents of O 0. § 198 
 
 Note 1. — Point is the point which was called the In-center of the 
 triangle in § 169. The reason is clear now. 
 
 Note 2. — A circle can be constructed which is 
 tangent to the sides AB and AC prolonged and to 
 BG as in the adjoining figure. It is called an 
 Escribed Circle and its center is called an Ex- 
 center of the triangle. 
 
 There are three ex-centers for each triangle. 
 
 Ex. 92. Consti-uct a triangle and construct 
 Its three escribed circles. 
 
 Ex. 93. If is the center of the circumscribed circle of A ABC and 
 OD is drawn perpendicular to BC, prove Z BOD — /.A. 
 
MEASUREMENT OF ANGLES AND ARCS 123 
 
 Proposition XXIY. Problem 
 
 227. I. Construct a tangent to a circle at a point 
 on the circle. 
 
 Given O and point A on it. 
 
 Required to construct a tangent to O at A. 
 
 Construction indicated in the figure. 
 
 [Description and proof to be given by the pupil.] 
 
 II. Construct a tangent to a circle from a 2^oint 
 outside the circle. 
 
 Given. O and point A outside O 0. 
 
 Required to construct a tangent to O from point A. 
 
 Construction. 1. Draw AO. 
 
 2. Construct a O on ^0 as diameter intersecting O at J5 
 and C. 
 
 3. Draw AB and AC. 
 Statement. AB and AC are both tangent to O O. 
 
 [Proof to be given by the pupil.] 
 Suggestion. — Draw OB and OC. Prove LB and Z. C are rt. A. 
 
124 PLANE GEOMETRY — BOOK II 
 
 LOCI 
 
 228. Illustrative Problem 1. — Where are all jioints J 
 in. from 0.? 
 
 Evidently the place of points \ in. from 
 is the circle with center and radius 
 iin. 
 
 Instead of using the word " place '^ it is 
 customary to use the word locus — a Latin 
 word meaning place. So the preceding 
 sentence becomes 
 
 The locus of points \ in. from is the circle with center 
 and radius \ in. 
 
 It is evident that : 
 
 (a) Every point "i in. from 0'' is on the circle. 
 
 (6) Every point on the circle is " \ in. from 0.'' 
 
 " ^ in. from '' is the condition which the points satisfy. 
 
 Ex. 94. Draw the locus of points which are 2 in. from a given point. 
 
 Ex. 95. Draw the locus of the end of a pump handle which is 33 in. 
 long from its end to the point about which it turns, if the handle may be 
 moved through an angle of 100°. (Let 1 in. represent 11 in.) 
 
 Ex. 96. Draw any line of indefinite length. 
 
 (a) Locate freehand three points above the line which are 1 in. from 
 the line. 
 
 (6) Draw the line which contains all points which are 1 in. from the 
 line and lie above the line. 
 
 (c) Are there any other points which are 1 in. from the line ? 
 
 (d) Draw the line showing where they are to be found. 
 
 Ex. 97. Where are, that is, what is the locus of all points on this 
 page which are ^ in. from the left-hand edge of the page ? 
 
 Ex. 98. A rectangular lot is 100 ft. wide and 300 ft. long. Shrubs 
 are to be planted 5 ft. from the lot line along the two sides and the back 
 of the lot. What is the locus of the shrubs ? 
 
 Ex. 99. Draw two parallel lines. 
 
 (a) Locate freehand three points which are equidistant from the two 
 parallels. 
 
 (&) Draw the locus of all points which are equidistant from the 
 parallels. 
 
LOCI 125 
 
 Ex. 100. (a) Draw a line AB and locate on it a point C. Construct 
 three circles, all tangent to AB at C 
 
 (6) What is the locus of the center of a O which will be tangent to a 
 given line at a given point ? 
 
 Ex. 101. Draw a circle with radius 1 in. 
 
 Draw five radii of the circle. On each radius locate a point J in. from 
 the circle. Draw the locus of all such points. 
 
 Ex. 102. Draw the locus of all points outside a circle with 1 in. 
 radius and \ in. from the circle. 
 
 229. Def. If a single geometrical condition is given, the 
 Locus of Points satisfying that condition is the line or group 
 of lines such that : 
 
 (a) Every point in the line (or lines) satisfies the condition. 
 (6) Every point which satisfies the condition lies in the 
 line or group of lines. 
 
 230. Problem. Determine the locus of points equidistant from 
 tico given points. .^^,„ 
 
 Solution. (a) 1. E is located so that ^'^ 
 EA=:EB. Similarly, S and Tare located. ^ ^ 
 
 2. Their position suggests that the locus of 
 such points is the _L bisector of AB. ^"-'^ 
 
 (6) 1. Assume that CD, the ± bisector ABy ^:a 
 
 is the locus of points equidistant from A and B. 
 
 2. Is every point on CD equidistant from A 
 
 and B? 
 
 Yes, by § 118, I. 
 
 3. Is every point equidistant from A and B in line CD? 
 
 Yes, by § 118, II. 
 
 (c) .-. Tlie locus of points equidistant from two given points is 
 the perpendicular-bisector of the segment between the points. 
 
 Note. — In solving a locus problem, first locate three or more points 
 satisfying the condition ; then decide what you think the locus is ; then 
 try to prove that the supposed locus is the required locus. In doing this 
 last, prove theorems (a) and (b) of § 229, as is done in part (6) of the 
 solution in § 230. 
 
 A- 
 
126 
 
 PLANE GEOMETRY — BOOK II 
 
 Ex. 103. Locate two points Xand F which are 2 in. apart. 
 
 Construct the locus of points equidistant from X and Y. 
 
 Ex. 104. What is the locus of the vertex of an isosceles triangle 
 which has a given base ? 
 
 Ex. 105. What is the locus of the center of a circle which will pass 
 through two given points ? 
 
 231. Problem. Determine the locus of points within an angle 
 
 which are equidistant from the sides of the angle. 
 
 [The solution is to be given by the pupil.] 
 
 Suggestions. — Model your solution after that for the problem in § 230. 
 Recall § 120, I and II. 
 
 At the end of your solution complete the following sentence : 
 The locus of points within an angle which are equidistant from the 
 sides of the angle is . . . . 
 
 Ex. 106. Construct the locus of points equidistant from the sides of 
 a right angle and within the angle. 
 
 Ex. 107. What is the locus of the center of a circle which is tangent 
 to the sides of a given angle and lies within the angle ? 
 
 Note. — For additional discussion of loci see § 238. 
 
 CONSTRUCTION OF TRIANGLES 
 
 232. In a A ABC, the sides op- 
 posite angles A, B, and C are marked 
 by the small letters a, b, and c, re- 
 spectively. 
 
 The letter h denotes an altitude. 
 h^ (read h — sub — a) denotes the alti- 
 tude to side a. Similarly, there are the altitudes h^, and h^. 
 
 The letter m denotes a median. The medians to sides a, b, 
 and c are denoted by m^, m^,, and m^ respectively. 
 
 The letter t is used to denote the length of the bisector of an 
 angle between the vertex and the opposite side. The bisectors 
 of angles A, B, and C are denoted by t^, t^, and tg. 
 
 Note. — This notation was introduced by Euler (1707-1783). 
 
 233. A triangle is determined when three independent parts 
 are known. 
 
CONSTRUCTION OF TRIANGLES 127 
 
 General Suggestions 
 
 1. Draw freehand a triangle which represents the desired Jig- 
 iwe, marking with heavy lines the parts which correspond to the 
 given parts. Use this figure as a guide in constructing the desired 
 triangle. It is rwt the desired triangle, and the parts marked 
 are not necessarily equal in size to the given parts. 
 
 2. Make the construction, using the given parts. 
 
 3. Prove that the resulting triangle has all the given parts, and 
 is the kind of triangle specified. 
 
 4. Discuss the construction, determining whether there are con- 
 ditions under which it may he impossible to construct a triangle 
 having the given parts. {See Prop. IV, Book I, Discussion.) 
 
 234. The following seven problems are the fundamental con- 
 struction problems for triangles : 
 
 Ejc. 108. Review Proposition IV', Book I. 
 
 Ex. 109. Construct a triangle having given two of its sides and the 
 inchided angle. 
 
 Suggestion. — Let a and 6 be two given segments and Z C a given Z, — 
 drawn at random ; then construct the triangle. 
 
 Ex. 110. Construct a triangle having given two of its angles and the 
 included side. 
 
 Discussion. — Can the triangle be constructed always : 
 
 (a) If both A are acute ? (&) If both zi are rt. zl ? (c) If both A are 
 
 obtuse ? (d) If one Z is obtuse and one is acute ? 
 
 Ex. 111. Construct a right triangle having given its hypotenuse and 
 
 a leg. 
 
 Ex. 112. Construct a triangle having given a side, the opposite angle, 
 and another angle. 
 
 Suggestion. —If «, Z^, and Z A are given, then Z C may be determined 
 by subtracting AA+ /.B from 180°. Then A ABC can be constructed. 
 
 Ex. 113. Construct a right triangle having given a leg and the op- 
 posite acute angle. 
 
 Ex. 114. Construct a right triangle having given the hypotenuse 
 and an acute angle. 
 
 Note. — For further discussion of construction of figures see § 235 
 and § 241. 
 
128 
 
 PLANE GEOMETRY — BOOK II 
 
 Note. — 
 be omitted 
 
 SUPPLEMENTARY TOPICS 
 
 The rest of Book II is supplementary material and may safely 
 
 235. Triangles may be constructed when numerous other 
 combinations of three independent parts are given, besides 
 those mentioned already in § 234. 
 
 Illustkative Problem. — Construct a triangle having given 
 an angle, the length of its bisector, and the length of the alti- 
 tude drawn from its vertex. 
 
 Given 
 
 <^A 
 
 Hence t\ABE can be con- 
 
 Required to construct A ABC. 
 
 Analysis. 1. Let A^^C, with AD ± 
 BC and AE bisecting A A represent the 
 required figure. 
 
 2. The known parts are marked with 
 heavy lines, including Z.BAE = AEAG 
 = l/.A. 
 
 3. A ADE is a rt. A with a known leg 
 {—ha) and known hypotenuse {=1^). 
 structed. (Ex. 111.) 
 
 4. B is on DE extended and A BAE = J Z A 
 
 5. G is on ED extended and Z EA C = ^ZA. 
 Construction. 1." Construct rt. A ADE with 
 
 leg = ha and hypotenuse = Ia- 
 
 2. Extend DE in both directions. 
 
 3. Bisect ZA, and construct AB, making 
 Z EAB = ^ Z ^ ; let AB meet DE extended at B. 
 
 4. Construct AC, making Z EAC = ^ ZA, and meeting DE extended 
 at C. 
 
 5. Then A ABC is the required triangle. 
 
 Proof. 1. Z BAC = given Z A, since it equals 2(| Z A). Const. 
 
CONSTRUCTIONS 129 
 
 2. AD = given ha and is an altitude since Z D = rt. Z. Const. 
 
 3. AE = given tj, and is the bisector of Z A. Const. 
 Discussion. The construction is impossible if t^ < ha. 
 
 Note. — Observe that the final triangle may appear quite different 
 from the triangle drawn for the first step in the analysis. 
 
 236. Analysis of Construction Problems. 
 
 1. Draw a figure which represents the desired figure. 
 
 (a) Make this figure general. For example, if a triangle is to be 
 drawn, do not draw a right or an isosceles triangle unless such a triangle 
 is specified. 
 
 (b) Remember that this is not the final figure and that the parts in it 
 are not necessarily the given parts. 
 
 2. IVIark with heavy lines or with colored lines the parts 
 which are known and also those which may be readily deter- 
 mined from the known parts by fundamental constructions. 
 
 For example, if a known line is bisected, then each of the halves is 
 known. 
 
 3. Try to determine some part of the figure which can be 
 constructed by known methods. Usually this is a triangle. 
 This part can usually be made the basis for the rest of the 
 construction. 
 
 In this connection, remember the first five fundamental triangle con- 
 structions given in § 234. 
 
 4. Try to determine how the remaining parts can be obtained 
 from the figure constructed in step 3. 
 
 5. Make the construction, following the points rioted in steps 
 3 and 4. 
 
 6. Prove that the resulting figure satisfies the conditions of 
 the problem. 
 
 7. Discuss the resulting figure, determining, in particular, 
 whether the construction is or is not always possible. 
 
 Note. — Systematic use of this form of analysis is attributed to Plato. 
 The method of analysis has been described as one of the four great steps 
 in mathematics. Plato also introduced the restriction that constructions 
 should be made by ruler and compass alone. 
 
130 
 
 PLANE GEOMETRY — BOOK II 
 
 Illustrative Problem 1. — Construct A ABC having given 
 Z B, h^, and the radius r of the circumscribed circle. 
 Given 
 
 ^B= 
 
 Required to construct A ABC. 
 
 Analysis. 1. Let the adjoining figure repre- 
 sent the required figure. 
 
 2. A ABD is a rt. A, with known leg ( = ha) 
 and known acute angle {^B). .-. it can be con- 
 structed by Ex, 113. 
 
 3. Point is equidistant from A and B^ a dis- 
 tance equal to r. Hence O can be located. 
 
 4. The circle can then be drawn, and BD^ 
 extended, will meet the circle at C. 
 
 Construction left to the pupil. Follow up the steps 2, 3, and 4. 
 
 1. Construct a rt. ^ with side = ha and opposite Z = Z J5. 
 
 2. Locate the point O and draw the circle. 
 
 3. Extend BD and thus determine point C. 
 Proof and Discussion left to the pupil. 
 
 Illustrative Problem 2. 
 tangents of two circles. 
 
 Analysis. 1. Let the circles 
 unequal. Let the adjoining 
 
 Construct the common external 
 
 be assumed 
 figure represent 
 the desired figure. 
 
 2. Evidently ABCO' is a O. 
 
 3. .•.AB=CO'. 
 
 4. .. A0= OB- CO'. 
 
 5. Also, AO' and A' 0' are tangents to circle AA'. 
 
 Construction. 1. Construct a circle with radius equal to the difference 
 between the radii of the two circles, and concentric with the larger circle. 
 
 2. Draw tangents to this circle from the center of the smaller circle, 
 meeting the constructed circle at points A and A'. 
 
 3. Draw OA and OA' meeting the large circle at B and B'. 
 Complete the construction. 
 
 Give the Proof and the Discussion- 
 
CONSTRUCTION OF TRIANGLES 
 
 131 
 
 Ex. 115. Construct the common inter- 
 nal tangents of two unequal circles. 
 
 Constmct the triangle ^5C having given : 
 
 Ex. 116. b, c, he. 
 
 Ex. 117. a, c, and wig. 
 
 Ex. 118. a, 6, and hg. 
 
 Ex. 119. 6, he, B. 
 
 Ex. 120. Construct an isosceles triangle having given one base angle 
 and the altitude to the base. 
 
 Ex. 121. Construct an isosceles triangle having given one side and 
 the altitude to one of the sides. 
 
 Note. — Supplementary Exercises 31-49, p. 287, can be studied now. 
 
 Proposition XXV. Problem 
 
 237. Ujoon a given segment as chord, constricct on 
 arc of a circle such that every angle inscribed in it 
 shall equal a given angle. 
 
 \ N 
 
 Given segment AB and Z T. 
 
 Eequired to construct an arc upon AB as chord such that 
 every angle inscribed in the arc shall equal Z T. 
 
 Construction. 1. Construct Z BAC= Z T. 
 
 2. Construct DE J_ AB at its mid-point. 
 
 3. Construct AFl. AC at A, intersecting DE at 0. 
 
 4. Construct a circle with center and OA as radius. 
 
 Statement. AMB is the required arc. 
 
 Proof. (The pupil should now prove that any Z AGE = Z T.) 
 
132 PLANE GEOMETRY — BOOK II 
 
 FURTHER DISCUSSION OF LOCI* 
 238. Method of Attacking a Locus Problem. 
 
 1. Locate either freehand or by construction three or more 
 points which satisfy the given condition. These points 
 should suggest the probable locus. 
 
 2. Draw the probable locus and try to prove that it is the 
 real locus. To do this, try to prove either (a) and (6) below, 
 or else (a) and (c) : 
 
 (a) Every point on the locus satisfies the given condition. 
 (6) Every point which satisfies the given condition lies on 
 the locus. 
 
 (c) Every point not on the locus does not satisfy the given 
 condition. 
 
 Note. — (6) is the converse of (a) and (c) is the opposite of (a). The 
 opposite of (&) is : 
 
 (d) every point which does not satisfy the given condition does not lie 
 on the locus. 
 
 When (a) and (6) are known, then (c) and (d) can be proved by the 
 indirect method ; when (a) and (c) are known, then (6) and (d) can be 
 proved in the same manner. 
 
 Illustrative Problem. — Determine the locus of the ver- 
 tex of the right angle of a right triangle having a given seg- 
 ment as hypotenuse. ^ q 
 
 Solution. 1. Let A ABC be right tri- 9y^/^^\^<A 
 angles having the hypotenuse AB and rt. ,^^^^^^^^^\V 
 
 z at a ^^— -^=^-« 
 
 2. This figure suggests that the points C 
 lie on a circle having AB as diameter. 
 
 3. Assume that the locus is the circle hav- 
 ing AB as diameter. 
 
 (a) Is every AAXB, where X is any 
 point on the circle, a rt. A? 
 
 Yes, since Z AXB is a rt. Z, by § 218. 
 
 * Review at this time § 229 and the two loci discussed in § 230 and § 231. 
 
LOCI 133 
 
 (b) Is every A A YB, where Y is any point not on the cir- 
 cle, an oblique A ? 
 
 Yes, since every Z A YB is either acute or obtuse, according as Y lies 
 outside or inside of the circle. (Easily proved.) 
 
 4. Hence the locus of the vertex of the right angle of a right 
 triangle having a given segment as hypotenuse is a circle draivn 
 on the hypotenuse as diameter. 
 
 239. Summary of Fundamental Loci. 
 
 1. The locus of points at a given distance d from a given 
 point is the circle drawn with as center and d as radius. 
 (§ 228.) 
 
 2. The locus of points at a given distance d from a fixed 
 line I (of indefinite length) is the pair of parallels to I at the 
 distance d from it. (Ex. 96.) 
 
 3. The locus of points equidistant from two parallel lines 
 is the line parallel to them and midway between them. (Ex. 99.) 
 
 4. The locus of points equidistant from two given points is 
 the perpendicular bisector of the segment joining the points. 
 (§ 230.) 
 
 5. The locus of points equidistant from the sides of an angle 
 and within the angle is the bisector of the angle. (§ 231.) 
 
 Cor. The locus of points equidistant from two intersecting 
 straight lines is the set of bisectors of their included angles. 
 (These bisectors form two straight lines.) 
 
 6. The locus of the vertex of the right angle of a right 
 triangle which has a given hypotenuse is a circle drawn upon 
 the hypotenuse as diameter. (§ 238.) 
 
 7. If A and B are any two fixed points and X is a point 
 such that Z AXB is a given angle, the locus of X is the arc 
 of a circle constructed upon AB as chord such that every angle 
 inscribed in it equals the angle given. (§ 237.) 
 
 Caution. — 1. Remember that " what is the locus of ? " means " what 
 is the place of?" 2. Be certain that you know what " the given con- 
 dition" is in each locus theorem. 
 
134 
 
 PLANE GEOMETRY — BOOK II 
 
 240. Intersection of Loci. — Sometimes it is specified that 
 a point shall satisfy each of two given conditions. Each con- 
 dition determines a locus for the point. The point then must 
 lie at the intersection of the two loci. 
 
 Illustrative Problem. — Find all points which are equi- 
 distant from two intersecting lines and also equidistant from 
 two fixed points. 
 
 Given intersecting lines AB and CD and points B and S. 
 
 Required to find all points which are equidistant from AB and CD and 
 also equidistant from B and S. 
 
 Solution. 1. The locus of 
 points equidistant from AB and 
 CD is the set of bisectors of the 
 angles included by them. (Lines 
 ^i^and GH.) 
 
 2. The locus of points equi- 
 distant from B and S is the 
 perpendicular bisector of BS. 
 (Line TW.) 
 
 3. The required points will be 
 at the intersection of TW with 
 EF and GH. 
 
 Discussion. 1. Usually there 
 are two points ; as Xi and X2. 
 
 2. There may be only one point, however, for TW ma,y be, parallel to 
 one of the lines, EF and GH. 
 
 3. There must always be at least one point, for T W cannot be parallel 
 to both EF and GH. 
 
 4. There may be a whole line full of points, for TW may coincide with 
 EF or GH. 
 
 Ex. 122. In a given line, find all points which are equidistant from 
 two given points. 
 
 Ex. 123. In a given line, find all points which are equidistant from 
 two given intersecting lines. 
 
 Ex. 124. In a given circle, find all points which are equidistant from 
 two given parallel lines. 
 
 Ex. 125. Find all points which are equidistant from two given points 
 and also at a given distance from a given point. 
 
LOCI 135 
 
 Ex. 126. Find all points which are equidistant from two given points 
 and also at a given distance from a given line. 
 
 Ex. 127. Find all points which are equidistant from two given points 
 and also equidistant from two given parallels. 
 
 Ex. 128. Find all points which are equidistant from two given 
 parallels and also at a given distance from a given point. 
 
 Ex. 129. Find all points which are at a given distance from a given 
 line and also at another given distance from a given point. 
 
 Ex. 130. Find all points which are equidistant from two parallels 
 and also equidistant from two intersecting lines. 
 
 Ex. 131. Find all points which are equidistant from two intersecting 
 lines and at a given distance from a given point. 
 
 Ex. 132. What is the locus of the vertex of a triangle whose base 
 lies in a given straight line if the altitude to the base is a given segment ? 
 
 Ex. 133. What is the locus of the center of a circle which shall be 
 tangent to a given line and have a given radius ? 
 
 Ex. 134. What is the locus of points at a given distance from a 
 given circle ? 
 
 (The distance is measured along a line between the point and the 
 center of the circle.) 
 
 Ex. 135. What is the locus of the center of a circle which has a 
 given radius and passes through a given fixed point ? 
 
 Ex. 136. What is the locus of the center of a circle which shall be 
 tangent to each of two parallel lines ? 
 
 Ex. 137. What is the locus of the mid-points of all chords of a circle 
 that have a given length ? 
 
 Ex. 138. What is the locus of the points such that the tangents from 
 the points to a given circle shall have a given length ? 
 
 Ex. 139. What is the locus of the mid-points of all parallel chords 
 of a circle ? 
 
 Ex. 140. What is the locus of the mid-points of all segments drawn 
 from one vertex of a triangle and terminated by the opposite side ? 
 
 Ex. 141. A line AB of fixed length moves so that A is constantly on 
 one side of a given right angle and B is on the other side of the angle. 
 What is the locus of the mid-point of the segment AB ? (Recall Ex. 175, 
 Book I.) 
 
136 
 
 PLANE GEOMETRY — BOOK II 
 
 241. Construction of Figures by Intersection of Loci. 
 
 Illustrative Problem. — Construct A ABC having given 
 c, h^, and Z C. 
 
 Given 
 
 he 
 
 from c. (Lo- 
 
 Required to construct A ABC. 
 
 Analysis. 1. Let A ABC represent the de- 
 sired triangle, the known parts being marked 
 by heavy lines. 
 
 2. Line c can be drawn, thus locating defi- 
 nitely points A and B. 
 
 3. Point C is at the distance h^ from c. It 
 therefore lies on one of two parallels to c at the distance 
 cus2, §239.) 
 
 4. Point C is such that ZACB must equal Z C. It therefore lies 
 on the arc constructed on ^B as chord, 
 the inscribed angles of which equal Z C. 
 (Locus 7, §239.) 
 
 Construction is made so as to obtain 
 the loci mentioned in steps 3 and 4. The 
 circle cuts the line BS at two points (7i 
 and 02. A ACiB and A AC2B each sat- 
 isfy the given conditions. 
 
 Proof and Discussion left to the pupil. 
 
 Ex. 142. Given the base and altitude of an isosceles triangle, construct 
 the triangle. 
 
 Ex. 143. Construct an isosceles triangle having given the base and 
 the radius of the circumscribed circle. 
 
 Ex. 144. Construct a rhombus having given its base and altitude. 
 
 Ex. 145. Construct a right triangle having 
 given the hypotenuse and the length of the altitude 
 upon it. 
 
 Ex. 146. Construct an isosceles triangle having 
 given the base and the angle opposite the base. 
 
LOCI 
 
 137 
 
 Ex. 147. Construct a triangle having given the base, the altitude, and 
 the radius of the circumscribed circle. 
 
 Ex. 148. Construct a triangle having given a side, an adjacent angle, 
 and the radius of the circumscribed circle. 
 
 Ex. 149. Through a given point construct a circle with a given radius 
 which shall be tangent to a given line. 
 
 Analysis. 1. Let the circle with the center C pass 
 through P and be tangent to line ?. 
 
 2. O is r distant from P. Hence it must lie on a 
 circle having P as center and r as radius. 
 
 3. C is r distant from I. Hence C must lie on one 
 of two parallels to I at the distance r from I. 
 
 4. C must be at the intersection of these two loci. 
 Construction, Proof, and Discussion left to the pupil. 
 
 Ex. 150. Construct a circle with a given radius which shall be tan- 
 gent to each of two intersecting lines. 
 
 Ex. 151. Construct a circle which shall be tangent to each of two 
 intersecting lines, tangent to one of them at a given point. 
 
 Ex. 152. Construct a circle through a given point not in a given line 
 which shall be tangent to the given line at a given 
 point in the line. 
 
 Ex. 153. Construct a circle having a given 
 radius which shall be tangent to each of two given 
 circles. 
 
 Ex. 154. A circular cylinder head 12 in. in diameter is to have holes 
 bored in it for 12 1-in. bolts, equally spaced around the edge, with their 
 centers 1^ in. from the edge. Make a scale drawing of the cylinder head 
 (\" = 1") and mark the centers for the 12 bolt holes. 
 
 Ex. 155. Determine how to construct the unit which is repeated in 
 the design below. Construct it in a circle of 3-in. radius. 
 
 Note. — Supplementary Exercises 50 to 63, p. 288, can be studied now, 
 
138 
 
 PLANE GEOMETRY — BOOK II 
 
 Ex. 156. Determine how to 
 construct the left-hand figure. 
 Notice that it is the basis for the 
 artistic design at the right. 
 Construct the left-hand figure 
 in a circle of 4- in. diameter. 
 
 Ex. 157. The adjoining figure indicates a form of mansard roof. 
 The chords AB, CD, CE, and FG are all ^ 
 
 equal. ^^^^f^<:^^^ 
 
 Construct such a roof outline for a build- ^"C^' ^^\e 
 
 mg in which AQ is 30 ft. and the distance 
 IM is 10 ft. (Let 1 in. = 5 ft.) 
 
 1&AI = IC= CJ = JG'^ 
 
 Is thelineZ/l|yl(y? 
 
 Ex. 158. Construct between two parallel lines a set of circular rings 
 like those in the design below. 
 
 7rV^/^rV/ 
 
 MMM%B 
 
 Design for Ornamental 
 Stonework on a Bridge 
 
 Ex. 159. The adjoining design is a panel for 
 ornamental ironwork on a bridge. 
 
 Determine how to construct the fundamental units 
 of the design, units (a) and (6). 
 
 Unit (a) 
 
 Unit (&) 
 
 Ex. 160. Construct a tangent to a circle which will 
 be parallel to a given line. 
 
 Suggestion. — Make an analysis based on the adjoining 
 figure. 
 
MISCELLANEOUS EXERCISES 139 
 
 Miscellaneous Exercises 
 
 Ex. 161. AOB is a diameter of O 0. C is any point of AB. D is 
 the mid-point oi BC and E is the mid-point of AC. Prove Z DOE is a 
 right angle. 
 
 Suggestion. — Draw CO. 
 
 Ex. 162. Points A and B are on the diameter XY of circle at 
 equal distances from 0. CA and DB are perpendicular to XF, meeting 
 the semicircle at C and D respectively. Prove ABDC 
 is a rectangle. . 
 
 Ex. 163. If a circle is inscribed in a right tri- cl/' 
 
 angle, the sum of its diameter and the hypotenuse is /J Or- 
 
 equal to the sum of the legs of the triangle. / V i 
 
 E 
 
 Ex. 164. If AB is a common external tangent of twc^ circles which 
 touch each other externally at (7, prove Z ACB is a right angle. 
 
 Suggestion. — Draw the common tangent of the © at C, meeting AB at D. 
 
 Ex. 165. Prove that the bisector of the angle betw^een two tangents 
 to a circle passes through the center of the circle. 
 
 Suggestions. — Draw radii to the points of contact. Recall § 120. 
 
 .,„..„. ».,....„..™. .,...,.. ,^1\ 
 
 sides of an isosceles triangle as diameter bisects the I / i Nl 
 
 Ex. 167. Two circles are tangent externally at C. In one circle 
 A ABC is inscribed, having one vertex at the point of contact 6f the 
 circles. ^Cand BC are extended through C, meeting the other circle 
 at D and E respectively. Prove DE II AB. 
 
 Suggestion. — I)ra.w the common tangent through point C. 
 
 Ex. 168. If a straight line be drawn through the point of contact of 
 two circles which are tangent externally, terminating in their circumfer- 
 ences, the tangents at its extremities are parallel. 
 
 Suggestion. — Draw the common internal tangent of the circles. 
 
 Ex. 169. If AB and AC are the tangents from point A to the circle 
 O, Z BAC = 2Z0BC. 
 
 Suggestions. — 1. Draw OA. "What relation does it bear to BC? 
 2. Compare ZBAO with Z OB C. 
 
140 PLANE GEOMETRY — BOOK II 
 
 Ex. 170. Euclid's construction for the tangent to a circle with, 
 center M from a point A outside of it is as follows : 
 
 1. Draw the circle with center M and radius MA. 
 
 2. Draw MA intersecting the given O at B. 
 
 3. Draw BC ± MA at B, meeting the larger O at G. 
 
 4. Draw MC^ intersecting the given O at D. 
 
 Statement. AD is tangent to the given O. 
 Make the construction and give the proof. 
 
 Ex. 171. Given a side and the diagonals of a parallelogram, con- 
 struct the parallelogram. 
 
 Ex. 172. Through a given point within a circle, construct a chord 
 equal to a given chord. . 
 
 Is there any restriction on the location of the point ? 
 
 A 
 Ex. 173. Construct a parallel to the side BC of 
 A ABC meeting AB and AC &t D and E respec- 
 tively, so that DE will equal EC. 
 
 B' 
 
 Ex. 174. If point B bisects arc AC of a circle, then ZA of A ABC 
 equals Z C. 
 
 Ex. 175. Prove that the bisectors of the angles of a circumscribed 
 quadrilateral pass through a common point. 
 
 Ex. 176. Prove that two chords which are perpendicular to a third 
 chord at its extremity are equal. 
 
 Ex. 177. If XA = YG and BG = AB, prove 
 A AXB ^ A YCS. 
 
 Ex. 178. In the figure for Ex. 177 draw A C cutting 
 XB at M and YB at N. Prove A AXM^ A YCJ^. 
 
 Ex. 179. A carpenter has a tool called a gauge which illustrates and 
 applies one of the fundamental loci theorems. 
 
 The shaded rectangle represents the end of a board ; the tool is upon 
 the right-hand side of the board. P is a marking point which extends to 
 the under side of the tool. AB is & movable part which can be fixed at 
 any short distance from P by means of a p ^ 
 
 screw at A. By moving the gauge so that '■ ^^^^^^gLgg;:^:^;-::^^^ J 
 AB is constantly against the edge of the 
 board, the point P traces upon the upper side of the board a line 
 parallel to the edge of the board. Why is this so ? 
 
 lY 
 
 B 
 
BOOK III 
 PROPORTION — SIMILAR POLYGONS 
 
 242. The Eatio of one number to another is the quotient of 
 the first divided by the second. 
 
 Thus, the ratio of a to & is - ; it is also written a: b. 
 h 
 
 The numerator is called the Antecedent and the denomina- 
 tor is called the Consequent. 
 
 Since a ratio is a fraction, it is subject to the usual rules 
 for operations with fractions. 
 
 243. The ratio of two concrete quantities of the same kind is 
 the ratio of their measures in terms of a common unit. (§ 212.) 
 
 Thus, the ratio of 350 lb. to 2 tons is ^^ or ^. 
 
 Ex. 1. Express the following ratios in their simplest form. 
 
 (a) 3 to 9. (c) 6x to 2 x. (e) f to y\. (g) 25 to 375. 
 
 (6) 12 to 2. (d) 6 a2 to 15 a^. (/) ^ to ]. {h) a^ - b'^ to a^ - b\ 
 
 Ex. 2. A line 15 in. long is divided into two parts which have the 
 ratio 2 : 3. Find the parts. 
 
 Suggestion. — If the short part contains x in. and the long part (15 — a;) in., 
 
 X 2 
 
 then = -• Complete the solution. 
 
 15 — a; 3 
 
 Ex. 3. Divide a line 63 in. long into two parts whose ratio is 3 : 4. 
 
 Ex. 4. Divide 36 into two parts such that the ratio of the greater 
 diminished by 4 to the less increased by 3 will be 3 : 2. 
 
 Ex. 5. The ratio of the height of a tree to the length of its shadow 
 on the ground is 17 : 20. Find the height of the tree if the length of 
 the shadow is 110 feet. 
 
 Ex. 6. What is the ratio of: (a) a right angle to a straight angle? 
 (6) a right angle to the perigon ? (c) one angle of an equilateral tri- 
 angle ^to the sum of all the angles of the triangle ? (d) one side of a 
 square to the perimeter of the square ? 
 
 141 
 
142 PLANE GEOMETRY — BOOK III 
 
 244. A Proportion is a statement that two ratios are equal j 
 
 as, - = -, ov a:o = c: a. 
 b d 
 
 This proportion is read " a is to 6 as c is to d." 
 
 1 5 
 Thus, 1, 3, 5, and 15 form a proportion since - = ^. 
 
 S 15 
 
 This means that 1 bears to 3 the same relation that 5 bears to 15. 
 
 The first and fourth terms of a proportion are called the 
 Extremes, and the second and third terms, the Means. 
 
 In the proportion a : b = c : d, a and d are the extremes and b and c 
 are the means ; a and c are the antecedents, and b and d are the con- 
 sequents. 
 
 Ex. 7. Select four numbers which form a proportion like the arith- 
 metical illustration in § 244. 
 
 Ex.8. (a)Is^ = ^? (b)ls^ = ^? (c)Is?zz:A? 
 
 Ex. 9. Find the value of the literal number in each of the following 
 proportions. 
 
 ^^48 ^M6 8 ^^2 3 
 
 (^,) L0 = 2. 1^3. 3+^ ^5. 
 
 Ex. 10. Find the value of x in each of the following proportions. 
 
 ^ ^ b X ^ ^ 36 2c ^ s t P X 
 
 245. Proportion is used in a great variety of ways. 
 
 Example. — The cost of a number of articles of a given 
 kind is ''proportional" to the number of articles. 
 
 Thus, the cost of seven books is to the cost of three books of the same 
 
 kind as 7 is to 3. Hence, if 3 books cost .§1.35, the cost of 7 books maj'- 
 
 X 7 
 
 be determined from the proportion = - • 
 
 ^ ^ 1.35 3 
 
 Ex. 11. Determine by proportion the cost of 13 yd, of cloth if the 
 
 cost of 5 yd. of the same cloth is 70^. 
 
 Ex. 12. Determine by proportion the distance an automobile, will 
 travel in one hour if it travels 2 mi. in 5 minutes. 
 
PROPORTION 143 
 
 Ex. 13. If a girl makes 3 14.25 profit from 15 hens in one year, what 
 profit can she expect from 60 hens, assuming the same average profit per 
 lien ? 
 
 246. The Fourth Proportional to three numbers a, 6, and c 
 is the number x in the proportion a: b = c: x. 
 
 2 4 
 Thus, the fourth proportional to 2, 3, and 4 is the number x in - = - . 
 
 3 X 
 .-. 2x = 12, orx = 6. 
 
 Note. — The numbers must be placed in the proportion in the order in 
 which they are given as in the illustrative example. 
 
 Ex. 14. Find the fourth proportional to : 
 
 (a) 2, 3, and 4. (d) 25, 15, and 10. 
 
 (6) 3, 2, and 4. (e) 3 a, 2 6, and c. 
 
 (c) 4, 3, and 2. (/) r, rs, and s. 
 
 247. The Third Proportional to two numbers a and b is the 
 
 number x in the proportion a: b = b : x. 
 
 2 3 
 Thus, the third proportional to 2 and 3 is a; in - = - • 
 
 • 9 ^ ^ 
 
 .-. 2x = 9, orx = -. 
 
 2 
 
 Ex. 15. Find the third proportional to : 
 
 (a) 3 and 5. (&) 2 and 5 t. (c) 5 and 10. 
 
 248. A Mean Proportional between two numbers a and b is 
 the number x in the proportion a: x = x: b. 
 
 Example. — A mean proportional between 2 and 3 is a; in : 
 
 2^x 
 X Z' 
 .'. a;2 = 6, orx=±V6. 
 
 There are two mean proportionals between any two num- 
 bers. The positive one is implied when "the" mean propor- 
 tional is specified. 
 
 Ex. 16. Find the mean proportional between : 
 
 (a) 75 and 12. (c) 2 r^t and IS rt. 
 
 (6) 3 a and a. (d) 6f and |. 
 
144 PLANE GEOMETRY — BOOK III 
 
 FUNDAMENTAL THEOREMS OF PROPORTION 
 
 249. The mean proportional between two numbers is the square 
 root of their product. 
 
 For the mean proportional between a and 6 is a; in : 
 a _x 
 X b 
 .'. x'^ = ab; or ic = Vab. 
 
 250. In a proportion, the product of the extremes is equal 
 to the product of the means. 
 
 If - = -, then ad = be. 
 
 b d 
 
 Suggestion, — Multiply both members of the proportion by bd. 
 
 2 6 
 Example.— Since - = -,2x9 should equal 3x6. Does it?- 
 
 251. If three terms of one proportion are equal respectively to 
 the three corresponding terms of another proportion, the fourth 
 terms also are equal. 
 
 If - = - and - = -,thena: = 2/. 
 
 b x by 
 
 Suggestion. — Determine x from the first proportion and y from the second. 
 
 Example. — If - = - and - = -, then x should equal y. 
 3 X 3 y 
 
 252. If the product of two numbers is equal to the pi^oduct of 
 two other numbers, one pair may be made the means of a propor- 
 tion having the other pair as the extremes. 
 
 If mn = xy, then — = ^ , 
 
 X n 
 
 Proof. Dividing both members of the given equation by xn, 
 
 mn _xy m _y_ 
 xn xn X n 
 
 3 4 
 
 Example. — Since 3x8 = 6x4, - should equal - . Does it? 
 
 6 8 
 
PROPORTION 145 
 
 Ex. 17. If mn = xy, prove : 
 
 (a) ^=^. (Divide by yn.) (6) ^ = ^'. (c) ^ = i^. 
 
 ^y?i my X m 
 
 Ex. 18. Since 4 x 5 = 2 x 10, write four proportions which involve 
 4, 6, 2, and 10. 
 
 253. I7i any proportion, the terms are in proportion by Alter- 
 nation ; that is, the first term is to the third as the second is to 
 the fourth. 
 
 If ^ = ^,then«=^. 
 
 b d c d 
 
 Proof. 1. Since - = -, then ad = be. § 250 
 
 b d 
 
 2. Since ad = be, then - = - . § 252 
 
 c d 
 
 Example. — Since - = — - , then - should equal — . Does it ? 
 6 12' 4 ^12 
 
 Ex. 19. Write each of the following proportions by alternation : 
 
 ^ ^ S 15' ^^5 r' ^ s y 
 
 254. Ill any proportion, the terms are in proportion by In- 
 version ; that is, the second term is to the first as the fourth is to 
 the third. 
 
 If « = ^,then * = 5?. 
 
 b d a c 
 
 Proof. 1. Since - = - , then ad = be. Why ? 
 
 b d 
 
 2. Since ad = be, then - = - • Why ? 
 
 a c 
 
 O A fi "I O 
 
 Example. — Since - = — , then - should equal — . Does 
 it? 6 12' 2 ^4 
 
 Ex. 20. Write each of the three proportions of Ex. 19 by inversion. 
 
 Ex. 21. Write the proportion f = -, by inversion, and then write 
 
 h y 
 
 the resulting proportion by alternation. 
 
146 PLANE GEOMETRY — BOOK III 
 
 255. lyi any proportion, the terms are in proportion by Com- 
 position ; that is, the sum of the first two terms is to the second, 
 as the sum of the last two terms is to the fourth. 
 
 If « = :?,then^i^ = ^+^. 
 
 b d b d 
 
 Proof. 1. Since - = -, then ^+1=^ + 1. Why ? 
 
 b d b d 
 
 2. .*. — ^^ = -^ — . Algebraic addition. 
 
 b d 
 
 Example. Since -= — , then -^ — should equal — ^t — ^^ 
 6 12' 6 ^ 12 
 
 Does it ? 
 
 Ex. 22. Write the three proportions of Ex. 19 by composition. 
 
 Ex. 23. Write the proportion - = - : 
 
 b y 
 
 (a) By composition and the result by inversion. 
 
 (6) By inversion and the result by composition. 
 
 (c) By composition and the result by alternation. 
 
 (d) By alternation and the result by composition. 
 
 256. In any proportion, the terms are in proportion by Divi- 
 sion; that is, the first term minus the second is to the second, as 
 the third minus the fourth is to the fourth. 
 
 Tu a c,i a — b c — d 
 
 If - = -, then = 
 
 b d b d 
 
 Proof. 1. Since - = -, then ^-1 = ^-1. AVhy ? 
 
 b d b d 
 
 (Complete the proof.) 
 Ex. 24. Write the three proportions of Ex. 19 by division. 
 
 Ex. 25. Write the proportion - = - : 
 
 b y 
 
 (a) By division and the result by alternation. 
 
 (&) By alternation and the result by division. 
 
 (c) By inversion and the result by division. 
 
 {d) By division and the result by inversion. 
 
PROPORTION 147 
 
 257. In any proportion^ the terms are in proportion by Com- 
 position ayid Division ; that is, the first term plus the second is to 
 the first term minus the second, as the third term plus the fourth 
 is to the third term minus the fourth. 
 
 If ? = «,then^Jl^ = <l±^. 
 
 h a a — b c — d 
 
 Proof. 1. Since « = ^, then ^^i^ = ^^t_^-. §255 
 
 b d b d 
 
 2. Also a-±^c^-_d^ ^ 256 
 
 b d 
 
 3 • c^4-& , a — b _ c-\- d . c — d Why? 
 
 " b ' b d ' d ' ^ ' 
 
 J, a -\-b b c-\-d d 
 
 or 
 
 a — b d 
 
 a+ b _ c +d 
 a— b c—d 
 
 ^ o- 10 15 ^, 10+2 15 + 3 T^ .,., 
 
 Example. — Since -— = — , then ^^ = ^^ ' . Does it .'* 
 2 3 ' 10 - 2 15 - 3 
 
 Ex. 26. Write the proportion - = - : 
 
 h y 
 
 (a) By composition and division and the result by inversion. 
 
 (6) By inversion and tiie result by composition and division. 
 
 (c) By composition and division and the result by alternation. 
 
 (d) By alternation and the result by composition and division. 
 
 258. In any jyroportion, like powers or like roots of the terms 
 are in proportion. 
 
 If ^ = ^, then ^ = ^, and also ^5 = ^. 
 b d' b^ d-' yi ^ 
 
 The first of these conclusions follows from raising both members of the 
 given proportion to the nth power, and the second from taking the nth 
 roots of both members. 
 
 Ex. 27. n« = ^, prove: (a) m=,n^. a) r^ = ^. 
 
 b d b d mb d 
 
148 PLANE GEOMETRY — BOOK III 
 
 259. The preceding paragraphs give some of the essential 
 facts about the subject of proportion. Remember that the 
 terms in every case are numbers. We shall be dealing with 
 ratios and proportions of geometrical magnitudes. However, 
 as explained in § 243, we replace the magnitudes themselves 
 by their measures in terms of common units so that the terms 
 are again numbers. The consequence is that the theorems 
 about proportions all apply to the proportions which we shall 
 encounter. 
 
 Thus, if AB, CD, EF, and OH are four segments such that 
 
 4^ = RZ then ABx GH=EFx CD. 
 CD OH' 
 
 This means that the product of the numerical measures of AB and 
 dr equals the product of the numerical measures of ^i^'and CD. 
 
 A similar interpretation must be given to all applications of § 246 to 
 § 258 inclusive. 
 
 PROPORTIONAL LINE-SEGMENTS 
 
 260. Introduction. If AJEJ = EB , a E 
 
 and CF=FD, then ^=^ since ^ // f ^ 
 
 EB FD ^ ' • ^ 
 
 each ratio equals 1. Again, in the same figure, if G bisects 
 
 AC 1 PIT 1 
 
 AE and H bisects CF, then ^=i and also ~^ = ^', hence 
 
 GB 3 HD 3 
 
 AO OTT 
 
 '^=-^ = — — This means that AG bears to GB the same relation 
 
 GB HD 
 
 that CH bears to HD. G and H are said to divide AB and 
 CD proportionally. 
 
 Def. Two line-segments are divided proportionally when the 
 segments of one have the same ratio as the corresponding seg- 
 ments of the other. 
 
 Ex. 28. Draw a A ABC, having AB = 2 in., AC = 3 in., and BC 
 = 4 in. Place X on AB, so that AX= .5 in. Draw from X a parallel 
 io BC meeting ^C at Y. 
 
 (a) Measure A Y and YC, and determine the ratio of A Y to YC. 
 
 (6) What is the ratio of ^Xto XB ? 
 
 (c) Do the sides appear to be divided proportionally ? 
 
PROPORTION 
 
 Proposition I. Theorem 
 
 149 
 
 261. A parallel to one side of a triangle, intersecting 
 the other two sides, divides the other two sides pro- 
 portionally. 
 
 Hypothesis. In A ABC, DE II BC, meeting AB at D and 
 AC at E. 
 
 Conclusion. ^ = ^. 
 
 DB EC 
 
 Case I. Suppose that AD and DB are commensurable. 
 
 §211 
 Proof. 1. Let AF be a common measure contained 4 times 
 in AD and 3 times in DB. 
 
 " DB 3 * 
 
 2. Draw lis to BC through the points of division on AB. 
 Then AC will be divided into equal segments, of which 4 are 
 in AE and 3 are in EC. § 147 
 
 3 . AE^^ 
 
 "EC 3* 
 
 AD AE 
 
 4. Then, from steps (1) and (3), 
 
 DB EC 
 
 Ax. 1, § 51 
 
 Case II. Suppose that segments AD and DB are incom- 
 mensurable. § 211 
 
 The proof given for Case I will not apply, as no common 
 measure with which to divide both AD and DB can be found. 
 
 The theorem is true hoivever for the incommensurable case 
 also. The proof is given in § 424 and, if desired, may be read 
 at this time. 
 
150 
 
 PLANE GEOMETRY — BOOK III 
 
 262. Cor. 1. Since AD : DB 
 
 AD 4- DB AE + EC AB 
 
 ' — ' or 
 
 Why 
 
 AE : EC, then 
 
 AC 
 
 DB EC ' ' DB EG 
 
 That is, one side is to its lower segment as the other side 
 is to its lower segment. 
 
 A 
 
 AD 
 
 ^, then m 
 EC AD 
 
 EG 
 
 ae' 
 
 AB AC 
 
 or = 
 
 Why? 
 Why? 
 
 263. Cor. 2. Since 
 
 DB 
 
 DB -\- AD ^ EC + AE 
 "AD AE ' AD AE 
 
 That is, one side is to its upper segment as the other side 
 is to its upper segment. 
 
 264. Numerous other proportions may be derived from the 
 proportions obtained in §§ 261-2G3 by making allowable 
 changes in them. 
 
 From §261, AD: AE =^ DB: EC. Why? 
 
 Erom § 262, BD:AB = EC: AC Why ? 
 
 From § 263, AD:AB = AE : AC Why ? 
 
 Note. — In every case, corresponding segments occur in the proportion 
 in the same manner. 
 
 265. For convenience, reference may be made to any of the 
 proportions developed in §§ 261-264 by quoting the authority: 
 
 A parallel to one side of a triangle divides the other two sides 
 proportionally. 
 
 Ex. 29. If, in the figure of § 261, AD is i of BD, what is the ratio of 
 AE to EC? 
 
 Ex. 30. liAD = S in., DB = 6 in., and EC = 6 in., find AE. 
 
 Ex. 31. If AB = 12 in., AC = 15 in., and AE = 6 in., find AD. 
 
PROPORTION 151 
 
 Proposition II. Problem 
 
 266. Construct the fourth proportio'nal to three given 
 segments. 
 
 Given segments m, n, and p. 
 
 Required to construct the fourth proportional to m, w, and p. 
 Analysis. 1. Let x represent the fourth proportional. 
 Then m : n =p : x. 
 
 2. This suggests the following construction. 
 Construction. 1. On side AB of a convenient angle, Z BAC, 
 take AD = m, and DE = w ; on ^C, take AF=2). 
 
 2. Draw DF and construct EG \\ DF, meeting AC at G. 
 Statement. Then FG is the fourth proportional to m, n, 
 
 and p. 
 
 [Proof to be given by the pupil.] 
 
 267. Cor. Construct the third proportional to m and n. 
 Analysis. 1. Let x represent the third proportional to m 
 and 71. 
 
 2. Then m:n = n: x. 
 
 [Construction and proof to be given by the pupil.] 
 
 Ex. 32. Construct the fourth proportional to segments which are 2 in., 
 1 in., and 3 in. in length. Measure the resulting segment. Verify your 
 work by computing the fourth proportional to 2, 1, and 3, as in Ex. 14. 
 
 Ex. 33. Let OB be any line within ZAOC and Fand V any two 
 points on OB. Let FXand yX' be perpendiculars to OA, and TZ and 
 Y'Z' be perpendiculars to OC. Prove that OX : OX' = OZ : OZ'. 
 
 Ex. 34. A line drawn parallel to the bases of a b c 
 
 trapezoid and intersecting the non-parallel sides, A \ 
 
 divides the non-parallel sides proportionally. ^^ — '-^ ^^ 
 
 Prove BE:EA=CF: FD. 
 
 A^ k 'D 
 
152 PLANE GEOMETRY — BOOK III 
 
 Proposition III. Theorem 
 
 268. A line ivhich divides tivo sides of a triangle pro- 
 portionally is parallel to the third side. 
 
 A 
 
 Hypothesis. In A ABC, DE intersects AB and AC so that 
 
 AB^AC 
 
 AD Ae' 
 Conclusion. DE \\ BC. 
 
 Proof. 1. Assume DF \\ BC, meeting AC at F. 
 
 4. .-. AF=AE. § 251 
 
 5. .*. i^ coincides with E, and DE with DF. Why ? 
 
 6. .'. DE \\ BC Step 1 
 
 Ex. 35. If ^D = 3 in., AB = 12 in., A0= 10 in., and AE = 2.5 in., 
 is DE li BC? 
 
 Ex. 36. It AD = 6 in., BD = 10 in., AE = 6 in., and EC = 11 in., is 
 DE II BC? 
 
 Ex. 37. What relation is tliere between Prop. I and Prop. Ill ? 
 
 269. Def. If P is a point of segment 
 
 AB, then P divides AB internally into two a £ b 
 
 segments AP and PB. 
 
 Ex. 38. Construct a A ABC having AB = 2 in., BC = 4 in., and 
 -4(7 = 4.5 in. Let the bisector of Z J5 meet ^O at D. Measure AD and 
 DC. Compare the ratio of AB to BC with the ratio of AD to DC. 
 
 Note. — Supplementary Exercises 1 to 3, p. 289, can be studied now. 
 
SIMILAR POLYGONS 
 
 153 
 
 Proposition IV. Theorem 
 
 270. In any triangle, the bisector of an interior angle 
 divides the opposite side internally into segments pro- 
 portional to the adjaceiit sides of the triangle. 
 
 Hypothesis. 
 Conclusion. 
 
 2. Then, in A EBC, 
 
 AD bisects Z A oi A ABC, meeting BC at D. 
 
 BD^BA 
 
 DC AC 
 
 Proof. 1. Draw BE II DA, meeting CA extended at E. 
 
 BD EA ^, o 
 
 _ = _. Why? 
 
 Prove now that BA = EA and substitute it for EA in step 2. 
 
 Suggestions. — (1) Recall § 123. (2) Compare Z 1 aud Z 3 with Z 5 and Z 4 
 respectively, and use the hypothesis. 
 
 Ex. 39. The sides of a given triangle are 10, 20, and 12 inches re- 
 spectively. Find the segments of the side of length 12 in. made by the 
 bisector of the angle opposite it. 
 
 Ex. 40. The sides of a triangle are 6, 7, and 8 inches respectively. 
 Find the segments of each side made by the bisector of the opposite angle. 
 Note. — Supplementary Exercises 4 to 5, p. 289, can be studied now. 
 
 SIMILAR POLYGONS 
 
 271. Introduction. The triangles below are similar tri- 
 angles. Notice that they appear to have the same shape. 
 
154 PLANE GEOMETRY — BOOK III 
 
 Ex. 41. Construct a A ABC having AB = 2 in., Z^ = 50°, and 
 ZB = 80°. Also construct a A A'B' C, having A'B' = iin., ZA' = 50°, 
 and ZB' = 80°. 
 
 (a) Do the triangles appear to be similar, in the sense that they have 
 the same shape ? 
 
 (&) Determine the lengths of BG, B'C, AC, and A'C. Then deter- 
 mine the approximate values of the ratios: ABiA'B'; BC:B'C'i 
 AC: A'C. 
 
 (c) Do the ratios appear to be about equal ? 
 
 272. Def. Two polygons are similar (~) if: 
 
 (1) Their homologous angles are equal ; 
 
 (2) Their homologous sides are proportional. 
 
 E' D' 
 
 Thus, ABCDE^A'B'C'D'E' if : 
 
 (1) ZA=ZA'', ZB = ZB'', ZC^ZO', etc. and 
 
 (2)^=^ = ^ = ...etc. 
 ^ ^ A'B' B'C CD' 
 
 The ratio of any two homologous sides of two similar poly- 
 gons is called the Ratio of Similitude of the polygons. 
 
 Note. — Two polygons may have their homologous angles equal and 
 still fail to be similar ; as a square and a rectangle. 
 
 Ex. 42. Are two squares similar ? Why ? 
 
 Ex. 43. Are two equilateral triangles similar ? Why ? 
 
 Ex. 44. Are two rectangles necessarily similar ? 
 
 Ex. 45. The sides of one triangle are 1 in., 1.5 in., and 2 in. respec- 
 tively. The shortest side of a similar triangle is 2 in. What are the other 
 sides of the second triangle ? Construct the two triangles. 
 
 Ex. 46. The sides of one pentagon are 3, 4, 5, 8, and 11 in. respec- 
 tively. The shortest side of a similar pentagon is 9 in. How long are the 
 remaining sides of the second pentagon ? 
 
SIMILAR POLYGONS 
 
 155 
 
 Proposition V. Theorem 
 
 273. Two triangles are similar if they are mutually 
 equiangular. 
 
 Hypothesis. In A ABC and AXFZ: 
 
 /:a = z.X', zb^^Y', zc=z.z. 
 
 Conclusion. A ABC ^ AXYZ. 
 
 Plan. We must prove the homologous sides proportional. 
 
 Proof. 1. Place AXYZm the position ADE, Z. X coincid- 
 ing with Z Ay and vertices y and Z falling at D and E respec- 
 tively. 
 
 2. Since Z ADE= Z B, then DE |j BG. Why ? 
 
 .-. 4^ = ^^ Why? 
 
 AD AE ^ 
 
 AB^AC 
 
 XY XZ 
 
 5. By placing A XZY so that Z Y coincides with its equal 
 Z B, it may be proved that 
 
 AB BC 
 
 3. 
 
 4. That is 
 
 XY YZ 
 
 6. From steps (4) and (5), 11= 11= ||. 
 
 7. 
 
 AABC^AXYZ. 
 
 Why? 
 , §272 
 
 274. Cor. 1. Tivo triangles are similar if two angles of one 
 are equal respectively to two angles of the other. 
 
 Suggestion. — Recall § 111. 
 
 275. Cor. 2. Two right triangles are similar if an acute 
 angle of one is equal to an acute angle of the other. 
 
156 
 
 PLANE GEOMETRY — BOOK III 
 
 276. Cor. 3. Two triangles are similar if their sides are 
 parallel each to each. 
 
 Hypothesis. In A ABG and (either) A A'B'C : 
 AB II ^'i3'; BG II B^O ; AC \\ A'C. 
 
 Conclusion. A ABC ^ A A'B'C. 
 
 Suggestion. — Recall § 105. 
 
 277. Cor. 4. Two triangles are similar if their sides are 
 perpendicular each to each. ^ 
 
 Hypothesis. In A ABC and A A'B'C ; 
 
 AB±A'B'; BC±B'C'', AC ± A'C. 
 Conclusion. A ABC ~ A A'B'C, ^ 
 
 Ex. 47. Construct any triangle ABC. Upon a segment XY which 
 equals 2 AB construct a triangle similar to A ABC. (§ 274) 
 
 Ex. 48. If X and Y are any two points on the side ^ ^^ 
 
 BC ot acute angle ABC and XW and YZ are perpen- 
 diculars to AB, then A BX W ^ hBYZ. ^ 
 
 w z 
 
 Ex. 49. If chords AB and CD of a circle intersect at E within the 
 circle, then A AED ~ A BEC. 
 
 Ex. 50. If AD and CE are the altitudes drawn from A and O respec- 
 tively in A ABC, then AABD-^A CBE. 
 
 Ex. 51. In the figure for Ex. 50, if AD intersects CE at 0, prove 
 A ^£"0 is similar to A CDO. 
 
 Ex. 52. If a line be drawn parallel to the base of a triangle intersect- 
 ing the other two sides, the triangle formed is similar to the given triangle. 
 
 278. Homologous sides of similar triangles are proportional. 
 
 Homologous sides lie opposite equal angles. The following 
 exercise illustrates a device for selecting the homologous sides and 
 for forming the three equal ratios. 
 
SIMILAR POLYGONS 
 
 157 
 
 Illustrative Exercise 
 
 Hypothesis. In A ABC: 
 
 AE±BC', 
 BD±AC. 
 
 n 1 . BD BC 
 
 Conclusion. — — = -— 
 
 AE AC 
 
 Proof. 
 
 CD 
 EC 
 
 1. InAAECsLTidABDC: 
 Z3 = Z4; ZC=ZG', 
 
 2. 
 
 .-. A AEC - A BDC, 
 
 3. 
 
 In A AECsiud A BDC: 
 
 
 AC 
 
 Z3 = Z4 
 
 BC 
 
 
 AE 
 
 Z(7=Z(7 
 
 BD 
 
 
 EC 
 
 Z2 = Z1 
 
 DC 
 
 A 
 
 
 AC AE 
 
 EC 
 
 Why? 
 Why? 
 
 Read 
 
 Note 1 
 
 now. 
 
 4. 
 
 BC BQ DC 
 
 §278 
 
 Note 1. — (a) Below A AEC^ write its three angles and to the left of 
 them write the sides which are opposite them in A AEC. 
 
 (6) Opposite the angles of A AEC write the equal angles of A BDC. 
 
 (c) Beside the angles of A BDC, write the sides opposite them in 
 A BDC. 
 
 (d) Then ^Cand BC are horaologoua sides ; also AE and BD', also 
 EC and DC. 
 
 Note 2. -^ Sometimes only two equal ratios are wanted. In that case, 
 cross out in step 4 the ratio that is not wanted. 
 
 279. Principle IV. Four segments can be proved propor- 
 tional by proving them homologous sides of similar triangles. 
 
 Ex. 53. If ^5 ±^C and DC ±50, prove that Ap^^ 
 
 AB^BO^AO^ J^"-"--^o c 
 
 DC CO do' ^d 
 
 Ex. 54. If the altitudes AD and CE oiAABC intersect at F, prove 
 that^F: CF=EF'.DF. 
 
 Ex. 55. In the figure for Ex. 54, prove that AD : CE = AB : CB. 
 Ex. 56. Prove that the diagonals of a trapezoid divide each other 
 so that the corresponding segments are proportional. 
 
 Note. — Supplementary Exercises 6 to 13, p. 290, can be studied now. 
 
158 
 
 PLANE GEOMETRY — BOOK III 
 
 Proposition VI. Theorem 
 
 280. Two triangles are similar if an ajigle of one 
 equals an ayigle of the other and the sides including 
 these angles are proportional. 
 
 Hypothesis. In A ABC and A DEF : 
 
 ' DE DF 
 
 Conclusion. 
 
 A ABC -^ A DEF. 
 
 Proof. 1. Place A DEF in the position AXY, Z D coincid 
 ing with its equal Z. A, E falling at X and F at Y. 
 AB^AC 
 AX AY 
 .-. XY II BO. 
 .: A AXY ^ A ABC. 
 .'. A DEF '^ A ABC. 
 
 2. Then 
 
 3. 
 4. 
 
 5. 
 
 Hyp. and step 1 
 
 Why? 
 
 Give the full proof. 
 
 Why? 
 
 Ex. 57. Two segments AOB and COD intersect at so that AG — 
 3 OB and CO = S CD. Prove AC = 8 BD. 
 
 Ex. 58. ZAoi A ABC is a right angle. From E, any point of AC^ 
 ED is drawn perpendicular to BC^ meeting it at Z>. (a) Examine the 
 figure to discover a pair of similar triangles ; (&) prove the triangles 
 similar ; (c) from these triangles determine the three equal ratios of 
 sides of the triangle. 
 
 Ex. 59. Z ABC is an acute angle. CD is perpendicular to AB 
 and AF is perpendicular to BC. (a) Discover a pair of similar tri- 
 angles ; (&) prove the triangles similar ; (c) write down three equal 
 ratios of sides of these triangles. 
 
 Ex. 60. The shadow of a chimney is 36 yd. long. At the same time 
 the shadow of a stake 2 yd. long is 1.5 yd. in length. How high is the 
 chimney ? 
 
SIMILAR POLYGONS 
 
 159 
 
 Proposition VII. Theorem 
 
 281. Two triangles are similar if their homologous 
 sides are proportional. 
 
 Hypothesis 
 
 Conclusion. 
 Proof. 1. 
 Draw XY. 
 
 2. Then 
 
 3. 
 4. 
 
 5. 
 
 6. But 
 
 In A ABC and A DEF: 
 
 AB^AC^BC 
 
 DE DF EF 
 
 A ABC '^ A DEF. 
 
 On AB, take AX = DE : on AC take A Y= DF. 
 
 AB 
 AX 
 
 .'. XT 
 .'. AAXY 
 AB^BC 
 AX 
 
 AB 
 DE 
 
 , or 
 
 XY DE 
 
 AC 
 AY 
 BC. 
 
 A ABC Give the 
 AB^BC 
 
 xy' 
 
 BC 
 
 ef' 
 
 ~T^ = -^p^- By hyp. and step 1 
 
 Why? 
 full proof. 
 
 Why? 
 Why? 
 
 Then 
 
 .-. XY=EF. 
 AAXY^ADEF. 
 . A DEF ^ A ABC. 
 
 Note. — Notice that A DEF is not superposed on A ABC ; 
 A AXY is constructed, and is proved similar to A ABC and 
 A DEF. 
 
 §251 
 
 Give the full proof. 
 
 Why? 
 
 that, rather, 
 congruent to 
 
 Ex. 61. Construct any scalene triangle. Then construct a triangle 
 whose sides are double the corresponding sides of the first triangle. Are 
 the two triangles similar ? 
 
 Ex. 62. Determine three segments which shall bear to the sides of a 
 given triangle the ratio 3:2. Then construct the triangle having the new 
 segments as sides. Are the two triangles similar ? 
 
160 
 
 PLANE GEOMETRY — BOOK III 
 
 Proposition YIII. Theorem 
 
 282. Homologous altitudes of similar triangles have 
 the same ratio as any two homologous sides, 
 
 4 X 
 
 Hypothesis. AABC^/^XYZ. 
 
 AD and XTTare homologous altitudes. 
 
 Conclusion. ^ = ^=^ = ^. 
 
 XW XY YZ XZ 
 
 Proof. 1. In rt. A ABD and rt. A XYW: 
 2. .'./\ABD^/\XYW. 
 
 " XW XY 
 
 4. But^ = ^ = ^. 
 
 XY YZ XZ 
 
 g . AD ^AB ^ BO ^AG 
 
 YZ ' 
 
 XW XY 
 
 XZ 
 
 §272 
 Why? 
 
 Why? 
 
 Why? 
 
 Ax. 1, § 51 
 
 Note. — It can be proved that any two homologous lines of similar 
 triangles are proportional to any two homologous sides. 
 
 Ex. 63. The base and altitude of a triangle are 5 ft. and 3 ft. re- 
 spectively. If the homologous base of a similar triangle is 7 ft., find its 
 homologous altitude. 
 
 Ex. 64. Prove that the bisectors of homologous angles of similar tri- 
 angles have the same ratio as any two homologous sides. 
 
 Suggestion. — The length of the bisector is the length of the segment of 
 the bisector between the vertex of the angle and the opposite side of the 
 triangle. 
 
 Ex. 65. Prove that two homologous medians of two similar triangles 
 have the same ratio as any two homologous sides of the triangles. 
 
 Suggestion. — Use § 280. 
 
SIMILAR POLYGONS 161 
 
 Proposition IX. Theorem 
 
 283. If tivo chords are draion through a fixed point 
 within a circle, the product of the segments of one is 
 equal to the product of the segments of the other. 
 
 Hypothesis. AB and CD are any two chords of O inter- 
 secting at point P. 
 
 Conclusion. APPB^DP- PC. 
 
 Analysis. 1. ltAP'PB = DP- PC, then AP : PC = DP : PB. 
 
 §252 
 2. .-. try to prove A APD ~ A PBC. § 279 
 
 Proof. 1. Draw AD and BC. 
 
 2. A APD ~ A PBC. Give the full proof. 
 
 3. ...^=^. Why? 
 
 PC PB ^ 
 
 4. .-. AP'PB = PC' DP. Why ? 
 
 284. Principle V. To prove that the product of two seg- 
 ments equals the product of two other segments, first derive 
 from the equation a proportion by § 252 and then try to pro- 
 ceed as in § 279. 
 
 This principle is illustrated in the proof of § 283. 
 
 Ex. 66. Two chords of a circle intersect so that the segments of one 
 are 4 in. and 5 in. respectively. If the shorter segment of the other is 
 6 in., what is the longer segment ? 
 
 Ex. 67. In a circle whose diameter is 16 in., a chord 14 in. long is 
 drawn through a point which is 4 in, from the center. What are the two 
 segments of the chord ? (Represent one segment by x.) 
 
162 
 
 PLANE GEOMETRY — BOOK III 
 
 285. If a secant PA is drawn to a circle 
 from a point P, cutting the circle at point 
 B, then PA is called the whole secant, PB 
 the external segment, and AB the internal 
 segment. 
 
 Propositio:n" X. Theorem 
 
 286. If any two secants are drawn through a fixed 
 point outside a circle^ the product of one and its ex- 
 ternal segment equals the product of the other and its 
 external segment. 
 
 Hypothesis. ABP and CDP are two secants of O 0. 
 Conclusion. AP'BP= CP- DP. 
 
 Suggestion. — Make an analysis and a proof similar to that for Proposition 
 IX. 
 
 Ex. 68. From a point P, a secant 18 in. long is drawn to a circle ; 
 the external segment is 3 in. The external segment of a second secant 
 from the same point is 6 in. long. How long is the whole secant ? 
 
 Ex. 69. A secant is drawn from point P to a circle. The external 
 segment is 4 in. and the internal segment is 6 in. How long must a 
 second secant be in order that its internal segment shall be 3 in. ? 
 
 Ex. 70. If from an exterior point P, any number of secants be 
 drawn, the product of the whole secant and the external segment is 
 constant. 
 
 Note. — The conclusion means that the product of the whole secant 
 and the external segment is the same for each secant. 
 
 Ex. 71. Prove that the product of the segments of one diagonal of 
 an inscribed quadrilateral is equal to the product of the segments of the 
 other diagonal. 
 
SIMILAR POLYGONS 163 
 
 Proposition XI. Theorem 
 
 287. If a secant and a tangent are draioii to a circle 
 from the same point outside a circle, the square of the 
 tangent is equal to the product of the whole secant and 
 its external segment. 
 
 H3rpothesis. PC is a tangent to O ; secant PA intersects 
 the O at ^ and A. 
 Conclusion. CP"^ = AP - BP. 
 
 (Analysis and proof left to the pupil.) 
 Note. — Proposition XI maybe stated: If a secant and a tangent are 
 drawn to a circle from the same point outside the circle, the tangent 
 is the mean proportional between the whole secant and its external 
 segment. 
 
 For, when CP = AP   BP, AP : CP = CP : BP. Why ? 
 
 Ex. 72. The length of the tangent to a circle from a point outside 
 is 12 in. What must be the length of a secant from the same point in 
 order that the external segment will be 8 in. ? 
 
 Ex. 73. If altitudes AD and BE of A ABC intersect at F, prove 
 that the product of the segments of one is equal to the product of the 
 segments of the other. (Apply § 284.) 
 
 Ex. 74. If AD and BE are two altitudes of A ABC, then AD   BC 
 = BE.AC. 
 
 Ex. 75. If Z J. of A ABC is a right angle and ED is drawn perpen- 
 dicular to CB from any point E of AB, meeting CB at D, then EB   AB 
 = CB . DB. 
 
 Ex. 76. If altitudes AD and BE of A ABC intersect at F, then : 
 (a) BE • EF= AE'EC; (6) AD   DF = BD • DC. 
 
 Note. — Supplementary Exercises 14 to 18, p. 290, can be studied now. 
 
164 
 
 PLANE GEOMETRY — BOOK III 
 
 Proposition^ XII. Theorem 
 
 288. If the altitude he draivn to the hypotenuse of a 
 right triayigle : 
 
 I. The altitude is the mean proportional hetiveen the 
 segments of the hypotenuse ; 
 
 II. Each leg is the mean propo7'tional hetioeen the 
 whole hypoteiiuse and the adjacent segment. 
 
 Hypothesis. In A ABC, Z (7 is a rt. Z. 
 CD ± AB, 
 
 Conclusion. I. ^^ = ^. 
 
 CD DB 
 
 Proof. 1. Z 3 and Z 1 are each complements of Z 2. 
 
 Why? 
 
 .-. Z 1 = Z 3. Why ? 
 
 2. .-. A ADC ~ A DCB, Give the full proof. 
 
 "CD db' 
 
 See § 278 
 
 Conclusion. II. (a) 
 
 Plan. For («) prove A ABC '^ A ACD. 
 
 AB^AC, ... AB^BC 
 AC AD' ^ ^ BC db' 
 
 Cor. If a perpendicular be drawn 
 from any point on a circle to a diameter : 
 
 (a) The perpendicular is the mean propor- 
 tional between the segments of the diameter. 
 
 (b) The chord joining the point to either 
 extremity of the diameter is the meayi proportional between the 
 whole diameter and the segment of it adjacent to the chord. 
 
SIMILAR POLYGONS 165 
 
 Hx. 77. Find the altitude drawn to the hypotenuse of a right tri- 
 angle if it divides the hypotenuse into two segments whose lengths are 
 3 in. and 12 in. respectively. Find each leg of the riglit triangle. 
 
 Ex. 78. The hypotenuse of a right triangle is 20 in. and the perpen- 
 dicular to it from the opposite vertex is 8 in. Find the segments of the 
 hypotenuse, and the two legs of the triangle. 
 
 Ex. 79. C and D are respectively the mid-points of a chord AB and its 
 subtended arc. If J.i> = 12 and CZ>=8, what is the diameter of the circle ? 
 Suggestion. — DC extended passes through the center of the circle. 
 
 Ex. 80. A chord of a circle is 20 in. in length. Its mid-point is 5 in. 
 from the mid-point of its arc. Find the diameter of the circle. 
 
 Proposition XIII. Theorem 
 
 290. Construct the mean proportional between two 
 given segments. j) 
 
 A'- 
 
 \ 
 
 1 
 
 B'' C 
 
 ^.4_n_-...l_._._E 
 
 Given segments m and n. 
 
 Required to construct the mean proportional between m and n. 
 
 Construction. 1. On line AE take AB = m and BC = n. 
 
 2. Construct semicircle ADC on AC as diameter. 
 
 3. Construct DB _L AC, meeting the semicircle at D. 
 Statement. DB is the mean proportional between m and n. 
 
 (The proof Ls to be given by the pupil. See § 289.) 
 
 Ex. 81. Construct a segment equal to aV3 where a is any segment 
 whatever. 
 
 Analysis. 1. Let a; = a V3. Then x^ = 3 a^. Why ? 
 
 2. .-. 3a : X =a;:a. § 262 
 
 3. .-. X is the mean proportional between a and 3 a. 
 
 (Construction is to be given by the pupil.) 
 
 Ex. 82. Construct a segment equal to VZab where a and b are any 
 given segments. 
 
166 PLANE GEOMETRY — BOOK III 
 
 • Proposition XIV. Theorem 
 
 291. The square of the hypotenuse of a right triangle 
 is equal to the sum of the squares of the legs. 
 
 
 Hypothesis. In A ABC, Z (7 is a right angle. 
 
 Conclusion. c^ = a^ + b\ 
 
 Proof. 1 . Draw CD ± AB. Let AD = r, and DB = s. 
 
 2. 
 
 Then 
 
 as h r 
 
 § 288, II 
 
 3. 
 
 
 .'. a^ = cs and ¥ = cr. 
 
 Why? 
 
 4. 
 5. 
 
 
 .-. a^ -\- b"^ = cs -\- cr 
 
 = c(s-^r). 
 ... ct2 + 62 = c • c, or a2 + 62 ^ c^. 
 
 
 Note. — This theorem is called the Pythagorean Theorem, after Pythag- 
 oras, who formulated it. The theorem was evidently known even to the 
 Egyptians. This proof of the theorem is attributed to Hindu mathema- 
 ticians. In Book IV, we shall study Euclid's proof of the theorem — a 
 strictly geometric proof, whereas this is more an algebraic one. 
 
 292. Cor. TJie square of either leg of a right triangle is 
 equal to the square of the hypotenuse minus the square of the other 
 leg. 
 
 Ex. 83. How long must a rope be to run from the top of a 12-foot 
 tent pole to a point 16 ft. from the foot of the pole ? 
 
 Ex. 84. The diameters of two concentric circles are 14 in. and 60 in., 
 respectively. Find the length of a chord of the greater circle which is tan- 
 gent to the smaller. 
 
 Ex. 85. A baseball diamond is a square whose sides are each 90 ft. 
 Ions;. What is the distance from " first" to " third" ? 
 
SIMILAR POLYGONS 167 
 
 Ex. 86. Find the formula for the diagonal of a square whose side is s. 
 By this formula determine the diagonal when : (a) s = 10 ; (b) s = 15. 
 
 Ex. 87. The equal sides of an isosceles trapezoid are each 10 in. long. 
 One of the bases is 30 in., and the other is 42 in. in length. What is the 
 altitude of the trapezoid ? 
 
 Ex. 88. Find the length of the altitude of an equilateral triangle if 
 each side is 10 in. 
 
 Ex. 89. Derive the formula for the length of the alti- 
 tude of an equilateral triangle if each side of the triangle is s. 
 By this formula determine the altitude when : , ^ 
 
 {a) s = 6in. ; (6) 8 = 13 in. 
 
 Ex. 90. A piece of silk 27 in. wide is folded " on the 
 bias " along the line AB. How long is AB ? 
 
 Ex. 91. Find the length of each side of a rhombus if the diagonals 
 are 6 in. and 8 in. respectively. 
 
 Ex. 92. If AD is the perpendicular from A to BC of A ABC, prove 
 
 AB^ - 'AO^ = DB^ - CD^. 
 
 Plan. Find an expression for Alf and AC^ ; then subtract the latter 
 from the former. 
 
 Note. — This might be called a "common sense" plan. After form- 
 ing in this manner the left member, if the right member is not obtained at 
 once, then proceed to form in the same manner the right member, and 
 afterwards try to prove the two values obtained are equal. 
 
 Ex. 93. If D is any point in the altitude from A to side BC ot 
 A ABC, prove that A^ - AC^ = ~DB^ - DC\ 
 
 Suggestion. — Read the note under Ex. 92. 
 
 Ex. 94. If a parallel to hypotenuse AB of right triangle ABC 
 meets AC and BC at D and E respectively, prove that 
 
 'AE'^ 4- BD^ ='AB^ + DE^. 
 
 Ex. 95. If perpendiculars PF, PD, and PE be 
 
 drawn from any point P within an acute-angled tri- 
 angle ABC to sides AB, BC, and CA respectively, F 
 prove that 
 
 AF^ -^ BB^ -\-CE^ = AE^ +^F^ + CJD^. " D 
 
 Note. — Supplementary Exercises 19 to 40, p. 291, can be studied now. 
 
168 
 
 PLANE GEOMETRY — BOOK III 
 
 Proposition XY. Theorem 
 
 293. Two polygons are similar if they are composed 
 of the same number of triangles, similar each to each, 
 and similarly placed. 
 
 Hypothesis. A AEB ^ A A'E'B' ; A EBD ~ A E'B'D' ; 
 ABCD^AB'C'D'. 
 The triangles are similarly placed. 
 
 Conclusion. Polygon ABODE ~ polygon A'B'C'D'E'. 
 
 Analysis. The homologous A must he proved equal, and the homolo- 
 gous sides must he proved proportional. § 272 
 
 Proof. 1. Z1 = Z1'; Z2 = Z2'; Z3 = Z3'. Why? 
 
 2. .-. ZjB = Z5'. Why? 
 
 3. Similarly, Z Z> = Z Z)' ; ZE = ZE'', ZA^ZA'; 
 
 z c=za. 
 
 BE 
 
 4. 
 5. 
 
 6. Similarly, 
 
 7. Also 
 8. 
 
 AE_^BE^ ^^^ ED^^ 
 
 A'E' B'E'' E'D' B'E' 
 
 AE ED 
 
 Prove it. 
 Why? 
 
 A'E' E'D' 
 ED ^ CD 
 E'D' CD' ' 
 
 ^ = ^,and^ 
 A'B' A'E" B'C 
 
 CD 
 
 AB 
 
 AE ED CD 
 
 CD' 
 
 ^ ^ ^ ^ BC 
 
 A'B' A'E' E'D' CD' B'C' 
 9. .-. Polygon ABCDE ^ polygon A'B' CD' E'. 
 
 Ax. 1, § 51 
 
 Why? 
 
 Ax. 1, § 51 
 Why? 
 
SIMILAR POLYGONS 
 
 169 
 
 Proposition XVI. Problem 
 
 294. Upon a given segment ^ homologous to a given 
 side of a given polygon, construct a polygon similar to 
 the given polygon. 
 
 Given polygon ABCDE and segment A'B'. 
 
 Required to construct upon A'B' as side homologous to AB 
 a polygon similar to ABCDE. 
 
 Construction. 1. Divide ABCDE into triangles by draw- 
 ing EB and BD. 
 
 2. Construct A A'B'E' similar to A ABE, by making Z A' 
 = ZA, andZl' = Zl. 
 
 3. Construct A E'B'D' similar to A EBD. How ? 
 
 4. Construct A D'B'C similar to A DBC.   How ? 
 Statement, polygon A'B'CD'E' ~ polygon ABCDE. 
 
 Why? 
 Ex. 96. ABCD is the shape of an irregular 
 piece of groimd. Make a figure similar to ABCD 
 such that each side of the resulting figure shall be 
 three times as long as the corresponding side of the 
 given figure. 
 
 Ex. 97. An ordinary shed roof is said to have a 
 the distance AB is one third of the distance CD. 
 
 A carpenter wishes to order some " two by fours "" for 
 the rafters AE of a garage which is to be 24 ft. wide and 
 have a one-third pitch. He allows \ ft. for cutting at 
 the end A^ and wants the rafters to project beyond the 
 wall at C so that CE will be 2 ft. What length of " two 
 by fours " must he order if they can be obtained only in even lengths ? 
 
 one-third pitch 
 
170 
 
 PLANE GEOMETRY — BOOK III 
 
 Pkoposition XVII. Theorem 
 
 295. Tivo similar polygons can he divided into the 
 same number of triangles, similar each to each, and 
 similarly placed. 
 
 Hypothesis. Polygon ABODE ~ polygon A'B'C'D'E', ho- 
 mologous vertices being indicated by corresponding letters. 
 
 Conclusion. The polygons can be divided into the same 
 number of triangles, similar each to each and similarly 
 placed. 
 
 Construction. Draw BE, BD, B'E', and B'D'. 
 Statement. A ABE ~ A A'B'E' ; A EBD ~ A E'B'D' ; 
 A BCD ~ A B'C'D'. 
 
 AB AE 
 
 Proof. 1. 
 
 2. 
 3. 
 
 4. 
 
 ZJ. = Zyl'and , ^, , 
 
 A'B' A'E' 
 
 .-. A BAE ~ A B'A'E'. 
 
 ZE = ZE' Sind Z 4: = Z 4'. 
 
 .-. Z5 = Z5'. 
 
 BE AE -, T ED AE 
 
 B'E' 
 
 AE . 1 ED 
 
 — ; — -, and also -^ 
 
 A'E'' E'D' 
 
 A'E' 
 
 BE 
 
 ED 
 
 5. 
 
 6. Similarly 
 
 B'E' E'D' 
 .'. A EBD ~ A E'B'D'. 
 A BCD ~ A B'C'D'. 
 
 Why? 
 
 Why? 
 Why? 
 Why? 
 
 Why? 
 
 Why? 
 Why? 
 
 Note. — If X and T are any two points of one polygon and X' and 
 Y' are the homologous points of a similar polygon, then XZand XT' 
 are homologous segments and XY : X'Y' equals the ratio of similitude. 
 
SIMILAR POLYGONS l7l 
 
 296. Fundamental Theorem about Equal Ratios. 
 
 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. 
 
 If -=- = - = 1 
 
 b d f h' 
 
 then ^±_^+l+^ = « = ^ = etc. 
 
 h-{-d-\-f-^h b d 
 
 Proof. 1. Let r = - and hence br = a. 
 b 
 
 2. .-. dr = c, fr = e, hr = g. Prove it. 
 
 3. .'. br -\- dr -\- fr -^ hr = a -\- c -^ e + g. Why? 
 
 4. ...,.=:!L±A+i+Lg. Prove it. 
 
 b + d+f-hh 
 
 5. Hence ^Jl^_±l+5 = « = £ = etc. 
 
 b + d-{-f-{-h b d 
 
 Proposition XVIII . Theorem 
 
 297. JTie perimeters of two similar polygons have the 
 same ratio as any tioo homologous sides. 
 
 B 
 
 b' 
 
 D E U 
 
 Hypothesis. ABCDE and A'B'C'D'E' are similar polygons 
 with homologous vertices indicated by corresponding letters. 
 
 Conclusion. 
 
 AB + BC+CD + DE-^EA ^ AB ^ BO ^ 
 A'B' + B'C + CD' + D'E' + E'A' A'B' B'C ^ ^' 
 
 Proof. 1. -^ = ^ = -^ = etc. 
 
 A'B' B'C CD' 
 
 (Complete the proof. Apply §296.) 
 Note. —Supplementary Exercises 41 to 44, p. 293, can be studied now. 
 
172 PLANE GEOMETRY — BOOK III 
 
 SUPPLEMENTARY TOPICS 
 
 Five groups of theorems follow, — all of which appear in 
 modern geometries. It is not necessary, — in fact, it may be 
 unwise to study all of them in every class. The teacher 
 should feel free to select the group or groups which appear to 
 be of most value to the class. 
 
 Each group is independent of each of the others. 
 
 None of these theorems is required as an authority in the main 
 lists of theorems of -succeeding Books. 
 
 Group A, — Scales and Scale Drawing. 
 
 Group B. — Trigonometric Ratios and their Application. 
 
 These two groups are interesting and valuable applications of 
 Book III. 
 
 Group C. — Proportional Division of a Segment. 
 Group D. — External and Harmonic Division of a Segment. 
 Group ^. — Numerical Relations among Segments of a 
 Triangle. 
 
 These last three groups have long appeared in geometries. 
 
 A. Scales and Scale Drawing 
 
 298. Scale drawings are a common and 
 useful application of similar polygons. 
 
 The adjoining figure represents a lot 150' x 
 276'. It is drawn to the scale of 1" to 200 ' ; that 
 is, AB, I" in length, represents 150' and jBC, If" in 
 length, represents 275'. If the corners of the lot 
 itself are denoted by J.', B', C, and D' respectively, 
 then AB : A'B' =-. 1 : 2400, and 5(7 : 5' C = 1 : 2400. 
 
 (1" to 200' is 1" to 2400" or 1 : 2400.) 
 
 ABCD is similar to A'B' CD', for the two figures 
 are mutually equiangular (being rectangles) and 
 their homologous sides are proportional (the ratio of similitude being 
 1:2400). 
 
 Since, in similar figures, the ratio of any two homologous sides equals 
 the ratio of similitude (Note, § 295), it is possible to determine from 
 ABCD the approximate length of any segment on the field itself. 
 
SCALES AND SCALE DRAWING 173 
 
 299. Scales. The construction and use of scale drawings 
 are made easy by the construction in advance of the scale it- 
 self, unless such a scale is already at hand. 
 
 Example. — Below is the scale of 1" to 100' to measure 
 350 ft. 
 
 Scale r = 100 
 
 The segment extending from the zero mark to any division 
 point represents the number of feet indicated above that point. 
 Notice that the left-hand section is divided into ten equal 
 parts. 
 
 To determine the number of feet represented by a given 
 segment according to the given scale : take the segment on the 
 dividers ; place one point of the dividers on a division mark 
 at the right of the zero mark, so that the other point of the 
 dividers will fall on the scale either at the zero mark or to the 
 left of it. The length represented by the segment may then 
 be read to the nearest 5 feet. 
 
 Thus, the segment a below represents 235 ft. if drawn to the scale of 
 1" to 100'. 
 
 Ex. 98. Determine the length represented by each of the following 
 segments, assuming that they are drawn to the scale of 1" to 100'. 
 
 Ex. 99. Determine the approximate number of feet represented by 
 the diagonal ^C in the figure of § 298. 
 
 Ex. 100. Determine the approximate distance of the tree, T, from 
 each of the comers of the lot ABCD in the figure of § 298. 
 
 Ex. 101. Construct the scale of 1" to 1' to measure 5 ft., having 
 the left-hand section show the segments corresponding to 2", 4", etc., 
 to 12". 
 
 What length, in feet and inches, do segments a, 6, and c of Ex. 98 
 represent if it is assumed that they are drawn to the scale of 1" to 1' ? 
 
174 
 
 PLANE GEOMETRY — BOOK III 
 
 Ex. 102. Construct the scale of 1" to 25', to measure 100 ft., having 
 the left-hand section show segments corresponding to 5', 10', etc., to 25', 
 
 What length do segments a, 6, c of Ex. 98 represent if it is assumed 
 that they are drawn to the scale of 1" to 25' ? 
 
 Ex. 103. Draw to the scale of 1" to 25' the ad- 
 joining figure. From the figure so drawn, determine 
 the approximate height of AB. 
 
 Ex. 104. Draw to the scale of 1" to 25' a 
 figure similar to the adjoining one. From the 
 resulting figure, determine the approximate dis- 
 tance represented by AB. 
 
 Ex. 105. Draw to the scale of 1" to 25' a figure 
 similar to the adjoining one. From the resulting 
 figure, determine the approximate length of AB, if 
 CD =60', ZACB = QQ°, ZADC=ZO'', said Z ABC 
 = 90°. 
 
 Ex. 106. The perimeters of two similar polygons are 119 and 68 ; if a 
 side of the first is 21, what is the homologous side of the second ? (§ 297.) 
 
 Ex. 107. Draw to the scale of 1" to 100' a 
 figure similar to the adjoining one. From the 
 resulting figure, determine the approximate perime- 
 ter of the field having the dimensions indicated. 
 
 B. Trigonometric Ratios and their 
 Application 
 
 300. Sine of an Angle. Let Z ABO be any angle, 
 on BC any points Pi and Pg- Draw per- 
 pendiculars PiEi and P2R2 to AB. 
 
 Then ABP^R.^ABPA- (Why?) 
 
 . R\P\ -K2P2 
 
 ** 'BPi~'BP,' 
 
 That is, the ratio of the perpendicular BP to the distance 
 BP is the same, regardless of where P^ and P^ are located 
 onjBC 
 
TRIGONOMETRIC RATIOS 
 
 175 
 
 This constant ratio is called the Sine of Z B. (Sin B.) 
 When the angle is acute, its sides and the perpendicular 
 form a right triangle. In this triangle, 
 
 sine of acute angle = side opposite -f- hypotenuse. 
 
 The sine of a given angle may be computed as in the fol- 
 lowing 
 
 Example. — Let Z B= 60°. Determine sin 60°. 
 
 Solution. 1. Draw Pi? ±^A DrawPT^P^. 
 
 2. Then APBT^A PBB, and Z T = 60°. 
 
 (Why?) 
 
 3. .-. A PP7'is equilateral, and BB = ^ BP. 
 
 (Why ?) 
 
 4. Let BP = 2 m and hence BB = m. 
 
 5. In A BPB, BP^ = 4 m^- m2 = 3 m2._ (Why ?) 
 
 6. .-. i?P = »nV3. 
 
 ...sin 60° = ^=??^ =1^232 
 BP 2 m 2 
 
 .866+. 
 
 Ex. 108. Determine as in the example the value of sin 45° and of 
 
 sin 30°. 
 
 Ex. 109. Construct a figure similar to the ad- 
 joining one making : /. ABC = 35° ; Z ABD = 50° ; 
 Z ABE = 65° ; and Z ABF = 75°. Draw the per- 
 pendiculars from C, D, E, and F to AB. Measure 
 these perpendiculars and also the radius. Then com- 
 pute the approximate values of the sine of each of 
 the angles indicated ; that is, of sin 35°, sin 50°, 
 sin 65°, and sin 75°. 
 
 (If you have a metric scale, make AB = 100 mm. and measure the 
 perpendiculars in mm. ; if you do not have a metric scale, make AB = 3| 
 in., and measure the perpendiculars in sixteentlis of an inch. If you use 
 metric measures, your sines should be approximately correct to the 
 second decimal place ; if you use the English scale, the values should 
 be correct to the first decimal place. Keep your figure for use in a later 
 exercise.) 
 
 Ex. 110. Construct the acute angle : 
 
 (a) Whose sine is J ; (6) whose sine is f. 
 
 Ex. 111. BC, 40 mm. long, is on one side of Z ABC, whose sine is ^. 
 How long is CA, the perpendicular to side AB ? 
 
i?j /?2 
 
 176 PLANE GEOMETRY— BOOK III 
 
 301. Cosine and Tangent of an Angle. 
 
 It is easily proved that the ratio -~ ^'^^ 
 
 is constant for all positions of P on J5(7, 
 as in § 300 ; and also that is constant. „ 
 
 ~ is called the Cosine of Z B, (Cos 5.) 
 
 ^ is called the Tangent of Z B. (Tan B^ 
 BR 
 
 In the right triangle formed when Z 5 is an acute angle : 
 
 cosine of acute angle = adjacent side -^ hypotenuse ; 
 
 tangent of acute angle = opposite side -v- adjacent side. 
 
 Example. —When Z5 = 60°, if BP=2m, then BR = m 
 and EP = m V3. (See Example, § 300.) 
 
 Hence, Cos 60° = ^^ = - = .500. 
 
 2m 2 
 
 Tan 60° =V12^ = Vs = 1.732. 
 m 
 
 Ex. 112. Compute the cosine and the tangent of 45° and 30° respec- 
 tively. 
 
 Ex. 113. In the figure constructed for Ex. 109, measure BM, BN, 
 BB, and BS. Then compute the approximate values of : 
 (a) cos 85° ; cos 50° ; cos 65° ; cos 75° ; 
 (6) tan 35°; tan 50°; tan 65°; tan 75°. 
 
 302. Table of Values of Trigonometric Ratios. The values 
 of the sine, cosine, and tangent of certain angles have been 
 computed. See the table opposite. 
 
 To determine the sine of 37° from the table : in the first 
 column find 37° ; on the same line with it, and in the column 
 headed by the word Sin, is found .602. This is the sine of 37°. 
 
 Similarly the cosine or tangent of a given angle may be 
 determined from the table. 
 
TRIGONOMETRIC RATIOS 
 Table of Values of Trigonometric Ratios 
 
 177 
 
 Anolr 
 
 Sin 
 
 Cos 
 
 Tan 
 
 Angle 
 
 Sin 
 
 Cob 
 
 Tan 
 
 10^ 
 
 .174 
 
 .985 
 
 .176 
 
 45° 
 
 .707 
 
 .707 
 
 1.000 
 
 11° 
 
 .191 
 
 .982 
 
 .194 
 
 46° 
 
 .719 
 
 .696 
 
 1.036 
 
 12° 
 
 .208 
 
 .978 
 
 .213 
 
 47° 
 
 .731 
 
 .682 
 
 1.072 
 
 13° 
 
 .226 
 
 .974 
 
 .231 
 
 48° 
 
 .743 
 
 .609 
 
 1.111 
 
 14° 
 
 .242 
 
 .970 
 
 .249 
 
 49° 
 
 .755 
 
 .656 
 
 1.150 
 
 16° 
 
 .259 
 
 .966 
 
 .268 
 
 60° 
 
 .766 
 
 .643 
 
 1.192 
 
 16° 
 
 .276 
 
 .961 
 
 .287 
 
 61° 
 
 .777 
 
 .629 
 
 1.236 
 
 17° 
 
 .292 
 
 .956 
 
 .306 
 
 62° 
 
 .788 
 
 .616 
 
 1.280 
 
 18° 
 
 .309 
 
 .951 
 
 .326 
 
 63° 
 
 .799 
 
 .602 
 
 1.327 
 
 19° 
 
 .326 
 
 .946 
 
 .344 
 
 64° 
 
 .809 
 
 .588 
 
 1.376 
 
 20° 
 
 .342 
 
 .940 
 
 .364 
 
 66° 
 
 .819 
 
 .674 
 
 1.428 
 
 21° 
 
 .358 
 
 .934 
 
 .384 
 
 66° 
 
 .829 
 
 .659 
 
 1.483 
 
 22° 
 
 .376 
 
 .927 
 
 .404 
 
 67° 
 
 .839 
 
 .646 
 
 1.640 
 
 23° 
 
 .391 
 
 .921 
 
 .424 
 
 68° 
 
 .848 
 
 .630 
 
 1.600 
 
 24° 
 
 .407 
 
 .914 
 
 .446 
 
 69° 
 
 .867 
 
 .516 
 
 1.664 
 
 26° 
 
 .423 
 
 .906 
 
 .466 
 
 60° 
 
 .866 
 
 .600 
 
 1.732 
 
 26° 
 
 .438 
 
 .899 
 
 .488 
 
 61° 
 
 .876 
 
 .486 
 
 1.804 
 
 27° 
 
 .454 
 
 .891 
 
 .610 
 
 62° 
 
 .883 
 
 .4t)9 
 
 1.881 
 
 28° 
 
 .469 
 
 .883 
 
 .532 
 
 63° 
 
 .891 
 
 .454 
 
 1.963 
 
 29° 
 
 .485 
 
 .875 
 
 .654 
 
 64° 
 
 .899 
 
 .438 
 
 2.050 
 
 30^ 
 
 .500 
 
 .866 
 
 .677 
 
 66° 
 
 .906 
 
 .423 
 
 2.144 
 
 31° 
 
 .515 
 
 .857 
 
 .601 
 
 66° 
 
 .914 
 
 .407 
 
 2.246 
 
 32° 
 
 .530 
 
 .848 
 
 .625 
 
 67° 
 
 .921 
 
 .391 
 
 2.356 
 
 33° 
 
 .545 
 
 .839 
 
 .649 
 
 68° 
 
 .927 
 
 .376 
 
 2.476 
 
 34° 
 
 .559 
 
 .829 
 
 .675 
 
 69° 
 
 .934 
 
 .358 
 
 2.606 
 
 36° 
 
 .674 
 
 .819 
 
 .700 
 
 70° 
 
 .940 
 
 .342 
 
 2.747 
 
 36° 
 
 .588 
 
 .809 
 
 .727 
 
 71° 
 
 .946 
 
 .326 
 
 2.904 
 
 87° 
 
 .602 
 
 .799 
 
 .754 
 
 72° 
 
 .961 
 
 .309 
 
 3.078 
 
 38° 
 
 .616 
 
 .788 
 
 .781 
 
 73° 
 
 .956 
 
 .292 
 
 3.271 
 
 39° 
 
 .629 
 
 .777 
 
 .810 
 
 74° 
 
 .961 
 
 .276 
 
 3.487 
 
 40° 
 
 .643 
 
 .766 
 
 .839 
 
 76° 
 
 .966 
 
 .259 
 
 3.732 
 
 41° 
 
 .656 
 
 .756 
 
 .869 
 
 76° 
 
 .970 
 
 .242 
 
 4.011 
 
 42° 
 
 .669 
 
 .743 
 
 .900 
 
 77° 
 
 .974 
 
 .226 
 
 4.331 
 
 43° 
 
 .682 
 
 .731 
 
 .933 
 
 78° 
 
 .978 
 
 .208 
 
 4.706 
 
 44° 
 
 .696 
 
 .719 
 
 .966 
 
 79° 
 
 .982 
 
 .191 
 
 6.146 
 
 48° 
 
 .707 
 
 .707 
 
 1.000 
 
 80° 
 
 .986 
 
 .174 
 
 6.671 
 
 Ex. 114. Obtain from the table : (a) sin 40*^ 
 (b) cos 37° ; cos 76° ; cos 64° ; (c) tan 29° 
 
 sin 63° ; sin 26° ; 
 tan 68° : tan 71°. 
 
 Ex. 115. What is the angle x if : 
 
 (a) sin X = .454 ; sin a; = .829 ; sin x = .978 ; 
 (6) cos X = .966 ; cos a; = .719 ; cos x = .368 ; 
 (c) tan X = .424 ; tan x = 1.111 ; tan x = 3.732. 
 
178 PLANE GEOMETRY — BOOK III 
 
 303. Application of Trigonometric Ratios. 
 
 Example 1. — Assume that at point B 
 ZCBA = 34° and that BC=125 ft. How 
 high above ground is point A? 
 
 Solution. 1 . ^ = tan 34°, ot AC = BC xta,B. 34°. ^ ' 225^ 
 
 2. .-. ^C= 125 X .68=85ft. 
 
 Note. — Z CBA is called the Angle of Elevation of A at point B. 
 
 Example 2. — AB represents a lighthouse j).. 
 250 ft. high. DA is an imaginary line paral- 
 lel to BO. C represents the position of a 
 ship. Z DAC= 31°. How far from B is C? 
 
 Solution. 
 
 1. ZBCA = S1°. 
 
 2. 
 
 T&nSl" -"^^^ or BC-AB : tan 31°. 
 BC 
 
 3. 
 
 .•.BC=: 250 - .601 = 415.9 ft. 
 
 Why? 
 
 That is, BCis about 416 ft. 
 
 Note. — ZDACis called the Angle of Depression of C at A. 
 
 Ex. 116. In the adjoining figure, ii AC ± CD, CD 
 = 100 ft., and ZD = 62°, what are the distances AC 
 and AD ? 
 
 Ex. 117. rind the angle of elevation of the sun 
 when a monument whose height is 360 ft. casts a shadow 
 400 ft. in length. 
 
 Ex. 118. Determine the length of one side of an equilateral polygon 
 having nine sides which is inscribed in a circle of radius 10 ft. 
 
 Determine the distance of each of the sides from the center of the 
 circle. 
 
 Ex. 119. A boy flying a kite knows that he has 500 ft. of twine. 
 Another boy stations himself approximately below the kite. The angle 
 made by the string with an imaginary line running from the first boy to 
 the second is approximately 50°. Determine the approximate height of 
 the kite. 
 
 Ex. 120. At a point 169 ft. from the foot of a tower surmounted by 
 a pole, the angle of elevation of the top of the tower is 35° and that of 
 the top of the pole is 47°. Find the length of the pole. 
 
DIVISION OF A SEGMENT 179 
 
 C, Proportional Division of a Segment 
 Proposition XIX. Theorem 
 
 304. Parallel lines intercept proportional segments on all 
 transversals. 
 
 X 
 
 Hypothesis. Parallels AB, CD, EF, and GH intercept seg- 
 ments a, b, and c, on XY and d, e, and/ on ZW, respectively. 
 
 Conclusion. - = - = -. 
 
 d e f 
 
 Suggestions. — 1. Draw TK II Z W. Let the lU intercept the segments 
 r, s, and t on TK. 
 
 2. Compare r, s, and t, with d, e, and /, respectively. 
 
 3. In A TKL, compare a : r with TL : TK. 
 
 4. In A TKL, compare c : t with TL : TK 
 
 5. In A TMX, compare a : r with b : s. 
 
 6. Then compare a :r, c :t, and b : s. 
 
 Complete tlie proof, using the facts obtained in step 2. 
 
 Ex. 121. Divide a segment into parts proportional to any number of 
 given segments. 
 
 Given segment AB, and segments w, n, and p. 
 Required to divide AB into segments x, y, and z, so that 
 
 — = ^ = -. 
 m n p 
 
 Suggestion. — Base the construction on Proposition XIX. 
 
180 PLANE GEOMETRY — BOOK III 
 
 Ex. 122. Divide a segment 6 in. in length into segments proportional 
 to 2, 3, and 4. 
 
 Ex. 123. Construct a triangle similar to a given triangle, having 
 given its perimeter. 
 
 Ex. 124. A line parallel to the bases of a trapezoid, passing through 
 the intersection of the diagonals, and terminating in the non-parallel 
 sides, is bisected by the diagonals. 
 
 Note. — Supplementary Exercises 45 to 46, p. 293, can be studied now. 
 
 D. External and Harmonic Division of a 
 Segment 
 
 305. External Division of a Segment. If P is a point on 
 line AB but not located between A and B, then P divides AB 
 externally into segments AP and PB. 
 
 FA B A^ B P 
 
 Fig. 1 
 
 The following justification is given for calling AP and PB segments 
 of AB. Direction on a segment may be indicated by reading the seg- 
 ment from its beginning point to its end point ; thus 
 
 AB =A >- B, and BA = A -^ B. 
 
 Direction from left to right is regarded as positive and from right to 
 left as negative. Hence BA =— AB and AB + BA = 0. 
 
 Clearly then the algebraic sum of JP and PB in Fig. 1 equals AB, 
 for AP+ PB=AP+ PA-\- AB and AP + PA ^ 0. 
 
 The consequence is that the algebraic sum of AP and PB is AB, no 
 matter where P is located on the line AB. (See figures below.) 
 
 A. P B p A B A B P 
 
 Fig. 2 
 Note. — Prove that AP + PB = AB in each case in Fig. 2. 
 
 ;^ BCD E^- 125. What segment represents the alge- 
 
 ' •"• braic sum of : (a) AB + J5C? (b) AD -{■DC'} 
 
 (c) BA^- AG -\- CB'} 
 
DIVISION OF A SEGMENT 
 Proposition XX. Theorem 
 
 181 
 
 306. In any triangle, the bisector of an exterior angle at any 
 vertex divides the opposite side externally into two segments whose 
 ratio equals the ratio of the two adjacent sides of the triangle. 
 
 D B 
 
 Hypothesis. AD bisects ext. Z BAE of A ABC, meeting CB 
 
 extended at D. 
 Conclusion. BD:DC= BA : AC. 
 
 Proof. 1. Draw BF II DA, meeting AC at F. 
 2. Then BD : DC = FA: AG. 
 
 It remains to prove that FA = BA. Try to prove that Z3=Z4, 
 using the hypothesis and construction. Proof left to the pupil. 
 
 Note. — The converse of Proposition XX is also true. It may be 
 proved by laying off AF = AB. 
 
 Ex. 126. The sides of a triangle are AB = 5, BC = 7, and CA = 8 ; 
 find the segments into which side 8 is divided by the bisector of the ex- 
 terior angle at the opposite vertex. 
 
 307. A segment is divided harmonically if it is divided inter- 
 nally and externally into seg- j_ X B Y 
 ments having the same ratio. i \ I 
 
 Thus. MAB=\,AX= -, and JIF = % then ^ = ^, since each 
 4 2 JlB YB 
 
 ratio equals -. Hence Xand F divide AB harmonically. 
 
 Ex. 127. Prove that the bisector of an interior angle of a triangle 
 and the bisector of the exterior angle at the same vertex divide the op- 
 posite side harmonically. 
 
 Ex. 128. If Xand F divide AB harmonically, then A and B divide 
 Xr harmonically. (Use § 263. Verify afterwards for the figure in § 307. ) 
 
182 
 
 PLANE GEOMETRY — BOOK III 
 
 E. Numerical Relations among Segments of a 
 
 Triangle ./> 
 
 308. The Projection of a Point upon a 
 given line is tlie foot of the perpendicu- 
 lar drawn from the point to the line. j ^. ^^^ 
 
 Thus, B is the projection of P on AB. 
 
 309. The Projection of a Segment 
 
 upon a given line is the distance between 
 the projections of its end-points. 
 
 Thus, the projection of AB on line CD is 
 A'B'. 
 
 The symbol " jo^^ " is read "the projec- 
 tion of AB on CDr 
 
 C A 
 
 B D 
 
 Ex. 129. Draw a segment AB and also four straight lines not paral- 
 lel to AB but also not crossing AB. 
 
 (a) Determine the projection of AB on each of the straight lines. 
 
 (b) Are the projections all of the same length ? 
 
 Ex. 130. Draw an acute scalene triangle. Show by means of a 
 drawing the projection of the shortest side upon each of the other sides. 
 
 Ex. 131. Repeat the preceding exercise for an obtuse triangle. 
 
 Ex. 132. Draw an obtuse triangle. Obtain the projection of the 
 longest side upon each of the other sides. 
 
 Ex. 133. If AB, extended, makes an acute angle with a line m, 
 prove that p^^ is less than AB. 
 
 Ex. 134. If AB II ?7i, how does p^ compare with segment AB ? 
 
 Ex. 135. If AB ± m, what is the length oip'l^ ? 
 
 Ex. 136. What part of the base of an isosceles triangle is the pro- 
 jection upon the base of one of the equal sides ? 
 
 Ex. 137. If the equal sides of an isosceles trapezoid be projected 
 upon the lower base, the projections are equal. 
 
 Ex. 138. Draw a right triangle and draw the median to the hypotenuse. 
 Prove that the projection of the median upon either leg of the triangle is 
 one half of that leg. 
 
NUMERICAL RELATIONS AMONG SEGMENTS 183 
 
 Propositiox XXI. Theorem 
 
 310. 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, 
 minus twice the product of one of these sides and the projection of 
 the other upon it. 
 
 Fig. 1 
 
 Hypothesis. In A ABC, Z 5 is acute. 
 Conclusion. 6^ = a^ + c'^ — 2 a • ^^. 
 
 Proof. 1. Draw^Z)±50. Then ^i>=i9^=i). 
 
 2. In Fig. 1, ?>2 = /i2 4- DC". Why ? 
 
 3. But i)0 = a - p, and /i^ = c^ - p\ Why ? 
 
 4. .-. 62 = c2 - />2 -f (a - j))\ Ax. 2, § 51 
 
 (Complete the proof.) 
 
 Note 1. — A similar proof may be given from Fig. 2. In Fig. 2, DC = 
 p — a. 
 
 Note 2. — The conclusion of Proposition XXI is a formula connecting 
 the three sides of a triangle with the projection of one side upon one of the 
 otlier two sides. Altogether four different numbers are involved. Hence, 
 when three of these numbers are known, the fourth may be determined by 
 substituting in the formula and solving the resulting equation. In the 
 right member, there appear the squares of two sides and the projection of 
 one of these upon the other ; in the left member, there appears the square 
 of the third side. 
 
 Ez. 139. Determine : 
 
 (a) ;)« when a = 13, & = 14, and c = 15. 
 (6) 6 when a = 10, c = 12, and jo« = 9. 
 
 (c) c when a = 11, 6 = 16, and p^ = 7. 
 
 (d) a when 6 = 18, c = 12, andp^ = 4. 
 
184 PLANE GEOMETRY — BOOK III 
 
 Proposition XXII. Theorem 
 
 311. In any triangle having an obtuse angle, the square of the 
 side opposite the obtuse angle is equal to the sum of the squares of 
 the other two sides, plus twice the product of one of these sides and 
 the projection of the other side upon it. 
 
 A 
 
 Hypothesis. In A ABC, Z (7 is an obtuse Z. 
 Conclusion. c^=^a''+b^-\-2 a-pl 
 
 Proof. 1. Draw AD ± BC extended. Then p^c =CD=p. 
 
 2. Then in A ABD, ^ = li^ + Bff- Why ? 
 
 3. But 7i2 = ?>2 _p2 . and BD = a+p. 
 
 4. .-. c' = (b''-p'')^{a+pf. 
 (Complete the proof.) 
 
 312. Cor. If a, b, and c are the sides of a triangle : 
 Z ^ is acute if a^ < b'^^c}\ 
 Z.Ais obtuse if a^ > 6^ + c^ ; 
 Z A is Si right Z if a^ = 6^ + c\ 
 (Proof is indirect in each case.) 
 
 Ex. 140. Is the greatest angle of the triangle whose sides are 8, 9, 
 and 12 acute, right, or obtuse ? 
 
 Ex. 141. Is the greatest angle of the triangle whose sides are 12, 35, 
 and 37 acute, right, or obtuse ? 
 
 Ex. 142. Prove that the sum of the squares of the diagonals of a 
 parallelogram equals the sum of the squares of the sides of the parallelo- 
 gram. (Use § 310 and § 311.) 
 
 Ex. 143. If AB and A C are the equal sides of an isosceles triangle, 
 and BD is drawn perpendicular to AC, prove 
 2AC X CD = BC^. 
 Note. — Supplementary Exercises 47 to 51, p. 293, can be studied now. 
 
NUMERICAL RELATIONS AMONG SEGMENTS 185 
 
 313. When the three sides of a triangle are known, the 
 altitude to each side can be computed. 
 
 A 
 
 1. Assume AD = /i„, and Z ^ to be an acute angle. 
 
 2. 
 3. 
 4. 
 5. 
 
 6. 
 
 7. 
 8. 
 
 62=;a2 + c2-2a .p^ or ¥ = a^ -\' c^ - 2 ap. 
 
 .-. p = 
 
 2a 
 
 /l2 = c2-p2=(c4-p)(C-p). 
 
 K= 
 
 ;+ 2a r ^^~"J 
 
 2 a JL 2 a J 
 
 4a2 
 
 ^ (g -h c 4- &) (q + c - 6)(& 4- g - c) (& - g H- c) 
 
 4g2 
 
 9. Let g4-6 + c = 2s. 
 
 10. .-. g + 6-c = 2s-2c = 2(s-c). 
 
 Similarly, 6-|-c — g = 2(s — g); and c -f- g — 6 = 2 (s — 6). 
 
 -,. . ^2 _ 2/?. 2 (8 -6) '2. (a-c) '2{s-a) 
 J. J.. . . /i^ — 
 
 4g* 
 
 _ 4 8(8 — g)(/t — 6)(8 — c) 
 
 12. .-. ^„ = -Vs(8-g)(s-6)(s-c) 
 g 
 
 Similarly, 
 and 
 
 K = -y/s{s-a) (s - 6) (s - c) ; 
 
 
 
 o 
 
 h^ = - Vs(« — g) (s — 6) (s — c)- 
 c 
 
 Ex. 144. Find the three altitudes of the triangle whose sides are 13, 
 14, and 16, getting the results correct to one decimal place. 
 
186 PLANE GEOMETRY — BOOK III 
 
 Proposition XXIII. Theorem 
 
 314. In any triangle, the sum of the squares of two sides 
 equals twice the square of half the third side plus twice the square 
 of the median drawn to that side. 
 
 O 
 
 L .^..A _J 
 
 Hypothesis. In A ABC, CD is the median to side AB. 
 Conclusion. a^-{-¥ = 2 ^^ Y + 2 mf. 
 
 Note. — A ABC is either a rt. Z, an acute Z, or an obtuse Z. When 
 it is a rt. Z, the proof is quite easy. 
 
 Proof. 1. Assume that Z ABC is obtuse and hence Z BBC 
 is acute. 
 
 2. Draw CE ± AB, so that BE=pll. 
 
 (Complete the proof.) 
 
 Suggestion. — Determine a^ from ABCDhj^ 310 ; 62 from AACDhy^ 311 ; 
 then add, so as to obtain a^-{- b^. 
 
 Note. — By Proposition XXIII, it is possible to determine the three 
 medians of a triangle when the three sides of the triangle are known. 
 
 315. Cor. The difference between the squares of tzvo sides of 
 a triangle equals tivice the product of the third side and the pro- 
 jection of the media}! upon that side. 
 
 Con. b^-a^ = 2G' p'^c. 
 
 Suggestion. — Determine 6^ and a^ and then subtract the value of a^ from 
 that of 62. 
 
 Ex. 145. Determine m^ when & = 12, c = 16, and a = 20. Deter- 
 mine also m^ and rric. 
 
 Ex. 146. Prom the conclusion of § 314 derive a formula for m^ in 
 terms of a, 6, and c. 
 
NUMERICAL RELATIONS AMONG SEGMENTS 187 
 
 Proposition XXIV. Theorem 
 
 316. In any triangle, the product of two sides equals the product 
 of the diameter of the circumscribed circle and the altitude upon 
 the third side. 
 
 Hjrpothesis. A ABC is inscribed i n O ; AD is a diameter 
 of O 0; AE = h,. 
 
 Conclusion. b - c = d -h^. 
 
 Analysis and proof left to the pupil. See analysis of § 283. 
 
 2 
 
 317. By § 313, h, = - Vs(s - a)(s - b)(s - c). 
 
 2d 
 
 Hence, by § 316, b - c = — Vs(s — a){s — b){s — c). 
 
 a 
 
 .'. 2 dVs(s — a){s — b)(s—c) = abCj 
 d= ^^^ 
 
 2Vs(s-a)(s-b){s-c) 
 Hence, when the sides of a triangle are known, the diameter of 
 the circumscribed circle can be computed. 
 
 Ex. 147. Determine the diameter of the circle circumscribed about 
 the triangle whose sides are 13, 14, and 15. 
 
 Ex. 148. If two adjacent sides and one of the diagonals of a parallelo- 
 gram are 7, 9, and 8 respectively, find the other diagonal. (§ 314.) 
 
 Ex. 149. The sides AB and ^C of A ABC are 16 and 9 respectively, 
 and the length of the median drawn from C is 11. Find side BC. (§ 314.) 
 
 Ex. 150. If the sides oi £\ABC are 10, 14, and 16, find the lengths of 
 the three medians. Determine also the diameter of the circumscribed 
 circle. 
 
188 
 
 PLANE GEOMETRY — BOOK III 
 
 Proposition XXV. Theorem 
 
 318. In any triangle, the product of any two sides is equal to 
 the product of the segments of the third side formed by the bisector 
 of the opposite angle, plus the square of the bisector. 
 
 ■~-v^ 
 
 Hypothesis. AD bisects Z J. of A ABC, meeting BC at D. 
 (Let BD=r, and DC = p.) 
 
 Conclusion. b-c^tl + r-p. 
 
 Proof. 1. Circumscribe a circle about A ABC. 
 
 Extend AD to meet the circle at E. Draw CE. 
 
 2. 
 
 A ABD ~ A AGE. 
 
 
 Give full proof. 
 
 3. 
 
 •••;^=i'°^'"'= 
 
 AE ' t^. 
 
 Give full proof. 
 
 4. 
 5. 
 6. 
 
 7. 
 
 ••• bc=^t\'(tj, + s). 
 .:bc = t\ + tj,'S. 
 But t^- s = r 'p. 
 
 .'. bc = t\-\-r'p. 
 
 
 {AE = t^ + s) 
 §283 
 
 Note. — This proposition makes it possible to compute tlie bisectors of 
 the three angles of a triangle when the three sides of the triangle are 
 known. 
 
 Ex. 151. If c =: 4, 6 = 5, and a = 6, find Ia- 
 
 Suggestions. — 1. It is necessary to find r and p first. This may be done 
 by using § 270. 
 
 r ^4 
 6-r 5 
 2. Then substitute in the conclusion of § 318 
 
 Whence r=? and p = 6 — r = ? 
 mclusion of § 318. 
 Note. — Supplementary Exercises 52 to 56, p. 294, can be studied now. 
 
MISCELLANEOUS EXERCISES 189 
 
 Miscellaneous Exercises 
 Ex. 152. The vertices of quadrilateral ABCD are joined to a point 
 lying outside the quadrilateral. Points A', B' , C, and D' are taken 
 on OA, OB, OC, and OD, respectively, so that A'B' II AB, B'C II BC\ 
 and CD' II CD. Prove A'D' II AD. 
 
 Ex. 153. Two circles are tangent externally at point C. Through 
 C, a straight line is drawn, meeting the first circle at A and the second at 
 D ; another straight line through C meets the first circle at B and the 
 second at E. Prove AC: CD = BC: CE. 
 
 Suggestion. — Draw the common tangent at C, and also chords AB and 
 ED. 
 
 Ex. 154. If P and 8 are two points on the same side of line OX 
 such that the perpendiculars PB and /ST drawn to OX have the same 
 ratio as OB and OT, then points 0, P, and SWq in a straight line. 
 
 Suggestion. — Frove L ROP = L TOS by proving A OPR ~ A OST. 
 
 Ex. 155. If two parallels are cut by three or 
 more straight lines passing through a common 
 point, the corresponding segments are propor- 
 tional. 
 
 Prove ^-=.^ = ^^. -y 
 
 A'B' B'C CD' /' 
 
 Ex. 156. If three transversals intercept proportional lengths on two 
 parallels, the transversals meet at a point. 
 
 Suggestion. — Let A' A and B'B meet at and draw OC and OC"; then 
 prove A OBC&nd OB'C similar. (Fig. Ex. 155.) 
 
 Ex. 157. Derive a formula for the altitude to the base of an isosceles 
 triangle if the base is b and the equal sides are each a. By means of the 
 formula determine the altitude when : (^a) a = 12 and b =6; (b) a = 
 15 and 6=7. 
 
 Ex. 158. Find the length of the common external tangent to two 
 circles whose radii are 11 and 18, if the distance between their centers 
 is 25. 
 
 Suggestion. — See the figure for Problem 2, § 236. 
 
 Ex. 159. If BE and CF are the medians drawn from the extremities 
 of the hypotenuse of right triangle ABC, prove 4 BE^ + 4 CY'^ = 5 BC'\ 
 
 Ex. 160. Prove that the projections of two parallel sides of a paral- . 
 lelogram upon either of the other sides are equal. 
 
190 PLANE GEOMETRY — BOOK III 
 
 Ex. 161. BC is the base of an isosceles triangle ABC inscribed in 
 a circle. If a chord AD is drawn, cutting BG at E, prove AB^ = AE^ + 
 BE X CE. 
 
 Suggestions. — 1. The proof is like that for § 318. 
 2. Prove A ABD ~ A ABE. 
 
 Ex. 162. Prove that the non-parallel sides of a trapezoid and the line 
 joining the middle points of the parallel sides, if extended, meet in a com- 
 mon point. 
 
 Review Questions 
 
 1. What is meant by taking a proportion by (a) inversion? (b) 
 composition ? (c) alternation ? 
 
 . 2. Complete the following theorem : "If the product of two numbers 
 equals the product of two other numbers, one pair, .- " 
 
 3. Define: (a) mean proportional; (6) fourth proportional; (c) 
 similar polygons ; (d) ratio of similitude. 
 
 4. Are mutually equiangular triangles similar ? 
 Are mutually equiangular polygons similar ? 
 
 5. State all of the theorems, by which two triangles can be proved 
 similar. 
 
 6. How do you select the homologous sides of similar triangles ? 
 What do you know about them ? 
 
 7. What do you know about the ratio of homologous altitudes of 
 similar triangles ? 
 
 What about the ratio of the perimeters of similar triangles ? 
 What about the ratio of the areas of similar triangles ? 
 
 8. What is the Pythagorean theorem ? 
 
 9. Find the mean proportional between 5 and 15. 
 Construct the mean proportional between segments r and s. 
 
 10. What are the two conclusions which follow from the hypothesis 
 that the altitude is drawn to the hypotenuse of a right triangle ? 
 
BOOK IV 
 
 AREAS OF POLYGONS 
 
 319. A polygon, being a closed line (§ 7), incloses a limited 
 portion of the plane. 
 
 In measurement theorems, the words "rectangle," "parallelogram," 
 *' polygon," etc., mean the surface within the figure mentioned. 
 
 320. The Area of the surface within a closed line is the 
 ratio of the surface to the unit of surface measure. 
 
 Thus, in the adjoining figure, if the unit of surface 
 is one small square, the area of the rectangle is 30. 
 
 It has become customary, when speaking of the 
 area of a figure, to mention at once the unit of sur- 
 face ; thus, in the foregoing example, it is customary 
 to say that the area is 30 small scjuares. Remember, 
 however, that the area is 30. -^ 
 
 321. The usual Unit of Surface is a square whose side is 
 some linear unit : as, a square inch or a square centimeter. 
 In this text, it will be assumed that the unit always is such a 
 square unit. 
 
 Ex. 1. In the following figures, assume that the unit of surface is a 
 small square, (a) What is the exact area of Figs. 1 and 2 ? 
 
 i._.H.._4 — J.-_4--- 
 
 1...4.— ^--4._.^._- 
 — -i— T— I-— f-- -f— 
 
 Fig. 1 
 
 FiQ. 2 
 
 191 
 
192 
 
 PLANE GEOMETRY — BOOK IV 
 
 
 
 2 7 - /^-. ^" ""—^ 
 
 L t / ^^ t A 
 
 i 7 V ^^ V -^ 
 
 ' Jl f ^ ^^Jli -^ ^ 
 
 
 
 Fig. 3 
 
 Fig. 4 
 
 Fig. 5 
 
 (?)) What is the approximate area of Figs. 3, 4, and 5 ? 
 (Include the square in the area if half or more than half of it lies 
 within the figure ; do not include it otherwise.) 
 
 322. Two limited portions of a plane are Equal (=) if their 
 areas are equal when they are measured by the same unit. 
 
 Since the test of the equality of two figures is the equality of two 
 numbers, the usual axioms apply when equal figures are added or sub- 
 tracted, or when they are multiplied or divided by the same number. 
 
 Thus, if equal figures are added to equal figures, the sums are equal ; 
 also halves of equal figures are equal. 
 
 Ex. 2. Of the figures in Ex. 1, are any two or more equal ? 
 
 323. Two congruent figures are necessarily equal, hut two 
 equal figures are not necessarily congruent. 
 
 Also, two figures which consist of parts which are respectively con- 
 gruent are equal. 
 
 Thus, the parallelogram and the kite- 
 shaped figure made from it by placing mm^^^ 
 the two triangles together as hi the fig- //////////§i 
 ure adjoining are equal. 
 
 Ex. 3. If E is the mid-point of one of the non- 
 parallel sides of trapezoid ABCD, and a parallel to 
 AB drawn through E meets BC extended at F and 
 AD at G, prove that parallelogram ABFG is equal ^ 
 to trapezoid ABGD. 
 
 Suggestion. — Fvoye A CEF ^ A GED, and apply § 323. 
 
 Ex. 4. In the adjoining figure, D and E are 
 the mid-points of ^B and ^C; AJA.BG; BF 
 ± DE extended at F; CG ± DE extended at G. ^. ,^ 
 Prove th at A ^^ O equals un BFGC. 
 
 Suggestion. — Prove A BDF ^ A DAH, and 
 A CEG ^/\AEH. 
 
 \L 
 
AREAS OF POLYGONS 
 
 193 
 
 Ex. 5. In the adjoining figure, E and F are 
 the mid-points of sides AB and CD of trapezoid 
 ABCD ; XY and Z \V are drawn througli E and 
 F respectively ± AD, meeting BC extended at X 
 and irrespectively. VvovQih2Lt XYWZ=ABCD. "^ .^ " 
 
 Ex. 6. Let K be the mid-point of side 5Cand if the mid-point of 
 side AD of CD ABCD ; let FE, drawn through the mid-point G of KH, 
 intersect BC and AD at F and E respectively. Prove that FE divides 
 ABCD into two equal quadrilaterals. 
 
 Note. — Supplementary Exercises 1-3, p. 294, can be studied now. 
 
 MEASUREMENT OF RECTANGLES 
 
 324. The Dimensions of a rectangle are the Base and 
 Altitude. 
 
 325. Area of a Rectangle. (Informal treatment.) If the 
 base of a rectangle measures 6 and its altitude 5 linear units, 
 the area is evidently 6 x 5 or 30 surface units. 
 
 If the base measures 6 units and the alti- 
 tude measures 3^ units, the area is evidently 
 6 X 3.5 or 21 surface units. 
 
 These two examples suggest the theorem : 
 
 The number of surface units in the area of a rectangle is the 
 product of the number of linear units in its base and the number 
 in its altitude. More 'briefly, this theorem is expressed : the 
 area of a rectangle is the product of its base and its altitude. 
 
 The theorem is proved in the following three propositions. 
 
 T w 
 
 — I— r----|-T 
 J_.L4.J__I 
 
 .4-1-^-1-4 
 
 X 
 
 326. Comparison of Rectangles. Rectangles may be com- 
 pared without computing their areas. 
 
 Two rectangles having equal bases and altitudes are equal, for 
 it is evident that they can be made to coincide by superposition. 
 
194 
 
 PLANE GEOMETRY — BOOK IV 
 
 Pkoposition I. Theokem 
 
 327. Two rectangles having equal altitudes are to 
 each other as their bases. 
 
 K 
 
 c 
 
 D 
 
 E 
 
 G 
 
 H 
 
 Hypothesis. Rectangles ABCD and EFGH have equal 
 altitudes AB and EF, and bases AD and EH, respectively. 
 
 Conclusion. ABCD ^ AD ^ 
 
 EFGH EH 
 
 Case I. Assume that AD and EH are commensurable. 
 
 §211 
 Proof. 1. Let AK, a common measure of AD and EH, be 
 contained in AD 5 times and in EH 3 times. Draw Js to AD 
 and EH at the points of division. 
 
 2. Then ABCD and EFGH are divided into equal rec- 
 tangles. Why ? 
 (Complete the proof.) 
 
 Suggestions. — What is the value of ^-^ ? of - ? Then compare these 
 
 ratios. ^^ ^^^H 
 
 Case IL When AD and EH are incommensurable, the 
 theorem is still true. The proof is given in § 425. 
 
 328. Cor. Two rectangles having equal bases are to each 
 other as their altitudes. 
 
 Ex. 7. Construct a rectangle which will be three times a given rec- 
 tangle ; ' also one which will be three fourths a given rectangle. 
 
 Ex. 8. Two rectangles M and T have equal bases b and altitudes 
 r and s respectively. What is the ratio of J^f to 2'? 
 
 Note. — Supplementary Exercises 4 to 5, p. 295, can be studied now. 
 
AREAS OF POLYGONS 195 
 
 Proposition II. Theorem 
 
 329. Two rectangles are to each other as the products 
 of their bases by their altitudes. 
 
 M 
 
 B 
 
 h' 
 
 Hypothesis. Rectangle M has base h and altitude a ; rec- 
 tangle N has base V and altitude a'. 
 
 Conclusion . — = -— • 
 
 N a'b' 
 
 Proof. 1. Let rectangle R have base 6' and altitude a. 
 
 2. .-. — = — , since M and E have equal altitudes. Why? 
 
 7? n 
 
 3. Also — = — , since R and iVhave equal bases. Why? 
 
 , M ^R ab ^ M ah .^ttj. ^ 
 
 4. ••• — X -T^=— :t-., or — = -^rT- Why? 
 
 . =-x- = — or^=-^ 
 " R N a'b'' N a'b' 
 
 Ex. 9. What is the ratio of rectangles B and S if their dimensions 
 are as follows ? 
 
 (A 
 
 ) 
 
 
 
 (B) 
 
 
 
 
 (C) 
 
 
 
 
 Ji 
 
 s 
 
 
 
 li 
 
 s 
 
 
 
 It 
 
 *s- 
 
 Altitude 
 
 k 
 
 X 
 
 Altitude 
 
 4 
 
 6 
 
 Altitude 
 
 12 
 20 
 
 15 
 18 
 
 Base 
 
 w 
 
 y 
 
 Base 
 
 10 
 
 16 
 
 Base 
 
 Ex. 10. If i?, 8, T, and X are rec- 
 tangles having the dimensions indicated in 
 the adjoining table, determine the ratio of 
 each rectangle to each of the others. 
 
 (Thus, determine B : 8, B : T, Qiz.) 
 
 
 Altitl-de 
 
 Base 
 
 B 
 
 10 
 
 6 
 
 8 
 
 5 
 
 12 
 
 T 
 
 10 
 
 8 
 
 X 
 
 10 
 
 12 
 
196 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition III. Theorem 
 
 330. The area of a rectangle is the product of its 
 base and altitude. 
 
 N 
 
 Hypothesis. Eectangle M has altitude a and base b. 
 
 Conclusion. Area oi M=ab. 
 
 Proof. 1. Let square ^be the unit of surface measure. 
 
 2. Then area of M= the ratio of Mto N. § 320 
 
 M_ ab 
 
 N~lxl' 
 4. .-. area of M= ab. 
 
 3. 
 
 Why? 
 
 Note. — Remember that this theorem means that the number of 
 square units in the area equals the product of the number of linear units 
 in the base by the number in the altitude. A similar interpretation 
 must be given for each of the measurement theorems of this Book. 
 
 Ex. 11. A business corner 50 ft. x 120 ft. is valued at .$9000. What 
 is the value per square foot ? 
 
 Ex. 12. The area of a square is 590.49 sq. ft. Find its perimeter. 
 
 Ex. 13. A rectangle has the dimensions 30 ft. and 120 ft. Compare 
 its perimeter with that of an equal square. 
 
 Ex. 14. An ordinary eight-room house costs approximately $4.75 
 per square foot of ground covered by it. What is the approximate cost 
 of a house 27 ft. x 36 ft.? 
 
 Ex. 15. The area of a rectangle is 147 sq. ft. Its base is three times 
 its altitude. What are its dimensions ? 
 
 Ex. 16. What are the dimensions of a rectangle whose area is 168 sq. 
 ft. and whose perimeter is 58 ft ? 
 
 Suggestion. — Let the base = x and the altitude = y. Form two equations 
 and complete the solution algebraically. 
 
 Ex. 17. What is the length of the diagonal of a rectangle whose area 
 is 2640 sq. ft., if its altitude is 48 ft.? 
 
AREAS OF POLYGONS 
 
 197 
 
 Proposition IV. Theorem 
 
 331. TJie area of a parallelogram equals the product 
 of its base and altitude. 
 
 E B 
 
 Hypothesis. ABCD is a parallelogram. 
 
 Its altitude DF= a : its base AD 
 
 b. 
 
 Conclusion. Area of ABCD = ah. 
 
 Proof. 1. Draw AE \\ DF, meeting BC extended at E. 
 
 2. .-. AEFD is a rectangle. Why ? 
 
 3. A AEB = A FCD. Give the full proof. 
 
 4. .-. O ABCD = □ AEFD. Why ? 
 
 5. But area of □ AEFD = ah. Why ? 
 
 6. .-. area of ABCD = ah. Ax. 1, § 51 
 
 332. Corollaries. Let O P^ have base ftj and altitude a/; 
 and O P., have base &2 and altitude ag. 
 
 (1) Parallelograms having equal hases and equal altitudes are 
 equal. 
 
 (2) Two parallelograms are to each other as the products of 
 their hases hy their altitudes. 
 
 For, since O Pi = ai6i and O r^ = a-rbi, then ^^ = ^ • 
 
 O P2 0262 
 
 (3) Parallelograms having equal altitudes are to each other as 
 their hases. 
 
 For, in (2), if ai = ao, then O Pi : O P2 = 61 : 62. 
 
 (4) Parallelograms having equal bases are to eojch other as 
 their altitudes. 
 
198 
 
 PLANE GEOMETRY — BOOK lY 
 
 Ex. 18. What is the area of OB, of O S, and ofUT? 
 
 (a) O B has altitude 4 in. and base 9 in. 
 
 (b) JU S has altitude 15 ft. and base 20 ft. 
 
 (c) CJ T has altitude 3 a; in. and base 11 y in. 
 
 Ex. 19. What is the altitude of a parallelogram whose area is 56 
 sq. in., if its base is 14 in. ? 
 
 Ex. 20. Construct a parallelogram equal to twice a given parallelo- 
 gram. 
 
 Ex. 21. Construct a rectangle equal to two thirds a given parallelo- 
 gram. 
 
 Ex. 22. Divide a parallelogram into two equal parallelograms ; into 
 four equal parallelograms. 
 
 Ex. 23. What is the ratio oi O P to O B \i the base of each is 10 
 in. and the altitudes are 5 in. and 8 in. respectively ? 
 
 Ex. 24. Construct a CJ ABCD having AB = 3 in. and BG = 4 in., 
 and having : (a) ZB = 30^; (6) ZB = 45°. (c) Determine the area of 
 each of the parallelograms. 
 
 Ex. 25. The base of A ABC is 10 and the alti- a d 
 
 tude is 5. What is the area of A ABC ? 
 
 Suggestion. — Draw AD \\ EC and CD \\ AB to form 
 O ABCD. Compare A ABC with EJ ABCD. Then 
 determine the area of O ABCD and finally ot AAB G. 
 
 Propositiq]^ v. Theorem 
 333. The area of a triangle equals one half the prod- 
 
 uct of its base and altitude. 
 
 Hypothesis. A ABC has altitude AE = a and 
 Conclusion. Area of A ABC= ^ ah. 
 
 [Proof to be given by the pupil.] 
 Suggestion. — Construct CJ ABCD and proceed as in Ex. 25. 
 
 BC = h. 
 
AREAS OF POLYGONS 199 
 
 334. Corollaries. By proofs similar to those given in § 332, 
 it follows that : , 
 
 (1) Triangles having equal bases and equal altitudes are^equal. 
 
 (2) Tioo triangles are to each other as the products of their 
 bases by their altitudes. 
 
 (3) Triangles having equal altitudes are to each other as their 
 bases. 
 
 (4) Triangles having equal bases are to each other as their 
 altitudes. 
 
 (5) A triangle is one half a parallelogram having the same base 
 and altitude. 
 
 Ex.26, (a) CompsLTeCJ ABCD with A BCE. a x d 
 
 (6) Compare A BOX with A BCE. 
 (c) If Xis the mid-point of AD, compare A ABX 
 
 with A XCZ>; also compare A XCD with A 5 O-E. 
 
 Ex. 27. Determine the area of an isosceles right triangle whose leg 
 is 9 in. 
 
 Ex. 28. Determine the area of an equilateral triangle whose side is 
 10 in. 
 
 Ex. 29. (a) Prove that the area of an equilateral triangle whose 
 
 side is s is — v3. (Memorize this formula.) 
 4 
 (&) Using the formula developed in (a), obtain the area of an equi- 
 lateral triangle whose side is: (1) 12 in. ; (2) 15 in. 
 
 Ex. 30. Find the area of the front of the garage whose K('y\ lo' 
 dimensions are indicated in the adjoining figure. 
 
 Ex. 31. What is the area of the rhombus whose diago- 
 nals are 10 and 16 respectively ? 
 
 Ex. 32. What is the length of the side of a square whose area equals 
 that of a triangle whose base is 24 and whose altitude is 12 ? 
 
 Ex. 33. If BD is the median to side AC of A ABC, prove that 
 A ABD = ABDC. (Draw the altitude BF to side AC. ) 
 
 Ex. 34. Prove that the diagonals of a parallelogram divide the par- 
 allelogram into four equal triangles. 
 
 Ex. 35. If segments are drawn from two opposite vertices of a quad- 
 rilateral to the mid-point of the diagonal joining the other two vertices, 
 the broken line so formed divides the quadrilateral into two equal parts. 
 
200 PLANE GEOMETRY — BOOK IV 
 
 Ex. 36. Through the vertex ^ of A ABC, draw a line MN parallel 
 to BC. On MN^ take any point X and prove that A XBC = A ABC. 
 
 Ex. 37. Construct a triangle twice as large as a given triangle : 
 
 (a) having the same base as the given triangle ; 
 
 (&) having the same altitude as the given triangle. 
 
 Ex. 38. Construct a rectangle equal to a given triangle. 
 
 Ex. 39. Construct a triangle equal to a given rectangle. 
 
 Ex. 40. Construct a right triangle equal to a given triangle and hav- 
 ing the same base as the triangle. 
 
 Ex. 41. Construct an isosceles triangle equal to a given triangle and 
 having the same base as the given triangle. 
 
 335. The Area of a Triangle can be expressed in Terms of 
 its Sides. 
 
 Solution. 1. If a, 5, and carethe sidesof A^iiC, ands = ^(a + & + c), 
 it can be proved that the altitude drawn to side a is given by the formula : 
 
 ha=-Vs{s-a){s-b){s-c) §313 
 
 a 
 
 Note. — The proof may be read if desired. Often in mathematics, we 
 use provable formulae which we may not have proved ourselves. 
 
 2. Areaof A^^C = ia- /i„. 
 
 3. .-. areaof A^i?(7 = - a- - >/s(s - a)(s- b)(s-c), 
 
 2 a 
 
 4. or area of A ^j5C = vs(s — a)(s— 6) (s — c). 
 
 Eemember that s is one half the perimeter of the triangle. 
 Example. — Find the area of the triangle whose sides are 
 13, 14, and 15. 
 
 Solution. 1. Let a = IS, 6 = 14, and c = 15. 
 
 2. .-. 8 = 1(13 + 14+15) =21. 
 
 3. . •. area = V2I x 8 x 7 xli ■= V3x7x2x4x7x3x2 
 
 4. • = 3 X 7 x 2 X 2 = 84. 
 
 Ex. 42. The sides of the lots A 
 and B in the adjoining figure have the 
 lengths indicated. Find the area of 
 each of the lots. 
 
 Note. — Supplementary Exercises 6 to 30, p. 295, can be studied now. 
 
AREAS OF POLYGONS 201 
 
 Proposition VI. Problem 
 336. Construct a triangle equal to a given polygon. 
 
 G D 
 Given polygon ABODE. 
 Required to construct a A = ABODE. 
 
 1. Change ABODE into an equal quadrilateral. 
 Construction. 1. Draw diagonal AO, cutting off A ABO. 
 
 2. Draw BF II AO^ meeting DC extended at F. Draw AF. 
 Statement. AFDE = ABODE. 
 
 Proof. 1. A ABO and A AOF have the same base, AO, 
 and equal altitudes, — the distance between the lis AO and BF. 
 
 2. .: AABO=^AOF. Why? 
 
 3. AFDE = AODE + A AOF-, 
 
 and ABODE = AODE -h A ABO. Ax. 3, § 51 
 
 4. .-. AFDE = ABODE. Ax. 7, § 51 
 II. OJiange AFDE into an equal triangle. 
 
 Construction. 1. Draw AD ; draw OE II ADy meeting FD 
 extended at 6?; draw AG, 
 
 Statement. A AFG = quadrilateral AFDE. Prove it. 
 
 Ex. 43. (a) Make a reasonably large pentagon, and construct a tri- 
 angle equal to the pentagon. (6) Measure the base and altitude of the 
 triangle, and compute the area of the triangle, (c) What is the area of 
 the pentagon ? D 15 C -e 
 
 Ex. 44. Determine the area of the trapezoid /j ""^-^ \ J 
 
 ABCD whose dimensions are indicated in the ad- /l'^ ^"^^ \! 
 
 joining figure. 
 
 A F B 
 
 Suggestions. — Area of A ABD = ? Area of A BCD = ? 
 
202 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition^ VII. Theorem 
 337. Tlie area of a trapezoid equals one half its alti- 
 
 tude multiplied hy the sum of its bases. 
 
 A 
 
 D b' 
 
 
 1 
 
 
 \i 
 
 B 
 
 Hypothesis. Trapezoid ABCD has its altitude DE = a, its 
 base AB = h, and its base CD = b\ 
 
 Conclusion. Area ABCD = ^a(b + b'). 
 
 Proof. 1. Draw BD and altitude BF of A DBC. 
 Complete the proof. See Ex. 44. 
 
 338. Cor. Tlie area of a trapezoid equals the product of its 
 altitude and its median. (Recall § 153.) 
 
 Ex. 45. Determine the area of the trapezoids A and B whose dimen- 
 sions are given in the table below : 
 
 
 Altitude 
 
 Lower Hase 
 
 Upper Base 
 
 Trapezoid A 
 Trapezoid B 
 
 10 in. 
 
 20 in. 
 
 9 in. 
 
 15 ft. 
 
 30 ft. 
 
 20 ft. 
 
 Ex. 46. Find the lower base of a trapezoid whose area is 675 sq. ft., 
 upper base 35 ft., and altitude 15 ft. 
 
 Ex. 47= The non-parallel sides, AB and CD, 
 of a trapezoid are each 25 in., and the sides AD 
 and BC are 33 in. and 19 in., respectively. Find the 
 area of the trapezoid. A 
 
 Suggestions. — Draw through jB a 1| to CD, and al 
 to AD. 
 
 Ex. 48. Prove that the straight line joining the mid-points of the 
 bases of a trapezoid divides the trapezoid into two equal trapezoids. 
 
AREAS OF POLYGONS 
 
 203 
 
 Ex. 49. How many square feet of wood will be re- 
 quired for 100 waste-paper boxes like the one pictured in 
 the adjoining figure, — allowing 15% extra for wood 
 wasted in cutting ? 
 
 Note. — Assume that each side is an isosceles trapezoid 
 having the dimensions indicated in the figure. 
 
 Ex. 50. The adjoining figure represents the end of 
 a barn. If the barn is 85 ft. in length, determine the 
 expense of painting its sides, its end, and its roof at 4 j^ 
 per square foot. 
 
 Ex. 51. The longest diagonal AD of pentagon 
 ABODE is 44 in., and the perpendiculars to it from 
 B, C, and E are 24, 16, and 15 in. respectively. 
 If AB = 25 in. and CZ) = 30 in., what is the area 
 of the pentagon ? 
 
 Ex. 52. Find the lower base of a trapezoid whose area is 9408 sq. 
 ft., whose upper base is 79 ft., and whose altitude is 96 ft. 
 
 Ex. 53. Construct a triangle equal to a given trapezoid and having 
 the same altitude as the trapezoid. 
 
 Ex. 54. Draw through a given point in one side of a parallelogram 
 a straight line, dividing the parallelogram into two equal parts. 
 
 Ex. 55. Construct a parallelogram equal to a given trapezoid, having 
 the same altitude as the trapezoid. 
 
 Ex. 56. If AD is the median to side BC of A ABC and E is the 
 mid-point of AD, then A BEC = ^ A ABC. 
 
 Ex. 57. If E and F are the mid-points of sides AB and AC respec- 
 tively of A ABC, and D is any point in side BC, prove quadrilateral 
 AEDF=^AABC. 
 
 Ex 58. If E is any point in side BC oi CJ ABCD, then A ABE -H 
 AECD^^/DABCD. 
 
 Ex. 59. Draw a straight line perpendicular to the bases of a trape- 
 zoid which will divide the trapezoid into two equal parts. 
 
 339. The following Proposition is an alternative demon- 
 stration of the Pythagorean Theorem given in Proposition 
 XIV of Book III. 
 
204 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition VIII. Theorem 
 
 340. The square upon the hypotenuse of a right tri- 
 angle is equal to the sum of the squares upon the two 
 legs of the triangle. 
 
 Hypothesis. Z C of A ABC is a right angle. 
 
 ABEF, ACGB, and BGKL are squares. 
 Conclusion. Area ABEF = area ACGH+ area BCKL. 
 Proof. 1. Draw CD ± AB and extend it to meet FE at M. 
 
 2. Draw BH and CF. 
 
 3. • A ACF ^ A ABH. (Give the full proof.) 
 
 4. BCO is a st. line and parallel to AH. § 40. 
 
 5. .-.A ABH and D ACGH have the same base, AH, and 
 eqiial altitudes, — the distance between the lis BO and AH. 
 
 6. .-. area ACGH = 2 area A ABH. Why ? 
 
 7. Similarly area ^Z)3fi^= 2 area A ACF. 
 
 (Give the full proof.) 
 
 8. .-. area ACGH = area ADMF. Why ? 
 
 [From steps 6 and 7.] 
 
 9. Similarly it can be proved that 
 
 area BCKL = area BDME. 
 10. .-. area ACGH + area BCKL = area ABEF. 
 
 [From steps 8 and 9.] 
 
AREAS OF POLYGONS 
 
 205 
 
 Note. — Many other proofs of this important theorem can be given. 
 The proof suggested in Ex. 00, which follows, is an extremely suggestive 
 one ; the one in Ex. 61 has of course special interest. i 
 
 Ex. 60. Prove the Pythagorean Theorem, us- 
 ing the adjoining figure. (Note. Square AH is 
 " turned in " over A ABC.) 
 
 Prove [JAD = UBF-\-U AH. 
 
 Suggestions. — 1. Draw J5^X" and prove HKE is a 
 St. line, by proving LAKE is a rt. Z. 
 
 2. Prove [J AH = CJ AXYE, by comparing each 
 with A ABE. 
 
 3. FroyeOBF = CJCXYD. 
 Note. — The Pythagorean Theorem can be proved 
 
 from figures obtained by " turning in " any of the squares, one at a time, two 
 at a time, or all three of them. 
 
 ~~^E 
 
 Ex. 61. Garfield's Proof of the Pjrthagorean a 
 Theorem. 
 
 Hyp. In AABC,ZB =90^". 
 Con. b' = (fi + c2. 
 
 Suggestions. — 1. Extend BC to D, making CD = 
 An. Draw DEIBD at D, making DE = BC. 
 Draw CE and AE. 
 
 2. Prove ABDE is a trapezoid. 
 
 3. Express the area of ABDE in terms of a and c. 
 
 4. Prove £2 = 90°, and that CE = b. 
 
 5. Express the area of A ABC, CDE, ACE in terms of a, b, and c. 
 
 6. Form an equation based on the fact that the trapezoid consists of the 
 three triangles. 
 
 7. Complete the proof algebraically.' 
 
 Ex. 62. Prove C, H, and L lie in a st. line. (Fig. § 340.) 
 (Draw CH &nd CL, and prove Z HCL = 1 st. Z.) 
 
 Ex. 63. Prove AG || BK. 
 
 Ex. 64. Prove that the sum of the Js from H and L to AB extended 
 equals AB. 
 
 Suggestion. — Compare AD and DB with Jfe. 
 
 1 A number of alternative proofs of the Pythagorean Theorem, and other 
 interesting theorems, are given in Heath's Mathematical Monographs, Num- 
 bers 1-4. Published by D. C. Heath & Co., Boston, New York, Chicago. 10/* 
 each. 
 
206 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition IX. Problem 
 
 341. Construct a square equal to the sum of two 
 given squares. 
 
 N 
 
 A B 
 
 Given squares M and N. 
 
 Required to construct a square equal to the sum of M and N. 
 Construction. 1. Construct ^(7 _L^-B, making ^C= n and 
 AB = m. Draw BC. 
 
 Statement. The square constructed on J50 as side ==M-{-N. 
 [Proof to be given by tlie pupil.] 
 
 342. Cor. Construct a square equal to the difference between 
 two given squares. 
 
 A yC D 
 
 Given squares M and N. 
 
 Required to construct a square equal to M— N. 
 Construction and proof to be given by the pupil. 
 [Construction suggested by the figure.] 
 
 Ex. 65. Construct a square equal to the sum of three given squares. 
 
 Ex. 66. The area of an isosceles right triangle is equal to one fourth 
 the area of the square described upon the hypotenuse. 
 
 Suggestion.— Com^2iYe the right triangle with the square on one leg. 
 
 Ex. 67. In the figure for § 340, prove that A AFH, BEL, and CGK 
 each equals A ABC. 
 
AREAS OF POLYGONS 
 
 207 
 
 Proposition X. Theorem 
 
 343. The areas of two similar triangles are to each 
 other as the squares of any- two homologous sides. 
 
 Hypothesis. AB and A^B' are homologous sides of similar 
 A ABC and A'B'C respectively. 
 
 Conclusion. 
 
 A ABC AB' 
 
 AA'B'C JT^i'' 
 
 Proof. 1. Draw altitudes CD and CD'. 
 
 A ABC lAB'CD AB - CD 
 
 AA'B'C 
 
 ^AB'CD 
 . A'B' ' CD' 
 
 A ABC ^( ^B\ 
 AA'B'C \A'B'J 
 
 But 
 . A ABC 
 
 CD\ 
 
 CD')' 
 
 CD 
 
 A'B' . CD' 
 
 § 333; Ax. 6, § 51 
 
 An algebraic change 
 
 AB 
 
 CD' A'B' 
 
 AB AB AB^ 
 
 A A'B'C A'B' A'B' 
 
 §282 
 Ax. 2, § 51 
 
 A'B'' 
 
 Note. — Since the ratio of two homologous lines of two similar tri- 
 angles equals the ratio of any two homologous sides, the areas of two 
 similar triangles are to each other as the squares of any two homologous 
 lines. 
 
 Ex. 68. A ABC ~ A A'B'C and AB = 2 A'B'. 
 
 (a) Compare the area of A ABC with the area of A A'B'C. 
 
 (b) Draw a figure to illustrate the correctness of your result. 
 
 Ex. 69. What is the ratio of A ABC to A A'B'C, if they are similar, 
 and: 
 
 (a) itAB = S A'B' ? (b) if AB = A'B' ? (c) it AB = ^ A'B' ? 
 
 Note. — Supplementary Exercises 32 to 35, p. 297, can be studied now. 
 
208 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition XI. Theorem 
 
 344. The areas of two similar polygons are to each 
 other as the squares of any two homologous sides. 
 
 Hypothesis. AB and A'B' are homologous sides of similar 
 polygons AC and A'C. 
 
 Conelusion. Area of polygon AO ^ AW^ 
 
 Area of polygon A'C A'B'^ 
 Proof. 1. Draw the diagonals EB, EC, E'B', and E'C. 
 2. Then A ABE ~ A A'B'E' ; A BCE ^ A B'G'E' ; etc. 
 
 _ §295 
 
 . A ABE AB" 
 
 A A'B'E' 
 A BCE 
 
 A'B'' 
 ^ BCf ^ AB" 
 
 AB'C'E' B^'"" A'W 
 
 [Since AB__BC_^ 
 L A'B'~ B'C J 
 
 3. 
 
 4. Similarly 
 
 5. Similarly 
 
 6. 
 
 A A'B'E' A B'OE' A C'D'E' 
 
 Complete the proof, applying § 296. 
 
 §343 
 
 ACDE 
 
 CD" 
 
 A C'B'E' c'j)'^ 
 A ABE A BCE 
 
 AW 
 
 JJb'^ 
 
 A CDE 
 
 Why 
 
 345. Since the perimeters of two similar polygons have the 
 same ratio as any two homologous sides (§ 297), then the areas 
 of two similar polygons must have the same ratio as the squares 
 of their perimeters. 
 
AREAS OF POLYGONS 
 
 209 
 
 Proposition XII. Theorem 
 
 346. Two triangles having an angle of one equal to 
 an angle of the other are to each other as the jproducts 
 of the sides including these angles. 
 
 Hypothesis. A ^IBC and A AB'C have Z A common. 
 
 Conclusion. AABO^ABxAC^ 
 
 A AB'C AB'xAC 
 
 Proof. 1. Draw B'C; also draw CD±AB'. 
 
 2. A ABC and A AB'C have the common altitude CD. 
 
 3. . AABC^AB^^ Cor. 3, §334 
 
 A AB'C AB' 
 
 4. A AB'C and A AB'C have as common altitude the ± 
 from B' to AC. 
 
 5. ,AAB!C^AC ^, 
 
 AC ^ 
 
 A AB'C 
 
 6. Multiplying the equations of steps 3 and 5, 
 A ABC A AB'C ^ AB x AC A ABC ^ AB x AC 
 A AB'C ' A AB'C AB' x AC' °^ A AB'C AB' x AC' 
 
 Ex. 70. If the area of a poly^fon, one of whose sides is 16 in., is 375 
 sq. in., what is the area of a similar polygon whose homologous side is 
 18 in. ? 
 
 Ex. 71. The longest sides of two similar polygons are 18 and 3 in. 
 respectively. How many polygons, each equal to the second, will form a 
 polygon equal to the first ? 
 
 Note. — Supplementary Exercises 36 to 41, p. 298, can be studied now. 
 
210 
 
 PLANE GEOMETRY — BOOK IV 
 
 SUPPLEMENTARY TOPICS 
 
 Three groups of supplementary material follow. This 
 material appears in some form in most geometries. All of it 
 is interesting and instructive mathematically; none of it is 
 strictly necessary in subsequent parts of geometry. 
 
 The teacher should feel free to select the group or groups 
 which best meet the needs of the class. 
 
 Group A. — Constructions based upon Algebraic Analysis. 
 This group is especially instructive and interesting. 
 
 Group B. — Constructions without Formal Analysis. 
 Group C. — Miscellaneous Problems. 
 
 The first two problems of this group are usually studied. Teachers 
 often omit the remaining ones. 
 
 A. Constructions based on Algebraic Analysis 
 Proposition XIII. Problem 
 347. Construct a square equal to a given parallelogram. 
 
 1 
 1 
 
 1 
 
 s 
 
 
 —Ir ^ 
 
 Given O ABCD having base h and altitude a. 
 Required to construct a square equal to O ABCD. 
 
 Analysis. 1. Let x — the side of the required square. 
 
 2. Then x^ = the area of the required square, Why ? 
 and ah = the area of the given parallelogram. Why ? 
 
 3. .'.x'^ah. Why? 
 
 4. .'.a:x = x:h. § 252 
 
 5. .-. X is the mean proportional between a and 6, and can 
 be constructed by § 290. 
 
CONSTRUCTIONS 2 1 1 
 
 Construction. 1. Construct a;, the mean proportional between 
 a and b. § 290 
 
 2. On X as side, construct the square S, 
 
 Statement. Square S = CJ ABCD. 
 
 Proof. 1. Area of S = x^y and area of O ABCD = ab. 
 
 2. a : X = X : by OT x^ = ab. Why ? 
 
 3. .-. area of /S = area of O ABCD, 
 
 Discussion. The construction is always possible, for the 
 mean proportional between a and b can always be found. 
 
 Note 1. — The analysis gives the pupil an idea of how such a construc- 
 tion is discovered. In many cases the proof of the correctness of the 
 resuking construction is ratlier trivial after such an algebraic analysis, — 
 and in such cases the teacher may decide to omit the proof. 
 
 Note 2. — The algebraic solution of such a problem as that proposed 
 in §347 would duplicate the analysis as far as step 3. Then the 4th 
 step would be: .-. x =Vab, After x had been computed, the square 
 would be constructed upon a line of the length determined. 
 
 Theoretically the geometric solution is preferable, for x as constructed 
 actually equals the mean proportional between a and 6, so that the 
 square on side x actually equals the parallelogram ; whereas, in the case 
 of the algebraic solution, the value of x is determined only approximately 
 (in most cases) when the square root is found, and hence the square will 
 be only approximately equal to the parallelogram. 
 
 348. Cor. Construct a square equal to a given triangle. 
 
 Analysis. 1. Let x = the side of the required square, b = the base of 
 the given triangle, and h = the altitude of the triangle. 
 
 2. .•.x^=ihb. 
 
 3. .'. \ b : X = X . h, or X is the mean proportional between ^ b and h. 
 Construction to be given by the pupil. 
 
 Ex. 72. Construct a square equal to twice a given triangle. 
 Ex. 73. Construct a square equal to twice a given square. 
 Ex. 74. Construct a square which will be twice a given parallelogram. 
 Ex. 75. Construct a square which will be three times a given triangle. 
 Ex. 76. Construct a square which will be two thirds a given rectangle. 
 Ex. 77. Construct a square which will be equal to a given pentagon. 
 (First construct a triangle equal to the pentagon, and then a square 
 equal to the triangle.) 
 
212 PLANE GEOMETRY — BOOK IV 
 
 Ex. 78. Construct a parallelogram which will equal a given rectangle 
 and have a given segment as base. 
 
 Analysis. 1. Let a — the altitude and h = the base of the given rec- 
 tangle, and let c = the given base of the parallelogram. Let x — the re- 
 quired altitude of the parallelogram. 
 
 2. Then ex = ah. Why ? 
 
 3. .'. c : a = & : X, or X is the 4th proportional to c, a, and 6. Why ? 
 Construction left to the pupil. 
 
 Suggestion. — Construct the fourth proportional x and then construct the ZZ7. 
 
 Ex. 79. Construct a rectangle equal to a given rectangle, having a 
 given segment as base. (Analyze as in Ex. 78.) 
 
 Ex. 80. Construct a triangle equal to a given triangle, having a given 
 segment as base. 
 
 Suggestion. — Determine the altitude as in Ex. 78, then construct the triangle. 
 How many such triangles can be constructed ? 
 
 Ex. 81. Construct a line parallel to the base of a triangle dividing 
 the triangle into two equal parts. 
 
 Analysis. 1. Assume 5' C to be the required line : ht 
 
 let AB' =x. / ^\ 
 
 2. ,^AABC^2^^^AABC^AB\ ,3 b/ \c' 
 
 Complete the analysis and then make the construction. That is, de- 
 termine X first and then draw B'C at the distance x from A on AB. 
 
 Ex. 82. Construct a rectangle having a given base and equal to f 
 a given square. (Analyze as in Ex. 78.) 
 
 Ex. 83. Construct a triangle having a given base and equal to a 
 given parallelogram. 
 
 Ex. 84. Construct a parallelogram having a given altitude and equal 
 to a given triangle. 
 
 Ex. 85. Construct a parallelogram having a given altitude and equal 
 to a given square. 
 
 Ex. 86. Construct a parallelogram having a given altitude and equal 
 to a given trapezoid. 
 
 Ex. 87. — Construct a triangle having a given altitude and equal to a 
 given trapezoid. 
 
 Ex. 88. Construct a square equal to a given trapezoid. 
 
CONSTRUCTIONS 213 
 
 B. Construction without Formal Analysis 
 
 349. Clearly, if MN\^ a line parallel a X 
 
 to BO through Aj and X is any point /^^SxT' '\ 
 
 on MN, then A XBC = A ABC. /-""^'^^^^V 
 
 In fact, it is clear that : 
 
 The locus of the vertex of a triangle equal to A ABC and hav- 
 ing the base BC is a pair of lines parallel to BC at the distance 
 of A from BC. 
 
 This fact aids in making numerous constructions. 
 
 Ex. 89. (a) Construct a triangle equal to a given triangle, having 
 the same base BC but having the Z XBC = 60^. 
 (6) Make a similar construction if Z XBC = 45°. 
 (c) Make a similar construction if Z XBC = 30°. 
 
 Ex. 90. Construct a A XBC equal to a given A ABC, having tlie 
 same base BC and side XB equal to a given segment d. 
 
 Ex. 91. Construct a A XBC equal to a given A ABC, and having 
 the median from Xto BC equal to a given segment m. 
 
 Ex. 92. Construct a parallelogram XBCYeqnaA to a given O ABCD^ 
 having the same base BC : 
 
 (a) having Z XBC = a given angle ; 
 (6) having side XB = a given segment ; 
 (c) having diagonal YB = a given segment. 
 
 Ex. 93. Construct a triangle equal to a given triangle and having 
 two of its sides equal to given segments m and n. 
 
 Suggestion. — Select m as base, and determine the altitude to m as in 
 Ex. 78. Continue as in § 349. 
 
 Ex. 94. Construct a triangle equal to a given triangle and having one 
 side equal to a given segment m, and one angle adjacent to that side 
 equal to a given angle, Z T. 
 
 Ex. 95. Construct a triangle equal to a given square, having given 
 its base and an angle adjacent to the base. 
 
 Ex. 96. Construct a triangle equal to a given square, having given 
 its base and the median to the base. 
 
 Ex. 97. Construct a rhombus equal to a given parallelogram, having 
 one of its diagonals coincident with a diagonal of the parallelogram. 
 
214 PLANE GEOMETRY — BOOK IV 
 
 C. Miscellaneous Supplementary Problems 
 
 Proposition XIV. Problem 
 
 350. Construct a rectangle equal to a given square, having 
 the sum of its base and altitude equal to a given segment. 
 
 C D,. ~ 
 
 -"F 
 
 M 
 
 N 
 
 A E 
 
 Given square M and segment AB. 
 
 Required to construct a rectangle = M, having the sum of 
 its base and altitude = AB. 
 
 Construction. 1. On AB as diameter construct semicircle 
 ADB. 
 
 2. Draw AC 1.AB, making AC = side of M. 
 
 3. Draw CF il AB, intersecting arc ADB at D. 
 
 4. Draw DE ± AB. 
 
 5. Construct □ N, having its base = BE and its altitude 
 = AE. 
 
 Statement. Rectangle N= square M. 
 
 Proof. 1. AE:DE = DE: BE. § 289 
 
 2. .-. DE^ = AEx BE. Why ? 
 
 3. .-. area oi M= area of N. Why ? 
 Discussion. The construction is impossible when the side 
 
 of the square is more than i AB. Why ? 
 
 Note. — § 350 suggests a geometrical solution of a quadratic of the form 
 x'^-tx + m^ = 0. 
 
 From this equation, 7/1"^ = x(t — x). Clearly, w corresponds to a side 
 of the square, x(t — x) corresponds to the area of the rectangle equal to 
 the square, and t corresponds to the given segment, for x + (t — x) =^ t. 
 
 Solve ic^ — 10 X + 16 = geometrically and check the solution alge- 
 braically. 
 
CONSTRUCTIONS 215 
 
 Proposition XV. Theorem 
 
 351. Construct a rectangle equal to a given square, having the 
 difference between its base and altitude equal to a given segment. 
 
 
 kC 
 
 M 
 
 i ')p, ..^ 
 
 A\ \ 1 
 
 D 
 
 N 
 
   E 
 
 Given square M and segment AB. 
 
 Required to construct a rectangle equal to Jf, having the 
 difference between its base and altitude equal to AB. 
 
 Constmction. 1. On AB as diameter, construct O ADB. 
 
 2. Draw ACl. AB, making AC=2i side of M. 
 
 3. Through the center 0, draw CO, intersecting the O at 
 D and E. 
 
 4. Construct □ N, having its base = CE and its altitude 
 = CD. 
 
 Statement. □ -A^ is the required rectangle. 
 
 Proof. 1. CE-CD = DE = AB', 
 
 that is, the base of ^— the altitude of -^= AB. 
 
 2. AC is tangent to the circle. Why ? 
 
 3. .-. CExCD= C7l\ § 287 
 
 4. .-. area of ^= area of M, Why ? 
 
 Discussion. The construction is always possible, since a 
 secant can always be drawn through the center of the circle 
 from an exterior point. 
 
 Note. — § 351 suggests a geometrical solution of a quadratic of the form 
 x^ ■}- tx — m^ = 0, for this equation may be written, m^ = x(t + x). 
 
 Solve the equation x^ + Sa; — 9 = geometrically, and check the solu- 
 tion algebraically. 
 
216 
 
 PLANE GEOMETRY — BOOK IV 
 
 Proposition XVI. Problem 
 
 352. Construct a square having a given ratio to a given 
 square. 
 
 R 
 
 i 
 
 a 
 
 X 
 
 
 ^ 
 
 
 S 
 
 ^a 
 
 b 
 
 
 '^>C}\ 
 
 
 V 
 
 
 o 
 
 
 Given square R and the segments a and h. 
 
 Required to construct a square S such that S : B = a : b. 
 
 Analysis. 1. Let x = one side of the required square. 
 2. . •. x^ : r'^ = a: b. 
 
 3. 
 
 4. .-. r : x = x 
 
 .'. bx'==ar\ ovx' = [~]xr. 
 
 ar 
 
 r and 
 
 b 
 5. Let 
 
 Algebra 
 
 or x is the mean proportional between 
 
 Why? 
 
 ar 
 y = —,OT by = ar. 
 
 
 
 6. .'.b :a = r :y, ov y is the fourth proportional to b, a, 
 and r. Why ? 
 
 Construction. 1. Construct y as determined in step 6. 
 
 2. Construct x as determined in step 4. 
 
 3. Construct square S having x as side. 
 
 Statement. 
 
 S:R = a:b. 
 
 
 
 Proof. 1. 
 
 S:R = x'':r^ 
 
 
 
 2. 
 
 But x"^ = ry, and 2/ = 
 
 ar ^ 
 
 b ' 
 
 Construction 
 
 3. 
 
 « '(f) 
 
 R- r' ' 
 
 
 Algebra 
 
 4. 
 
 , S_ar^ l_a 
 " R b ' r^ b' 
 
 
 
CONSTRUCTIONS 217 
 
 Proposition XVII. Problem 
 
 353. Construct a polygon similar to a given polygon and hav- 
 ing a given ratio to it. 
 
 V 
 
 A B 
 
 Given polygon AC and segments a and 6. 
 
 Required to construct a polygon A'O similar to AC and 
 p olygon A C a 
 
 such that —, yrvy, = T • 
 
 polygon A'C o 
 Analysis. 1. Let x = the side homologous to AB. 
 
 2. Then yolygon AC ^AW_ ^ 344 
 
 polygon A'C x' 
 
 o AW a 
 
 Complete the analysis as in Prop. XVI, thus determining x 
 in terms of AB^ a, and h. Then construct upon x as side 
 homologous to AB a polygon similar to polygon AC, by § 294. 
 This will be the required polygon. 
 
 Note. — Notice that § 362 is a special case of § 363, for all squares are 
 similar. 
 
 Ex. 98. Construct a rectangle similar to a given rectangle and having 
 to it the ratio 2:1. 
 
 Ex. 99. Construct an equilateral triangle equal to a given triangle. 
 
 Suggestion. — Determine the side s (Ex. 29, Book IV) as in § 348. Recall 
 Ex. 81, Book III. 
 
 Ex. 100. Construct a triangle equal to the sum of two given triangles. 
 
 Suggestion. — First construct squares equal to the given triangles. 
 
 Ex, 101. Construct a triangle equal to the difference of tvv^o given 
 triangles. 
 
 Ex. 102. Draw a line parallel to the ba.se of a triangle which will 
 divide the triangle into two parts which will have the ratio 1 : 2, 
 
 Suggestion. — A.xxq\jz% as hi Ex. 81, Book IV. 
 
218 PLANE GEOMETRY — BOOK IV 
 
 Proposition XYIII. Problem 
 
 354. Construct a polygon similar to one of two given polygons 
 and equal to the other. 
 
 m 
 
 Given polygons iWand N. 
 
 Required to construct a polygon similar to M and equal to N. 
 
 Analysis. 1. Let m= the side of the square equal to J/, 
 and n = the side of the square equal to N. Let x= that side 
 of the required polygon P which is homologous to AB. 
 
 2. ...^=^. Why? 
 
 3. But ^,= ~ = —• Since P = N. 
 
 Why 
 
 5. r.- = ^^-. Why? 
 n X 
 
 6. Hence x is the fourth proportional to m, n, and AB. 
 Construction. 1. Construct the squares equal to M and N, 
 
 thus determining segments m and n. See Ex. 77 
 
 2. Construct x as determined in step 6. 
 
 3. Upon X as side homologous to AB, construct a polygon 
 P similar to Jf. § 294 
 
 Statement. P ^ M, and P=N. 
 
 Proof. 1. P-^M. Why? 
 
 2. P: Jf=a;2:ZB2. Why? 
 
 3. But X = ^^^^ . Construction 2 
 
 m 
 
 4. .*. P : Jf = 71^ : ml Substituting in step 2 
 
 5. .-. P: il^= iV': J/, or P = iV: 
 
 ■•£= 
 
 M 
 
 ^. Since P 
 m^ 
 
 m^ 
 
 X^ 
 
 AB" 
 
 m 
 
 a; 
 ^5 
 
 
 m 
 n 
 
 ^5 
 
 
MISCELLANEOUS EXERCISES 219 
 
 Miscellaneous Exercises 
 
 Ex. 103. A road CO ft. wide runs from one corner to the opposite 
 corner of a square field measuring 500 ft. on a side, the diagonal of the 
 field running along the center of the road. What is the area of that 
 portion of the field occupied by the road ? (Carry out the results to two 
 decimal places.) 
 
 Ex. 104. What is the length of the side of an equilateral triangle 
 equal to a square whose side is 15 in.? 
 
 Suggestion. — Recall Ex. 29, Book IV. 
 
 Ex. 105. From one vertex of a parallelogram, draw lines dividing the 
 parallelogram into three equal parts. 
 
 Ex. 106. The sides AB, BC, CD, and DA of quadrilateral ABCD are 
 10, 17, 13, and 20 respectively, and the diagonal ^O is 21. Find the 
 area of the quadrilateral. 
 
 Ex. 107. If diagonals ^Cand BD of trapezoid ABCD intersect at E, 
 then AAEB = A DEC. {BC and AD are the bases of ABCD.) 
 
 Suggestion. — Compare A ABD and A ACD. 
 
 Ex. 108. If Xis any point in diagonal AC of O ABCD, then 
 AABX=AAXD. 
 
 Suggestion. — Draw the altitudes from B and D to base AX. 
 
 Ex. 109. If E and F are the mid-points of sides AB and ^C of 
 A ABC, then A AEF ;= J A ABC. 
 
 Ex. 110. If E is any point within O ABCD, then A ABE + A CDE 
 equals I the parallelogram. 
 
 Suggestion. —Throngh E draw a line parallel to AB. 
 
 Ex. 111. If Z ^ of A ABC is 30°, prove that the area of A ABC = 
 \ABx AC. 
 
 Suggestion. — Draw BD 1 AC. Recall Ex. 128, Book I. 
 
 Ex. 112. Prove that the area of a rhombus is one half the product of 
 its diagonals. 
 
 Ex. 113. If E is the mid-point of CD, one of the non-parallel sides of 
 trapezoid ABCD, prove that ABE = ^ ABCD. 
 
 Suggestion. — Through E, draw a line parallel to AB. 
 
 Ex. 114. Construct an isosceles triangle equal to a given triangle, 
 having given one side of length m. 
 
 Suggestion. — Use m as the base. Determine the altitude to m as in Ex. 78, 
 Book IV. Then follow § 241. 
 
220 PLANE GEOMETRY — BOOK IV 
 
 Ex. 115. Draw through a given point in one base of a trapezoid a 
 straight line which will divide the trapezoid into two equal parts. 
 
 Ex. 116. If the diagonals of a quadrilateral are perpendicular, the 
 sum of the squares on one pair of opposite sides of the quadrilateral 
 equals the sum of the squares on the other pair. 
 
 Note. — Supplementary Exercises 42 to 46, p. 298, can be studied now. 
 
 Review Questions 
 
 1. Define area of a plane figure. 
 
 2. Distinguish between congruent, similar, and equal figures. 
 
 3. State the rule for determining the area of : 
 
 (a) a rectangle ; (c) a triangle ; 
 
 (6) a parallelogram ; (d) a trapezoid. 
 
 4. State the formula for the area of any triangle in terms of its sides 
 a, 6, and c, and the number s. 
 
 What is the number s ? 
 
 5. State the formula for the area of an equilateral triangle in terms of 
 its side s. 
 
 6. State the corollaries by which the areas of two rectangles are 
 compared : 
 
 (a) If the rectangles have equal altitudes. 
 
 (&) If the rectangles have equal bases. 
 
 (c) When no known relation exists between the altitudes or the bases. 
 
 7. State the corresponding corollaries for two parallelograms. 
 
 8. State the corresponding corollaries for two triangles. 
 
 9. State a theorem connecting the areas of a triangle and a parallel- 
 ogram having equal bases and equal altitudes. 
 
 10. State a theorem connecting the areas of two similar polygons. 
 
BOOK V 
 
 REGULAR POLYGONS. MEASUREMENT 
 OF THE CIRCLE 
 
 355. Review the definitions given in § 125, § 128, and § 178. 
 
 356. A Regular Polygon is a polygon which is both equi- 
 lateral and equiangular. 
 
 The figures below illustrate some uses of regular polygons : 
 
 Two Linoleum Patterns 
 Notice the regular triangles, hexagons, squares, and octagons. 
 
 Ex. 1. Prove that the exterior angles at the vertices of a regular 
 polygon are equal. 
 
 Ex. 2. What is the perimeter of a regular pentagon one of whose 
 sides is 7 in. ? of a regular octagon one of whose sides is 6 in. ? 
 
 Ex. 3. In § 154, we have proved that the sum of the angles of any 
 polygon having n sides is (n — 2) st. A. 
 
 IIow large is each angle of a regular polygon having : (a) 3 sides ? 
 (6) 4 sides ? (c) 6 sides ? (d) 6 sides ? (e) 8 sides ? (/) 10 sides ? 
 
 Ex. 4. (a) Four square tile can be used to cover the space around 
 a point. (Why?) 
 
 (6) In the shape of what other regular polygon can tile be made in 
 order that the surface around a point can be completely covered by 
 using tile of the same shape ? 
 
 357. Each angle of a regular polygon having n sides is 
 
 Cv^) 
 
 A. (See Ex. 3.) 
 
 221 
 
222 
 
 PLANE GEOMETRY — BOOK V 
 
 Proposition I. Theorem 
 
 358. A circle can he circumscribed about any regu- 
 lar polygon. 
 
 Hypothesis. ABODE is a regular polygon. 
 Conclusion. A circle can be circumscribed about ABODE. 
 Proof. 1. A O can be constructed through A, B, and O. 
 Let be its center and OA, OB, and 00 be radii of it. 
 2. It can now be proved that this circle passes through D 
 by proving OD — OA. (Draw OD.) 
 
 Suggestions. — 1. Compare Z. AB C and L B CD ; Zl and Z 2 ; then Z 3 and Z4. 
 
 2. Prove A AOB ^ A OCD, and then OD = OA. 
 
 3. Hence the passes through D. 
 
 3. Similarly the circle can be proved to pass through E. 
 
 4. Hence a O can be circumscribed about ABODE. 
 
 359. Cor. 1. A circle can be inscribed in any regular polygon. 
 
 Proof. 1. AB, BO, OD, etc. are equal 
 chords of the circle which can be cir- 
 cumscribed about ABODE. 
 
 2. Hence these sides are equidistant 
 from 0. Why ? 
 
 3. Hence a circle can be drawn 
 tangent to each of the sides of ABODE. 
 
 § 198 
 
REGULAR POLYGONS 223 
 
 360. The Center of a regular polygon is the common center 
 of the circumscribed and inscribed circles ; as 0. 
 
 The Eadius of a regular polygon is the distance from its 
 center to any vertex ; as OA. 
 
 The Apothem of a regular polygon is the distance from its 
 center to any side ; as OF. 
 
 The Central Angle of a regular polygon is the angle between 
 the radii drawn to the ends of any side; as Z AOB, 
 
 The Vertex Angle of a regular polygon is the angle between 
 two sides of the polygon. 
 
 q/»AO 
 
 361. Cor. 2. The central angle of a regular n-gon is . 
 
 n 
 
 362. Notation. The following notation will be employed : 
 (a) s^, Sg, or s„ will denote one side of a regular inscribed 
 
 polygon of 4, 6, or n sides respectively. 
 
 (6) a^j ttg, or a„ will denote the apothem of a regular in- 
 scribed polygon of 4, 6, or ri sides respectively. 
 
 (c) P4j Pq, or p„ will denote the perimeter of a regular in- 
 scribed polygon of 4, 6, or n sides respectively. 
 
 (d) 7Ci, k^ or k^ will denote the area of a regular inscribed 
 polygon of 4, 6, or n sides respectively. 
 
 To denote the corresponding quantities for a regular circum- 
 scribed polygon, a capital letter with the appropriate subscript 
 will be employed. Thus, 
 
 S^ = one side of the regular circumscribed pentagon. 
 
 A^ = the apothem of the regular circumscribed pentagon. 
 
 Pg = the perimeter of the regular circumscribed pentagon. 
 
 /ig = the area of the regular circumscribed pentagon. 
 
 Ex. 5. Find the number of degrees in the central angle and in the 
 vertex angle of a regular polygon of : (a) 3 sides ; (6) 4 sides ; ^(c) 5 sides ; 
 (d) 6 sides ; (e) 8 sides ; (/) 10 sides. 
 
 Find also the sum of the central angle and the vertex angle in each case. 
 Do the results suggest any theorem ? 
 
 Ez. 6. Prove that any radius of a regular polygon bisects the angle to 
 whose vertex it is drawn. 
 
 Note. — Supplementary Exercises 1 to 2, p. 299, can be studied now. 
 
224 
 
 PLANE GEOMETRY — BOOK V 
 
 Pkoposition II. Theorem 
 
 363. The area of a regular polygon is equal to one 
 half the product of its apothem and its perimeter. 
 
 A F B 
 Hypothesis. The perimeter of regular polygon AC is p and 
 
 the apothem OF is r. 
 Conclusion. Area of ABODE = | rp. 
 
 Plan. From the center of polygon AC, draw the radii 
 OA, OB, OC, etc. forming A having the common altitude r. 
 Determine the area of each triangle and add the results. 
 
 Proposition III. Theorem 
 
 364. If a circle he divided into any number of equal 
 arcs, the chords of these arcs form a regular inscribed 
 polygon of that number of sides. 
 
 Hypothesis. AB=BC= CD=BE^EA in O 0. 
 Conclusion. ABODE is a regular pentagon. 
 
 [Proof to be given by the pupil.] 
 Note. — Supplementary Exercise 3, p. 299, can be studied 
 
REGULAR POLYGONS 
 
 225 
 
 365. Cor. 1. If from the mid-point of 
 each arc subtended by a side of a regular 
 polygon lines be drawn to its extremities^ a 
 regular inscribed polygon of double the num- 
 ber of sides is formed. 
 
 366. Cor. 2. An equilateral polygon 
 inscribed in a circle is regular. 
 
 Note. — Supplementary Exercises 4 to 5, p. 299, can be studied now. 
 
 Proposition IV. Theorem 
 
 367. If a circle is divided into any number of equal 
 arcs, the tangents at the points of division form a regu- 
 lar circumscribed polygon of that number of sides. 
 
 Hypothesis. O ACD is divided into five equal arcs, AB, BCj 
 etc. X T, YZ, etc. are tangent to O ACD at A, B, etc., form- 
 ing pentagon XTZWK. 
 
 Conclusion. X YZ WK is a regular pentagon. 
 
 Suggestions. 1. Draw ^B, ^C, (7Z>, etc. 
 
 2. Prove ^ AXE, A YB, etc. congruent isosceles A. 
 
 3. Prove AX = AY= BY= BZ, etc. 
 
 4. Prove XY = YZ = ZW, etc. 
 Recall § 356. Complete the proof. 
 
 Ex. 7. Prove that any apothem of a regular polygon bisects the side 
 to which it is drawn. 
 
 Ex. 8. Prove that the diagonals drawn from one vertex of a regular 
 hexagon divides the angle at the vertex into 4 equal angles. 
 
226 
 
 PLANE GEOMETRY — BOOK V 
 
 Corollaries to Proposition. IV 
 
 368. Cor. 1. Tarigents drawn to the cir- 
 cle at the mid-points of the arcs iiicladed be- 
 tween two consecutive points of contact of a 
 regular circumscribed polygon form, with the 
 sides of the original circumscribed polygon, 
 a regular circumscribed polygon having 
 double the number of sides. 
 
 YoT the circle is divided into double the number of equal 
 arcs. The theorem follows by § 367. 
 
 369. Cor. 2. Tangents drawn to the 
 circle at the mid-points of the arcs sub- 
 tended by the sides of a regular inscribed 
 polygon form a regular circumscribed 
 polygon of the same number of sides. 
 
 If ABODE is regular, and F, G, H, 
 K, 3ind L are the mid-points of -arcs 
 
 AB, BC, etc., then LF = FG, etc. 
 Hence A'B'C'D'E' is regular. 
 
 Note. — Supplementary Exercise 6, p. 299, can be studied now. 
 
 370. Construction of Regular Polygons is based upon Prop- 
 ositions III and IV. In order to divide a circle into any number 
 of equal parts, it is sufficient to be able to divide the total angle 
 around the center into that same number of equal angles. 
 
 Ex. 9. By using your compass, ruler, and protractor, draw a regu- 
 lar inscribed pentagon within a circle of radius 2.5 in. 
 
 Ex. 10. Draw a regular polygon of 9 sides within a circle of 2.5 in. 
 as in Ex. 9. 
 
 371. It is customary in geometry, however, to use only the 
 comjjass and straightedge in making constructions. It is in- 
 teresting therefore to inquire what regular polygons can be 
 made by using only these tools. 
 
REGULAR POLYGONS 
 
 227 
 
 Proposition V. Problem 
 372. Inscribe a squai^e in a given circle. 
 
 Given circle 0. 
 
 Required to inscribe a square in circle 0. 
 
 Construction. 1. Draw AG and BD, perpendicular diameters. 
 2. Draw chords AB, BC, CD, and AD. 
 
 Statement. ABCD is the required square. 
 
 Proof. 1. AB=BG=CD = DA. Why? 
 
 2. .-. ABCD is an inscribed square. . § 364 
 
 373. Cor. 1. Regular inscribed x>olygons of 8, 16, 32, etc. 
 sides can he constructed. § 365 
 
 Note. — Hence, by § 372 and § 373, regular inscribed polygons the 
 number of whose sides is a number of the form 2" where n is an integer 
 ^ 2, can be constructed by ruler and compass alone. 
 
 Thus, when n = 2, 2" = 4 ; when n = 3, 2'» = 8 ; etc. 
 
 Ex. 11. Construct within a circle having a 3-in. 
 radius an eight-pointed star like the one which forms 
 the central unit of the adjoining linoleum pattern. 
 
 Ex. 12. A designer wishes to make a pattern for 
 the octagonal top of a taboret whose longest diag- 
 onal is to be 18 in. Make a scale drawing of the 
 octagon, letting 1 in. represent 3 in. 
 
 Ex. 13. A square is inscribed in a circle of radius 10 in. Compute S4, 
 04, 7)4, and kj^ to two decimal places. (See § 362.) 
 
 Ex. 14. A square is circumscribed about a circle of radius 10 in. 
 Compute /S'4, P4, and K^. 
 
 Note. — Supplementary Exercises 7 to 14, p. 299, can be studied now. 
 
228 PLANE GEOMETRY — BOOK V 
 
 Proposition VI. Problem 
 374. Inscribe a regular hexagon in a circle, 
 -^.^^ ^ C 
 
 1^^- ^^^E 
 
 Given circle 0. 
 
 Required to inscribe a regular hexagon in circle 0. 
 
 Analysis. The central angle of a regular hexagon is 60°. 
 
 Construction. Draw any radius OA. With A as center 
 and Oxi as radius draw an arc cutting the at B. 
 
 Statement. AB = \ oi the O, and may be applied 6 times to 
 the circle. The chords of these arcs form the regular inscribed 
 
 ^ ' [Proof to be given by the pupil.] 
 
 Suggestions. — Prove Z AOB = 60° and that AB = | of the circle. 
 
 375. Cor. 1. Chords joining the alternate vertices of a regu- 
 lar inscribed hexagon, starting with any vertex, form a regular 
 inscribed triangle. 
 
 376. Cor. 2. Regular inscribed polygons of 12, 24, JfS, etc., 
 sides can be constructed. (§ 365.) 
 
 Note. — By §§ 374, 375, and 376, regular inscribed polygons the number 
 of w^hose sides is a number of the form 3 • 2« can be constructed with ruler 
 and compass alone, where n is an integer ^ 0. 
 
 (What is 3 . 2^ when w = 0?l?2?3?) 
 
 Ex. 15. Prove that the diagonals joining alternate vertices of a regu- 
 lar hexagon are equal. 
 
 Ex. 16. Prove that the radii of a regular inscribed hexagon divide it 
 into six congruent equilateral triangles. 
 
REGULAR POLYGONS 
 
 229 
 
 Ex. 17. Prove that diagonals AD, BE, and CF of a regular hexagon 
 ABCDEF are the diameters of its circumscribed circle. 
 
 Ex. 18. Prove that the opposite sides of a regular hexagon are 
 parallel. 
 
 Ex. 19. Prove that the diagonal FC of regular hexagon ABCDEF is 
 parallel to sides AB and DE. 
 
 Ex. 20. Prove that the diagonal AD of regular hexagon ABCDEF is 
 perpendicular to diagonal BF and bisects it. 
 
 Ex. 21. A regular hexagon is inscribed in a circle of radius 10 in. 
 Compute the lengths of se, «6» P^^ and k^. (See § 362.) 
 
 Ex. 22. A regular triangle is inscribed in a circle of radius 10 in. 
 Compute the lengths of as, Ss, Pa, and kz. 
 
 Ex. 23. Find the area of a regular hexagon whose apothem is 6 in. 
 
 Ex. 24. Construct one of the following designs : 
 
 AB = 6" 
 CD = 4" 
 
 AB = b" 
 CD = 3" 
 
 CD = 6" 
 EF = 2" 
 
 Ex. 25. Construct a pattern for a doily like the one 
 adjoining, making the dimension AB = 9 in. and having 
 the radius of the outer arcs ^ in. longer than the radius 
 of the inner arcs of the scallops. 
 
 Note. — Supplementary Exercises 15 to 31, p. 301, can be studied now. 
 
 377. A segment is divided by a given point in Extreme and 
 Mean Ratio when one part is the mean proportional between 
 the whole segment and the other part. 
 
 Thus G divides AB internally in extreme ^ 
 and mean ratio if C 
 
 AB:AC= AC: CB. 
 
 Notice that in this case the whole segment is to its longer part as 
 the longer part is to the shorter part. 
 
230 PLANE GEOMETRY — BOOK V 
 
 Proposition YII. Problem 
 
 378. Divide a given segment in extreme and mean 
 ratio. ^ ^ 
 
 A' 5 'B A^ e^-J^ 
 
 Fig. 1 Fig. 2 
 
 Given segment AB — m. 
 
 Required to divide AB in extreme and mean ratio. 
 
 Analysis. 1. Let x = AQ and .•. m — x= CB. 
 
 2. .'. m:x = x: {m — x). 
 
 3. .*. x^ = m{m — x). 
 
 4. .*. a;2 + mx = m^. 
 
 6. ,.(.+|J = ^.+f^ 
 
 2^ 
 
 X =w,^4-f'- 
 
 7. ,'.x-\ is the hypotenuse of a right triangle whose 
 
 base is m and whose altitude is — . 
 
 z 
 
 Construction. (Fig. 2.) 1. Draw EB ± AB, making AB = m, 
 and^^ = ^. 
 
 7)1 
 
 2. Draw ^^ and on it take EF=EB = ^. 
 
 3. On AB, take yl e = AF. 
 Statement. AB : AC = AC : CB. 
 
 Note. — If the equation of step 6 of the Analysis is solved for x : 
 
 ^ 4 2 2 > 2 
 
 .•.i» = -(V5-l)=- (2.236-1) =-(1.230)=. 6 m. 
 
 2 ^ ji 
 
REGULAR POLYGONS 231 
 
 Proof.* 1. Complete the circle with center E and radius 
 EB, and extend AEj cutting the circle at G. 
 
 2. AB is tangent to O BFG. Why ? 
 
 3. ... :^=:^. §287 
 
 AB AF 
 
 4. ... ^ = ^. Why? 
 
 AB AC _ 
 
 5 . AG-AB ^ AB-AG ^.^ 
 
 " AB AC ^' 
 
 6. But AB=: 2 EB = FG. Const. 
 
 7. .-. ^G - yl£ = ^(? - i^G^ = AF= AC. 
 g . AC^CB 
 
 ' ' AB AC 
 (Substituting in step 5.) 
 
 9. ...4^ = ^. Why? 
 
 AC CB ^ 
 
 Note 1. — A point D divides a segment AB externally (see § 305) in 
 extreme and mean ratio if AB : AD = AD : DB. This form of division is 
 not used in this text. 
 
 Note 2. — The Greeks called this method of division of a segment Golden 
 Section. It represented to them the most artistic division of a segment 
 into two parts. 
 
 Ex. 26. Find AC and CB in § 378 algebraically if AB = 10 in. 
 (Let AC =x, and hence CB = 10 — x. Continue algebraically.) 
 
 Ex. 27. (a) What is the relation between the area of the inscribed 
 and of the circumscribed squares of a given circle ? 
 
 (6) What is the relation between the perimeter of the inscribed and 
 of the circumscribed squares of a given circle ? 
 
 Ex. 28. Prove that the opposite sides of a regular octagon are 
 parallel. 
 
 Ex. 29. Prove that diagonals HC and DG of regular octagon 
 ABCDEFGH are parallel . 
 
 Ex. 30. Prove that figure ACDF of regular octagon ABCDEFGH 
 is an isosceles trapezoid. 
 
 * This proof may be omitted if desired. 
 
232 PLANE GEOMETRY — BOOK V 
 
 Proposition VIII. Problem 
 379. Inscribe a regular decagon in a given circle. 
 
 Given o ACD. 
 
 Required to inscribe a regular decagon in O ACD. 
 
 Construction. 1. Draw any radius OA and divide it internally 
 at M in extreme and mean ratio so that 
 
 OA: 0M== OM.AM. 
 
 2. With A as center and OM (the longer segment) as radius, 
 draw an arc cutting the given circle at B. 
 
 Statement. AB is -^^ of the circle, and AB is the side of 
 the regular inscribed decagon. 
 
 Proof. 1. Draw OB and BM. 
 
 2. In A OAB and A ABM: 
 
 Z.A=AA; 
 
 OA:AB = AB: AM. 
 
 [Substituting AB for OM in the proportion of step 1, Construction.] 
 
 3. .-. A OAB ~ A ABM § 280 
 
 4. .'. Z1 = ZA0B. Why? 
 
 5. A OAB is isosceles, and hence A ABM is isosceles. 
 
 6. .-. 0M= AB = BM, or A 0MB is isosceles. Why ? 
 
 7. .:ZA0B=Z2. 
 
 8. Z 4 = Z 2 + Z AOB, or Z 4 = 2 • Z AOB. Why? 
 
 9. .'. Z3 = 2'ZAOB. Why? 
 
 10. .: ZAB0 = 2'ZA0B. Why? 
 
 11. ZAOB + Z3 + ZABO=1SO°. Why? 
 
REGULAR POLYGONS 233 
 
 12. .',5-^A0B = 180°. Why ? 
 
 13. ..ZA0B=S6°. 
 
 14. Hence AB = y^^ of the circle and AB is one side of the 
 regular inscribed decagon. 
 
 Note. — This construction is attributed to Pappus. 
 
 380. Cor. 1. Chords joining the alternate vertices of a regu- 
 lar inscribed decagon, starting ivith any vertex, form a regular 
 inscribed jyentagon. 
 
 381. Cor. 2. Regular inscribed polygons of 20, Jf.0, 80, etc., 
 sides can be constnicted with ruler and compass alone. Why ? 
 
 Note. — By §§ 879, 380, and 381, regular inscribed polygons the num- 
 ber of whose sides is a number of the form 5 • 2" can be constructed with 
 ruler and compass alone. 
 
 ( What is 6 . 2» when ?iisO? 1? 2? 3? etc.) 
 
 Ex. 31. In a circle having a 3 in. radius inscribe a regular decagon 
 making all of the constructions, 
 
 (Keep the resulting figure for use in later exercises.) 
 
 Ez. 32. Construct the adjoining figure, having the 
 points A 2 1 in. from the center and the points B 3 in. 
 from the center. 
 
 (From Ex. 31, obtain the arcs which are ^^ of the 
 larger circle.) 
 
 Ex. 33. Prove that the diagonals of a regular pentagon are equal. 
 
 [Construct the pentagon in a circle of radius 3 in., using the arcs ob- 
 tained in Ex. 31. J 
 
 Ex. 34. Prove that diagonal AC oi regular pentagon ABODE is 
 parallel to side BE. (Circumscribe a O about the pentagon.) 
 
 Ex. 35. If ABODE is a regular inscribed pentagon in circle O, prove 
 that a diameter perpendicular to side DE passes through B and is also 
 the perpendicular-bisector of the diagonal AG. 
 
 Ex. 36. If the diagonals AC and BE of a regular inscribed pentagon 
 ABODE intersect at F, prove that A ABF is isosceles. 
 Prove also that A AEF is isosceles. 
 
 Ex. 37. Construct by ruler and compass alone an angle of 36° ; also 
 an angle of 18°. 
 
 Note. — Supplementary Exercises 32 to 37, p. 301, can be studied now. 
 
234 PLANE GEOMETRY — BOOK V 
 
 Proposition IX. Problem 
 
 382. Inscribe a regular pentadecagon (15-^ow) in a 
 circle. 
 
 Given O MN. 
 
 Required to inscribe in O MN a pentadecagon. 
 
 Analysis. 1. The central Z of a pentadecagon = ^-— = 24°. 
 
 2. But 24° = 60° -36°. ^^ 
 
 3. This suggests a combination of the constructions of § 374 
 and § 379. 
 
 Construction. 1. Draw chord AB, a side of a regular in- 
 scribed hexagon, and chord AC, a side of a regular inscribed 
 decagon. 
 
 2. Draw chord BC. 
 
 Statement. BC is one side of the regular inscribed penta- 
 decagon. 
 
 Proof. 1. BC = {\ — yV) or tt of the circle. Const. 
 
 383. Cor. Regular polygons of 30, 60, etc., sides can he in- 
 scribed in a circle by ruler and compass alone. 
 
 Note. — Regular polygons, the number of whose sides is a number of 
 the form 15 • 2", where n is an integer, can be constructed with ruler and 
 compass alone. (§§382 and 383.) 
 
 384. Combining the results of §§ 373, 376, 381, and 383, it 
 can be said that regular polygons of 2", 3 • 2% 5 • 2", and 15 • 2" 
 sides (n = an integer) can be inscribed in a circle by ruler and 
 compass alone. 
 
 Eac. 38. How large is the vertex angle of a regular pentadecagon ? 
 
 Ex. 39. What regular polygons having a number of sides less than 
 100 can be constructed by ruler and compass alone ? 
 
REGULAR POLYGONS 
 
 235 
 
 Proposition X. Theorem 
 
 385. Regular polygons of the same number of sides 
 are similar. 
 
 Hypothesis. ABODE and A'B'C'iyE^ are regular polygons 
 of 5 sides. 
 
 Conclusion. ABODE - A'B'O'D'E', 
 Proof. 1. The polygons are mutually equiangular. 
 [Since each Z of each polygon is (^)° or 108°.] 
 
 2. Since AB = 5(7 = OD, etc., and A'B' = B'C = O'D', etc. 
 
 .,^^ = ^=_^,etc. Ax. 6, §51 
 
 A'B' B'O' CD'' 
 
 3. .-. the polygons have their homologous sides proportional. 
 
 4. .-. ABODE - A'B'O'D'E'. Why ? 
 
 Ex. 40. Construct a square having given one of its diagonals. 
 Ex. 41. A square is inscribed in a circle of radius B. Prove : 
 
 (a) s, = EV2', (c) a4 = |V2; 
 
 (6) p4 = 4:By/2; (d) *4 = 2 i?^. 
 
 Ex. 42. In the figure for § 369, prove that the apothem of the in- 
 scribed polygon becomes, when extended, the apothem of the circum- 
 scribed polygon. 
 
 Ex. 43. In the figure for § 367, prove that the radius drawn to any 
 vertex Y is the perpendicular bisector of the side AB of the inscribed 
 polygon. 
 
 Ex. 44. Prove that the sides of a regular x)olygon circumscribed about 
 a circle are bisected by the points of tangency. 
 
236 
 
 PLANE GEOMETRY — BOOK V 
 
 Proposition XI. Theorem 
 
 386. The perimeters of two regular polygons of the 
 same number of sides have the same ratio as their 
 radii, or as their apothems. 
 
 F' B 
 
 Hypothesis. P and P' are the perimeters, R and P' are 
 the radii, and r and r' are the apothems respectively of the 
 regular polygons AC and A'C of the same number of sides. 
 
 Conclusion. -^ = ^ = L, 
 
 P' M' / 
 
 Proof. 1. Let and 0' be the centers of polygons AC and 
 A^C respectively. 
 
 Draw radii OA, OB, O'A', and O'B', and apothems OF and 
 O'F'. 
 
 Polygon AC ^ polygon A'C\ 
 
 ' ' P' A'B'' 
 A OAB - A O'A'B'. 
 
 AB' 
 
 6. 
 
 AB _R 
 ~ R" 
 P 
 P' 
 
 and also 
 
 AB 
 A'B' 
 
 R' 
 
 §385 
 §297 
 §295 
 
 Why? 
 
 Why? 
 
 Ex. 45. The perimeters of regular inscribed polygons of 6 and 12 
 sides respectively inscribed in a circle of diameter 2 are approximately 
 6 in. and 6.21 in. respectively. What are the perimeters of regular in- 
 scribed polygons of 6 and 12 sides respectively in a circle of diameter 4 ? 
 of diameter 7 ? of diameter 10 ? 
 
REGULAR POLYGONS 237 
 
 387. Cor. The areas of two regular polygons of the same 
 number of sides have the same ratio as the squares of their radii 
 or as the squares of their apothems. 
 
 1. Let K and K' represent the areas of the polygons AC 
 
 and A'C respectively. 
 
 Then 4^=4^. §344 
 
 K 
 K' 
 
 AB" 
 A'B'' 
 
 K 
 K' 
 
 
 2. j^ = fi- = :!_. See step 5, § 386 
 
 Eac. 46. The perimeters of regular polygons of 4 and 8 sides respec- 
 tively circumscribed about a circle of diameter 2 in. are 8 in. and 6.63 in. 
 respectively. What are the perimeters of regular circumscribed polygons 
 of 4 and 8 sides respectively circumscribed about a circle of diameter 6 ? 
 of diameter 5 ? 
 
 Ex. 47. The area of a regular hexagon inscribed in a circle of radius 
 3 in. is 23.38 sq. in. What is the area of a regular hexagon inscribed in a 
 circle of radius 6 in. ? of one in a circle of radius 1 in. ? 
 
 Ex. 48. The area of a regular octagon circumscribed about a circle 
 of radius 1 in. is 1.656 sq. in. What is the area of a regular octagon cir- 
 cumscribed about a circle of radius 2 in. ? of radius 5 in. ? 
 
 Ex. 49. Prove that the square inscribed in a semi- 
 circle is equal to two fifths the square inscribed in the 
 entire circle. 
 
 Sugrjestions, — Let R = the radius of the circle. Com- 
 pute the areas of the two squares. 
 
 Ex. 50. Prove that the perimeter of any regular inscribed polygon is 
 less than the perimeter of the regular circumscribed polygon of the same 
 number of sides. 
 
 Ex. 51. In a circle of 2 in. radius, inscribe a square, a regular octa- 
 gon, and also a regular 16-gon. Prove thatp4 < p% <pi6. 
 
 Ex. 52. About a circle of 2 in. radius circumscribe a square, a regu- 
 lar octagon, and a regular 16-gon. Prove P4 > Pg > Pie- 
 Ex. 53. (a) Prove that p4, pg, Pie, etc., are each less than P4. 
 (6) Prove that P4, Pg, Pie, etc., are each greater than ^4. 
 
 Suggestions. — 1. Compare pi^ with Pig. See Ex. 50. 
 2. Compare Pie with P4. 
 
238 
 
 PLANE GEOMETRY — BOOK V 
 
 MENSURATION OF A CIRCLE. INFORMAL TREATMENT 
 
 388. Length of a Circle. We have defined the length of a 
 straight line segment as the ratio of that segment to the unit of 
 linear measure, — another straight line segment. Clearly we 
 cannot define the length of a circle, in that manner, because we 
 cannot lay off the linear unit of measure along a circle. In 
 defining the length of a circle therefore, an entirely new pro- 
 cedure is necessary. The treatment which follows, while in- 
 formal, involves nevertheless the ideas which underlie the 
 formal treatment of this same topic given in § 401 to § 413 
 inclusive. 
 
 (a) In the adjoining circle are inscribed 
 a square and a regular octagon ; imagine 
 that the regular inscribed polygons of 16, 
 32, etc., sides also are drawn. 
 
 The perimeters of these polygons have 
 been denoted by p^, p^, piQ, etc. (§ 362). 
 
 We have proved (Ex. 51) tliSit p8> p^; 
 that piQ > ps ; that ^32 > PiQ ; etc. In other 
 words, the perimeters of the regular inscribed polygons increase 
 as the number of sides increases. 
 
 (6) About the adjoining circle there are 
 circumscribed the regular polygons of 4 
 and 8 sides. Imagine that those of 16, 
 32, etc., sides also are drawn. 
 
 The perimeters of these polygons have 
 been denoted by P4, Pg? Pm etc. (§ 362). 
 
 We have proved (Ex. 52) that P,> Ps; 
 that Pg > P16 ; that Pig > P32 ; etc. In other words, the perim- 
 eters of the regular circumscribed polygons decrease as the 
 number of sides increases. 
 
 (c) From the figure it is evident that the successive inscribed 
 polygons come closer and closer to the circle ; likewise that the 
 successive circumscribed polygons come closer and closer to the 
 circle. 
 
MEASUREMENT OF THE CIRCLE 239 
 
 It is evident also that the length of the circle is greater than 
 the perimeter of any inscribed polygon and that the length of the 
 circle is less thay the perimeter of any circumscribed polygon. 
 
 It is natural therefore to regard the successive perimeters of 
 the regular inscribed polygons and also of the regular circum- 
 scribed polygons as better and better approximations to the 
 length of the circle. 
 
 (d) By careful computation it has been found that when the 
 diameter of a circle is 1 : 
 
 p, =2.82843. P, =4. 
 
 ^8 =3.06147. Pg =3.31371. 
 
 j9i6 = 3.12145. Pi6 = 3.18260. 
 
 ^32 = 3.13655. P32 = 3.15172. 
 
 p^ = 3.14033. P64 = 3.14412. 
 
 i)i28 = 3.14128. Pi28 = 3.14222. 
 
 jP26e = 3.14151. P256 = 3.14175. 
 
 2)512 = 3.14157. P512 = 3.14163. 
 
 Apparently when the diameter of a circle is 1, the length of 
 the circle is approximately 3.1416. If we let C=the length 
 of the circle and d = the length of the diameter, then C -i-d — 
 3.1416. 
 
 (e) By Proposition XI, § 386, the perimeters of regular poly- 
 gons of the same number of sides have the same ratio as their 
 radii, and hence as their diameters, and also as their apothems. 
 
 If we double the diameter of the circle considered in 
 part (d), then we shall obtain for the successive perimeters of 
 the inscribed and of the circumscribed polygons exactly double 
 the lengths given in part (d). Evidently then the length of a 
 circle of diameter 2 is approximately double that of a circle of 
 diameter 1 ; that is, C = 2 x 3.1416 = 6.2832. Again, C-^d = 
 3.1416. 
 
 Similarly the length of a circle of diameter 5 is approxi- 
 mately 5 X 3.1416, or 15.7080. Again G^d = 3.1416. 
 
 389. The relation derived in parts (d) and (e) of § 388 is 
 not only apparently true but can be proved to be true. We 
 
240 PLANE GEOMETRY ~ BOOK V 
 
 shall assume it for the present. It amounts to assuming that 
 the length of a circle bears to the length of its diameter a con- 
 stant ratio. This fact is proved in § 415. 
 
 The Greek letter tt (pi) is used to denote this constant ratio. 
 
 That is, C^d=7r, or C=7rd. 
 
 Two useful approximations of tt are 3.1416 and 3^. 
 
 The length of a circle is called the Circumference of the circle. 
 
 Note. — The determination of the value of ir and of what sort of num- 
 ber TT is has been one of the most famous problems of mathematics. 
 
 The Egyptians early recognized that C -^ d i^ constant, and obtained 
 for this ratio a value which corresponds to 3.1605. 
 
 The Babylonians and Hebrews were content with the much less accurate 
 value, TT = 3. (See I Kings, vii. 23.) 
 
 The method employed in this text was introduced by Antiphon (469- 
 399 B.C.), improved by Bryson (a contemporary, probably), and finally 
 carried out arithmetically in a remarkable manner by Archimedes (287- 
 212 B.C.) in a pamphlet on the mensuration of the circle. Antiphon 
 suggested the use of inscribed regular polygons of 4, 8, etc., sides as a 
 means of approximating the length of the circle, and Bryson suggested 
 using at the same time the corresponding circumscribed regular polygons. 
 Archimedes employed inscribed and circumscribed regular polygons hav- 
 ing 3, 6, .-.96 sides in his computation, and showed that tt >3^^ and 
 
 <^. 
 
 The methods employed by Archimedes remained for a long time the 
 standard procedure in efforts to compute tt. As mathematical skill in- 
 creased, formulse for ir were derived, particularly in trigonometric form, 
 which enabled diligent computers to obtain the value to more and more 
 decimal places. 
 
 Vieta (1540-1603) was the first to derive a formula for tt (not, however, 
 a trigonometric one). He gave for tt the value 3.141529653. Others car- 
 ried out the computation to as many as 700 decimal places. 
 
 A Holland mathematician, Huygens (1629-1695), at the age of twenty- 
 five, proved some theorems which made it possible to improve greatly on 
 the methods of Archimedes, He was able to obtain from a regular hexagon 
 as accurate a value for tt as Archimedes obtained from the regular 96-gon. 
 
 Mathematicians were particularly interested in determining what kind 
 of number tt is. In 1766-1767, Lambert proved that it is not rational ; 
 I.e., that it cannot be expressed as the quotient of two integers. In 1882, 
 through methods introduced by Hermite in 1873, Lindeman proved that 
 
MEASUREMENT OF THE CIRCLE 241 
 
 IT is a transcendental number ; i.e. , that it cannot be the root of an ordinary 
 algebraic equation. This was the goal toward which previous efforts had 
 been directed, and thus completely solved a problem to which many of the 
 great mathematicians had given some attention. 
 
 390. Cor. 1. The circumference of a circle equals 2 ttt, where 
 r equals the number of linear units in the radius. 
 
 391. Cor. 2. The circumferences of tivo circles have the same 
 ratio as their diameters or as their radii. 
 
 Proof. Let r„ cZj, and Oj be the radius, diameter, and circum- 
 ference of one circle ; and let r^, d^, and Cg be the radius, diam- 
 eter, and circumference of another circle. 
 
 2. Then C\ = ttc^i = 2 vTi, and Cj = 7rc?2 = 2 Trrg. 
 
 o (7, Trdi 2 7rr. Ci f7i r, 
 
 O. .•.—! = i = 4, or — i=:— i = -^« 
 
 Ca TTfZz ^Trrg C2 di r^ 
 
 Note. — Remember that this proof is based on an informal treatment. 
 For the customary formal treatment of this theorem, read, if it seems 
 desirable, § 414. 
 
 Ex. 54. Find the circumference of a circle whose diameter is 5 in. ; 
 8 in. ; 10 in. 
 
 Ex. 55. How long is the piece of rubber for the tire of a buggy 
 wheel 4 feet in diameter ? 
 
 Ex. 56. If the diameter of a circle is 48 in., what is the length of an 
 arc of 85° ? 
 
 Ex. 57. How long must the diameter of a circular table be in order 
 to seat 20 people, allowing 30 in. to each person ? 
 
 (Express the result correct to the nearest inch.) 
 
 Ex. 58. A fly wheel in an engine room has a diameter of 10 feet. 
 Through how many feet does a point on its outer rim move in a minute 
 if the wheel makes 100 revolutions per second ? 
 
 Ex. 59. (a) What is the diameter of a circular race track whose 
 length along its inside edge is one mile ? 
 
 {h) If the track is 100 feet wide, determine the distance around it in 
 the middle of the track. 
 
 Ex. 60. Draw any circle. Construct the circle : 
 
 (a) Whose circumference is 3 times that of the given circle. See § 391. 
 
 (6) Whose circumference is \ that of the given circle. 
 
242 PLANE GEOMETRY — BOOK V 
 
 392. Area of a Circle. In the adjoining circle are inscribed 
 a square and a regular octagon. The area of the square is one 
 half the product of its apothem and its 
 perimeter. In symbols (§ 362) : 
 
 Similarly the area of the regular octa- 
 
 And the area of the regular inscribed 
 16-gon would be ^ 
 
 "'16 = yPi6 X Ct-16- 
 
 It is evident that the surface within each successive polygon 
 is more nearly equal to the surface within the circle. On the 
 other hand, it is clear that each successive apothem is more 
 nearly equal to the radius and that the length of the polygon 
 is more nearly equal to the length of the circle. (See § 388.) 
 
 It is reasonable therefore to conclude that the area of a circle 
 is one half the product of its radius and its circumference. 
 
 Letting K represent the area of the circle, then 
 
 393. Cor. 1. Since (7 = 2 Trr, then 7^ = 1 r x 2 ttt- = ttt^. 
 
 394. Cor. 2. Since C = nd, and r = -, then 
 
 K=lxix7rd = ^7rd\ 
 
 2 2 4 
 
 395. Cor. 3. Tlie areas of two circles have the same ratio as 
 the squares of their radii or of their diameters. 
 
 Letting K^ and K^ represent the areas of the circles whose 
 diameters are d^ and c?2j ^^id whose radii are ri and rg respec- 
 tively, then ^, ^ ^ ^ i^.^ ^^. ^ ^ ^ ^ ^_ 
 ^2 Trrj^ \Trd^^ 7^2 r^ d^ 
 
 Note. — This theorem was proved by Hippocrates (450-400 b.c. ). Look • 
 up his history. 
 
MEASUREMENT OF THE CIRCLE 
 
 243 
 
 396. A Sector of a Circle is the portion 
 of the interior of a circle which is within 
 a given central angle. The central angle 
 is called the angle of the sector. 
 
 397. Cor. 4. Tlie area of a sector is one 
 half the i)Toduct of the radius and the length 
 of the arc intercepted by its angle. 
 
 Let c = the length of the arc and k = the area of the sector 
 of a circle whose area, circumference, and radius are K, C, and 
 r, respectively. 
 
 Prove that 
 
 k = ^r X c. 
 
 Proof. The area of a sector has the same ratio to the area of the circle 
 that the length of its arc has to the circumference ; that is, 
 
 ^=^, orJfc 
 
 But, since 
 Substituting in (1), 
 
 K 
 K 
 
 '"^ C 
 
 (1) 
 
 ^-•f=i^- 
 
 k = c X ^r, or ^r x c. 
 
 398. A Segment of a Circle is that portion of the interior of 
 a circle which is between a chord of the circle 
 and its subtended arc ; as segment AXB, in- 
 dicated by the shaded part of the adjoining 
 figure. 
 
 The area of a segment AXB may be deter- 
 mined by subtracting the area of A AOB 
 from the area of sector OAXB. 
 
 Ex. 61. Find the circumference and area of a circle whose diameter 
 is 6 in. ; 8 in. ; 10 in. 
 
 Ex. 62. Find the radius and area of a circle whose circumference is 
 
 20 TT in. ; 38 tt in. ; 15 tt in. 
 
 Ex. 63. Find the radius and circumference of a circle whose area is 
 64 T sq. in. ; 81 tt sq. in. ; 225 tt sq. in. ; 289 tt sq. in. 
 
 Ex. 64. Find the side of a square equivalent to a circle whose diame- 
 ter is 12 in. 
 
244 PLANE GEOMETRY — BOOK V 
 
 Ex. 65. The diameters of two circles are 6 and 8 respectively. 
 
 (a) What is the ratio of their areas ? 
 
 (6) What is the ratio of their circumferences ? 
 
 Ex. 66. The radii of three circles are 3, 4, and 12, respectively. 
 What is the radius of a circle equal to their sum ? 
 
 Ex. 67. rind the area of a segment having for its chord a side of a 
 regular inscribed hexagon, if the radius of the circle is 10. See § 398 
 
 Ex. 68- If the radius of a circle is 4, what is the area of a segment 
 whose arc is 120°? 
 
 Ex. 69. Draw any circle. 
 
 (a) Construct the circle whose area is four times that of the given 
 circle. § 395 
 
 (6) Construct the circle whose area is i that of the given circle. 
 
 Ex. 70. Two pulleys in a machine shop are 
 connected by a belt. One has a radius of 9 in. 
 and the other a radius of 1 in. For each revolution 
 of the large pulley how many revolutions will the 
 small pulley make ? 
 
 Ex. 71. What is the area of the ring between two concentric circles 
 whose radii are 8 in. and 10 in. respectively ? 
 
 Ex. 72. A circular grass plot, 100 ft. in diameter, is surrounded by 
 a walk 4 ft. wide. Find the area of the walk. 
 
 Ex. 73. How many tulip bulbs will be required for a circular flower 
 bed 6 feet in diameter, allowing 16 sq. in. to eacli bulb ? 
 
 Ex. 74. In a steam engine having a piston 20 in. in diameter, the 
 pressure upon the piston is 90 lb. to the square inch. What is the total 
 pressure upon the piston ? 
 
 Ex. 75. A woman had a number of potted plants with which to 
 plant a circular flower bed. She planned to make the bed 4 feet in 
 diameter and found that she used up in that way just one half of her 
 plants. Approximately how large should she make the bed to use up all 
 of her plants ? 
 
 Ex. 76. Prove that the area of the ring included be- a /^IIA^Xc 
 
 tween two concentric circles is equal to the area of a f 'yC \ ^^\\ 
 
 circle whose diameter is that chord of the outer circle j i n j j 
 
 which is tangent to the inner. V V J j 
 
 (To prove area of ring = \ irAC'-^-) \.^^^^ 
 
MEASUREMENT OF THE CIRCLE 
 
 245 
 
 Ex. 77. In erecting a hot air furnace for dwellings, certain pipes are 
 installed for carrying the air to the various rooms of the house, and one 
 or more other pipes are put in to convey cold air to the furnace. The 
 cross section area of tlie cold air supply pipes must equal approximately 
 tlie sum of the cross section areas of the warm air pipes. 
 
 A house is to have four warm air pipes 9 in. in diameter, and three 12 in. 
 in diameter. One cylindrical cold air duct is to be installed. How large, 
 approximately, must its diameter be ? 
 
 Ex. 78. Prove that the area of a circle is equal to four times the area 
 of the circle described upon its radius as a diameter. 
 
 Ex. 79. In the adjoining semicircular arch con- 
 structed about center 0, the distance AB is 10 ft. If 
 the arch is to be constructed of 13 stones of equal size, 
 how long is each of the arcs like arc DE ? 
 
 Ex. 80. The adjoining figure represents a segmen- 
 tal arch. The method of construction and the dimen- 
 sions are indicated in the figure. If the arch is made 
 of 11 stones of equal size, what is the length of the arc 
 Xr? What is the height of the arch ? 
 
 Note. — Supplementary Exercises 38 to 58, p. 302, 
 can be studied now. 
 
 SUPPLEMENTARY TOPICS 
 
 Four groups of supplementary topics follow. Each is inde- 
 pendent of the others. Teachers should feel free to select the 
 group or groups which appear to meet the needs of the class. 
 
 Group A. — Inscription of Circles within Regular Polygons 
 and within Circles. 
 
 A topic of considerable interest because of its frequent application in 
 artistic design. 
 
 Group B. — Variables and Limits together with the Formal 
 Treatment of the Mensuration of the Circle and of the Incom- 
 mensurable Cases. 
 
 The treatment of this topic is scientifically correct, but is nevertheless 
 as elementary and pedagogical as the nature of the subject renders possible. 
 
 Group C. — Symmetry in Plane Figures. 
 
 Group D. — Maxima and Minima of Plane Figures. 
 
246 PLANE GEOMETRY — BOOK V 
 
 Group A 
 
 399. Inscription of Circles within Regular Polygons and 
 within Circles is a characteristic feature of art wIdcIow and 
 other designs. 
 
 Many of the necessary constructions are based upon the fol- 
 lowing illustrative problem or may be discovered by means of 
 an analysis similar to that employed in this problem. 
 
 Illustrative Problem. — Inscribe a circle in a given sector 
 of a circle. 
 
 Analysis. Let O XTZ be tangent to 
 
 radius OB at Z, to OA at Y, and to arc /j\ 
 
 AXB at X. Let CD be the common tan- / | \ 
 
 gent to arc AXB and O XYZ at X. /H"~^ 
 
 2. Then O XFZ is inscribed in A OOZ). YJ i Az 
 
 3. Hence the center P of Q XYZ lies A J:j )\ 
 on the bisectors of Z COD and Z OCD. ^,/<< I ^JJ^ 
 
 The radius is the distance from P to X. O X ^ 
 
 The construction is evident at once. 
 
 Rs. 81. Construct a circle with radius 2 in. ; and within it construct 
 a sector whose angle is 90°. 
 
 (a) Within this sector inscribe a circle. 
 
 (?)) Compare the area of this circle with the area of the sector itself 
 when the radius of the given circle is r instead of 2. 
 
 Ex. 82. Construct six equal circles within a circle of 
 radius 2 in., each tangent internally to the given circle 
 and tangent externally to two of the inner circles. 
 
 Ex. 83. (a) If the radius of the given circle in Ex. 82 
 is r, what is the radius of the inscribed circles ? 
 
 (&) Compare the circumference of one of the inscribed circles with the 
 circumference of the given circle. 
 
 (c) Compare the total area of the inscribed circles with the area of the 
 given circle. 
 
 Ex. 84. Construct a circle which will be tangent to each of the con- 
 structed circles in Ex. 82. 
 
 How does the radius of this circle compare with the radius of the six 
 inscribed circles ? 
 
INSCRIBED CIRCLES 
 
 247 
 
 Ex. 85. Construct six equal circles tangent externally 
 to a circle of radius ^ in. such that each circle is also 
 tangent externally to two of the constructed circles. 
 
 Ex. 86. If the radius of the given circle in Ex. 85 
 is r: 
 
 (a) What is the radius of the escribed circles ? 
 
 (6) Compare the circumference of the given circle with the circumfer- 
 ence of one of the escribed circles. 
 
 (c) Compare the area of the given circle with the area of one of the 
 escribed circles. 
 
 Ex. 87. In an equilateral triangle inscribe three equal 
 circles, each tangent to two sides of the triangle and tan- 
 gent externally to the other two circles. 
 
 Ex. 88. In a regular hexagon inscribe six equal cir- 
 cles, each tangent to two sides of the hexagon and tangent 
 externally to two of the circles. 
 
 Ex. 89. In a regular octagon inscribe eight equal circles, each tangent 
 to two sides of the octagon and also tangent externally to two of the 
 circles. 
 
 Ex. 90. In a regular hexagon inscribe six equal circles, each tangent 
 to one side of the hexagon and tangent externally to two of the circles. 
 
 Ex. 91. Inscribe in a regular hexagon three equal circles 
 each tangent to two sides of the hexagon and tangent exter- 
 nally to two circles. 
 
 Ex. 92. In a regular octagon inscribe four equal circles 
 each tangent to two sides of the octagon and also tangent ex- 
 ternally to two circles. 
 
 Ex. 93. The adjoining design appears in a floor 
 pattern in a corridor of the new Congressional Li- 
 brary. Construct such a figure, making a 5-in. square, 
 the radius of the inner O 1.5 in., and the radius of the 
 concentric O 1.76 in. 
 
 Ex. 94. The adjoining curve is a trefoil. 
 
 (a) Construct such a figure based upon an equilateral 
 triangle whose side is 2 in. long. 
 
 (b) What is the length of the trefoil if the side of the 
 equilateral triangle is s inches ? 
 
 (c) What is the area within the trefoil if the length 
 of the side of the equilateral triangle is s inches ? 
 
 Note. — Supplementary Exercises 59 to 63, p. 304, can be studied now. 
 
248 PLANE GEOMETRY — BOOK V 
 
 Group B. Mensuration of a Circle 
 
 400. The formal treatment of the mensuration of a circle 
 involves the use of certain ideas which are fundamental in 
 mathematics. 
 
 401. Variable, Constant, and Limit. 
 
 Example 1. — Consider the numbers 1, -, -, -, ... 
 
 ' 2' 4' 8' 
 
 Each number is one half the preceding; while each number 
 
 is greater than zero, the numbers ultimately become very small. 
 
 Imagine a literal number x which has these values successively. 
 
 Then ultimately x — becomes less than an}^ small positive 
 
 number, and thereafter remains less than that number. Thus, 
 
 ultimately, a; — becomes and remains less than , or 
 
 1 •^' _ 1000' 
 
   , or any other small positive number. 
 
 1,000,000 -^ ^ 
 
 This is clear, since x takes successively the values 
 
 .1111111 1 1 1 
 
 J _ _ . etc 
 
 ' 2' 4' 8' 16' 32' 64' 128' 256' 512' 1024' 
 
 Example 2. — Consider the numbers 1, 1^, 1|, l-J, 1||, ... 
 Although each number is less than 2, the numbers are constantly 
 increasing. 
 
 Imagine a literal number x which has these values succes- 
 sively. Then ultimately 2 — x becomes and remains less than 
 any small positive number ; thus 2 — x becomes and remains 
 less tinan ■yq o o7' 
 
 (Write down enough of the successive values of x to make 
 certain of this last statement.) 
 
 A Variable is a number which assumes different values 
 during a particular discussion. Thus, in Example 1, a; is a 
 variable ; it is a decreasing variable. In Example 2, x is an in- 
 creasing variable. 
 
 Note. — Variables do not either always increase or always decrease. 
 Thus, the variable which takes the values 1, — |, + \^ — |, etc., al- 
 ternately decreases and increases. 
 
VARIABLES AND LIMITS 249 
 
 A Constant is a number which has a fixed value tliroughout a 
 particular discussion. Thus, in Example 2, 2 is constant ; in 
 Example 1, is constant. 
 
 A Limit of a Variable is a constant such that the numerical 
 value of the difference between the constant and the variable 
 becomes and remains less than any small positive number. 
 
 We say that a variable approaches its limit. 
 
 Not every variable has a limit. 
 
 402. Axiom of Limits. If an increasing variable is always 
 less than some constant^ then it. approaches a limit which is less 
 than or equal to that constant. Jf a decreasing variable is always 
 greater than some constant, then it approaches a limit ivhich is 
 greater than or equal to that constant. 
 
 We may represent the foregoing definitions and axiom geo- 
 metrically as follows : 
 
 X 
 
 A z (J B 
 
 Let the distance from A to X as X moves toward B represent 
 a variable x. Let it be agreed that X never passes beyond B. 
 Then ^X must approach a limit such as AC which is less than 
 or equal to AB. (In the figure, AC is made less than AB.) 
 This means that ultimately point X comes and remains as close 
 to C as we please, possibly even coinciding with C. 
 
 403. Two Limits Theorems. 
 
 (a) If a variable x approaches a finite limit I, then ex, where c 
 is a constant, approaches the limit d. 
 
 For cl — cx = c(l — x). As X approaches the limit Z, the numerical 
 value of Z — X becomes and remains less than any small positive number. 
 Hence the numerical value of cl — ex becomes and remains less than any 
 small positive number. Therefore ex approaches the limit cl by definition. 
 
 (b) If two variables are constantly equal and each approaches a 
 finite limit, then their limits are equal. 
 
250 PLANE GEOMETRY — BOOK V 
 
 Let X approach the limit I and y approach the limit m. If x 
 always equals y, then / must equal m. 
 
 A ^ ? ^ L 
 
 ^- y P ^ 
 
 Let the distance AX represent the variable x, approaching 
 AL. Let the distance B Y represent the variable y, approach- 
 ing the limit BM. 
 
 Point X ultimately must come and remain close to the point 
 L ; point Y ultimately must come and remain close to the point 
 M. If AL were greater than BM, then AX would ultimately 
 become greater than BY. But this is impossible, for the vari- 
 able X must always equal the variable y. Similarly if AL were 
 less than BM. Hence the limit I equals the limit m. 
 
 404. Sequences of Regular Polygons. Let AB be a side of 
 
 a regular inscribed polygon in a circle of radius ^ -^ 
 
 r; let AC be a side of the regular polygon /^ \ 
 
 having double the number of sides of the (?) 
 
 first; let AD be a side of the regular polygon V /j J 
 
 having double the number of sides of the ^X>^i ^^^ 
 second; etc. 
 
 Such regular inscribed polygons, the number of whose sides 
 is successively doubled, will be called a sequence of regular 
 inscribed polygons. 
 
 Similarly, we shall have occasion to speak of sequences of 
 regular circumscribed polygons. 
 
 405. In a sequence of regular inscribed polygons, the length of 
 the side of the polygon is a decreasing variable which approaches 
 zero as limit, as the number of sides increases indefinitely. 
 
 For 3^rc AC = I arc AB (see Fig. § 404) ; arc AD = J arc AB ; etc. 
 
 Hence the arcs decrease indefinitely, approaching zero as limit. (See 
 Ex. 1, §401.) 
 
 The chords are less than the corresponding arcs. Hence the chordg 
 decrease indefinitely, approaching zero as limit. 
 
VARIABLES AND LIMITS 
 
 251 
 
 Evidently, also in a sequence of regular circumscribed poly- 
 gons the length of the side is a decreasing variable which 
 approaches zero as limit as the number of sides increases 
 indefinitely. 
 
 406. In a sequence of regular inscribed polygons, the length 
 of the apothem is an increasing variable which approaches the 
 radius of the circle as limit. 
 
 If ^5 is a side of any regular inscribed polygon and 
 00 is the apothem of the polygon, then 
 
 AO- 0C< AC, or AO- 00 <^AB. (§160.) 
 
 As the number of sides increases indefinitely, AB 
 decreases indefinitely in numerical value. Hence 
 AO — 00 must also decrease indefinitely in numerical 
 value. Therefore, 00 must approach AO as limit by definition. 
 
 LENGTH OF A CIRCLE 
 
 407. Consider the sequences of regular inscribed and circum- 
 scribed polygons having 4} S, 16, etc., sides in a circle of radius r. 
 
 Let p denote the variable perimeter of the inscribed poly- 
 gon, assuming the values p^, pg, pi^, etc. ; let P denote the 
 variable perimeter of the circumscribed polygon, assuming the 
 values of P^, Pg, P^,, etc. 
 
 i 
 
 ' . ^ 
 
 >- 
 
 -r^v 
 
 r 
 
 ^ 
 
 X 
 
 •0 z 
 
 V 
 
 wy 
 
 K 
 
 408. The perimeter p approaches a limit as the jiumber of 
 sides increases indefinitely. 
 
 Proof. 1. Pi <ps <pi6, etc. See Ex. 51 
 
 2. 1>4, P8, Pi6, etc. are all less than P4. See Ex. 63, (a) 
 
 3. .-.p approaches a limit as the number of sides increases indefinitely. 
 Call this limit ^4. § 402 
 
252 PLANE GEOMETRY — BOOK V 
 
 409. The j^erimeter P (see § 407) approaches a limit as the 
 number of sides increases indefinitely. 
 
 Proof. 1. P4 > Pg > P16, etc. See Ex. 52 
 
 2. P4, Pg, P16, etc. are each greater than jt)4. See Ex. 53, (6) 
 
 3. .'. P approaches a limit as the number of sides increases indefi- 
 nitely. Call this limit L^. § 402 
 
 410. The perimeters p and P of the sequences of regular in- 
 scribed and circumscribed polygons described in § 407 approach 
 one and the same limit ; that is, the limit l^ equals the limit L^. 
 
 Proof. 1. Let AB = a side of one of the inscribed polygons, and AD 
 and DB halves of two consecutive sides of the circum- 
 scribed polygon having the same number of sides. OD 
 and OA are the radii of these polygons, p and P de- 
 note their perimeters. 
 
 2. The two polygons are similar. 
 
 3. .-. P:p= OD: OA. 
 
 4. .-. (P-p) :p=(OD- OA): OA. 
 
 5. ,',P-p = J^(^OD- OA) . By Algebra 
 
 OA 
 
 6. .'.P-p<^x AD. 
 
 r 
 
 (Since J9 < P4 ; r = OA; smd OD - 0A< AD.) 
 
 7. Successively double the number of sides, letting the inscribed and 
 the circumscribed polygons always have the same number of sides. The 
 length of each side of the polygons will decrease, approaching the Umit 
 zero ; in particular, AD will approach as limit. 
 
 8. .'. -i X AD will approach as limit, since -^ is constant. 
 
 ^ ^ § 403, (a) 
 
 9. P — p will approach as limit. Def., § 401 
 
 10. .-. X4 = Z4. 
 
 For suppose that X4 > ^4, and that ^4 — ^4 = m, a number > 0. Ulti- 
 mately,' P differs but little from X4 and p but little from h ; hence P — p 
 ultimately differs but little from m. But P — p approaches the limit 0. 
 
 Similarly if ^4 < ^4. 
 
 11. Let C represent the common value of L\ and l^. 
 
 Then, as the number of sides increases indefinitely, the perimeters p 
 and P of the regular inscribed and circumscribed polygons respectively 
 approach the limit C. 
 
VARIABLES AND LIMITS 253 
 
 411. If now, instead of starting with the sequences of regu- 
 lar polygons having 4, 8, 16, etc. sides, we start with any other 
 sequences of regular inscribed polygons and circumscribed 
 polygons, such as those having 3, 6, 12, etc. sides, it can be 
 proved that the perimeters p and P again approach this same 
 limit C obtained in § 410. 
 
 This fact justities the following definition. 
 
 412. The Length of a Circle is the limit of the perimeter of 
 any regular inscribed polygon as the number of sides is in- 
 definitely increased. 
 
 Remember that the length of a circle is called the circum- 
 ference of the circle. 
 
 413. The perimeter of any regular circumscribed polygon ap- 
 proaches the circumference of the circle as limit if the number of 
 sides is indefinitely increased. §§ 410, 411, and 412 
 
 Ex. 95. What is a constant ? 
 Ex. 96. What is a variable ? 
 Ex. 97. What is the limit of a variable ? 
 
 Ex. 98. If a "sequence" of regular inscribed polygons be formed 
 (§ 404) in a circle : 
 
 (a) What magnitude is constant ? 
 
 (h) What magnitudes are decreasing variables, and what are their 
 limits ? 
 
 (c) What magnitudes are increasing variables, and what are their 
 limits ? 
 
 Ex. 99. What limit is approached by the variable which assumes the 
 values given in the note at the bottom of page 248 ? 
 
 Ex. 100. Suppose a variable assumes the values 1, — 1, 1, — 1, .... 
 
 Does it approach a limit ? 
 
 Ex. 101. What is the limit of — as n assumes the values 1, 2, 3, 
 
 3n ' ' ' 
 
 4...? 
 
254 PLANE GEOMETRY — BOOK V 
 
 Proposition XII. Theorem 
 
 414. Tlie circumferences of two circles have the same ratio as 
 their radii or their diameters. 
 
 Hypothesis. Ci and 64 are the circumferences of two circles 
 whose radii are r^ and r^ and whose diameters are d^ and c?2 
 respectively. 
 
 Conclusion. O, ^ n ^ d^ 
 
 Proof. 1. Inscribe in the circles regular polygons having 
 the same number of sides. Let p^ and p2 be the perimeters 
 of the polygons inscribed in the circles whose radii are 7\ and 
 rg respectively. 
 
 2. The polygons are similar. § 385 
 
 3. Pi = h. §386 
 
 P2 ^2 
 
 4. .-. p^ X r2 = P2 X n. Why ? 
 
 5. Let the number of sides of each polygon be successively 
 doubled, the two polygons continuing to have the same num- 
 ber of sides. 
 
 6. Then pi x 7*2 will approach the limit 0^ X ?*2 
 
 and P2 X Ti will approach the limit C^ X r^. 
 (§411, §403, (a)) 
 
 7. .-. Ci X ^2 = (72 X ?'i. § 403, (6) 
 
 8. •••^=-- §252 
 
 62 ^2 
 
 9. .-. ^l^^Ti^di^ ^^^^ 
 
 O2 ^ ^2 ^2 
 
VARIABLES AND LIMITS 
 
 255 
 
 415. Cor. 
 
 Since ^^ = ^,then ^=^^ 
 
 di d^ 
 
 That is, the ratio of the circumference of a circle to the length 
 of the diameter of the circle is constant for all circles. 
 Note. —This proves the fact assumed in § 389. 
 
 We recall that the constant value — is denoted by tt. 
 
 d 
 
 416. Area of a Circle. The formal treatment of this topic 
 is exactly like that for the length of a circle. 
 
 Consider the sequences of regular inscribed and circum- 
 scribed polygons in a circle of radius r, having 4, 8, 16, etc. 
 sides. (§ 407.) 
 
 A\ ^ - ^ 
 
 E 
 \B 
 
 L 
 
 Let k denote the variable area of the inscribed polygon, and 
 K the variable area of the circumscribed polygon as the num- 
 ber of sides increases. 
 
 The following theorems can then be proved : 
 
 (a) The area k approaches a limit as the number of sides in- 
 creases indefinitely. Call this limit i^. 
 
 (b) The area K approaches a limit as the number of sides 
 increases indefinitely. Call this limit I^. 
 
 (c) The limit ii = the limit I^; that is, the areas of the regu- 
 lar inscribed and circumscribed polygons approach the same 
 limit as the number of sides increases indefinitely. Call this 
 area S. 
 
 (d) If any other sequence of regular inscribed or circum- 
 scribed polygons of the same circle be formed, the areas of the 
 inscribed and of the circumscribed polygons approach the 
 limit S obtained in the theorem (c). 
 
256 PLANE GEOMETRY — BOOK V 
 
 417. The Area of a Circle is the limit of the area of a 
 regular inscribed polygon as the number of sides increases 
 indefinitely. 
 
 418. Tlie area of any regular circumscribed polygon ap- 
 proaches the area of the circle as limit as the number of sides is 
 increased indefinitely. This follows at once from theorem (d) 
 of § 416. 
 
 Proposition XIII. Theorem 
 
 419. The area of a circle is the product of one half its radius 
 and its circumference. 
 
 Hypothesis, r is the radius, C is the circumference, and S 
 is the area of the circle. 
 
 Conclusion. zS = | r x C. 
 
 Proof. 1. Circumscribe about the circle any regular polygon. 
 Let P denote its perimeter and ^its area. 
 
 2. Then, since its apothem is r, 
 
 /f=irxP. Why? 
 
 3. Let the number of sides of the circumscribed polygon be 
 indefinitely increased. 
 
 Then /^will approach the limit 8\ § 418 
 
 P will approach the limit (7; § 411 
 
 i rP will approach the limit | rO. § 403, (a) 
 
 4. .:S = ^r-C. §403,(5) 
 
 420. The Value of tt. In § 389, the approximate value 
 3.1416 for TT was given. This value was derived from a table 
 of values of perimeters of regular inscribed and circumscribed 
 polygons. The following proposition provides a means of 
 computing such tables of perimeters. 
 
LIMITS AND VARIABLES 257 
 
 Proposition XIV. Problem 
 
 421. Given p^ and P„, the perimeters of the regular inscribed 
 and of the regular circumscribed polygons having n sides ; find 
 ;)2„ and Po^j ^^^^ perimeters of the regular inscribed and the 
 regular circumscribed polygons having double the number of 
 sides. 
 
 a! M F V p' 
 
 A. 
 
 yB 
 
 O 
 (a) To find P^^. . 
 Solution. 1. Let ^B be one side of the regular inscribed 
 polygon having n sides and F the. mid-point of arc AB, 
 
 2. Draw the tangent to the circle at F, meeting OA and OB 
 extended at A' and B^ respectively. 
 
 Then A'B^ is one side of the regular circumscribed polygon 
 having n sides. Also A'F= ^ A'B', and hence 2nA'F= P„. 
 
 3. Let tangents to the circle at A and B meet A'B' at 3f 
 and ^respectively. Then MNis one side of the regular cir- 
 cumscribed polygon having 2 n sides; MF=^MN2ind hence 
 
 §270 
 
 nMF=P^,. 
 
 
 4. 
 
 A'M OA' 
 MF OF 
 
 
 (Since Oilf bisects Z ^' OF. ) 
 
 5. 
 
 But ^-^■ 
 
 6. 
 
 . P,_A'M 
 
 rr 
 
 P,+p, A'M+MF 
 
 
 P., MF 
 
 8. 
 
 P.. + P. A'F 
 
 
 2K MF 
 
 §386 
 Why? 
 Why? 
 Why? 
 
258 PLANE GEOMETRY — BOOK V 
 
 A' M F N B' 
 
 A'\ 
 
 Pn+Pn_^nA'F_2P^ 
 
 or 
 
 Pn 
 
 4:nMF 
 
 I Multiplying num. and denom. of ^^ by 4 w."| 
 
 10. 
 
 .•.i^2„(P.+K)=2PA. 
 
 11. 
 
 
 1. 
 
 2. 
 
 (b) To find p2,.. 
 A ABF and A AFM are isosceles triangles 
 ZABF:=ZAFM. 
 
 3. 
 
 .-. A ABF ^ A AFM. 
 
 4. 
 
 AF:AB = MF:AF. 
 
 5. 
 
 .'. AF' = AB X MF. 
 
 6. 
 
 But AF- ^'- ; AB-^'^; and MF = ^^^ . 
 2n n An 
 
 . 
 
 [Steps 2 and 3, part (a)] 
 
 P2^V_ 
 jP2„ 
 
 4/^2 
 
 P_n 
 
 n 
 4tn^ 
 
 
 ••• P2„ = yp„P2, 
 
 Why? 
 
 Algebra 
 
 Why? 
 Why? 
 Why? 
 Why? 
 Why? 
 
 Ax. 2, § 51 
 
 Algebra 
 Algebra 
 
 Note. — The formulae P2„ = ^^ and ^2n = y/pnPin are quite re- 
 
 Pn + Pn 
 
 markable. If the perimeters of the regular inscribed and regular circum- 
 scribed polygons of say 4 sides are known, then the perimeters of the 
 regular circumscribed and regular inscribed polygons of 8 sides can be 
 computed by mere substitution ; then those of 16 sides ; and so on. 
 
VARIABLES AND LIMITS 
 
 259 
 
 Proposition XV. Problem 
 422. Compute an approximate value o/tt. 
 
 Solution. 1. If the diameter of a circle is 1, the side of an 
 inscribed square is ^ V2, and hence the perimeter of the square 
 is 2 V2, or p4 = 2.82843. 
 
 2. The side of a circumscribed square is 1, and P^ = 4. 
 
 q p_ 2P 4 Xi?4 . 
 tJ. /-g— -^- 
 
 „ 2 X 4 X 2.82843 
 
 §421 
 
 Hence 
 
 4. 
 
 Hence 
 
 4 + 2.82843 
 
 3.31371. 
 
 P% = Vi)4 X Pg- 
 
 §421 
 
 Ps = V2.82843 X 3.31371 = 3.06147. 
 
 5.- Similarly P^^ 
 
 2P,xps _ 2 X 3.31371 x 3.06147 
 
 Pg + ^8 3.31371 + 3.06147 
 
 = 3.18260. 
 And p,8 = Vj98 X P16 = V3.06147 x 3.18260 = 3.12145. 
 
 6. In this manner, we compute the following table ; 
 
 No. OP 
 
 Perimeter of 
 
 Perimeter of 
 
 Sides 
 
 Reg. Giro. Polygon 
 
 Reg. Inso. Polygon 
 
 4 
 
 4. 
 
 2.82843 
 
 8 
 
 3.31371 
 
 3.06147 
 
 16 
 
 3.18260 
 
 3.12145 
 
 82 
 
 , 3.15172 
 
 3.13655 
 
 64 
 
 3.14412 
 
 3.14033 
 
 128 
 
 3.14222 
 
 3.14128 
 
 256 
 
 3.14176 
 
 3.14151 
 
 512 . 
 
 3.14163 
 
 3.14157 
 
 7. The last results show that the circumference of the 
 circle whose diameter is 1 > 3.14157 and < 3.14163. 
 
 Hence an approximate value of tt is 3.1416, correct to the 
 fourth decimal place. 
 
260 PLANE GEOMETRY — BOOK V 
 
 THE INCOMMENSURABLE CASES* 
 
 Proposition XVI. Theorem 
 
 423. In the same circle or in equal circles^ central angles have 
 the same ratio as their intercepted arcs. (When the angles are 
 incomm ensurable.) 
 
 Hypothesis. In O ABC, Z AOB and Z BOC are two in- 
 commensurable central angles intercepting the arcs AB and 
 BC respectively. 
 
 Z BOC arc BC 
 
 Conclusion. 
 
 ZAOB arc ^5 
 
 Proof. 1. Divide Z AOB into two equal parts and let one 
 of these be applied as unit of measure to Z BOC. 
 
 2. Since Z AOB and Z BOC are incommensurable, a certain 
 number of angles equal to J- Z AOB will equal Z BOX^, leav- 
 ing a remainder Z X^OC which is less than the unit of 
 measure. 
 
 3. Z AOB and Z BOX^ are commensurable. 
 
 ^ Z BOX^ ^ arc BX , 
 
 Z AOB arc AB 
 
 *Note. — There are three incommensurable cases. (§423-§425.) 
 These propositions complete the proofs of the theorems given in § 213, 
 § 261, and § 327 respectively. If it is desired to read § 423 when studying 
 § 213, then it will be necessary to read also § 401 to § 403 inclusive, which 
 give an introduction to the theory of limits. 
 
VARIABLES AND LIMITS 261 
 
 5. Take now as unit of measure \ Z AOB. This measure 
 will be contained an integral number of times in Z AOB and 
 also in /. BOX^^ further, the unit of measure may be con- 
 tained once in Z XiOC, leaving a remainder Z X^OC which is 
 less than the new unit of measure. 
 
 Again Z^OX, ^arc^X,. 
 
 6. Continue in this manner to decrease indefinitely the unit 
 of measure. The remainder Z XOC, being always less than 
 the unit of measure, will approach the limit 0. 
 
 Using the symbol = to express "approaches the limit,'' 
 
 Z BOX = Z BOC, and hence yTS^ZJ^' ^ ^^^' ^^^ 
 
 arc BX = arc BC, and hence ^I^_BX^^c^C ^ ^^^ , . 
 
 arc^S arc^B ' ^ ^ 
 
 7 ZBOX . SLTcBX . ,, , . , , 
 
 '• -^ — 7w^ ^-^d -— are variables which are always 
 
 ZAOB arc AB ^ 
 
 equal. 
 
 o . Z BOC are BC ,, .^o /i.x 
 
262 PLANE GEOMETRY — BOOK V 
 
 Proposition XVII. Theorem 
 
 424. A parallel to one side of a triangle divides the other two 
 sides proportionally, when the segments of one side are incom- 
 mensurable. 
 
 A 
 
 Hypothesis. In A ABC, segments AD and BD are incom- 
 mensurable ; DE II BC, meeting AC at E. 
 
 Conclusion. ^^9^. 
 
 AD AE 
 
 Proof. 1. Divide AD into any number of equal parts (say- 
 two), and apply one of these parts to BD as unit of measure. 
 
 2. Since AD and BD are incommensurable, a certain num- 
 ber of segments equal to the unit of measure will extend from 
 D to Xj, leaving a remainder XiB which is less than the unit 
 of measure. 
 
 3. Draw X^Y^ II BC, meeting AC at Y^. 
 
 Then :5^ = ^S. §261 
 
 AD AE 
 
 [Since AD and DX^^ are commensurable.] 
 
 4. Take now as unit of measure \ AD. This measure will 
 be contained an integral number of times in AD and also in 
 i>Xi ; further, the unit of measure may be contained once in 
 XiB, leaving a remainder X.^B which is less than the new 
 unit of measure. | Draw X2Y2 11 BC, meeting ^Oat Y^. 
 
 Then DX^^^^Ii. 
 
 AD AE 
 
 5. Continue in this manner to decrease the unit of measure 
 indefinitely. The remainder XB, being always less than the 
 unit of measure, will also approach as limit. 
 
VARIABLES AND LIMITS 
 
 Then DX = BB, amd hence ^ = ^ 
 
 AD AD 
 
 Also EY= EC, and hence 
 6. 
 
 7. 
 
 EY^EC 
 AE AE 
 
 263 
 
 § 403, (a) 
 § 403, (a) 
 
 and are variables which are always equal. 
 
 AD AE . ^ ^ 
 
 DB^EG 
 AD AE 
 
 403, (5) 
 
 Proposition XVIII. Theorem 
 
 425. Two rectangles having equal altitudes are to each other as 
 their bases, when the bases are incommensurable. 
 
 Hypothesis. Rectangles ABCD and EFGH have equal al- 
 titudes AB and EF, and incommensurable bases BC and FG. 
 
 Conclusion. 
 
 EFGH^FG 
 ABCD BC' 
 
 Proof. 1. Divide BC into any number of equal parts (say 
 two), and apply one of these parts to FG as unit of measure. 
 
 2. Since BC and FG are incommensurable, a certain num- 
 ber of segments equal to the unit of measure will extend from 
 Fto Xi, leaving a remainder XiG which is less than the unit 
 of measure. 
 
 3. Draw X^Yi±FG, meeting EH at Y^. Then rectangles 
 EFXi Yi and ABCD have equal altitudes and commensurable 
 bases. 
 
 , . EFX, Y, ^ FXi 
 
 ' " ABCD BC' 
 Complete the proof. 
 Suggestion. — Model the proof after that for § 424. 
 
264 
 
 PLANE GEOMETRY — BOOK V 
 
 Group C. Symmetry in Plane Figures 
 
 426. Two points are symmetrical with respect to a third 
 point, called the Center of Symmetry, when the latter bisects 
 the segment which joins them. 
 
 Thus, if is the mid-point of segment AB, points A and B are sym- 
 metrical with respect to as center. A O B 
 
 427. Two points are symmetrical with 
 respect to a straight line, called the Axis of 
 Symmetry, when the latter bisects at right 
 angles the segment which joins them. c- 
 
 B 
 
 Thus, if CD bisects segment AB at right angles, 
 points A and B are symmetrical with respect to 
 CD as an axis. 
 
 428. A figure is symmetrical with 
 respect to a center when every straight 
 line drawn through the center cuts the 
 figure in two points which are symmetri- 
 cal with respect to that center. 
 
 429. A figure is symmetrical with respect to 
 
 an axis when every straight line perpendicular /'^^V j 7/^ 
 
 to the axis cuts the figure in two points which ^- Mui//^ ^ 
 are symmetrical with respect to that axis. \fF^^ 
 
 Ex. 102. Does a circle have a center of symmetry ? 
 
 Does it have an axis of symmetry ? 
 
 Does it have more than one axis of symmetry ? 
 
 Ex. 103. (a) Locate upon a sheet of paper a point and four other 
 points X, r, Z, and W. 
 
 (b) Construct the points X', Y', Z', and W, which are symmetrical 
 respectively to X, Y, Z, and W, with respect to as center. 
 
 Ex. 104. (a) Draw any straight line AB of indefinite length and 
 upon one side of it locate at random points X, Y, and Z. 
 
 (b) Construct the points X, Y', and Z', which are respectively sym- 
 metrical to X, Y, and Z, with respect to AB as axis. 
 
SYMMETRY 265 
 
 430. Theorem. Tioo segments wliicli are symmetrical with 
 respect to a center are equal and parallel. 
 
 .." "O" 
 
 Hypothesis. Segments AB and A'B' are symmetrical with 
 respect to center 0. 
 
 Conclusion. AB and A'B' are equal and parallel. 
 
 Proof. 1. Draw lines AA' and BB' intersecting at 0; draw 
 AB' and A'B. § 428 
 
 2. bisects AA' and BB'. Prove it. 
 
 3. .-. AB'A'B is a O. Prove it. 
 
 4. .'. ^5 and A'B' are equal and parallel. 
 
 Ex. 105. (a) Draw a figure something like the adjoining 
 one. 
 
 (Let ^B and FG be perpendicular to AG, and let BCDEF 
 be a curved line. ) 
 
 (6) Construct the figure symmetrical to ABCDEFG with e\ 
 respect to u4(t as axis. 
 
 F 
 
 Ex. 106. Prove that two segments which are equal and parallel are 
 symmetrical with respect to a center. 
 
 Ex. 107. Prove that the bisector of the vertical angle of an isosceles 
 triangle is an axis of symmetry of the triangle. 
 
 Ex. 108. How many axes of symmetry does an equilateral triangle 
 have ? 
 
 Ex. 109. Prove that the intersection of the diagonals of a parallelo- 
 gram is the center of symmetry of the parallelogram. 
 
 Ex. 110. Does a rhombus have a center of symmetry ? 
 Ex. 111. Does the rhombus have an axis of symmetry ? 
 Ex. 112. Does a rectangle have an axis of symmetry ? 
 Does it have a second axis of symmetry ? 
 Does it have a center of symmetry ? 
 
266 
 
 PLANE GEOMETRY — BOOK V 
 
 431. Theorem. If a figure is symmetrical with respect to 
 each of two perpendicular axes, it is symmetrical with respect to 
 their intersection as center. 
 
 
 
 c 
 
 
 A 
 
 / 
 
 /--, 
 
 \p 
 
 It 
 D 
 
 H 
 
 > 
 
 Q / 
 
 < 
 
 
 
 B 
 
 
 o 
 
 F 
 
 
 Hypothesis. Figure AE is symmetrical with respect to 
 axes XX' and YY\ XX' is perpendicular to YY' at 0. 
 
 Conclusion. AE is symmetrical with respect to as center. 
 
 Proof. 1. From P' any point of AE draw P'Q A. to XX' at 
 Q, and meeting AE again at P. Then PQ= P'Q. 
 
 2. From P draw PR± YY' meeting YY' at R and meet- 
 ing AE again at P". Then PR = RP". 
 
 3. Draw P'P". 
 
 4. YY' II PP' and bisects PP". 
 
 .-. YY' passes through the mid-point of P'P". Why ? 
 
 5. Similarly, XX' passes through the mid-point of P'P". 
 
 6. Hence O, the intersection of XX' and YY', must be 
 the mid-point of P'P". 
 
 7. In the same manner, for any other point like P' of AE 
 there is a corresponding point P" of AE, such that P'P" passes 
 through and is bisected by it. 
 
 8. .-. AE is symmetrical with respect to as center. 
 
 Ex. 113. 
 
 Ex. 114. 
 
 in Ex. 112. 
 
 Ex. 115. 
 
 Ex. 112. 
 
 Ex. 116. 
 
 Answer for a square the questions proposed in Ex. 112. 
 Answer for an isosceles trapezoid the questions proposed 
 
 Answer for a regular hexagon the questions proposed in 
 
 Answer for a regular pentagon the same questions. 
 
MAXIMA AND MINIMA 
 
 267 
 
 Group D. Maxima and Minima of Figures 
 
 432. Two figures are Isoperimetric when they have equal 
 perimeters. 
 
 433. If one geometric magnitude of a number which satisfy 
 certain given conditions has a value greater than that of any 
 of the others, it is called the Maximum ; if it has a value less 
 than that of any of the others, it is called the Minimum. 
 
 Thus, of all segments drawn from a given point to a given line the 
 perpendicular is the minimum ; again, of all chords of a circle^ the diame- 
 ter is the maximum. 
 
 434. Theorem. Of all triangles ivith two given sides, that in 
 which these sides are perpendicular is the maximum. 
 
 Hypothesis. In A ABC and A A'BC; 
 
 AB = A'B ; and AB ± BC. 
 
 Conclusion. Area of A ABC > area of A A'BC. 
 Proof. 1. Draw A'D ± BG, 
 
 2. A'B > A'D. 
 
 3. .-. AB > A'D. 
 
 4. .-. AB'BC> A'D ' BC 
 [Multiplying both members by BG.'\ 
 
 6. But area A ABC= \ AB • BC, 
 
 and area A A'BC =\A'D.BG. 
 6. .-. area A ABC > area A'BC 
 
 Why? 
 Why? 
 
 Ex. 117. Of all parallelograms having two given adjacent sides, that 
 is the maximum in which these sides include a right angle. 
 
268 
 
 PLANE GEOMETRY — BOOK V 
 
 435. Theorem. Of isoperimetric triangles having the same 
 base J that which is isosceles is the maximum. 
 
 Ay 
 
 --j:: 
 
 r 
 
 Hypothesis. A ABC and A A!BQ are isoperimetric and 
 have the common base BQ\ A ABQ is isosceles. 
 
 Conclusion. Area of A ABQ > area of AA'BC. 
 
 Analysis. We must prove the altitude AD > altitude A'D/ 
 
 Proof. 1. Extend BA to E, making AE = BA. Draw EO. 
 
 2. Z BGE is a rt. Z, for it can be inscribed in a semicircle 
 whose center is A and whose radius is AB. 
 
 3. Extend J5;(7 to X. Btslw A'E' = A'C] dvsiW BE\ Con- 
 struct A'F' and AF both perpendicular to EE'. 
 
 4. BA' + A'E' = BA' -f A'C. 
 
 5. BA + AE = BA-{- AC. 
 
 6. ' But BA'+ A'C = BA + AC. 
 
 7. .-. BA' + A'E' ==BA + AE. 
 
 8. .-. BA' + A'E' = SE-. 
 
 9. But BA' + ^'^' > jB^'. 
 
 10. .-. BE > J5^'. 
 
 11. «•. (7^ > (7J5;'. 
 
 [Since BC±EE', and 5^ > BE'.^ 
 
 12. .'.CF>CF'. 
 
 [Since Oi^= i 0^, and C2^' = i CE'.^ 
 
 13. .% ^D > A'D'. 
 
 (Prove it.) 
 
 14. .-. area of A ABC > area of A A'BC. 
 
 Why? 
 
 Why? 
 
 Hyp. 
 
 Why? 
 Why? 
 Why? 
 Why? 
 
 Why? 
 
MAXIMA AND MINIMA . 269 
 
 436. Theorem. Of isoperimetric polygons having the same 
 number of sideSy the maximum is equilateral. 
 
 Hypothesis. ABCDE is the maximum of polygons having 
 the given perimeter and the same number of sides as ABCDE. 
 
 Conclusion. ABCDE is equilateral. 
 
 Proof. 1. Assume that AB and BC are unequal. Draw 
 AC. 
 
 2. Let A AB'C be the isosceles triangle on base AC having 
 its perimeter equal to that of A ABC. 
 
 3. Then the area of A AB'C > area of A ABC. § 435 
 
 4. Then the area of AB'CDE > area of ABCDE. 
 
 5. But this is impossible for ABCDE is the maximum of all 
 polygons having the same perimeter and the same number of 
 sides as ABCDE. 
 
 6. .-. AB and BC cannot be unequal. 
 
 7. Similarly BC = CD ^ DE, etc. 
 
 8. .-. ABCDE is equilateral. 
 
 437. Cor. Of all isoperimetric triangles, the maximum is 
 equilateral. 
 
 Ex. 118. A parallelogram and a rhombus each have a perimeter of 
 40 in. Which has the greater area ? 
 
 Ex. 119. A man is planning for himself a house. He has a rectan- 
 gular plan, the dimensions of which are 30 ft. and 20 ft., making the 
 perimeter of the base 100 ft. 
 
 Will such a house cover a greater or a less number of square feet than 
 a square house whose perimeter also is 100 ft. ? 
 
270 . 
 
 PLANE GEOMETRY — BOOK V 
 
 438. Theorem. Of isoperimetric equilateral polygons of the 
 same number of sides, the maximum is equiangular. 
 
 H 
 
 .'^D 
 
 Fig. 1 
 
 Fig. 2 
 
 Fig. 3 
 
 Hypothesis. Consider an equilateral polygon of which AB, 
 BC, and CD are any three consecutive sides, and whose re- 
 maining part is denoted by P. Assume that this polygon is 
 the maximum, of all equilateral polygons isoperimetric with 
 the given polygon, equal in area to it, and having the same 
 number of sides as it. 
 
 Conclusion. This polygon is equiangular. 
 Plan. We shall assume that Z ABC > Z BCD. We shall 
 consider the three cases : Case I. AB II CD. Case II. AB 
 meets CD at H. Case III. AB meets CD at K. 
 
 Proof. Case I. (Fig. 1.) 1. Let E be the mid-point of BC. 
 Draw EF meeting AB prolonged at F, making EF = BE. 
 Then EF extended will meet CD at G. 
 
 2. Then A BEF ^ A ECG. Prove it. 
 
 .-. BF = CG, and EF = EG. Why ? 
 
 .-. EG = BC Why ? 
 
 .-. AB-\-BF+FG-\-GD=AB-\-BC+CG-\-GD. Why? 
 Hence the polygon composed of AFGD and P has the 
 same perimeter as the given polygon, composed of ABCD 
 and P. 
 
 7. Also AFGD and ABCD have the same area. Step 2 
 
MAXIMA AND MINIMA 271 
 
 8. Hence the polygon composed of AFOD and P has the 
 same area as the given polygon, composed of ABCD and P. 
 
 9. But the given polygon was the maximum of polygons 
 having the given number of sides. Hence the polygon com- 
 posed of AFGD and P is equal to the maximum of polygons 
 having that number of sides. 
 
 10. Hence the polygon composed of AFOD and P is 
 equilateral. § 436 
 
 11. But this is impossible since AF > DO. 
 • 12. Hence Z ABC cannot be > Z BCD. 
 
 Case II. (Fig. 2.) Assume that AB meets CD at H, 
 
 1. Let HE bisect Z BHC, meeting BC at E. 
 
 2. Revolve A BCH on HE as axis until it takes the position 
 oiAFGH. 
 
 3. Then FG=BC\ BF= CO ; and A BEF=A CEO. Why ? 
 
 4. .'. AB+BF-\-FG-\-GD=AB-{-BC-^CG-hGD. Why? 
 Complete the proof as in steps 6 to 12 inclusive, of Case I. 
 
 Case III. (Fig. 3.) Assume that AB and CD meet at K. 
 
 1. Let KE bisect Z BKC, meeting BC at E. 
 
 2. Revolve A BCK on KE as axis until it takes the position 
 of A FOK 
 
 3. ThenFO^BC; BF=CO', 2indABEF=ACEO. Why? 
 
 4. .: AB-^BF-^FG-\-GD=AB + BC-i-CO+GD. Why? 
 Complete the proof as in steps 6 to 12 inclusive, of Case I. 
 
 It follows from Cases I, II, and III that Z ABC cannot be 
 > Z BCD. 
 
 In the same manner, it can be proved Z ABC cannot be 
 < Z BCD. 
 
 Hence Z ABC = Z BCD. 
 
 Since these are any two consecutive angles of the given 
 polygon, then the given polygon must be equiangular. 
 
 439. Cor. Of isoperimetric polygons having the same num- 
 ber of sides, the maximum is regular. (§ 436 and § 437) 
 
272 
 
 PLANE GEOMETRY — BOOK V 
 
 440. Theorem. Of two isoperimetric regular polygons, that 
 which has the greater number of sides has the greater area. 
 
 Hypothesis. ABQ is an equilateral triangle, and M is an 
 isoperimetric square. 
 
 Conclusion. Area of Jf > area of A ABC. 
 
 Proof. 1. Let D be any point in side AB of A ABC. 
 
 2. Draw DO, and construct upon it as base isosceles A CDE 
 isoperimetric with A BCD. 
 
 3. Area of A CDE > area of A BCD. 
 
 4. .-. area of ADEC > area of A ABC 
 
 5. But ADEC and square M are isoperimetric, and hence 
 area of Jf > area of ADEC § 439 
 
 6. .-. area of M> area of A ABC 
 
 In like manner it can be proved that the area of a regular 
 pentagon is greater than that of an isoperimetric square ; etc. 
 
 441. Cor. The area of a circle is greater than the area of 
 any polygon having an equal perimeter. 
 
SUPPLEMENTARY EXERCISES 273 
 
 SUPPLEMENTARY EXERCISES 
 BOOK I 
 
 Ex. 1. Two quadrilaterals are congruent if three sides and the two 
 included angles of one are equal respectively to three sides and the two 
 included angles of the other. 
 
 Suggestion. — Prove by superposition. 
 
 Ex. 2. Two quadrilaterals are congruent if three angles and the two 
 included sides of one are equal respectively to three angles and the two 
 included sides of the other. 
 
 Ex, 3. Prove that the base angles of an isosceles triangle are equal, 
 using the following construction. 
 
 Hypothesis. AB = AC. 
 
 Conclusion. ZABC = ZA CB. 
 
 Construction. Extend AB to D. Extend AC to 
 E, making CE = BD. Draw DC and BE. 
 
 Plan. 1. Prove A ADC ^ A ABE in order to 
 prove DC= BE. 
 
 2. Prove ADBC^ABCE. 
 
 3. Prove ZDBC = ZBCE. 
 
 4. Prove ZABC=ZACB. 
 
 Ex. 4. If AB and ^C are two equal chords of the circle whose center 
 is O, then the radius OA bisects ZBAC. 
 
 Ex. 5. Books for carpenters give the follow- 
 ing method of bisecting an angle by means of 
 the " square " alone. 
 
 Make OD and OC of equal length. Place 
 the square so that DP= CP. Then OP bisects 
 ZAOB. 
 
 Prove that the method is correct. 
 
 Ex. 6. Prove that the bisectors of homologous angles of congruent 
 triangles are equal. 
 
 Suggestions. — 1. Recall § 66. 
 
 2. Remember that the homologous sides and angles of two congruent 
 triangles are equal. 
 
 Ex. 7. Construct the angle which is double ZJ5 of Ex. 61, Book I. 
 
 Ex. 8. Construct the angle which is the sum of Z ^ and Z B of Ex. 61. 
 
 Ex. 9. Construct an isosceles triangle having its equal sides 3 in. in 
 length and the angle included by them equal to ZB given in Ex. 61, 
 Book I. 
 
274 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 10. If a diagonal of a quadrilateral ABCD bi- 
 sects two of its angles, it is perpendicular to the other 
 diagonal and bisects it. 
 
 Suggestions. — 1. Let AC bisect Z.A and ^C; prove 
 AC 1 BD and AC bisects BD. 
 
 2. Try to prove AB = AD and BC=DC. 
 
 Ex. 11. In the adjoining figure, if J.0, BO, 
 
 and CO are extended to Z, F, and X respectively, 
 
 so that AO=OZ, BO=OY, and C0= OX, then 
 
 CiABC^/\ZYX. 
 
 (First prove AB = ZY, BC = XY, and AC=XZ.) 
 
 After proving /\ABC '^ A XYZ, vi^hat angle does /.BCA equal ? 
 
 Ex. 12. Prove that homologous medians of congruent triangles are 
 equal. 
 
 Suggestion. — Read the suggestions for Ex. 6, p. 273. 
 
 Ex. 13. If two triangles have two sides and the median to one of 
 them equal respectively to two sides and the corresponding median of the 
 other, the triangles are congruent. 
 
 Suggestion. — Read the note following § 77. 
 
 Ex. 14. Construct the perpendicular-bisector of a segment taken 
 along the lower edge of the paper. 
 
 Ex. 15. Draw any angle and construct its bisector. Through its 
 vertex construct a line perpendicular to the bisector. Prove that this last 
 line makes equal angles with the sides of the given angle. 
 
 Ex. 16. Construct a line through a given point 
 within a given acute angle, which will form with the 
 sides of the angle an isosceles triangle. 
 
 Suggestion. — If A ABC represents the desired triangle, 
 and AH bisects Z CAB, then CB 1 AH. Try now to work 
 toward this figure if only I. FAG and point D within it 
 are given. 
 
 Ex. 17. Prove Cor. 1 (§ 96) if Z 3 = Z 7. 
 
 Ex. 18. Prove Cor. 3 (§98) ifZ3 + Z5 = 
 1 St. Z. 
 
 Ex. 19. Prove that AB \\ CD (Fig. § 98) if 
 Z4 + Z7 = lst. Z. 
 
 Ex. 20. If ^5 = CZ) and Z 1 = Z 2 in the ad- 
 joining figure, prove ABW CD and also AGW BD. 
 
 Suggestion. — UecSiW § 95. 
 
SUPPLEMENTARY EXERCISES 
 
 275 
 
 
 
 
 
 tread -S 
 
 A<^^ 
 
 y^ 
 
 tread 
 
 < 
 
 ^ 
 
 1 
 
 
 Ex. 21. In building stairs two or more 
 stringers are required. To make a stringer 
 having a 9-in. tread and a 6-in. riser, a car- 
 penter uses his square as in the figure below. 
 For each step he places his square so that the 
 9-in. mark and the 0-in. mark fall along the 
 edge of the board from which he is cutting 
 the stringer. 
 
 Will the treads all be parallel ? Why ? 
 
 Will the risers all be parallel ? Why ? 
 
 Will each riser be perpendicular to its tread ? 
 Why? 
 
 Ex. 22. One triangle used by draughtsmen has an angle of 90° and 
 an angle of GO"". Why should it be called a " 60-30 " triangle ? 
 
 Ex. 23. The other triangle used by draughtsmen has an angle of 90=^ 
 and an angle of 45°. How large is the remaining angle ? 
 
 Ex. 24. In a A ABC, it ZA = 90°, and Z B = Z C, how large are 
 ZB and ZC? 
 
 Ex. 25. Find the three angles of a triangle if the second is four times 
 the first, and the third is seven times the first. (Algebraic solution.) 
 
 Ex. 26. Find the three angles of a triangle if the second exceeds the 
 first by 40°, and the third exceeds the second by 40°. 
 
 Ex. 27. The vertical angle of an isosceles triangle is n degrees. Ex- 
 press each of the base angles. 
 
 Ex. 28. One base angle of an isosceles triangle is n degrees. Ex- 
 press each of the other angles of the triangle. 
 
 Ex. 29. Determine by construction the angle C of a A ABC if ZA 
 and Z B are the angles given in Ex. 61, Book I. 
 
 Ex. 30. Prove that two isosceles triangles are congruent when the 
 vertical angle and the base of one are equal respectively to the vertical 
 angle and the base of the other. 
 
 Suggestion. — Prove the homologous base angles also are equal. 
 
 Ex. 31. If one acute angle of a right triangle is 35°, how large is the 
 other acute angle ? 
 
 Ex. 32. If perpendiculars be drawn from any point in the base of an 
 isosceles triangle to the equal sides, they make equal angles with the base. 
 
 Suggestion. —The proof i.s based on § 37 and § 109. 
 
 Ex. 33. Prove that the altitude drawn to the hypotenuse of a right 
 triangle divides the right angle into two parts which are equal respectively 
 to the acute angles of the right triangle. 
 
276 SUPPLEMENTARY EXERCISES 
 
 Ex. 34. If two opposite angles of a quadrilateral are equal and if the 
 diagonal joining the other two angles bisects one of them, then it bisects 
 the other also. 
 
 Ex. 35. If two triangles have two angles and the bisector of one of 
 these angles equal respectively to two angles and the corresponding bisector 
 of the other, the triangles are congruent. 
 
 Suggestion. — Recall the note following § 77. 
 
 Ex. 36. If two triangles have two sides and the altitude drawn to 
 one of them equal respectively to two sides and the corresponding altitude 
 of the other, the triangles are congruent. 
 
 Suggestion. — Read the note following § 77. 
 
 Ex. 37. Construct a pattern for the pointed end of 
 
 a belt, assuming that the belt material is 2 in. wide, and ^ '2' kj^ 
 
 that the point is to project 1 in. beyond the square end '^ 
 
 of the belt. 
 
 Ex. 38. Draw any straight line of indefinite length and select two 
 points not in it. Find the point in the line which is equidistant from the 
 two given points. 
 
 Ex. 39. Find a point in one side of a triangle which is equidistant 
 from the other two sides of the triangle. ^^ 
 
 Ex. 40. Prove that either exterior angle at the 
 base of an isosceles triangle is equal to the sum of a 
 right angle and one half the vertical angle. A^ 
 
 E- 
 
 Ex. 41. If from the vertex of one of the equal angles 
 of an isosceles triangle a perpendicular be drawn to the 
 opposite side, it makes with the base an angle equal to 
 one half the vertical angle of the triangle. 
 
 Suggestion. — Construct the bisector of Z C. 
 
 Ex. 42. A ABC is an equilateral triangle. BP, the bisector of Z B, 
 meets ^C at P; CM, the bisector of exterior angle ACB, meets BP 
 extended at M. MNis perpendicular to CB. Prove MN = BP. 
 
 Ex. 43. If the equal sides of an isosceles triangle be extended be- 
 yond the base, the bisectors of the exterior angles so formed form with 
 the base another isosceles triangle. 
 
 Ex. 44. If A ABC and A ABD are two tri- 
 angles on the same base and on the same side of 
 it, such that AC = BD and AD = BC, and AD 
 and BC intersect at 0, then A OAB is isosceles. 
 
SUPPLEMENTARY EXERCISES 
 
 277 
 
 Ex. 45. If CD is the bisector of Z C of A ABC, 
 and DF be drawn parallel to ^O meeting BC 2it E 
 and the bisector of the angle exterior to C at F, 
 prove DE-EF. B- 
 
 Sugrjestion. — Compare BE and EF with EC. .^ 
 
 Ex. 46. If equiangular triangles be constructed upon the sides of any 
 triangle, the lines drawn from their outer vertices to the opposite vertices 
 of the given triangle are equal. 
 
 Suggestion. — Recall § 124. ^D 
 
 Ex. 47. If -4C be drawn from the vertex of the right 
 angle to the hypotenuse of right A BCD so as to make 
 Z ACD = ZZ>, it bisects the hypotenuse. 
 
 Suggestion. — Prove ^B = Z.ACB hy ^ 109 and § 37. 
 
 Ex. 48. If the angle at the vertex of isosceles 
 A ABC is equal to twice the sum of the equal angles 
 B and C, and if CD is perpendicular to BC, meeting 
 BA extended at 2), prove A ACD is equilateral. 
 
 Suggestion. — Determine the number of degrees in each angle of A ABC. 
 
 aC 
 
 Ex. 49. If the bisectors of the equal angles of an 
 isosceles triangle meet the equal sides at D and E re- 
 spectively, prove that DE is parallel to the base of the 
 triangle. a^ :ab 
 
 Suggestions. — 1. Compare Z CED + L CDE with Z ^ + Z ^ (§ 106) . 
 
 2. Is LCED=^CDE? 3. Is ZC£Z> = Z^? 
 
 Ex. 50. Prove that two parallelograms are congruent if two sides 
 and the included angle of one are equal respectively to two sides and the 
 included angle of the other. 
 
 Suggestion. — Prove by superposition. Recall § 132. 
 
 Ex. 51. Prove that the sum of the perpendiculars 
 drawn from any point within an equilateral triangle to 
 the sides of the triangle is equal to the altitude of the 
 triangle. 
 
 Prove OR -\- 0F+ 0D = BG. 
 
 Suggestions. — 1. Let KM be II ^C and KE 1 AB. 
 
 2. Compare EK and BL. 3. Prove 0/J + 0F= EX. 
 
 Ex. 52. An ironing board is supported on each "^ 
 side as shown in the adjoining figure. If ^0 = OB 
 and DO = OC, prove that AC is always parallel to 
 the floor DB. " 
 
 G R 
 
278 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 53. What angle is formed by the bisectors of two consecutive 
 angles of : (a) a rectangle ? (6) an equilateral triangle ? (c) a parallel- 
 ogram ? B^ _,C 
 
 Ex. 54. Prove that the bisectors of the in- 
 terior angles of a parallelogram form a rectangle. 
 
 Ex. 55. Construct a rhombus whose sides are each 3 in. and whose 
 acute angles are each 45°. Draw and measure its diagonals. 
 
 Ex. 56. Construct a rhombus, having given one side and one diagonal. 
 
 Ex. 57. Prove that the two altitudes of a rhombus are equal. 
 
 Ex. 58. If on the diagonal BB of square ABCD a distance BE is 
 taken equal to AB, and if EF is drawn perpendicular to BD meeting AD 
 at F, then AF = EF = ED. 
 
 Suggestion. — What kind of angle is angle EDF? 
 
 Ex. 59. If AD and BD are the bisectors of the exterior angles at the 
 ends of the hypotenuse AB of right triangle ABC, and DE and Z>jPare 
 perpendicular respectively to CA and CB extended, prove CEDE is a 
 square. 
 
 Suggestion. —Ilecall § 143. Prove DE = DF, using § 120, I. 
 
 Ex. 60. Prove that the bisectors of the angles of ^ 
 
 a rectangle form a square. 
 
 Suggestions. — 1. Make a plan based upon § 143. 
 
 2. To prove EF = EH, prove AF=BH and AE = ^^ 
 
 BE. 
 
 Ex. 61. If the non -parallel sides of an isosceles trapezoid are extended 
 until they meet, they form with the base an isosceles triangle. 
 
 Ex. 62. If the line joining the mid-points of the bases of a trapezoid is 
 perpendicular to the bases, the trapezoid is isosceles. 
 
 Ex. 63. If the bisectors of the interior angles of a 
 trapezoid do not meet at a point, they form a quad- 
 rilateral, two of whose angles are right angles. 
 
 Suggestion.— Frove Z.FEH and ^FGH are right A' 
 angles. 
 
 Ex. 64. If D is the mid-point of side ^C of isosceles 
 A ABC, and DE is perpendicular to base BC, then EC 
 is \ BC. 
 
 Suggestion. — Draw DF parallel to AB. 
 
SUPPLEMENTARY EXERCISES 279 
 
 Ex. 65. A BCD is a trapezoid whose parallel 
 sides AD and BC are perpendicular to CD. If Eis 
 the inid-point of AB, prove EC = ED. 
 
 Suggestion. — Br&w EF parallel to AD. Recall A 
 § 149. 
 
 Ex. 66. The following method of dividing a segment into equal seg- 
 ments may be used. 
 
 To divide AB into five equal parts. "H,,—^''^ 
 
 1. Draw ^ C making with ^5 any con- X-^'X''' \ \ 
 venient angle. a .--''^" \ \ \ ,}-^ 
 
 2. Draw BD parallel to AC. \ \ \ ^,,^—'''0" 
 8. Lay off on 40 five equal segments, \ ^ "■-){"" ^ 
 
 and on BD five other segments of the same ^ 
 length. 
 
 4. Connect the points of division as in the figure. 
 
 Prove now that AB is divided into five equal segments. 
 
 Ex. 67. If the base of an isosceles triangle be trisected, the lines 
 joining the points of trisection to the vertex of the triangle are 
 equal. 
 
 Ex. 68. Prove that the line which joins the mid-points of two sides of 
 a triangle bisects any segment drawn to the third side from the opposite 
 vertex. 
 
 Ex. 69. If E and F are the mid-points of BC and AD respectively 
 of parallelogram ABCD, prove that AE and CF 
 trisect BD. 
 
 Suggestion. — Frove AE || FC, by proving AECF / ^^ 
 
 is a parallelogram. Then prove that AE bisects BH a 
 aud CF bisects GB. 
 
 Ex. 70. If E and i?'are the mid-points of sides AB and AC respec- 
 tively of t^ABC, and AD is the perpendicular from A\.o BC^ prove 
 /.EDF^AEAF. 
 
 Suggestion. —Recall Ex. 175, Book I. 
 
 Ex. 71. If ^and G are the mid-points of AB 
 and CD respectively of quadrilateral ABCD, and 
 /rand L are the mid-points of diagonals AC and 
 BD respectively, prove that EKGL is a parallelo- 
 gram. 
 
 Suggestion. — ^cdkW § 161. 
 
280 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 72. Prove that the lines joining the 
 mid-points of the opposite sides of a quadri- 
 lateral and the line joining the mid-points 
 of the diagonals of the quadrilateral meet in 
 a point. 
 
 Suggestion. — Recall Ex. 71. 
 
 Ex. 73. Prove that the line joining the ^ 
 
 mid-points of the diagonals of a trapezoid is parallel to the bases and 
 equal to | their difference. 
 
 Suggestions. — 1. Dravr EG II AD meeting CD at G. 
 
 2. Prove EG passes through point F. 
 
 3. Compare EG with AD and FG with BC. 
 
 Ex. 74. If the perpendiculars AE, BF, CG, 
 and DH be drawn from the vertices of parallelo- 
 gram ABCD to any line in its plane not intersecting 
 its surface, prove that AE + CG = BF -»- DH. 
 
 Suggestion. — See adjoining figure. Apply § 153. 
 
 Ex. 75. Prove that the sum of any three sides of a quadrilateral is 
 greater thau the fourth side. 
 Suggestion. — Draw a diagonal. 
 
 Ex, 76. Prove that the sum of the lines drawn from any point within 
 a triangle to the vertices is less than the sum of the three sides. 
 Suggestion. — 1. Let within A ABC be joined to A, B, and C 
 
 2. OA-\-OB<AC+BC. (Ex. 188, Book I.) 
 
 3. Similarly express OB -\- OC'and also OC -\- OA. 
 
 4. Add these inequalities and divide by 2. 
 
 Ex. 77. In triangle ABC, if D is any point on ^C so that AD = AB, 
 then BC> DC. 
 
 Suggestion. — Compare BC-\r AB with AC. 
 
 Ex. 78. Prove that each of the equal sides of an isosceles triangle is, 
 greater than one half the base. 
 
 Ex. 79. If is any point within triangle ABC, then AO-\-BO-\-GO 
 > \ perimeter. 
 
 Suggestions. — 1. Apply § 159 (a) to each side of the triangle. 
 2. Add the inequalities and divide by 2. 
 
 Ex. 80. Prove that any side of a triangle is less than one half the 
 perimeter of the triangle. 
 
 Suggestions. — 1. Apply § 159 (a) to one side. 
 
 2. Add that side to both members of the inequality. 
 
SUPPLEMENTARY EXERCISES 
 
 281 
 
 Ex. 81. Prove that the median to any side of a triangle is l^s than 
 one half the perimeter of the triangle. 
 
 Suggestion. — The median lies in each of two As. 
 
 Apply § 159 (a) to the median in each A, and add. 
 
 Ex. 82. Prove that the median to any side of a triangle is less than 
 one half the sum of the other two sides of the triangle. 
 
 Suggestion. — 1. Extend the median its own length, through the side of the 
 triangle. Connect the end of the new segment with one of the other vertices 
 of the triangle. 
 
 Ex. 83. Prove that the median to any side of a triangle is greater 
 than one half the sum of the other two sides diminished by the* side to 
 which it is drawn. 
 
 Ex. 84. Prove that the sum of the medians to the sides of a triangle 
 is greater than one half the perimeter of the triangle. 
 
 Suggestion. — Apply Ex. 83 to each of the medians. 
 
 Ex. 85. The bisectors of the exterior angles at 
 two vertices, and the bisector of the interior angle at 
 the third vertex of a triangle are concurrent. 
 
 Suggestion. — The proof is like that for § 169. 
 
 Ex. 86. If two medians of a triangle are equal, 
 isosceles. 
 
 Ex. 87. Prove that the line joining the ortho- 
 center of a triangle to the circum-center of the tri- 
 angle passes through the center of gravity (§178) of 
 the triangle. 
 
 Suggestions. — 1. Draw AR, and try to prove that K 
 is the center of gravity, by proving that AK = 2 KR. 
 . 2. Recall § 152, and Ex. 190, Book I. 
 
 Ex. 88. If is the point of intersection of the 
 medians AD and BE of equilateral triangle ABC, and 
 OF is drawn parallel to AC, meeting BC at F, prove 
 that DF ial BC. 
 
 Suggestion. — Let G be the mid-point of OA, and 
 draw GH || AC; also imagine a line through D II AC. 
 Apply § 147. 
 
 Ex. 89. If the exterior angles at the vertices A and J5 of A ABC are 
 bisected by lines which meet at D, prove Z D = ^ /IB -\- ^ZA. 
 
282 
 
 SUPPLEMENTARY EXERCISES 
 
 Proof. 1. ZD = lSO- ZDAB- ZABD. Why? 
 
 2. ZDAB=lZEAB = l(iZC+ZABG). Why? 
 
 3. Similarly Z ABD = ? 
 
 4. 180° = ZBAC + ZC + ZABG. Why? 
 
 5. Substitute in step 1, and complete the proof. 
 Note. — This proof is typical of many that involve" ^^ 
 
 numerical relations among angles of a figure. In triangles, the facts in 
 §§ 106, 109, and 110 are used frequently. 
 
 Ex. 90. Prove that the exterior angle at the base of an isosceles 
 triangle equals the angle between the bisectors of the base angles. 
 
 Ex. 91. D is any point in the base BG ol isosceles 
 triangle ABG. The side ^C is extended from G to E, 
 so that GE equals CD, and DE is drawn, meeting AB 
 at F. Prove Z AFE = 3 Z AEF. 
 
 Suggestions. — l.Z AFE is exterior to A BDF. 
 
 2. ZB = ZA CD, which is exterior to A ODE. 
 
 Ex. 92, If GD is the altitude to the hypotenuse 
 AB of right triangle ABC^ and E is the mid-point of 
 AB, prove ZDCE=ZA-ZB. 
 
 Suggestions. — 1. ZDCE is the complement of ZDEC. ^ 
 Why ? 
 
 2. Express Z DEC. 3. Recall Ex. 175, Book I. 
 
 Ex. 93. If GD is the altitude to the hypotenuse 
 AB of right triangle ABC, and GE is the bisector of 
 Z C, meeting AB at E, then ZDGE=\ (ZA- ZB). 
 
 Suggestions. — 1. ZDCE = Z ACE - ZACD. 
 
 2. ZACE=^h90°. 3. 90°= ZA-\-ZB 
 
 Ex. 94. If ZB oi A ABG is greater than Z C, and BD is drawn to 
 AG making AD equal to AB, prove 
 
 Z ADB = 1{ZB + Z G), and Z GBD = l(ZB-Z<y). 
 
 Suggestions. — 1. First apply § 110. 
 
 2. ZDBC= ZB-ZABD= ZB- ZADB. Why? 
 
 Ex. 95. If Z> and E are the mid-points of sides q 
 BG and AG respectively, of l^ABG, and AD hQ 
 extended to F and BE to O, making DF -. AD 
 and EG = BE, prove that GGF is a straight line 
 and that GG = GF. 
 
 Suggestion. — 'Recsill Ex. 146, Book I, and § 90. 
 
 Why? 
 Why? 
 
 
 
 C 
 
 
 '^N 
 
 \ 
 
 A 
 
 
 \ 
 
 'w 
 
 / ^ 
 >< 
 
 ><D / 
 
 
 A 
 
 
 
SUPPLEMENTARY EXERCISES 
 
 283 
 
 Ex. 96. If the median drawn from any vertex of a triangle is greater 
 than, equal to, or less than one half the opposite side, the angle at the 
 vertex is acute, right, or obtuse respectively. (§ 161.) 
 
 Ex. 97. The perpendicular from the inter- 
 section of the medians of a triangle to any straight 
 line in the plane of the triangle, not intersecting 
 its surface, is equal to one third the sum of the 
 perpendiculars from the vertices of the triangle to 
 the same line. (§ 163.) 
 
 G Q K L M H 
 
 Ex. 98. Prove that the diagonals of an oblique-angled parallelogram 
 are unequal, the one joining the acute angles being the greater. 
 
 Ex. 99. Define: 
 
 (a) parallelogram (/) isosceles trapezoid 
 
 (b) rectangle (^g) altitude of a parallelogram 
 
 (c) square (h) altitude of a trapezoid 
 
 (d) rhombus (i) median of a trapezoid 
 
 (e) trapezoid 
 
 Ex. 100. What are the important facts known about every parallelo- 
 gram ? 
 
 Ex. 101. State four theorems by which a quadrilateral can be proved 
 a parallelogram. 
 
 Ex. 102. State facts known about a rectangle. 
 
 Ex. 103. State facts known about a square. 
 
 Ex. 104. State facts known about a trapezoid. 
 
 Ex. 105. State facts known about an isosceles trapezoid. 
 
 Ex. 106. State methods for proving two segments equal. 
 
 Ex. 107. State methods for proving two angles equal. 
 
 Ex. 108. State methods for proving two lines are parallel. 
 
 BOOK II 
 
 Ex. 1. If AB is one of the non-parallel sides of a trapezoid circum- 
 scribed about a circle whose center is 0, prove ZAOB is a right angle. 
 Suggestion. — Recall Ex. 38 (6) , p. 106. 
 
 Ex. 2. The straight line joining the raid-points of th^ non-parallel 
 sides of a circumscribed trapezoid is equal to \ the perimeter of the 
 trapezoid. 
 
 Suggestion. — Recall Ex. 39, p. 106. 
 
284 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 3. If tangents are drawn to a circle at the extremities of any pair 
 of diameters which are not perpendicular to each other, the figure formed 
 is a rhombus. (Recall Ex. 39, p. 106.) 
 
 Ex. 4. If the angles of a circumscribed quadrilateral are right angles, 
 the figure is a square. 
 
 Ex. 5. A, B, C, and Z> are four points in a 
 straight line, B lying between C and D ; EF is a 
 common tangent to the circles drawn upon AB and 
 CD as diameters. Prove Z BAE = Z DCF. 
 
 Ex. 6. If ABCD is a quadrilateral circumscribed 
 about a circle whose center is 0, prove that ZAOB + 
 Z COD = 180°. 
 
 SuggesHo7i.~Com]^8iTe ZEOB and ZBOF; LEO A 
 and ZAOH, etc. 
 
 Ex. 7. Construct a figure like the one adjoin- 
 ing, using for the equal circles from which it is 
 constructed the radius 1 in. Notice that the cir- 
 cles are tangent circles. 
 
 Ex. 8. A very small triangular 
 piece of ground AB G lies in the inter- 
 section of three streets. Make a 
 drawing to scale (1" =20'). Then 
 construct corners which will be both 
 more useful and more artistic than 
 the sharp corners. Indicate on 
 your drawing the radii of the circles 
 you construct in the corners. 
 
 Ex. 9. Prove that an inscribed angle whose intercepted arc is less 
 than a semicircle is an acute angle ; and one whose arc is greater than a 
 semicircle is an obtuse angle. 
 
 Ex. 10. If any number of equal angles are inscribed in an arc, their 
 bisectors pass through a common point. 
 
 Ex. 11. If the diagonals of an inscribed quadrilateral intersect at the 
 center of the circle, the figure is a rectangle. 
 
 Ex. 12. Prove that a parallelogram inscribed in a circle is a rectangle. 
 
SUPPLEMENTARY EXERCISES 285 
 
 Ex. 13. If AB and AF are tangents to the circle whose center is 
 and E is any point in major arc Z>jP, then, Z DEF = 90° — \Z.A. 
 Suggestions. — 1. Draw OD and OF. 
 2. Compare LE \\\ih. LDOF, and ADOF with LA. 
 
 Ex. 14. If AB and AC are tangents to a circle whose center is O 
 from a point A, touching the circle at B and C respectively, and D is any 
 point on the minor arc BC, then ZBDC = 90'' -h^ZA. (Ex. 71, p. 117.) 
 
 Ex. 15. ABCD is a quadrilateral inscribed in a circle. Another 
 circle is drawn upon AB as chord, meeting AB and CB at JS^ and F re- 
 spectively. Prove chords BC and ^F parallel. (Ex. 71, p. 117.) 
 
 Ex. 16. If the opposite angles of a quadrilateral ABCB are supple- 
 mentary, a circle can be circumscribed about the quadrilateral. 
 
 Suggestions. — 1. Assume that D falls outside the O through A, B, and 
 C, and that the cuts CD at E. 
 
 2. Derive two contradictory facts about LD and Z.AEC, using the 
 hypothesis and Ex. 71, p. 117. 
 
 3. Next, assume that D falls inside the ABC and complete the indirect 
 proof. 
 
 Ex. 17. If a right triangle has for its hypotenuse the side of a square 
 and lies outside the square, the straight line drawn from the center of the 
 square to the vertex of the right angle bisects the right angle. 
 
 Suggestio7i. —The on the hypotenuse as diameter must pass through 
 the center of the square and also through the vertex of the right angle. 
 (Ex. 16, p. 285.) Draw this circle. 
 
 Ex. 18. The perpendiculars drawn from the vertices of a triangle to 
 the opposite sides are the bisectors of the angles of the triangle formed 
 by joining the feet of the perpendiculars. 
 
 Suggestions. — 1. © can be circumscribed about 
 quadrilaterals BDOF, CDOE, and AEOF. (Ex. 
 16, p. 285.) 
 
 2. Compare Z ODF with L OBF, and Z ODE 
 with Z.OCE. 
 
 3. Compare Z OBF with Z OCE, by connecting 
 each with Z^^C 
 
 4. Then AD bisects Z EDF. 5. Similarly for Z DEF and Z DFE. 
 
 Ex. 19. Construct the triangle having given the feet of the perpen- 
 diculars from the vertices to the opposite sides. (Recall Ex. 18, p. 286.) 
 
 Ex. 20. If sides AB and BC of inscribed hexagon ABCDEF are 
 parallel to sides 2)^ and EF respectively, prove side AF parallel to side 
 CD. 
 
286 SUPPLEMENTARY EXERCISES 
 
 Ex. 21. If a circle be drawn upon the radius of another circle as 
 diameter, any chord of the greater circle passing through the point of 
 contact of the circles is bisected by the smaller circle. 
 
 Suggestion. — Recall § 148. 
 
 Ex. 22. Prove Prop. XXI, Book II, by drawing through B a chord 
 parallel to CD. (Recall § 208. ) 
 
 Ex. 23. If sides AB and BC of inscribed quadrilateral ABCD 
 subtend arcs of 69'' and 112° respectively, and Z AED between the 
 diagonals is 87°, how many degrees are there in each angle of the 
 quadrilateral ? 
 
 Suggestions. — 1. Let x = AD and y = I)C. Determine these arcs alge- 
 braically. 
 
 2. Then determine the size of each of the required angles. 
 
 Ex. 24. Prove Prop. XXII, Book II, by drawing through B a chord 
 parallel to (7i>. (Recall §208.) 
 
 Ex. 25. Prove that the measure of the angle between two tangents 
 is the supplement of the measure of the smaller of the two intercepted 
 arcs. 
 
 Suggestion. — After obtaining the measure of the angle, substitute in it 
 for the larger arc the value of that arc in terms of the smaller arc. 
 
 Ex. 26. If sides AB., BC, and CD of an inscribed quadrilateral sub- 
 tend arcs of 99°, 106°, and 78° respectively, and sides BA and CD 
 extended meet at E^ and sides AD and 5(7 at F., find the number of 
 degrees in Z AED and Z AFB. 
 
 Ex. 27. li A A., B, and C of circumscribed quadrilateral ABCD are 
 128°, 67°, and 112°, respectively, and sides AB, BC, CD, and DA are 
 tangent to the circle at points E, F, G, and H respectively, find the 
 number of degrees in each angle of the quadrilateral EFGH. 
 
 Ex. 28. If AB and AC are the tangents to a circle from a point A, 
 and D is any point on the major arc subtended by chord BC, prove that 
 Z ABD + Z ^ CD is constant. 
 
 Suggestions. — !. AABD -{- AACD = 360° - /lA - /.D. Why? 
 
 2. Substitute for Z.A and Z.D their measures. 
 
 Ex. 29. If ABCD is a circumscribed quadrilateral, prove that the 
 angle between the lines joining the opposite points of contact equals 
 \{/.A-\- Z.C) or is supplementary to it. 
 
 Suggestion. — Find the measure of each of the angles. Add the measure 
 of /^and ZC 
 
SUPPLEMENTARY EXERCISES 287 
 
 Ex. 30. ABCD is a quadrilateral inscribed 
 in a circle. If sides AB and DC extended 
 intersect at JF, and AD and BC extended inter- 
 sect at F, prove that the bisectors of Z E and H^ 
 Z F are perpendicular. 
 
 Suggestions. — 1. AM+ AH+ KC ■\-(jL must aN 
 = 180°. 
 
 2. ^3/ enters in the measure of Z AEM, and 
 u4^in that of LAFH. Express these measures and add the results. 
 
 This will give a start on the proof. 
 
 Construct the A ABC having given : 
 
 Ex. 
 
 31. 
 
 a, K, he. 
 
 Ex. 
 
 32. 
 
 a, he, to. 
 
 Ex. 
 
 33. 
 
 A, C, to. 
 
 Ex. 
 
 34. 
 
 c, he, nic' 
 
 Ex. 
 
 39. 
 
 Construct a 
 
 Ex. 
 
 35. 
 
 a, 
 
 &, 
 
 A. 
 
 Ex. 
 
 36. 
 
 A, 
 
 B, 
 
 he. 
 
 Ex. 
 
 37. 
 
 b, 
 
 ma, 
 
 C. 
 
 Ex. 
 
 38. 
 
 A, 
 
 tA, 
 
 ha. 
 
 Construct a right triangle having given the altitude upon 
 the hypotenuse and one of the legs of the triangle. 
 
 Ex. 40. Construct a right triangle having given the altitude upon the 
 hypotenuse and one of the acute angles. 
 
 /^ 
 Ex. 41. Construct a triangle having given the / Nv 
 
 mid-points of its sides. ^;" /v 
 
 Suggestion. —Kow does C'B' compare with BC? / \/ \ ^ 
 
 Ex. 42. Construct a tangent to an arc of a circle at a given point of 
 the arc without using the center of the circle. 
 
 Ex. 43. Construct a tangent to a given circle which will be perpen- 
 dicular to a given straight line. 
 
 Ex. 44. Given the mid-point of a chord of a circle, construct the 
 chord. 
 
 Ex. 45. Construct a parallel to side BC of 
 A ABC meeting AB and ^C at D and E respec- 
 tively, so that DE will equal the sum of BD and CE. 
 
 Ex. 46. Inscribe a square within a given right 
 triangle having one of its angles coincident with the 
 right angle of the triangle and the opposite vertex lying on the hypotenuse. 
 
 Suggestion. — In the analysis figure, draw the diagonal from the vertex of 
 the right triangle. 
 
288 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 47. Construct a rhombus within a given tri- 
 angle, having one angle coincident with an angle of the 
 triangle, and the opposite vertex lying on the opposite 
 side of the triangle. 
 
 Ex. 48. Construct a square which will have its ver- 
 tices on the sides of a given rhombus. 
 
 Suggestion. — Make an analysis based upon the adjoining 
 figure. 
 
 Ex. 49. Construct two tangents to a given circle which will make a 
 given angle with the circle. 
 
 Suggestion. — Draw the given angle at the center of the circle. 
 
 Ex. 50. Given an angle of a triangle and 
 
 the segments of the opposite side made by the 
 
 altitude drawn to that side. Construct the tri- 
 angle. 
 
 Ex. 51. Construct a A ABC having given 
 c, 6, and Wo. Make an analysis based upon the 
 adjoining figure. 
 
 Ex. 52. Through a given point outside a circle, construct a secant 
 whose internal and external segments will be equal. 
 
 Suggestion. — For the analysis figure, connect the center of the circle with 
 the given point, and also with the points of intersection of the secant and the 
 circle. Recall Ex. 51, p. 288. 
 
 Ex. 53. Given the base, an adjacent acute angle, 
 and the difference between the other two sides of the 
 triangle, construct the triangle. 
 
 Suggestions. — 1. LetCD=CB. Then AD=AC-BC. 
 2. Then A ABB can be made the basis of the con- 
 struction. 
 
 Ex. 54. Given the base of a triangle, an adjacent 
 angle, and the sum of the other two sides, construct 
 the triangle. 
 
 Suggestions. — 1. Let AD = AC+ CB. Draw CB 1 
 
 BD. 
 
 2. A ABB can be made the basis of the construction. 
 
 /^' 
 
SUPPLEMENTARY EXERCISES 289 
 
 Ex. 55. Construct full size the pattern for the 
 faces of a mission lamp as shown in the adjoining 
 figure, using the dimensions indicated. 
 
 om 
 
 Ex. 56. Construct a circle tangent to a given line and having its 
 center at a given point not on the line. 
 
 Ex. 57. Construct a circle which will be tangent to each of two 
 parallels and will pass through a given point lying between the parallels. 
 
 Ex. 58. Construct a circle having its center in a given line, and pass- 
 ing through two points not in the line. 
 
 Ex. 59. Construct a circle with given radius which will be tangent 
 to a given circle and will pass through a given point outside of the circle. 
 
 Ex. 60. Construct a circle with given radius which will be tangent to 
 a given circle and pass through a given point inside of the circle. 
 
 Ex. 61. Construct a circle with a given radius which will be tangent 
 to a given line and also to a given circle. 
 
 Ex. 62. Construct a circle which will be tangent to a given circle at 
 a given point on it and also tangent to a given straight line. 
 
 Ex. 63. Construct a circle which will be tangent to a given circle at 
 a given point on it and also pass through a given point outside of the circle. 
 
 BOOK III 
 
 Ex. 1. Prove the theorem of § 268 on the hypothesis that AD : DB 
 = AE : EC. 
 
 Suggestion. — Use the same construction. Write the hypothesis by com- 
 position, and use § 262. 
 
 Ex. 2. Let P be any point not in line AB and B any point in AB. 
 Let ^ be a point in segment PB, such that FS : PB = 1:3. Suppose 
 that B moves along AB. What is the locus of point S ? 
 
 Ex. 3. XY is parallel to the side AB of A OAB, meeting OA at X 
 and OB at Y. Point C is taken between X and A of OA, and ^C is 
 drawn. XZ is drawn parallel to BC, meeting YB at Z. Prove CY\\ AZ. 
 
 Suggestion. — Try to prove OC • OZ = OA   OY; then use § 252. 
 
 Ex. 4. State and prove the converse of Prop. IV, § 270. (Fi§. of 
 Prop. IV. Prove Z BAD=Z CAD. Extend CA to E^ making AE=AB.) 
 
290 SUPPLEMENTARY EXERCISES 
 
 Ex. 5. The sides of a triangle are a, &, and c, respectively. Derive 
 formulae for the segments of side c made by the bisector of Z G. 
 
 Ex. 6. AB is the hypotenuse of right A ABC. If perpendiculars be 
 drawn to AB at A and B^ meeting AC extended at 2>, and BG extended 
 at jE', prove ^ ACE and BCD similar. 
 
 Ex. 7. If altitudes AB and GE of A ABC intersect at F, prove 
 AF:AB = EF.BD. 
 
 Ex. 8. ^^ is a chord of a circle, and CE is any chord drawn through 
 the middle point C of arc AB, cutting chord ^^ at D. 
 Prove ^O is a mean proportional between CD and CE. 
 
 Ex. 9. Two circles are tangent internally at C. GA is drawn meet- 
 ing the smaller circle at B and the larger at A ; CE is drawn meeting the 
 smaller circle at Z> and the larger at E. Prove GB : GA = CD : GE. 
 
 Ex. 10. The diagonals of a trapezoid, whose bases are AD and BG, 
 intersect at E. If AE =9, EC= 3, and BB = 16, find BE and ED. 
 
 Suggestion. — Fvoye AE : EC= DE : EB. 
 
 Ex. 11. Let ^C be the hypotenuse of right A ABC, and E and F be 
 any points on AB and BG respectively ; let JE'i> and FO be perpendic- 
 ulars to AC, meeting ylC at Z> and G respectively. Prove AE : FC 
 = ED : GC. (Recall § 109.) 
 
 Ex. 12. Z ^ of A ABC is a right angle. DEFG is a square having E 
 and F on BG, D on AC, and G on AB. Prove GE: EF= EF : FB. 
 
 Suggestion. — Compare A (7i)jE; and A BFG. 
 
 Ex. 13. AB and ^C are the tangents to a circle from point A. If 
 CD is drawn perpendicular to OB produced at D, then AB : OB = BD : 
 CD. 
 
 Suggestion. — Draw OA and BG. Prove OA 1 BG. 
 
 Ex. 14. AABCis^Lu isosceles triangle. If the perpendicular to AB 
 at A meets base BG, extended if necessary, at E, and D is the mid-point 
 of BE, then AB is the mean proportional between BG and BD. 
 
 Suggestion. — Recall §284 and Ex. 175, Book I. 
 
 Ex. 15. Let r be the radius of a circle and c be the distance from the 
 center of the circle to a point P outside the circle. Express the length of 
 the tangent to the circle from P, in terms of r and c. 
 
 Ex. 16. What is the length of the tangent to a circle whose diameter 
 is 16, from a point whose distance from the center is 17 ? 
 
SUPPLEMENTARY EXERCISES 291 
 
 Ex. 17. Prove that the tangents to two inter- 
 secting circles from any point in their common 
 chord produced are equal. (Figure adjoining.) 
 
 Ex. 18. If two circles intersect, their common 
 chord produced bisects their common tangents. 
 
 Ex. 19. If the altitude be drawn to the hypot- 
 enuse of a right triangle, the segments of the hypotenuse have the same 
 ratio as the squares of the adjacent legs. 
 
 Ex. 20. What is the length of a chord of a circle which is 6 in. from 
 the center, if the radius is 10 in. ? 
 
 Ex. 21. The equal angles of an isosceles triangle are each 30°, and the 
 equal sides are each 8 in. in length. What is the length of the base ? 
 Suggestion. — Recall Ex. 128, Book I. 
 
 Ex. 22. Find the altitude to the base of an isosceles triangle if the 
 base is 8 inches and the sides are each 10 inches in length. 
 
 Ex. 23. If the equal sides of an isosceles right triangle are each 18 
 in. in length, what is the length of the median drawn from the vertex of 
 the right angle ? 
 
 Ex. 24. One of the non-parallel sides of a trapezoid is perpendicular 
 to the bases. If the length of this side is 40, and of the parallel sides 31 
 and 22, respectively, what is the length of the other side ? 
 
 Ex. 25. If the length of the common chord of two intersecting circles 
 is 16, and their radii are 10 and 17, what is the distance between their 
 centera ? 
 
 Ex. 26. If BC is the hypotenuse of right triangle ABC, prove 
 (a + 6-|-c)2 = 2(a + c){a + b). 
 
 Ex. 27. If the diagonals of a rhombus are m and n respectively, 
 derive a formula for the perimeter of the rhombus. 
 
 Ex. 28. The diameter which bisects a chord 12 in. long is 20 in. in 
 length. Find the distance from either extremity of the chord to the 
 extremities of the diameter. 
 
 Suggestions. — 1. Let x represent one segment of the diameter made by the 
 chord. 2. Recall § 289. 
 
 Ex. 29. The radius of a circle is 16 in. Find the length of the chord 
 which joins the points of contact of two tangents, each 30 in. in length, 
 drawn to the circle from a point outside the circle. 
 
 Suggestions. — 1. Draw the radii to the points of contact. 2. Recall § 288. 
 
292 SUPPLEMENTARY EXERCISES 
 
 Ex. 30. Two parallel chords on opposite sides of the center of a circle 
 are 48 in. and 14 in. long, respectively, and the distance between their 
 mid-points is 31 in. What is the diameter of the circle ? 
 
 Suggestion. — Let x represent the distance from the center to the middle 
 point of one chord, and 31 — x the distance from the center to the middle point 
 of the other. Then the square of the radius may be expressed in two ways in 
 terms of x. 
 
 Ex. 31. The parallel sides, AD and BC, of a circumscribed isosceles 
 trapezoid are 18 and 6 respectively. Find the diameter of the circle. 
 
 Suggestions. — 1. Recall Ex. 35, Book II. 
 
 2. Through B, draw BE II CD, meeting AD at E. 
 
 Ex. 32. The diameters of two circles are 12 and 28, respectively, and 
 the distance between their centers is 29. Find the length of the common 
 internal tangent. 
 
 Suggestion. — Find the 1 drawn from the center of the smaller to the 
 radius of the greater extended through the point of contact. 
 
 Ex. 33. Prove that the square of the common tangent to two circles 
 which are tangent to each other externally is equal to 4 times the product 
 of their radii. 
 
 Ex. 34. If D is the mid-point of leg BC of right triangle ABC, prove 
 that the square of the hypotenuse AB exceeds 3 times the square of CD 
 by the square of AD. 
 
 Ex. 35. If AB is the base of isosceles triangle ABC and AD is 
 perpendicular to BC, prove AB^ + BC^ + AC^ = 3 AD^ + 2 CD^ + BD^ 
 
 Ex. 36. If D is the mid-point of leg BC of right triangle ABC, and 
 DE is drawn perpendicular to hypotenuse AB, prove AE — BE = ACT. 
 
 Ex. 37. If in right triangle ABC, acute angle B is double acute 
 angle A, prove AC^ = S BC^. 
 
 Suggestion. — Recall Ex. 128, Book I. 
 
 Ex. 38. Prove that the sum of the squares of the 
 distances of any point on a circle from the vertices of 
 an inscribed square is equal to twice the square of the 
 diameter of the circle. 
 
 Ex. 39. If ABC and ADC are angles inscribed 
 in a semicircle, and AE and CF are drawn perpen- 
 dicular to BD extended, prove E 
 
 BE^ + BF'^ = DE^ + DF\ 
 
SUPPLEMENTARY EXERCISES 293 
 
 Ex. 40. If lines be drawn from any point P to 
 the vertices of rectangle ABCD^ prove that 
 
 PA^ + PC^ = P& + Plf. 
 
 A 
 
 t 
 
 Ex. 41. Inscribe in a given circle a triangle similar to a given triangle. 
 
 Suggestion. — Circumscribe about the given A a 0, and draw radii to the 
 vertices. Recall § 293. 
 
 Ex. 42. Construct a right triangle having given its perimeter and an 
 acute angle. 
 
 Suggestion. — Any right triangle containing the given acute angle will be 
 similar to the required triangle. The sides of the required triangle can be de- 
 termined by § 297. 
 
 Ex. 43. The perimeter of one of two similar polygons is 153 in. ; 
 the shortest side of this polygon is 18 in. The shortest side of a similar 
 polygon is 24 in. ; what is the perimeter of the second polygon ? 
 
 Ex. 44. The adjoining figure is similar to the „ 
 
 boundary of an irregular field of a farm ; the ratio 
 of similitude of the figure and the boundary of the 
 field is 1 : 2400. Determine the perimeter of the 
 field itself by first finding the perimeter of the ad- 
 joining figure and then applying § 297. 
 
 Ex. 45. If JE7 is the mid-point of one of the parallel sides BC, of 
 trapezoid ABCD, and AE and DE extended meet DC and AB extended 
 at i^and G respectively, then FO is parallel to BC. 
 
 Suggestion. — GF II BC if GB :GA = FE: FA. 
 
 Ex. 46. A ABC and yl'i?C have their vertices KT" ----^A' 
 
 A and A' in a line parallel to their common base d \ ^v — y ' / ^' 
 BC. If a parallel io BC cuts AB at i) and ^O at \ ^X^ / 
 E, A'B at D' and ^'C at E\ then DE = D'E'. q^ — -^ 
 
 Suggestion. — Prove DE : BC = D'E' : BC. 
 
 Ex. 47. If AB and CD are equal and parallel segments, prove that 
 p^^ equals p^^, where m is any line. 
 
 Ex. 48. If AD and BE are the perpendiculars from vertices A and B, 
 respectively, of acute-angled triangle ABC to the opposite sides, prove 
 ACxAE + BCxBD = AB\ 
 
 Suggestion. — Find 2ACx AE by § 310, and in like manner find 2BCx BD 
 Then add. 
 
294 SUPPLEMENTARY EXERCISES 
 
 Ex. 49. In triangle ABC, if angle C equals 120°, prove 
 
 AB^^ = BC^ + AC^ + BCx AC. 
 Suggestion. — Recall § 311. 
 
 Bx. 50. If a line be drawn from vertex C of isosceles triangle ABC, 
 meeting base AB extended at D, prove CB^ — CB^ = AD x BD. 
 Suggestion. — Apply § 311 in A BCD. 
 
 Ex. 51. From the conclusion of § 311, derive a formula for pj in terms 
 of a, 6, and c. 
 
 Ex. 52. In any triangle, the product of any two 
 
 sides is equal to the product of the segments of the F>^ '-'P^'^'n 
 
 third side formed by the bisector of the. exterior ^^^>fC; \ '^' 
 
 angle at the opposite vertex, minus the square of /^'^J\\ \ ; 
 
 the bisector. q ^ X — — 4'q 
 
 Prove AB X AC = DB X DC - AD\ ^ "~ " 
 
 Suggestions. — 1. The solution is similar to that of § 318. 
 
 2. Firstprove A^£Z)~ A^C^. 
 
 Ex. 53. DEFG is a square having its vertices D and E on sides AB 
 and BC respectively of triangle ABC and its vertices F and O on side AC. 
 Let BH be II to AC, meeting AE extended at G ; let UK he ± AC and 
 BT±AC. Prove BHKT is a square, 
 
 Ex. 54. In a given triangle, construct a square which shall have two 
 vertices lying on one side of the triangle and having its other two vertices 
 on the other two sides of the triangle, one on each side. 
 
 Ex. 55. Construct a square which will have two of its vertices on a 
 diameter of a given circle, and the remaining two vertices on the semicircle 
 constructed on this diameter. 
 
 Ex. 56. Circumscribe about a given circle a triangle similar to a given 
 triangle. 
 
 Suggestion. — Inscribe in the given triangle a circle and draw radii to the 
 points of tangency. 
 
 BOOK IV 
 
 Ex. 1. The sides of a triangular field are 10 rd., 8 rd., and 9 rd. 
 respectively. Make a scale drawing of the boundary of the field on coor- 
 dinate paper, and estimate the area of the field. 
 
 Ex. 2. Angle B of A ABC is a right Z. D and E are the mid-points 
 of AB and AC respectively. CF, perpendicular to BC at C, meets DE 
 extended at F. Prove A J-jSC = □ BCFD. 
 
SUPPLEMENTARY EXERCISES 295 
 
 Ex. 3. E is any point on diagonal ^C of 
 O ABCD. Tlirough E, parallels to AD and 
 AB are drawn, meeting AB and CD at F and 
 irrespectively, and BC and AD at G and K re- 
 spectively. 
 
 Prove CJFBGE = CD EHDK. 
 
 Ex. 4. All the lots of a certain city block are rectangular and 125 ft. 
 in depth (from front to back). Compare two lots A and B if the frontage 
 of Lot A is 40 ft. and that of Lot J5 is 60 ft. (Do not obtain their areas.) 
 
 Ex. 5. Two rectangles B\ and B<2 have equal altitudes. 
 
 (a) What part of B-z is B\ if the base of Bi is 5 and the base of 7^2 is 8 ? 
 
 (6) What is the ratio of ^i to B2 if the bases are 25 and 10 respec- 
 tively ? 
 
 Ex. 6. Divide a given triangle into three equal parts by lines drawn 
 through one of its vertices. 
 
 Ex. 7. Determine the area of the triangle whose sides are 25, 17, 
 and 28. 
 
 Ex. 8. If h is the base and s is one of the equal sides of an isosceles 
 triangle, prove that the area is ^ 6 V4 s^ — ft^. 
 
 Ex. 9. The area of an isosceles right triangle is 81 sq. in. Deter- 
 mine its hypotenuse. 
 
 Suggestion. — I^iet x represent one of the sides. Determine x and then de- 
 termiue the hypotenuse. 
 
 Ex. 10. The area of an equilateral triangle is 9V3. Determine its 
 side. 
 
 Suggestion. — Use the formula proved in Ex. 29, Book IV. 
 
 Ex. 11. The altitude of an equilateral triangle is 3. Determine its 
 area. 
 
 Suggestion. — Let x represent one side. Determine x and then determine 
 the area. 
 
 Ex. 12. The area of an equilateral triangle is 16 V3. Determine its 
 altitude. 
 
 Ex. 13. The area of a rhombus is 240 sq. in. and its side is 17 in. 
 Find its diagonals. 
 
 Suggestion. — Represent the diagonals by 2 a; and 2 y. Proceed algebraically. 
 
 Ex. 14. One diagonal of a rhombus is five thirds the other; the dif- 
 ference of the diagonals is 8 in. Determine the area of the rhombus. 
 
 Ex. 15. The segments of the hypotenuse of a right triangle made by 
 the altitude drawn to the hypotenuse are 5^ and 9f respectively. Deter- 
 mine the area of the triangle. (§ 288.) 
 
296 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 16. The sides of A ABO are AB = 13, BC = 14, and AC = 15. 
 Bisector AD ot Z A meets BG ?itD. Find tlie areas of ^ ABD and ACD. 
 
 Suggestion. — 1. Compute the altitude Aa. (§313.) 
 2. Determine BD and i>(7 by § 270. 
 
 Ex. 17. If Z> and E are the mid-points of sides BC and AC respec- 
 tively of A^^O, prove A ABD =. A ABE. 
 
 Suggestion. — Compare the altitudes to AB from D and E, ' 
 
 Ex. 18. If diagonal AC of quadrilateral ABCD bisects diagonal BD^ 
 thenAJJ5a = A^D(7. 
 
 Ex. 19. Two equal triangles have a common 
 base, and lie on opposite sides of it. Prove that 
 the base, extended if necessary, bisects the line 
 joining their vertices. (Prove CD = CD.) 
 
 Ex- 20. If EF is any straight line drawn 
 through the point of intersection of the diagonals 
 oi O ABCD, meeting sides AD and BC at E 
 and F respectively, then A BEE = A CED. 
 
 Suggestion. — Does BF = ED ? 
 
 Ex. 21. If E, F, G, and ^ are the mid-points of 
 sides AB, BC, CD, and DA, respectively, of quad- 
 rilateral ABCD, prove ^i?'(y/i" a parallelogram equal 
 to one half ABCD. A 
 
 Ex. 22. Prove that the sum of the perpendiculars 
 from any point within an equilateral triangle to the 
 three sides is equal to the altitude of the triangle. 
 
 Suggestions. — -i. A J5PC+A BPA-^A APC=A ABC. 
 
 2. Express the area of each triangle and substitute in 
 this equation. 
 
 Ex. 23. If E is any point in side BCoi CJ ABCD, and DE is drawn, 
 meeting AB extended at F, prove A ABE equals A CEF. 
 
 Suggestion.— Comp2kTe AFCD with O ABCD. 
 
 Ex. 24. Prove that the area of a triangle is equal 
 to one half the product of its perimeter by the radius 
 of the inscribed circle. 
 
 De 
 
SUPPLEMENTARY EXERCISES 297 
 
 Ez. 25. A circle whose diameter is 12 is inscribed in a quadrilateral 
 whose perimeter is 50. Find the area of the quadrilateral. 
 
 Ex. 26. If the sides of a triangle are 15, 41, and 52, determine the 
 radius of the inscribed circle. 
 
 Suggestions. — 1. Find the area of the triangle. 
 2. Make use of the fact proved in Ex. 24. 
 
 Ex. 27. If Z> is the mid-point of side BC of A ABC, E the mid-point 
 of AD, F of BE, and G of CF, then AABC = %A EFG. 
 Suggestion. — Draw EC. 
 
 Ex. 28. If BE and CF are medians drawn from ^ 
 
 vertices B and C of A ABC, intersecting at D, prove / Nv 
 
 A BCD equals quadrilateral AEDF. p/ ..\e 
 
 Suggestion. — Compare A ABE with A BEC and /^-''^^^D^^"«v^\ 
 
 with A BFC. B^=^— — - — -^^c 
 
 Ex. 29. Any quadrilateral ABCD is equivalent to ^^F 
 
 a triangle, two of whose sides are equal to diagonals -'"'y \ 
 
 AC and BD, respectively, and include an angle equal ^f\/^ '• \ 
 
 to either of the angles between AC and BD. / y^s. \ / 1 
 
 Prove A EFG = ABCD, where EF=AC, and J/ X\[ I 
 
 EG = BD. D\J^ 
 
 G 
 Suggestion. — ComjpSiTe A DFG with A BEF and then with A ABC. 
 
 Ex. 30. Prove that two triangles are equal if two sides of one equal 
 respectively two sides of the other and the included angles are supple- 
 mentary. 
 
 Suggestion. — Place the triangles so that the supplementary angles are 
 adjacent and so that one pair of equal sides coincide. 
 
 Ex. 31. On coordinate paper, draw the pentagon whose vertices are 
 ^ = 0, ; B = 5, ; C = 8, 3 ; Z> = 4, 9 ; ^ = 0, 6. 
 
 Determine its approximate area as in Ex. 1, and Ex. 2, Book IV. 
 Construct a A equal to the pentagon. Then construct its base and alti- 
 tude, and compute its approximate area. 
 
 Ex. 32. If, in the figure of Prop. X, AB = 9 in., A'B' = 7 in., and 
 the area of A A'B'C is 147 sq. in., find the area of A ABC. 
 
 Ex. 33. The area of a certain triangle is f the area of a similar 
 triangle. If the altitude of the first is 4 ft., what is the altitude of the 
 second ? 
 
298 SUPPLEMENTARY EXERCISES 
 
 Ex. 34. If, in §343, area of AA'B'C =147 sq. in., AB = 9 in., 
 and A'B' = 3 in., find the area of A ABC. 
 
 Ex. 35. The sides AB and ^C of A ABC are 15 and 22, respectively. 
 From a point D in AB^ a parallel to BC is drawn meeting AC a.t E, and 
 dividing the triangle into two equal parts. Find AD and AE. 
 
 Ex. 36. If similar polygons be drawn upon the legs of a right tri- 
 angle as homologous sides, the polygon drawn upon the hypotenuse is 
 equal to the sum of the polygons drawn upon the legs. 
 
 Suggestions. — 1. Compare the polygon on each leg with the one on the 
 hypotenuse by § 344. 
 
 2. Add the resulting equations and simplify. 
 
 Ex. 37. Construct a triangle similar to two given similar triangles 
 and equal to their sum. 
 
 Ex. 38. Two similar triangles have homologous sides of 8 in. and 
 15 in. respectively. Find the homologous side of a similar triangle equal 
 to their sura. 
 
 Ex. 39. Construct a triangle similar to two similar triangles, and 
 equal to their difference. 
 
 Ex. 40. If the area of a polygon, one of whose sides is 15 in., is 375 
 sq. in., what is the area of a similar polygon whose homologous side is 
 10 in. ? 
 
 Ex. 41. If the area of a polygon, one of whose sides is 36 ft., is 648 
 sq. ft,, what is the homologous side of a similar polygon whose area is 
 392 sq. ft. ? 
 
 Ex. 42. Construct a rectangle having a given altitude and equal to 
 a given parallelogram. 
 
 Suggestion. — Recall Ex. 78, Book IV. 
 
 Ex. 43. Construct a parallelogram equal to a given parallelogram 
 and having two adjacent sides equal to given segments m and n respec- 
 tively. 
 
 Ex. 44. Construct a parallelogram equal to a given parallelogram 
 and having one side equal to a given segment m, and one diagonal equal 
 to a given segment n. 
 
 Ex. 45. Construct a right triangle equal to a given square, having 
 given its hypotenuse. 
 
 Suggestion. — Determine the altitude to the hypotenuse as in Ex. 78, Book 
 IV ; then construct the triangle, using the methods of § 241. 
 
 Ex. 46. Construct a right triangle equal to a given triangle, having 
 given its hypotenuse. 
 
SUPPLEMENTARY EXERCISES 299 
 
 BOOK V 
 
 Ex. 1. Prove that the diagonals drawn from one vertex of a regular 
 polygon having n vertices to each of the other vertices divides the angle at 
 that vertex into (n — 2) equal parts. 
 
 Ex. 2. Prove that the central angle of any regular polygon is the 
 supplement of the vertex angle of the polygon. 
 
 Ex. 3. Prove that the sum of the perpendiculars drawn from any 
 point within a regular polygon to the sides of the polygon is equal to the 
 apothera multiplied by the number of sides of the polygon. 
 
 Suggestions. — Connect the point with each vertex. Notice that the sum 
 of the triangles so formed equals the polygon. Express the area of each tri- 
 angle and form an equation. 
 
 Ex. 4. In the figure for § 365 prove that : 
 (a) S4 > S8 > «i6, etc. (See § 362. ) 
 (6) a4 < as < aie, etc. 
 (c) ki < ^8 < A:i6, etc. 
 
 Ex. 5. Prove that an equiangular polygon inscribed in a circle is reg- 
 ular if the number of sides is odd. 
 
 Ex. 6. Prove that an equiangular polygon circumscribed about a cir- 
 cle is regular. 
 
 Suggestions. — 1. Draw the chords joining the points of tangency. 
 2. Prove the resulting As : 
 
 (a) are isosceles; (6) are mutually equiangular; (c) that XY= YZ, etc. 
 See diagram in § 3G7. 
 Complete the proof. 
 
 Ex. 7. Repeat Ex. 14, p. 227, for a regular octagon circumscribed 
 about a circle of radius 10. 
 
 Ex. 8. Prove that diagonal AE oi regular octagon ABCDEFGH is 
 the perpendicular-bisector of diagonal BH. 
 
 F 
 
 Ex. 9. In the adjoining figure, ABCD and /O'y^ '< ^v\ 
 EFOH are squares inscribed in the circle, such (/\ ^^\!y' Y\\ 
 
 that AF= FB = BO, etc. Is BSTUVWXY a ^t\r"7,'^;"""'br 
 regular octagon ? \Nc' ' 
 
 H 
 
300 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 10. Construct a Maltese cross having the dimen- 
 sions indicated. 
 
 Ex. 11. Prove that the construction indicated in 
 the adjoining figure serves to inscribe a regular octa- 
 gon in the square. 
 
 Ex. 12. A regular octagon is inscribed in a circle of radius 10. Com- 
 pute S8, i?8, «8, and ^'8. 
 
 Ex. 13. Prove that for a regular octagon inscribed in a circle of 
 
 radius i? : 
 
 (a) SB = -BV2-V2 
 
 B 
 
 (c) a8=^V2+V2; 
 2 
 
 (&) i)8 == 8 i? V2 - V2 ; {d) ki=2 i?2v'2. 
 
 Ex. 14, Construct a regular octagon having its sides 1 inch long. 
 
 Ex. 15. What is the relation between the area of the inscribed and of 
 the circumscribed equilateral triangles of a given circle ? 
 
 Ex. 16. What is the relation between the perimeter of the inscribed 
 and of the circumscribed equilateral triangles of a given circle ? 
 
 Ex. 17. A regular hexagon is inscribed in a circle of radius r. Prove : 
 
 (a) s^ = r) (6) ae = 
 
 rVS 
 
 (c) j96 = 6 r ; (d) Tc^ = 
 
 3rV3 
 
 2 ' ^ ' ^" ' " ' " 2 
 
 Ex. 18. A regular triangle is inscribed in a circle of radius r. Prove 
 
 3rVS; (d) kz 
 
 (a) S3 = rVS ; (&) as=-r; (c) ps 
 2 
 
 3r2V3. 
 
 Ex. 19. Prove that the apothem of an equilateral triangle is one 
 third the altitude of the triangle. 
 
 Ex. 20. (a) In a circle of radius 2.5 in., inscribe a regular hexagon. 
 
 (b) Also inscribe in the same circle a regular triangle and a regular 
 12-gon. 
 
 (c) Prove that S3 > se > S12, etc. 
 
 (d) Prove that as < ae < «i2, etc. 
 
 (e) Prove thatps <pe<pi2, etc. 
 (/) Prove that ks < ke < ki2, etc. 
 
SUPPLEMENTARY EXERCISES 301 
 
 Ex. 21. What is the perimeter and area of a regular circumscribed 
 hexagon about a circle of radius 10 ? 
 
 Ex. 22. Repeat the foregoing exercise for a circle of radius B. 
 
 Ex. 23. What is the perimeter and the area of a regular triangle 
 circumscribed about a circle of radius 10 ? 
 
 Ex. 24. Repeat the foregoing exercise for a circle of radius B. 
 
 Ex. 25. Prove that the diagonals AC^ BD^ CE^ etc., of regular hex- 
 agon ABCDEFiorm another regular hexagon. 
 
 Suggestion. — ^^Prove that a circle can be inscribed in the inner hexagon. 
 
 Ex. 26. Prove that the area of the inner B^^;^ — ^ 
 
 hexagon of the foregoing exercise is one third the 
 area of ABC DBF. 
 
 Suggestion. — Express the area of each polygon 
 in terras of the radius OB of ABC DBF. 
 
 Ex. 27. Prove that the area of a regular in- 
 scribed hexagon is a mean proportional between 
 the areas of an inscribed and of a circumscribed equilateral triangle/ 
 
 Suggestion. — Express the areas of each in terms of the radius. 
 
 Ex. 28. In a given equilateral triangle, inscribe a regular hexagon 
 having two of its vertices lying on each side of the triangle. 
 
 Ex. 29. Construct a regular hexagon having given one of the diag- 
 onals joining two alternate vertices. 
 
 Ex. 30. A square is inscribed in an equilateral triangle whose side 
 is a, having two vertices in one side of the triangle, and one in each of 
 the other sides. Compute the area of the square. 
 
 Ex. 31. A regular 12-gon is inscribed in a circle of radius B. Prove : 
 (a) Si2 = i? V2 - V3 ; (c) pi^ = 12 i? V2 - Vs ; 
 
 (6) ai2 = ^V2Wl; id)k,, = SB^. 
 
 Z 
 
 Ex. 32. If the diagonals AC and BE of regular pentagon ABCDE 
 intersect at F, prove that BE = AE -\- EF. 
 
 Ex. 33. Prove that the figure FOHKL formed by parts of the diag- 
 onals of regular inscribed petagon ABCDE is also a regular pentagon. 
 (See figure on page 302.) 
 
302 SUPPLEMENTARY EXERCISES 
 
 Suggestions. — Prove that a circle can be inscribed 
 in FGHKL. 
 
 Ex. 34. — In the figure of Prop. VIII, Book V, f"^ ^ 
 prove that OM is the side of a regular pentagon in- \ l 
 scribed in the circle which can be circumscribed about 
 /\OBM. 
 
 Suggestion. — How large is Z OBM ? 
 
 Ex. 35. Construct a regular pentagon having given one of its sides. 
 
 Ex. 36. Construct a regular pentagon having given one of its diago- 
 nals. 
 
 Ex. 37. If B represents the radius of the circle circumscribed about 
 a regular decagon, prove : 
 
 (a) sio=_(V5-l); (c) pio = 5i?(\/5 - 1) ; 
 
 (6) aio= — Vl0 + 2\/5; (d) A^io = ^^ VlO - 2 V5. 
 
 4 4 
 
 Ex. 38. Find the area of the circle inscribed in a square whose area 
 is 25. 
 
 Ex. 39. If the radius of a circle is SVS, what is the area of the sector 
 whose central angle is 150° ? 
 
 Ex. 40. Find the radius of the circle equal to a square whose side is 
 10. 
 
 Ex. 41. Find the radius of the circle whose area is one half the area 
 of the circle whose radius is 15. 
 
 Ex. 42. Find the area of the square inscribed in the circle whose area 
 is 196 IT sq. in. 
 
 Ex. 43. The area of one circle is ^ the area of another. Find the 
 radius of the second if the area of the first is 15. 
 
 Ex. 44. The side of a square is 8. Find the circumference of its in- 
 scribed and circumscribed circles. 
 
 Ex. 45. The side of an equilateral triangle is 6. Find the area of its 
 inscribed and circumscribed circles. 
 
 Ex. 46. The area of a regular hexagon inscribed in a circle is 24 VS. 
 What is the area of the circle ? 
 
 Ex. 47. If the apothem of a regular hexagon is 6, what is the area of 
 its circumscribed circle ? 
 
SUPPLEMENTARY EXERCISES 303 
 
 Ez. 48. Two plots of ground, one a square and one a circle, each 
 contain 70686 sq. ft. How much greater is the perimeter of the square 
 than the length of the circle ? 
 
 Ex. 49. The perimeter of a regular hexagon circumscribed about a 
 circle is 12 VS. What is the circumference of the circle ? 
 
 Ex. 50. The length of the arc subtended by the side of a regular in- 
 scribed 12-gon 18 -IT in. What is the area of the circle ? 
 ° 3 
 
 Ex. 51. If the length of a quadrant is 1, what is the diameter of the 
 circle ? 
 
 Ex. 52. Prove that the area of the square inscribed in 
 a sector whose central angle is a right angle, is equal to one 
 half the square on the radius. 
 
 Ex. 53. If a circle is circumscribed about a right 
 triangle, and on each of the legs of the triangle as 
 diameters semicircles are drawn, exterior to the 
 triangle, the sum of the areas of the crescents thus 
 formed equals the area of the triangle. 
 
 Prove area AECG + area BFCH= area A ABC. 
 
 Suggestion. —From the sum of A ABC and the semicircles on ^Cand BC, 
 subtract the semicircle on AB. Express each area in terms of sides a, b, and 
 c of the triangle. 
 
 Ex. 54. Construct three equal circles having the vertices of an equi- 
 lateral triangle as their centers and for their radii one half the side of the 
 triangle. Compute the area of that part of the interior of the triangle 
 which is exterior to each of the circles, if the length of the side of the 
 triangle is s. 
 
 Ex. 55. Upon a segment AC draw a semicircle. Upon AC locate a 
 point B, not the center of AC. Upon AB and BC a,s diameters draw 
 semicircles within the one drawn upon ^C as diameter. Prove that the 
 area of the surface lying within the largest semicircle and exterior to the 
 smaller ones equals the area of the circle drawn upon BD as diameter, 
 where BD is the perpendicular to AC at B meeting the largest semicircle 
 at D. (Due to Archimedes.) 
 
 Ex. 56. With the vertices of an equilateral triangle as centers and 
 the side of the triangle as radius, three equal circles are drawn. Deter- 
 mine the area of that figure which is common to the three circles. 
 
304 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 57. Express in terms of the radius B the area of the segment of 
 a circle whose chord is a side of the inscribed square. 
 
 Ex. 58. Repeat Ex. 67 if the chord is the side of the equilateral 
 inscribed triangle. 
 
 Ex. 59. The arch ABC is a lancet arch. It consists of two arcs with 
 equal radii, drawn from centers Ci and d out- 
 side the span BG. Within the arch are two 
 other lancet arches. 
 
 Let JSO = 2 a ; let CiCa = s ; let J5Z) = BC. 
 
 (a) Determine the height h of the arch BAC. 
 
 (b) What is the length of the radius of the arc 
 XD? 
 
 (c) What is the height of the arch BXD ? 
 
 (d) What is the radius of the circle indicated as tangent to the arches ? 
 
 (e) What is the area and circumference of the circle ? 
 
 Ex. 60. In a given circle, inscribe three equal 
 circles, tangent to each other and to the given 
 circle. 
 
 Ex. 61. In a given equilateral triangle, in- 
 scribe three equal circles, tangent to each other 
 and each tangent to one and only one side of the 
 triangle. 
 
 Ex. 62. The figure below at the left is a quatrefoil. 
 
 (a) Construct such a figure based upon a square whose side is 2 in. 
 (6) What is the length of the curved line if AB = s inches ? 
 
SUPPLEMENTARY EXERCISES 
 
 305 
 
 (c) What is the area within the curved line if AB = s inches ? 
 
 (d) Notice that the quatrefoil is used in the adjoining design. 
 
 Ex. 63. Construct a figure like Fig. 1, below, upon a square of side 
 2 in. 
 
 Fig. 1 
 
 Fig. 2 
 
 (a) What is the length of the curved line when the side of the square 
 is s inches ? 
 
 (b) What is the total area within the curved line when the side of the 
 square is s inches ? 
 
 (c) Notice that the curved line of Fig. 1 is the fundamental unit of the 
 adjoining window design. 
 
SOLID GEOMETRY 
 
 BOOK VI 
 
 LINES AND PLANES — POL YEDKAL 
 ANGLES 
 
 442. Surfaces. No satisfactory elementary definition of 
 surface in general can be given. The surface of a physical 
 object is that part of the object which can, in general, be 
 touched; it separates the portion of space occupied by the 
 object from surrounding space. 
 
 The surface of a small pond on a calm day is approximately a, plane 
 surface. 
 
 The surface of a croquet ball or of a billiard ball is a spherical surface. 
 The surface of a " round " marble column is a cylindrical surface. 
 
 Ex. 1. If two points on the surface of a ball were joined by a straight 
 line, where would the line lie ? 
 
 Ex. 2. Are there any two points on the surface of a ball such that the 
 straight line through them lies upon the surface of the ball ? 
 
 Ex. 3. Are there two points on the surface of a cylindrical column 
 such that the straight line joining them lies on the surface of the column ? 
 
 Ex. 4. Does the straight line joining every pair of points on the sur- 
 face of a cylindrical column lie upon the surface of the column ? 
 
 443. A Plane is a surface such that the straight line join- 
 ing any two points of it lies wholly in it. 
 
 A plane is represented to the eye by a quadri- D 
 
 lateral like the figure adjoining. / 
 
 The plane may be referred to as plane ABCD, /, 
 
 as plane AC, or as plane M. A^ — 
 
 307 
 
308 SOLID GEOMETRY — BOOK VI 
 
 Pupils will find it convenient at times to represent a plane by a thin 
 card. 
 
 Ex. 5. In plane geometry, it was agreed that a straight line is in- 
 definite in extent. Do you think we should agree that a plane is indefinite 
 in extent ? Why ? 
 
 Ex. 6. Let MN represent a thin card of which AB is an edge. Let 
 C represent the " point " of a pencil lying above 
 MN. Let MN be turned about AB as an axis in 
 
 the direction indicated by the arrow\ ^"^^ q 
 
 A 
 
 Will the card eventually come in contact with / 
 
 point C? / 
 
 If AB and C are kept stationary, will the card B ' — ^ 
 
 come in contact with C in more than one position of the card ? 
 
 444. A plane is determined by a combination of points and 
 lines if it is the only plane which contains those points and 
 lines. 
 
 Points and lines lying in the same plane are said to be 
 Co-planar. 
 
 445. Postulate. A plane can he extended indefinitely. 
 
 446. Axiom. A plane is determined by three non-collinear* 
 points. 
 
 Ex. 7. Do two straight lines drawn at random necessarily lie in a 
 plane ? Illustrate by holding two pencils. 
 
 Ex. 8. Do four points usually lie in a plane ? Select four in the school- 
 room that do and four that do not. 
 
 Ex. 9. Are two straight lines in space which do not meet no matter 
 how far they are extended parallel ? 
 
 Ex. 10. Why is a tripod used as mounting for a camera or a sur- 
 veyor's instrument ? 
 
 Ex. 11. Why does a stool with three legs stand firmly whereas one 
 with four legs cannot always be made to stand firmly ? 
 
 Ex. 12. Prove that a plane and a straight line not lying in the plane 
 can have only one common point. 
 
 * Nou-collinear points are points which do not all lie in one straight line. 
 
LINES AND PLANES 309 
 
 Proposition I. Theorem 
 
 447. A plane is determined by 
 
 I. A straight line and a point outside the line, 
 
 II. Two intersecting straight lilies. 
 
 III. Two parallel straight lines. 
 
 1. Hypothesis. Point C lies outside st. line AB. 
 Conclusion. C and AB determine a plane. 
 
 Proof. 1. Points A, B, and C lie in one and only one 
 
 plane. § 446 
 
 2. AB lies in that plane. § 443 
 
 3. .*. AB and C determine a plane. § 444 
 
 II. Hypothesis. AB and BO are intersecting st. lines. 
 
 Conclusion. AB and BC determine a plane. 
 
 Proof. 1. AB and point C determine a plane. § 447, I 
 
 2. BC lies in that plane. § 443 
 
 3. .*. AB and BC determine a plane. 
 
 III. Hypothesis. AB and CD are parallel straight lines. 
 
 Conclusion. AB and CD determine a plane. 
 
 Proof. 1. AB and CD lie in a plane. § 89 
 
 2. AB and CD cannot lie in more than one plane, for, if they 
 did, points A, B, and C would lie in more than one plane, which 
 is impossible. Why ? 
 
 3. .-. AB and CD determine a plane. 
 
310 
 
 SOLID GEOMETRY — BOOK VI 
 
 448. The Intersection of two surfaces or of a surface and a 
 line consists of all points common to the surfaces, or to the 
 surface and. the line. 
 
 449. If a straight line intersects a plane, the point of inter- 
 section of the line and the plane is called the Foot of the line. 
 
 450. Axiom. If two planes intersect, they have at least two 
 common points. 
 
 Proposition II. Theorem 
 451. The intersection of tioo planes is a straight line. 
 
 Hypothesis. A and B are two points common to planes MN 
 and PQ. 
 
 Conclusion. The intersection of MN and PQ is a straight 
 line. 
 
 Proof. 1. Draw straight line AB. 
 
 2. AB lies in plane MN and also in plane PQ. Why ? 
 
 3. No point outside AB can be in both MN and PQ, for, if 
 there were, MN and PQ would coincide. § 447, I 
 
 4. Hence the complete intersection of MN and PQ is the 
 straight line AB. ' § 448 
 
 452. A line is perpendicular to a plane if it is perpendicular 
 
 to every line in the plane passing through its foot. 
 The plane is also perpendicular to the line. 
 
LINES AND PLANES 
 
 311 
 
 Proposition III. Theorem 
 
 453. If a line is per2)endicular to each of two inter- 
 secting lines at their intersection, it is perpendicular to 
 their plaiie. 
 
 Hypothesis. AD and AC intersect at A^ determining plane 
 RS. 
 BALAD^xi^BAl^AC. 
 Conclusion. BA ± plane RS. 
 
 Proof. 1. Let AE be any other straight line in R8 through A. 
 Let DQ be a straight line in R8 intersecting AD^ AE, and 
 AC, at Z), E, and (7, respectively. 
 
 2. Extend Bx\ to B\ making B'A = BA. 
 Draw BD, BE, BC, B'D, B'E, and B'C. 
 
 3. AD and AC are ± bisectors of BB'. Why ? 
 
 4. .-. ^Z) = i^'i> and BC = B'C. Prove it. 
 6. .-. A iJZ>C ^ A 5'Z)a Prove it. 
 
 6. Revolve A B'DC on DC as axis until ^' falls on B. 
 
 7. Then B'E = BE. Why ? 
 
 8. .'.A and JE7 are both equidistant from B and B'. 
 
 9. .-. ^^ ± BB', or 5B' ± ^^. § 77 
 
 10. But ^£7 is any st. line in iJ-S' through A. 
 
 11. .*. ^B ± every st. line in RS through A, and hence 
 
 AB± plane RS. 
 
 §452 
 
 Ex. 13. If a line is perpendicular to a line of a plane, is it perpendic- 
 ular to the plane ? 
 
312 
 
 SOLID GEOMETRY — BOOK VI 
 
 454. Cor. 1. Through a point of a 
 
 line a plane can he drawn perpendicular 
 
 to the line. 
 
 Suggestion. — Draw BC and BD, any two per- 
 pendiculars to AB. 
 
 Note. — Through a point of a line only one plane can he drawn per- 
 pendicular to the line. If ME and BE were both ± 
 to AB at J5, a plane AD through AB would intersect 
 ME and BE in two lines BF and jBZ>, each ± to ABT 
 at B. But this is impossible, for, in a plane {AB)., 
 only one line can be drawn perpendicular to a given 
 line at a point of the line. 
 
 455. Cor. 2. Through a point outside a line, ^ 
 a plane can he drawn perpendicular to the line. (See Fig. § 454.) 
 
 Suggestion. — Draw CB 1 AB from C; then draw BD 1 AB at D. 
 
 Note. — Through a point outside a line, only one 
 plane can he drawn perpendicular to the line. If 
 planes BT and 8T through C were both ± to AB., 
 the plane ABC determined by AB and C, would 
 intersect BT and ST in lines XC and YC, each ± 
 to AB. But this is impossible, for, in a plane., only 
 one line can be drawn perpendicular to a given line 
 from a point outside the line. 
 
 456. Cor. 3. At a point in a plane, a straight line can he 
 drawn perpendicular to the plane. 
 
 Construction. 1. Draw CD any line in 
 iifiNT through O. 
 
 2. Draw plane BB ± to CD at O, meet- 
 ing MN in line AB. 
 
 3. Draw EO in plane BB, ± to AB at O. 
 
 Statement. EO ± plane MN at 0. 
 
 Note. — At a point of a plane, only one straight line 
 can be drawn perpendicular to the plane. If EO and 
 E'O were both ± to plane MN at O, they would de- 
 termine a plane BB which would intersect MN in 
 line AB. EO and E'O would both be ± to AB at 0, 
 and that is impossible. Why ? 
 
 I 
 
LINES AND PLANES 
 
 313 
 
 457. Cor. 4. Any point in the plane 
 ichich is perpendicular to a segment at its 
 mid-point is equidistant from the ends of 
 the segment. 
 
 Ex. 14. Each of three concurrent lines is 
 perpendicular to each of the other two. Prove 
 that each is perpendicular to the plane of the other two. 
 
 Ex. 15. If two oblique lines, drawn to a plane from a point in a 
 perpendicular to the plane, cut oflE equal distances from the foot of the 
 perpendicular, they are equal. 
 
 Ex. 16. State and prove the converse of Ex. 15. 
 
 Ex. 17. If two oblique lines, drawn to a plane 
 from a point in a perpendicular to the plane, cut off 
 unequal distances from the foot of the perpendicular, 
 the more remote is the greater. 
 
 If BE > BD, prove AE > AD. 
 Suggestion. — Take BF = BD, and draw AF. Recall § 165 
 
 Ex. 18. State and prove the converse of Ex. 17. 
 Suggestion. —Give an indirect proof, basing it upon Ex. 15 and 17. 
 
 Ex. 19. If a circle be drawn in a plane and at its center a perpen- 
 dicular to the plane be erected, any point in this perpendicular is equi- 
 distant from the points of the circle. 
 
 Ex. 20. A line segment of fixed length, having one extremity at a 
 fixed point lying outside a plane, has its other extremity in the plane. 
 What is the locus of the extremity which lies in the plane ? 
 
 Ex. 21. IIow many different planes can be passed through one straight 
 line '? 
 
 Ex. 22. How many different planes are determined by : 
 (a) Three concurrent lines which do not all lie in one plane ? 
 (6) Three parallel lines which do not all lie in one plane ? 
 (c) Two intersecting lines and a point which does not lie in their plane? 
 {d) Four points, no three of which are collinear and which do not all 
 lie in one plane ? 
 
 Ex. 23. Prove that two parallels and any transversal of them are 
 co-planar. 
 
 Ex. 24. How many lines of intersection are determined, in general, 
 by three planes ? 
 
314 
 
 SOLID GEOMETRY — BOOK VI 
 
 Proposition IV. Theorem 
 
 458. All the perpendiculars to a straight line at a 
 point of the line lie in a plane perpendicular to the line 
 at the point. 
 
 Hypothesis. AC, AD, and AE are any three Js to AB at A. 
 Conclusion. AC, AD, and AE lie in a plane A. to AB at A. 
 Proof. 1. Let AC and AD determine plane MN. 
 
 2. .-. xiB A. plane MN. Why ? 
 
 3. Let AB and AE determine plane ABE, intersecting MN 
 in AE\ 
 
 4. .-. AB ± AE, since AE^ is in MN. § 453 
 
 5. .-. AE and AE, both in plane BE, must coincide. 
 
 [/)i a plane, only one _L can be drawn to a line at a point in the line.] 
 
 §81 
 
 6. .-. AE must lie in MN. Why ? 
 
 7. Hence all Js to AB at J. must lie in MN. Why ? 
 
 3f. 
 
 459. Cor. 1. Any point equidistant from 
 the eyids of a seginent lies in the plane per- 
 pendicular to the segment at its midpoint. 
 
 460. Cor. 2. TJie locus of points in space 
 equidistant from the ends of a segment is the 
 plane perpendicular to the segment at its mid-point. 
 
 D 
 
 y-. 
 
 Suggestion.— Reyiew, if necessary, § 229 of the Plane Geometry and apply 
 §§ 457 and 459. 
 
LINES AND PLANES 
 
 315 
 
 Proposition V. Theorem 
 
 461. If through the foot of a perpendicular to a 
 plane a line he drawn at right angles to any line in 
 the plane, the line drawn from its intersection with this 
 line to any point in the perjyendicidar will he perpen- 
 dicular to the line in the plane. 
 
 Hypothesis. AB ± plane MN\ CD is any line in MN\ 
 AE J_ CD ; BE is drawn from any point B of AB to E. 
 Conclusion. BE A. CD. 
 
 Suggestions. — 1. Take CE = DE, and draw BD, BC, AD, and AC. 
 2. Compare ^C and ^Z). 3. Compare 5Z) and 56'. 
 
 462. Cor. 1. From a point outside a 
 plane, a straight line can^be drawn per- 
 pendicular to the plane. 
 
 Construction. 1. Draw DE, any st. line in 
 MX. 
 
 2. Draw AF ± to DE at F, and BF, in 
 Jf iV, ± to DE at F. 
 
 3. Draw AB ± to BF. 
 Statement. AB ± plane MN from A. 
 Proof. 1. Draw BE. 
 
 2. EF ± the plane determined by AF and BF. Why ? 
 
 3. .'.BE±AB. §461 
 [Since BF, through the foot of EF, is ± to AB in plane ABF.^ 
 
 4. .•.AB±3IN. 
 
 [See step 3 of the Construction and of the Proof.] 
 
 Note. — From a point outside a plane only one straight line can he 
 drawn perpendicular to the plane. If AC and AB were both ± to plane 
 J/iVfrom A, A ABC would have two right angles in it. 
 
316 SOLID GEOMETRY — BOOK VI 
 
 463. Cor. 2. The perpendicular is the shortest segmerit that 
 can be drawn from a point to a plane. 
 
 464. The Distance from a point to a plane is the length of 
 the perpendicular from the point to the plane. 
 
 PARALLEL LINES AND PLANES 
 
 465. A straight line is parallel to a plane if it does not meet 
 the plane however far they are extended. 
 
 Two planes are parallel if they do not meet however far 
 they are extended. 
 
 Proposition VI. Theorem 
 
 466. If a line outside a plane is parallel to a line of 
 ^ the plane, it is parallel to the plane. 
 
 Hypothesis. AB II CD. 
 
 Plane MN contains CD but not AB. 
 Conclusion. AB II plane MN. 
 
 Proof. 1. AB and CD lie in a plane AD. § 89 
 
 2. This plane intersects MN in line CD. Why ? 
 
 3. If AB were to intersect MN, the point of intersection 
 would be in plane MN and also in plane AD, and therefore 
 in CD.   
 
 4. Hence AB would intersect CD. 
 
 5. But, AB cannot meet CD. Why ? 
 
 6. .*. AB cannot meet JOT" and hence AB II MW. 
 
PARALLEL LINES AND PLANES 
 
 317 
 
 467. Cor. 1. TJirough a given straight line, a plane can be 
 drawn parallel to any other straight 
 line. 
 
 Prove a plane can be drawn through 
 CD II AB. 
 
 Suggestion. — Bra.w CE \\ AB. 
 
 Note. — Discuss the solution when 
 AB II CD and when AB is not II CD. 
 
 
 A 
 
 B 
 
 
 M 
 
 
 
 ^C 
 
 T1 / 
 
 / 
 
 / '^ 
 
 ^^^^ / 
 
 
 
 N 
 
 468. Cor. 2. Through a given point, a plane can he drawn 
 parallel to each of two given straight 
 lines in space. 
 
 Prove a plane can be drawn through 
 point A, parallel to straight lines BC 
 and DE. 
 
 Suggestion. — DTa\r AG \\ DE and AF 
 BC. 
 
 B- 
 
 D\ 
 
 c 
 
 M 
 
 
 ^^E 
 
 A ^ / 
 
 y 
 
 \ 
 
 \(? / 
 
 
 
 N 
 
 Note. — Discuss the solution when BC II 2)^ and when BC is not II DE. 
 
 Ex. 25. Can more than one perpendicular be drawn to a line at a 
 point of the line in a plane ? in space ? 
 
 Ex. 26. Through a given point outside a line, a plane can be drawn 
 parallel to the line. How many such planes can be drawn ? 
 
 Ex, 27. I*rove that a straight line and a plane, 
 both perpendicular to the same straight line, are 
 parallel. 
 
 AB±AC; plane MN'± A C. 
 
 Hyp. 
 Con. 
 
 AB II plane MN. 
 
 7 
 
 Suggestion. — Let the plane determined by AB and 
 AC intersect MN in line DE. 
 
 Ex. 28. Through a given point outside a plane, a line can be drawn 
 parallel to the plane. Can more than one line be drawn parallel to the 
 plane ? 
 
 Ex. 29. Through a given point, a line can be drawn parallel to each 
 of two intersecting planes. 
 
318 SOLID GEOMETRY — BOOK VI 
 
 Pkoposition VII. Theorem 
 
 469. If a straight line is parallel to a plane, the 
 intersection of the plane tvith any plane draivn through 
 the line is parallel to the line. 
 
 Hypothesis. AB li plane MJSf. 
 
 Plane BCj through AB, intersects JfiV in CD. 
 
 Conclusion. AB II CD. 
 
 Proof. 1. AB and CD lie in the same plane BC. 
 
 2. AB and CD cannot intersect, for if they did, AB would 
 intersect plane MN, which is impossible. 
 
 3. .-. AB II CD. § 89 
 
 470. Cor. If a Uyie and a plane are parallel, a parallel to the 
 line through any point of the plane lies in the plane. 
 
 Hyp. AB II plane MN. 
 
 C is any pt. in MN. -^I 1^ 
 
 CI) II AB. / / 
 
 Con. CD lies in MN. i + — ^/ 7^" 
 
 Proof. 1. The plane determined by AB and / ^ ^ / 
 
 G intersects MN in a line CE., through C, -^^ 
 parallel to AB. Why ? 
 
 2. But CD, through O, II AB. Why ? 
 
 3. .'.CD and CE coincide. Why ? 
 
 4. .'. CD lies in plane MN. 
 
 Ex. 30. Prove that three non-concurrent straight lines, each of which 
 intersects the other two, lie in a plane. 
 
. PARALLEL LINES AND PLANES 319 
 
 Proposition VIII. Theorem 
 
 471. If two parallel planes are cut hy a third plane, 
 the intersections are parallel. 
 
 31 
 
 n 
 
 
 // 
 
 ^\ 
 
 ^ 
 
 P ^ 
 
 \ 
 
 V 
 
 
 
 c 
 
 Q 
 
 Hypothesis. Plane MN li plane PQ. 
 
 Plane AD intersects MN in AB and PQ in CD. 
 
 Conclusion. AB II CD. 
 
 Suggestion. — Recall § 89. 
 
 Ex. 31. Prove that parallel segments between parallel planes are equal. 
 
 Ex. 32. If two planes are parallel, a line parallel to one of them 
 through any point of the other lies in the other. (Fig. of Prop. VIII.) 
 
 Suggestion. — Given parallel planes MN and P Q, and AB through any 
 point A of MN \\ PQ. Prove that^J? lies in MN. Through AB pass a plane 
 intersecting MN in a line AB' and PQ in line CD. Consider the relation of 
 lines AB' and CD, and also of AB and CD. 
 
 Ex. 33. From a line parallel to a plane, two parallels are drawn to 
 the plane and terminated by the plane. Prove that the segments are equal. 
 
 Review Exercises 
 
 Ex. 34. When is a line perpendicular to a plane ? 
 
 Ex. 35. Where do all lines lie which are perpendicular to a given 
 line at a given point of the line ? 
 
 Ex. 36. What is true about two lines in the same plane which are 
 perpendicular to the same line ? 
 
 Ex. 37. Line AB is perpendicular to plane iHf.V at A. A line is 
 drawn from B meeting any line CB of plane MN at E. If line BE is 
 perpendicular to CD, prove AE perpendicular to CD. (Fig. Prop. V.) 
 
320 
 
 SOLID GEOMETRY — BOOK VI 
 
 Peopositiois' IX. Theorem 
 
 472. Two lines perpendicular to the same plane are 
 parallel. 
 
 A 
 
 \ 
 
 G 
 
 M 
 
 
 
 
 
 / / 
 
 B 
 
 / 
 
 D / 
 
 Hypothesis. AB^MN^\.B', CDI.MN2XD. 
 Conclusion. AB II CD. 
 
 Proof. 1. Draw AD from any pt. A of AB. Draw BD. 
 In plane MN, draw FD ± BD. 
 
 2. CD1.FD. Why? 
 
 3. AD ± FD. § 461 
 
 4. .-. AD, BD, and CD lie in a plane ABDG. § 458 
 
 5. .'. AB and CD lie in the same plane, ABDC. 
 
 6. But AB A. BD and CD ± BD. Why ? 
 
 7. .*. ^J5 II CD. § 97 
 
 473. Cor. 1. If one of two parallel lines is perpendicular to a 
 plane, the other is also perpendicular to the plane. 
 
 A O 
 
 Hyp. 
 
 Con. 
 
 AB II CD. 
 
 AB ± plane MJST. 
 
 CD ± MN. 
 
 Suggestions. — 1. Assume CE 1 MN. 
 2. Prove CE \\AB. 
 
 474. Cor. 2. If each of two straight 
 lines is parallel to a third straight line, 
 they are parallel to each other. 
 
 Hyp. ^J5II CD; EFW CD. 
 
 Con. AB II EF. 
 
 Suggestion. — Let plane MNhe 1 CD. 
 
 I \ /I / ^ 
 
 / BED/ 
 M'- ' 
 
PARALLEL LINES AND PLANES 321 
 
 Ex. 38. Through a given line which is parallel to a given plane, a 
 number of planes are passed intersecting the given plane. Prove that 
 the lines of intersection with the given plane are all parallel. 
 
 Ex. 39. If two parallel planes intersect two other parallel planes, the 
 four lines of intersection are parallel. 
 
 Ex. 40. Prove that a line parallel to a plane is everywhere equidis- 
 tant from it. 
 
 Ex. 41, If two points are equidistant from a plane, and on the same 
 side of it, they determine a line parallel to the plane. 
 
 Ex. 42. If two points lie on opposite sides of a plane and equidistant 
 from it, the segment joining them is bisected by the plane. 
 
 Ex. 43. If one of two parallel lines is parallel to a plane, the other is 
 also, unless it lies in the plane. 
 
 Suggestion. — Through the line which is 1| to the plane, pass a plane inter- 
 secting the given plane. Use § 466. 
 
 Ex. 44. Prove that the lines joining in order the mid-points of the 
 sides of a quadrilateral in space form a parallelogram. 
 
 Proposition X. Theorem 
 
 475. Two planes perpendicular to the same straight 
 line are parallel. 
 
 My 
 
 N 
 
 Hypothesis. Planes MN and PQ are _L to AB, 
 Conclusion. MN \\ PQ. 
 
 Suggestion. — Prove it by the indirect method, using § 455. 
 
 Ex. 45. Are lines in space which are perpendicular to the same line 
 necessarily parallel ? 
 
322 
 
 SOLID GEOMETRY — BOOK VI 
 
 Propositiois' XI. Theorem 
 
 476. If each of tivo intersecting lines is j^ctrallel to a 
 plancj their plane is pai^allel to the given plane. 
 
 Hypothesis. AB II plane PQ ; AC 11 plane PQ, 
 AB and AO determine plane MN. 
 Conclusion. MN II PQ. 
 
 Proof. 1. . From A draw AD ± PQ. 
 
 2. AC and AD determine a plane which intersects PQ in a 
 line DF parallel to AC. Why ? 
 
 3. Similarly, DE II AB. 
 
 4. AD A. DF and also AD ± DE. Why ? 
 
 5. .-. ^Z> ± AC and also J.Z> ± AB. Why ? 
 Complete the proof, using § 475. 
 
 Proposition^ XII. Theorem 
 
 477. A straight line perpendicular to one of tioo 
 parallel planes is pferpendicular to the other also. 
 
 Hypothesis. Plane MN II plane PQ. AD A. PQ. 
 
 (Fig. § 476.) 
 Conclusion. AD A. MN. 
 
 Proof. 1. Through AD pass two planes intersecting MN 
 in AB and ACy and PQ in DE and DF, respectively. 
 
 2. .-. AB II DE and AC II J9i<^. Why ? 
 
 3. ^D J. i)£; and also AD ± Di^. Why ? 
 
 • Complete the proof. 
 
/ 
 
 -/ 
 
 
 
 JS 
 
 p 
 
 
 
 / 
 
 ' / 
 
 
 
 y 
 
 PARALLEL LINES AND PLANES 323 
 
 478. The distance between two parallel planes is the length 
 of the segment perpendicular to them and 
 lying between them. 
 
 479. Cor. 1. Two parallel planes are 
 everywhere equally distant. 
 
 480. Cor. 2. Through a given point, a 
 plane can be drawn parallel to a given plane. 
 
 Note. — TTirough a given point only one plane can he draion parallel 
 to a given plane. For, if two planes through A were parallel to PQ, 
 each would be ± to AB at A, and that is impossible. 
 
 Proposition XIII. Theorem 
 
 481. If two angles not in the same plane have their 
 sides parallel and extending in the same direction, they 
 are equal, and their planes are parallel. 
 
 M 
 
 
 / iH\ 
 
 / 
 
 / ^iT" 
 
 1. / 
 
 
 .V 
 
 A \ 
 
 \ 
 
 /\^ 
 
 -^\ / 
 
 / y-^ 
 
 ^c'X 
 
 
 Q 
 
 Hypothesis. Z BACis in plane MN; Z B'A!C' is in plane PQ. 
 
 AB II A'B^ and extends in the same direction. 
 
 AC II A'C and extends in the same direction. 
 Conclusion. Z BAC = Z B'A'C ; MN II PQ. 
 Proof. 1. AB II plane PQ and AC I! PQ. § 466 
 
 2. .-. MN II PQ. 
 
 3. Lay off AB = AB', and AC = A'C ; draw BC, B'C, 
 AA', BB', and GO. 
 
 4. ABB:A' i^d. O, QXidi. '. BBf ^AA^iTidiBB' WAA'. Why? 
 6. Similarly, CC = AA and CC II ^^.' 
 
 6. .-. BB'C'C is a O. Prove it. 
 
 Complete the proof. 
 
324 
 
 SOLID GEOMETRY — BOOK VI 
 
 Proposition XIV. Theorem 
 
 482. If two straight lines are cut by three or more 
 parallel planes, the corresponding segments are pro- 
 portional. 
 
 Hypothesis. Planes MN, PQ, and RS are parallel. 
 AC intersects the planes at A, B, and C respectively. 
 A^C intersects the planes at A\ B', and C respectively. 
 
 Conclusion. 4^ = 4^. 
 
 BC B'C 
 
 Proof. 1. Draw AC cutting PQ at D, 
 
 2. Plane CAC intersects PQ at BD and RS at CC 
 Plane AC A' intersects PQ at DB' and MN Sit AA'. 
 
 3. .-. BD II CC and also BB' \\ AA', Why ? 
 Complete the proof, using § 265. 
 
 Ex. 46. Discuss Prop. XIII : (a) if BA and B'A' extend in opposite 
 directions and CA and C'A.' in the same direction ; 
 (&) if both pairs extend in opposite directions from 
 their vertices. 
 
 Ex. 47. If each of two intersecting planes be 
 cut by two parallel planes, not parallel to their in- 
 tersection, their intersections with the parallel 
 planes include equal angles. 
 
 Suggestion. — FroYB Z ABC= Z DBF. 
 
 Ex. 48. If two planes are parallel to a third plane, they are parallel 
 to each other. (§ 477 and § 475.) 
 
DIEDRAL ANGLES 
 
 325 
 
 ZDZ7 
 
 DIEDRAL ANGLES 
 
 483. It is evident that a straight line divides a plane into 
 two parts, each indefinite in extent. A ^ 
 
 part of the plane, like BCD, is called a 
 Half-plane. BG is called the Edge of the ^, 
 
 half-plane. ^ ^ 
 
 484. A Siedral Angle is the figure formed by two half-planes 
 which have a common edge. 
 
 The common edge is the Edge and the two 
 half-planes are the Faces of the diedral angle. 
 
 Thus, half-planes BF and BD form the diedral angle 
 whose edge is BE, and whose faces are FBE and DBE. 
 
 The diedral angle may be read : 
 
 diedral Z BE, or diedral Z ABEC. 
 
 485. Two diedral angles are Adjacent when 
 they have the same edge and a common face be- 
 tween them : as Z ABEC and Z CBED. 
 
 Two diedral angles are Vertical when the 
 faces of one are the extensions of the faces of 
 the other. 
 
 486. A Plane Angle of a diedral angle is the angle formed 
 by two straight lines one in each plane, drawn perpendicula.r 
 to the edge at the same point. 
 
 Thus, if lines AB and ^O be drawn in faces DE and 
 DF respectively, of diedral angle DG, perpendicular 
 to DO at A, Z BAC is a plane angle of the diedral 
 angle DG. 
 
 487. Cor. 1. All p/ane angles of a given die- 
 dral angle are equal. 
 
 488. Cor. 2. A plane loerpendicular to the edge of a diedral 
 angle intersects the faces of the angle in lines which form the 
 plane angle of the diedral angle. 
 
 489. Two diedral angles are equal if they can be made to 
 coincide. 
 
326 
 
 SOLID GEOMETRY — BOOK VI 
 
 Proposition XY. Theorem 
 
 490. Tioo diedral angles are equal if their plane 
 angles are equal. 
 
 Hypothesis. Z ABC and Z A'B'C are the plane A of die- 
 dral A BD and B'D' respectively ; Z ABC = Z A'B'C 
 
 Conclusion. A BD = A B'D'. 
 
 Proof. 1. Apply A B'D' to Z BD so that A'B' will coincide 
 with AB and 5'(7' will coincide with BC. 
 
 2. 5i) ± plane ABC, and JB'D' ± plane A'B'C. Why? 
 
 3. .-. B'D' coincides with BD. Note, § 456 
 
 4. .-. ^'D' coincides with AD, and 0'i>' with CD. § 447, ll 
 
 5. .-. Z J5'Z>' = ABD. § 489 
 
 491. Cor. 1. i/* tico diedral angles are equal, their plane angles 
 are equal. 
 
 For the diedral angles can be made to coincide. Then a plane, per- 
 pendicular to the common edge, will intersect in each its plane angle, and 
 evidently these plane angles coincide. 
 
 492. Cor. 2. If two planes intersect, the vertical diedral angles 
 are equal. 
 
 Suggestion. — Compare their plane angles. 
 
 Ex. 49. Prove that a plane can be drawn bisecting a diedral angle. 
 
 493. A diedral angle is right, acute, or obtuse according as 
 its plane angle is right, acute, or obtuse. Two diedral angles 
 are supplementary or complementary according as their plane 
 angles are supplementary or complementary. 
 
DIEDRAL ANGLES 
 
 327 
 
 Ex. 50. If one plane meets another plane, the adjacent diedral angles 
 formed are supplementary. 
 
 Ex. 51. If two parallel planes are cut by a third plane, the alternate- 
 interior diedral angles are equal. 
 
 Suggestion. — Prove the plane A of the alt. -int. diedral A equal. 
 
 p 
 
 494. Two planes are perpendicular if the 
 
 diedral angles formed are right diedral 
 angles. 
 
 Proposition XVI. Theorem 
 
 495. If a straight line is perpendicular to a plane, 
 every plane drawn through the line is perpendicular to 
 the plane. 
 
 Hypothesis. AB _L plane MN. 
 
 PQ is any plane through AB. 
 
 Conclusion. PQ ± MN. 
 
 Proof. 1. Let PQ and MN intersect in line QR. 
 
 2. 
 
 3. 
 
 4. 
 
 6. 
 
 Draw C'BC in MN± QR at B. 
 
 AB±QR. Why? 
 
 .-. Z ABC is the plane angle of the diedral Z PQRN. Def. 
 But Z ABC is a rt. Z. Why ? 
 
 Complete the proof, recalling §§ 494 and 493. 
 
 Ex. 52. (a) Prove that a plane can be drawn through a point per- 
 pendicular to a given plane. 
 
 (b) How many such planes can be drawn ? 
 
 Ex. 53. Prove that a plane perpendicular to the edge of a diedral 
 angle is perpendicular to the faces of the angle. 
 
328 SOLID GEOMETRY — BOOK VI 
 
 Propositiois^ XVII. Theorem 
 
 496. If two planes are perpendicular to each other, a 
 straight line drawn in one of them perpendicular to 
 their intersection is perpendicular to the other. 
 
 M^ 
 
 C'A 
 
 Why? 
 Why? 
 
 Q N 
 
 Hypothesis. Plane PQ ± plane MN. PQ intersects MN 
 
 in QR. Line AB in PQ is ± QR. 
 Conclusion. AB ± MN. 
 
 Proof. 1. Draw OBC in plane MN 1. QR. 
 
 2. .'. Z ABC is the plane Z of diedral Z PQRN. 
 
 3. But ZPQi^ 2V^ is art. diedral Z. 
 
 Complete the proof. 
 
 497. Cor. 1. If two planes are perpendicular to each other, a 
 perpendicular to one of them at a7iy point of their intersection lies 
 in the other. 
 
 Hyp. Plane PQ ± plane MN, intersecting it in QB. 
 
 AB, drawn from any pt. B of QB, is ± to MN. 
 Con. AB lies in PQ. 
 
 Suggestions. — 1. A line in PQ 1 Q7? at £ is 1 MN. Why ? 
 
 2. Prove that it and AB coincide. 
 
 498. Cor. 2. If two planes are perpendicular to each other, a 
 perpendicular to one of them from any poi7it of the other lies, in 
 the other. 
 
 Hyp. Plane PQ ± plane MN, intersecting it in QB ; 
 
 AB, drawn from any pt. A of PQ, is ± to MN. 
 Con. AB must lie in PQ. 
 
 Suggestions. — In PQ draw a 1 to QR from A. Prove that AB must coin- 
 cide with this perpendicular. 
 
DIEDRAL ANGLES 
 
 329 
 
 Proposition XVIII. Theorem 
 
 499. A plane perpendicular to each of tivo intersect- 
 ing planes is perpendicular to their intersection. 
 
 Hypothesis. Plane MN A. planes RS and PQ. 
 RS intersects PQ in line AB. 
 
 Conclusion. MN ± AB. 
 
 Suggestion. — Assume a line 1 MN from A. Where will this line lie ? 
 
 (§498.) 
 
 Ex. 54. Are two planes which are perpendicular to the same plane 
 necessarily parallel ? 
 
 Ex. 55. If a plane is perpendicular to a line of a plane, it is perpen- 
 dicular to the plane. ^ 
 
 Ex. 56. If a straight line is parallel to a plane, 
 any plane perpendicular to the line is perpendicular 
 to the plane. 
 
 Hyp. AB II plane MN. 
 
 Plane PB ± AB at C. 
 
 Con. PB ± MN. 
 
 Suggestions. — 1. Draw line CD in PR 1 QR. 
 
 2. Let the plane determined by CB and CD intersect MNin line DE. 
 
 3. Prove CD 1 MN. 
 
 Ex. 57. Prove that any point in the 
 plane through the bisector of an angle and 
 perpendicular to the plane of the angle is 
 equidistant from the sides of the angle. j^- 
 
 Suggestions. — I. Draw Pi? 1 OC, RT 1 OA, and RS L OB. Draw PT 
 and PS. 
 
 2. Prove PR 1 MN (§ 496), PT 1 OA (§ 461), PS 1 OB. 
 
330 
 
 SOLID GEOMETRY — BOOK VI 
 
 Pkoposition XIX. Theorem 
 
 500. Every point in the plane hisecting a diedral 
 angle is equidistant from the faces of the angle. 
 
 Hypothesis. Plane BE bisects diedral Z ABDC. 
 P is any point in plane BE. 
 PM ± plane AD', PN ± plane DC. 
 Conclusion. PM = PN. 
 
 Proof. 1. The plane determined by PM and PN intersects 
 planes AD, BE, and CD in lines FM, FP, and FN respectively. 
 
 2. Plane PMFJSf ± AD and also ± CD. Why ? 
 
 3. .-. PMFN ± BD. Why ? 
 
 4. .-. APFMsiiid PEN are the plane A of diedral AABDE 
 and CBDE. Why ? 
 
 5. .'.APFM=APFN. Why? 
 
 Complete the proof. 
 
 501. Cor. Any point loithin a diedral angle and equidistant 
 from its faces lies in the plane bisecting the diedral angle. 
 
 Suggestions. — 1. Let BE be the plane determined by P and BD. 
 2. Prove that BE bisects Z ABDC by proving A PFM and PEN are the 
 plane angles of the diedral angles and are equal. 
 
 Ex. 58. If perpendiculars are drawn to the faces 
 of a diedral angle from any point within the angle, 
 they lie in a plane perpendicular to the edge of the 
 diedral angle and form an angle which is the supple- 
 ment of the plane angle of the diedral angle. 
 
 Suggestion. — Wha,t is the sum of the angles of a 
 plane quadrilateral? 
 
DIEDRAL ANGLES 
 
 331 
 
 Proposition XX. Theorem 
 
 502. Tlirough a given straight line not jjerpendicular 
 to a given 2^lcine, one and only one plane can he drawn 
 peypendicular to the given plane. 
 
 AB is not ± to plane MN. 
 A plane can be drawn tlirough AB ± MN, and 
 
 Hypothesis. 
 Conclusion. 
 
 only one. 
 
 Proof. 1. Draw AC 1. MN, from point A. 
 
 2. AC and AB determine a plane ± MN. Why ? 
 
 3. If a second plane through AB were J_ MN, their inter- 
 section AB would also be X MN. Why ? 
 
 4. But AB is not ± MN. 
 
 5. .*. Only one plane can be drawn through AB A. MN. 
 
 Note 1. — If AB lies in MN, the theorem is still true. 
 Note 2. — If AB ± MN, an infinity of planes can be drawn through 
 AB±MN{^495). 
 
 503. The projection of a point on a plane is the foot of the 
 perpendicular drawn from the point to the 
 plane. 
 
 The projection of a given line on a plane 
 is the line which contains the projections 
 of all the points of the given line. 
 
 Thus, A'B'C is the projection of ABC on MN. 
 
 504. Cor. The projection of a straight line on a plane is a 
 straight line. (Fig. § 502.) 
 
 Suggestions. — 1. Through AB pass a plane AD 1 MN. 
 2. Prove that the feet of the J2 to MN from AB lie in CD. 
 
 (§ 498.) 
 
332 SOLID GEOMETRY — BOOK VI 
 
 Proposition XXI. Theorem 
 
 505. The acute angle between a straight line and its 
 projection on a plane is the least angle ivhich it makes 
 ivith any line drawn in the plane through its foot. 
 
 Hypothesis. BC is the projection of AB on plane MN. 
 
 BD is any other line in MN through B. 
 Conclnsion. Z ABC < A ABB. 
 
 Suggestions. — 1. Take -BZ) = £C. 2. Compare J. Z) and ^C. 
 3. Then compare A ABC and ABD, recalling § 167. 
 
 506. The angle between a line and a plane is the acute angle 
 made by the line with its projection on the plane. This angle 
 is called the Inclination of the line to the plane. 
 
 Ex. 59. If two equal segments are drawn to a plane from a 
 point outside the plane, they make equal angles with the plane. 
 
 Ex. 60. If two parallels meet a plane, they make 
 equal angles with it. 
 
 Suggestion. — Giwen AB || CD; A A' 1 JfiV, and 
 CC 1 'm2^. Prove Z ABA' = ACDC'. 
 
 Ex. 61. Prove that a straight line and its projection upon a plane lie 
 in a plane which is perpendicular to the given plane. 
 
 Ex. 62. If a straight line-segment is parallel to a plane, it is parallel 
 to its projection upon the plane, and is equal to it. 
 
 Ex. 63. If two parallel lines are oblique to a plane, their projections 
 upon the plane are parallel. (§ 481 and § 471.) 
 
 Ex. 64. Prove that the ratio of two parallel line-segments is the same 
 as the ratio of their projections upon a given plane. 
 
 u 
 
 K 
 
 /f 
 
 // 
 
 >/. 
 
 Jc/ 
 
 / ^ 
 / ^ 
 
 D 
 
 / 
 
 
 
 -^N 
 
POLYEDRAL ANGLES 
 
 333 
 
 Ex. 65. Can the projection upon a plane of a curved or broken line 
 be a straight line ? 
 
 Ex. 66. If the projection upon a plane of a given figure is a straight 
 line, then the figure lies in a plane. 
 
 Ex. 67. Let the projection upon a given plane M of a segment I be de- 
 noted by v. What is tlie relation between I and V if : 
 
 (a) Z ± itf ? {h) IWM? (c) I and Jf form an angle of 45° ? 
 
 Note. — Supplementary Exercises 1-17, p. 454, can be studied now. 
 
 POLYEDRAL ANGLES 
 507. The figures below represent polyedral angles. 
 
 
 B C 
 
 A 0-ABCDE 
 
 Notice that each is formed of portions of three or more inter- 
 secting planes ; these planes are the Faces of the polyedral 
 angle. The faces intersect in one common point; this is the 
 Vertex of the polyedral angle. Each face has two edges which 
 pass through the vertex of the angle ; these are the Edgesl of 
 the polyedral angle. On each face, the two edges form an 
 angle, whose vertex is also the vertex of the polyedral angle ; 
 these angles are the Face Angles of the polyedral angle. Each 
 pair of consecutive faces intersect in an edge, forming a diedral 
 angle; these diedral angles are the Diedral Angles of the 
 polyedral angle. 
 
 The edges and the faces are unlimited in extent. It is con- 
 venient to indicate the polygon which results if a plane is 
 drawn, not through the vertex, but intersecting all the faces ; 
 this polygon aids in picturing the number of faces of the 
 polyedral angle. 
 
334 SOLID GEOMETRY — BOOK VI 
 
 508. A Triedral Angle is a polyedral angle having three 
 faces. 
 
 A Tetraedral Angle is a polyedral angle having four faces. 
 
 Ex. 68. If a polyedral angle has 4 faces, how many vertices, edges, 
 face angles, and diedral angles does it have ? 
 
 Ex. 69. Name the edges, face angles, and the diedral angles of triedral 
 angle O-ABC. (Fig. § 507.) 
 
 509. Two polyedral angles are vertical if the edges of one 
 are the prolongations of the edges of the other. 
 
 510. Two polyedral angles are congruent if they can be made 
 to coincide. 
 
 Ex. 70. (a) Construct two triedral angles which have the face angles 
 of one equal respectively to the face angles of the other and in the same 
 order. Determine whether they can be made to coincide. 
 
 (b) Construct a third triedral angle whose face angles are equal to 
 those of the triedrals of part (a), but arrange them in order opposite to 
 that in part (a). (See Fig. §511.) Can this triedral angle be made 
 to coincide with either of those in part (a)? 
 
 511. Two polyedral angles are symmetrical if the face angles 
 and the diedral angles of one are equal respectively to the face 
 angles and the diedral angles of the other, provided these parts 
 occur in opposite orders. 
 
 Thus, if face AAOB, BOC, and COA are equal respectively to face 
 AA'O'B', B'O'C, and CO' A', and 
 
 diedral A OA, OB, and 00 to die- O' 
 
 dral A O'A', O'B', and O'C, triedral /A /v\ 
 
 A O-ABC and O'-A'B'C are sym- // \ / \\ 
 
 metrical, since the parts in Z 0'- / \ / \ \ / 
 
 A'B'C occur in opposite order to the A<^-h---^C Ci^:^^--Ar--^A' 
 equal parts of Z O-ABC; that is, to jb^ ^B^^ 
 
 pass from OA to OB to OC, one 
 
 moves from left to right, whereas, to pass from O'A' to O'B' to O'C, one 
 moves from right to left. 
 
 It is evident that, in general, two symmetrical polyedrals 
 cannot be placed so that their faces will coincide. 
 
POLYEDRAL ANGLES 
 
 335 
 
 Proposition XXII. Theorem 
 512. Tioo vertical polyedral angles are symmetricaL 
 
 H3rpothesis. 0-ABC and O-A'B'C are vertical triedral 
 angles. (Fig. 1.) 
 
 Conclusion. A 0-ABC and O-A'B'C are symmetrical. 
 
 Proof. 1. Face A AOB, BOC, etc. equal respectively face 
 AA'OB', B'Oa, etc. Why ? 
 
 2. Diedral A BOAC and B'OA'C are vertical ; for AOB and 
 A' OB' are parts of the same plane, as also are AOC and A' 00'. 
 
 Similarly, A OB and OB' are vertical, etc. 
 
 3. .•. Diedral A OA, OB, etc. equal respectively diedral 
 angles OA', OB', etc. § 492 
 
 4. The parts of 0-ABC occur in opposite order to the equal 
 parts of Z O-A'B'C. 
 
 This may be understood by moving O-A'B'C parallel to itself to the 
 right, and then revolving it about an axis through (as shown in Fig. 2) 
 until face OA'C comes into the same plane as before. OB' is then in 
 front of plane C OA' instead of back of that plane as in Fig. 1. Now, in 
 Fig. 1, to pass from ^0 to OB to OC, one moves from left to right; in 
 Fig. 2 to pass from OA' to OB' to OC , one moves from right to left. 
 
 5. .-. A 0-ABC and O-A'B'C are symmetrical. § 511 
 
 Note. — The theorem may be proved for any two vertical polyedral 
 angles in the same manner. 
 
336 SOLID GEOMETRY — BOOK VI 
 
 Proposition XXIII. Theorem 
 
 513. The sum of any tioo face angles of a triedral 
 angle is greater than the third. 
 
 Note. — The theorem requires proof only in the case when the third 
 face angle is greater than either of the others. 
 
 Hypothesis. In triedral Z 0-ABC 
 
 Z. AOC > Z AOB, and also Z AOC > Z BOC 
 Conclusion. Z AOC <ZAOB+Z BOG. 
 Proof. 1. In face AOO, draw OD = OB, making 
 
 ZAOD = ZAOB. 
 2. Through B and D pass any plane cutting the faces of the 
 triedral Z in AB, BC, and CA, respectively. 
 3. 
 4. 
 
 5. 
 6. 
 
 7. 
 
 8. 
 9. 
 
 Ex. 71. Prove that any face angle of a polyedral angle is less than 
 the sum of the remaining face angles. 
 
 Suggestion. — Divide the polyedral Z into triedral A by passing planes 
 through any lateral edge, and apply § 513. 
 
 A AOB ^ A AOD. 
 
 Prove it. 
 
 .'. AB = AD. 
 
 Why? 
 
 In A ABC 
 
 
 AB + BO >AD-{- DO. 
 
 Why? 
 
 .'.BC>DC. 
 
 § 158, Ax. 18 
 
 .'. in A ^OC and A COD 
 
 
 ZBOC>ZCOD. 
 
 §167 
 
 AOB + Z BOC > Z AOD + Z COD. Why ? 
 
 .-. Z AOB + Z BOC > Z AOC. 
 
 
POLYEDRAL ANGLES 337 
 
 Proposition XXIV. Theorem 
 
 514. The sum of the face angles of any convex poly e- 
 dral angle is less than four right angles. 
 
 Hypothesis. 0-ABCDE is any convex polyedral Z. 
 
 Conclusion. AAOB+ A BOC + etc. < 4 rt. A. 
 
 Proof. 1. Pass a plane cutting the faces in the polygon 
 ABCDE. 
 
 Let 0' be any point within ABCDE, and draw OA, O'B, 
 O'G, O'D, and O'E. 
 
 2. Then, in triedral Z A-EOB, 
 
 Z OAE + Z OAB >ZEAO' -\-Z O'AB. § 513 
 
 3. Similarly Z OB A + Z OBC > Z ABO' + Z O'BC', etc. 
 
 4. Adding these inequalities, the sum of the base A of the 
 A whose common vertex is is greater than the sum of the 
 base A of the A whose common vertex is 0'. § 158, Ax. 19 
 
 5. The sum of all the A of the A with vertex equals the 
 sum of all the A of the A with vertex 0'. ^Vhy ? 
 
 6. .*. the sum of the A a^t is less than the sum of the A 
 at 0'. § 158, Ax. 20 
 
 7. .-. the sum of the z^ at < 4 rt. A. Why ? 
 
 Note. — The pupil's understanding of this theorem will be increased if 
 a pasteboard model of the figure of § 514 is at hand when this theorem is 
 first studied. 
 
 Notice that the inequality in step 2 does not mean that ZEAO 
 >ZEAO' and Z OAB > Z O'AB ; rather, the sum of Z EAO diud Z OAB 
 is greater than the face angle EAB of triedral Z A-EOB. 
 
338 
 
 SOLID GEOMETRY — BOOK VI 
 
 Proposition XXV. Theorem 
 
 515. If two triedral angles have the face angles of 
 one equal respectively to the face angles of the other, 
 their homologous diedral angles are equal. 
 
 C'C 
 
 Fig. 1 
 
 Fig. 2 
 
 Fig. 3 
 
 Hypothesis. In triedral A 0-ABC and O'-A'B'C 
 ZAOB = ZA'0'B'', ZBOC=ZB'0'C'', Z 00A = Z C'O'A'. 
 
 Conclusion. Diedral Z OA = diedral Z OA'; etc. 
 
 Proof. 1. Lay off OA, OB, 00, O'A', O'B', and O'C all of 
 the same length, and draw AB, BO, OA, A'B', B'C, and C'A'. 
 
 2. .-. A OAB ^ A O'A'B' and AB = A'B'. Prove it. 
 
 3. Similarly BO = B'C and AO = A'O'. 
 
 4. AlsoAABC^AA'B'C'sindZBAO=ZB'A'C'. Why? 
 
 5. On OA and O'A', take AD = A'D' ; draw DE in face 
 OAB±OA, D'E' in A'0'B'±A'0', DF in A0G1.A0, D'F' 
 in A'0'0' ± A'O', EF in face ABC, and E'F' in face A' B'C. 
 
 6. Then A ADE ^ A A' D'E'. Prove it. 
 .-. AE = A'E' and DE = D'E'. 
 
 . 7. Also AF = A'F' and DF = D'F'. Prove it. 
 
 8. Then A AEF ^ A A' E'F' and EF = E'F'. Prove it. 
 
 9. Then A DEF ^ A D'E'F'. Prove it. 
 
 10. .-. Z EDF = Z ^'i)'i^'. Why ? 
 
 11. .-. diedral Z ^0 = diedral Z ^'0'. Why ? 
 
 Note. — The above proof holds for Fig. 3 as well as for Fig. 2. In 
 Figs. 1 and 2, the equal parts occur in the same order, and in Figs. 1 and 
 3 in the reverse order. 
 
SUPPLEMENTARY TOPICS 339 
 
 516. Cor. If two triedrcU angles have the face angles of one 
 equal respectively to the face angles of the other ^ 
 
 1. They are congruent if the equal parts occur in the same 
 order. 
 
 2. They are symmetrical if the equal parts occur in the reverse 
 order. 
 
 SUPPLEMENTARY TOPICS 
 
 The following topics, theorems, and exercises of Book VI are 
 supplementary. Each group is independent of each of the others. 
 None of this material is required in the main parts of subse- 
 quent Books. 
 
 Group A. — Analogy between Triedral Angles and Triangles. 
 
 Group B. — Loci in Space. 
 
 Group C. — Consists of two supplementary theorems usually 
 given in texts. 
 
 Group A. Analogy between Triedral Angles 
 AND Triangles 
 
 517. The analogy between triangles and triedral angles is 
 very striking. Many propositions of plane geometry about tri- 
 angles may be changed into propositions about triedral angles 
 by substituting for the word angle of the former diedral angle, 
 and for the word side the words /ace angle. 
 
 Ex. 72. Two triedral angles are congruent when a face angle and the 
 adjacent diedral angles of one are equal respectively to a face angle and 
 the adjacent diedral angles of the other, if the parts are arranged in the 
 same order. 
 
 Suggestion. — Prove by superposition. 
 
 Note. — If the parts of one are arranged in reverse order to the parts 
 of the other, the triedral angles are symmetrical. 
 
 Ex. 73. Two triedral angles are congruent if two face angles and the 
 included diedral angle of one are equal respectively to two face angles and 
 the included diedral angle of the other, if the parts are arranged in the 
 same order. 
 
340 SOLID GEOMETRY — BOOK VI 
 
 Ex. 74. If two face angles of a triedral angle are equal, the opposite 
 diedral angles are equal. 
 
 Suggestion. — Recsdl the proof of § 69. 
 
 Ex. 75. An exterior diedral angle of a triedral angle is greater than 
 either remote interior diedral angle. 
 
 Suggestion. — Model the proof after that in § 87. 
 
 Ex. 76. — If two triedral angles have a face angle, the opposite diedral 
 angle and another diedral angle of one equal respectively to the corre- 
 sponding parts of the other, they are congruent if the parts are in the 
 same order. 
 
 Suggestion. — Superpose the equal face angles, so that the equal diedral 
 angles adjacent to the faces superposed also coincide. Prove that the faces 
 opposite these diedral angles also coincide by an indirect proof, based upon 
 Exercise 75. 
 
 Note. — If the parts are in reverse order, the figures are symmetrical. 
 
 Ex. 77. State and prove the converse of Ex. 74. 
 Suggestion. — The proof is based upon Ex. 76, Note. 
 
 Group B. Loci in Space 
 
 518. The following more general definition of locus will be 
 employed in solid geometry. The locus of points satisfying a 
 given condition consists of all points which satisfy the condi- 
 tion, and. of no other points. 
 
 The points which constitute a locus may form one (or more) 
 lines, or one or more surfaces. 
 
 To prove that a particular assumed locus is actually the locus 
 satisfying a given condition, or conditions, prove either (a) and 
 (b) below or else (a) and (c). 
 
 (a) Every point of the locus satisfies the conditions. 
 
 (h) Every point not of the locus does not satisfy the conditions. 
 
 (c) Every point which does satisfy the conditions lies m the 
 locus. 
 
 Ex. 78.' As a consequence of § 457 and § 459, what is the locus of 
 points equidistant from the extremities of a line ? 
 
 Ex. 79. As a consequence of § 500 and § 501, what is the locus of 
 points equidistant from two intersecting planes ? 
 
SUPPLEMENTARY TOPICS 341 
 
 Ex. 80. What is the locus of points in space at a given distance from 
 a given plane ? 
 
 Ex. 81. What is the locus of points in space equally distant from 
 two parallel planes ? 
 
 Ex. 82. What is the locus of points in space equidistant from the 
 points of a given circle ? 
 
 Ex. 83. What is the locus of points in space equidistant from the 
 vertices of a given triangle ? 
 
 519. Intersection Loci. — When two conditions are imposed 
 upon a point in space, each condition determines a locus for 
 the point, and the desired point lies in the intersection of the 
 loci. It is often inadvisable to attempt to draw the two loci, 
 for that demands considerable skill in drawing. The following 
 form of solution brings out all the mathematical value of such 
 a problem quite as well, if not better, than if a figure were 
 drawn. 
 
 Illustrative Problem. — What is the locus of points in 
 space at a given distance from a given plane and equidistant from 
 two given points ? 
 
 Solution. 1. The locus of points at a given distance from a given plane 
 consists of two planes parallel to the given plane and at the given distance 
 from it. Call this Locus 1. 
 
 2. The locus of points equidistant from two given points is the plane 
 perpendicular to and bisecting the segment between the two points. 
 Call this Locus 2. 
 
 3. The desired locus is the intersection of Locus 1 and Locus 2. 
 Discussion. 1. Generally the plane of Locus 2 will intersect the planes 
 
 of Locus 1 in two straight lines. 
 
 2. Locus 2 may be parallel to the planes of Locus 1. In this case, 
 there will not be any points satifying the given conditions. 
 
 3. Locus 2 may coincide with one of the planes of Locus 1. In this 
 case, the plane common to Loci 1 and 2 will be the desired locus. 
 
 Ex. 84. What is the locus of points in a given plane equidistant from 
 two parallel planes ? 
 
 Ex. 85. What is the locus of points in a given plane at a given dis- 
 tance from another given plane ? 
 
342 SOLID GEOMETRY — BOOK VI 
 
 Ex. 86. What is the locus of points in a given plane equidistant 
 from two given points not in the plane ? 
 
 Ex. 87. What is the locus of points in a given plane equidistant 
 from two intersecting planes ? 
 
 Ex. 88. What is the locus of points equidistant from two given points 
 and also equidistant from two parallel planes ? 
 
 Ex. 89. What is the locus of points equidistant from two given 
 points and also equidistant from two intersecting planes ? 
 
 Ex. 90. What is the locus of points equidistant from two parallel 
 planes, and also equidistant from two intersecting planes ? 
 
 Ex. 91. What is the locus of points equidistant from two intersect- 
 ing planes, and also at a given distance from a given plane ? 
 
 Ex. 92. Find all points which are at a given distance from a given 
 plane, equidistant from two other parallel planes, and equidistant from 
 two given points. 
 
 Ex. 93. Find all points which are equidistant from two given inter- 
 secting planes, equidistant from two parallel planes, and equidistant from 
 two given points. 
 
 Ex. 94. Prove that the three planes bisecting the diedral angles of 
 a triedral angle meet in a common straight line. o 
 
 Snggestion. — Planes OAD and OBE intersect in / i<t\ 
 
 line OG. Prove OG is in plane OCF. / //|1\\ 
 
 Ex. 95. Prove that the three planes deter- / / liW \ 
 mined by the edges of a triedral angle and the ^^^^-:t-V-V::Y-^C 
 bisectors of the opposite face angles intersect in a f^^ m /^ 
 
 line. ^B 
 
 Suggestions. — 1. In a figure like that for Ex. 94, assume OB, OF, and 
 OE bisect the angles BOO, BOA, and AOC, respectively. 
 
 2. Take OA=OB=OC, and draw AB, BC, and AC 
 
 3. Prove AD, CF, and EB are concurrent. 
 
 4. Prove planes AOD, BOE, and COF meet in a line. 
 
 Ex. 96. Prove that the locus of points equidistant from the sides of 
 an angle is the plane through the bisector of the angle, and perpendicular 
 to the plane of the angle. 
 
 Suggestion. — Recall § 460 and Ex. 57, Book VI. 
 
 Ex. 97. Prove that the planes through the bisectors of the face 
 angles of a triedral angle, perpendicular to the planes of the faces, meet 
 in a line, whose points are equidistant from the edges of the triedral 
 angle. 
 
SUPPLEMENTARY THEOREMS 
 
 343 
 
 Group C. Supplementary Theorems 
 
 Proposition XXVI. Theorem 
 
 520. Tim straight lines not in the same plane have one com- 
 mon perpendicular, and only one ; and this segment is the shortest 
 segment that can be drawn between them. 
 
 Hypothesis. AB and CD do not lie in the same plane. 
 
 Conclusion. One and only one common ± to AB and CD 
 can be drawn ; also, this ± is the shortest segment between 
 AB and CD. 
 
 Proof. 1. Through CD draw plane MN II AB. § 467 
 
 2. Through AB, draw plane AH ± plane MN, intersecting 
 MN in line OH. § 502 
 
 3. .'.QHWAB. Why? 
 
 4. .-. GH intersects CD at a point G. 
 
 [If CD II GH, CD would be II to AB, which is impossible.] 
 
 5. In plane AH, draw AG ± GH at G. 
 
 6. Then AG ± AB and also to CD. Prove it. 
 
 7. Assume KE also ± to both AB and CD. 
 
 8. Draw EF II AB, and KL in plane AH ± GH. 
 
 9. ^ii^ is in MN. 
 
 10. EK±EF. 
 
 11. .'.EKA.MN. 
 
 12. But this is impossible for /li ± 307". 
 
 13. Hence ^G^ is the only common X to AB and CD. 
 
 14. ^^ > /O:. 
 
 15. .-. EK > AG. 
 
 16. .-. AG is the shortest segment from AB to CD. 
 
 §470 
 Why? 
 Why? 
 
 §496 
 Why? 
 Why? 
 Why? 
 
344 
 
 SOLID GEOMETRY — BOOK VI 
 
 Proposition XXVII. Theorem 
 
 521. Two diedral angles have the same ratio as their plane 
 angles. 
 
 Case I. When the plane angles are commensurable. 
 
 c 
 
 Ill 
 
 1 1 1 
 1 1 1 
 
 ^ 
 
 [^ 
 
 Hypothesis. A ABC and A'B'C, the plane A of diedral A 
 ABDC and A'B'D'C respectively, are commensurable. 
 
 Conclusion. = • 
 
 Z ABCD Z ABO 
 
 Proof. 1. Let Z ABE, a common measure of Z ABC and 
 Z A'B'C, be contained 4 times in Z ABC and 3 times in 
 
 Z A'B'C. 
 
 Z. A'B'C 3 
 4* 
 
 2. Then 
 
 ZABC 
 
 3. Passing planes through the edges BD and B'D', and the 
 division lines of ZABC and Z A'B'C, respectively, diedral 
 Z ABCD is divided into 4 parts, and Z A'B'D'C into three 
 parts, all of which are equal. W^hy ? 
 
 Z A'B'D'C 3^ 
 4* 
 
 5. 
 
 ZABDC 
 Z A'B'D'C Z A'B'C 
 
 Why 
 
 Z ABDC Z ABC 
 
 Case II. When the plane angles are incommensurable. 
 
 Hypothesis. Z ABC and Z A'B'C, plane A of diedral 
 A ABDC and A'B'D'C, respectively, are incommensurable. 
 
Conclusion. 
 
 SUPPLEMENTARY THEOREMS 
 Z A'B'D'C Z A'B'C 
 
 345 
 
 Proof. 1. Let Z x\BC be divided into any number of equal 
 parts, and let one of these parts be applied to Z A'B'C as unit 
 of measure. 
 
 Since Z ABC and Z A'B'O' are incommensurable, the unit 
 will not be contained exactly in Z A'B'C 
 
 A certain number of equal parts will extend from A'B' to 
 B'E, leaving the remainder Z EB'C less than the unit of 
 measure. 
 
 2. Pass a plane through B'D' and B'E. Then 
 
 Z A'B'D'E Z A'B'E 
 
 ZABDC A ABC 
 
 Case I 
 
 3. Now let the number of subdivisions of Z ABC be in- 
 definitely increased ; then the unit of measure will be indefi- 
 nitely decreased, and consequently the remainder Z EB'C will 
 approach the limit 0. § 401 
 
 Z A'B'D'E . Z A'B'D'C 
 
 4. 
 
 6. Also 
 6. 
 
 ZABDC ZABDC 
 
 [" =," means " approaches the limit."] 
 
 Z A'B'E . Z A'B'C 
 
 ZABC 
 Z A'B'D'C 
 
 ZABC 
 Z A'B'C 
 
 ZABDC ZABC 
 
 § 403, (a) 
 
 § 403, (a) 
 § 403, (h) 
 
BOOK YII 
 
 POLYEDEA 
 
 522. Surfaces. So far only plane surfaces have been con- 
 sidered. 
 
 A Curved Surface is a surface no part of which is plane. 
 
 The surface of a spherical object is a familiar example of a curved sur- 
 face. 
 
 It is evident that there may be surfaces 
 of which part is plane and part is curved ; 
 as surface S. 
 
 Also there are surfaces consisting of two 
 or more parts each of which is plane ; as 
 surface T. 
 
 523. Closed Surfaces. Let a plane ABC intersect the facfes 
 of triedral angle 0-XYZ, and consider the sur- 
 face consisting of triangles OAB, GAG, GBG, 
 and ABC, and the portions of planes within 
 them. This surface incloses a finite portion 
 of space. Such a surface is a closed surface. 
 A closed surface is such that the intersection 
 of it made by evei-y intersecting plane is a 
 closed line. 
 
 524. A closed surface is convex if the intersection with it of 
 every intersecting plane is a convex closed line. 
 
 It will be assumed that all closed surfaces considered in this text are 
 convex. 
 
 525. A Solid is the finite portion of space inclosed by a 
 closed surface. The surface is called the surface of the solid, 
 and is said to bound the solid. 
 
 346 
 
POLYEDRA 
 
 347 
 
 In the remaining part of solid geometry a detailed study is 
 made of certain common solids. 
 
 526. A Polyedron is a solid bounded by portions of planes, 
 called the Faces of the polyedron. The faces intersect in 
 straight lines, called the Edges of the polyedron. The edges 
 intersect in points, called the Vertices of the polyedron. The 
 straight line joining any two vertices of the polyedron which 
 do not lie in the same face is a Diagonal of the polyedron. 
 
 527. The least number of planes that can form a polyedral 
 angle is three. Then the least number of planes that can form 
 a polyedron is four. 
 
 A polyedron of four faces is a Tetraedron ; one of six faces 
 is a Hexaedron ; one of eight faces is an Octaedron ; one of 
 twelve faces is a Dodecaedron ; and one of twenty is an 
 Icosaedron. 
 
 The cube is a familiar hexaedron. 
 
 Tetraedron 
 
 Pentaedron 
 
 Hexaedron 
 
 Ex. 1. Verify for a tetraedron, a peDtaedron, and a hexaedron, the 
 
 foniiula 
 
 F-\- V=E + 2, 
 
 where F is the number of faces, V is the number of vertices, and E is the 
 number of edges. 
 
 Note. — This theorem is due to the mathematician Leonard Euler. 
 
 A 
 
 Ex. 2. If E, F, G, and H are the mid-points of 
 edges BD, BC, AC, and AD, respectively, of tetraedron 
 ABCD, prove EFGII a. parallelogi'am. D 
 
348 
 
 SOLID GEOMETRY — BOOK VII 
 
 Ex. 3. If ABCD is a tetraedron, the section made by a plane parallel 
 to each of the edges AB and CD is a parallelogram. 
 
 Note. — Remember that by § 468 a plane can he drawn parallel to each 
 of two straight lines in space. 
 
 Ex. 4. The lines joining, by pairs, the mid-points 
 of opposite edges of any tetraedron, intersect in a com- 
 mon point. 
 
 Ex. 5. In tetraedron ABCD, a plane is drawn 
 through edge CD perpendicular to AB, intersecting 
 faces ABC and ABD in CE and ED, respectively. If 
 the bisector of Z CED meets CD at F, prove 
 
 CF : DF = area ABC : area ABD. 
 
 Suggestion. — Recall § 270. 
 
 PRISMS AND PARALLELOPIPEDS 
 
 528. A Prism is a polyedron, two of whose faces lie in 
 parallel planes, and whose remaining 
 faces, in order, intersect in parallel lines. 
 The parallel faces are the Bases; the 
 other faces are the Lateral Faces; the 
 edges which are not sides of the bases are 
 the Lateral Edges ; the perpendicular be- 
 tween the bases is the Altitude ; the sum 
 of the areas of the lateral faces is the 
 Lateral Area. 
 
 If a plane is perpendicular to the lateral edges of a prism, 
 its intersection with the prism is called a Right Section of the 
 prism. 
 
 529. The following Important Facts about a Prism should be 
 proved by the pupil : 
 
 I. TJie lateral faces of a prism are inclosed by parallelograms. 
 
 Prove BCHG is a O, using § 471. Fig. § 528 
 
PRISMS AND PARALLELOPIPEDS 
 
 349 
 
 II. The lateral edges of a prism are parallel and equal. 
 
 III. The bases of a prison are inclosed by congruent polygons. 
 Suggestion. — Recall § 481. 
 
 IV. Sections of the lateral surface of a 
 prison made by two parallel planes cutting 
 all the lateral edges are congruent polygons. 
 
 Suggestion. — Let planes CF and C'F' he paral- 
 lel. 
 
 Prove CDEFG ^ C'D'E'F'G'. 
 
 V. A section of a prism made by a 
 plane parallel to the base is congruent to 
 the base. 
 
 530. Kinds of Prisms. A prism is triangular, quadrangulary 
 etc., according as its base is triangular, quadran- 
 gular, etc. 
 
 A Right Prism is a prism whose lateral edges 
 are perpendicular to its bases. 
 
 An Oblique Prism is a prism whose lateral 
 edges are not perpendicular to its bases. 
 
 A Regular Prism is a right prism whose base 
 is inclosed by a regular polygon. 
 
 A Truncated Prism is that portion of a prism 
 bounded by the base and a plane not parallel to 
 the base, cutting all the lateral edges. 
 
 Ex. 6. Prove that every pair of lateral edges of a prism determines a 
 plane which is parallel to each of the other lateral edges of the prism. 
 
 Ex. 7. Prove that the lateral edges of a right prism are equal to the 
 altitude. 
 
 Ex. 8. Prove that the lateral faces of a right prism are inclosed by 
 
 rectangles. n' c 
 
 r~- — yf~~-^H 
 Ex. 9. Prove that the section of a prism made by ^- ^ ^ 
 
 a plane parallel to a lateral edge is inclosed by a par- 
 allelogram. 
 
 Suggestion.— Jjet plane EF' be || AA', a lateral edge 
 of prism AC. 
 
 .-L., 
 
350 SOLID GEOMETRY — BOOK VII 
 
 Proposition I. Theorem 
 
 531. The lateral area of a prism equals the perimeter 
 of a right section multiplied by the length of a lateral 
 edge. 
 
 Hypothesis. DEFGH is a rt. section of prism AO. 
 
 P= perimeter of DEFGH', L = length of A A' ; 
 
 S = the lateral area. 
 Conclusion. S = LP. 
 
 Proof. 1. Area of O AB' = Lx DE. Why ? 
 
 2. Similarly, area of CJ BC' = Lx EF. Why ? 
 
 Complete the proof. 
 
 532. Cor. TJie lateral area of a right prism equals the perim- 
 eter of the base multiplied by the length of the altitude. 
 
 Ex. 10. Find the lateral area of a regular hexagonal prism, each side 
 of whose base is 3 and whose altitude is 9. 
 
 Ex. 11. There are upon a porch six columns having the form of regu- 
 lar octagonal prisms. If the side of the base is 4 inches and the altitude 
 of the column is 7 feet, find the total of the lateral areas of the columns. 
 
 533. The Volume of a Solid is a number which indicates the 
 measure of that solid in terms of a unit of solid measure ; it is 
 the ratio of the solid to the unit of solid measure. 
 
 534. Two solids are equal if they have equal volumes. Evi- 
 dently two congruent solids are equal ; likewise two solids 
 which can be divided into parts which are respectively congru- 
 ent are equal. 
 
PRISMS AND PARALLELOPIPEDS 351 
 
 Proposition II. Theorem 
 
 535. An oblique prism equals a right prism which 
 has for its base a right section of the oblique prism, and 
 for its altitude a lateral edge of the oblique prism. 
 
 Hypothesis. Fir is a right prism. Its base FKis a right 
 section of oblique prism AD'. 
 
 Its altitude FF' = AA\ a lateral edge of AD'. 
 Conclusion. FK' = AD'. 
 
 Proof. 1. ABCDE^A'B'C'D'E', 
 
 and FGHKL ^ F'O'H'K'L'. § 529, IV 
 
 2. AF= A'F' ; BG = B'G' ; CH = G'H' ; etc. Prove it. 
 
 3. Slide polyedron AK upward, letting AF, BG, etc., move 
 along lines AF', BG', etc., until ABODE coincides with 
 A'B'C'D'E'. 
 
 4. Then F, G, H, etc., fall upon F', G', H', etc. Step 2 
 
 5. .-. FGHKL will fall upon F'G'H'K'L'. 
 
 6. .-. polyedron AK ^ polyedron A'K'. 
 
 7. .♦. prism AD' = prism FK'. § 534 
 
 536. Cor. Two right prisms are congruent, and hence equal, 
 when they have congruent bases and equal altitudes. 
 
 For the bases can be made to coincide. Then the lateral edges of one 
 will coincide with the homologous edges of the other by § 530 and Note, 
 § 456. Then the upper bases must coincide. 
 
352 
 
 SOLID GEOMETRY — BOOK VII 
 
 537. A Parallelepiped is a prism whose base is inclosed by 
 a parallelogram. As a consequence, all the 
 faces of a parallelopiped are inclosed by par- 
 allelograms. 
 
 A Eight Parallelopiped is a parallelopiped 
 whose lateral edges are perpendicular to its 
 bases. 
 
 A Rectangular Parallelopiped is a right par- 
 allelopiped whose base is inclosed by a rec- 
 tangle. Consequently all the faces are inclosed 
 by rectangles. (See Ex. 8, p. 349.) 
 
 A Cube is a rectangular parallelopiped whose six faces are 
 inclosed by squares. 
 
 Ex. 12. How many faces of a right parallelopiped are rectangles ? 
 Ex. 13. Is a cube a prism ? 
 
 Proposition III. Theorem 
 
 538. The opposite lateral faces of a parallelopiped 
 are congruent and parallel. 
 
 
 Hypothesis. AC and A'C are the bases of parallelo- 
 piped AC. 
 
 Conclusion. Eaces AB' and DC are congruent and II. 
 
 Suggestions. — 1. To prove AB' ^ DC , prove them mutually equiangular 
 and mutually equilateral. 
 
 2. To prove AB' \\DC', recall § 481. 
 
 539. Cor. Any pair of opposite faces of a parallelopiped may 
 he taken as its bases. 
 
PRISMS AND PARALLELOPIPEDS 353 
 
 Proposition IV. Theorem 
 
 540. The plane through tivo diagonally opposite 
 edges of a parallelopiped divides it into two equal tri- 
 angular prisms. 
 
 Hypothesis. Plane AC passes through, edges AA' and CC 
 of parallelopiped A'C. 
 
 Conclusion. Prism ^'-^B(7= prism ^'-^CZ>. 
 
 Proof. 1. Let EFGH be a right section of the parallelo- 
 piped, intersecting plane AA'C'C in EG. 
 
 2. EFGH is a O. Prove it. 
 
 3. .-. AEFG^AEGH. Why? 
 
 4. Prism A'- ABO = a right prism with base EFG and 
 altitude AA\ 
 
 Prism A'-ACD = Si right prism with base EGH and 
 altitude AA'. § 535 
 
 6. But these right prisms are equal. § 536 
 
 6. .-. A'-ABC = A'-AGD. 
 
 Ex. 14. Prove that the diagonals of a rectangular parallelopiped are 
 equal. 
 
 Ex. 15. Prove that the diagonals of a parallelo- / yz^^f^ 
 piped bisect each other. \/^y'~'^^\^ 
 
 B 
 Suggestion. — Prove that each of the other diagonals bisects BG. 
 
 Note. — The point Of intersection of the diagonals of a parallelopiped is 
 called the center of the parallelopiped. 
 
354 
 
 SOLID GEOMETRY — BOOK VII 
 
 Ex. 16. Prove that any line drawn through the center of a parallele- 
 piped, terminating in a pair of opposite faces, is bisected at that point. 
 
 Ex. 17. Prove that the line joining the center of a parallelepiped to 
 the center of any face is parallel to any edge of the parallelopiped which 
 intersects that face, and is equal to one half of it. 
 
 Ex. 18. Prove that the centers of two opposite faces of a parallelo- 
 piped and the center of the parallelopiped are collinear. 
 Suggestion. — Recall Ex. 17 and § 90. 
 
 Ex. 19. If the four diagonals of a quadrangular prism pass through a 
 common point, the prism is a parallelopiped. 
 
 Plan. — Prove BCHE is a O. 
 
 Suggestion. — DE and ^Zf determine a plane (why?) 
 which intersects planes DF and BH in lines AD and ^ 
 HE, respectively. Compare AD and HE; also compare 
 AD 2iX\dBC. Then compare BC and HE. 
 
 Ex. 20. Recall that the diagonals of a square are equal, bisect each 
 other, and are perpendicular to each other. 
 
 What questions about the diagonals of a cube do these facts suggest ? 
 Prove the answers to your questions. 
 
 541. The Dimensions of a rectangular parallelopiped are the 
 lengths of its three edges which meet at any vertex. 
 
 The Volume of a rectangular parallelopiped is readily deter- 
 mined when the three dimensions are multiples of the linear 
 unit of measure. 
 
 Thus, if the dimensions of P are 5 units, 4 units, 
 and 3 units, respectively, the solid can be divided into 
 60 unit cubes. In this case 60, the number which 
 expresses the area, is the product of 3, 4, and 5, the 
 three dimensions. 
 
 /-/-/--/-/-A 
 
 A-/--/--/--/- 
 
 -A\. 
 
 -l-j-l-i- 
 
 
 1 1 ' 1 
 
 ___ — I — , — ) — 
 
 A 1 A 
 
 1 ! 1 i 
 
 V 
 
 The following three propositions prove 
 that this is the correct formula for determining the volume of 
 any rectangular parallelopiped. Before taking up these propo- 
 sitions, certain necessary new ideas are introduced.* 
 
 * The class may have studied limits in Plane Geometry (§ 401 and follow- 
 ing) . In that case the following two paragraphs constitute a review of those 
 paragraphs of the Plane Geometry. 
 
PRISMS AND PARALLELOPIPEDS 355 
 
 542. a. A Variable is a number which assumes different 
 values during a particular discussion. 
 
 Thus a number x which assumes successively the values 1, ^, 
 \j \y ••• is a decreasing variable, assuming ultimately values 
 which differ by as little as we please from zero. 
 
 A number y which assumes successively the values 1, 1^, 1|, 
 1 J, ••• is an increasing variable^ assuming ultimately values which 
 differ V)y as little as we please from 2. 
 
 h. A Constant is a number which has a fixed value through- 
 out a particular discussion. 
 
 c. A Limit of a Variable is a constant such that the numeri- 
 cal value of the difference between the constant and the varia- 
 ble becomes and remains less than any small positive number. 
 We say that a variable approaches its limit. If a variable has a 
 limit, it has only one limit. The symbol = means " approaches 
 the limit." 
 
 Thus, X above = 0, and y = 2. 
 
 543. Limits Theorems. It can be proved that : 
 
 a. If a variable x approaches a finite limit I, then ex, where 
 c is a constant, approaches the limit cl. § 403, a 
 
 b. If two variables are constantly equal and each approaches 
 a finite limit, then their limits are equal. § 403, b 
 
 c. If a variable x approaches a finite limit a, and a variable 
 y approaches a limit 6, then : 
 
 (x ± y) approaches the limit a±b\ 
 xy approaches the limit a • 6 ; 
 
 and - approaches the limit -, provided b is not zero. 
 y & 
 
 Ex. 21. Find the length of the diagonal of a rectangular parallelo- 
 piped whose dimensions are 8, 9, and 12. 
 
 Ex. 22. Prove that the square of a diagonal of a rectangular paral- 
 lelopiped is equal to the sum of the squares of its edges. 
 
 Ex. 23. Determine the length of the diagonal of a cube whose edge is 
 of length s. 
 
 Note. — Supplementary Exercises 18-22, p. 456, can be studied now. 
 
356 
 
 SOLID GEOMETRY — BOOK VII 
 
 Proposition Y. Theorem 
 
 544. TiDO rectangular parallelopipeds having congru- 
 ent bases have the same ratio as their altitudes. 
 Case I. When the altitudes are commensuraUe, 
 
 />-- 
 
 ■A 
 
 
 
 Q 
 
 
 / 1 
 
 / / 1 
 
 A 
 
 / A 
 
 
 
 -/-7 
 
 / i- 
 
 B 
 
 / 1 
 
 
 
 'A 
 
 y 1 
 
 / J 
 
 
 V 
 
 / 
 
 v 
 
 Hypothesis. P and Q are rectangular parallelopipeds with 
 congruent bases, and commensurable altitudes AA! and BB'. 
 
 Conclusion. - = 4^!- 
 
 P AA 
 
 Proof. 1. Let AC^ a common measure of AA and BB'^ be 
 
 contained 4 times in AA and 3 times in BB'. 
 
 BB' J^ 
 
 " AA 4 * 
 
 2. Through the points of division of AA and BB' draw 
 planes ± AA and BB' , respectively. 
 
 3. Then P is divided into 4, and Q is divided into 3, parts 
 which are all congruent. § 536 
 
 4. • 5-5. 
 
 P 4 
 Q^BB' 
 P AA 
 
 Why? 
 
 Case II (Fig. p. 857). 
 cojumensurahle* 
 
 When the altitudes are in- 
 
 Hypothesis. P and Q are rectangular parallelopipeds with 
 congruent bases and incommensurable altitudes AA' and BB'. 
 * This proof may be omitted if desired. 
 
PRISMS AND PARALLELOPIPEDS 
 
 357 
 
 Conclusion. 
 
 P AA' 
 
 A 
 
 
 P 
 
 A 
 
 / 
 
 L 
 
 
 A 
 
 / 
 
 / 
 
 A / 
 
 A 
 
 // 
 
 / 1 
 
 / 
 
 
 / 
 
 / 
 
 Proof. 1. Divide AA! into any number of equal parts, and 
 apply one of these parts to BB! as unit of measure. Since A A 
 and BB' are incommensurable, a certain number of these parts 
 will extend from B to O, leaving a remainder CB' less than the 
 unit of measure. 
 
 2. Draw plane CD J_ BB', and let parallelopiped BD be 
 denoted by Q'. 
 
 Then % = ^.' Case I 
 
 P AA 
 
 [Since A A and BC are commensurable.] 
 
 3. Let the number of subdivisions in AA! be indefinitely in- 
 creased. The length of each subdivision will then diminish 
 indefinitely, and hence CB' will approach the limit 0. § 542, c 
 
 Also 
 
 and 
 4. 
 
 Qt Q 
 
 -^ will approach the limit -^, 
 
 - will approach the limit — —   
 AA 
 
 AA' 
 
 Q^BB' 
 P AA' 
 
 § 543, a 
 
 § 543, a 
 § 543, b 
 
 545. Cor. If tivo rectangular parallelopipeds have two dimen- 
 sions of one equal respectively to two dimensions of the other j their 
 volumes have the same ratio as their third dimensions. 
 
 Ex. 24. Compare the volumes of two rooms having the same floor 
 space if their heights are 8' 6" and 9' respectively. 
 
358 
 
 SOLID GEOMETRY — BOOK VII 
 
 Proposition VI. Theorem 
 
 546. Two rectangular parallelopipeds having equal 
 altitudes have the same ratio as their bases. 
 
 p 
 
 
 
 
 Q 
 
 
 n 
 
 A 
 
 c 
 
 A) 
 
 
 /\ 
 
 c 
 
 A 
 
 / 
 
 1 
 
 ! 
 
 1 
 
 / 
 
 
 
 
 } 
 
 C 
 
 X 
 
 
 J 
 
 
 Hypothesis. Eectangular parallelopipeds P and Q have equal 
 
 altitudes c and bases with dimensions a, b, and a', b', respec- 
 tively. 
 
 n^^ -.!«-: P Ct X b 
 
 
 
 
 Q a'xb' 
 
 
 
 
 Proof. 1. Let i2 be a 
 sions c, a', and b. 
 
 rectangular parallelopiped 
 
 with dimen- 
 
 2. 
 
 Then 
 
 
 P a 
 R-a" 
 
 
 
 §545 
 
 and 
 
 
 
 R b 
 Q b'' 
 
 
 
 Why? 
 
 3. 
 
 Then, 
 
 multiplying 
 
 the equations 
 P ab 
 Q a'b' 
 
 of step 2, 
 
 Ax. 
 
 5, §51 
 
 Note. — Two rectangular parallelopipeds having a dimension of one 
 equal to one dimension of the other, have the same ratio as the products 
 of their other two dimensions. 
 
 Ex. 25. Two rectangular parallelopipeds, with equal altitudes, have 
 the dimensions of their bases 6 and 14, and 7 and 9, respectively. Find 
 the ratio of their volumes. 
 
 Ex. 26. Compare each of the following rectangular parallelopipeds 
 with each of the others : i2, having dimensions 5, 7, and 9 ; ^, having 
 dimensions 9, 5, and 4 ; T, having dimensions 4, 0, and 7. 
 
PRISMS AND PARALLELOPIPEDS 
 
 359 
 
 Proposition VII. Theorem 
 547. Two rectangular parallelopipeds have the same 
 ratio as the products of their three dimensions. 
 
 F 
 
 
 
 
 R 
 
 A > 
 
 
 Q 
 
 
 /V /\ 
 
 /\ 
 
 / 
 
 /\ / 
 
 
 /\ 
 
 / 
 
 i 
 ] — 
 
 / 
 
 c 
 
 
 / 
 
 
 
 c 
 
 
 
 
 V 
 
 / 
 
 A) 
 
 
 / 
 
 Hypothesis. P and Q are rectangular parallelopipeds having 
 dimensions a, 6, c, and a', 6', c', respectively. 
 
 Conclusion. ^=-^- 
 
 q a'b'c' 
 
 Suggestions. — 1. Let ^ be a rectangular parallelepiped with dimensions 
 a', b', and c. 
 
 2. Find - and - by § 546 and § 545. Then multiply - by -• 
 R Q R Q 
 
 Proposition VIII. Theorem 
 548. If the unit of solid measure is the cube ivhose 
 edge is the linear unit, the volume of a rectangular par- 
 allelopiped equals the product of its three dimerisions. 
 
 /\ 
 
 / 
 
 1 
 
 1 
 
 / 
 
 c 
 
 7 
 
 A 
 
 / 
 
 1 
 >— 
 
 /i 
 
 Hypothesis, a, 6, and c are the dimensions of rectangular 
 parallelopiped P, and Q is the unit of measure. 
 Conclusion. Volume of P = a x 6 x c. 
 
 Proof. 1. 
 
 Volume of P = 
 
 Complete the proof, applying § 647. 
 
 533 
 
360 SOLID GEOMETRY — BOOK VII 
 
 549. Cor. 1. Tlie volume of a cube is equal to the cube of its 
 edge. 
 
 550. Cor. 2. The volume of a rectangular parallelopiped is 
 
 equal to the product of its base and altitude. 
 
 Note. — Corollaries 1 and 2 are expressed in their commonly abbreviated 
 form. Expressed more accurately, the second would be " the volume of a 
 rectangular parallelopiped is equal to the product of the area of the base 
 and the length "of the altitude." The brief form of statement will be 
 employed in the remaining theorems of solid geometry. 
 
 Remember that volume of a solid means the number of cubic units in it. 
 In all succeeding theorems relating to volumes, it is understood that the 
 unit of solid is the cube whose edge is the linear unit, and the unit of 
 surface the square whose side is the linear unit. 
 
 Ex. 27. Find the ratio of the volumes of two rectangular parallele- 
 pipeds whose dimensions are 8, 12, and 21, and 14, 15, and 24, respectively. 
 
 Ex. 28. Find the volume and the area of the entire surface of a cube 
 whose edge is 4 in. 
 
 Ex. 29, (a) If the edge of a cube is e, express by formulae the 
 total area, the volume, and the length of a diagonal. (6) Using the 
 proper formula, determine the edge when the diagonal is 12. (c) Using 
 the proper formula, determine the edge when the total area is 150 sq. in. 
 
 Ex. 30. Find the altitude of a rectangular parallelopiped, the dimen- 
 sions of whose base are 21 and 30, equal to a rectangular parallelopiped 
 whose dimensions are 27, 28, and 35. 
 
 Ex. 31. What must be the height of a tank having the form of a 
 rectangular parallelopiped whose base has the dimensions 5 ft. and 8 ft., 
 in order that the tank will contain 1800 gal. of water when the water rises 
 to within one foot of the top ? (One cu. ft. of water = 7^ gal. approxi- 
 mately. ) 
 
 Ex. 32. How many barrels of water will run into a cistern during a 
 ^ in. fall of rain from the roof of a barn whose total roof area is 800 sq. ft. 
 (One cu. ft. of water = 7| gal. ) 
 
 Ex. 33. (a) How many cubic yards of concrete are required for the 
 foundation walls of a house 25 ft. x 35 ft., if the walls are 10 in. thick and 
 are 8 ft. high ? (6) How many bags of cement are required if the mixture 
 contains 4 bags of cement to one yard of gravel ? 
 
 Note. — Supplementary Exercises 23-28, p. 456, can be studied now. 
 
PRISMS AND PARALLELOPIPEDS 
 
 361 
 
 Proposition IX. Theorem 
 
 551. TJie volume of any parallelopiped is equal to 
 the product of its hose and altitude. 
 
 H3rpothesi8. ^C is an oblique parallelopiped. 
 Let the length of altitude AE be H, the area of base ABCD 
 be B, and the volume of AC be V. 
 Conclusion. V=BH. 
 
 Proof. 1. Extend edges AB, A'B', D'C, and DC. 
 
 On AB extended, take FG = AB. 
 Draw planes FK' and GH' 1. FG, forming right parallelo- 
 piped FH'. 
 
 2. .-. prism FH' = prism AC. § 535 
 
 3. Extend edges HG, H'G', K'F, and KF. 
 
 On HG extended, take NM=HG; draw planes NP and 
 ML' 1. NM, forming rectangular parallelopiped LN^. 
 
 4. .-. prism L'N= prism FH'. 
 
 5. .'. prism L'N= prism AC. 
 
 6. But volume L'N = LMNP x 30f . 
 
 7. .-. volume AC = LMNP x MM. 
 
 8. But the length of MM' = H 
 and area LMNP = area ABCD = B. 
 
 9. .-. vol. AC = BH. 
 
 Why? 
 Why? 
 
 Note, p. 362 
 
 , Note. — The student's understanding of this theorem will be increased 
 greatly if a model of the above figure is at hand. 
 
362 SOLID GEOMETRY — BOOK VII 
 
 Note 1. — Proof that LN' (step 3, § 551) is a rectangular parallelopiped. 
 
 1. Since FG ± plane GH', .-. plane LH± plane MH'. § 495 
 
 2. Since MM' ± MN, .-. MM' ± plane LH. § 496 
 
 3. .-. Z LMM' = a rt. Z. 
 
 4. .'. LM' is a rectangle. § 141 
 
 5. .-. LN' is a rectangular parallelopiped. § 537 
 Note 2. — Proof that LMNP = ABCD. 
 
 LMNP = FGHK, since Os having equal bases and equal altitudes 
 are equal. Similarly FGHK= ABCD, and .-. LMNP = ABCD. 
 
 Proposition X. Theorem 
 
 552. The volume of a triangular prism is equal to 
 the product of its base and altitude. 
 
 Hypothesis. C'-ABC is a triangular prism. 
 Length of altitude AE = H ; area of A ABC = B ; volume 
 oi C'-ABC =V. 
 Conclusion. F= BH. 
 
 Suggestions. — 1. Consider the parallelopiped D'-ABCD, having its edges 
 parallel to AB, BC, and BB' respectively. 
 
 2. Compare volume of C'-ABC with that of D'-ABCD. 
 
 3. Express the volume of D'-ABCD, and then of C'-ABC. 
 
 Note. — At this point, the pupil should memorize the following formulae 
 if they are not already known. 
 
 1. Area of a A = V s(s - a)(^s -h){s- c), § 335 
 where the sides are a, 6, and c, and s = |(a + & + c). 
 
 2. Area of an equilateral A of side s = ^^-^. (Ex. 29, p. 199, Book IV.) 
 
 4 
 
 sV3 
 
 3. Altitude of an equilateral A of side s = 
 
PRISMS AND PARALLELOPIPEDS 363 
 
 Ex. 34. Find the volume of a regular triangular prism the side of 
 whose base is 6 and whose altitude is 10. 
 
 Ex. 35. Derive a formula for the volume of a regular triangular 
 prism the side of whose base is s and whose altitude is h. 
 
 Ex. 36. Find the lateral area and volume of a right triangular prism, 
 having the sides of its base 4, 7, and 9, respectively, and the altitude 8. 
 
 Suggestion. — To determine the area of the base, use the first formula in 
 the note of § 552. 
 
 Ex. 37. Prove that the volume of a right triangular prism is equal to 
 the product of the area of any face and one half the altitude to that face. 
 
 Ex. 38. The volume of a triangular prism is QQVE, and one side of 
 
 its base is 8. Find its lateral area. 
 
 Proposition XI. Theorem 
 
 553. The volume of any prism is equal to the prod- 
 uct of its base and altitude. 
 
 Hjrpothesis. Let H = the length of altitude AMj B = the 
 area of base FGHJK, and V= the volume of prism AJ. 
 Conclusion. F= BH. 
 
 Suggestions. — 1. Through edge AF mid diagonals FH and FJ of the base, 
 pass planes AFHC and AFJD dividing prism P into triangular prisms P\y 
 P2, and Ps, whose base areas are Bi, B^, and B^, resj)ectively, and whose com- 
 mon altitude is of length H. 
 
 2. Express the volumes of Pj, P^, and P^. Add the results and simplify, 
 thus obtaining an expression for the volume P. 
 
364 SOLID GEOMETRY — BOOK VII 
 
 554. Cor. 1. Two prisms having equal bases and equal alti- 
 tudes are equal. 
 
 Suggestion. — Let prisms P and P' have equal bases B and B' respec- 
 tively, and equal altitudes H and H' . Prove P = P' . 
 
 555. Cor. 2. Tico prisms haviiig equal altitudes have the 
 same ratio as their bases. 
 
 556. Cor. 3. Two prisms having equal bases have the same 
 ratio as their altitudes. 
 
 557. Cor. 4. Two prisms have the same ratio as the products 
 of their bases and altitudes. 
 
 Ex. 39. Find the volume of a regular hexagonal prism the side of 
 whose base is 4 in. and whose altitude is 9 in. 
 
 Ex. 40. Express by formulae the total area and the volume of a 
 regular hexagonal prism whose base edge is e and whose height is h. 
 
 Ex. 41, A contractor agreed to dig a cellar at 35^ per cubic yard. 
 The lot was located upon a hillside so that the depth of the cellar at the 
 back was 9 ft. and in front 5 ft. If the cellar was 38 ft. from the front 
 to back, and was 25 ft. wide, how much did the contractor receive ? 
 
 Ex. 42. How many cubic yards of concrete are re- 
 quired for a retaining wall 2 ft. thick whose dimensions 
 are indicated on the adjoining figure ? " ' W 
 
 Note. — Supplementary Exercises 29-33, p. 457, can be studied now. 
 
 PYRAMIDS 
 
 558. A Pyramid is a polyedron bounded by three or more 
 triangular faces wliicli have a common vertex, and one other 
 plane face, the Base, which intersects each of 
 
 the triangular faces. 
 
 The common vertex of the triangular faces is 
 the Vertex of the pyramid ; the triangular faces 
 are the Lateral Faces ; the edges terminating at 
 the vertex are the Lateral Edges ; the sum of the 
 areas of the lateral faces is the Lateral Area; 
 the perpendicular from the vertex to the plane of the base is 
 the Altitude. 
 
PYRAMIDS 
 
 365 
 
 559. A pyramid is called triangular, quadrangular, etc., ac- 
 cording as its base is triangular, quadrangular, etc. 
 
 A Regular Pyramid is a pyramid whose base 
 is inclosed by a regular polygon, and whose 
 vertex lies in the perpendicular to the base at 
 the center of the base. 
 
 A Truncated Pyramid is the part of a pyramid 
 included between its base and a plane cutting 
 all the lateral edges. 
 
 The base of the pyramid and the section of the cutting plane 
 are called the bases of the truncated pyramid. 
 
 560. A Frustum of a Pyramid is a truncated 
 pyramid whose bases are parallel. 
 
 The Altitude of a frustum is the perpendic- 
 ular between the planes of the bases. 
 
 561. The following important facts 
 about pyramids should be proved by the 
 pui)il : 
 
 I. The lateral edges of a regular pyramid 
 are equal. 
 
 II. The lateral faces of a regular pyramid 
 are inclosed by congruent isosceles triangles. 
 
 III. TJie lateral faces of a frustum of any pyramid 
 closed by trapezoids. 
 
 IV. The lateral faces of a frustum of a 
 regular pyramid are inclosed by congruent 
 trapezoids. 
 
 Suggestion. — Superpose A OAB on A OBC. 
 Prove ABB' A' ^ BCC'B'. 
 
 V. TJie lateral edges of a frustum of a 
 regular pyramid are equal. 
 
 Note. — It will be assumed that the boundary of the base of the 
 is a convex polygon. 
 
366 SOLID GEOMETRY — BOOK VII 
 
 Proposition^ XII. Theorem 
 
 562. If a pyramid is cut hy a plaiie parallel to the 
 base : 
 
 I. The lateral edges and the altitude are divided 
 proportionally. 
 
 II. The section is sir)iilar to the base. 
 
 B 
 
 Hypothesis. Plane A'C^, parallel to the base of pyramid 
 0-ABCD, intersects faces OAB, OBC, OGD, and ODA in 
 lines A'B', B'C, CD', and D'A', respectively, and altitude OP 
 at P'. 
 
 Conclusion. I. M'^ M'= 0^= OP'. 
 OA OB on OP 
 
 II. A'B'O'D' ~ ABCD. 
 
 Proof of I. 1. Through 0, pass plane MN || ABCD. 
 
 2 . OA^^OB^^^qc^qD^^qp^ ..^^ 
 
 " OA OB OC OD op' 
 Proof of n. 1. Z A'B'C = Z ABC, Z B'CD' = Z BCD, 
 etc. Prove it. § 481 
 
 ^ A^^OAi B^C^OB: ^^^ 
 
 AB OA' BC OB' ^ 
 
 ^ . A^^B^C^OD^^D^A^^ ^^., 
 
 " AB BC CD DA ^ ' 
 
 4. .'.A'B'C'D'^ABCD. Why? 
 
PYRAMIDS 
 
 367 
 
 563. Cor. 1. Tfie area of a section of a pyramid parallel to 
 the base is to the area of the base as the square of its distance 
 from the vertex is to the square of the altitude of the pyramid. 
 
 Proof. 1. 
 
 2. But 
 
 ArQ2L A' B' C D' A'B'' 
 
 3. 
 
 Area ABCD 
 
 AB' 
 
 A'B> 
 AB 
 
 OA' 
 OA 
 
 OP' 
 OP 
 
 Area, A' B' CD' 
 
 OP'^ 
 
 §344 
 
 Area ABCD 
 
 OP 
 
 564. Cor. 2. If two pyramids have equal altitudes and equal 
 basesi sections parallel to the bases at equal distances from the 
 vertices are equal. 
 
 Hypothesis. Pyramids 0-ABC and O'-A'B'C have the common 
 altitude //, and equal bases, ABC and A'B'C. 
 
 DEF and D'E'F' are sections of the pyramids parallel to the bases at 
 the distance h from the vertices O and 0', respectively. 
 
 Conclusion. DEF= D'E'P. 
 
 Area DEF h^ _„^ Area. D' E' F^ h^ 
 
 Proof. 1. 
 
 ^ and Area D'E'Ff 
 
 §563 
 
 Area ABC m Area A'B'C 
 
 Complete the proof. 
 
 Ex. 43. Prove that the areas of sections of a pyramid made by 
 planes parallel to the base have the same ratio as the squares of the dis- 
 tances to the planes from the vertex. 
 
 Ex. 44. The altitude of a pyramid is 12 in., and its base is a square 
 9 in. on a side. "What is the area of a section parallel to the base, 
 whose distance from the vertex is 8 in. ? 
 
 Ex. 45. What part of the area of the base of a pyramid is the area of a 
 section made by a plane which is parallel to the base and bisects the altitude ? 
 
 Note. — Supplementary Exercises 34-37, p. 457, can be studied now. 
 
368 
 
 SOLID GEOMETRY — BOOK VII 
 
 565. The slant height of a regular pyramid is the altitude of 
 
 any lateral face. (See II, § 561.) 
 
 The slant height of a frustum of a regular pyramid is the 
 altitude of any lateral face. (See III, § 561.) 
 
 Ex. 46. Prove that the perimeter of the mid-section of a frustum of 
 a pyramid is one half the sum of the perimeters of the bases. 
 
 Ex. 47. What is the slant height of a regular quad- 
 rangular pyramid whose altitude is 12 and the side of 
 whose base is 4 ? 
 
 Suggestions. — 1. Let ABV be one face, VC be the alti- 
 tude, and (7 the center of the base. 
 
 2. Determine the length of AC, and of DC; then of VD. 
 
 Ex. 48. What is the slant height of a regular hexagonal pyramid 
 whose altitude is 10 and whose base edge is 4 ? 
 
 Proposition XIII. Theorem 
 
 566. The lateral area of a regular pyramid is equal 
 to the perimeter of its base multiplied by one half its 
 slant height. 
 
 Hypothesis. 0-ABCDE is a regular pyramid. 
 P = perimeter of its base ; L = length of its slant height 
 S= the lateral area. 
 
 Conclusion. 
 Proof. 1. 
 
 S=iPL. 
 
 iLx AB. 
 
 Area of A OAB 
 
 Complete the proof. 
 
 Why? 
 
PYRAMIDS 369 
 
 567. Oor. Tlie lateral area of the frustum of a regular pyra- 
 mid is equal to one half the sum of the perimeters of the bases 
 multiplied by the slant height. 
 
 Hypothesis. AD' is a frustum of a regular pyramid. 
 L = the length of the slant height ; p and P = the perimeters of the 
 upper and lower bases respectively ; and S = the lateral area. 
 
 Conclusion. S=^L(^p + P). 
 
 Proof. 1. Aresiof ABB' A' = ^L(AB + A' B'). Why? 
 
 Complete the proof. 
 
 Ex. 49. What is the slant height of a regular triangular pyramid 
 whose altitude is 10 and whose base edge is 4 ? 
 Suggestion. — Recall § 172. 
 
 Ex. 50. Express the lateral area of a regular pyramid in terms of 
 the length of the slant height and the perimeter of the section midway 
 between the base and the vertex. 
 
 Ex. 51. Prove the lateral surface of any pyramid greater than its 
 base, when the perpendicular from the vertex to the base falls within the 
 base. 
 
 Suggestion. — From the foot of the altitude draw lines to the vertices of 
 the base ; each A formed has a smaller altitude than the corresponding 
 lateral face. 
 
 Ex. 52. Determine the lateral area of each of the pyramids in Exer- 
 cises 47 to 49. 
 
 Ex. 53. In each of the Exercises 47 to 49 pass a plane parallel to the 
 base at a distance of 5 in. from the vertex. Determine the lateral areas 
 of the resulting frustums. 
 
 Ex. 54. Determine the total areas of the pyramids of Exercises 47 to 
 49. 
 
 Ex. 55. The edges of the bases of a frustum of a regular square 
 pyramid are 5 in. and 10 in. respectively, and the altitude is 6 in. Deter- 
 mine the slant height and then the lateral area. 
 
370 
 
 SOLID GEOMETRY — BOOK VII 
 
 Pkoposition XIV. Theokem 
 568. Two triangular pyramids having equal altitudes 
 
 and equal bases are equal. 
 
 Hypothesis. 0-ABC and O'-A'B'C have equal altitudes 
 and equal bases ABC and A'B'C. 
 
 
 Conclusion. 
 
 0-ABC = O'-A'B'C. 
 
 Proof. 1. Place the pyramids with their bases in the same 
 plane, and let H represent their common altitude. 
 
 Divide H into 3 equal parts. 
 
 Through the points of division pass planes II to the plane of 
 the bases, cutting 0-ABC in sections DEF and GHK, and 
 O'-A'B'C in sections D'E'F' and G'H'K'. 
 
 .'. DEF = D'E'F' 
 and GHK= G'H'K'. 
 
 §564 
 
 3. With ABC, DEF, and GHK as lower bases, construct 
 prisms X, Y, and Z with their lateral edges equal and II to AD ; 
 with D'E'F' and G'H'E^ as upper bases, construct prisms F' 
 and Z', with their lateral edges equal and II to A'D'. 
 
 4. 
 
 .-. prism Y= prism F' 
 and prism Z = prism Z'. 
 
 Why? 
 
PYRAMIDS 371 
 
 6. Hence the sum of the prisms circumscribed about 
 0-ABC exceeds the sum of the prisms inscribed in O-A'B'O 
 by prism X. 
 
 6. Evidently, 0-ABO <X+Y+Z, 
 
 and 0-ABC > Y + Z\ 
 Likewise O'-A'B'C <X+Y-\-Z and > F' -h Z\ 
 
 7. .-. 0-ABC and O'-A'B'C differ by less than the differ- 
 ence between the sum of the circumscribed prisms and the sum 
 of the inscribed prisms ; 
 
 i.e. 0-ABC and O'-A'B'C differ by less than the lower 
 prism X. 
 
 8. By increasing indefinitely the number of subdivisions 
 of H, the volume of X can be made less than any assigned 
 number, however small. 
 
 9. Suppose now that the volume of 0-ABC and 0- 
 A'B^O differ by any amount k. 
 
 Since X > the difference between 0-ABC and 0'-A'B'C\ 
 then X would be > k. 
 
 10. But this contradicts step 8. 
 
 11. .-. O'-A'B'C and 0-ABC cannot differ at all ; 
 
 i.e. O'-A'B'C = 0-ABC. 
 
 Note. — An interesting and instructive exercise at this point is that of 
 proving the equality of two triangles which have equal bases and alti- 
 tudes, by a proof like that given for Proposition XIV. In fact, it aids in 
 understanding Proposition XIV if the exercise proposed is studied before 
 taking up § 668. 
 
 The two triangles are compared with two sets of parallelograms, one 
 inscribed in one triangle, the other circumscribed about the other triangle. 
 The resulting figures are like the triangles AOB and A'O'B' of the figure 
 of § 668. 
 
372 SOLID GEOMETRY — BOOK VII 
 
 Proposition XV. Theorem 
 
 569. TJie volume of a triangular pyramid is equal to 
 one third the product of its base and altitude. 
 
 Hypothesis. 11= the length of the altitude ; 5= the area 
 of the base ; and V= the volume of pyramid 0-ABC. 
 
 Conclusion. V^l HB. 
 
 Proof. 1. Let EOD-ABC be the triangular prism having 
 base ABC, and its lateral edges equal and parallel to OB. 
 
 2. Prism EOD-ABC is composed of pyramid 0-ABC and 
 pyramid 0-ACDE. 
 
 3. Divide 0-ACDE into two triangular pyramids, 0-ACE 
 and 0-CDE by passing a plane through E, 0, and C. 
 
 4. In pyramids 0-ACE and 0-ECD : 
 
 The altitudes are common. Why ? 
 
 Base ACE = base ECD. Why ? 
 
 .-. 0-ACE = 0-ECD. Why ? 
 
 5. Pyramid 0-ECD is the same as pyramid C-EOD. 
 
 6. In pyramids 0-ABC and C-EOD : 
 
 The altitudes are equal. Why ? 
 
 Base ABC = base EOD. Why ? 
 
 .-. 0-ABC = C-EOD. Why ? 
 
 7. .-. O-ABC = 0-ECD = 0-ACE. 
 
 8. .-. 0-ABC = i prism EOD-ABC. 
 
 9. The altitude of prism EOD-ABC = H, and base = B. 
 
 10. .-. vol. EOD-ABC = HB. Why ? 
 
 11. .-. vol. 0-ABC = i HB. 
 
PYRAMIDS 
 
 373 
 
 Ex. 56. If the base of a pyramid is a parallelogram, the plane deter- 
 mined by the vertex of the pyramid and a diagonal of the base divides the 
 pyramid into two equal triangular pyramids. 
 
 Ex. 57. Determme the ratio to a given parallelopiped of the pyramid 
 whose lateral edges are the three edges of the parallelopiped which inter- 
 sect at any one corner. 
 
 Ex. 58. Each side of the base of a regular triangular 
 pyramid is 6, and its altitude is 4. Find its lateral 
 edge, lateral area, and volume. 
 
 Suggestion. — In the figure, C is the center of the base. 
 
 Ex 59. Find the area of the entire surface and the 
 volume of a triangular pyramid, each of whose edges 
 is 2. 
 
 Proposition XVI. Theorem 
 570. The volume of any pyramid is equal to one third 
 
 the product of its base and altitude. 
 
 B C 
 
 The proof is like that for § 553. 
 Proof to be given by the student. 
 
 571. Cor. 1. Any two pyramids having eqttal bases and equal 
 altitudes are equal. 
 
 Cor. 2. Two pyramids having equal altitudes have the same 
 ratio OS their bases. 
 
 Cor. 3. Two pyramids having equal bases have the same ratio 
 as their altitudes. 
 
 Cor. 4. Any two pyramids have the same ratio as the products 
 of their bases and altitudes. 
 
374 SOLID GEOMETRY — BOOK VII 
 
 Proposition XYII. Theorem 
 
 572. The volume of a frustum of any pyramid is 
 equal to one third of its altitude multiplied hy the sum 
 of its upper base, its lower base, and the mean pro- 
 portional between its bases. 
 
 Hypothesis. F=the volume, B = the area of the lower 
 base, b = the area of the upper base, and H = the length of 
 the altitude of AC, a frustum of any pyramid 0-AC. 
 
 Conclusion. V=iH(B-\-b-{- VBb). 
 
 Proof. 1. Draw altitude OP, cutting A'C at Q. 
 
 2. V= vol. 0-AC - vol. O-A'C 
 
 = i^B{H+oq)-\h{oq) 
 
 = iHB-^iOQ(B-b). 
 
 3. But B:b= OP: OQ^ § 563 
 
 4. .-. VB :Vb= OP: OQ. Algebra 
 
 5. .-. (VB-VI) : V6= (OP - OQ) :OQ = H: OQ. 
 
 § 256 
 
 6. .-. OQ (VB - Vb) = HVb. 
 
 7. Multiplying both members by V-B -f- V6, 
 
 ... OQ {B-b) = H(VBb + b). 
 
 8. Substituting in step 2 for OQ(B — b) its value from 
 step 7, 
 
 V=\HB + \H{-\/Bb^+b) 
 
 = i H{B + ?> + V56). 
 
 1 
 
PYRAMIDS 375 
 
 Ex. 60. Find the volume of a regular quadrangular pyramid each 
 side of whose base is 3, and whose altitude is 6. 
 
 Ex. 61. Find the volume of a regular hexagonal pyramid each side of 
 whose base is 4, and whose altitude is 9. 
 
 Ex. 62. The slant height and lateral edge of a regular quadrangular 
 pyramid are 25 and \/674, respectively. Find its lateral area and volume. 
 
 Ex. 63. Prove that the lines joining the center of a cube to the four 
 vertices of one face are the edges of a regular quadrangular pyramid whose 
 volume is ^ that of the cube. 
 
 Ex. 64. Express the volume of a pyramid in terms of its altitude and 
 the area of its mid-section parallel to the base. 
 
 Ex. 65. Find the lateral area and volume of a regular quadrangular 
 pyramid, the area of whose base is 100, and whose lateral edge is 18. 
 
 Ex. 66. Find the area of the base of a regular quadrangular pyramid, 
 whose lateral faces are equilateral triangles, and whose altitude is 5. 
 
 Suggestion. — Represent the lateral edge and the side of the base by x. 
 
 Ex. 67. Find the volume of a frustum of a regular quadrangular 
 pyramid, the sides of whose bases are 9 and 5, respectively, and whose 
 altitude is 10. 
 
   Ex. 68. Find the volume of a frustum of a regular triangular pyra- 
 mid, the sides of whose bases are 18 and 6, respectively, and whose alti- 
 tude is 24. 
 
 Ex. 69. Find the volume of a frustum of a regular hexagonal pyramid, 
 the sides of whose bases are 8 and 4, respectively, and whose altitude is 12. 
 
 Ex. 70. A monument is in the form of a frustum of a regular quad- 
 rangular pyramid 8 ft. in height, the sides of whose bases are 3 ft. and 
 2 ft., respectively, surmounted by a regular quadrangular pyramid 2 ft. 
 in height, each side of whose base is 2 ft. What is its weight, at 180 lb. 
 to the cubic foot ? 
 
 Ex. 71. The areas of the bases of a frustum of a pyramid are 12 and 
 75 respectively, and its altitude is 9. What is the altitude of the pyramid ? 
 
 Suggestion. — Let the altitude of the pyramid D' ^C ' 
 
 = z ; then a; — 9 is the 1 from its vertex to the /; ^1__^/A 
 
 upper base of the frustum ; then use § 564. f^ Y ' /^] ^c \ \ 
 
 Ex.72. The lateral edge of a frustum of / Dil..|__A.l.V--Ac 
 
 a regular hexagonal pyramid is 10, and the / / j\ [ \[ \/ 
 
 sides of its bases are 10 and 4, respectively. 1/ V/C^ \/^ 
 Find its lateral area and volume. A p g 
 
 Note. — Supplementary Exercises 38-42, p. 457, can be studied now. 
 
376 
 
 SOLID GEOMETRY — BOOK VII 
 
 SUPPLEMENTARY TOPICS 
 
 Five groups of supplementary material follow. None of this 
 material is needed for subsequent parts of solid geometry. 
 Each group is independent of each of the others. The groups 
 are arranged in order of importance and of interest. The 
 teacher should select from this material such parts as seem 
 best to meet the needs of the class. 
 
 Group A. Prismatoids 
 
 573. A Prismatoid is a polyedron bounded by two parallel 
 faces called Bases, and by a number of lateral faces which are 
 bounded by either triangles, trapezoids, or parallelograms. 
 
 The Altitude of a prismatoid is the perpendicular between 
 the bases. 
 
 The Mid-section of a prismatoid is the section of the plane 
 parallel to the base and midway between them. 
 
 Proposition XVIII. Theorem 
 
 574. . If the areas of the lower and upper bases of a prisma- 
 toid are B and b, respectively, the area of the mid-section is m, 
 the length of the altitude is H, and the volume is V, then, 
 
 V=\H{B^b-\-4.m). 
 
 Proof. 1. Through any point Fof the mid-section and each 
 edge of the prismatoid, pass planes. These planes divide the 
 
PRISMATOIDS 377 
 
 prismatoid into pyramid V-ABC, pyramid V-DEFO, pyramids 
 like V-BCFy and polyedra like V-ABED. 
 
 (a) Volume V-ABC =^\'\ H h = \ Ub. 
 
 (b) Similarly, volume V-DEFG = | IIB. 
 
 (c) To compute the volume of V-BCF : 
 
 1. Draw VE and VS^ thus forming A VRSy which is a part 
 of the mid-section. 
 
 2. Draw plane VBS. 
 
 3. V-BCF = V-RSF + V-BRS + V-BCS. 
 
 4. V-RSF=\-^H'AVRS = \H'AVR1S. Prove it. 
 
 5. V-BRS = \-^H'AVRS = \H'AVRS. Prove it. 
 
 6. V-BCS=^2 . V-BRS, for A BGS=2 • A BRS. § 571, Cor. 2. 
 
 .-. F-J5C>S'=f ^.AFie.S'. 
 
 7. .-. V-BCF =iH'A VRS. 
 
 Similarly, the volume of any triangular pyramid with vertex 
 V and as base a triangular lateral face is equal to |^ ^ multi- 
 plied by that part of m which is in the triangular pyramid. 
 
 (d) V-ABED can be divided into two triangular pyramids 
 by passing a plane through V, A, and E. Hence its volume 
 can be obtained as in part (c). 
 
 (c) Hence, the sum of all pyramids with vertex F, whose 
 bases are lateral faces of the prismatoid, is ^ H^m. 
 
 (/) .-. volume of the prismatoid = ^ HB -f ^ Hb + J Hm 
 
 = iH(B-^b-\-4.m). 
 
 Note. — This Proposition is particularly interesting not alone because 
 it enables us to determine the volumes of many irregularly shaped figures, 
 but because it includes many previous propositions as special cases. 
 
 Ex. 73. Is a prism a special case of a prismatoid ? In a prism, what 
 relation is there between J5, 6, and m ? Does the formula of § 674 reduce 
 to the usual formula for the volume of a prism ? 
 
 Ex. 74. Answer the same questioiw for a pyramid. 
 
378 SOLID GEOMETRY — BOOK VII 
 
 Group B. Truncated Prisms 
 Proposition XIX. Theorem 
 
 575. A truncated triangular prism is equal to the sum of 
 three pyramids having as common base the lower base of the 
 given prism, and having as their vertices the three vertices of the 
 upper base of the truncated prism. p 
 
 Hypothesis. DEF-ABG is a truncated triangular prism. 
 
 Conclusion. DEF-ABG = E-ABC + D-ABC + F-ABO. 
 
 Proof. 1. Pass planes through A, E, and O, and through 
 D, E, and C, thus dividing DEF-ABQ into E-ABC, E- 
 ADC, and E-DFC. 
 
 2. E-ABC is one of the required pyramids. 
 
 3. E-DAC= B-DAC. Cor. 1, § 571 
 [For the altitudes from B and E to ADC are equal.] 
 
 -Bvit B-DAC = D-ABC 
 .-. E-DAC = D-ABC, the second required pyramid. 
 
 4. E-DFC=B-AFC. §571 
 
 [For A DFC = A AFC, and the altitudes from E and B to DFC and 
 AFC, respectively, are equal.] 
 
 But B-AFC = F-ABC 
 .'. E-DFC = F-ABC, the third required pyramid. 
 
 5. .-. DEF-ABC = E-ABC + D-ABC + F-ABC. 
 
 576. Cor. 1. The volume of a truncated right triangular 
 prism is equal to one third the base multiplied by the sum of the 
 lateral edges. 
 
 
TRUNCATED PRISMS 
 
 379 
 
 577. Cor. 2. Tlie volume of any tnincated triangular prism is 
 equal to the product of one third the area of a ^f 
 
 right section by the sinn of the lateral edges. jy^ 
 
 Suggestions. — 1. Let XYZ be the right section whose 
 area is r. 
 
 2. Apply § 676, to DEF-XYZ, and to ABC-XYZ. 
 
 3. Prove DEF-ABC = l^r {AD -\- BE + CF) . 
 
 Ex. 75. Find the volume of a truncated right triangular prism, the 
 sides of whose base are 5, 12, and 13, and whose lateral edges are 3, 7, 
 and 5, respectively. 
 
 Ex. 76. Find the volume of a truncated right triangular prism 
 whose lateral edges are 11, 14, and 17^ having for its base an isosceles 
 triangle whose sides are 10, 13, and 13, respectively. 
 
 Ex. 77. Find the volume of a truncated regular quadrangular prism, 
 a side of whose base is 8, and whose lateral edges, taken in order, are 2, 
 C, 8, and 4, respectively. 
 
 Suggestion. — Pass a plane through two diagonally opposite lateral edges, 
 dividing the solid into two truncated right triangular prisms. 
 
 Ex. 78. If ABCD is a rectangle, and EF is any 
 line not in its plane parallel to AB, the volume of the 
 solid bounded by figures ABCD, ABFE, CDEF, ADE, 
 and BCF, is 
 
 Ih X AD x(2AB-\- EF), 
 
 where h is the perpendicular from any point of EF to 
 ABCD. § 577 
 
 Ex. 79. If ABCD and EFQH are rectangles lying in parallel 
 planes, AB and BC being parallel to EF and FQ, respectively, the solid 
 bounded by the figures ABCD, EFGH, ABFE, 
 BCGF, CDHQ, and DAEH, is called a rec- 
 tangular prismoid. 
 
 ABCD and EFGHa.Te called the bases of the 
 rectangular prismoid, and the perpendicular dis- 
 tance between them, the altitude. 
 
 Prove the volume of a rectangular prismoid 
 equal to the sum of its bases, plus four times A B 
 
 a section equidistant from the bases, multiplied by one sixth the altitude. 
 
 Suggestion. —Pass a plane through CD and EF, and find the volumes of 
 the solids ABCD-EF and EFGH-CD by Ex. 78. 
 
 Note. — Supplementary Exercises 43-47, p. 468, can be studied now. 
 
 
 
 — 
 
 f 
 
 e/- 
 
 1 
 
 <F 
 
 \ 
 
 h' 
 
 -'"n" 
 
 \ 
 
 ^ 
 
 kA 
 
 
 \ 
 
 
 // 
 
 V- 
 
 
380 
 
 SOLID GEOMETRY — BOOK VII 
 
 Group C. Miscellaneous Theorems 
 
 The following theorems about tetraedra are very much 
 like certain theorems about triangles and about triedral angles. 
 
 Proposition XX. Theorem 
 
 578. Two tetraedra having a triedral angle of one equal to 
 a triedral angle of the other, have the same ratio as the products 
 of the edges including the equal triedral angles. 
 
 Hypothesis. V and F' are 
 
 volumes of tetraedra 0-ABC 
 and O-A'B'C, respectively, 
 having the common triedral 
 ZO. 
 
 Conclusion. 
 
 V ^ OAx OBx 00 
 
 V OA' X OB' X 00'' 
 
 Proof. 1. Draw lines OP and O'P' ± to face OA'B'. 
 
 2. Let their plane intersect face OA'B' in line OPP'. 
 
 3. Now, OAB and OA'B' are the bases, and OP and O'P' 
 the altitudes, of triangular pyramids 0-0 AB and 0'-OA'B\ 
 respectively. 
 
 4 . F^ area OAB x OP 
 
 " V 
 
 area OA'B' x O'P' 
 
 But 
 
 _ area OAB 
 " area OA'B' 
 area OAB 
 
 X 
 
 OP 
 O'P'' 
 OA X OB 
 
 area OA!B' OA x OB' 
 
 Also A OOP and 00' P' are rt. A. 
 Then A OOP and 00' P' are similar. 
 
 ' ' O'P' 00' ' 
 Substituting from steps 5 and 8 in step 4, 
 
 Why? 
 
 (1) 
 
 §346 
 
 Why? 
 Why? 
 
 Why? 
 
 V 
 
 OAx OB 00 OAx OB X 00 
 
 OA' X OB' 00' OA' X OB' x 00' 
 
MISCELLANEOUS THEOREMS 
 
 381 
 
 Ex. 80. State the theorem of plane geometry about triangles which 
 corresponds to Proposition XX. 
 
 Note. — For each of the following exercises, state also the correspond- 
 ing theorem about triangles. 
 
 Ex. 81. Prove that two tetraedra are congruent if a diedral angle 
 and the adjacent faces of one are congruent, respectively, to a diedral 
 angle and the adjacent faces of the other, if the congruent parts are 
 arranged in the same order. 
 
 Suf/gestion. — Prove by superposition. 
 
 Ex. 82. Two tetraedra are congruent if three faces of one are con- 
 gruent, respectively, to three faces of the other, if the congruent parts are 
 arranged in the same order. 
 
 Suggestion. — Recall § 515. 
 
 Ex. 83. Prove that the three planes passing 
 through the lateral edges of a triangular pyramid, 
 bisecting the sides of the base, meet in a common 
 straight line. 
 
 Ex. 84. Prove that the six planes through the 
 edges of a tetraedron bisecting the opposite edges 
 meet in a common point. 
 
 Suggestion. — By Ex. 83, three planes meet in line 
 VO. Let plane XBC intersect VO at Y. Prove Y lies 
 in the remaining two planes. 
 
 Note. — The common point is the center of gravity 
 of the tetraedron. 
 
 Ex. 85. Prove that the center of gravity of a 
 tetraedron divides the line drawn from any vertex 
 to the center of gravity of the opposite face in the ratio 3:1. 
 
 Ex. 86. Prove that the six planes bisecting the diedral angles of a 
 tetraedron meet in a common point. 
 
 Note. — Pupils will find it interesting to attempt to make up other 
 theorems about tetraedra which are suggested by theorems about 
 triangles. 
 
 Group D. Regular Polyedra 
 
 679. A Regular Polyedron is a polyedron whose faces are 
 congruent regular polygons, and whose polyedral angles are all 
 congruent. 
 
382 SOLID GEOMETRY — BOOK VII 
 
 Proposition XXI. Theorem 
 
 580. Not more than Jive regular convex poly edr a are possible. 
 A convex polyedral Z. must have at least three faces, and the 
 sum of its face A must be < 360°. § 514 
 
 1. With equilateral triangles. 
 
 Since each Z of an equilateral A is 60°, we may form a con- 
 vex polyedral Z. by combining either 3, 4, or 5 equilateral A- 
 
 Not more than 5 equilateral A can be combined to form a 
 convex polyedral Z. § 514 
 
 Hence not more than three regular convex polyedra can 
 be bounded by equilateral A. 
 
 2. With squares. 
 
 Since each Z of a square is 90°, we may form a convex poly- 
 edral Z by combining 3 squares. 
 
 Not more than 3 squares can be combined to form a convex 
 polyedral Z. 
 
 Hence not more than one regular convex polyedron can be 
 bounded by squares. 
 
 3. With regular pentagons. 
 
 Since each Z of a regular pentagon is 108°, we may form a 
 convex polyedral Z by combining 3 regular pentagons. 
 
 Not more than 3 regular pentagons can be combined to form 
 a convex polyedral Z. Why ? 
 
 Hence not more than one regular convex polyedron can be 
 bounded by regular pentagons. 
 
 4. With other regular polygons. 
 
 Since each Z of a regular hexagon is 120°, no convex poly- 
 edral Z can be formed by combining regular hexagons. Why ? 
 
 Hence no regular convex polyedron can be bounded by reg- 
 ular hexagons. 
 
 In like manner, no regular convex polyedron can be bounded 
 by regular polygons of more than six sides. 
 
 Therefore, not more than five regular convex polyedra are 
 possible. 
 
REGULAR POLYEDRA 
 
 383 
 
 Ex. 87. Prove that the following construction produces a regular 
 tetraedron having a given side. 
 
 Construction. — 1. With given side AB, construct 
 equilateral ABC. 
 
 2. At its center E, draw ED 1 ABC. 
 
 3. Take D on DE so that AD = AB, and draw AD, 
 BD, and CD. 
 
 Statement! — AB CD is a regular tetraedron. 
 Prove its faces are inclosed by congruent equilateral 
 As and that its triedral angles are congruent. 
 
 581. Models of the five regular polyedra may be made by 
 drawing upon cardboard figures like the following. 
 
 Cut out the figures along the outer line ; cut only halfway 
 through the cardboard on the inner lines; bring the edges 
 together and fasten them with gummed paper. 
 
 Tbtbaedbon 
 
 Hexaedron 
 
 OCTAEDEON 
 
 DODECAEDBON 
 
 ICOSAEDBON 
 
384 
 
 SOLID GEOMETRY — BOOK VII 
 
 Ex. 88. Make a model of at least one of the regular polyedra. 
 
 Ex. 89. What is the sum of the face angles at any vertex of each of 
 the regular polyedra ? 
 
 Ex. 90. Find the volume and the total area of a regular tetraedron 
 whose edge is 10. 
 
 Ex. 91. Find the volume and the total area of a regular tetraedron 
 whose edge is a. 
 
 Ex. 92. The sum of the perpendiculars 
 drawn to the faces from any point within a 
 regular tetraedron is equal to its altitude. 
 
 Suggestion. — Divide the tetraedron into trian- 
 gular pyramids, having the given point for their 
 common vertex. Find the volume of each and of b 
 the whole pyramid and form an equation. 
 
 Ex. 93. Prove that the volume of a regu- 
 lar octaedron is equal to the cube of its edge 
 multiplied by |Vii. 
 
 Group E. Similar Polyedra 
 
 582. Two polyedra are similar when they have the same 
 number of faces similar each to each and similarly placed, and 
 have their homologous polyedral angles congruent. 
 
 Ex. 94. Prove that the ratio of any two homologous edges of two 
 similar polyedra is equal to the ratio of any other two homologous 
 edges. 
 
 Ex. 95. Prove that any two homologous faces of two similar polye- 
 dra are to each other as the squares of any two homologous edges. 
 
 Ex. 96. Prove that the entire surfaces of two similar polyedra 
 are to each other as the squares of any two homologous edges. 
 
SIMILAR POLYEDRA 385 
 
 Proposition XXII. Theorem 
 
 583. Two tetraedra are similar when three face triangles 
 including a triedral angle of one are similar, respectively, to 
 three face triangles including a triedral angle of the other, and 
 similarly placed. 
 
 Hypothesis. In tetraedra ABCD and A'B'C'D' 
 
 A ABC - A A'B'C, A ACD - A A' CD', and 
 AADB'^AA'D'B'. 
 
 Conclusion. ABCD ~ A'B'C'D. 
 
 Proof. 1. From the given similar A, we have 
 
 BC fAC\ CD fAD\BD ™ . 
 
 B'C \A'G') CD' \A'D'J B'D' ^ ' 
 
 2. Hence, A BCD - A B',C'D'. Why ? 
 
 3. Again, A BAC, CAD, and DAB are equal, respectively, 
 
 to A B'A'C, C'A'D', and D'A'B'. Why ? 
 
 4. Then, triedral A A-BCD and A'-B'C'D' are congruent. 
 
 §516 
 6. Similarly, any two homologous triedral A are congruent. 
 
 6. Therefore, ABCD and A'B'C'D' are similar. § 582 
 
 Ex. 97. Two tetraedra are similar when a diedral angle of one is 
 congruent to a diedral angle of the other, and the face triangles including 
 the congruent diedral angles are similar each to each, and similarly placed. 
 
 Ex. 98. If a tetraedron be cut by a plane parallel to one of its faces, 
 the tetraedron cut off is similar to the given tetraedron. 
 
386 
 
 SOLID GEOMETRY — BOOK VII 
 
 Proposition XXIII. Theorem 
 
 584. Two similar tetraedra have the same ratio as the cubes 
 of any two homologous edges. 
 
 Hjrpothesis. Fand V^ are the volumes of similar tetraedra 
 ABCD and A'B'C'D', respectively, A and A' being homologous 
 vertices. 
 
 Conclusion. 
 Proof. 1. 
 
 2. 
 3. 
 
 V^ AB 
 
 Since ABCD ~ A'B'C'D', 
 triedral Z.A = triedral Z A'. 
 AB xACx AD 
 
 V 
 
 A'B' X A'C X A'D' 
 
 V A'B' 
 
 AB ^ AC ^^ AD 
 
 But 
 and 
 
 A'C" A'D' 
 AC AB 
 
 A'C A'B' 
 AD AB 
 
 A'D' A'B' 
 5. Substituting from step 4 in step 3, 
 
 V AB ^ AB ^ AB 
 
 6. 
 
 A'B' A'B' A'B' 
 
 V aS" 
 
 A^'' 
 
 §578 
 
 Why? 
 
SIMILAR POLYEDRA 387 
 
 Note 1. It can be proved that any two similar polyedra can be 
 divided into the same number of tetraedra, similar each to each and sim- 
 ilarly placed. Consequently, it can be proved that any two similar 
 polyedra have the same ratio as the cubes of any two homologous edges. 
 
 Note 2. It is interesting to notice that in similar figures 
 
 (a) two homologous lines have the same ratio as any two homologous 
 sides ; 
 
 (6) the areas of two homologous limited or bounded surfaces have the 
 same ratio as the squares of any two homologous sides ; 
 
 (c) the volumes of two homologous solid parts have the same ratio as 
 the cubes of any two homologous sides. 
 
 Ex. 99. The volume of a pyramid whose altitude is 7 in. is 686 cu. 
 in. Find the volume of a similar pyramid whose altitude is 12 in. 
 
 Ex. 100. If the volume of a prism whose altitude is 9 ft. is 171 cu. ft., 
 find the altitude of a similar prism whose volume is 50f cu. ft. 
 
 Ex. 101. Two bins of similar form contain, respectively, 375 and 648 
 bushels of wheat. If the first bin is 3 ft. 9 in. long, what is the length of 
 the second ? 
 
 Ex. 102. A pyramid whose altitude is 10 in. weighs 24 lb. At what 
 distance from its vertex must it be cut by a plane parallel to its base so 
 that the frustum cut off may weigh 12 lb. ? 
 
 Ex. 103. An edge of a polyedron is 66, and the homologous edge of a 
 similar polyedron is 21. The area of the entire surface of the second 
 polyedron is 135, and its volume is 162. Find the area of the entire sur- 
 face and the volume of the first polyedron. 
 
 Ex. 104. The area of the entire surface of a tetraedron is 147, and its 
 volume is 686. If the area of the entire surface of a similar tetraedron is 
 48, what is its volume ? 
 
 Suggestion. — Let z and y denote the homologous edges of the tetraedra. 
 
 Ex. 105. The area of the entire surface of a tetraedron is 75, and its 
 volume is 500. If the volume of a similar tetraedron is 32, what is the 
 area of its entire surface ? 
 
 Ex. 106. The homologous edges of three similar tetraedra are 3, 4, 
 and 6, respectively. Find the homologous edge of a similar tetraedron 
 equivalent to their sum. 
 
 Suggestion. — Represent the edge by z. 
 
BOOK VIII 
 
 THE CYLINDER AND THE CONE 
 
 585. Generating a Surface. If a m 
 
 straight line I moves so that it con- 
 stantly intersects a straight line b 
 and is constantly parallel to a 
 straight line a which intersects b, it 
 can be proved that / constantly lies 
 in the plane M determined by a and 
 b ; also it can be proved that every point in M lies in line I at 
 some time during its period of movement. 
 
 Line I is said to generate the plane M. 
 
 By suitable movement of a straight line, the straight line 
 can be made to generate various curved surfaces as well as a 
 plane surface. 
 
 586. A Cylindrical Surface is the surface generated by a 
 moving straight line which constantly intersects a given plane 
 curve and which is constantly parallel 
 to another given straight line not in 
 the plane of the curve. 
 
 Thus, if AB moves so as constantly to in- 
 tersect plane curve AD and to be constantly -^ 
 parallel to MN^ not in the plane of AB^ then 
 AB generates the cylindrical surface BD. 
 
 The moving line is called the Generatrix ; the curved line is 
 called the Directrix. Any position of the generatrix, as EF, 
 is called an Element of the cylindrical surface. 
 
 388 
 
THE CYLINDER AND THE CONE 389 
 
 Ex. 1. How many elements does a cylindrical surface have ? 
 Ex. 2. Consider any two elements of a cylindrical surface. What 
 kind of lines are they ? 
 
 Ex. 3. Can more than one cylindrical surface be generated by use of 
 a given directrix ? 
 
 587. If the directrix is a closed plane 
 curve, the cylindrical surface separates an 
 infinite part of space from surrounding 
 space. 
 
 The cylindrical surfaces considered in this 
 text always have closed convex directrices. 
 
 588. A Cylinder is the solid (§ 525) bounded by a portion 
 of a cylindrical surface and by portions of two ^ — — ,^^ 
 parallel planes which intersect all the elements j^ j 
 of the surface. /^ -^ 
 
 The portions of the parallel planes are the / / 
 
 Bases of the cylinder ; and the portion of the I,.- - -..^ / 
 
 cylindrical surface between the planes is the Lat- f J 
 
 eral Surface of the cylinder. 
 
 The Total Surface of a cylinder consists of its lateral surface 
 and its bases. 
 
 The perpendicular between the two parallel planes is the 
 Altitude of the cjdinder. The segment of an element of the 
 cylindrical surface which lies between the bases is an Element 
 of the cylinder. 
 
 A Section of a cylinder is the part of the cylinder common 
 to it and a plane cutting all the elements of the cylinder. 
 
 If a section of a cylinder is made by a plane perpendicular 
 to the elements, it is a Right Section of the cylinder. 
 
 Ex. 4. The elements of a cylinder are equal and parallel. 
 
 589. Kinds of Cylinders. A Right Cylinder is a cylinder 
 whose elements are perpendicular to its bases. 
 
 A Circular Cylinder is a cylinder whose base is inclosed by a 
 circle. 
 
390 SOLID GEOMETRY — BOOK VIII 
 
 Proposition I. Theorem 
 
 590. If a plane jJCtsses through an element of a cylin- 
 der and through at least one other point of the surface 
 of the cylinder, the intersection ivith the total surface 
 of the cylinder is a parallelogram. 
 
 Hypothesis. Plane M, passing through element AB of cylin- 
 der AF, intersects the lateral surface of ^i^in point C, not in JLB. 
 
 Conclusion. The section of AF made by plane M is inclosed 
 by a parallelogram. 
 
 Proof. 1. Through C draw CD II AB, 
 
 2. .-. CD is an element of surface AF, intersecting the bases 
 at C and D respectively. § 586 
 
 3. CD is also in M. § 447, I and III 
 
 4. .-. CD is in the intersection of AF and M. 
 
 5. BC and AD are il straight lines, lying in both M and the 
 bases of AF. Why ? 
 
 6. .-. ABCD is the intersection of ^i^and M. 
 
 7. Also ABCD is a O. Why ? 
 
 Ex. 5. If a rectangle revolves about one of its sides 
 
 as an axis, it generates a right circular cylinder. 
 
 Suggestion. — Prove that C describes a circle, that BC 
 generates a cylinder, and that the cylinder is a right cylin- 
 der. 
 
 591. As a consequence of Ex. 5, a right cir- 
 cular cylinder is also called a cylinder of revolution. 
 
THE CYLINDER AND THE CONE 391 
 
 Proposition II. Theorem 
 592. The bases of a cylinder are congruent 
 
 JE^ ^ 
 
 Hypothesis. ABf is any cylinder, with bases A'B' and AB. 
 
 Conclusion. Base A'B' ^ base AB. 
 
 Proof. 1. Let E' and F' be two particular points of curve 
 A'B' and G' any other point. Draw elements EE'j FF'j and 
 GG' ; also draw EF, FG, EG, E'F', F'G', and E'G'. 
 
 2. .-. A EFG ^ A E'F'G'. Prove it. 
 
 3. .*. base AB^ may be superposed on base AB so that E\ 
 F\ and & will fall upon E, F, and G respectively. 
 
 4. But G' was any point of AB' except E' and F\ 
 
 5. .-. E\ F, and every other point of curve AB^ will fall 
 upon a corresponding point of curve AB. 
 
 6. .-. base AB ^ base AB. Why ? 
 
 Note. — It is obvious that every point of curve AB can be proved to 
 fall upon a corresponding point of A'B'. 
 
 Ex. 6. Prove that a section of a circular cylinder made by a plane 
 parallel to the base is inclosed by a circle. 
 
 Note. — A section of a circular cylinder made by a plane not parallel 
 to the base is inclosed by a curve called an ellipse. 
 
 Ex. 7. Prove that a line drawn parallel to the 
 elements of a circular cylinder from the center of one 
 base intersects the other base at its center. 
 
 Suggestions. — 1. Let A' be a particular point and 
 B' be any other point of the bounding circle of the 
 upper base whose center is C . 
 
 2. Draw C'C II B'B, and draw element A' A. 
 
 3. Prove that CB = CA, and that C is the center of curve AB. 
 
392 
 
 SOLID GEOMETRY — BOOK VIII 
 
 593. The Axis of a circular cylinder is a straight line drawn 
 between the centers of its bases. 
 
 Ex. 8. Prove that the axis of a circular cylinder is parallel to the 
 elements of the lateral surface. 
 
 Ex. 9. Prove that the axis of a circular cylinder passes through the 
 centers of all sections parallel to the bases. 
 
 MEASURING THE CYLINDER 
 
 594. In measuring the cylinder, difficulties are encountered 
 which are like those met in measuring a circle. For example, 
 the ratio of the cylindrical surface to the customary unit of 
 surface measure cannot have meaning in the ordinary sense 
 since they are not magnitudes of the same kind, one being a 
 plane and one a curved surface. 
 
 595. Application of Limits to the Circle. 
 
 (a) The circumference of a circle, i.e. the length of a circle, 
 is defined to be the limit of the perimeter of any regular 
 inscribed polygon as the number of sides is indefinitely 
 increased. § 412 
 
 (h) It is proved that the perimeter of any regular circum- 
 scribed polygon (as well as inscribed polygon) approaches the 
 circumference of the circle as limit if the number of sides is 
 indefinitely increased. § 413 
 
 (c) It can be proved that the perimeter of any inscribed or cir- 
 cumscribed polygon approaches the circumference of the circle 
 as limit if the number of sides is increased indefinitely in such 
 manner that the length of every side approaches the limit zero. 
 
MEASURING THE CYLINDER 393 
 
 (d) The area of a circle is defined to be the limit of the 
 area of any regular inscribed polygon as the number of sides 
 increases indefinitely. § 417 
 
 (e) It is proved that the area of any regular circumscribed 
 polygon approaches the area of the circle as limit as the num- 
 ber of sides is increased indefinitely. § 418 
 
 (/) It can be proved that the area of any inscribed or cir- 
 cumscribed polygon approaches the area of the circle as limit 
 if the number of sides is increased indefinitely in such manner 
 that the length of each side approaches the limit zero. 
 
 Note. — Remember that = is the symbol for " approaches the limit.'* 
 
 596. A prism is inscribed in a cylinder when its lateral edges 
 are elements of the cylinder and its bases are ^^^-^•^^ 
 in the planes of the bases of the cylinder. J^'^^,' ^^ 
 The polygons bounding the bases of the J^^^y^^^^^ 
 prism are inscribed in the boundaries of the / / / / 
 bases of the cylinder. / / / / 
 
 597. Application of Limits to a Cylinder. h-^''t^>J 
 
 Inscribe in a circular cylinder a prism \x/ ..■';^ 
 having its base inclosed by a regular poly- 
 gon ; then inscribe a second prism whose base is inclosed by 
 a regular polygon having double the number of sides ; imagine 
 that this process is continued indefinitely. Pass a plane form- 
 ing right sections of the cylinder and the prisms. 
 
 It will be assumed evident that the prisms come nearer and 
 nearer to occupying the same space as the cylinder. As a 
 consequence : 
 
 (a) The Volume of the Cylinder is defined to be the limit of 
 the volume of the inscribed prism as the number of faces in- 
 creases indefinitely. 
 
 (h) The Lateral Area of a Cylinder is defined to be the limit 
 of the lateral area of the inscribed prism as the number of 
 faces increases indefinitely. 
 
 (c) The length of a right section of the lateral surface of 
 the cylinder is defined to be the limit of the length of the right 
 
394 SOLID GEOMETRY — BOOK VIII 
 
 section of the lateral surface of the inscribed prism as the 
 number of faces increases indefinitely. 
 
 (d) It is evident that the edge and altitude of the inscribed 
 prism equal respectively the element and altitude of the 
 cylinder. 
 
 Proposition III. Theorem 
 
 598. The lateral area of a circular cylinder is equal 
 to the perimeter of a right section multiplied by the 
 length of an element. 
 
 Hypothesis. 8 = the lateral area, P = the perimeter of a 
 right section, E = the length of an element of a circular 
 cylinder. 
 
 Conclusion. S = E x F. 
 
 Proof. 1. Inscribe in the cylinder a prism whose base is 
 inclosed by a regular polygon. 
 
 Let S' = the lateral area and P' = the perimeter of a right 
 section. 
 
 2. Then S' = E x P'. Why? 
 
 3. Let the number of faces of the prism increase indefinitely. 
 
 Then 
 
 S' = S, and P' = P. § 597, 6 and c ; also Note, § 595, 
 
 4. .'. ExF^ExP. § 543, a 
 
 5. .: S = E xP. § 543, 6 
 
 599. Cor. 1. The lateral area of a right circular cylinder is 
 equal to the circumference of its base multiplied by the length of 
 the altitude. 
 
MEASURING THE CYLINDER 395 
 
 600. Cor. 2. If R = the length of the radius of the base, 11 = 
 the length of the altitude^ S = the lateral area, and T = the total 
 area of a right circular cylinder, then 
 
 (a) S = 2irRH. 
 
 (6) T=27rE^ + 27rRH=27rR(E + H). 
 
 Ex. 10. Find the lateral area of a right circular cylinder whose alti- 
 tude i.s 16 and the diameter of whose base is 18. 
 
 Ex. 11. Find the total area of a cylinder of revolution whose altitude 
 is 15 and the radius of whose base is 5. 
 
 Ex. 12. Determine the cost at 15^ per square yard of painting the 
 vertical surface and top of a gas holder whose diameter is 30 ft. and 
 wliose height is 20 ft. 
 
 Ex. 13. How many square feet of tin are required to make 30 sec- 
 tions of hot-air furnace pipe 10 in. in diameter and 30 in. in length ? 
 
 Proposition IV. Theorem 
 
 601. The voluine of a circular cylinder is equal to the 
 area of its base multiplied by the length of its altitude. 
 
 Hypothesis. V = the volume, B = the area of the base, and 
 // = the length of the altitude of a circular cylinder. 
 
 Conclusion. V= H x B. 
 
 Proof. 1. Inscribe in the cylinder a prism having its base 
 inclosed by a regular polygon. Let F'= the volume and J5' = 
 the area of the base. 
 
 2. .-. F' = //X i?'. Why? 
 
 3. Increase indefinitely the number of faces of the prism. 
 Then V^ = V, and 2J' = B. 
 
 4. .'.HxB' = HxB, §543, a. 
 
 5. .'.V=HxB. §543,6. 
 
 602. Cor. If V = the volume, H = the length of the altitude, 
 and R = the radius of the base of a circiUar cylinder, then 
 
 F= wR'H. 
 
396 SOLID GEOMETRY — BOOK VIII 
 
 Ex. 14. What is the cost of digging a dry well 5 ft. in diameter and 
 15 ft. deep at 50 f per cubic yard ? 
 
 Ex. 15. What is the capacity in gallons of a water tank 12 ft. in 
 length and 36 in. in diameter, estimating 1\ gal. of water to a cubic foot ? 
 
 Ex. 16. How many cubic feet of metal are there in a hollow cylin- 
 drical tube 18 ft. long, whose outer diameter is 8 in. and whose thickness 
 is 1 in. ? 
 
 Ex. 17. Determine the number of cubic yards of concrete required 
 for the wall and floor of a circular cistern 8 ft. in outside diameter, and 
 12 ft. deep, if the walls and floor are 8 in. thick. 
 
 Ex. 18. Determine the diameter of a 2-bbl. water reservoir having 
 the form of a right circular cylinder if the length is 4 ft. (2 bbl.=63 gal. ; 
 1 cu. ft. contains 7| gal.) 
 
 Note. — Supplementary Exercises 48-58, p. 459, can be studied now. 
 
 THE CONE 
 
 603. A Conical Surface is the surface generated by a moving 
 straight line, which constantly intersects a given plane curve 
 and constantly passes through a given point 
 not in the plane of the curve. 
 
 Thus, if line OA moves so as constantly to in- 
 tersect plane curve ABC, and constantly passes 
 through point O, not in the plane of the curve, it 
 generates a conical surface. 
 
 The moving line is called the Generatrix, 
 and the curve the Directrix. 
 
 The given point is called the Vertex, and any position of the 
 generatrix, as OB, is called an Element of the surface. 
 
 If the generatrix be supposed to be indefinite in length, it 
 will generate two conical surfaces of indefinite extent, 0- 
 A'B'C and 0-ABC. 
 
 These are called the upper and lower nappes, respectively, of 
 the conical surface. 
 
 It will be assumed that the directrix is a closed plane curve 
 so that each nappe of the surface separates an infinite portion 
 of space from surrounding space. 
 
THE CONE 397 
 
 604. A Cone is a solid bounded by a portion of one nappe 
 of a conical surface and that part of a plane cutting all the 
 elements of the surface which lies within the 
 surface. 
 
 The plane is called the Base of the cone, and the 
 conical surface the Lateral Surface. 
 
 The Altitude of a cone is the perpendicular from 
 the vertex to the plane of the base. 
 
 605. Kinds of Cones. 
 
 A Circular Cone is a cone whose base is inclosed by a circle. 
 
 The Axis of a circular cone is a straight line drawn from the 
 vertex to the center of the base. 
 
 A Right Circular Cone is a circular cone whose axis is per- 
 pendicular to its base. 
 
 A Frustum of a Cone is a portion of a cone included between 
 the base and a plane parallel to the base. 
 
 The base of the cone is called the lower hose, and the section 
 made by the plane the upper base, of the frustum. 
 
 The altitude of a frustum is the perpendicular between the 
 planes of the bases. 
 
 Ex, 19. Prove that the elements of a right circular cone are equal. 
 
 Ex. 20. Prove that the elements of a frustum of a right circular cone 
 are equal. 
 
 Ex. 21. If a right triangle be revolved about one of its legs as an axis, 
 it generates a right circular cone. 
 
 606. As a consequence of Ex. 19, the distance from the ver- 
 tex to any point of the circle bounding the base of a right cir- 
 cular cone is called the Slant Height of the right circular cone. 
 
 As a consequence of Ex. 20, the portion of the slant height 
 of a right circular cone between the base and a plane parallel 
 to the base is called the Slant Height of the frustrum of the 
 right circular cone. 
 
 As a consequence of Ex. 21, a right circular cone is also 
 called a cone of revolution. 
 
398 SOLID GEOMETRY — BOOK VIII 
 
 Pkoposition y. Theorem 
 
 607. If a plane passes through an element of a cone 
 and through at least one other point of the surface of 
 the cone, the intersection with the total surface of the 
 cone is a triangle. 
 
 Hypothesis. Plane M, passing through element OC of cone 
 0-AB, intersects the surface again in point D, not in OC. 
 Conclusion. Section OCD is a A. 
 
 Proof. 1. CD is a straight line. Why ? 
 
 2. Since the base AB is inclosed by a closed line, line CD 
 intersects it in two points C and D which lie in the conical 
 surface. 
 
 3. Lines OC and OD are elements of the surface. Def . 
 
 4. Also OC and OD lie in the plane OCD. Why ? 
 
 5. .'. the complete intersection of the surface and the plane 
 is the triangle OCD. 
 
 Ex. 22. Find, correct to one decimal, the slant height of a right cir- 
 cular cone whose altitude is 9 in. and the radius of whose base is 3 in. 
 
 Ex. 23. Find the slant height of a right circular cone whose altitude 
 is h and the radius of wjiose base is r. 
 
 Ex. 24. The radii of the upper and lower bases of the frustum of a 
 right circular cone are 3 in. and 5 in. respectively; the altitude of the 
 frustum is 6 in. Determine the slant height correct to one decimal place. 
 
 Ex. 25. Find the altitude of a right circular cone whose slant height 
 is 13 and the radius of whose base is 5. 
 
 Ex. 26. Kepeat Ex. 24 when the radii are r and B respectively and 
 the altitude is if. 
 
THE CONE 
 
 399 
 
 Proposition VI. Theorem 
 
 608. TJie section of the lateral surface of a circular 
 cone made by a plane 2Mrallel to the base is a circle. 
 
 Hypothesis. A'B'C is the section of circular cone S-ABC, 
 made by a plane II base ABC whose center is O. 
 
 Conclusion. A'B'C is a ©. 
 
 Proof. 1. Draw axis O/S, intersecting A'B'C at O'. 
 
 2. Let A' and B' be any two points in curve A'B'C. Let 
 planes A'O'S and B'O'S intersect the base in radii OA and 
 OB, the cutting plane in lines O'A' and O'B', and the lateral 
 surface in lines SA'A and SB'B respectively. 
 
 3. SA'A and SB'B are straight lines. 
 
 4. A SA'O' ~ A SAO, and A SB'O' - A SBO. 
 
 5 . O'^' ^ O'B' 
 
 " OA OB' 
 
 .-. O'A' = O'B'. 
 
 § 607 
 Prove it. 
 
 Prove it. 
 Prove it. 
 
 6. 
 
 7. Since A' and B' are any two points of curve A'B'C and 
 are equidistant from O', curve A'B'C is a circle and 0' is its 
 center. 
 
 609. Cor. The axis of a circular cone passes through the 
 center of every section parallel to the base. 
 
 Ex. 27. Prove that the radii of the upper and lower bases of a frus- 
 tutn of any cone have the same ratio as the distances of the bases from 
 the vertex of the cone. 
 
400 SOLID GEOMETRY — BOOK VIII 
 
 MEASURING THE CONE 
 
 610. A pyramid is inscribed in a cone when y 
 its lateral edges are elements of the cone and /I 
 the bounding polygon of the base of the pyra- / / 
 mid is inscribed in the boundary of the base y^-'-'---y'''''^ 
 of the cone. The vertex and altitude of the f \ / \ j 
 pyramid coincide with the vertex and the ^"^-^^^U:::!':::^^^^ 
 altitude of the cone. 
 
 611. A frustum of a pyramid is inscribed in a frustum of a 
 cone when its lateral edges are elements of 
 
 the frustum of the cone, and the boundaries / S^ ^"T k 
 
 of the bases of the frustum of the pyramid /k \ M J . 
 
 are inscribed in the boundaries of the bases // / / / / 
 
 of the frustum of the cone. The altitude //- -/ ,. / / / 
 
 of the frustum of the pyramid coincides with /' i / \ k/ 
 
 the altitude of the frustum of the cone. \l/ '{/ 
 
 612. Application of limits to the Cone. 
 
 Inscribe in a circular cone a pyramid having a base inclosed 
 by a regular polygon ; then inscribe a second pyramid whose 
 base is inclosed by a regular polygon having double the num- 
 ber of sides ; imagine that this process is continued indefinitely. 
 It will be assumed evident that the pyramids come nearer 
 and nearer to occupying the same space as the cone. As a 
 consequence : 
 
 (a) The Volume of a Cone is defined to be the limit of the 
 volume of a regular inscribed pyramid as the number of faces 
 is increased indefinitely. 
 
 (6) The Lateral Area of a Cone is defined to be the limit of 
 the lateral area of a regular inscribed pyramid as the number 
 of faces is increased indefinitely. 
 
 (c) It can be proved that the slant height of a right circular 
 cone is the limit of the slant height of a regular inscribed pyra- 
 mid as the number of faces is increased indefinitely. 
 
 (d) The volume and lateral area of a frustum of a cone are 
 
MEASURING THE CONE 401 
 
 defined to be the limits of the volume and area respectively 
 of the frustum of a regular inscribed pyramid, having its base 
 inclosed by a regular polygon, as the number of faces is in- 
 creased indefinitely ; also, .the slant height of a frustum of a 
 right circular cone can be proved to be the limit of the slant 
 height of the frustum of a regular inscribed pyramid as the 
 number of faces is increased indefinitely. 
 
 Proposition YII. Theorem 
 
 613. The lateral area of a right circular cone is equal 
 to the circumference of its base multiplied by one half 
 its slant height. 
 
 Hypothesis. S = the lateral area, C = the circumference of 
 the base, and L = the slant height of a right circular cone. 
 
 Conclusion. >S — ^ CL. 
 
 Proof. 1. Inscribe in the cone a pyramid whose base is 
 inclosed by a regular polygon. Let S' = the lateral area, P'= 
 the perimeter of the base, and L' = the slant height of the 
 pyramid. 
 
 2. Then S' = ^ P'lJ. Why ? 
 
 3. Let the number of faces of the pyramid increase indefi- 
 nitely, keeping the pyramid always a regular pyramid. Then 
 
 S' = iS, P' = C, and V ^ L. § 612, 6, c 
 
 4. .-. \P'L' ^\CL. § 543, c 
 
 5. .'. S=^\CL § 542, 6 
 
402 SOLID GEOMETRY — BOOK VIII 
 
 614. Cor. If S denotes the lateral area, T the total area, L 
 the slant height, and R the radius of the base of a right circular 
 cone, then 8 = 2 .R^^^ L = .RL. 
 
 Also, T = ttRL + irR'' = irR^L + R). 
 
 Ex. 28. Find the lateral area and the total area of a right circular 
 cone, the radius of whose base is 7 in. and whose slant height is 25 in. 
 
 Ex. 29. How many square yards of canvas are required for a circus 
 tent having the form of a right circular cylinder, surmounted by a right 
 circular cone, if the diameter of the tent is 100 ft., the height of the 
 vertical wall 15 ft., and the height of the highest point of the tent 50 ft. ? 
 
 Ex. 30. The diameter of the base of a right circular cone is equal to 
 its altitude. Determine its lateral and total area. 
 
 Proposition YIII. Theorem 
 
 615. The volume of a circular cone is equal to the 
 area of its base multiplied hy one third the length of 
 its altitude. 
 
 Hypothesis. F'= the volume, ^ = the area of the base, and 
 H = the altitude of a circular cone. 
 Conclusion. V= J BH. 
 
 Proof left to the pupil. 
 Suggestion. — Model the proof after that in § 601 . Use Fig. § 613 
 
 616. Cor. // V denotes the volume, H the altitude, and R 
 the radius of the base of a circular cone, 
 
 F= i ttR'H. 
 
 Ex. 31. Determine the volume of a right circular cone, the radius of 
 whose base is 7 in. and whose slant height is 25 in. 
 
 Ex. 32. Determine the volume of the solid generated when a right 
 triangle of base b and altitude h 
 
 (a) revolves about its side h; (h) revolves about its side b. 
 
 Ex. 33. Determine the ratio of a circular cone and a circular cylin- 
 der having the same base and altitude. 
 
MEASURING THE CONE 403 
 
 Proposition IX. Theorem 
 
 617. The lateral area of a frustum of a right circu- 
 lar cone is equal to the sum of the circumferences of its 
 bases, multiplied by one half its slant height. 
 
 Hypothesis. S = the lateral area, C and c the circumfer- 
 ences of the lower and upper bases respectively, and L = the 
 slant height of a frustum of a right circular cone. 
 
 Conclusion. S = ^L(0-\-c). 
 
 Proof. 1. Let aS" = the lateral area, C and c' the circum- 
 ferences of the lower and upper bases, and L' = the slant 
 height of the frustum of a regular pyramid inscribed in the 
 frustum of the cone. 
 
 2. .'.S' = iL'(C+c'), Why? 
 
 3.. Let the number of faces of the frustum of the pyramid 
 be increased indefinitely, keeping the pyramid always a regular 
 pyramid. Then 
 
 S' = S, C'=0, c'=c, L'=L. Why? 
 
 4. .-. J L\C' H- cO = 4 i>(0 -h c). § 543, c 
 
 5. .'.S=\L{C + c). §543,6 
 
 618. Cor. 1. If S denotes the lateral area^ L the slant height^ 
 and R and r the radii of the bases of a frustum of a right cir- 
 cular cone, /S = (2 7ri2 -f 2 irr) X 4 -L = Tr(R + r)L. 
 
 619. Cor. 2. Jlie lateral area of a frustum of a right drcidar 
 cone is equal to the circumference of a section midway between the 
 bases, multiplied by the slant height. 
 
404 SOLID GEOMETRY — BOOK VIII 
 
 Proposition X. Theorem 
 
 620. The volume of a frustum of a circular cone is 
 equal to the stem of its bases and the mean proportional 
 between them, multiplied hy one third the length of its 
 altitude. 
 
 Hypothesis. V= the volume, B and h = the areas of the 
 lower and upper bases respectively, and II = the length of 
 the altitude of a frustum of a circular cone. 
 
 Conclusion. V= -J H(B + 6 + V^). 
 
 Proof. 1. Inscribe in the frustum of the cone a frustum of 
 a pyramid having its base inclosed by a regular polygon. Let 
 V = the volume, and B' and b' = the areas of the lower and 
 upper bases respectively of the frustum of the pyramid. 
 
 2. Then F' =i H(B' + 6' + VW). § 572 
 
 Complete the proof as in § 601 and § 615. 
 
 621. Cor. If V denotes the volume, H the altitude, and R 
 and r the radii of the bases of a frustum of a circular cone, then 
 
 V=\iT{B^+r'+Rr)H. 
 
 Ex. 34. Find the lateral area, the total area, and the volume of a 
 frustum of a cone of revolution, the diameters of whose bases are 16 in. 
 and 6 in., and whose altitude is 12 in. 
 
 Ex. 35. Determine the contents in quarts of a water pail having the 
 form of a frustum of a cone of revolution, if the diameters of the bottom 
 and top are 9 in. and 12 in. respectively, and the height of the pail is 14 
 in. (One quart occupies about 231 cu. in.) 
 
 Note. — Supplementary Exercises 59-71, p. 459, can be studied now. 
 
TANGENT PLANES 405 
 
 SUPPLEMENTARY TOPICS 
 
 A. Planes Tangent to a Cylinder or a Cone 
 
 622. A plane is tangent to a circular cylinder or to a circular 
 cone when it contains one and only one element of the cylinder 
 or of the cone. 
 
 Proposition XI. Theorem 
 
 623. A plane drawn throtigh an element of a circular cylinder 
 and a tangent to the base at its extremity is tangent to the cylinder. 
 
 Hypothesis. AA' is an element of the lateral surface of cir- 
 cular cylinder AB', line CD is tangent to the base AB at Aj 
 and plane CD' is drawn through AA' and CD. 
 
 Conclnsion. CD' is tangent to the cylinder. 
 
 Proof. 1. Let E be any point in plane CD', not in AA', and 
 draw through E a plane II to the bases, intersecting CD' in line 
 EF and the cylinder in FII. 
 
 2. Draw axis 00' ; then 00' is II AA'. Ex. 8, p. 392. 
 
 3. Let the plane of 00' and A A' intersect the planes of AB 
 and FH in radii OA and OF, respectively. 
 
 4. Then, GF II OA and FE II AD. Why ? 
 6. .-. Z GFE = Z OAD. Why ? 
 6. .'. FE± OF, and tangent to O FH. Prove it. 
 
406 SOLID GEOMETRY — BOOK VIII 
 
 7. Whence, point E lies outside the cylinder. 
 
 8. .-. all points of CD', not in AA', lie outside the cylinder, 
 and CD' is tangent to the cylinder. 
 
 Proposition XII. Theorem 
 
 624. A plane determined by an elemerit of the lateral surface 
 of a circular cone and a tangeyit to the base at its extremity, is 
 tangent to the cone. 
 
 
 
 Hypothesis. OA is an element of the lateral surface of cir- 
 cular cone OAB, line CD is tangent to base AB at A, and 
 plane OCD is drawn through OA and CD. 
 
 Conclusion. OCD is tangent to the cone. 
 
 (Prove that E lies outside the cone.) 
 
 Suggestion. — Model the proof after that of § 623. 
 
 B. Similar Cylinders and Cones of 
 Revolution 
 
 625. Similar cylinders of revolution are right circular cylin- 
 ders generated by the revolution of similar rectangles about 
 homologous sides as axes. 
 
 Similar cones of revolution are right circular cones generated 
 by the revolution of similar right triangles about homologous 
 sides as axes. 
 
SIMILAR CYLINDERS AND CONES 
 
 407 
 
 Proposition XIII. Theorem 
 
 626. The lateral or total areas of two siinilar cylinders of 
 revolution are to each other as the squares of their altitudes, or as 
 the squares of the radii of their bases ; and their volumes are to 
 each other as the cubes of their altitudes, or as the cubes of the 
 radii of their bases. 
 
 \H 
 
 .""» 
 
 Hypothesis. S and s are the lateral areas, T and t are the 
 total areas, V and v are the volumes, // and h are the altitudes, 
 and M and r the radii of the bases, of two similar cylinders 
 of revolution. 
 
 S T H' R' _,, V IP^^ 
 
 h^ r^* 
 
 Conclusion. 
 
 Proof. 1. 
 
 2. 
 
 3. 
 
 - = -— = -—, and — 
 t h^ r^ V 
 
 Since the generating rectangles are similar. 
 
 Why? 
 
 II 
 
 h 
 
 r 
 
 h '' 
 
 II + R 
 
 h + r' 
 
 § 255, § 253 
 
 2z^ = 5x^ = ^ = ^'. Whv^ 
 
 2 7rrh r r 7^ h^ ^ ' 
 
 6. 
 
 T_ 2 7rR(H-\-R) ^R^^R_R_II^ 
 
 t 2 irr{h -f- r) r r r^ h^ ' 
 
 V ttRH^R^R^R^IP 
 
 TtrVi 
 
 h^ 
 
 Why? 
 Why? 
 
408 PLANE GEOMETRY — BOOK VIII 
 
 Proposition XIV. Theorem 
 
 627. The lateral or total areas of two similar cones of revolu- 
 tion are to each other as the squares of their slant heights, or as 
 the squares of their altitudes, or as the squares of the radii of 
 their bases; ayid their volumes are to each other as the cubes of 
 their slant heights, or as the cubes of their altitudes, or as the cubes 
 of the radii of their bases. 
 
 Hypothesis. S and s are the lateral areas, T and t the total 
 areas, "Fand v are the volumes, L and I are the slant heights, 
 H and h are the altitudes, and R and r the radii of the bases, 
 of two similar cones of revolution. (§ 625.) 
 
 Conclusion. ^=2'=^ = ^ = «^,a„di: = ^ = ^=^. 
 
 s t P h^ r^ V P h' 7^ 
 
 The proof is left to the pupil ; model it after that of § 626. 
 
 Ex. 36. At what distance from the vertex of a right circular cone 
 with altitude H must a plane parallel to the base be passed so that the 
 lateral area will be bisected ? 
 
BOOK IX 
 
 THE SPHERE 
 
 628. A Spherical Surface is a closed surface all points of 
 which are equidistant from a point within, called the Center. 
 
 629. A Sphere is the solid bounded by a spherical surface. 
 
 630. A Radius of a sphere, or of its surface, is the straight 
 line drawn from its center to any point of its surface. 
 
 A Diameter of a sphere, or of its surface, is the straight line 
 drawn through the center having its extremities in the surface. 
 
 Proposition I. Theorem 
 
 631. The intersection of a spherical surface and a 
 plane is a circle. 
 
 
 ~~^\ 
 
 
 0'"'.:!\n 
 
 /^l^."-"'' 
 
 "''•^.S^^^ 
 
 '^^«s 
 
 &'■' 
 
 Hypothesis. ABC is the intersection of the spherical surface 
 whose center is and a plane. 
 
 Conclusion. Curve ABC is a O 
 
 Proof. 1. Draw 00' ± plane of ABC. 
 
 2. Let A and B be any two points in curve ABC. Draw 
 OA, OB, O'A, and O'B. 
 
 Suggestion. — Now prove that O'A = O'B, and ABC is a O. 
 
 409 
 
410 SOLID OPTOMETRY — BOOK IX 
 
 632. A Great Circle of a sphere is the in- 
 tersection of its surface and a plane passing /^ 
 through its center ; as O ABQ. A,-— - 
 
 A Small Circle of a sphere is the intersec- J-t;:;;;;^ 
 tion of its surface and a plane which does \ ^ 
 not pass through its center. ^^J_-^-^ 
 
 The Axis of a circle of a sphere is the 
 diameter of the sphere which is perpendicular to the plane of 
 the circle ; as axis POP\ 
 
 The Poles of a Circle of a sphere are the extremities of the 
 axis of the circle. 
 
 633. Cor. 1. Tlie axis of a circle of a sphere passes through 
 the center of the circle. 
 
 634. Cor. 2. All great circles of a sphere are equal. 
 
 635. Cor. 3. Every great circle bisects the sphere and its 
 surface. 
 
 For if the portions of the sphere formed by the plane of the great circle 
 be separated, and placed so that their plane surfaces coincide, the spheri- 
 cal surfaces falling on the same side of this plane, the two spherical sur- 
 faces will coincide throughout ; for all points of either surface are equally 
 distant from the center. 
 
 636. Cor. 4. Any tivo great circles bisect each other. 
 
 For the intersection of their planes is a diameter of the sphere, and 
 therefore a diameter of each circle. 
 
 637. Cor. 5. Between any tivo points on the surface of a 
 sphere, not the extremities of a diameter, one and only 07ie arc of 
 a great circle, less than a semicircle, can be draion. 
 
 For the two points, with the center of the sphere, determine a plane 
 which intersects the surface of the sphere in the required arc. 
 
 638. Two spheres are equal ivhen their radii are equal. 
 
 All radii and diameters of the same sphere or equal spheres 
 are equal. 
 
 639. A spherical surface may be generated by the revolu- 
 tion of a semicircle about its diameter as an axis. 
 
 For all points of such a surface are equally distant from the center of 
 the circle. 
 
THE SPHERE 411 
 
 640. Through two points of a spherical surface, an infinity 
 of small circles of the sphere can be drawn. Why ? 
 
 Through three points of a spherical surface, one and only 
 one small circle of the sphere can be drawn. Why ? 
 
 Ex. 1. Prove that a great circle of a sphere which passes through one 
 pole of a circle must pass through the other pole also. 
 
 Ex. 2. How many great circles can be passed through two points 
 which are the extremities of a diameter of a sphere ? 
 
 Ex. 3. What kind of circles of the earth are the parallels' of latitude ? 
 
 Ex. 4. What kind of circles of the earth are the meridians ? 
 
 Ex. 5. What kind of circle of the earth is the equator ? 
 
 Ex. 6. Speaking strictly, is it accurate to speak of the North Pole of 
 the earth ? Of what circle or circles is it a pole ? 
 
 Ex. 7. Into how many parts do two great circles of a sphere divide 
 the surface of the sphere ? 
 
 Ex. 8. Into how many parts do three great circles of a sphere divide 
 the surface of the sphere, if they do not all have a common diameter ? 
 
 Ex. 9. Prove that all circles of a sphere made by parallel planes have 
 the same axis and the same poles. 
 
 Ex. 10. Given a point of a spherical surface. Prove that it is the 
 pole of one and only one great circle. 
 
 Ex. 11. Prove that circles of a sphere made by planes equidistant 
 from the center of the sphere are equal. 
 
 Suggestion. — Use the Pythagorean Theorem. (§ 291.) 
 
 Ex. 12. State and prove the converse of Ex. 11. 
 
 Ex. 13. Prove that circles of a sphere made by planes unequally dis- 
 tant from the center of the sphere are unequal, the more remote being 
 the smaller. 
 
 Ex. 14. State and prove the converse of Ex. 13. 
 
 Ex. 15. In how many points can two straight lines intersect ? 
 
 In how many points on one hemisphere can two great circles intersect ? 
 
 641. The great circles of a sphere in sphencal geometry cor- 
 respond to the straight lines of a plane in plane geometry. 
 § 637 and Ex. 15 are two instance's pointing to this similarity 
 of great circles and straight lines ; others will appear in the 
 remaining paragraphs of Book IX. 
 
412 SOLID GEOMETRY — BOOK IX 
 
 642. It can be proved that the length of the arc of the great 
 circle, less than a semicircle, between two points of a spherical 
 surface is less than the length of any other curved line on the 
 surface between the two points. Consequently, the distance 
 between two points on the surface of a sphere^ measured on the 
 surface, is defined to be the arc of the great circle, less than a 
 semicircle, drawn between them. 
 
 Proposition II. Theorem 
 
 643. All points in a circle of a sphere are equidistant 
 from each of its poles. 
 
 •p' 
 
 Hypothesis. P and P' are the poles of O ABC of sphere M. 
 
 Conclusion. All points of O ABC are equidistant from P, 
 and also from P'. 
 
 Proof. 1. Let A and B be any two points of O ABC, and 
 draw great circle arcs PA and PB. Draw axis PMP', intersect- 
 ing the plane of ABC at 0. Draw OA, OB, PA, and PB. 
 
 2. .'.PA = PB. Prove it. 
 
 3. .♦. PA = PB. Prove it. 
 
 4. Since A and B are any two points of O ABC, .'. all 
 points of O ABC are equidistant from P. 
 
 5. Similarly all points of O ABC are equidistant from P\ 
 
 644. The Polar Distance of a circle of a sphere is the dis- 
 tance from the nearer of its poles to the circle, or from either 
 pole if they are equally near. 
 
 Thus, in the figure of Proposition II, the polar distance of O ABC is 
 arc PA. 
 
THE SPHERE 
 
 413 
 
 645. Cor. All points of a great circle of a sphere are at a 
 quadranfs distance from either of its poles. 
 
 Note. — The term quadrant in Spherical Geom- 
 etry, usually signifies a quadrant of a great circle. 
 
 Hypothesis. P is a pole of great circle ABC of 
 sphere AFC; B is any point in QABC, and PB is 
 an arc of a great O. 
 
 Conclusion. Arc PB = a quadrant. 
 
 Swjgestion. — Draw radii OA, OB, and OP. 
 
 Note. — An arc of a circle may be drawn on the surface of a sphere by 
 placing one foot of the compasses at the nearer pole of the circle, the dis- 
 tance between the feet being equal to the chord of the polar distance. 
 
 Proposition III. Theorem 
 
 646. A point on the surface of a sphere at a quad- 
 rant's distance from each of two points^ not the extremi- 
 ties of a diameter of the sphere, is a pole of the great 
 circle through those points. 
 
 Hypothesis. P is on the surface of the sphere whose center 
 is 0. AB is an arc of great O ABC, not a semicircle. PA 
 and PB are quadmnts. 
 
 Conclusion. P is a pole of AB. 
 
 Suggestions. — 1. Recall the definition of " pole of a 0." 
 2. Draw PO, AO, OB, and prove PO 1 plane ABC. 
 
 "Ex. 16. If a point lies at a quadrant^s distance from the ends of a 
 diameter of a sphere, is it necessarily a pole of the great circle through 
 those points ? 
 
414 
 
 SOLID GEOMETRY — BOOK IX 
 
 647. The angle between two intersecting curves is the angle 
 formed by the tangents to the curves at the point of inter- 
 section. 
 
 A Spherical Angle is the angle between two intersecting arcs 
 of great circles. 
 
 Propositiois^ IV. Theorem 
 
 648. A spherical angle is measured by an arc of a 
 great circle having its vertex as a pole, included be- 
 tween its sides extended if necessary. 
 
 Hypothesis. ABO and AB'C are arcs of great (D on the 
 surface of the sphere whose center is ; lines AD and AD' 
 are tangent to ABC and AB'C, respectively, and BB' is an arc 
 of a great O having ^ as a pole, included between arcs ABC 
 and AB'C. 
 
 Conclusion. Z BAB' is measured by arc BB'. 
 
 Proof. 1. Draw diameter AGO and radii OB and OB'. 
 
 2. Arcs AB and AB' are quadrants. Why ? 
 
 3. .-. A AOB and AOB' are rt. A. Why ? 
 
 4. OB II AD and OB' II AD'. Why? 
 
 5. .-. Z DAD' = Z BOB'. § 481 
 
 6. But Z BOB' is measured by arc BB'. Why ? 
 
 7. Then, Z DAD' is measured by arc BB'. Why ? 
 
 8. .-. Z £^5' is measured by arc BB'. § 647 
 
THE SPHERE 415 
 
 649. Cor. 1. Tlie angle between two arcs of great circles is 
 equal to the diedral angle formed by their planes. 
 
 650. Cor. 2. An arc of a great circle drawn to another great 
 circle from the latter^ s pole is popendicular to that great circle. 
 
 Suggestions.— 1. What Z does OA form with plane BOB'? 
 2. What kind of Z is diedral Z AOB'B? (§ 495.) 
 
 Ex. 17. If a spherical blackboard can be had, construct a spherical 
 angle and measure it. 
 
 Ex. 18. The chord of the polar distance of a circle of a sphere is 6. 
 If the radius of the sphere is 6, what is the radius of the circle ? 
 
 Ex. 19. What is the locus of points in space at a given distance d from 
 a fixed point, and equidistant from two given points ? 
 
 SPHERICAL POLYGONS 
 
 Note. — Recall at this point the definition of a polygon in plane geom- 
 etry as a closed broken line lying in a plane. (§ 125.) 
 
 Recall also § 641, calling attention to the similarity in the rdles of the 
 straight line in plane geometry and the great circle in solid geometry. 
 
 651. A Spherical Polygon is a closed line on the surface of a 
 sphere consisting of arcs of three or more 
 great circles ; as polygon A BCD. 
 
 Just as in plane geometry we considered 
 only convex polygons, so we shall consider 
 only convex spherical polygons. (See § 126.) 
 
 It will be assumed as evident that a 
 simple spherical polygon iucloses a por- 
 tion of the surface of the sphere. The 
 bounding arcs are the Sides of the polygon ; they are usually 
 measured in arc-degrees. (§ 214.) 
 
 The angle formed by two consecutive sides of a polygon is 
 an Angle of the spherical polygon, and its vertex is a Vertex of 
 the polygon. 
 
 A Diagonal of a spherical polygon is the arc of the great 
 circle joining two non-consecutive vertices of the polygon, and 
 lying withia the polygon. 
 
416 SOLID GEOMETRY — BOOK IX 
 
 652. A Spherical Triangle is a spherical polygon having 
 three sides. A spherical triangle is Isosceles when it has two 
 equal sides ; it is Equilateral when it has three equal sides ; it 
 is Right-angled when one of its angles is a right angle. 
 
 653. The planes of the sides of a spherical polygon form a 
 polyedral angle, whose vertex is the center of the sphere, and 
 whose face angles are measured by the sides of the spherical 
 polygon. 
 
 Thus, in the figure of § 651, the planes of the sides of the spherical 
 polygon form a polyedral, Z 0-ABCD, whose face angles AOB, BOC, 
 etc., are measured by arcs AB, BC, etc., respectively. 
 
 Ex. 20. Prove that the angles of a spherical polygon have the same 
 measures as the diedral angles of the corresponding polyedral angle. 
 
 654. If great circles be drawn with the vertices of a spheri- 
 cal triangle as poles, they divide the surface of the sphere into 
 eight parts whose boundaries are tri- 
 angles. 
 
 Thus, if circle B'C'B" be drawn with 
 vertex A of spherical A ABC as a pole, 
 circle A'C^A" with ^ as a pole, and circle 
 A'Bf'A"B' with (7 as a pole, the surface 
 of the sphere is divided into eight spheri- 
 cal A ; namely, A'B'C, A'B"Q', A'B'C, 
 and A'B"C' on the hemisphere represented in the figure, and 
 four others on the opposite hemisphere. 
 
 Of these eight spherical A, one is called the Polar Triangle 
 of ABO, and is determined as follows : 
 
 Of the intersections. A! and A' of circles drawn with B and 
 C as poles, that which is nearer to A, i.e. A', is a vertex of the 
 polar triangle ; and similarly for the other intersections. 
 
 Thus, A'B'C is the polar A of ABC. 
 
 Ex. 21. The polar distance of a circle of a sphere is 60°. If the di- 
 ameter of the circle is 6, find the diameter of the sphere, and the distance 
 of the circle from its center. 
 
 Suggestion. — Represent the radius of the sphere by 2 x. (§ 288.) 
 
SPHERICAL POLYGONS 417 
 
 Proposition V. Theorem 
 
 655. If one spherical triangle is the polar triangle 
 of another, then the second sjjherical triangle is the 
 polar triangle of the first. 
 
 Hypothesis. A'B'C is the polar A of the spherical A ABC; 
 Af B, and C are the poles of arcs B'C\ O'A', and A'B', 
 respectively. 
 
 Conclusion. ABC is the polar A of spherical A A'BfC. 
 
 Proof. 1. A' is at a quadrant's distance from B. § 645 
 (Since B is the pole of arc A'C.) 
 
 2. A' is at a quadrant's distance from C. Why ? 
 
 3. .-. A' is a pole of the great O arc BC. Why ? 
 
 4. Similarly Bf and C are the poles oi AC and AB, 
 respectively. Prove it. 
 
 5. .-. A ABC is the polar A of A A'BfC. 
 
 (For of the two intersections of the great (D having B^ and 
 C, respectively, as poles, A is nearer to A' ; similarly for B 
 and C. § 654 
 
 Note. — Two spherical triangles, each of which is the polar triangle of 
 the other, are called polar triangles. 
 
 Ex. 22. How many degrees are there in the polar distance of a circle 
 whose plane is 6V2 units from the center of the sphere, the diameter of 
 the sphere being 20 units ? 
 
 Suggestion. — The radius of the is a leg of a rt. A, whose hypotenuse is 
 the radius of the sphere, and whose other leg is the distance from its center 
 to the plane of the 0. 
 
418 SOLID GEOMETRY — BOOK IX 
 
 Proposition VI. Theorem 
 
 656. In two polar triangles, each angle of one has the 
 same measure as the supplement of that side of the other 
 of which it is the pole. 
 
 Hypothesis. A ABC and A'B'C are polar A, point A being 
 
 the pole of B^', etc. 
 
 Let a, a', etc. be the measures in degrees of BO, B'C, etc., 
 respectively ; let A, A', etc. be the measures in degrees of 
 A A, A', etc. 
 
 Conclusion. ^ = 180 - a' ; JB = 180 - &' ; (7 = 180 - c'. 
 ^' = 180 - a ; B'== 180 -b ; C = 180 - c. 
 
 Proof. 1. Extend AB and AC to meet B^ at D and E, 
 respectively. 
 
 2. Since B' is the pole of AE, B'E = 90°. Why ? 
 
 3. Similarly, CD = 90°. 
 
 4. .-. B'E + CI) = 1S0°. 
 
 5. .'. m) + DE-^CD = 180°, OY DE-{-B^' = 1S0°. 
 
 6. But DE is the measure of Z ^. § 648 
 
 7. .-. A-\-a' = 180, or 
 
 ^=180 -a'. 
 
 8. Similarly for each of the other angles of either triangle. 
 
 Ex. 23. Prove on the figure for § 656, that A' = 180 — a. 
 
 Ex. 24. If the sides of a spherical triangle are 77°, 123^, and 95°, 
 how many degrees are there in each angle of its polar triangle ? 
 
 Ex. 25. If the angles of a spherical triangle are 86°, 131°, and 68°, 
 how many degrees are there in each side of its polar triangle ? 
 
SPHERICAL POLYGONS 
 
 419 
 
 Proposition VII. Theorem 
 
 657. Tlie sum of the sides of a convex spherical poly- 
 (jon is less than 360°. 
 
 Hypothesis. ABCD is a convex spherical polygon. 
 
 Conclusion. AB + BC+ CD -\- DA < 360°. 
 
 Proof. 1. Let polyedral angle 0-ABCD be the polyedral 
 angle which corresponds to spherical polygon ABCD. 
 
 2. Then the measure of AB equals the measure of central 
 angle AOB. Why ? 
 
 3. Similarly for arcs BC, CD, and DA. 
 
 4. But Z AOB + Z BOC + Z COD + Z DO A < 360°. 
 
 _ ^ ^ ^ §614 
 
 5. .-. AB + BC-^CD-\-DA< 360°. 
 
 Note I. — The pupil should recall at this point that one arc-degree is ^^^ 
 of a circle. Since arcs AB, BC, CD, and DA are arcs of great circles, 
 Proposition VII means that the sum of the sides of any spherical polygon 
 is less than 300 arc-degrees of a great circle — i. e. is less than a great 
 circle. 
 
 Note 2. — Proposition VII is one of many theorems about spherical 
 polygons which can be formulated from corresponding theorems about 
 polyedral angles by replacing in the latter the words "face angle" and 
 "diedral angle" by "side" and "angle" respectively. 
 
 Ex. 26. If two sides of a spherical triangle measure 80° and 70° respec- 
 tively, between what two values must the remaining side lie ? 
 
 Ex 27. What is the greatest possible length for the sum of the sides 
 of a convex spherical polygon on a circle of radius 12 inches ? 
 
420 SOLID GEOMETRY — BOOK IX 
 
 Proposition VIII. Theorem 
 
 658. Tlie sum of the angles of a spherical triangle is 
 greater than 180° and less than 540°. 
 
 Hypothesis. A, B, and C are tlie measures in degrees of 
 the A of spherical A ABC. 
 
 Conclusion. A + B + C> 180° and < 540°. 
 
 Proof. 1. Let A A'B'C be the polar triangle of spherical 
 
 A ABC, A being the pole of B^, B of A^, and C of A^'. 
 
 Let the measures in degrees of B'C, C'A', and A'B' be a" 
 b', and c', respectively. 
 
 2. .-. ^ = 180 - a', 5 = 180 - 6', and = 180 - c'. Why ? 
 
 3. .♦. ^ + ^H-C=540-(a'+6' + c'). Why? 
 
 4. .-. ^-h^+C<540°. 
 
 5. But a' + b' + c' < 360°. § 657 
 
 6. .'.A + B+C> 180°. Ax. 20, § 158 
 
 659. Cor. Tlie sum of the angles of a spherical polygon of n 
 sides is greater than (n — 2) x 180°. 
 
 Consider ABCD a spherical quadrilateral. 
 
 Draw great O arc BD, dividing the quadrilateral into two spherical i^, 
 ABD and BBC. The sum of the angles of the triangles equals the sum 
 of the angles of the quadrilateral. In each triangle the sum of the angles 
 is > 180°. Hence for the quadrilateral, the sum of the angles is greater 
 than 2 X 180^ ; that is, the sum >(4 - 2) x 180°. 
 
 In like manner, if there are n sides, the polygon can be divided into 
 (n— 2) spherical triangles, in each of which the sum of the angles is 
 gi-eater than 180°. Therefore, in the polygon the sum of the angles is 
 greater than (n — 2) x 180°. 
 
SPHERICAL POLYGONS 421 
 
 660. The Spherical Excess of a spherical triangle, measured 
 in degrees, is the diii'erence between the sum of its angles and 
 180°. 
 
 The Spherical Excess of a spherical polygon of n sides, meas- 
 ured in degrees, is the difference between the sum of its angles 
 and (n - 2) X 180°. 
 
 Note. — In each case, the spherical excess is the amount by which the 
 sum of the angles of the spherical polygon exceeds the sum of the angles 
 of a plane polygon of the same number of sides. 
 
 Ez. 28. Prove that a spherical triangle may have one, two, or three 
 right angles, or one, two, or three obtuse angles. 
 
 661. A spherical triangle having two right angles is called 
 a Bi-rectangular Triangle, and one having three right angles 
 a Tri-rectangular Triangle. 
 
 Ex. 29. Prove that the sum of the angles of a spherical hexagon is 
 greater than 8, and less than 12, right angles. 
 
 Ex. 30. What is the spherical excess of a triangle whose angles are 
 100°, 95°, and 65^ respectively ? 
 
 Ex. 31. What is the spheiical excess of a tri-rectangular triangle ? 
 
 Ex. 32. Prove that the spherical excess of a bi-rectangular triangle 
 is the measure of the remaining angle of the triangle. (§ 048.) 
 
 Ex. 33. What is the spherical excess of a triangle if the sides of its 
 polar triangle measure 80°, 85°, and 95° ? 
 
 Ex. 34. What relation exists between a tri-rectangular spherical 
 triangle and its polar ? 
 
 Ex. 35. Prove that the sides opposite the equal angles of a bi-rectaii- 
 gular triangle are quadrants. 
 
 Suggestion. — Recall § 499 and the definition of "pole." 
 
 Ex. 36. Prove that each side of a tri-rectangular triangle is a quadrant. 
 
 Ex. 37. Prove that in a bi-rectangular spherical triangle, the third 
 angle has the same measure as the side opposite it. 
 
 662. If a ^lane or a line has only one point in common 
 with the surface of a sphere, it is said to be Tangent to the 
 Sphere. The sphere is said to be tangent to the plane or line. 
 The point common to the plane or line and the spherical sur- 
 face is called the Point of Contact or Tangency. 
 
422 
 
 SOLID GEOMETRY — BOOK IX 
 
 Proposition IX. Theorem 
 
 663. A plane perpendicular to a radius of a sjjJiere 
 at its outer extremity is tangent to the sphere. 
 
 
 
 ^^^^ 
 
 \y 
 
 f 
 
 > / y 
 
 y^\ 
 
 ,' 
 
 \ y X 
 
 / / 
 
 
 ~— L-^^ / 
 
 / ^ 
 
 
 A y 
 
 M 
 
 Hypothesis. Plane MN~ ± radius OA at A. 
 Conclusion. Plane MN is tangent to the sphere. 
 
 Suggestion. — Let B be any poiut of MN except A. 
 Prove that B lies outside the sphere. 
 
 664. Cor. A plane tangent to a sphere is perpendicular to the 
 radius drawn to the point of contact. (Fig. of Prop. IX.) 
 
 665. Two Spheres are Tangent to each other when each is 
 tangent to the same plane at the Same point. 
 
 Ex. 38. Prove that a straight line perpendicular to a radius of a 
 sphere at its outer extremity is tangent to the sphere. 
 
 Ex. 39. How many straight lines can be tangent to a sphere at a 
 point of the sphere ? 
 
 Ex. 40. If two straight lines are tangent to a sphere at the same point, 
 their plane is tangent to the sphere. 
 
 Ex. 41. Prove that all lines tangent to a sphere at a point of the 
 sphere lie in the plane tangent to the sphere at that point. 
 
 Ex. 42. How many straight lines can be tangent to a sphere from a 
 point outside the sphere ? Compare the lengths of these tangents. 
 
 Ex. 43. Prove that the points of contact of all lines tangent to a 
 sphere from an exterior point lie in a circle. 
 
 Ex. 44. Is the circle in Ex. 43 a great circle or a small circle ? 
 
 Ex. 45. If two spheres are tangent to the same plane at the same 
 point, the straight line joining their centers passes through the point of 
 contact. 
 
MEASUREMENT OF SPHERICAL POLYGONS 423 
 
 MEASUREMENT OF SPHERICAL POLYGONS 
 
 666. A Zone is the portion of a spherical surface included 
 between two parallel planes. 
 
 The circles which bound the zone are its baseSj and the dis- 
 tance between their planes is its dUitude. 
 
 A zone of one base is a zone lying between one plane and 
 a parallel plane tangent to the sphere. 
 
 667. If semicircle ACEB be revolved 
 about diameter AB as an axis, and CD and 
 EF are lines ± AB, then arc CE generates 
 a zone whose altitude is DF, and arc AC a. 
 zone of one base whose altitude is AD. 
 
 668. Application of Limits to Zones. 
 
 Let be the center of AGB, and OMhe any diameter of the 
 circle. Let AA' and BB' be _L OM. Let C bisect arc AB, and 
 draw broken line ACB. If arc AB is ,, 
 
 revolved about OM as axis, it gener- 
 ates a zone, and broken lines AC and 
 CB generate curved surfaces. 
 
 It will be assumed as evident that 
 the sum of the areas of the surfaces 
 generated by AC and CB is less than 
 
 the area of the zone generated by ACB. 
 
 Assume arcs AC and CB to be bisected at X and Y", and 
 imagine the broken line AXCYB. It will be assumed as evi- 
 dent that the area of the surface generated by line AXCYB is 
 greater than that generated by ACB but is still less than that 
 of the zone generated by AB. If the process of subdividing 
 arc AB by successively halving the subdivisions of arc AB be 
 continued indefinitely, it will be assumed evident that the sur- 
 face generated by the resulting broken line approaches as limit 
 the zone AB ; also, as the chords like AC decrease indefinitely 
 in length, their distance from center increases and approaches 
 the radius of arc AB as limit. 
 
424 
 
 SOLID GEOMETRY — BOOK IX 
 
 Proposition X. Theorem 
 
 669. The area of the surface generated hy the revolu- 
 tion of a straight line about a straight line in its plane, 
 not parallel to a7id not intersecting it, as an axis, is 
 equal to its projection on the axis, multiplied hy the cir- 
 cumference of a circle, whose radius is the perpendicu- 
 lar erected at the mid-point of the line and terminating 
 in the axis. 
 
 Hypothesis. Straight line AB is revolved about straight line 
 FM in its plane, not ± to and not intersecting it, as an axis ; 
 lines AC and BD ± FM, and EF is the ± erected at the mid- 
 point of AB terminating in FM. 
 
 Conclusion. Area AB'^=CDy.2 irEF. 
 
 Proof. 1. Draw line AG ± BD, and line EH ± CD. 
 
 2. The surface generated by AB is the lateral surface of a 
 frustum of a cone of revolution, whose bases are generated by 
 AC and BD. 
 
 3. .-. area AB=ABx2 ttEH 
 
 4. A ABG and EFH are similar. 
 g . AB^EF 
 
 ' ' AG EH 
 6. .\ABxEH = AG xEF 
 
 = CDx EF. 
 1. Substituting in Step 3, 
 
 area AB=CDx2 ttEF. 
 
 §619 
 Prove it. 
 
 Why? 
 
 Why? 
 Why? 
 
 *The expression 
 generated by AB. 
 
 area AB " is used to denote the area of the surface 
 
MEASUREMENT OF SPHERICAL POLYGONS 425 
 
 Proposition XI. Theorem 
 
 670. The area of a zone is equal to its altitude multi- 
 plied by the circumference of a great circle. 
 
 Hypothesis. AB is revolved about diameter OM as axis ; 
 AA' and BB' ± OM', R is the radius of AB. 
 
 Conclusion. Area of zone generated by AB = ^'jB' x 2 irR. 
 
 Proof. 1. Bisect ^^ at C; draw AG and GB] also draw 
 CC J. 03f and OELAG. 
 
 2. Revolve the ligure about OiHf as axis. 
 
 3. .•.area^a = ^'C" x2 7rO^. §669 
 
 area GB = O'jB' x 2 ttOE. Why ? 
 
 4. Adding, the surface generated by AGB 
 
 = (A'G' + CB')x2 7rOE. 
 
 5. Continue to bisect the subdivisions of AB, indefinitely. 
 
 6. Then, the area of the surface generated by revolving the 
 inscribed broken line approaches as limit the area of the zone 
 generated by AB, and 
 
 OE=R. 
 '. area of zone = A'B' x2 wR. 
 
 §668 
 §403* 
 
 Note. — The proof of § 670 holds for auy zone which lies entirely on the 
 surface of a hemisphere ; for, in that case, no chord is II OM, and § 669 is 
 applicable. 
 
 Since a zone which does not lie entirely on the surface of a hemisphere 
 may be considered as the sum of two zones, each of which does lie entirely 
 on the surface of a hemisphere, the theorem of § 670 is true for any zone. 
 
426 
 
 SOLID GEOMETRY — BOOK IX 
 
 671. Cor. 1. If Z denotes the area of a zone, h its altitude, 
 ayid R the radius of the sphere^ 
 
 Z=2 irBh. 
 
 672. Cor. 2. The area of a spherical surface equals the 
 square of its radius multiplied by 4 tt. 
 
 Proof. A spherical surface may be regarded as a zone whose altitude 
 is a diameter of the sphere. Letting S represent the area of the spherical 
 surface, 
 
 673. Cor. 3. The area of the surface of a sphere equals the 
 area of four great circles of the sphere. 
 
 674. Cor. 4. The areas of two spherical surfaces have the 
 same ratio as the squares of their radii or the squares of their 
 diameters. 
 
 675. A Lune is the portion of a spherical surface bounded 
 by two semicircles of great circles ; as ACBD. 
 
 The Angle of a Lune is the angle between 
 its bounding arcs. 
 
 It is evident that two lunes on the same 
 sphere or equal spheres are congruent if their 
 angles are equal. 
 
 676. It is evident that lunes on the same 
 sphere or on equal spheres may be added by 
 placing them so that their angles become adja- 
 cent angles ; thus lune ACBE + lune AEBF 
 = lune ACBF. 
 
 When two lunes are added, the angle of the 
 sum equals the sum of the angles of the given 
 iunes. 
 
 If Lj^ is used to denote the lune whose angle is Z X, then 
 
 Lj[-{- Ly= L(^x+r) 5 
 
 i.e. lune of Z X + lune of Z F= lune of Z(X + Y). 
 
 Note. — It is very important that the symbol Lx he understood and 
 remembered. 
 
MEASUREMENT OF SPHERICAL POLYGONS 427 
 
 Proposition XII. Theorem 
 
 677. Two lunes on the same sphere or equal spheres 
 ham the same ratio as their angles. 
 
 Case I. When the angles are commensurable. 
 
 Hjrpothesis. ACBD and ACBE are lunes on sphere AB, 
 having their A CAD and CAE commensurable. 
 
 Conclusion. ACBD^Z^CAD^ 
 
 ACBE Z CAE 
 
 Proof. 1. Let Z CAa be a common measure of Z CAD and 
 CAE, and let it be contained 5 times in Z CAD, and 3 times 
 in Z CAE. 
 
 2 . ZC^D ^5 ... 
 
 ' ' Z CAE 3* ^ ^ 
 
 3. Extending the arcs of division of Z CAD to JB, lune 
 
 ACBD will be divided into 5 parts, and lune ACBE into 3 
 
 parts, all of which parts will be equal. Why ? 
 
 . ACBD^5 
 
 "ACBE 3* 
 
 4. 
 
 6. From (1) and (2), 
 
 ACBD ^ Z CAD 
 ACBE Z CAe' 
 
 (2) 
 Why? 
 
 Note. — The theorem may be proved in a similar manner when the 
 given lunes are on equal spheres. 
 
 Case II. Wheii the angles are incommensurahle. 
 
 Suggestion. — Model the proof after that in § 544. 
 
428 SOLID GEOMETRY — BOOK IX 
 
 678. The surface of a hemisphere may be regarded as a lune 
 of angle 180° and the surface of the sphere, a lune of angle 360°. 
 
 679. Cor. 1. The surface of a lune is to the surface of the 
 sphere as the measure of its angle in degrees is to 360. 
 
 680. Cor 2. If the radius of the sphere is R, and the angle of 
 the lune, measured in degrees, is A, and the area of the lune is de- 
 noted hy L^, then 
 
 ^ 360 90 
 
 Ex. 46. Prove that the areas of two zones on the same sphere, or 
 equal spheres, are to each other as their altitudes. 
 
 Ex. 47. Determine the area of a zone whose altitude is 13, if the 
 radius of the sphere is 16. 
 
 Ex. 48. Prove that the area of a zone of one base is equal to the area 
 of the circle whose radius is the chord of its generating arc. (§ 288.) 
 
 Ex. 49. Determine the area of the surface of a sphere whose radius 
 is 12. 
 
 Ex. 50. If the radius of a sphere is i?, what is the area of a zone of 
 one base, whose generating arc is 45° ? 
 
 Ex. 51. Find the radius of a sphere whose surface is equivalent to the 
 entire surface of a cylinder of revolution, whose altitude is lOi, and ra- 
 dius of base 3. 
 
 Ex. 52. What is the area of a lune whose angle is 40° on the surface 
 of a sphere whose radius is 15 in. ? 
 
 Ex. 53. What part of the surface of the earth is included between 
 the 30th and 35th meridians ? 
 
 Ex. 54. The area of a lune is 28f . If the area of the surface of the 
 sphere is 120, what is the angle of the lune ? 
 
 Ex. 55. Prove that the surface of a sphere is equal to two thirds the 
 entire surface of the right circular cylinder circumscribed about it. 
 
 Ex. 56. Compare the surface of a sphere with the lateral surface of 
 the right circular cylinder circumscribed about the sphere. 
 
 Ex. 57, What circles of the earth bound the North Temperate Zone ? 
 What part of the earth's surface lies within that zone ? 
 
 Ex. 58. What zones of the earth are zones of one base ? 
 
MEASUREMENT OF SPHERICAL POLYGONS 429 
 
 681. Two spherical polygons, on the same or equal spheres, 
 are Symmetrical when the sides and angles of one are equal, re- 
 spectively, to the sides and angles of the other, if the equal 
 parts occur in opposite orders. 
 
 Thus, if spherical ^ABC and A'B'C\ on 
 the same or equal spheres, have sides AB^ BC, 
 and CA equal, respectively, to sides A'B'^ B'C\ ^^ ^ 
 
 and C'A', and A A, B, and C to^ A', B', and C", and the equal parts 
 occur in the opposite orders the A are symmetrical. 
 
 Proposition XIII. Theorem 
 
 682. TJie spherical triangles corresponding to a pair 
 of vertical triedral angles are symmetrical. 
 
 Hypothesis. AOA', BOB', and COC are diameters of the 
 sphere with center 0; the planes determined by them inter- 
 sect the spherical surface in® ABA' B', AC A' C, and BCB'C. 
 
 Conclusion. Spherical A ABC and A'B'C are symmetrical. 
 
 Suggestions. — 1. Prove A' B' = AB ; B'C = BC, etc. 
 
 2. Prove Z BCA = L B'C'A' ; Z BAG = Z B'A'C ; etc. 
 
 3. Prove that the parts of A ABC occur in C' 
 
 opposite order to those of A A'B'C . y'"'^^/' /*^\ 
 
 The adjoining figure will aid in doing h\^,jL..A^,\B'\ 
 
 this. A'B'C has been slid around the ^ ^J!^/_.__ir::ir.llVJ^' 
 
 sphere until it occupies the position in- \l7h~~~/- — / 
 
 dicated in this figure. Determine the \ l / / 
 
 direction from A to B to C ; and also \\ / y^ 
 
 the direction from A' to B* to C. p" 
 
430 SOLID GEOMETRY — BOOK IX 
 
 Pkoposition XIY. Theorem 
 
 683. Two symmetrical spherical triangles, of lohich 
 one is isosceles, are congruent and hence equal. 
 
 1 
 
 Hypothesis. A ABC is symmetrical to AA'B^C^-, that is, 
 
 AB = A^, AC = A^y BC=B^, ZB = ZB', ZC==ZG', 
 Z A= ZA', with the equal parts arranged in opposite orders in 
 
 the triangles ; also AB = AC. 
 
 Conclusion. A ABC ^ A A'B'C 
 
 Proof. 1. Since AB = A^, and AB= AC, 
 
 .-. A^= AG. 
 
 2. In like manner A^C = AB. 
 
 Complete the proof by superposing AA'B'C on A ABC, 
 making A'C coincide with AB, with point A' on point A. 
 
 684. Cor. If one of two symmetrical spherical triangles is 
 isosceles, the other is also. 
 
 Proposition XV. Theorem 
 
 685. Two spherical triangles corresponding to a pair 
 of vertical triedral angles are equal. 
 
 Hypothesis. AOA', BOB', and COC are diameters of sphere 
 0; also, the planes determined by them intersect the surface 
 in arcs AB, BC, AC, A'B', B'C, and A'C. (Fig. p. 431.) 
 
MEASUREMENT OF SPHERICAL POLYGONS 431 
 Conclusion. Area of A ABC = area of triangle A'B'C. 
 
 A 
 
 Proof. 1. Let P be the pole of the small circle passing through 
 points Aj B, and C ; draw arcs of great circles PA, PB, and PC. 
 
 2. ..PA = PB=PC. §643 
 
 3. Draw PP\ a diameter of the sphere, and P'A', P'B', and 
 PC arcs of great CD ; then spherical A PAB and P'A'B' are 
 symmetrical. § 682 
 
 4. But spherical A PAB is isosceles. 
 
 5. .-. A PAB = A P'A'B', Why ? 
 
 6. In like manner, 
 
 A PBC = A P'B'C and A PC A = A PC A'. 
 
 7. Then the sum of the areas of A PAB, PBC, PAC equals 
 the sum of the areas of A P'A'B', P'B'C, and P'CA'. 
 
 8. .-. area A ABC = area A A'B'C. 
 
 686. Cor. Two symmetrical triangles on the same or equal 
 spheres are equal. 
 
 1. ljQtAA"B"C" be symmetrical to /S ABC', let A A'B'C 
 be the spherical triangle on the same sphere as A ABC, such 
 that A', B', and C are diametrically opposite to A, B, and C, 
 respectively. 
 
 2. Then A A'B'C is also symmetrical to A ABC, and equal 
 to A ABC. 
 
 3. .-. the parts of A A"B"C" and A A'B'C are equal and are 
 arranged in the same order. Hence A A"B"C" ^ A'BC". 
 
 4. .'.AABC==AA"B"C". 
 
432 SOLID GEOMETRY — BOOK IX 
 
 Proposition XVI. Theorem 
 
 687. A spherical triangle of a sphere equals one half j 
 a lune of that sphere whose angle in degrees equals the 1 
 spherical excess of the triangle. 
 
 Hypothesis. A, B, and C are the measures in degrees of the 
 angles of spherical A ABC. E represents the spherical excess 
 of the triangle. 
 
 Conclusion. A ABC = \Le ; 
 
 that is, A ABC = one half a lune whose angle is E. 
 Proof. 1. Complete the ® ABA'B', BCB'C, smd AGA'C, 
 and draw diameters AA', BB', and CC. 
 
 2. A ABC -^AACB' = lune of Z B, or Lb- 
 
 3. A ABC -irAA'CB = lune of Z A, or Lj,. 
 
 4. A ABC + A ABC = lune of Z (7, or Lc. 
 
 But A A'B'C = A ABC, so § 685 
 
 5. A ABC -f- A A'B'C = lune of Z C, or X^. 
 
 6. Adding the equations of steps 2, 3, and 5, 
 
 2 A ^5(7 + (A ABC + A ACB' ^AA'CB-\-A A'B'C) 
 
 7. But A ^50+ A ^05' + A J[' 05 + A ^'^'0 
 
 = surface of hemisphere 
 
 = Aso- § 6^^ 
 
 8. .-. 2 A ABC-]- Li8o = i^^+^+c. 
 
 9. .*. 2 A ^5(7 =i^+5+c' — Aso =-^ (^+5+0-180' 
 
 10. But J + 5 + (7- 180 = E. 
 
 11. .-. 2 A yl^C = X^, or A ABC = i L^. 
 
 i 
 
MEASUREMENT OF SPHERICAL POLYGONS 433 
 
 688. Cor. 1. If the radius of the sphere is Rand the sjjherical 
 excess of A ABC in degrees is E, then 
 
 689. Cor. 2. The area of any spherical polygon whose excess 
 
 IS E ts 
 
 180 
 
 Suggestion, — Divide the polygon into A by drawing diagonals from one 
 vertex. Express the area of each triangle. Remember that the excess of the 
 polygon equals the sum of the excesses of the triangles, when the polygon is 
 divided as suggested. 
 
 Note. — A spherical degree may be defined as being a bi-rectangular 
 spherical triangle whose third angle is one spherical angular degree. The 
 area of a spherical triangle in spherical degrees can be proved to equal its 
 spherical excess in degrees. 
 
 Ex. 59. Determine the area of a spherical triangle whose angles are 
 125°, 133°, and 156°, on a sphere whose radius is 10 in. 
 
 Ex. 60. What is the ratio of the areas of- two spherical triangles on the 
 same sphere whose angles are 94°, 135°, and 146°, and 87°, 105°, and 118°, 
 respectively. 
 
 Ex. 61. Determine the area of a spherical triangle whose angles are 
 103°, 112°, and 127° on a sphere whose area is 160. 
 
 Ex, 62. Find the area of a spherical hexagon whose angles are 120°, 
 139°, 148°, 155°, 162°, and 167°, on a sphere whose radius is 12. 
 
 Ex. 63. The sides of a spherical triangle on a sphere whose radius is 
 15 in. are 44°, 63°, and 97°. Find the area of its polar triangle. 
 
 Ex. 64. Determine the part of the area of the surface of a sphere in- 
 tercepted by a triedral angle whose face angles are 89°, 55°, and 100°. 
 
 Ex. 65. Express the ratio of a spherical triangle to the surface of the 
 sphere in terms of the spherical excess E of the triangle, when (a) the 
 excess is measured in degrees ; (b) the excess is measured in right angles. 
 
 Ex. 66. The area of a spherical pentagon, four of whose angles are 
 112°, 131°, 138°, and 168°, is 27. If the area of the surface of the sphere 
 is 120, what is the other angle ? 
 
 Ex. 67. What part of the surface of a sphere is a tri-rectangular 
 triangle of the sphere ? 
 
 Ex. 68. Compare the area of a tri-rectangular spherical triangle of a 
 sphere whose radius is 10 in. with the area of the plane triangle formed by 
 the chords of the sides of the spherical triangle. 
 
434 SOLID GEOMETRY — BOOK IX 
 
 VOLUME OF A SPHERE 
 
 690. If a semicircle be revolved about its diameter as an 
 axis, the solid generated by any sector of the semicircle is 
 called a Spherical Sector. 
 
 A 
 
 Thus if semicircle ACDB be revolved about 
 diameter AB as an axis, sector OCD generates a 
 spherical sector. 
 
 The zone generated by the arc of the circu- 
 lar sector is called the base of the spherical 
 sector. 
 
 Note. — In the following pages, the expression "Vol. OCD'^ will be 
 used to denote the volume of the solid generated by revolving the portion 
 of the plane within OCD around some axis specified. 
 
 691. Application of Limits to Spherical Sectors. Let O be 
 
 the center of arc AB, and OM be any diameter of the circle 
 whose center is O. Let G bisect arc AB, and 
 draw radii OA, OB, and 00. Draw OE ± 
 AC; draw broken line ACB. 
 
 If the sector OAB is revolved about OM 
 as an axis, it generates a spherical sector. 
 The portion of the plane bounded by polygon 
 OACB generates a solid which is less than the spherical sector 
 generated by circular sector OAB. 
 
 If arcs AG and GB are bisected at D and F respectively 
 and broken line ADGFB is drawn, then when the figure is 
 revolved about OM, the part of the plane bounded by polygon 
 OADGFB generates a solid more nearly equal to the spherical 
 sector. If the process of halving the arcs be continued indefi- 
 nitely, it will be assumed evident that the solid generated by 
 the part of the plane bounded by the inscribed polygon ap- 
 proaches the spherical sector as limit. 
 
 Notice that the surface generated by broken line ADGFB 
 approaches as limit the zone generated by arc AGB. (§ 668.) 
 
VOLUME OF A SPHERE 
 
 435 
 
 Proposition XVII. Theorem 
 
 692. If a portion of a plane inclosed hy an isosceles 
 triangle he revolved about a straight line in its plane 
 as axis, ivhich passes through its vertex ivithout inter- 
 secting its surface and ivithout being parallel to its 
 base, the volume of the solid generated is equal to the 
 area of the surface generated by its base multiplied by 
 one third its altitude. 
 
 Hypothesis. Isosceles A OAB, and the surface inclosed, are 
 revolved about straight line OF in its plane ; OF is not II base 
 AB; OC±AB. 
 
 Conclusion. Vol. OAB = area AB x ^ 00. 
 
 Proof. 1. Draw AD ± OF and BE A. OF; extend BA to 
 meet OF at F. 
 
 2. Vol. OBF= vol. OBE -f vol. BEF 
 
 = \TrB^ X 0^ + i ttBE" X EF § 616 
 
 = \7rBE'{0E 4- EF)=^7rBE X BE x OF. 
 BE X OF = OCX BF. Prove it. 
 
 3. 
 4. 
 
 5. 
 6. 
 7. 
 8. 
 9. 
 10. 
 
 11. 
 
 But 
 
 But 
 
 •. vol. OBF= \ ttBE X OC X BF. 
 
 area BF x ^ OC. 
 vol. OAF = area AF x ^ OC. 
 
 ttBE X BF is the area BF. 
 .;. vol. OBF 
 Similarly 
 Subtracting 9 from 8, 
 
 vol. OAB = (area BF - area AF) x ^ OC. 
 = area AB X ^ OC. 
 
 §614 
 
436 SOLID GEOMETRY — BOOK IX 
 
 Pkoposition XYIII. Theorem 
 
 693. The volume of a spherical sector is equal to the 
 area of the zone which forms its base, multiplied hy 
 one third the radius of the sphere. 
 
 Hypothesis. Sector OAB of O is revolved about di- 
 ameter OM as an axis ; E is the radius of the sphere. 
 
 Conclusion. Vol. of the spherical sector generated by cir- 
 cular sector OAB = area of zone generated by AB X ^B. 
 
 Proof. 1. Let C bisect^. Draw AC, CB, OC \ and draw 
 OEl.Aa 
 
 2. Vol. OAC = area AG x \ OE. § 691 
 
 3. Vol. OCB = area CB x \ OE. 
 
 4. Adding, vol. 0^05 = (area AC + area CB) x \ OE 
 
 = area ACB x \ OE. 
 
 5. Let the subdivisions of AB be bisected indefinitely. 
 
 6. Then vol. OACB = vol. generated by sector OAB, 
 and area ACB = area of zone generated by AB 
 and OE = R. 
 
 7. .*. vol. generated by sector OAB 
 
 = area of zone generated by AB x ^B. § 403 
 
 694. Cor. 1. If V denotes the volume of a spherical sector, h 
 the altitude of the zone which forms its base, and R the radius of 
 the sphere, 
 
 V = 27rRh xiR = lirR'h. § 671 
 
 695. Cor. 2. The sphere may be regarded as a spherical 
 sector whose base is the entire surface of the sphere. Letting 
 
VOLUME OF A SPHERE 437 
 
 V denote the volume of the sphere, and M its radius, 
 
 The volume of a sphere is equal to the cube of its radius multi- 
 plied by j^TT. 
 
 696. Cor. 3. The volumes of two spheres have the same ratio 
 as the cubes of their ludii. 
 
 Ex. 69. Find the volume of the sphere whose radius is 12. 
 
 Ex. 70. Determine the volume of metal in a spherical shell 10 in. in 
 diameter and 1 in. thick. 
 
 Ex. 71. A spherical cannon ball 9 in. in diameter is dropped into a 
 cubical box filled with water, whose depth is 9 in. How many cubic 
 inclies of water will be left in the box ? 
 
 Ex. 72. If a sphere in. in diameter weighs 351 oz., what is the 
 weight of a sphere of the same material whose diameter is 10 in.? 
 
 Ex. 73. The outer diameter of a spherical shell is 9 in., and its thick- 
 ness is 1 in. What is the weight, if a cubic inch of the metal weighs one 
 third pound ? 
 
 Ex. 74. Find the area of the surface and the volume of the sphere 
 inscribed in a cube the area of whose surface is 486 sq. in. 
 
 Ex. 75. Find the radius and the volume of a sphere, the area of 
 whose surface is 324 ir sq. in. 
 
 Ex. 76. Prove that the volume of a sphere is two thirds the volume 
 of its circumscribed cylinder. 
 
 Ex. 77. "Within a sphere of radius E is inscribed a right circular 
 cylinder whose altitude equals the diameter of its base. 
 
 (a) Determine its lateral area and compare the result with the area of 
 the surface of the sphere. 
 
 (b) Compare its volume with the volume of the sphere. 
 
 Ex. 78. Given a spherical surface of radius R and its circumscribed 
 right circular cylinder. From the center of the sphere, draw lines to the 
 points of the circles bounding the bases of the cylinder, thus forming the 
 two right circular cones. Compare the volume of the sphere with the dif- 
 ference between the volume of the cylinder and the sum of the volumes 
 of the two cones. 
 
 Ex. 79. A cylindrical vessel, 8 in. in diameter, is filled to the brim 
 with water. A ball is immersed in it, displacing water to the depth of 2 J 
 in. Find the diameter of the ball. 
 
 Note. — Supplementary Exercises 72-85, p. 460, can be studied now. 
 
438 SOLID GEOMETRY — BOOK IX 
 
 SUPPLEMENTARY TOPICS 
 
 Group A. Construction of Spheres 
 
 Proposition XIX. Theorem 
 
 697. Through four points not in the same plane, one and only 
 owe spherical surface can be passed. 
 
 Hypothesis. A, B, O, and D are four points, not in the same 
 plane. 
 
 Conclusion. One and only one spherical surface can be passed 
 through A, B, C, and Z>. 
 
 Proof. 1. Every point equidistant from O and D must lie 
 in a plane EKFl. CD at its mid-point K\ and conversely. 
 
 §§457; 459 
 
 2. Every point equidistant from B and G must lie in a plane 
 IJFJ^BG 2it its mid-point J-, and conversely. 
 
 3. These planes intersect in a line HF, which is ± plane 
 BCD at F, the circumcenter of A BCD. § 499 
 
 Also, by steps 1 and 2, every point equidistant from B, G, 
 and D must lie in HF-, and conversely. 
 
 4. Similarly, every point equidistant from A, G, and D lies 
 in line OE, which is ± plane ACD at E, the circumcenter of 
 AACD. 
 
 5. Since E and F are in plane EKF, then OE and HF lie 
 in plane EKF. § 497 
 
CONSTRUCTION OF SPHERES 439 
 
 6. .-. OE intersects HF at a point 0. 
 
 7. .'. is one and the only point which is equidistant from 
 A, B, C, and D. Steps 3 and 4 
 
 8. .-.a sphere with center and radius OA is the one and 
 only sphere through Af B, C, and D. 
 
 698. Cor. A sphere may he circumscribed about any tetror 
 edron. 
 
 Ex. 80. What is the locus of the center of a sphere which will have 
 a given radius r and will pass through a given point P ? 
 
 Ex. 81. What is the locus of the center of a sphere which will pass 
 through each of two given points ? 
 
 Ex. 82. What is the locus of the center of a sphere which will pass 
 through each of three given points ? 
 
 Ex. 83. What is the locus of the center of a sphere which will pass 
 through all the points of a given circle ? 
 
 Ex. 84. Is it possible for a sphere to pass through all the points of a 
 given circle and also through a given point outside the plane of the circle ? 
 If so, tell how to determiue its center and its radius. 
 
 Ex. 85. Prove that a sphere can be circumscribed about a cube. 
 
 Ex. 86. What is the locus of the center of a sphere which is tangent 
 to a given plane at a given point ? 
 
 Ex. 87. What is the locus of the center of a sphere which will have 
 a given radius r and will be tangent to a given plane ? 
 
 Ex. 88. What is the locus of the center of a sphere which will be 
 tangent to each of the faces of a given diedral angle ? 
 
 Ex. 89. Is it possible for a sphere to be tangent to each of the faces 
 of a given triedral angle ? If there is more than one such sphere, what 
 is the locus of the center ? 
 
 Ex. 90. Find the area of the spherical surface 
 passing through the vertices of a regular tetraedron 
 whose edge is 8. 
 
 Suggestion. — Draw DOE and AOF perpendicular 
 to &^ ABC and BCD respectively. 
 
 C 
 Ex. 91. Prove that a sphere can be inscribed in a given tetraedron. 
 
440 solid geometry — book ix 
 
 Group B. General Theorems of Spherical 
 Geometry 
 
 699. Special interest attaches to the following theorems 
 because of their similarity to certain theorems of plane geom- 
 etry. In each case, the pupil should recall the corresponding 
 theorem of plane geometry, if there is one, or should note the 
 difference between the theorem of spherical geometry and the 
 corresponding situation in plane geometry. 
 
 Proposition XX. Theorem 
 
 700. The intersection of two spherical surfaces is a circle, whose 
 center is in a straight line joining the centers of the spheres and 
 whose plane is perpendicular to that line. 
 
 Hypothesis. and 0' are the centers of two intersecting 
 
 spherical surfaces. 
 
 Conclusion. The intersection of the surfaces is a circle 
 whose center is in line 00' and whose plane is A. 00'. 
 
 Proof. 1. Through and 0' and any point A of the inter- 
 section, pass a plane. This plane cuts the two surfaces in two 
 intersecting great (D. Let AB be the common chord of these 
 two (D, intersecting 00' at C. 
 
 2. .'. 00' bisects AB at right angles. § 207 
 
 3. If the entire figure is revolved about 00' as an axis, the 
 (D will generate the spherical surfaces whose centers are 
 and 0'. 
 
 Point A will generate a O whose center is C and radius 
 AC, which is common to the two spherical surfaces. 
 
 4. The plane of O ^C is ± 00'. § 458 
 
THEOREMS OF SPHERICAL GEOMETRY 441 
 
 6. No point outside O ACB can lie in both surfaces ; for, if 
 there were such a point, the two surfaces would necessarily 
 coincide. § 697 
 
 Ex. 92. What is the locus of points at the distance ri from a given 
 point Pi and at the distance rg from a given point Pa ? 
 
 Ez. 93. The distance betveeen the centers of two spheres whose 
 radii are 25 and 17, respectively, is 28. Find the diameter of their circle 
 of intersection, and its distance from the center of each sphere. 
 
 Suggestion. — Recall § 313. 
 
 Proposition XXI. Theorem 
 
 701. Any side of a spherical triangle is less than the sum of 
 
 the other two. 
 
 A 
 
 Hypothesis. AB is any side of spherical A ABO. 
 Conclusion. AB < AC 4- BC. 
 
 Suggestions. — ■[. Compare /L AOB with A AOC+ ABOC. 
 2. What is the relation of the measure of AB and that of /L AOB'i Of 
 iC'and AA0C1 etc. 
 
 Ex. 94. Prove that any side of a convex spherical pojygon is less 
 than the sum of the remaining sides. 
 
 Ex. 95. Prove that the sum of the arcs of great circles drawn from 
 any point within a spherical triangle to the extremities of any side, is less 
 than the sum of the other two sides of the triangle. 
 
 702. Two spherical polygons on the same sphere or equal 
 spheres are mutually equilateral or mutually equiangular when 
 the sides or angles of one are equal respectively to the sides or 
 angles of the other, whether taken in the same or in opposite 
 orders. 
 
442 SOLID GEOMETRY — BOOK IX 
 
 Proposition XXII. Theorem 
 
 703. If two spherical triangles on the same sphere, or equal 
 spheres, have two sides and the included angle of one equal respec- 
 tively to two sides and the included angle of the other, 
 
 I. They are congruent if the equal parts occ^ir in the same 
 order. 
 
 II. They are symmetrical if the equal parts occur in opposite 
 orders. 
 
 1. Hypothesis. ABC and DEF are spherical A on the same 
 sphere, or equal spheres, having AB = DE, AC = DF, and 
 /.A=/.Dy, and the equal parts occur in the same order. 
 
 Conclusion. A ABC ^ A DEF. 
 
 Suggestion. — Prove it by superposition as in § 63. 
 
 II. Hypothesis. ABC and D'E'i' are spherical A on the 
 same sphere, or equal spheres, having AB = D'E\ AC = D'F', 
 and AA = ZD'; and the equal parts occur in opposite orders. 
 
 Conclusion. ABC and D'E'F' are symmetrical A. 
 
 Proof. 1. Let DEF be a spherical A on the same sphere, 
 or an equal sphere, symmetrical to AD'E'F', having 
 
 DE = UE\ DF= UF', and Z Z> = Z D', 
 the equal parts occurring in opposite orders. 
 
 2. Then in spherical A ABC and DEF, 
 
 AB = DE, ACJ = DF, and Z A = Z D; 
 and the equal parts occur in the same order. 
 
 3. .-. A ABC ^ A DEF. 
 
 4. .-. A ABC is symmetrical to A D'E'F'. 
 
THEOREMS OF SPHERICAL GEOMETRY 443 
 
 Ex. 96. Prove that the arc of a great circle bisecting the vertical 
 angle of an isosceles spherical triangle is perpendicular to the base and 
 bisects the base, 
 
 Ex. 97. Prove that the angles opposite the equal sides of an isos- 
 celes spherical triangle are equal. 
 
 Proposition XXIII. Theorem 
 
 704. If two spherical triangles on the same sphere^ or on equal 
 spheres J have a side and two adjacent angles of one equal respec- 
 tively to a side and two adjacent angles of the other, 
 
 I. They are congruent if the equal parts occur in the same order. 
 
 II. They are symmetrical if the equal parts occur in opposite 
 orders. 
 
 The proof is left to the student. • 
 
 Proposition XXIV. Theorem 
 
 705. If two spherical triangles on the same sphere, or on equal 
 spheres, are mutually equilateral, they are mutually equiangular. 
 
 -"^0 
 
 
 Hypothesis. ABC and DEF are mutually equilateral 
 
 spherical A on the equal sphere ; BC and EF are homologous. 
 
 Conclusion. A ABC and DEF are mutually equiangular. 
 
 Suf/f/estioiis. — 1, Let O and 0' be the centers of the respective spheres, and 
 draw the radii to A,B, C, D, E, and F. Consider the two triedral angles 
 O- ABC And O'-DEF. 
 
 2. Compare the face angles of triedrals 0-ABC and O'-DEF. 
 
 3. Compare the diedral angles of O-ABC and O'-DEF. 
 
 4. Now compare the A A and D ; also the A B and /; ; also A C and F. 
 Note. — The theorem may be proved in a sinnlar manner when the 
 
 given spherical ^ are on the same sphere. 
 
444 SOLID GEOMETRY — BOOK IX 
 
 706. Cor, If two spherical triajigles on the same sjjhere, or on 
 equal spheres, are mutually equilateral, 
 
 I. They are congruent if the equal parts occur in the same 
 order. 
 
 II. Tliey are symmetrical if the equal parts occur in opposite 
 orders. 
 
 Ex. 98. Prove that the arc of a great circle drawn from the vertex 
 of an isosceles spherical triangle to the middle point of the base, is per- 
 pendicular to the base, and bisects the vertical angle. 
 
 Proposition XXV. Theorem 
 
 707. If two spherical triangles on the same sphere, or equal 
 spheres, are mutually equiaiigular, their polar triangles are 
 mutually equilateral. 
 
 i 
 
 Hypothesis. ABC and DEF are mutually equiangular 
 spherical A on the same sphere or equal spheres, A A and 
 D being homologous ; also, A A'B'C is the polar A of A ABC, 
 and A D'E'F' of A DEF, A being the pole of mj', and D of 
 E'F'. 
 
 Conclusion. A A'B'C and D'E'F' are mutually equilateral. 
 
 Suggestions. — 1. Compare the measure ot Z. A and B'C"; oi Z. D and 
 E^'. Then compare B^' and E^'. 
 
 2. Proceed similarly for the other pairs of homologous sides. 
 
 708. Cor. If two spherical triangles on the same sphere, or 
 equal spheres, are mutually equilateral, their polar triangles are 
 mutually equiangular. 
 
 Suggestion. — Model the proof after that of § 707. 
 
THEOREMS OF SPHERICAL GEOMETRY 445 
 
 Proposition XXVI. Theorem 
 
 709. If two sjjherical triangles on the same sphere, or on equal 
 spheres, are mutually equiangular, they are mutually equilateral. 
 
 D' 
 
 Hypothesis. ABC and DEF are mutually equiangular spher- 
 ical A on the same sphere or equal spheres. 
 
 Conclusion. A ABC and DEF are mutually equilateral. 
 
 Proof. 1. Let A A'B'C be the polar A of ABC, and D'E'F 
 of DEF. 
 
 2. Since A ABC and A DEF are mutually equiangular, 
 A A'B'C and A D'E'F' are mutually e(iuilateral. § 707 
 
 3. .'. A A'B'C and A D'E'F' are mutually equiangular. 
 
 4. .-. But A ABC is the polar A of A'B'C and DEF of 
 D'E'F. § 655 
 
 5. .*. A ABC and A DEF are mutually equilateral. Why ? 
 
 710. Cor. Iftv:o spherical triangles on the same sphere or on 
 equal spheres are mutually equiangular, 
 
 I. They are congruent if the equal parts are arranged in the 
 same order. 
 
 II. They are symmetrical if the equal paHs are arranged in 
 opposite orders. 
 
 Ex. 99. Compare the theorem of Proposition XXVI with the corre- 
 sponding theorem about two plane triangles. 
 
 Ex. 100. If three diameters of a sphere be drawn so that each is per- 
 pendicular to the other two, the planes determined by them divide the 
 surface of the sphere into eight congruent tri-rectangular triangles. 
 
 Note. — Recall at this point all the theorems by which two spherical 
 triangles can be proved congruent. 
 
446 
 
 SOLID GEOMETRY — BOOK IX 
 
 Proposition XXVII. Theorem 
 
 711. In an isosceles triangle, the angles opposite the equal sides 
 are equal. 
 
 D *C 
 
 Hypothesis. In spherical triangle ABC, AB = AG. 
 Conclusion. ZB= ZC. 
 
 Suggestion. — Let AD, an arc of a great circle, bisect BC Use § 705. 
 
 Proposition XXYIII. Theorem 
 
 712. If two angles of a spherical triangle are equal, the sides 
 opposite are equal. 
 
 Hypothesis. In spherical A ABG, Z.B= Z.C. 
 Conclusion. AB = AG. 
 
 Suggestions. — 1. Let A A'B'C be the polar A of ABC, B being the pole of 
 iX",and CoiA^'. 
 
 2. Compare A'C and A'B' by using § 656. 
 
 3. Compare Z B' and Z C, and from them determine the relation between 
 /b and Id. 
 
 Ex. 101. Prove that the great circle arcs drawn to the extremities of 
 an arc of a great circle from any point on the great circle perpendicular 
 to and bisecting it are equal. 
 
 Ex. 102. State and prove the converse of Ex. 101. 
 
THEOREMS OF SPHERICAL GEOMETRY 447 
 
 Proposition XXIX. Theorem 
 
 713. If tivo angles of a spherical triangle are unequal, the sides 
 opposite them are unequal, the side (ypposite the greater angle being 
 the greaier. 
 
 .'.BD = Da 
 
 Why? 
 
 Ab^Bb>AB. 
 
 Why? 
 
 \Ab-\-DC>AB. 
 
 
 .\AC>AB. 
 
 
 Hypothesis. In spherical A ABC, Z ABC > Z C 
 
 Conclusion. AC > AB. 
 
 Proof. 1. Let BD, an arc of a great circle, meeting AC at 
 D, make Z CBD =ZC 
 
 2. ^ ^ 
 
 3. 
 4. 
 5. 
 
 714. Cor. If turn sides of a spherical tnangle are unequal, 
 the angles opposite are unequal, the angle opposite the greater side 
 being the greater. 
 
 An indirect proof based upon §§ 712 and 713 is quite easy. 
 
 Bac. 103. What is the locus of points on the surface of a sphere which 
 are equidistant from the extremities of an arc of a great circle of that 
 sphere ? 
 
 Ex. 104. Prove that the great circle arcs perpendicular to and 
 hisecting the sides of a spherical triangle intersect in a point which is 
 equidistant from the vertices of the triangle. 
 
 Ex. 105. Prove that a circle can be circumscribed about a spherical 
 triangle. 
 
448 SOLID GEOMETRY — BOOK IX 
 
 Proposition XXX. Theorem 
 
 715. The shortest line on the surface of a sphere between tivo 
 given points is the arc of a great circle, not greater than a semi- 
 circle which joins the two points. 
 
 Hypothesis. Points A and B are on the surface of a sphere, 
 and AB is an arc of a great O, not greater than a semicircle. 
 
 Conclusion. AB is the shortest line on the surface of the 
 sphere between A and B. 
 
 Note. — The following proof is divided into four parts, (a), (6), (c), 
 and (d). 
 
 Proof, (a) 1. Let C be any point in AB. 
 2. Let DCF and ECG be arcs of small (D with A and B 
 respectively as poles, and AC and BC as polar distances. 
 
 (b) DCF and ECG have only point C in common. 
 
 1. For let F be any other point in DCF and draw AF and 
 BF, arcs of great circles. 
 
 2. .-. i?=iC. §643 
 
 3. But AF-\-BF>AC + BC. Why? 
 
 4. Subtracting AF from the first member of the inequality 
 and its equal AC from the second member, 
 
 BF>BC, or BF>BG. 
 
 5. .-. i?^lies outside small QECG, and DCF and ECG have 
 only point C in common. 
 
 (c) The shortest line on the surface of the sphere from A to 
 B must pass through C. 
 
THEOREMS OF SPHERICAL GEOMETRY 449 
 
 1. Let ADEB be any line on the surface of the sphere be- 
 tween A and B^ not passing through C, and cutting DCF and 
 ECG at D and E respectively. 
 
 2. Then, whatever the nature of line AD, it is evident that 
 an equal line can be drawn from A to C. 
 
 3. In like manner, whatever the nature of line BE, an equal 
 line can be drawn from B to C. 
 
 4. Hence a line can be drawn from AtoB passing through 
 O, equal to the sum of lines AD and BE, and consequently 
 less than ADEB by the part DE. 
 
 5. Therefore no line which does not pass through C can be 
 the shortest line between A and B. 
 
 (d) AB is the' shortest line from ^ to JB on the surface of 
 the sphere. 
 
 1. But C is any point in AB. 
 
 2. Hence the shortest line from ^ to i^ must pass through 
 every point of AB. 
 
 3. Then the great circle arc AB is the shortest line on the 
 surface of the sphere between A and B. 
 
 Ex. 118. Prove that any point in the arc of a 
 great circle bisecting a spherical angle is equally 
 distant (§ 573) from the sides of the angle. 
 
 (Prove PM = PN. Let ^ be a pole of arc AB, 
 and F of arc BC. Spherical A BPE and BFF ^^-' 
 are symmetrical by § 702, XL, and PE = PF.) 
 
 Ex. 119. Prove that a point on the surface of 
 a sphere, equally distant from the sides of a spherical angle, lies in the arc 
 of a great circle bisecting the angle. 
 
 (Fig. of Ex. 118. Prove Z ABP = Z CBP. Spherical ^ BPE and BPF 
 are symmetrical by § 706. ) 
 
 Ex. 120. What is the locus of points on the surface of a sphere 
 equally distant from the sides of a spherical angle? 
 
 Ex. 121. Prove that the arcs of great circles bisecting the angles of 
 a spherical triangb meet in a point equally distant from the sides of the 
 triangle. 
 
 Ex. 122. Prove that a circle may be inscribed in any spherical tri- 
 angle. 
 
450 
 
 SOLID GEOMETRY — BOOK IX 
 
 Gkoup C. Spherical Segments, Pyramids, 
 AND Wedges 
 
 716. A Spherical Segment is the portion of a sphere included 
 between two parallel planes which intersect the sphere. 
 
 The portions of the planes bounding the segment are the 
 bases of the segment ; the perpendicular between the planes is 
 the altitude of the segment. 
 
 A spherical segment of one base is the spherical segment 
 one of whose bounding planes is a tangent to the sphere. 
 
 If a semicircle ACEB be revolved about 
 diameter AB as an axis, and CD and EF are 
 perpendicular to AB, the portion of the 
 plane bounded by FECD generates a spheri- 
 cal segment whose altitude is DF, and whose 
 bases have radii CD and EF respectively ; 
 the portion ACD generates a spherical seg- 
 ment of one base wjiose altitude is AD. 
 
 lYJ. If r and r' are the radii of the bases, h the altitude, and 
 V the volume of a spherical segment, then 
 
 Let O be the center of ADB ; let AA' and BB' be ± to the 
 diameter OM; let AA' = v' , BB' = r, A'B' == h. Let the whole 
 figure revolve about OM as an axis. Let v = the volume of 
 the resulting spherical segment. 
 
 Solution. 1. Draw OA, OB, and AB ; draw OC ± AB, and 
 AE ± BB'. Let OA = R. 
 
SPHERICAL SEGMENTS AND WEDGES 451 
 
 2. Now, vol. ADBB'A' = vol. ACBD + vol. ABB' A'. (1) 
 
 3. Also, vol. ACBD = vol. OADB - vol. OAB. 
 
 4. vol. OADB = I Tram. § 694 
 
 5. And, vol. OAB = a^renAB x i 00 § 691 
 
 6. =hx2 ttOC X i OC § 668 
 
 7. =1 TrOO'/i. 
 
 8. .-. vol. ACDB = 1 7r727i - | ttOO'A 
 
 9. = J 7r(i22 - OC^)h. 
 
 10. But, Ii^-OC^ = AC^^ Why? 
 
 11. = a^)2 
 
 12. = \ Alf, 
 
 13. " .'.\ol ACDB = ^TrXiA&x h = iwA&h. 
 
 14. Now, Zb' = 5^' + 3E' 
 
 15. =(r-r')2+A2. 
 
 16. .-. vol. ^(7Z>i^ = i 7r[(r - r')2 + Ji^yi. 
 
 17. Also, vol. ABB' A' = J 7r(r2 + r"' -\- rr')h. § 621 
 
 18. Substituting in step 2, vol. ADBB'A' 
 
 = i 7r[(r - r'y + 7i2]/i + I ^(2 r2 + 2 r'2 + 2 rr')/* 
 
 19. = J 7r(r2 - 2 rr' + r'2 -\- h^ -^ 2 r^ -{- 2 r'^ -{- 2 rr')h 
 
 20. = -J 7r(3 7-2 + 3 r'2)/4 4- -^ nh^ 
 
 21. =i7r(r2+r'2)A + i7r/il 
 
 Ex. 123. Find the volume of a spherical segment, the radii of whose 
 bases are 4 and 5, and whose altitude is 9. 
 
 718. A Spherical Wedge is a solid bounded by a lune and 
 the planes of its bounding arcs. 
 
 Evidently, two spherical wedges in the same sphere, or 
 equal spheres, are congruent when their angles are equal. 
 
 Also, it is evident that two wedges in the same sphere can 
 be added by placing them so that they have one common 
 bounding plane. The angle of the sum is equal to the sum of 
 the angles of the wedges. 
 
 Note. — Review at this time § 675 and § 676, noting the analogy be- 
 tween wedges and lunes. 
 
452 
 
 SOLID GEOMETRY — BOOK IX 
 
 719. It can be proved as in § 677 that two wedges have the 
 same ratio as their angles. (Cf. § 677.) 
 
 720. A sphere may be regarded as a wedge whose angle is 
 360°. (Cf. § 678.) 
 
 Therefore, a wedge of a sphere whose radius is r, whose angle 
 contains A degrees has a volume v determined as follows : 
 
 or V = 
 
 TTT^A 
 
 irr 
 
 360' 270 
 
 721. Since the lune whose angle is A degrees, on a sphere 
 whose radius is r, has its area expressed by the formula ^^^-^ 
 
 (§ 680) 
 
 .*. the volume of a wedge equals one third the radius of the 
 
 sphere multiplied by the area of the lune which forms its base. 
 
 722. A Spherical Pyramid is a solid 
 bounded by the spherical polygon and the 
 planes of its sides ; as 0-ABGD in the ad- 
 joining figure. 
 
 The center of the sphere is the vertex of 
 the pyramid, and the spherical polygon is 
 its base. 
 
 Two spherical pyramids are congruent 
 when their bases are congruent, for they can be made to 
 coincide. 
 
 723. Two spherical pyramids whose bases are symmetrical 
 isosceles spherical triangles are congruent, for their bases are 
 congruent by § 683. 
 
 724. Two spherical pyramids corre- 
 sponding to a pair of vertical triedral 
 angles are equal. (Cf. § 685.) 
 
 Suggestions. — 1. Kecall the proof of § 685. 
 
 2. Compare spherical pyramids 0-APB, 
 0-BPC, and 0-CPA with spherical pyramids 
 0-A'P'B', O-B'P'C, and 0-C'P'A\ respectively. 
 
SPHERICAL PYRAMIDS AND WEDGES 453 
 
 725. Tlie volume of a triangular spherical pyramid equals one 
 half the volume of a spherical wedge whose angle is the spherical 
 excess of the base of the pyramid. 
 
 Tlie proof is exactly like that for § 687. 
 
 726. If the radius of the sphere is r and the excess of the base 
 of a triangular spherical pyramid is Ey and the volume of the 
 spherical pyramid is v, then 
 
 iTT^f^^^rr^. §720 
 
 2 270 540 
 
 727. The same formula may be employed to find the volume 
 of any spherical pyramid, with the understanding that E is 
 the spherical excess of the base of the pyramid, measured in 
 degrees. 
 
 728. In the case of any spherical pyramid, the area of the 
 
 base is — — "^ (§ 689). Hence the volume of any spherical pyra- 
 loO 
 
 mid is one third the area of its base multiplied by the radius 
 
 of the sphere. 
 
 Ex. 124. Find the volume of a triangular spherical pyramid the 
 angles of whose base are 92°, 119^, and 134°, if the volume of the sphere 
 is 192. 
 
 Ex. 125. Find the volume of a quadrangular spherical pyramid, the 
 angles of whose base are 107°, 118°, 134°, and 146°, if the diameter of the 
 sphere is 12. 
 
 Ex. 126. The volume of a triangular spherical pyramid, the angles 
 of whose base are 105°, 126°, and 147°, is 60^. What is the volume of the 
 sphere ? 
 
 Ex. 127. Find the volume of a pentagonal spherical pyramid the 
 angles of whose base are 109°, 128°, 137'', 163°, and 168°, if the volume of 
 the sphere is 180. 
 
 Ex. 128. The volume of a quadrangular spherical pyramid, the 
 angles of whose base are 110°, 122°, 136°, and 146° is 12 J. What is the 
 volume of the sphere ? 
 
 Ex. 129. What is the angle of the base of a spherical wedge whose 
 volume is Y" """^ i^ ^^6 radius of the sphere is 4 ? 
 
454 
 
 SOLID GEOMETRY 
 
 SUPPLEMENTARY EXERCISES 
 
 Ex. 1. Two planes DEF and GEF intersect in 
 line EF. A is any point in plane GEF. If ylC be 
 drawn perpendicular to EF^ and AB perpendicular to 
 plane DEF^ prove the plane determined hy AC and 
 BC perpendicular to EF. 
 
 Ex. 2. Prove that the line joining the mid-points of one pair of 
 opposite sides of a quadrilateral in space bisects the line joining the mid- 
 points of the other pair of sides. 
 
 Ex. 3. If two intersecting planes pass through two parallel lines, 
 their intersection is parallel to the parallel lines. 
 
 Ex. 4. If three planes intersect in pairs, the lines of intersection are 
 either parallel or concurrent. 
 
 Suggestions. — Case (a) 1. Assume that two lines of intersection meet at 
 a point P. 2. Prove that P is on the third line of intersection. Case (6) 
 1. Assume that two lines of intersection are parallel. 2. Prove that the third 
 line of intersection is parallel to the other two by an indirect proof. 
 
 Ex. 5. Prove that a line parallel to each of two inter- 
 secting planes is parallel to their intersection. 
 
 Hyp. AB is II planes PR and QS. 
 
 Con. AB II QB. 
 
 Suggestion. — Pass a plane through AB\\PR; then use ^ 
 §471. 
 
 Ex. 6. If a plane be drawn through a diago- 
 nal of a parallelogram the perpendiculars to it from 
 the extremities of the other diagonal are equal. 
 
 Hyp. ABCD is a O. 
 
 BG and DH 1. plane AECF. 
 
 Con. BG = DH. 
 
 Ex. 7. D is any point, in perpendicular AF from 
 A to side BC ot triangle ABC. If line I)E be drawn 
 perpendicular to the plane of ABC, and line GH be 
 drawn through E parallel to BC, prove line AE per- 
 pendicular to GH. 
 
 Suggestion. — FroYe BC 1 to plane AED and then GH 
 1 plane AED. 
 
SUPPLEMENTARY EXERCISES 
 
 455 
 
 Ex. 8. (a) Through a line which is parallel to a plane, a plane can 
 be drawn parallel to the given plane. 
 
 (6) Is the construction possible if the given line is not parallel to the 
 given plane ? 
 
 Ex. 9. If a plane, parallel to the edge of a diedral angle, intersects 
 ilie faces of the angle, its intersections with the faces are parallel to the 
 edge and to each other. 
 
 Ex. 10. If a straight line and a plane are both 
 perpendicular to a given plane, they are parallel, un- 
 less the line lies in the plane. 
 
 Hyp. CB ± plane PB ; plane MN 1. plane PB. 
 Con. CBWMN. 
 
 / 1 
 
 f- 
 
 -,E 
 
 / 
 
 B 
 
 , / 
 
 ^- 
 
 
 y 
 
 E 
 
 Suggestions. — \. Draw CD 1 RQ, and let the plane 
 
 determined by CB and CD meet MN in DE. R ^ 
 
 2. Prove CB \\ DE and hence || plane MN by § 466. 
 
 Ex. 11. If a plane is perpendicular to one of two perpendicular 
 planes, its intersection with the other plane is also perpendicular to the 
 first plane. 
 
 Ex. 12. From any point E within diedral angle 
 CABD., EF and EG are drawn perpendicular to faces 
 ABC and ABD^ respectively, and G^IZ" perpendicular to 
 face ABC at //. Prove i^^is perpendicular to AB. 
 
 Suggestion — Prove that FII lies in the plane of EF 
 and EG, by § 498; also consider the relation of AB and 
 plane GEF. 
 
 Ex. 13. If 7? C is the projection of line AB upon 
 plane MN, and BD and BE be drawn in the plane 
 making Z CBD = Z GBE, prove ZABD = ZABE. 
 
 Suggestions. — 1. Lay off BD = BE, and draw lines 
 AD, AE, CD, and CE. 
 
 2. Prove A ABD and ABE congruent. 
 
 Ex. 14. If a line is perpendicular to one of two intersecting planes, 
 its projection upon the other is perpendicular to the intersection of the 
 two planes. 
 
 Suggestion. — U EG 1 plane AD, and FII is the projection of EG on 
 plane BC, prove FH 1 AB. 
 
 Ex. 15. If a straight line intersects two parallel planes, it makes 
 equal angles with them. 
 
456 
 
 SOLID GEOMETRY 
 
 Ex. 16. The base of a rectangle ABCT) is 10, 
 and its altitude 8. Side 10 is parallel to plane QB. 
 Side 8 makes an angle of 60° with QB. Find the 
 area of the projection of O ABCD on plane QR, 
 correct to three decimal places. 
 
 Ex. 17. An equilateral A ABC, whose area is 25, 
 has its side BC parallel to a plane QB. The plane 
 of A ABC makes an angle of 45° with plane QB. 
 Find the area of the projection of A ABC on 
 plane BQ. 
 
 Ex. 18. Find the lateral area of a regular triangular prism each side 
 of whose base is 5 and whose altitude is 8. 
 
 Ex. 19. Prove that the upper base of a truncated parallelepiped is a 
 parallelogram. 
 
 Ex. 20. Prove that the sum of two opposite 
 lateral edges of a truncated parallelopiped is equal 
 to the sum of the other two lateral edges. 
 
 Suggestions. — 1. What kind of figure is 
 AA'C'C? 
 
 2. Compare AA' + CC with 00'. 
 
 Ex. 21. Prove that the perpendicular drawn 
 to the lower base of a truncated right triangular 
 prism from the intersection of the medians of the 
 upper base, is equal to one third the sum of the 
 lateral edges. 
 
 Suggestion. — Let P be the mid-point of DL, and 
 draw PQ 1 ABC', express LM in terms oi FQ and 
 GN. 
 
 Ex. 22. Prove that the sum of the squares of the 
 four diagonals of a parallelopiped is equal to the sum 
 of the squares of its twelve edges. 
 
 Suggestion. — 1^C2A\ Ex. 142, Book III, p. 184. 
 
 Ex. 23. Determine the approximate area of the base of a bin 6 ft. 
 deep that will hold 250 bu. of grain. (One bu. = 2150.42 cu. in.) 
 
 Ex. 24. Find the edge of a cube equivalent to a rectangular parallelo- 
 piped whose dimensions are 9 in., 1 ft. 9 in., and 4 ft. 1 in. 
 
SUPPLEMENTARY EXERCISES 457 
 
 Ex. 25. Find the volume of a rectangular parallelopiped, the dimen- 
 sions of whose base are 14 and 9, and the area of whose entire surface is 
 620. 
 
 Ex. 26. The diagonal of a cube is 8 VS. Find its volume, and the 
 area of its entire surface. 
 
 Suggestion. — Represent the length of the edge by x. 
 
 Ex. 27. Find the dimensions of the base of a rectangular parallelo- 
 piped, the area of whose entire surface is 320, volume 336, and altitude 4. 
 Suggestion. — Represent the dimensions of the base by x and y. 
 
 Ex. 28. Find the area of the entire surface of a rectangular parallelo- 
 piped, the dimensions of whose base are 11 and 13, and volume 858. 
 
 Ex. 29. A trench is 124 ft. long, 2\ ft. deep, 6 ft. wide at the top, and 
 5 ft. wide at the bottom. How many cubic feet of water will it contain ? 
 
 Ex. 30. Prove that the volume of any oblique prism is equal to the 
 product of the area of a right section by the length of a lateral edge. 
 
 Ex. 31. Prove that the volume of a regular prism is equal to its 
 lateral area multiplied by one half the apothem of the base. 
 
 Ex. 32. The volume of a right prism is 2310, and its base is a right 
 triangle whose legs are 20 and 21, respectively. Find its lateral area. 
 
 Ex. 33. The lateral area and volume of a regular hexagonal prism are 
 60 and 15 VS, respectively. Find its altitude, and one side of its base. 
 Suggestion. — Represent the altitude by x, and the side of the base by y. 
 
 Ex. 34. The altitude of a pyramid is 20 in., and its base is a rectangle 
 whose dimensions are 10 in. and 15 in. , respectively. What is the distance 
 from the vertex of a section parallel to the base, whose area is 54 sq. in.? 
 
 Ex. 35. At what distance from the altitude must a plane parallel to 
 the base be drawn so that the area of the section will be one half the base ? 
 
 Ex. 36. In Ex. 35, replace the fraction ^ by the fraction ^ and solve 
 the resulting exercise. 
 
 Ex. 37. In Ex. 35, replace the fraction ^ by the fraction ^ and solve 
 the resulting exercise. 
 
 Ex. 38. Prove that the volum/B of a regular pyramid is equal to its 
 lateral area, multiplied by one third the distance from the center of its 
 base to any lateral face. 
 
 Suggestion. — Pass planes through the lateral edges and the center of the 
 
458 
 
 SOLID GEOMETRY 
 
 Ex. 39. Find the lateral edge, lateral area, and volume of a frustum 
 of a regular quadrangular pyramid, the sides of 
 whose bases are 17 and 7, respectively, and whose 
 altitude is 12. 
 
 Suggestion. — Let ABB' A' be a lateral face of 
 the frustum, and and 0' the centers of the bases; 
 draw lines OC 1. AB, O'C 1 A'B', CD 1 OC, and 
 A'E 1 AB\ also lines 00' and CC. 
 
 Ex. 40. The bases of a frustum of a pyramid are rectangles, whose 
 sides are 27 and 15, and 9 and 5, respectively, and the line joining their 
 centers is perpendicular to each base. If the altitude of the frustum is 
 12, find its lateral area and volume. 
 
 Ex. 41. Find the lateral area and volume of a frustum of a regular 
 triangular pyramid, the sides of whose bases are 12 and 6, respectively, 
 and whose lateral edge is 6. 
 
 Ex. 42. The altitude and lateral edge of a frustum of a regular 
 triangular pyramid are 8 and 10, respectively, and each side of its upper 
 base is 2\/3. Find its volume and lateral area. 
 
 Ex. 43. Find the volume of the rectangular prismoid the sides of 
 whose bases are 10 and 7, and 6 and 5, respectively, and whose altitude 
 is 9. 
 
 Ex. 44. The volume of a triangular prism is equal to a lateral face, 
 multiplied by one half its perpendicular distance from any point in the 
 opposite lateral edge. 
 
 Suggestion. — Draw a rt. section of the prism, and apply § 577. 
 
 Ex. 45. Prove that the volume of a truncated parallelopiped is 
 equal to the area of a right section multiplied by 
 one fourth the sum of the lateral edges. 
 
 Ex. 46. Prove that a plane passed through 
 the center of a parallelopiped divides it into two 
 equal solids. 
 
 Ex. 47. The volume of a truncated parallelo- 
 piped is equal to the area of a right section, mul- 
 tiplied by the distance between the centers of the 
 
 Suggestion. — By Ex. 45, the distance between the centers of the bases 
 may be proved equal to one fourth the sum of the lateral edges. 
 
SUPPLEMENTARY EXERCISES 459 
 
 Ex. 48. How many square feet of heating surface are there in a hot- 
 water conducting pipe 9 feet long and 2 inches in outside diameter ? 
 
 Ex. 49. Determine the lateral area of the right circular cylinder 
 formed by revolving a rectangle, having base b and altitude h, 
 (a) about its base ; (6) about its altitude. 
 
 Ex. 50. The lateral area of a cylinder of revolution is 120 tt. The 
 area of the base is 36 tt. Find the altitude. 
 
 Ex. 51. The cross section of a tunnel, 2^ mi. in length, is in the form 
 of a rectangle yd. wide and 4 yd. high, surmounted by a semicircle 
 whose diameter is equal to the width of the rectangle ; how many cubic 
 yards of material were taken out in its constniction ? (tt = 3.1416.) 
 
 Ex. 52. What must be the length in inches of a 10-gal. gasoline tank 
 which is 10 in. in diameter ? 
 
 Ex. 53. Determine the volume generated when a rectangle of base b 
 and altitude h 
 
 (a) revolves about its side b ; (6) revolves about its side h. 
 
 Ex. 54. Two right circular cylinders have equal altitudes, but the 
 radius of the base of the one is double the radius of the base of the other. 
 Compare (a) their lateral areas ; (6) their volumes. 
 
 Ex. 55. A regular hexagonal prism is inscribed in a right circular 
 cylinder whose altitude is 10 in. and the radius of whose base is 3 in. 
 Determine the difference between the volumes of the prism and cylinder. 
 
 Ex. 56. Prove that the volume of a cylinder of revolution is equal to 
 its lateral area multiplied by one half the radius of its base. 
 
 Ex. 57. . Express by a formula the volume of a round cast-iron column 
 of length I ft., thickness t in., and outside diameter d in. 
 
 Ex. 58. Given the radius of the base B and the total area T of a 
 cylinder of revolution, find its volume. 
 
 (Find S'from the equation T=2 irEH + 2 -n-R^.) 
 
 Ex. 59. Given the diameter of the base D and the volume Fof a 
 cylinder of revolution, find its lateral area and total area. 
 
 Ex. 60. The volume of a circular cone is F. What is the effect 
 upon the volume : 
 
 (a) if the radius of the base is doubled ? 
 
 (6) if the altitude is doubled ? 
 
 (c) if both the radius and the altitude are doubled ? 
 
 Ex. 61. The altitude of a cone of revolution is 27 in., and the radius 
 of its base is 16 in. What is the diameter of the base of an equal cylin- 
 der, whose altitude is 16 in. ? 
 
460 SOLID GEOMETRY 
 
 Ex. 62. A plane is passed parallel to the base of a circular cone so 
 as to bisect the altitude. What is the ratio of the two parts into which 
 the given cone is divided ? 
 
 Ex. 63. Determine the lateral area of a right circular cone whose 
 volume is 320 tt cu. in., and whose altitude is 15 in. 
 
 Ex. 64. Determine the volume of a cone of revolution whose slant 
 height is 29 in., and whose lateral area is 580 tt sq. in. 
 
 Ex. 65. If the altitude of a cone of revolution is three fourths the 
 radius of its base, the volume is equal to its lateral area multiplied by one 
 fifth the radius of its base. 
 
 Ex. 66. Given the altitude H and the volume T of a right circular 
 cone. Derive the foi-mula for the lateral area in terms of V and H, 
 
 Ex. 67. Given the slant height L and the lateral area S of a right 
 circular cone. Derive the formula for its volume in terms of S and L. 
 
 Ex, 68. Find the lateral area of the frustum of a right circular cone, 
 whose altitude is 8 in., if the radii of its bases are 6 in. and 3 in., respec- 
 tively. 
 
 Ex. 69. A tapering hollow iron column, 1 in. thick, is 24 ft. long, 10 
 in. in outside diameter at one end and 8 in. in diameter at the other. 
 How many cubic inches of metal are there in it ? 
 
 Ex. 70. Prove that a frustum of a circular cone is equal to three 
 cones whose common altitude is the altitude of the frustum, and whose 
 bases equal the lower base, the upper base, and the mean proportional 
 between the bases of the frustum. 
 
 Ex. 71. The area of the entire surface of a frustum of ,a cone of 
 revolution is 306 ir sq. in., and the radii of its bases are 11 in. and 5 in., 
 respectively. Find the lateral area and the volume of it. 
 
 Ex. 72. The volume of a frustum of a right circular cone is 6020 tt 
 cu. in., its altitude is 60 in., and the radius of its lower base is 15 in. 
 Find the radius of the upper base and its lateral area. 
 
 Ex. 73. Find the diameter and the area of the surface of a sphere 
 whose volume is ^^^-^ir cu. in. 
 
 Ex. 74. The altitude of a frustum of a cone of revolution is 3^, and 
 the radii of its bases are 5 and 3 ; what is the diameter of an equal 
 sphere ? 
 
 Ex. 75. Find the area of the surface and the volume of a sphere cir- 
 cumscribing a cylinder of revolution, the radius of whose base is 9, and 
 whose altitude is 24. 
 
SUPPLEMENTARY EXERCISES 
 
 461 
 
 Ex. 76. A cone of revolution is inscribed in a sphere whose diameter 
 is ^ the altitude of the cone. Prove that its lateral surface and volume 
 are, respectively, f and /^ the surface and volume of the sphere. 
 
 Ex. 77. Given the area of the surface of a sphere S to find its 
 volume. 
 
 Ex. 78. Given the volume of a sphere V to find the area of its surface. 
 
 Ex. 79. A portion of a plane bounded by an eqiiilateral triangle, 
 whose side is 6, revolves about one of its sides as an axis. Find the area 
 of the entire surface, and the volume of the solid generated. 
 
 Ex. 80. A circular sector whose central angle is 45° and radius 12 
 revolves about a diameter perpendicular to one of its bounding radii. 
 Find the volume of the spherical sector generated. 
 
 Ex. 81. A portion of a plane bounded by a right triangle, whose 
 legs are a and 6, revolves about its hypotenuse as an axis. Find the area 
 of the entire surface, and the volume of the solid generated. 
 
 Ex. 82. A portion of a plane bounded by an equilateral triangle, 
 whose altitude is h, revolves about one of its altitudes as an axis. Find 
 the area of the surface, and the volume of the solid generated. 
 
 Ex. 83. A portion of a plane bounded by an equilateral triangle, 
 whose side is a, revolves about a straight line drawn 
 through one of its vertices parallel to the opposite side. 
 Find the area of the entire surface, and the volume of the 
 solid generated. 
 
 (The solid generated is the difference of the cylinder 
 generated by BCUG, and the cones generated by ABG 
 and ACH.) 
 
 Ex. 84. If a portion of a plane bounded by any triangle be revolved 
 about an axis in its plane, not parallel to its base, which passes through 
 its vertex without intersecting its surface, the volume of the solid gen- 
 erated is equal to the area of the surface generated by the base, multiplied 
 by one third the altitude. 
 
 Ex. 85. If a portion of a plane bounded by any triangle be revolved 
 about an axis which passes through its 
 vertex parallel to its base, the volume of 
 the solid generated is equal to the area of 
 the surface generated by the base, multi- 
 plied by one third the altitude. 
 
 Ex. 86. Find the volume of a spher- 
 ical sector, the altitude of whose base is 
 12, the diameter of the sphere being 25. 
 
IMPORTANT DEFINITIONS AND THEOREMS OF PLANE 
 GEOMETRY 
 
 § 8. {a) One and only one straight line can be drawn through two 
 points. 
 
 (h) A straight line can be extended indefinitely in each direction. 
 
 § 11, Two straight lines can intersect at only one point. 
 
 § 14. The straight line segment is the shortest line between two points. 
 
 § 16. A circle is a closed curved line (in a plane) all points of which 
 are equidistant from a point within called the center. 
 
 § 17. All radii of the same circle or of equal circles are equal. 
 
 § 20. An angle is the figure formed by two rays drawn from the same 
 point. 
 
 § 24. Adjacent angles are two angles that have a common vertex and 
 a common side between them. 
 
 § 26. If one straight line meets another straight line so that the adjacent 
 angles formed are equal, each of these angles is a right angle. 
 
 § 27. All right angles are equal. 
 
 § 29. Two lines are perpendicular if they form a right angle. 
 
 § 34. The sum of all the successive adjacent angles around a point on 
 one side of a straight line is one straight angle. 
 
 § 35. The sum of all the successive adjacent angles around a point is 
 two straight angles. 
 
 § 36. Two angles are complementary if their sum equals a right angle. 
 
 § 37. Complements of the same angle or of equal angles are equal. 
 
 § 38. Two angles are supplementary if their sum is equal to a straight 
 angle. 
 
 § 39. If two adjacent angles have their exterior sides in a straight line, 
 they are supplementary. 
 
 § 40. If two adjacent angles are supplementary, their exterior sides 
 are in a straight line. 
 
 § 41. Supplements of the same angle or of equal angles are equal. 
 
 462 
 
 i 
 
INDEX 
 
 Altitude, of a O, § 131 ; of a trape- 
 zoid, § 145 ; of a A, § 85. 
 
 Analysis, § 236, § 347. 
 
 Angle, § 20; acute, § 30; central, 
 § 180 ; diedral, § 484 ; inscribed, 
 § 216 ; intercepted, § 180 ; obtuse, 
 § 30 ; polyedral, § 507 ; right, § 26; 
 spherical, § 647 ; straight, § 31 ; 
 triedral, § 508 ; tetraedral, § 508 ; 
 between two intersecting curves, 
 § 047 ; between a line and a plane, 
 § 506 ; bisector of, § 23 ; cosine of, 
 § 301 ; measure of, § 28; sides of, 
 § 20 ; sine of, § 300 ; tangent of, 
 § 301 ; vertex of, § 20 ; bisect an, 
 § 74 ; construct an, § 75 ; meas- 
 ure an, by protractor, § 44 ; ex- 
 terior, of a A, § 86 ; of elevation, 
 § 303 ; of depression, § S03. 
 
 Angles, adj., § 24; alt. -int., § 92; 
 alt.-ext. , § 92 ; complementary, 
 § 30 ; corresponding, § 92 ; equal, 
 § 21 ; exterior of lis, § 92 ; homolo- 
 gous of /S^, § 65 ; interior of lis, 
 § 92 ; supplementary, § 38 ; supp.- 
 adj., § 39 ; vertical, § 42. 
 
 Antecedent, § 242. 
 
 Apothem, § 360. 
 
 Arc, § 179 ; minor, § 179 ; major, 
 § 179 ; measure of, § 214 ; inter- 
 cepted, § 180 ; subtended, § 184. 
 
 Area, § 320 ; of a O, § 392, § 416, 
 § 417 ; of a O, § 331 ; of a Q, 
 § 330 ; of a sector, § 397 ; of a 
 
 segment of a O, § 398 ; of a spher- 
 ical surface, § 672 ; of a trapezoid, 
 § 337, § 338 ; of a A, § 333, § 335 ; 
 lateral, of a prism, § 531 ; lateral, 
 of a regular pyramid, § 566 ; lat- 
 eral, of a rt. circ. cylinder, § 599 ; 
 lateral, of a rt. circ. cone, § 613. 
 
 Axiom, § 51 ; of congruence, § 61 ; 
 of limits, § 402 ; of lis, § 90. 
 
 Axioms, list of, § 51, § 61, § 90, 
 §158. 
 
 Axis, of a circle of a sphere, § 634 ; 
 of a circ. cone, § 605 ; of a circ. 
 cylinder, § 593; of symmetry, 
 § 427. 
 
 Base, of a O, § 131 ; of a A, § 68 ; 
 of an isosceles A, § 68. 
 
 Bases, of a O, § 131 ; of a trape- 
 zoid, § 145. 
 
 Bisect a segment, § 78 ; an angle, 
 §74. 
 
 Broken line, § 5. 
 
 Center, of a O, § 16; of gravity, 
 § 172 ; of a O, § 137 ; of a regu- 
 lar polygon, § 360 ; of symmetry, 
 § 426. 
 
 Chord, § 184. 
 
 Circle, § 16 ; arc of, § 179 ; area of, 
 § 392, § 416, § 417 ; center of, 
 § 16 ; chord of, § 16 ; circumfer- 
 ence of, § 389 ; diameter of, § 16 ; 
 interior of, § 174 ; radius of, § 16 ; 
 
 463 
 
464 
 
 INDEX 
 
 sector of, § 396 ; segment of, 
 § 398 ; circumscribed, § 178 ; 
 great, of a sphere, § 634 ; in- 
 scribed, § 225 ; small, of a sphere, 
 § 634. 
 
 Circles, concentric, § 176 ; equal, 
 § 17 ; tangent, § 204 ; tangent 
 internally, § 204; tangent ex- 
 ternally, § 204. 
 
 Circumcenter, § 170. 
 
 Circumference, § 389. 
 
 Circumscribed, circle, § 178; poly- 
 gon, § 225. 
 
 Commensurable, § 211. 
 
 Complement, § 36. 
 
 Conclusion, § 52. 
 
 Concurrent lines, § 168. 
 
 Cone, § 604 ; altitude of a, § 604 ; 
 axis of a circular, § 605 ; base of 
 a, § 604 ; circular, § 605 ; frustum 
 of a, § 605; lateral surface of a, 
 § 604; rt. circ, § 605. 
 
 Cones of revolutipn, § 625. 
 
 Congruence, § 59 ; axiom of, § 61. 
 
 Conical surface, § 603 ; directrix of 
 a, § 603 ; element of a, § 603 ; 
 generatrix of a, § 603 ; vertex of 
 a, § 604. 
 
 Consequent, § 242. 
 
 Constant, § 401, § 542. 
 
 Converse, § 104. 
 
 Co-planar, § 444. 
 
 Corollary, § 71. 
 
 Cosine, § 301. 
 
 Cylinder, § 588 ; altitude of a, § 588 ; 
 bases of a, § 588 ; circular, § 589 ; 
 elements of a, § 588 ; lateral area 
 of a, § 597 ; lateral area of a rt. 
 circ, § 599 ; lateral surface of a, 
 § 588 ; total surface of a, § 588 ; 
 right, § 589 ; volume of a, § 597, 
 §601. 
 
 Degree, of angle, § 28 ; of arc, § 214. 
 Diedral angle, § 484 ; edge of a, 
 
 § 484 ; faces of a, § 484 ; plane A 
 
 of a, § 486. I 
 
 Diedral angles, adj., § 485; equal, | 
 
 § 489 ; vertical, § 485. ^ 
 
 Dimensions of a rect. parallelopiped, 
 
 §541. 
 Directrix, of a conical surface, § 603 ; 
 
 of a cylindrical surface, § 586. 
 Distance, between II planes, § 478 ; 
 
 between points on the surface of 
 
 a sphere, § 641 ; from a point to 
 
 a line, § 84 ; polar, § 644. 
 Dodecaedron, § 527. 
 
 Edge, of a half plane, § 483. 
 Equal angles, § 21 ; ©, § 17 ; seg- 
 ment, § 13 ; surfaces, § 322. 
 Exterior angle, § 86. 
 Extreme and mean ratio, § 377. 
 Extremes, § 244. 
 
 Foot of a line on a plane, § 449. 
 
 Frustum, of a pyramid, § 560; of a 
 cone, §604 ; of a pyramid in- 
 scribed in a frustum of a cone, 
 §611. 
 
 Generatrix, of a conical surface, 
 § 603 ; of a cylindrical surface, 
 § 586. 
 
 Great circle, § 634. 
 
 Half-plane, § 483 ; edge of a, § 483. 
 
 Hexaedron, § 527. 
 
 Homologous parts, § 65 ; sides of 
 
 similar triangles, § 278. 
 Hypotenuse, § 107. 
 Hypothesis, § 52. 
 
 Icosaedron, § 527. 
 In-center, § 169, § 226. 
 
INDEX 
 
 465 
 
 Inclination of a line to a plane, § 50(). 
 Incommensurable, § 211 ; cases, 
 
 § 423, § 424, § 425. 
 Indirect method, § 94. 
 Inscribed, angle, § 216 ; polygon, 
 
 §178. 
 Interior of closed line, § 6. 
 Intersection, of lines, § 10 ; of two 
 
 surfaces, § 448. 
 Isoperimetric, § 432. 
 
 Lateral area, of a frustum of a reg. 
 pyramid, § 667 ; of a frustum of a 
 rt. circ. cone, §617; of a cylinder, 
 § 698 ; of a prism, § 631 ; of a 
 reg. pyramid, § 566 ; of a rt. circ. 
 cone, § 613. 
 
 Legs of a rt. triangle, § 107. 
 
 Length of a circle, § 388, § 407, 
 §412. 
 
 Limit, § 401. 
 
 Line, broken, § 5 ; curved, § 5 ; 
 II to a plane, § 464 ; ± to a plane, 
 § 452 ; straight, § 6 ; of centers, 
 § 206. 
 
 Lines, concurrent, § 168. 
 
 Loci, method of attacking, § 238 ; 
 intersection of, § 340 ; construc- 
 tion by, § 241. 
 
 Locus of points, § 229, § 518. 
 
 Lune, § 675 ; angle of a, § 675. 
 
 Maximum, § 433. 
 
 Means, § 244. 
 
 Measure, numerical, §210 ; common, 
 
 § 211 ; of arc, § 214 ; of central 
 
 angle, §216; of inscribed angle, 
 
 §217. 
 Median, of A, § 79 ; of trapezoid, 
 
 § 145. 
 Mid-point of a segment, § 16. 
 Minimum, § 433. 
 
 Octaedron, § 527. 
 Ortho-center, § 171. 
 
 Parallel lines, § 89 ; axiom of, § 90 ; 
 construction of, § 99. 
 
 Parallelogram, § 131 ; altitude of a, 
 § 131 ; bases of a, § 131. 
 
 Parallelopiped, § 537 ; rt., § 537 ; 
 rect., § 537 ; center of a, Ex. 15, 
 p. 353. 
 
 Perpendicular, § 29 ; construction 
 of, § 80, § 82. 
 
 Perpendicular-bisector, § 76 ; con- 
 struction of, § 78. 
 
 Plan of a proof, § 117. 
 
 Plane, § 443 ; determined, § 444 ; 
 Z of a diedral Z, § 486 ; ± to a 
 line, § 452 ; tangent to a cylinder 
 and to a cone, § 622. 
 
 Planes, parallel, § 465 ; perpendicu- 
 lar, § 494. 
 
 Point, § 4, note p. 27 ; mid-, § 15 ; 
 of contact, § 197 ; of tangency, 
 § 197 ; projection of a, § 308, 
 
 Points, of intersection, § 10 ; locus 
 of, § 228, § 229. 
 
 Polar distance, § 644. 
 
 Poles of a O of a sphere, § 634. 
 
 Polyedra, similar, § 582. 
 
 Polyedral angle, § 507 ; diedral A 
 of a, § 507 ; edges of a, § 507 ; 
 faces of a, § 607 ; face A of a, 
 § 507; vertex of a, § 507. 
 
 Polyedral angles, congruent, § 610, 
 symmetrical, § 511 ; vertical, § 509. 
 
 Polyedron, § 626 ; diagonal of a, 
 § 526 ; edges of a, § 526 ; faces of 
 a, § 526 ; regular, § 679 ; vertices 
 of a, § 526. 
 
 Polygon, § 126 ; .4 of, § 126 ; diag- 
 onal of, § 125 ; interior of, § 125 ; 
 perimeter of, § 126 ; sides of, 
 
466 
 
 INDEX 
 
 § 125 ; vertices of, § 125 ; circum- 
 scribed, § 225 ; concave, § 126 ; 
 convex, § 126 ; equiangular, 
 § 128 ; equilateral, § 128 ; in- 
 scribed, § 178 ; mutually equi- 
 angular, or equilateral, § 129 ; 
 regular, § 356 ; similar, § 272. 
 
 Postulate, § 56. 
 
 Postulates, list of, § 56, § 62. 
 
 Prism, § 528 ; altitude of a, § 528 ; 
 bases of a, § 528 ; lateral area 
 of a, § 528 ; lateral edges of a, 
 § 528 ; lateral faces of a, § 528 ; 
 oblique, § 530 ; regular, § 530 ; 
 right, § 530 ; truncated, § 630. 
 
 Prismatoid, § 573 ; altitude of a, 
 § 573 ; bases of a, § 573 ; mid- 
 section of a, § 573. 
 
 Problem, § 55. 
 
 Projection, of a point on a line, 
 § 308 ; of a point on a plane, 
 § 503 ; of a segment on a line, 
 § 309 ; of a line on a plane, § 503. 
 
 Proportion, § 244 ; by alternation, 
 § 253 ; by inversion, § 254 ; by 
 composition, § 255 ; by composi- 
 tion and division, § 257 ; by divi- 
 sion, § 256. 
 
 Proportional, fourth, § 246 ; third, 
 § 247 ; mean, § 248. 
 
 Proposition, § 57. 
 
 Protractor, § 43 ; use of, § 44, 
 §46. 
 
 Pyramid, § 58 ; altitude of a, § 658 ; 
 base of a, § 568 ; lateral area of 
 a, § 558 ; lateral edges of a, § 558 ; 
 lateral faces of a, § 568 ; vertex 
 of a, § 658 ; frustum of a, § 560 ; 
 regular, § 559 ; truncated, § 559 ,• 
 inscribed in a cone, § 610. 
 
 Quadrant, of circle, § 176. ] 
 
 Radius, of a O, § 16 ; of circum- 
 scribed O of a A, § 177 ; of in- 
 scribed O of a A, § 226 ; of a 
 regular polygon, § 360. 
 
 Ratio, § 212, § 242 ; of similitude, 
 § 272 ; extreme and mean, § 377. 
 
 Ray, § 19. 
 
 Rectangle, § 141. 
 
 Regular polyedron, § 579 ; pyramid, 
 §559. 
 
 Regular polygon, § 356 ; apothem 
 of, § 360 ; central A of, § 360; ra- 
 dius of, § 360 ; construction of, 
 § 370 ; vertex Z of, § 360. 
 
 Rhombus, § 142. 
 
 Scales, § 299. 
 
 Secant, § 194 ; whole, § 285. 
 
 Segment of a circle, § 398 ; of a line, 
 § 12 ; external, § 286 ; internal, 
 §386; spherical, §716; projec- 
 tion of a, § 309 ; divided exter- 
 nally, § 306 ; divided harmoni- 
 cally, § 307 ; divided internally, 
 §269. 
 
 Segments, equal, § 13 ; divided pro- 
 portionally, § 260. 
 
 Semicircle, § 176. 
 
 Sequence of polygons, § 404. 
 
 Sine of an angle, § 300. 
 
 Slant height, of a frustum of a reg. 
 pyramid, § 562 ; of a frustum of 
 a rt. circ. cone, § 606 ; of a reg. 
 pyramid, § 562 ; of a rt. circ. 
 cone, § 606. 
 
 Solid, § 529 ; volume of a, § 533. 
 
 Solids, equal, § 534. 
 
 Sphere, § 629 ; radius of a, § 630 ; 
 diameter of a, § 630. 
 
 Spheres, tangent, § 665. 
 
 Spherical, angle, § 647 ; excess, 
 § 660; pyramid, § 722; sector. 
 
INDEX 
 
 467 
 
 § «90 ; segment, § 716 ; surface, 
 § 027 ; triangle, § 662 ; wedge, 
 § 718. 
 
 Spherical polygon, § 651 ; angle of 
 a, § 051 ; diagonal of a, § 651 ; 
 sides of a, § 651. 
 
 Spherical polygons, symmetrical, 
 §681. 
 
 Square, § 143. 
 
 Subtend, § 184. 
 
 Superposition, § 62 ; postulate of, 
 §62. 
 
 Supplement, § 38. 
 
 Surface, § 442 ; closed, § 523 ; coni- 
 cal, § 603 ; convex, § 524 ; curved, 
 § 522 ; cylindrical, § 586 ; gen- 
 erating a, § 585 ; spherical, § 628. 
 
 Symmetrical spherical polygons, 
 §681. 
 
 Symmetry with respect to, a straight 
 line, § 429 ; a center, § 428. 
 
 Tangent, § 197 ; common, § 203 ; 
 
 external, § 203; internal, § 203; 
 
 of angle, § 301. 
 Tangent to a sphere, a plane, § 662. 
 Tetraedral angle, § 508. 
 Tetraedron, § 527. 
 Theorem, § 52 ; proved formally, 
 
 § 54 ; proved informally, § 53. 
 Transversal, § 92. 
 Trapezoid, § 145 ; altitude of, § 145 ; 
 
 area of, § 337 ; bases of, § 146 ; 
 
 isosceles, § 146; median of, 
 §146. 
 
 Triangle, § 47 ; altitude of, § 85 ; 
 base of, § 68; median of, § 79; 
 parts of, § 47 ; sides of, § 47 ; ver- 
 tical Z. of, § 68 ; vertices of, § 47 ; 
 bi-rectangular, § 661 ; circum- 
 scribed, § 225 ; equiangular, § 68 ; 
 equilateral, § 68 ; inscribed, § 178 ; 
 isosceles, § 68 ; isosceles right, 
 § 107 ; right, § 107 ; scalene, § 68 ; 
 tri-reetangular, § 661 ; construc- 
 tion of, § 232. 
 
 Triedral angle, § 508. 
 
 Trigonometric tables, § 302. 
 
 Trigonometry, § 300. 
 
 Unit, of surface, § 321 
 § 28 ; of arc, § 214. 
 
 of angle, 
 
 Variable, § 401, § 542 ; limit of a, 
 §542. 
 
 Vertical angle, of a A, § 68 ; of an 
 isosceles A, § 68. 
 
 Volume of a cone, § 612 ; of a rt. 
 circ. cone, § 615 ; of a frustum 
 of a rt. circ. cone, § 620 ; of a 
 cylinder, defined, § 597 ; of a 
 cylinder, § 601 ; of any parallele- 
 piped, § 651 ; of any prism, § 553 ; 
 of a solid, § 633; of a sphere, 
 
 Zone, § 666. 
 
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