UC-NRLF 
 
 SB E7fl 
 
 
IN MEMORIAM 
 FLORIAN CAJORI 
 
-y1 
 
PLANE GEOMETRY 
 
 BY 
 
 HERBERT E. HAWKES, PH.D. 
 
 PROFESSOR OF MATHEMATICS IN COLUMBIA UNIVERSITY 
 
 WILLIAM A. LUBY, M.A. 
 
 HEAD OF THE DEPARTMENT OF MATHEMATICS IN THE 
 JUNIOR COLLEGE OF KANSAS CITY 
 
 AND 
 
 FRANK C. TOUTON, PH.D. 
 
 SUPERVISOR OF HIGH SCHOOLS, STATE DEPARTMENT OF 
 PUBLIC INSTRUCTION, MADISON, WISCONSIN 
 
 GINN AND COMPANY 
 
 BOSTON NEW YORK CHICAGO LONDON 
 ATLANTA DALLAS COLUMBUS - SAN FRANCISCO 
 
COPYRIGHT, 1920, BY 
 
 HERBERT E. HAWKES, WILLIAM A. LUBY 
 AND FRANK C. TOUTON 
 
 A.LL RIGHTS RESERVED 
 321.3 
 
 CAJOR1 
 
 gfte fltftcnatum 
 
 GINN AND COMPANY PRO- 
 PRIETORS BOSTON ' U.S.A. 
 
PREFACE 
 
 Although the study of geometry is important from an 
 informational point of view, it is generally recognized that a 
 genuine mastery of the subject means real achievement in 
 the solution of original exercises. The chief aim of the authors 
 in preparing this text has been to give such assistance to 
 students as will stimulate insight and develop power to solve 
 exercises of gradually increasing difficulty. 
 
 The content and organization of the first book of geometry 
 are determining factors in the student's progress in the sub- 
 ject. The first ten or twenty theorems determine whether a 
 student will grasp quickly the general trend of the subject 
 or remain bewildered for an indefinite period. The simplicity 
 and directness of the early theorems and the order in which 
 they are presented are elements of the highest importance 
 in an effective introduction. These elements were given the 
 most careful consideration in the preparation of this text. If 
 the first theorem is used in proving the second and the sec- 
 ond in proving the third, and so on, the student will soon 
 see a reason for mastering the content of each theorem. But 
 if the initial theorem is used for the first time in the fifteenth 
 theorem, and the second is used next in the eighth, the first 
 month's work will give the impression that geometry deals 
 with unrelated facts which lead nowhere. Such an arrange- 
 ment of the theorems of geometry will hamper even the 
 strongest students, and will make progress almost impossible 
 for the less capable. 
 
iv PLANE GEOMETRY 
 
 In this text no time is wasted on side issues or on dis- 
 cussive explanations, but the student is brought by a direct 
 route to the theorems on angle sums of polygons, the theo=- 
 rems on parallel lines, and those on parallelograms. Before 
 reaching Theorem 13 he meets theorems on angle sums, the 
 one topic in Book I which has varied numerical applications. 
 Exercises based on angle sums are presented on page 35, at 
 which point the student begins his really independent work 
 upon numerical exercises. Having acquired some ability in 
 solving numerical exercises he is prepared to begin the de- 
 vising of general proofs involving congruent triangles. 
 
 The most important single method of elementary geom- 
 etry is the use of congruent triangles. The best experience 
 of teachers has shown that practice with this method should 
 begin early and last long. Superposition is used here only 
 when unavoidable, and the more difficult topics, such as in- 
 equalities and indirect methods, are deferred until near the 
 end of the book. 
 
 Attention is invited to certain general features of the 
 text. For example, axioms, postulates, and definitions are 
 given when needed, not before. The usual group of theo- 
 rems on proportion are distributed, each one being presented 
 at the proper point. An important feature is the placing of 
 a few exercises involving suitable numerical and general 
 applications of a theorem immediately after it. Many exer- 
 cises are presented which are designed to correlate geometry 
 with algebra, and to compel the student to use his knowl- 
 edge of fractions, equations, and radicals. A large number of 
 pertinent and stimulating queries are given which demand 
 only direct answers. Occasionally queries are used to develop 
 a specific topic such as loci, but usually they are designed 
 to broaden and perfect the student's knowledge of details 
 
PREFACE V 
 
 supplementary to the theorems and exercises. The parallel- 
 column arrangement of the demonstrations has been chosen 
 in the belief that it is an aid to clear thinking, and that it 
 places unusual and proper emphasis on the necessity of stat- 
 ing a reason for each assertion. Frequently, to encourage inde- 
 pendent thinking, reasons have been omitted and " Why ? " 
 has been inserted. This is done sparingly at first, but later 
 on with an increasing frequency graduated to the student's 
 growing knowledge of the subject. 
 
 Certain special features have received careful treatment. 
 Emphasis on right triangles having one angle 30, 45, or 60> 
 which begins with Theorem 24 of Book I, is continued over 
 into other books, especially Book III. This is justified by 
 the experience of every teacher of trigonometry and by the 
 specific recommendation of the Society for the Promotion of 
 Engineering Education. In fact, no theorem has been omitted 
 which any important educational body considers necessary. 
 In Book II particular attention is called to the emphasis 
 placed on constructions and loci and their interrelations. An 
 appreciation of these relations makes it clear why real con- 
 structional work should not come earlier in a course on 
 demonstrative geometry. If it does, the results must be 
 assumed without proof. It seems far more desirable to 
 assume, in the few cases needed, the existence of certain lines 
 and figures, and to defer the actual construction of them 
 until it is possible to prove that the method used is correct. 
 Full advantage has been taken of the unusual opportunity 
 offered by Book V for useful and necessary constructions. 
 The presentation of the matter of that book is one of the 
 unique features of the text. The graphical methods of solu- 
 tion given on pages 218 and 219 satisfy the most rigorous 
 criteria for practical problems. A further illustration of the 
 
vi PLANE GEOMETRY 
 
 importance of graphical methods is the work of page 281 on 
 areas. Attention is invited to Theorem 2, Book III, a simple 
 proof of which is given not involving the use of the theorem 
 of limits. In the other theorems usually proved by the 
 method of limits the meaning of the theorem has been 
 brought out by illustration and the student convinced of 
 its truth without any attempt at proof. This is in accord 
 with a widespread feeling that the method of limits is un- 
 profitable for the majority of American students when they 
 first study the plane geometry. The plan also avoids the 
 common error of designating as a proof a line of thought 
 which is not a proof. 
 
 One hundred and eighty additional exercises, grouped 
 according to the books on which they depend, will be found 
 at the end of the text. 
 
CONTENTS 
 
 PAGE 
 
 THE ORIGIN OF GEOMETRY . 1 
 
 BOOK I 3 
 
 BOOK II 89 
 
 BOOK III 155 
 
 BOOK IV . 214 
 
 BOOK V 249 
 
 SUPPLEMENTARY EXERCISES 286 
 
 Tii 
 
PLANE GEOMETRY 
 
 THE ORIGIN OF GEOMETRY 
 
 The elementary geometry of to-day is mainly due to the 
 genius of the Greeks. The foundation for their work, how- 
 ever, was obtained from the Babylonians and the Egyptians. 
 
 The Babylonian knowledge of geometry was developed, 
 in part at least, through the necessity of constructing figures 
 having religious significance. These figures involved triangles, 
 quadrilaterals, circles, and inscribed regular hexagons. Their 
 value 3 for IT was much less accurate than 3.1604, which was 
 used by the Egyptians probably as early as 3000 B.C. 
 
 The Egyptians were great builders. We read that Menes, 
 their first king, who built the temple of Ptah at Memphis, 
 also constructed a great reservoir and even changed the 
 course of the Nile. The ruins of many temples, the Sphinx, 
 and the great pyramids of Gizeh all attest considerable insight 
 into geometric relations. The geometry they developed was 
 applied to the calculation of the contents of granaries, to the 
 laying out of right angles, to obtaining north-and-south lines 
 for their temples and pyramids, and to carrying out land 
 surveys. Herodotus, the Greek historian, who traveled in 
 Egypt, says that the overflow of the Nile necessitated an 
 annual resurvey of the land along its banks for the double 
 purpose of determining ownership and of justly levying taxes. 
 The word geometry itself is of Greek origin and signifies 
 " earth measurement." 
 
 1 
 
2 PLANE GEOMETRY 
 
 Our information concerning the Egyptians goes back with 
 certainty to 1700 B.C. and with great probability to 3000 B.C. 
 We know that two thousand, perhaps three thousand, years 
 before our era they had formulas a few of which were incor- 
 rect for the areas of triangles, rectangles, parallelograms, and 
 trapezoids, and fairly accurate knowledge of the area of a 
 circle. They knew also that a triangle is a right triangle if its 
 sides are 3, 4, and 5 units respectively. Such geometry as they 
 had grew out of their practical needs and was embodied in 
 working rules. It was very useful, but vastly inferior to the 
 scientific geometry afterwards developed by the Greeks. In 
 judging the intellectual attainments of any people, however, 
 we must remember that first advances in any science are the 
 most difficult. Other peoples had the same practical needs as 
 the Egyptians but developed no geometry to meet them. 
 The Aztecs of Mexico, for example, were skillful artificers 
 in silver and gold, and they transported huge blocks of 
 stone from distant quarries to build their splendid temples, 
 but their imagination never led them to construct a vehicle 
 on wheels, one of the simplest applications of a circle, but 
 undoubtedly the most useful, which has ever been made. 
 
BOOK I 
 
 1. Introduction. Geometry is a science which treats of the 
 properties and measurements of flat figures, such as angles, 
 triangles, and circles, and of solid figures, such as cubes, cones, 
 and pyramids. The subject is not made up of entirely new 
 matter, for every student has had some experience with geo- 
 metric facts in his previous study. It will be found that geom- 
 etry uses arithmetic and algebra freely, though it is not 
 primarily concerned either with numbers or with the equation. 
 
 2. Fundamental ideas. Before taking up the part of geom- 
 etry which is entirely new, we must make more precise 
 our knowledge of some facts and relations already partially 
 known. Certain familiar terms must be carefully defined, cer- 
 tain ideas illustrated, and certain fundamental truths examined 
 and accepted as a basis for further work. 
 
 3. Definition. A definition is a statement explaining the 
 meaning of a word or phrase. 
 
 Every definition should be clear, complete, and in terms 
 more simple and familiar than the one defined. The essen- 
 tial qualities of the term defined should be made so clear that 
 whatever has these qualities, or does not have them, can 
 easily be recognized. 
 
 Not every term can be defined. The word space,, for ex- 
 ample, denotes a concept extremely hard to put in words. 
 Terms which are difficult to define accurately can usually be 
 made clear by description or by illustration or by both. This 
 method is necessary with the term straight line. Then there 
 will, of course, be various terms, such as length, breadth, and 
 
 3 
 
4 
 
 PLANE GEOMETRY 
 
 thickness, which need not be denned, as the ideas they convey 
 are already understood with sufficient clearness. 
 
 4. Solids. A board, a piece of chalk, or an iron bar is 
 an example of a physical solid. If a board be moved from 
 one place to another, the space it has occupied at any 
 instant is an example of a geometric solid. A physical solid 
 can be seen and touched; a geometric solid cannot it is 
 purely a mental concept. The science of physics is a study of 
 the nature and properties of the material of which physical 
 solids are constituted, while geometry is a study of the size 
 and shape of solids but not of the matter composing them. 
 
 5. Dimensions. The smooth block ABCD extends in three 
 principal directions ; that is, it has three dimensions, length 
 AB, breadth (or width) CD, and thickness (or height, depth, 
 or altitude) BC. In like manner 
 
 every solid has three dimensions. 
 
 6. Surfaces. Each of the flat 
 faces bounding ABCD is called a 
 surface. The boundaries of a geo- 
 metric solid are also called surfaces. A better illustration of a 
 geometric surface is one's shadow on a smooth sidewalk. 
 
 A surface has two dimensions, length and breadth. Even the 
 thinnest piece of paper is not a surface ; it is a solid, having length 
 and breadth and a small but measurable thickness. t 
 
 7. Lines. Any two surfaces of ABCD meet in an edge. In 
 the geometric solid the edges are called lines. We may think 
 of the intersection of any two surfaces as a line. 
 
 A geometric line has one dimension only, length. Of course any 
 line we may draw or engrave has breadth. Since over 20,000 par- 
 allel lines have been cut by a diamond point on a steel plate one 
 inch long, the breadth of such lines is very minute, yet they give 
 only an approximate idea of what is meant by a geometric line. 
 
BOOK I 5 
 
 8. Points. Any three edges of ABCD which meet form what 
 is called a corner. In the geometric solid this is called a- point. 
 We may think of the intersection of any. two lines as a point. 
 
 Any small dot which we can make on paper, such as the period at 
 the end of this sentence, has a measurable size. Such a point, how- 
 ever small, is only a crude representation of a geometric point. 
 
 9. Magnitudes. A magnitude is anything which is measur- 
 able or which may be thought of as measurable. 
 
 Anything of which we may ask, How much ? is a magnitude. 
 Lines, surfaces, and solids are examples of magnitudes. 
 
 10. The geometric magnitudes generated by motion. If a 
 point moves, it generates (describes or traces) a line ; if a 
 line moves (except along itself), it generates a surface ; if 
 a surface, such as a triangle or a square, moves (except along 
 itself), it generates a solid. 
 
 11. Straight line. If a weight is suspended by a fine 
 piano wire, the position taken by the wire is a good physical 
 representation of a straight line. 
 
 When the word line is used alone, usually a straight line 
 is meant. 
 
 A straight line is considered as extending indefinitely in both 
 directions. 
 
 A line beginning at a definite point and extending indefinitely 
 in one direction only is a ray. 
 
 A part of a line lying between any two of its points is called a 
 line-segment. Thus AB is a seg- 
 ment of KR. % 4 $ 
 
 12. Plane. A plane is a surface such that a straight line 
 joining any two of its points lies wholly within the surface. 
 
 In finishing up concrete walks a workman often lays a straight- 
 edge across the surface in several directions to test its flatness ; 
 that is, to determine if it is a plane. 
 
6 PLANE GEOMETRY 
 
 13. Geometric figures. A point, a line, a surface, a solid, 
 or any combination of them is called a geometrio t figure. 
 
 A plane figure is any geometric figure which lies in a plane 
 (flat) surface. 
 
 A triangle and a square are examples of 
 plane figures. 
 
 Plane geometry is concerned with the 
 study of plane figures only. 
 
 A rectilinear figure is a figure bounded by straight lines. 
 ABODE is a rectilinear figure. 
 
 14. Boundaries. The boundaries of a geometric figure are 
 the points, lines, curves, or surfaces which limit it. 
 
 Thus the boundaries of a line-segment are its two end points, 
 the boundaries of a surface are lines, and the boundaries of a 
 solid are surfaces. 
 
 15. Angle. A plane angle (symbol Z) is the figure formed 
 by two rays which meet. 
 
 The two rays are called the sides of the angle and the 
 point at which they meet is called the vertex. 
 
 Thus AB and CB are the sides of the angle 
 ABC and B is the vertex. B^ A 
 
 In naming an angle, the vertex letter is named 
 between the other two, thus : the angle ABC or the angle CBA . 
 It is also proper to name an angle by the vertex letter alone if 
 no other angle has the same vertex. Thus we may call angle CBA 
 angle B. In the adjacent figure, to speak of 
 the angle K would be indefinite, as any one of 
 three angles has the vertex K. Whenever two 
 or more angles have the same vertex we must 
 name each angle by three letters, or we can 
 write a small letter or number within the angle 
 near its vertex and name each angle by one character. Thus 
 angle RKH may be called angle 1 and angle HKL may be called 
 angle 2. 
 
BOOK I 7 
 
 16. Triangle. A triangle (symbol A) is a portion of a plane 
 bounded by three straight lines. 
 
 Every triangle is regarded as having six parts : three sides and 
 three angles. 
 
 17. Axiom. An axiom is a general statement which is 
 accepted as true without proof. 
 
 Axioms are truths so simple that we either cannot prove them 
 or do not care to do so. An example of a mathematical axiom is : 
 If the same number be added to each member of an equation, the 
 result is an equation. 
 
 18. Postulate. A postulate is a geometric statement which 
 is accepted as true without proof. 
 
 19. Postulate I. There is only one straight line through two 
 points. 
 
 20. Postulate II. Any geometric figure may be moved from 
 one place to another without changing its size or shape. 
 
 The postulates I and II are needed at once in Theorem I. 
 Other postulates and axioms will be stated as the need for 
 them arises. 
 
 21. Theorem. A theorem is a statement of a truth which 
 is to be proved. 
 
 An example of a theorem is : The square of the sum of two 
 numbers is the square of the first plus twice their product plus 
 the square of the second. 
 
 22. Proof. A proof, or a demonstration, is an argument or 
 an orderly arrangement of statements with reasons by which 
 we show the truth of an assertion. 
 
 In the course of a geometrical proof we proceed from inference 
 to inference, and we may appeal to any or all of the follow- 
 ing : to the definition of a term previously agreed on ; to an 
 axiom or a postulate already stated ; to a theorem already proved. 
 
8 PLANE GEOMETRY 
 
 Sometimes we appeal directly or indirectly to common sense, 
 to obvious facts, or to ideas or definitions which are so 
 universally accepted that a detailed statement. of them is not 
 required. 
 
 23. Equality of two geometric magnitudes. If two magni- 
 tudes of the same kind, like two angles or two distances, 
 are measured and their measures are expressed in terms of 
 a common unit by the same number, the two magnitudes 
 are equal. 
 
 In geometry, however, it is often desirable to prove two 
 magnitudes of the same kind (two angles, or two triangles, 
 or two other plane figures) equal without measuring them. 
 One method of doing this consists in proving that the 
 boundaries of the two magnitudes may be made to coincide. 
 
 Thus the line-segment AB is equal ^ ^ jt ^ 
 
 to the line-segment KR provided that 
 
 when A is placed on K, B at the same time can be placed on /?. 
 
 Similarly, the angle ABC is equal to the angle KRL if B oan 
 be placed on R and BA upon 
 RK, and if at the same time 
 BC can be placed upon RL. 
 
 Note that the size of an &</. r> **/ t 
 
 ^J \*f x\ ~~~ i^ 
 
 angle depends on the size of 
 
 the opening between its sides (see 15) and not on their 
 
 length. In showing that angle B equals angle R, BA, which 
 
 is represented as shorter than 
 
 RK, falls upon RK but does not 
 
 coincide with it. 
 
 Lastly, the triangle ABC is A B K R 
 
 equal to the triangle KRL if 
 
 when A is placed on K, B at the same time may be placed on R 
 and C on L. For then AB would coincide with KR, BC with 
 HL, and CA with LA"; that is, the boundaries of the triangle 
 ABC would coincide with the boundaries of the triangle KRL. 
 (See Postulate I.) 
 
BOOK I 9 
 
 24. Congruence. Two geometric magnitudes are congruent 
 if their boundaries can be made to coincide. 
 
 Congruent figures, therefore, are always equal, but equal 
 figures are not necessarily congruent. 
 
 Magnitudes having only one dimension, such as line-segments, 
 are congruent if they are equal. With all such magnitudes equal 
 and congruent mean the same thing. Magnitudes having two 
 or more dimensions may be equal without being congruent. 
 
 Thus the area of the triangle ABC (that is, the number of square 
 units in the surface) is equal to the area of the triangle KRL, but 
 the two triangles are not congruent. ^ 
 
 A triangle and a square, each 
 containing forty square inches, are 
 equal, but they are not congruent. 
 
 If two figures are equal they 
 
 have the same size but not necessarily the same shape, while 
 if two figures are congruent they have the same size and also 
 the same shape. 
 
 We shall now take up the first theorem and by reasoning 
 about it prove its truth. Other theorems depending on the 
 first will follow, and by their aid we shall prove still others. 
 Truth after truth will thus be demonstrated, all forming a 
 closely connected whole. Some of these truths without their 
 proof the student has met and used before. It may be of 
 assistance to mention a few of the theorems which are proved 
 in geometry. We shall prove that the sum of the angles 
 of any triangle is 180 ; that the square on the hypotenuse 
 of a right triangle is equal to the sum of the squares on the 
 other two sides ; that the area of any circle is equal to the 
 square of the radius multiplied by 3.1416 ; that the volume 
 of any pyramid is one third the area of the base times the 
 altitude ; and that the volume of any sphere is four thirds 
 of the product of the cube of the radius and 3.1416. 
 
10 
 
 PLANE GEOMETEY 
 Theorem 1 
 
 25. // two sides and the included angle of vne triangle 
 are equal respectively to two sides and the included angle 
 of another, the two triangles are congruent. 
 
 Given the triangles ABC and FGH, in which AB equals FG, AC 
 equals FH, and the angle A equals the angle F. 
 
 To prove that A ABC is congruent to AFGH. 
 
 Proof 
 
 STATEMENT 
 
 REASON 
 
 1. Place A ABC upon AFGH in 
 such a way that AB coincides 
 with its equal FG, and so that 
 point A falls on F and point B 
 on G. 
 
 2. AC will fall on FH. 
 
 3. Point C will fall on point H. 
 
 4. Therefore, since B falls on G 
 and (7'on H, EC will coincide 
 with GH. 
 
 5. AABC is congruent to A FGH. 
 
 1. 20. Postulate II. Any geo- 
 metric figure may be moved 
 from one place to another 
 without changing its size or 
 shape. 
 
 2. Z-A is given equal to /.F. 
 
 3. AC is given equal to FH. 
 
 4. 19. Postulate I. There is 
 only one straight line through 
 two points. 
 
 5. 24. The boundaries of the 
 triangles coincide. 
 
 26. Corresponding parts. If two triangles have the three 
 angles of one equal respectively to the three angles of the 
 other, any two equal angles (one chosen from each triangle) 
 are called corresponding or homologous angles, and any two 
 
BOOK I 11 
 
 sides which lie opposite equal angles are called corresponding 
 or homologous sides. 
 
 Thus, if Z.4 = Z F, Z = ZA, and ZC = ZZ, then Z.I and 
 ZF, Z and ZA, and ZC and ZZ are corresponding angles. 
 Further, BC and AZ, AC C Z 
 
 and F2T, yJ/J and FA, are 
 corresponding sides. 
 
 Also, if any two rec- 
 tilinear figures are con- 
 
 gruent, any angle or side of one and the equal and similarly 
 placed angle or side of the other are called corresponding parts. 
 
 27. Parts of congruent figures. From 26 it follows also that 
 Corresponding parts of congruent figures are equal. 
 
 It should be noted that in plane geometry the word part 
 refers either to lines or angles. 
 
 EXERCISES 
 
 1. Suppose A ABC congruent to AFGH. Name the other 
 corresponding parts if (&) /-A = Z.F and Z5 = ZG J ; 
 (b) Z.A = Z F, A C = FH, and AB = FG ; (c) AB = FG, 
 Z.4=Z J P,andZJ5=ZG J ; (d)AB=FG 
 
 and A C = FH. 
 
 2. In answering (a) to (d) the stu- 
 dent may have been guided by his 
 eye and not by a real understanding 
 of the definition of 26. To make it 
 
 emphatic that corresponding parts are similarly situated parts, 
 we shall now assume that A ABC and FGH are congruent, and 
 shall make some suppositions which the shape of the triangles 
 really contradicts. 
 
 Name two corresponding parts if (a) Z.A =ZFand ZC = . 
 ()ZB = Z#andZC = ZG Y ; (c) AC = FG and AB = FH. 
 
 QUERY 1. If two line-segments are equal, are they congruent? 
 QUERY 2. If two angles are equal, are they congruent ? 
 
12 PLANE GEOMETRY 
 
 28. Bisection. To bisect a magnitude means to divide it 
 into two equal parts. ^ 
 
 Thus, if AK equals KB, the point K bisects A 
 the line-segment AB. ^ 
 
 Also, if Z1 = Z2, the line BK bisects 
 the /.ABC. 
 
 For the present we assume as evident that 
 any line-segment or angle can be bisected. 
 Later it will be shown how to perform these constructions. 
 
 Theorem 2 
 
 29. If two sides of a triangle are equal, the angles 
 opposite them are equal. 
 
 K 
 
 Given the triangle ABC, in which AC equals EC. 
 To prove that 
 
 Proof 
 STATEMENT REASON 
 
 1. Let CK bisect Z.ACB and 1. 28. Any angle may be bi- 
 ineet AB at some point K. sected. 
 
 2. Then in AACK and BCK, 2. Construction. 
 
 Z1=Z2. 
 
 3. AC = BC. 3. Given. 
 
 4. CK = Ctf. 4. Identical, 
 
BOOK I IB 
 
 5. A A C K is congruent to A BCK. 5. 25. If two sides and the 
 
 included angle of one triangle 
 are equal respectively to two 
 sides and the included angle 
 of another, the two triangles 
 are congruent. 
 
 6. Z.I = Z7?. 6. 27. Corresponding parts of 
 
 ' congruent figures are equal. 
 
 NOTE. It should be observed as an implicit assumption of plane 
 geometry that any lines or points added to a figure are in the same 
 plane as the figure. 
 
 30. Isosceles triangle. An isos- 
 celes triangle is a triangle which 
 has two equal sides. 
 
 EXERCISES 
 
 3. State Theorem 2, using the word isosceles. 
 
 4. If the three sides of a triangle are equal, its three angles 
 are equal. 
 
 HINTS. Prove by using Theorem 2 twice. 
 
 Write down the several steps of the proof as in the proof of 
 Theorems 1 and 2 above. 
 
 31. Axiom I. If equals are added to equals, the results are 
 equal. 
 
 32. Axiom II. (1) Two numbers or magnitudes each equal 
 to a third are equal to each other. 
 
 This axiom is often needed in the following form : 
 
 (2) Two figures congruent to a third are congruent to each other. 
 
 One should quote the form (1) or (2) of Axiom II as the 
 case requires. 
 
14 
 
 PLANE GEOMETRY 
 
 Theorem 3 
 
 33. If the three sides of one triangle are equal respec- 
 tively to the three sides of another, the triangles are 
 congruent. 
 
 C H 
 
 Given the triangles ABC and FGH in which AB equals FG, 
 BC equals GH, and AC equals FH. 
 
 To prove that A ABC is congruent to A FGH. 
 
 Proof 
 
 Let AB and FG be the longest sides of the triangles respectively. 
 
 1. Suppose A FGH is placed ad- 
 jacent to AABC so that FG 
 coincides with its equal AB, 
 the point F falling on the 
 point A, the point G on the 
 point B, . and the point // 
 at K. 
 
 2. Draw CK forming angles 1 
 and 2 at C and 3 and 4 at K. 
 
 3. 
 
 AC=AK. 
 
 4. Therefore Z1= 
 
 5. Also 
 
 6. And 
 
 BC = BK. 
 Z2=Z4. 
 
 1. 20. Postulate II. Any figure 
 may be moved from one place 
 to another without changing 
 its size or shape. 
 
 2. 19. Postulate I. There is 
 only one straight line through 
 two points. 
 
 3. Given. 
 
 4. 29. If two sides of a tri- 
 angle are equal, the angles 
 opposite them are equal. 
 
 5. Given. 
 
 6. 29. 
 
BOOK I 15 
 
 7. Z1 + Z2 =Z3 + Z4 7. 31. Axiom I. If equals are 
 or Z.ACB/.AKB. added to equals, the results 
 
 are equal. 
 
 8. A ABC is congruent to 8. 25. If two sides and the 
 AABK. included angle of one tri- 
 angle are equal respectively 
 to two sides and the included 
 angle of another, the two tri- 
 angles are congruent. 
 
 9. Then A ABC is congruent to 9. 32. Axiom II. Two figures 
 AFGH. congruent to a third are con- 
 gruent to each other. 
 
 QUERY 1. Does the proof of Theorem 3 depend on Theorem 2 ? on 
 Theorem 1 ? If the proof of a theorem depends on Theorem 2, does it 
 depend on Theorem 1 ? 
 
 QUERY 2. If three angles of one triangle are equal respectively to 
 three angles of another, are the triangles congruent ? Illustrate. 
 
 QUERY 3. Draw a figure for Theorem 3 assuming AB and FG the 
 shortest sides respectively of the triangles. Would any difficulty then 
 arise in the proof ? 
 
 EXEKCISES 
 
 5. In a triangle ABC line CK is drawn from C to the middle 
 point of AB. If AC and BC are the same length, prove that 
 AACK is congruent to ABCK. 
 
 HINT. Show that Theorem 3 applies. 
 
 6. In a certain rectilinear four-sided figure the opposite sides 
 are equal. Prove that a line joining two opposite vertices forms 
 two equal triangles. . Q, 
 
 HINT. Use 33. 
 
 7. ABCDE is a rectilinear figure having. equal 
 sides and equal angles. Prove that AD equals A C. 
 
 HINT. Apply Theorem 1, then 27. 
 
 QUERY 1. Will three bars held together by one bolt at A, B, and C 
 respectively form a rigid figure ? 
 
16 
 
 PLANE GEOMETRY 
 
 QUERY 2. Will a quadrilateral formed by four bars and held to- 
 gether by one bolt at A, B, C, and D respectively form a rigid figure? 
 How can it be made rigid ? 
 
 QUERY 3. Will the gate (of the above figure) sag from its own, 
 weight ? If so, how can sagging be prevented ? 
 
 34. Adjacent angles. Two angles which have the same vertex 
 and a common side between them are called adjacent angles. 
 
 Thus angle ABK and angle KBC are adja- 
 cent angles. 
 
 35. Straight angle. A straight angle is 
 an angle whose sides extend in opposite 
 directions and form a straight line. 
 
 In the adjacent figures, if A KB is a straight 
 line, Zl 4- Z 2 = a straight angle and Z3 -f 
 Z 4 4- Z 5 = a straight angle. 
 
 36. Postulate III. All straight angles are 
 equal. 
 
 It follows that if two lines cross so that 
 any two of the adjacent angles are equal, all 
 four angles formed are equal, because each 
 is half a straight angle. Thus, if in the figure of 37 AC 
 and KR cross so that Z 1 = Z 3, then Z1 = Z2=Z3=Z CBR. 
 
 37. Perpendicular. If one straight line cuts another so as 
 to make any two adjacent angles equal, each K 
 
 line is perpendicular (symbol J_) to the other. 
 
 Thus, if ABC and KBR are straight lines, and 
 angle ABK equals angle KBC, each of the lines 
 A C and KR (or any portion of either) is perpendicular to the other, 
 
 1 2 
 
 A 3 
 
 B 
 R 
 
 C 
 
BOOK I 17 
 
 If one straight line cuts another so that there are no two 
 adjacent angles which are equal to each other, then the lines 
 would not be called perpendicular. 
 
 A corresponding remark holds regarding most of the definitions 
 in this and other texts. 
 
 38. Right angle. If one line is perpendicular to another, 
 the four angles between them are called right angles. 
 
 Thus, in the adjacent figures, if 
 KIl is _L to .1C, AC to BR, and EC 
 to BK respectively, the angles 1, 2, 
 3, and 4 are right angles. 
 
 39. Axiom III. If equals are 
 divided by the same number, the 
 results are equal. 
 
 40. Straight angle and right angle. From the definition of a 
 right angle it follows that a right angle is half a straight 
 angle, or a straight angle is twice as great as a right angle. 
 
 Hence the sum of the angles around a fixed point is four 
 right angles. 
 
 From 36 and 39 it immediately follows that 
 All right angles are equal. 
 
 41. In the figure of 37, if BK is perpendicular to AC, no 
 other line in the plane of the paper can be drawn through B 
 perpendicular to AC. This gives' 
 
 Postulate IV. At a given point of a line, one and only 
 one perpendicular can be drawn to the line. 
 
 Thus, if KR is perpendicular to AB, it bisects 
 the straight angle A KB. Obviously a second 
 J_ KL is impossible, because an angle cannot be 
 bisected by two different lines. 
 
 QUERY 1. In the figure of 34, do the angles .4 AT and ABC have the 
 same vertex and a common side ? Are they adjacent angles ? 
 
18 
 
 PLANE GEOMETKY 
 
 QUERY 2. In the figure of 37, if KB cuts AC, how many right 
 angles are formed? 
 
 QUERY 3. Can two angles have a common side and not be adjacent 
 angles ? Illustrate with a figure. 
 
 QUERY 4. Can two angles have a common vertex and not be adjacent 
 angles? Illustrate. 
 
 QUERY 5. If two perpendicular lines intersect, how many right angles 
 are formed ? 
 
 QUERY 6. If one line is perpendicular to a 
 second, is the second perpendicular to the first? 
 
 QUERY 7. If LKR is not a straight line, can the 
 sum of the angles 1 and 2 equal two right angles ? 
 If AKR is a straight line, what is the sum of Zl + Z 2 ? 
 
 Theorem 4 
 
 42. There is only one perpendicular from a point to 
 a line. 
 
 A 
 
 RA 3 
 
 P 
 
 K 
 
 B 
 
 N2 4 
 \ ! 
 
 Given the point P any point outside the line AB, PK a perpen- 
 dicular to AB at K, and PR any other line from P terminating 
 in AB. 
 
 To prove that PR is not _L to AB. 
 
 Proof 
 
 1. Let PK be extended to 1. 11. A straight line ex- 
 L, making KL =PK, and tends indefinitely. 
 
 draw LR. 
 
 2. Now in APKR and LKR 2. 40. All right angles are 
 
 Z3 = Z4. equal. 
 
 3. PK = LK. 3. Construction. 
 
BOOK I 
 
 19 
 
 KR=KR. 
 
 5. APKR is congruent to 
 &LKR. 
 
 6. 
 
 Z 1 = Z 2. 
 
 4. Identical. 
 
 5. 25. If two sides and the 
 included angle of one tri- 
 angle are equal respectively 
 to two sides and the in- 
 cluded angle of another, the 
 two triangles are congruent. 
 
 6. 27. Corresponding parts of 
 congruent figures are equal. 
 
 7. 19, Postulate I. There 
 is only one straight line 
 through two points. 
 
 8. 35. A straight angle is an 
 angle whose sides extend in 
 opposite directions and form 
 a straight line. 
 
 9. 40. A right angle is half 
 a straight angle. 
 
 10. Step 9. 38. If one line is 
 perpendicular to another, 
 "~the - four angles between 
 them are called right angles. 
 
 NOTE. If A, B, and C ai*e three theorems, and A is quoted to prove B 
 but B alone is quoted to prove C, then C depends directly upon B and 
 indirectly upon A. 
 
 QUERY. Does the proof of Theorem 4 depend directly or indirectly 
 on Theorem 3 ? on Theorem 2 ? on Theorem 1 ? 
 
 7. Since PKL is a straight line, 
 PRL is not a straight line. 
 
 8. Z 1 + Z 2, or Z.PRL, is not 
 a straight angle. 
 
 9. Zl, or half of Z.PRL, is 
 
 not a right angle. 
 10. PR is not _L to AB. 
 
 EXERCISES 
 
 8. Prove that two of the three angles of a triangle cannot 
 both be right angles. 
 
 NOTE. When unable to prove some property of a figure, one is 
 tempted to say, " Why, you can see that it is true." The following illus- 
 trations, however, will show that the eye may easily be deceived, and 
 that an appeal to the eye is not a proof. 
 
20 PLANE GEOMETRY 
 
 9. Which appears to be the longer in the adjacent figure, 
 a or b ? Measure each. Which is the >. / 
 
 longer ? /^ 
 
 10. Draw freehand a perpendicular <J^ ^ 
 
 to AB at its mid-point and equal to 
 
 AB. Measure the two lines. Are they // 
 
 equal ? 
 
 11. Which lines of the adjacent figure 
 are prolongations of the others ? Verify 
 your answer. 
 
 43. Parallel lines. Parallel lines are lines that lie in the 
 same plane and do not meet however far they are produced. 
 
 Theorem 5 
 
 44. Two lines perpendicular to the same line are parallel. 
 
 A 
 
 R 
 
 M 
 
 B 
 
 Given the lines KR and LMea.ch perpendicular to AB at K and 
 L respectively. 
 
 To prove that KR is II to LM. 
 
 Proof 
 
 1. Since KR and LM lie in the 1. No other possibility exists. 
 same plane, they either meet 
 
 or they do not meet. 
 
 2. If they meet, as at //, there 2. Hypothesis, 
 will be two perpendiculars 
 
 from that point to AB. 
 
BOOK I 21 
 
 3. But two perpendiculars from 3. 42. There is only one per- 
 the same point to a line are pendicular from a point to a 
 impossible. line. 
 
 4. Therefore KR cannot meet 4. Remaining alternative in 
 LM. statement 1. 
 
 5. Therefore KR is II to LM. 5. 43. 
 
 45. Postulate V. The postulate of parallels. Through a given 
 point outside a line, one line parallel to it exists, and only one. 
 
 Theorem 6 
 
 46. //' a line intersects one of two parallel lines, it 
 intersects the other also. 
 
 B 
 
 C 
 
 'R 
 
 Given the parallels AB and CD and the line KR cutting AB at 0. 
 To prove that KR cuts CD. 
 
 Proof 
 
 1. KR intersects CD or it does 1. No other possibility exists, 
 not intersect it. 
 
 2. If it does not, then KR and 2. 43. 
 CD are II. 
 
 3. Then AB and KR pass 3. 45. 
 through O and are II to CD. 
 
 But this is impossible. 
 
 4. Then if KR is not II to CD, 4. Remaining alternative in 
 it intersects it. statement 1. 
 
22 PLANE GEOMETRY 
 
 NOTE. The method of proof in which we suppose one geometric 
 figure placed upon another, as was done in Theorem 1, is called a proof 
 by superposition. \ 
 
 QUERY 1. Is any theorem from 2 to 6 proved by superposition? 
 
 QUERY 2. On what postulate must every proof by superposition 
 depend ? 
 
 QUERY 3. On what theorems from 1 to 4 does the proof of Theorem 5 
 directly depend? indirectly depend? 
 
 QUERY 4. On which of the five preceding theorems does the proof 
 of Theorem 6 directly depend? indirectly depend? 
 
 QUERY 5. Can three pencils be held so that two of them are per- 
 pendicular to the third at the same point? Illustrate. 
 
 QUERY 6. Can two pencils be held so that they are perpendicular 
 to the same line at different points, but are not parallel to each other ? 
 Illustrate. 
 
 NOTE. The answer to the preceding query illustrates the fact that 
 in the statements of the theorems of plane geometry it is implicitly 
 understood that all the points and lines referred to by the theorem 
 lie in one plane. This is important, for without this understanding 
 Theorem 5 as stated is not true. 
 
 Theorem 7 
 
 47. If a line is perpendicular to one of two parallel 
 lines, it is perpendicular to the other also. 
 
 M 
 
 Given the parallel lines AB and CD, with -XT perpendicular to 
 AB and intersecting it at L. 
 
 To prove that XY is _L to CD. 
 
BOOK I 
 
 Proof 
 
 1. XY intersects CD at some 
 point, as M. 
 
 2. CD is _L to XY at MOT it is 
 not J_ to XY. 
 
 3. Suppose CD is not _L to XY. 
 And let KR be drawn through 
 M _L to XY. 
 
 4. Then KR is II to AB. 
 
 5. Since KR and CD are II to 
 AB and both pass through 
 point M 9 they make but one 
 line, instead of being two dis- 
 tinct lines as indicated in the 
 figure. This means that KR, 
 when drawn through M _L to 
 XY, coincides with CD. 
 
 6. KR is _L to XY. 
 1. CD is _L to XY. 
 
 1. 46. If a line intersects one 
 of two parallel lines, it in- 
 tersects the other also. 
 
 2. No other possibility exists. 
 
 3. 41. 
 
 4. 44. Two lines perpendic- 
 ular to the same line are 
 parallel. 
 
 5. 45. Postulate V. Through 
 a given point outside a line, 
 one line parallel to it exists, 
 and only one. 
 
 6. Construction. 
 
 7. Since CD and KR are the 
 same line. 
 
 QUERY 1. On which of the preceding theorems does the proof of 
 Theorem 7 directly depend ? indirectly depend ? 
 
 QUERY 2. Can you hold three pencils so that two are parallel 
 and the third is perpendicular to one of them but not to the other? 
 
 QUERY 3. How many of the theorems 1 to 7 are true only 
 when the figure is supposed to lie in a plane ? Q 
 
 48. Right triangle. A right triangle is a triangle 
 in which one angle is a right angle. 
 
 49. Hypotenuse. The hypotenuse of a right tri- 
 angle is the side opposite the right angle. A 
 
24 
 
 PLANE GEOMETRY 
 
 Theorem 8 
 
 50. Two right triangles are congruent if the hypotenuse 
 and an adjacent angle of one are equal respectively to the 
 hypotenuse and an adjacent angle of the other. 
 
 C H 
 
 A B F ROM 
 
 Given the right triangles ABC and FGH in which the hypot- 
 enuse AC equals the hypotenuse FH and the angle A equals the 
 angle F. 
 
 To prove that A ABC is congruent to A FGH. 
 
 Proof 
 
 1. Place AABC upon. AFGH so 1. 20. 
 that the hypotenuse A C coin- 
 cides with its equal FH, and 
 
 so that the point A falls on 
 the point F and the point C 
 on the point II. 
 
 2. Now Z.4=ZK 2. 
 
 3. Hence AB will fall along FG. 3. 
 
 4. CB must fall in one of three 4. 
 positions : upon HG, to the left 
 
 as HR, or to the right as HM. 
 
 5. If CB falls in the position 5. 
 HR, then HR and HG would 
 form two _ls from // to FG, 
 which is impossible. 
 
 In like manner it can be proved 
 
 6. Therefore CB falls on HG. 6. 
 
 7. Therefore A ABC is con- 7. 
 gruent to A FGH. 
 
 Given. 
 
 Statements 1 and 2. 
 
 No other possibility exists. 
 
 42. There is only one per- 
 pendicular from a point to a 
 line. 
 
 that CB cannot fall on JI^f. 
 Remaining alternative in 4. 
 24. 
 
BOOK I 25 
 
 51. Axiom IV. Jf equah are subtracted from equals, the 
 results are equal. 
 
 52. Vertical angles. Two angles are vertical angles if the 
 sides of one are the prolongations of the sides of the other. 
 
 QUERY 1. In which of the first eight theorems did the proof depend 
 on no other one of the eight ? 
 
 QUERY 2. If two straight lines intersect, how many pairs of vertical 
 angles are formed? how many pairs of adjacent angles? 
 
 Theorem 9 
 
 53. If two straight lines intersect, the vertical angles 
 are equal. 
 
 D- 
 
 B C 
 
 Given the straight lines BA and DC intersecting at K. 
 To prove that Z 1 = Z 3 and Z 2 = Z BK C. 
 
 Proof 
 
 1. Z 1 + Z 2 = a straight angle. 1. 35. 
 
 2. Z 2 + Z 3 = a straight angle. 2. 35. 
 
 3. Z 1 + Z 2 = Z 2 -f Z 3. 3. 36. All straight angles are 
 
 equal. 
 
 4. Z 1 == Z 3. 4. 51. If equals are subtracted 
 
 from equals, the results are 
 equal. 
 
 5. Again Z 1 + Z 2 = a straight 5. Why ? 
 angle. 
 
 6. And Zl-f-Z BKC = a straight 6. Why ? 
 angle. 
 
 7. Hence Z 2 = Z BKC. 1. Why ? 
 
26 PLANE GEOMETKY 
 
 EXERCISE 12. Two straight lines KRL and GRH intersect so 
 that KR equals RH and GR equals RL. Draw KG and HL and 
 prove that the triangle KR G is congruent to the triangle HRL. 
 
 QUERY. In the figure of the preceding exercise, if the lines KH and 
 GL be drawn, are the triangles KRH and GRL congruent? Why? 
 
 54. Transversal. A transversal is a line that crosses (cuts 
 or intersects) two or more lines. 
 
 55. Angles made by a transversal. In the adjacent figure the 
 angles 1, 2, 7, and 8 are called exterior angles and angles 3, 4, 
 
 5, and 6 are called interior angles. 
 The two pairs of angles 3 and 
 
 6, 4 and 5, are called alternate- 
 interior angles ; and the two pairs 
 of angles 1 and 8, 2 and 7, are 
 called alternate-exterior angles. 
 
 The four pairs of angles 1 and 5, 
 2 and 6, 3 and 7, and 4 and 8 are called corresponding angles. 
 
 It should be noted that we now have applied the word corre- 
 sponding to certain angles of congruent figures and to certain 
 angles formed when a transversal cuts two other lines, which will 
 almost invariably be two parallel lines. Therefore, in using the 
 word corresponding we should always be careful to convey the 
 meaning . we intend, by saying either corresponding angles of 
 parallel lines or corresponding angles of congruent figures. 
 
 56. Degree. If a, right angle be divided into ninety equal 
 angles, each part is an angle of one degree. 
 
 EXERCISE 13. Draw a figure like that of 55, making Z 6 about 
 40 and AB parallel to CD. How many angles appear to contain 
 40? How many to contain 140? How many still another 
 number of degrees ? 
 
BOOK I 
 
 : 27 
 
 Theorem 10 
 
 57. If two parallel lines are cut by a transversal, the 
 alternate-interior angles are equal. 
 
 B 
 
 Given the two parallels AB and CD cut by the transversal 
 XY at H and K respectively, forming the pairs of alternate- 
 interior angles 1 and 4, 2 and 3. 
 
 To prove that ^1 = Z.4 and Z 2 = Z 3. 
 
 1. Let R be the mid-point of 
 HKj and let the line through 
 R which is _L to AB meet A B 
 in F and CD in G. Then F& 
 is _L to CD. 
 
 2. Z 5 = Z 6. 
 
 3.AKRG is congruent to 
 
 Proof 
 1. 
 
 4. 
 
 Z1 = Z4. 
 
 (1) 
 
 5. ButZl+ Z2 = Z3 + Z4. (2) 
 
 6. Z 2 = Z 3. 
 
 47. If a line is perpen- 
 dicular to one of two parallel 
 lines, it is perpendicular to 
 the other also. 
 
 2. 53. If two straight lines 
 intersect, the vertical angles 
 are equal. 
 
 3. 50. 
 
 4. 27. Corresponding parts of 
 congruent figures are equal. 
 
 5. 36. All straight angles are 
 equal. 
 
 6. (2)-(l), 51. If equals are 
 subtracted from equals, the 
 results are equal. 
 
28 PLANE GEOMETRY 
 
 58. Corollary. A corollary is a secondary, or a minor, theorem 
 dependent on a major theorem. Usually it is a theorem the 
 truth of which is apparent from the major theorem or which 
 follows from it with but little proof. 
 
 The theorems of SS 59 and 60 are corollaries to Theorem 10. 
 
 I . o o 
 
 The proqf of each is brief and depends on Theorem 10. 
 
 59. Corollary 1. If ttvo parallel lines are cut by a transversal, 
 the corresponding angles are equal. 
 
 K 
 
 A \2 B 
 
 5\6 
 
 C AS D 
 
 \R 
 
 Given the parallels AB and CD cut by the transversal KR, 
 forming four pairs of corresponding angles 1,5; 2, 6; 3, 7; 4, 8. 
 To prove that Z.l = ^5. 
 
 HINTS. Z1 = Z4. Why? 
 
 Z4=Z5. Why? 
 
 .-. Z1 = Z5. Why? 
 
 EXERCISES 
 
 14 . Prove in like manner (using the figure of 59) that Z 2 = Z 6. 
 
 15. Prove in like manner that Z3 =Z7. 
 
 16. Prove in like manner that Z 4 = Z 8. 
 
 17. If two parallel lines are cut by a transversal, the alternate- 
 exterior angles are equal. 
 
 Prove, using the figure of 59, that Zl =Z 8 and Z 2 =Z7. 
 HINTS. Z1=:Z4. Why? 
 
 Z4 = Z5. Why? 
 
 Z5=Z8. Why? 
 
 .-. Z1 = Z8. Why? 
 
 In like manner it can be proved that Z 2 = Z7. 
 
BOOK I 29 
 
 NOTE. In some of the proofs of the preceding theorems and corol- 
 laries the word " Why? " has been placed on the right of the page. Its 
 presence denotes the omission of a reason (see 22) at that point in the 
 demonstration. The purpose of the omission is to lead the student to 
 supply mentally the reason omitted and thus to do some of his own 
 thinking even in the proofs given in the text. He should in all such 
 cases pause long enough to decide in his own mind what the omitted 
 reason is. In oral or written demonstrations of these theorems and 
 corollaries all necessary reasons should be given in full, whether they 
 are given in the text or not. 
 
 60. Corollary 2. If two parallel lines are cut by a transversal, 
 the sum of the two interior angles on the same side of the trans- 
 versal equals two right angles. 
 
 Given the parallel lines AB and CD cut by the transversal KR. 
 To prove, using the figure of 59, that 
 
 Z.3+^5 = 2 rt. A, and Z4+Z6=2 rt A. 
 HINTS. Z3 + Z4 = 2 rt. A. (1) Why ? 
 
 Z4 = Z5. Why? 
 
 Substituting Z5 for Z4 in (1), Z3 + Z5 = 2 rt. /I. 
 In like manner it can be proved that 
 
 Z4 +Z6 = 2rt.zl. 
 
 61. Supplementary angles. One angle is the supplement of 
 another if their sum equals two right angles (or one hundred 
 and eighty degrees). 
 
 62. Corollary 3. If two angles have their sides parallel each 
 to each, they are equal or supplementary. 
 
 HINTS. What is the relation between Zl 3 
 
 and Z4? Z4 and Z2? What, then, is the 
 relation between Zl and Z2? What is the 
 
 relation between Z3 and Zl? What, then, is the relation between 
 Z3 and Z2? 
 
30 PLANE GEOMETRY 
 
 63. Converse. In the statement of every theorem there is a 
 conditional part (often expressed in an " if " clause) called 
 the hypothesis. The other part of the theorem ' is a conse- 
 quence of the hypothesis and is called the conclusion. In 
 any theorem, if the hypothesis and the conclusion be inter- 
 changed, the resulting theorem is called the converse of 
 the first. 
 
 For example, the converse of Theorem 1 is as follows : If 
 two triangles are congruent, two sides and the included angle of 
 one are equal respectively to two sides and the included angle 
 of the other. 
 
 EXERCISES 
 
 18. State the hypotheses of Theorems 1 to 10 inclusive. State 
 the conclusions. 
 
 19. State the converse of Theorems 2, 3, and 10 respectively. 
 
 20. State the converse of the corollaries in 59, 60, and 62 
 respectively. 
 
 It is important to note that the converse of a theorem is not 
 always true. 
 
 Thus : If a quadrilateral is a square, its diagonals intersect at 
 right angles. Conversely : If the diagonals of a quadrilateral 
 intersect at right angles, the quadrilateral 
 is a square. From the larger figure on the 
 right the first theorem appears to be true, 
 while the smaller quadrilateral shows that the 
 converse is false. Even outside of geometry 
 the converse of a theorem is not always true. 
 This will be apparent from stating the con- 
 verse of the following : If a man lives in 
 Chicago, he lives in Illinois. These illustrations make it clear 
 that though a theorem is true, its converse may not be true. Until 
 the converse of a theorem has been proved it must not be quoted 
 as a reason for any statement. 
 
BOOK I 
 
 31 
 
 Theorem 11 (Converse of Theorem 10) 
 
 64. If two straight lines are cut by a transversal, making 
 two alternate-interior angles equal, the lines are parallel. 
 
 B 
 
 D 
 
 R 
 
 Given AB and CD cut by the transversal .XT at L and # respec- 
 tively, making the alternate-interior angles 1 and LHD equal. 
 
 To prove that 
 
 AB is II to CD. 
 
 Proof 
 
 1. CD is II to AB OY it is not II 
 to AB. 
 
 2. Suppose CD is not II to AB. 
 Draw KR through H II to AB. 
 
 3. Then Z1 
 
 4. 
 
 5. 
 
 6. Hence Ktf.fi and CHD are 
 not two distinct lines, but 
 the same line. 
 
 7. But KHR, or KR, is II to <4JBL 
 
 8. Therefore CHD, or CD, 
 which coincides with KR, 
 is II to AB. 
 
 1. No other possibility exists. 
 
 2. 45. 
 
 3. 57. If two parallel lines 
 are cut by a transversal, the 
 alternate-interior angles are 
 equal. 
 
 4. Hypothesis. 
 
 5. 32. 
 
 6. 24. 
 
 7. Statement 2, 
 
 8. 45. 
 
32 PLANE GEOMETRY 
 
 EXERCISES 
 
 21 . If, in the adjacent figure, Z 1 = Z 8, prove that 4- B is II to CD. 
 
 22. IfZ5 = Z7,provethat,4 J BislltoC Y >. 
 
 23. If Z2+Z3 = 2rt.z, prove that ' 2/6 
 AB is II to CD. / 
 
 24. From the preceding exercises state ^ 473 "g 
 
 three corollaries to Theorem 11. / 
 
 65. Axiom V. A number may be substituted for its equal in 
 any operation on numbers. 
 
 NOTE. It should be observed that the axioms stated from time to 
 time apply to numbers. If, then, for example, we say subtract line AB 
 from line CD or substitute /.x for Za, it is the numerical measure of 
 the line or the angle respectively to which we refer and to which the 
 axiom applies. 
 
 Theorem 12 
 
 66. The sum of the angles of any triangle is two right 
 angles. 
 
 K 
 
 Given the triangle ABC. 
 
 Toprove that ZA +ZB+ZC= 2 rt. A. 
 
 Proof 
 
 1. Let KR be a line through 1. 45. 
 C II to AB, forming Zl and 
 
 Z3 respectively. 
 
 2. Then Z 1+ Z 2 + Z 3 = 2. 35, 40. 
 
 2rt.A (1) 
 
BOOK I 33 
 
 3. Z1=Z.4. 3. 57. If two parallel lines are 
 
 cut by a transversal, the alter- 
 .nate-interior angles are equal. 
 
 4. Also Z3=Z. 4. Why? 
 
 5. Z A -f- Z 2 -f Z B = 2 rt. Z , 5. Substituting Z ^ for Z 1 and 
 orZ4+Z J B + ZC = 2rt. A Z for Z3 in equation (1), 
 
 65. 
 
 67. Corollary 1. If two angles of one triangle equal respec- 
 tively two angles of another, the third angle of the first equals the 
 third angle of the second. 
 
 Given two triangles whose angles are denoted by a, 6, c, and 
 x, y, z respectively, where a = x and b = y. 
 
 To prove that c = z. 
 
 Proof 
 
 1. Nowa + b + c = x + y + z 1. 66. 
 
 = 2rt.Zs. (1) 
 
 2. a -f ^ = x -f y. (2) 2. Hypothesis and 31. 
 
 3. From (1) and (2), 3. 51. 
 
 c = z. 
 
 68. Exterior angle. An exterior angle of a triangle is the 
 angle between any side and the adjacent side produced. See 
 Z 2 of 69. 
 
 QUERY. How many exterior angles has a triangle? 
 
 69. Corollary 2. An exterior angle of a triangle is equal 
 to the sum of the two opposite interior 
 
 angles. 
 
 HINTS. Zl + Z2 = 2rt.A Why? 12 
 
 = 2rt.A Why? ^ B K 
 
 Hence Zl + Z2 = Zl + A + ZC. Why? 
 
 Therefore Z 2 = Z .4 + Z C. Why ? 
 
 70. Quadrilateral. A quadrilateral is a portion of a plane 
 bounded by four straight lines. 
 
34 PLANE GEOMETRY 
 
 71. Corollary 3. The sum of the angles of a quadrilateral is 
 four right angles. 
 
 Proof. Let A BCD be any quadrilateral. 
 Draw A C. 
 
 1. ThenZl4-Z4-fZD = 2rt.zl Why? 
 
 3. (Zl + Z 2) + (Z 4 + Z 3) + (Z B + ZD) = 4 rt. A Why ? 
 
 72. Complementary angles. One angle is the complement 
 of another if their sum is a right angle (or ninety degrees). 
 
 73. Acute angle. An acute angle is an angle less than a 
 right angle. 
 
 74. Obtuse angle. An obtuse angle is an angle greater than 
 a right angle and less than two right angles. 
 
 75. Corollary 4. The two acute angles of a right triangle are 
 complementary. 
 
 76. Corollary 5. If two angles have their sides perpendicular 
 each to each, they are equal or supplementary. t / 
 
 HINTS. Let the sides of Zl be _L respectively to L J-/2. 
 
 the sides of Z3. What is the relation between Z3 
 andZ4? Z4andZ5? Z5andZl? Conclusion? 
 
 What is the relation between ^2 and Z3? Z2 
 andZl? 5/ 
 
 77. Conjugate angles. One angle is the 
 
 conjugate of another if their sum is four right angles. 
 
 78. Polygon. A polygon is a plane figure bounded by three 
 or more straight lines. 
 
 79. Regular polygon. A regular polygon is a polygon all 
 of whose angles are equal and all of whose sides are equal. 
 
 Such a polygon is said to be equiangular and equilateral. 
 
BOOK I 35 
 
 80. Diagonal. A diagonal of a polygon is a line joining 
 any two nonconsecutive vertices. 
 
 81. Special polygons. A pentagon is a polygon having 
 five sides. A hexagon is a polygon having six sides. 
 
 ALGEBRAIC EXERCISES ON ANGLE RELATIONS 
 
 25. Solve for /.x, 20 +Z* = 38 30'. 
 
 26. Solve for Z.x, 3Zx +10 = 64 +Zjr. 
 
 27. Solve for x, 180 - x = f (90 - x). 
 
 28. What is the complement of 37 ? the supplement ? 
 
 29. What is the conjugate of 130 ? 
 
 30. One acute angle of a right triangle is four times the other. 
 Find the number of degrees in each. 
 
 HINT. Let 4 x and x represent the number of degrees in the two 
 angles respectively. 
 
 31. Two parallels are cut by a transversal. One of the two 
 interior angles on the same side of the transversal is five times 
 the other. Find the number of degrees in each. 
 
 32. In a certain quadrilateral whose opposite sides' are par- 
 allel, one of two adjacent angles is four times the other. Find the 
 number of degrees in each angle. 
 
 33. If x represents the number of degrees in one angle, rep- 
 resent its complement ; its supplement ; its conjugate. 
 
 34. The complement of an angle is three and one-half times 
 the angle. Find the angle. 
 
 35. The supplement of an angle is 21 more than twice the 
 angle. Find the angle. 
 
 36. One angle of a triangle is twice another, and the third is 
 15 more than four times the sum of the other two. Find each. 
 
 37. The complement of an angle is 52 less than two thirds 
 the supplement of the angle. Find the angle. 
 
36 PLANE GEOMETRY 
 
 38. In the triangle ABC, the angle B is twice the angle A. 
 The bisectors of these angles meet at Z), and BD prqduced meets 
 AC at R. The angle BRC equals 72. Find the angles of the 
 triangle ABC. 
 
 39. The side AB of the triangle ABC is extended to A". The 
 bisectors of the angles CBK and A meet at R. Assign a numerical 
 value to each angle of the triangle ABC. Then determine the num- 
 ber of degrees in the angle R. Compare this with the angle ('. 
 
 40. Assign other values to the angles of the triangle ABC and 
 repeat the work of the preceding exercise. 
 
 41. Find the sum of the angles of a pentagon. 
 HINT. From one vertex draw all the diagonals possible. 
 
 42. Find each angle of an equiangular pentagon. 
 
 43. Find the sum of the angles of an equiangular hexagon. 
 
 44. If a pentagon is regular, find the number of degrees in 
 the angle between two diagonals drawn from the same vertex. 
 
 45. A certain hexagon is equilateral and equiangular. Find 
 the angle between each adjacent pair of diagonals. 
 
 46. Assign unequal numerical values to the angles of a quadri- 
 lateral, making the opposite angles supplementary, and produce 
 the opposite sides of the quadri- R 
 
 lateral until they meet in K 
 
 and R. Then bisect the angles / 
 
 K and R and determine the 
 
 number of degrees in the angle _._ 
 
 between the bisectors. / ^13-- 
 
 47. Solve the preceding exer- 
 cise again, assigning other values to the angles of the quadri- 
 lateral than those used before. What is the angle between the 
 bisectors ? Is a conclusion possible here ? 
 
 48. ABCD is a quadrilateral with its diagonals AC and BD 
 intersecting at K. The bisectors of the angle . 4DB and the 'angle 
 A CB meet in R. Assign numerical values to the angle A DB, the 
 
BOOK I 37 
 
 angle ACB, and the angle AKD. Then determine the number of 
 degrees in the angle DRC. , 
 
 49. Solve the preceding exercise again, assigning still other 
 values to the angles. Is the relation between the angle DRC and 
 the sum of the angles DAC and DEC the same as before ? Is a 
 conclusion from this fact possible ? 
 
 Theorem 13 
 
 82. If a side and the two adjacent angles of one 
 triangle are equal respectively to a side and the two 
 adjacent angles of another, the triangles are congruent. 
 
 tC 
 
 Given the triangles ABC and FGHin which the side AB equals 
 the side FG, the angle A equals the angle F, and the angle B 
 equals the angle G. 
 
 To prove that A ABC is congruent to AFGH. 
 
 Proof 
 
 1. Superpose AFGH on AABC 1. 20. 
 so that FG falls on its equal 
 
 AB, the point F falling on the 
 point A and the point G on 
 the point B. 
 
 2. The side FH will fall on AC. 2. Z A = Z F. 
 
 3. The side GH will fall on BC. 3. Z G = Z.B. 
 
 4. The point H falls at the in- 4. It falls on the line A C and 
 tersection C of the lines AC also on the line BC. 
 
 and BC. 
 
 5. Therefore AABC is congru- 5. 24. 
 ent to AFGH. 
 
38 PLANE GEOMETRY 
 
 EXERCISES 
 Solve f OT x and y and check : 
 
 50. x-2y=8, 52. Solve for Z.A./.E, and ZC: 
 3x + 2y = 7. Z.4 -f Z5 + ZC = 180, 
 
 51. 5o;-3y = 2, Z.4 + 2Z =166, 
 15a;+12y = -5. Z.A + Z.B Z.C = 52. 
 
 53. The sum of two angles of a triangle is 105. The sum of 
 one of these and the third angle is 115. Find the number of 
 degrees in each angle of the triangle. 
 
 54. The difference of two angles of a triangle is 24. The sum 
 of one of these and the third angle is 132. Find each angle of 
 the triangle. 
 
 HISTORICAL NOTE. Theorem 13 is one of several theorems the dis- 
 covery and proof of which are attributed to Thales of Miletus (about 
 600 B.C.), one of the Seven Wise Men of Greece. He applied this 
 theorem to the measurement of the distances of ships from the shore. 
 Business affairs took him for a time to Egypt, where he learned what 
 the Egyptians knew of science and geometry, the study of which he 
 later introduced into Greece. He seems to have been the first man to 
 appreciate the necessity of a scientific proof in geometry. While yet 
 in Egypt his acute mind enlarged on what he learned and, according to 
 Plutarch, he soon excelled his Egyptian teachers and astonished King 
 Amasis by measuring the heights of the pyramids from the lengths of 
 their shadows. A more amazing performance, no doubt, was the predic- 
 tion by Thales of an eclipse of the sun for the year 585 B.C. In fact, 
 the scientific study of astronomy begins with him. It was Thales, also, 
 who first noticed that a piece of amber when rubbed becomes electrified. 
 Moreover, he expressed the conviction that there is some unifying prin- 
 ciple which links together all the physical phenomena and is able to 
 make them intelligible ; and that all matter is made of one primordial 
 element, the search for which should be the aim of natural science. 
 Modern physics has shown that an intimate relation exists between 
 heat, light, electricity, and magnetism, and it knows that electrons are 
 a constituent of the atoms of all substances. Thus we see that, in addi- 
 tion to doing fundamental work in geometry, Thales was the first to 
 describe the spirit which guided the growth of physical science in all 
 
BOOK I 
 
 39 
 
 ages and to express his belief in the ultimate nature of matter, which 
 the very latest work in physics comes near confirming. These facts 
 constitute a very remarkable tribute to the genius of the man. 
 
 83. Parallelogram. A parallelogram is a quadrilateral whose 
 opposite sides are parallel. 
 
 84. Rectangle. A rectangle is a parallelogram whose angles 
 are right angles. 
 
 Theorem 14 
 
 85. The opposite sides of a parallelogram are equal. 
 
 A B 
 
 Given the parallelogram ABCD. 
 To prove that AB = DC and AD 
 
 1. Draw the diagonal BD. 
 
 2. In AABD and .BCD, 
 
 3. Also Z1=Z4. 
 
 4. DB = DB. 
 
 5. A A BD is congruent to A B CD. 
 
 6. Therefore AB = DC 
 and AD = EC. 
 
 BC. . 
 
 Proof 
 
 1. 19. 
 
 2. 57. If two parallel lines 
 are cut by a transversal, the 
 alternate-interior angles are 
 equal. 
 
 3. Why ? 
 
 4. Why ? 
 
 5. 82. If a side and the two 
 adjacent angles of one tri- 
 angle are equal respectively 
 to a side and the two adja- 
 cent angles of another, the 
 triangles are congruent. 
 
 6. 27. Corresponding parts of 
 congruent figures are equal. 
 
40 PLANE GEOMETRY 
 
 86. Corollary. The opposite angles of a parallelogram are 
 equal. \ 
 
 HINTS. In the figure of 85, 
 
 Z2=Z3, (1) Why? 
 
 and Z1=Z4. (2) 
 
 Therefore, from (1) and (2), 
 
 (Z1 + Z2), orZABC = (Z3 + Z4),or ZADC. 31 and 65 
 Also ZA =ZC. Why? 
 
 EXERCISES 
 
 55. A diagonal of a parallelogram divides it into two congru- 
 ent triangles. 
 
 HINTS. In the proof of Theorem 14 it is shown that BD divides 
 the parallelogram into two congruent triangles. It may be proved in 
 like manner that A C does also. 
 
 56. If two lines are perpendicular to one of two parallels, the 
 portions of them included between the two parallels are equal. 
 
 HINTS. Let A and C be any two points on A C 
 
 one of the parallels. Let AB and CD be JL to 
 the other parallel. Zl=? Why? Z 2 = ? Why? 
 Is AB II to OD? Why? What 'is 457X7? Why? 
 
 Conclusion ? B D 
 
 . 57. If on_e angle of a parallelogram is a right angle, the figure 
 is a rectangle. 
 
 HINT. Use 60 and 84. 
 
 58. Parallel line-segments included between parallels are equal 
 HINTS. Given "A B II to CD, and KR 
 and LM two other parallels included A - 7 ~7 - B 
 
 between them. 
 
 What kind of figure is KLMR1 Why? / / 
 
 What relation exists between KR and C ^ - "^ - D 
 ZM? 'Why ? 
 
 : 59. .If the angles. of a quadrilateral are right angles, the figure 
 is tt rectangle. . ' . ; i-, . 
 
BOOK I 
 
 41 
 
 Theorem 15 (Converse of Theorem 14) 
 
 87. If the opposite sides of a quadrilateral are equal, the 
 figure is a parallelogram. 
 
 Given the quadrilateral ABCD in which AB equals DC and AD 
 equals BC. 
 
 To prove that ABCD is a parallelogram. 
 
 Proof 
 
 to 
 
 1. Draw the diagonal A C. 
 
 2. In the A ABC and ADC, 
 AB = DC and AD = B(\ 
 
 3. AC = AC. 
 
 4. A ABC is congruent 
 
 AADC. 
 
 5. 
 
 6. 
 
 7. Also 
 
 8. AD is II to BC. 
 
 9. From 6 and 8 it follows that 
 ABCD is a parallelogram. 
 
 DC is II to AB. 
 
 1. 19. 
 
 2. Hypothesis. 
 
 3. Identical. 
 
 4. 33. If the three sides of one 
 triangle are equal respectively 
 to the three sides of another, 
 the triangles are congruent. 
 
 5. 27. 
 
 6. 64. If two straight lines are 
 cut by a transversal, making 
 two alternate-interior angles 
 equal, the lines are parallel. 
 
 7. 27. 
 
 8. 64. 
 
 9. 83. A parallelogram is a 
 quadrilateral whose opposite 
 sides are parallel. 
 
42 
 
 PLANE GEOMETRY 
 
 Theorem 16 
 
 88, If two sides of a quadrilateral are < equal and, 
 parallel, the figure is a parallelogram. 
 
 Given the quadrilateral ABCD with AB equal to DC and 
 parallel to it. 
 
 To prove that the quadrilateral ABCD is a parallelogram. 
 
 Proof 
 
 1. Draw the diagonal ^4 C. 
 
 2. ^5 is II to DC. 
 
 3. Hence Z1 = Z4. 
 
 4. ^#=DC. 
 
 5. J.C = ^C. 
 
 6. A A EC is congruent to A CD A, 
 
 7. Z2 = Z3. 
 
 8. 
 
 AD is II to J5C. 
 
 9. ABCD is a parallelogram. 
 
 1. 19. 
 
 2. Hypothesis. 
 
 3. 57. If two parallel lines are 
 cut by a transversal, the 
 alternate-interior angles are 
 equal. 
 
 4. Hypothesis. 
 
 5. Identical. 
 
 6. 25. 
 
 7. Why ? 
 
 8. 64. If two straight lines 
 are cut by a transversal, 
 malting two alternate-interior 
 angles equal, the lines are 
 parallel. 
 
 9. 83. 
 
 89. Method of proof. Much of the student's ,work in the 
 exercises and theorems of Book I really consists in prov- 
 ing two angles equal or two lines equal. Sometimes only 
 
BOOK I 43 
 
 one theorem is needed to do this. Thus two angles can 
 frequently be proved equal by pointing out that they are 
 vertical angles, or that they are the base angles of an isosceles 
 triangle, or that they are alternate-interior or corresponding 
 angles of parallel lines; and two lines can often be 
 proved equal by showing that they are the equal sides of 
 an isosceles triangle, or that they are the opposite sides 
 of a parallelogram. Occasionally a proof requires the use of 
 only two theorems. 
 
 Among the various methods which require the use of 
 three or more theorems there is one which the student 
 must use so frequently that it will now be given special 
 emphasis. 
 
 To prove that two lines or two angles are equal : 
 
 1. Draw as accurately as possible a figure ivhieh the state- 
 ment of the theorem shows to be necessary. 
 
 2. Select two triangles which appear to be congruent and 
 which contain as parts the lines or angles to be proved equal. 
 
 It will often be found necessary to draw lines not mentioned 
 in the theorem in order to form "the two triangles needed. See, 
 for example, the proof of Theorems 14, 15, and 16. 
 
 3. Prove the two triangles congruent by the use of one of the 
 four following theorems : 1, 3, 8, and 13. 
 
 Later the student may use any one of six theorems, as Theo- 
 rems 19 and 20 also relate to congruent triangles. 
 
 4. State the conclusion : The required lines or angles are 
 equal because they are correspondiny parts of congruent triangles. 
 
 The student should study carefully the application of the 
 several steps of the preceding outline, especially in Theorems 
 14 and 15 and he should note their repeated application in 
 
44 PLANE GEOMETRY 
 
 Theorems 16, 17, and 18. The list of exercises which follows 
 96 will furnish material with which he can make a real begin- 
 ning in applying the method for himself. 
 
 QUERY 1. Iii which of the theorems from 1 to 15 was the method 
 described in 89 first used ? In which theorem was 
 it used next? / 
 
 QUERY 2. How many times has it been used in the / 
 theorems thus far? D f~ ~J C 
 
 QUERY 3. If A BCD is a parallelogram, does the p jr 
 
 point C lie within the angle KAR1 Why? 
 
 QUERY 4. If the point C lies within the angle KAR, will the 
 diagonal AC intersect the diagonal BD't 
 
 QUERY 5. Which of the first sixteen theorems were proved by 
 superposition ? 
 
 Theorem 17 
 
 90. The diagonals of a parallelogram bisect each other. 
 
 Given a parallelogram ABCD with its diagonals AC and BD 
 intersecting at K. 
 
 To prove that AK= KG and DK=KB. 
 
 HINTS. Show that a side and two adjacent A of AABK are equal 
 respectively to a side and two adjacent A of A CDK. 
 Then use 82 and 27. 
 
 QUERY 1. In Theorems 1, 3, and 13 change the word triangle to 
 parallelogram and then determine if the resulting statements are true. 
 
 QUERY 2. A diagonal is drawn in each of two quadrilaterals. If 
 the two triangles of the first are respectively congruent to the two tri- 
 angles of the second, are the quadrilaterals congruent? Explain. 
 
BOOK I 45 
 
 Theorem 18 (Converse of Theorem 17) 
 
 91. If the diagonals of a quadrilateral bisect each other, 
 the figure is a parallelogram. 
 
 Given the quadrilateral ABCD (using the figure of 90) in 
 which AK equals KC and DK equals KB. 
 
 To prove that the quadrilateral ABCD is a parallelogram. 
 
 HINTS. By the use of Theorem 1 show that 1\AKB is congruent to 
 &DKC. Then Z6 = Z2 by 27, and AB is II to CD by 64. In like 
 manner, using \AKD and &BKC, prove AD II ioBC. 
 
 92. Drawing of figures. In drawing the figures necessary 
 to prove any given theorem or exercise the student should 
 make it a rule to observe the following points. A little expe- 
 rience will soon demonstrate their value. 
 
 1. Draw the figures as accurately as possible. 
 
 Later on methods of constructing many figures with theoretical 
 exactness will be studied. Until then a sufficient degree of accu- 
 racy can be attained by the exercise of a little care and some 
 judgment, and by the use of a ruler and a sharp-pointed pencil. 
 A protractor may be used to lay off angles (see 280). 
 
 Accuracy is especially desirable because a properly drawn figure 
 often shows relations between its various lines and angles which 
 are of great help in discovering a proof of the theorem. These 
 relations are frequently obscured by a poorly drawn figure. 
 
 2. Make the figures as general as possible. 
 
 If the theorem refers to a triangle, draw one having the three 
 sides unequal, for that is the most general triangle possible. If 
 it calls for an isosceles triangle, do not draw one with three 
 equal sides. 
 
 In the most general kind of quadrilateral no two sides are 
 equal or parallel and no two angles are equal. Therefore, if the 
 theorem specifies a quadrilateral, do not draw a parallelogram or a 
 
46 PLANE GEOMETRY 
 
 quadrilateral having two sides parallel or equal. Similarly, if the 
 theorem refers to a parallelogram, do not draw a rectangle, or a 
 parallelogram all of whose sides are equal. 
 
 In drawing figures, then, avoid equal lines, parallel lines, equal 
 angles, or right angles, unless the statement of the theorem says 
 or implies that the figure should have them. And, furthermore, 
 if a point is to be chosen on a line, do not select the mid-point of 
 the line. 
 
 The student should form the habit of drawing general figures. 
 Otherwise he will frequently be disappointed to find that what he 
 thought was a valid proof is wholly incorrect, because it depends 
 on an unwarranted assumption which crept into his reasoning 
 without his noticing it. Very often he will find that he made the 
 assumption because his figure was a special one. 
 
 93. Order in naming the vertices of a polygon. In stating 
 exercises in which it is necessary to name a polygon of four 
 or more sides by the letters at its vertices, it is customary 
 to name the vertices in succession and in either order. 
 
 A B A D A B 
 
 FIG. 1 FIG. 2 FIG. 3 
 
 Thus the quadrilateral ABCD is correctly represented by either 
 Fig. 1 or Fig. 2 but not by Fig. 3. Unless this point is observed, 
 the figure which results may appear to make the theorem wholly 
 untrue or to give it a meaning other than the one intended. 
 
 94. Mutually equiangular and mutually equilateral poly- 
 gons. It should also be noted that when two polygons are 
 spoken of as being mutually equiangular the meaning is that 
 the angles of one are equal respectively to the angles of 
 the other. A like meaning holds for the phrase mutually 
 equilateral. 
 
BOOK I 47 
 
 95. Extending a line. In extending or producing a line it is 
 understood that if we say line AB is ^ K 
 produced to K, we mean that AB 
 
 is extended beyond B to K so that B lies between .4 and K. 
 
 96. On methods of study. In some of the following exer- 
 cises the method of 89 is not needed. In others it is 
 necessary, and if properly used will result in a solution. In 
 still others the method of 89 must be used first, followed 
 by the application of theorems other than those on congruent 
 triangles to complete the proof. 
 
 The necessity of application in the solution of the following 
 list of exercises cannot be made too emphatic, for an hour's study 
 daily at this point will go far to insure success and save many 
 hours later. If the student finds that he can solve few or none of 
 the exercises in the following list without help, it may be that 
 he has not applied himself properly to the work which precedes, 
 but has inferred that he could learn geometry by listening atten- 
 tively to proofs worked out by others. Success in geometry is 
 measured by the acquisition of power to solve original exercises. 
 This will come only with sufficient attentive study. 
 
 EXERCISES ON CONGRUENT TRIANGLES AND 
 PARALLELOGRAMS 
 
 60. In the triangle AB C the line KR, parallel to ,4.6, cuts AC 
 in K and EC in R. Prove that the angles of the triangle ABC 
 are equal respectively to the angles of the triangle K R C. 
 
 61. In the parallelogram ABCD lines are drawn from B and D, 
 the ends of the shorter diagonal, perpendicular to A C and meeting 
 it in K and R respectively. Prove that the angles of the triangle 
 BK C are equal respectively to the angles of the triangle DRA . 
 
 62. ABCD is a parallelogram. AB and CD are produced in 
 opposite directions the same distance to K and R respectively. 
 Prove that the triangle ADR is congruent to the triangle CBK. 
 
 63. The line joining the mid-points of two opposite sides of a 
 parallelogram divides it into two parallelograms. 
 
48 PLANE GEOMETRY 
 
 64. AC is the longer diagonal of the parallelogram A BCD. 
 From C and D perpendiculars are drawn to A B or A B produced, 
 meeting it in K and R respectively. Prove that the triangles BKC 
 and ADR are mutually equiangular. 
 
 65. A BCD is a parallelogram and O the mid-point of the diag- 
 onal AC. A line is drawn through cutting DC in K and AB 
 in R. Prove that KO equals OR. 
 
 66. K is the mid-point of BC in the parallelogram A BCD. 
 Line DA' produced intersects ^IZ* produced in H. Prove that the 
 triangle KCD is congruent to the triangle KBH. 
 
 67. ABCDE is a pentagon which is equi- 
 lateral and equiangular. Draw AC and AD 
 and prove that the triangle ABC is congruent 
 to the triangle AED and that the diagonals 
 AC and AD are equal. 
 
 68. In the quadrilateral A BCD the diago- 
 nals intersect at K so that KA equals KB and KC equals KD. 
 Prove that BC equals AD. 
 
 69. In Ex. 61 prove that BK equals DR. 
 
 70. In Ex. 64 prove that CK equals DR. 
 
 71. The mid-point of one side of a parallelogram is joined to 
 a vertex not adjacent to the side and the mid-point of the opposite 
 side is joined to the vertex opposite the first. 
 
 Prove that the two lines thus determined are 
 equal and parallel. 
 
 72. AC and BD, the diagonals of a parallelo- 
 gram, meet in K. Points F, G, H, and L are the mid-points of 
 AK, BK, CK, and DK respectively. Lines FG, GH, HL, and LF 
 are drawn. Prove that FGHL is a parallelogram. 
 
 73. If two opposite sides of a parallelogram are produced by 
 the same distance in opposite directions and their ends joined to 
 vertices so as to form a quadrilateral, it will be a parallelogram. 
 
 74. A BCD is a parallelogram. R is taken on AB and K on CD 
 so that BR equals DK. Prove that A KCR is a parallelogram. 
 
BOOK I 
 Theorem 19 
 
 49 
 
 97. Two right triangles are congruent if the hypotenuse 
 and another side of the. first are equal respectively to the 
 hypotenuse and another side of the second. 
 
 C G 
 
 12 
 
 B 
 
 K 
 
 Given the right triangles ABC and FGH, in which the hypot- 
 enuse AC equals the hypotenuse HG and the side CB equals the 
 side GF. 
 
 To prove that A AB C is congruent to A FGH. 
 
 Proof 
 
 1. 20. 
 
 1. Place A FGH adjacent to 
 A ABC in such a way that 
 GF coincides with its equal 
 CBj the point G falling on C 
 and the point F on B. The 
 point H will then fall at 
 some point K. 
 
 2. Zl=Z2 = a rt. Z. 2. Hypothesis. 
 
 3. Zl + Z2 = a st. Z. 3. 40. 
 
 4. ABK is a straight line. 4. 35. 
 
 5. ABKC is a triangle. 5. 16. 
 
 6. In AAKC, A C = CK. 6. Hypothesis. 
 
 7. Z.A = Z.K. 7. 29. 
 
 8. Hence A ABC is congruent 8. 50. 
 to ABCK. 
 
 9. Therefore A ABC is con- 9. 32. 
 gruent to A FGH. 
 
 QUERY. How could Theorem 19 have been proved if GH and Fll 
 had been given equal to CA and BA respectively? 
 
50 
 
 PLANE GEOMETRY 
 Theorem 20 
 
 98. Two right triangles are congruent if a 'side about 
 the right angle and an acute angle of the first are equal 
 respectively to a side about the right angle and a corre- 
 sponding angle of the second. 
 
 B C G H 
 
 Case I. When the acute angle is opposite the given side. 
 
 Given the right triangles ABC and FGH with the side EC 
 equal to the side GH, the angle B and the angle G right angles, 
 and the acute angle A equal to the corresponding acute angle F. 
 
 To prove that A ABC is congruent to A FGH. 
 
 1. In A ABC and FGH, 
 
 Proof 
 
 1. Hypothesis. 
 
 2. ZC=Z#. 2. Why? 
 
 3. EC = GH. 3. Hypothesis. 
 
 4. AABCiscongiuenttoAFGH. 4. 82. 
 
 Case II. When the acute angle is adjacent to the given side. 
 Given as in Case I except that the angle C equals the angle H. 
 HINT. Apply 82. 
 
 QUERY. What is the precise meaning of the word corresponding in 
 the statement of Theorem 20 ? Illustrate. 
 
 EXERCISE 75. Give in full the details of Case II. 
 
BOOK I 
 
 51 
 
 EXERCISES 
 
 76. Two adjacent angles of a parallelogram are supplementary. 
 
 77. The lines joining the mid-points of adjacent sides of a 
 parallelogram form a parallelogram. 
 
 78. The bisectors of two adjacent angles of a parallelogram 
 intersect each other at right angles. 
 
 79. Two parallelograms are equal if two sides and the included 
 angle of one are equal respectively to two sides and the included 
 angle of the other. 
 
 80. The image of an object A seen in a 
 
 plane mirror by the eye at E appears to be /' 
 
 at A 1 , as far behind the mirror MR as A is in 
 
 front of it and so that A A l is perpendicular 
 
 to MR. That is, rays of light from A strike 
 
 the mirror at L and are reflected to E. Show 
 
 that Z.ALR=Z.ELM. 
 
 Theorem 21 (Converse of Theorem 2) 
 
 99. If two angles of a triangle are equal, the sides 
 opposite those angles are equal. 
 
 L/ 
 
 Given the triangle ABC with the angle A equal to the angle B. 
 To prove that AC = BC. 
 
 HINT. Let CK be _L to AB. Then use 98. 
 
 NOTE. It is customary to call that side of an isosceles triangle which 
 is not equal to either of the other two the base. And frequently that 
 one of the three vertices of an isosceles triangle which lies opposite the 
 base is called the vertex. 
 
52 PLANE GEOMETRY 
 
 EXERCISES 
 
 81. In an isosceles triangle a line cutting two sides and parallel 
 to the third side forms another isosceles triangle. 
 
 82. Prove that the exterior angles at the base of an isosceles 
 triangle are equal. 
 
 QUERY. If a quadrilateral has two equal angles, lias it two equal 
 sides? 
 
 100. Equilateral triangle. An equilateral triangle is a tri- 
 angle which has three equal sides. 
 
 101. Equiangular triangle. An equiangular triangle is a 
 triangle which has three equal angles. 
 
 Theorem 22 
 
 102. If a triangle is equilateral, it is equiangular. 
 
 C 
 
 A B 
 
 Given the triangle ABC in which AB equals BC equals CA. 
 To prove that ZA = ZB = ZC. 
 
 HINT. Apply Theorem 2 twice. 
 
 EXERCISE 83. The sides AB, BC, and CA of the equilateral 
 triangle A BC are extended by the same length to K, R, and L 
 respectively. Prove that the triangle whose vertices are A', R, 
 and 7, is equiangular. 
 
BOOK I 
 
 53 
 
 Theorem 23 (Converse of Theorem 22) 
 
 103. If a triangle is equiangular, it is equilateral. 
 
 HINT. Use the figure of 102 and apply Theorem 21 twice. 
 
 104. Corollary. Each angle of an equilateral or an equi- 
 angular triangle is 60. 
 
 Theorem 24 
 
 105. If one acute angle of a right triangle is thirty 
 degrees, the side opposite is half the hypotenuse. 
 
 O 
 
 A B 
 
 Given the right triangle ABC in which the angle A is 30 and 
 AC is the hypotenuse. 
 
 To prove that BC = AC. 
 
 Proof 
 
 1. 30 + Z C = 90. 
 
 2. Hence Z C = 60. 
 
 1. 75 and hypothesis. 
 
 2. 51. 
 
 = 90. 
 
 3. Let BL be a line making 3. Z, 4 BC is greater than Z A. 
 
 Z.LBA, orZl, =30. 
 4. 
 5. 
 6. 
 7. 
 8. 
 9. 
 
 AL = BL. 
 Z 2 = 60. 
 Z3 = 60. 
 BL = LC = BC. 
 AL=LC = \A 
 BC = A C. 
 
 4. 99. 
 
 5. Zl = 30, and 
 
 6. 69. 
 
 7. From 2, 5, and 6, and 103. 
 
 8. 65. 
 
 9. Statements 7 and 8. 
 
 QUERY. What is the converse of Theorem 24? 
 
54 
 
 PLANE GEOMETRY 
 
 QUERY 1. If in Theorems 22 and 23 the word polygon replaces the 
 word triangle, would the resulting statement be true ? 
 
 QUERY 2. How many diagonals has a polygon of four sides? five sides? 
 six sides? seven sides? 
 
 EXERCISES 
 
 84. Prove that a line cutting two sides of an equilateral triangle 
 and parallel to the third side forms an equilateral triangle. 
 
 85. ABCDEF is a regular hexagon. Prove that the lines AC, 
 CE, and EA form an equilateral triangle. 
 
 86. Prove that the longer diagonals of a regular hexagon bisect 
 the angles at the vertices which they connect. 
 
 87. ABCDEFis a regular hexagon. Prove that if the diagonals 
 AD and BE are drawn, two equilateral triangles are formed. 
 
 88. In parallelogram ABCD the angle B is 120, and the diagonal 
 BD makes a right angle with A D. If DK is perpendicular to A B at K, 
 prove that BD is twice DK. 
 
 Theorem 25 (Converse of Theorem 24) 
 
 106. If one side of a right triangle is half the hypotenuse, 
 the angle opposite that side is thirty degrees. 
 
 Given the right triangle ABC, in which the side AB is equal 
 to one half the hypotenuse AC. 
 
 To prove that /.ACB = 30. 
 
BOOK I 
 
 55 
 
 Proof 
 
 1. Extend AB to A', making 
 BK=AB. 
 
 2. Then draw CK. 
 
 3. In AABC and KBC, 
 
 Z1=Z2. 
 
 4. BC = BC. 
 
 5. And ^5 = BK. 
 
 6. A ABC is congruent 
 A KBC. 
 
 7. AC = /fC. 
 
 8. ^5 = 5^=1^C. 
 
 9. (AB+BK)o?AK=AC= 
 
 10. Z^=Z, 
 
 11. Z3=Z4. 
 
 12. Z3 = l. 
 
 to 
 
 1. 95. 
 
 2. 19. 
 
 3. 40. 
 
 4. Why ? 
 
 5. By construction. 
 
 6. 25. 
 
 7. 27. 
 
 8. Construction and hypothesis. 
 
 9. Statements 7 and 8. 
 
 10. 102 and 106. 
 
 11. 27. 
 
 12. Statements 10 and 11. 
 
 107. Trapezoid. A trapezoid is a quadrilateral two and 
 only two of whose sides are parallel. , 
 
 EXERCISES 
 
 89. AK, perpendicular to the base EC of the equilateral tri- 
 angle ABC, meets EC in K ; and jO*, perpendicular io AB, meets 
 AB in R. Prove that AK is twice KR. 
 
 90. A BCD is a trapezoid with AB parallel to CD. If DK is 
 perpendicular to AB at K and makes A K equal to one half AD, find 
 the number of degrees in the angle A DC. 
 
 91. Using the adjacent figure and 
 a; different method from that given in 
 the text, prove Theorem 25 as follows. 
 Let BK be a line making Z1=Z/1. 
 
 = ? What, then, 
 
 B C 
 
 is the relation between Z C and Z 2 ? between A K and BK ? etc. 
 
56 PLANE GEOMETRY 
 
 108. Median. A median of a triangle is a line which joins 
 any vertex to the mid-point of the opposite side. f 
 
 NOTE. Among the more important elements of good writing or 
 speaking are brevity, completeness, and a logical order of arrange- 
 ment. The student's control of these three elements will be greatly 
 increased if his attitude toward geometry is correct and if his part of 
 the work involves the thinking out and the preparation of oral and 
 written demonstrations of exercises. In this work care should be taken 
 to avoid needless repetition, especially in the longer proofs. When 
 in the course of a demonstration two lines or two angles can be proved 
 equal just as two other lines or two other angles have been proved equal, 
 it is best to omit all details in the second case and say, "In like manner 
 it can be proved that AB KB, or that Z a Z r." On the other hand, 
 important details should not be left out. The student should state 
 clearly how any auxiliary lines are drawn and not assume that they are 
 drawn in a certain way. All vital reasons should be quoted. Omission 
 of a single one may ruin an otherwise perfect proof. Lastly the student 
 should avoid a haphazard arrangement of the matter of a proof. A 
 single statement not in its proper place may be fatal to a demonstration. 
 
 EXERCISES ON RIGHT, ISOSCELES, AND EQUILATERAL 
 TRIANGLES 
 
 92. If a line bisects the vertical angle of an isosceles triangle, 
 it bisects the base and is perpendicular to it. 
 
 93. The median from the vertex of an isosceles triangle is 
 perpendicular to the base and bisects the vertical angle. 
 
 94. If a line from the vertex of an isosceles triangle is perpen- 
 dicular to the base, it bisects the base and the vertical angle. 
 
 95. If any angle of an isosceles triangle is 60, the triangle is 
 equilateral. 
 
 96. The perpendiculars drawn from the middle point of the 
 base to the equal sides of an isosceles triangle and terminating 
 in them are equal. 
 
 97. The bisectors of the equal angles of an isosceles triangle 
 intersect, and form, with its base, an isosceles triangle. 
 
BOOK I 57 
 
 98. The medians drawn from the vertices of the equal angles 
 of an isosceles triangle to the opposite sides are equal. 
 
 99. The perpendiculars drawn from the extremities of the 
 base of an isosceles triangle to the opposite sides and terminating 
 in them are equal. 
 
 100. State and prove the converse of the theorem of Ex. 99. 
 
 101. If a triangle is isosceles, the lines from the extremities 
 of the base bisecting the equal angles and terminating in the 
 opposite sides are equal. 
 
 102. State the converse of the theorem of Ex. 101. 
 
 NOTE. The student will observe that he is not asked to prove the 
 theorem just stated. The theorem is true and its proof may appear 
 to be easy. It is really very difficult to prove it, however, even by the 
 indirect method of 152, if one is limited to the theorems of Book I. 
 A direct proof bv Ex. 116 of the Supplementary Exercises on page 29C 
 is not so difficult to obtain. 
 
 103. If the angles adjacent to the longer of the two parallel 
 sides of a trapezoid are equal, the iionparallel sides are equal. 
 
 HINT. What two lines forming two right triangles might be of use ? 
 or what single line forming a parallelogram ? 
 
 104. State the converse of the theorem of Ex. 103 and, if 
 true, prove it. 
 
 105. 'One angle formed by the bisectors of two angles of an 
 equilateral triangle is double the third angle of the triangle. 
 
 HINT. It is frequently helpful, as here, to determine the number of 
 degrees in each angle of the figure and to write it plainly within the 
 angle. Often the proof of the theorem will then be obvious. 
 
 106. ABC is an equilateral triangle. The bisectors of the angles 
 A and C meet at K. The line KR is drawn parallel to AB, meeting 
 A C at R, and KL is drawn parallel to BC, meeting A C at L. Prove 
 that AR, RK, RL, KL, and CL are equal. (See the hint to Ex. 105.) 
 
 107. The line through the vertex of an isosceles triangle parallel 
 to the base bisects the exterior angles at the vertex. 
 
58 PLANE GEOMETRY 
 
 108. Has the theorem of Ex. 107 a converse ? more than one ? 
 If so, state and prove them. 
 
 109. Under what conditions may a theorem have more than 
 one converse ? 
 
 110. ABC is any triangle. A KB and ARC are equilateral tri- 
 angles adjacent to the triangle ABC and having the sides AB and 
 AC respectively in common with it. 
 
 Show that CK equals BR. 
 
 111. The obtuse angle between the 
 
 bisectors of the equal angles of an s^x^^ y ^ ^*^^ C+2X 
 isosceles triangle is equal to an exte- 4 B 
 
 rior angle at the base of the triangle. 
 
 HINTS. This involves a geometrical identity and is best attacked 
 algebraically. Why can the angles be marked as in the figure ? Does 
 y + 2x = /.C + 2x + 2x1 Why? Conclusion? 
 
 Try to apply the method here illustrated wherever possible. 
 
 112. If each of the angles at the base of an isosceles triangle 
 is one fourth the vertical angle, every line perpendicular to the 
 base which does not pass through the vertex forms an equilateral 
 triangle with the other two sides, one being produced. (See hint 
 to Ex. 105.) 
 
 113. K is any point on the base EC of the isosceles triangle- 
 ABC. The side A C is produced from C to 72 so that CR equals 
 CK, and KR is drawn meeting AB at L. Prove that the angle ALR 
 equals three times the angle ARL. 
 
 HINTS. Let a be the number of degrees in the angle R. What other 
 two angles can also be marked a degrees? What two angles can be 
 marked 2 a degrees ? What angle 3 a degrees ? 
 
 114. In the triangle ABC the bisectors of the angle A and the 
 exterior angle at B intersect at K. Prove that the angle A KB is 
 one half of the angle C. 
 
 109. Square. A square is a rectangle whose sides are equal. 
 
 110. Rhombus. A rhombus is a parallelogram whose sides 
 are equal and whose angles are not right angles. 
 
BOOK I 
 
 Theorem 26 
 
 111. TJie diagonals of a rectangle are equal. 
 D 
 
 59 
 
 Given the rectangle ABCD with the diagonals AC and BD. 
 To prove that A C = BD. 
 
 HINT. Select two triangles, one having A C as a side and one having 
 BD as a side. Prove that these A are congruent by 25. Conclusion ? 
 
 112. Corollary. The diagonals of a square are equal. 
 
 Theorem 27 
 
 113. The diagonals of a rhombus or of a square 
 
 (1) meet at right angles; 
 
 (2) bisect the four angles of each. 
 
 DC D 
 
 A B A B 
 
 Given the rhombus ABCD and the square ABCD with diag- 
 onals AC and BD intersecting at K. 
 
 To prove that (1) Z 1 = Z 2 = a rt. Z. 
 
 (2) Z3=Z4, Z5-Z6, Z7 = Z8, Z9 =Z10. 
 
 HINTS. (1) Which A contains Zl? Which A contains Z2? Prove 
 these A congruent. Then is Zl equal to Z2? Why? Conclusion? 
 
60 PLANE GEOMETRY 
 
 114. Corollary. The diagonals of a square make an angle 
 of 45 with each of the four sides. 
 
 QUERY 1. Do the diagonals of a parallelogram meet at right 
 angles? Do they bisect the angles of the parallelogram? Are they 
 equal ? 
 
 QUERY 2. If the diagonals of a quadrilateral are equal, is the 
 figure necessarily a parallelogram? 
 
 EXERCISES 
 
 115. If the diagonals of a quadrilateral are equal and bisect 
 each other at right angles, the quadrilateral is a square. 
 
 116. If the diagonals of a quadrilateral bisect each other at 
 right angles and the shorter diagonal equals one side, the quad- 
 rilateral is a rhombus having one angle 120. 
 
 115. Mid-perpendicular. If a line is perpendicular to another 
 line at its middle point, the first line is called the mid- 
 perpendicular of the second. 
 
 Theorem 28 
 
 116. If a point is on the mid-perpendicular of a line, it 
 is equidistant from the ends of the line. 
 
 M 
 
 A P B 
 
 Given MP the perpendicular erected at the mid-point of AB, 
 K any point on MP, and the lines KA and KB. 
 
 To prove that KA = KB. 
 
 HINT. Prove &APK congruent to &BPK. 
 
BOOK I 61 
 
 EXERCISE 117. ABC and ABK are triangles on opposite sides 
 of the common base A B. If line AB is the mid-perpendicular of KC, 
 prove that the triangle ABC is congruent to the triangle ABK. 
 
 Theorem 29 (Converse of Theorem 28) 
 
 117. If a point is equidistant from the ends of a line, 
 it is on the mid-perpendicidar of the line. 
 
 Given the line KR and the point A so that AK equals AR. 
 To prove that A lies on the mid-perpendicular of KR. 
 
 Proof 
 
 1. Let AL be JL to KR. 1. 42. 
 
 2. Then in &AKL and ARL, 2. Why ? 
 
 AL = AL. 
 
 3. And AK = AR. 3. Why ? 
 
 4. A.I KL is congruent to A. 4 #Z. 4. Why ? 
 
 5. AY, = LR. 5. Why ? 
 
 6. AL is the mid-perpendicular 6. 115. 
 of KR. 
 
 118. Corollary. Two points each equally distant from the ex- 
 tremities of a line determine the mid-perpendicular of the line. 
 
 HINT. Apply 117 twice, followed by 41. 
 
 QUERY 1. Is the point of intersection of the mid-perpendiculars to 
 two sides of a triangle equally distant from the three vertices? 
 
 QUERY 2. If a point is equally distant from all three vertices of a 
 triangle, on how many of the mid-perpendiculars to the sides does it lie? 
 
62 PLANE GEOMETRY 
 
 119. Concurrent lines. Three or more lines which have one 
 point in common are said to be concurrent. 
 
 EXERCISE 118. Prove that the mid-perpendiculars to the three 
 sides of a triangle are concurrent. 
 
 120. Distance to a line. The distance from a point to a line 
 is the length of the perpendicular from the point to the line. 
 
 Theorem 30 
 
 121. If a point is on the bisector of an angle, it is 
 equally distant from the sides of the angle. 
 
 Given the angle ABC with BM its bisector, F a point on BM, 
 and FH and FG the distances from F to AB and CB respectively. 
 
 To prove that FH = FG. 
 
 Proof 
 
 1. In ABFH and J3FG, 1. Why ? 
 
 Z1=Z2. 
 
 2. Z 3 and Z 4 are rt. A. 2. 120. 
 
 3. BF = BF. 3. Why ? 
 
 4. ABFHis congruent to A BFG. 4. 50. 
 
 5. FH = FG. 5. Why ? 
 
 EXERCISE 119. CD is the shorter base of the trapezoid ABCD. 
 The bisectors of the angle C and the angle D intersect at A'. 
 Prove that K is the same distance from BC 9 CD, and DA. 
 
BOOK I 63 
 
 Theorem 31 (Converse of Theorem 80) 
 
 122. If a point is equally distant from the sides of an 
 angle, it is on the bisector of the angle. 
 
 A* 
 
 Given the angle ABC and the point K such that the distances 
 KR and KL are equal. 
 
 To prove that the point K is on the bisector of /.ABC. 
 
 Proof 
 
 1. Draw BK. 1. 19. 
 
 2. Then in &BKR and BKL, 2. Why ? 
 
 BK = BK. 
 
 3. KR = KL. 3. Why ? 
 
 4. Z 3 = Z 4 = 90. 4. 120. 
 
 5. ABKRiscongrueuttoABKL. 5. 97. 
 
 6. Z 1 = Z 2. 6. Why ? 
 
 7. A' bisects Z ABC. 7. 28. 
 
 QUERY 1. Is the point of intersection of the bisectors of two angles 
 of a triangle equidistant from the sides of the triangle ? 
 
 QUERY 2. If a point is equidistant from the three sides of a triangle, 
 on the bisectors of what angles does it lie ? 
 
 EXERCISE 120. Prove that the bisectors of the three angles of 
 a triangle are concurrent. 
 
 QUERY 3. Are the bisectors of the angles of a square concurrent? 
 QUERY 4. Are the bisectors of the angles of a parallelogram con- 
 current ? 
 
64 PLANE GEOMETRY 
 
 123. Convex and reentrant polygons. A polygon is called 
 convex if each of its angles is less than two rjght angles, 
 and reentrant if any one of its angles is greater than two 
 right angles. 
 
 QUERY 1. Does the following statement correctly define a convex 
 polygon ? If a straight line can cut the boundary of a polygon in but 
 two points, the polygon is convex. 
 
 QUERY 2. How can a reentrant polygon be defined without men- 
 tioning its angles? 
 
 124. Axiom VI. If equals are multiplied by equals, the results 
 are equal. 
 
 Theorem 32 
 
 125. If n is the number of sides of a convex polygon 
 and 8 is the sum of its interior angles, then s = (2 n 4) 
 right angles. 
 
 Given any convex polygon ABCDEF having n sides. 
 To prove that the sum of the interior angles of ABCDEF 
 is (2 n - 4) rt. A. 
 
 Proof 
 
 1. From Kj any point within the 1. 19. 
 polygon, draw a line to each 
 
 vertex forming n triangles. 
 
 2. The sum of the angles of each 2. 66. 
 triangle = 2 rt. A. 
 
 3. The sum of the angles of the 3. 124. 
 n triangles = 2 n rt. A. 
 
BOOK I 65 
 
 4. Angles about the point K 4. 40. 
 
 = 4rt. zi. 
 
 5. Interior angles of polygon = 5. 51. These angles at/f arein- 
 
 s = 2 n rt. A 4 rt. A. eluded in the 2 n rt. A, but 
 
 they are not included in the 
 angles of the polygon. 
 
 EXERCISE 121. Draw a reentrant polygon and determine 
 whether the method of proving Theorem 32 can be used to obtain 
 a demonstration of the theorem for such a polygon. Is Theorem 32 
 true for a reentrant polygon ? 
 
 126. Equiangular polygon. A polygon whose interior angles 
 are all equal is an equiangular polygon. 
 
 QUERY. If a polygon has n vertices, how many sides lias it? 
 Conversely ? 
 
 EXERCISES 
 
 122. If one interior angle of an equiangular polygon of n sides 
 
 2 n 4 
 
 is x rt. A. pTove that x = 
 
 n 
 
 123. A polygon has 18 sides. Find in degrees the sum of its 
 interior angles. 
 
 124. The sum of the interior angles of a polygon is 36 right 
 angles. How many sides has it ? 
 
 HINT. 36 = 2 n - 4, etc. 
 
 125. Find the number of sides of a polygon the sum of whose 
 interior angles is 2520. 
 
 126. A regular polygon has 10 sides. How many degrees in 
 each interior angle ? 
 
 127. One interior angle of a regular polygon contains 168. 
 How many sides has the polygon ? 
 
 168 2 n - 4 
 
 HINT. = , etc. 
 
 90 n 
 
66 PLANE GEOMETRY 
 
 Theorem 33 
 
 127. The sum, 8, of the exterior angles of any convex 
 polygon, counting one at each vertex, is four right angles. 
 
 Given the convex polygon ABCDE having n sides. 
 To prove that the sum of the exterior angles, one at each vertex, 
 is 4 rt. A. 
 
 Proof 
 
 1. Extend the sides, forming one 1. 68. 
 exterior angle at each vertex. 
 
 2. The sum of the exterior angle 2. 40. 
 and the interior angle at each 
 
 vertex is 2 rt. A For ex- 
 ample, Zl + Z7 = 2 rt. A. 
 
 3. Therefore the sum of the 3. Why ? 
 n interior angles and the n 
 
 exterior angles is 2 n rt. A. 
 
 4. But the sum of the interior 4. 125. 
 angles alone is (2 n 4) rt. A. 
 
 5. Therefore the sum, S, of the 5. 51. 
 exterior angles of any polygon, 
 expressed in right angles, is 
 S=2fc-(2tt-4) = 4. 
 
 EXERCISES 
 
 128. Is Theorem 33 true for reentrant polygons ? Explain. 
 
 129. How many degrees are there in each exterior angle of a 
 regular octagon ? a regular decagon ? 
 
BOOK I 67 
 
 130. Find the number of sides of a polygon if the sum of its 
 exterior angles equals the sum of its interior angles. 
 
 HINTS. Let n be the number of sides of the polygon. Then, by 125 
 and 127, 4 = 2 n - 4. 
 
 131. Find the number of sides of a polygon if the sum of its 
 exterior angles is twice the sum of its interior angles. 
 
 132. Find the number of sides of a polygon if the sum of its 
 interior angles is twice the sum of its exterior angles. 
 
 Theorem 34 
 
 128. If two lines are parallel to a third line, the two 
 lines are parallel to each other. 
 
 A B 
 
 Given the lines AB, CD, and XY in the same plane, with AB 
 and CD parallel to XY. 
 
 To prove that AB is II to CD. 
 
 Proof 
 
 1. The lines AB and CD either 1. No other possibility exists. 
 meet or do not meet. 
 
 2. If they meet, as at A', there 2. Hypothesis, 
 will then be two lines through 
 
 K II to AT. 
 
 3. But this is impossible. 3. 45. Postulate V. Through 
 
 a given point outside a line, 
 one line parallel to it exists, 
 and only one. 
 
 4. AB and CD cannot meet. 4. Remaining alternative from 
 
 statement 1. 
 
 5. AD is II to CD. 5. 43. 
 
68 PLANE GEOMETRY 
 
 Theorem 35 
 
 129. If three or more parallels intercept equal seg- 
 ments on one transversal, they intercept equal segments 
 on any other transversal. 
 
 / L 
 
 Given the parallels AF, BG y CH, and DI intercepting the equal 
 segments AB, BC, and CD on the transversal AD and intercept- 
 ing the segments FG, GH, and HI on another transversal. 
 
 To prove that FG= GH = HI. 
 
 Proof 
 
 1. Through A, B, and C draw Us 1. 45. 
 to FIj meeting BG, CH, and 
 
 DI in K, 7?, and L respectively. 
 
 2. Then AK, BU, and CL are II 2. 128. 
 lines. 
 
 3. In &ABK, BCR, and ODL, - 3. 59. 
 
 Z6 = Z5 = Z4. 
 
 4. Also Zl = Z2 = Z3. 4. 59. 
 
 5. AB = BC CD. 5. Hypothesis. 
 
 6. Therefore A ABK, BCR, and 6. Why ? 
 CJDL are congruent. 
 
 7. AK = BR = CL. 7. Why ? 
 
 8. But AK = FG, BR = GH, 8. 85. 
 Mild CL = HL 
 
 9. Therefore FG = GH = ///. 9. 32. 
 
BOOK I 69 
 
 130. Corollary. -If a line bisects one side of a triangle and 
 is parallel to another side, it bisects the third side. 
 
 HINTS. Let K be the mid-point of 
 A C and let KR be II to AB. Draw LM 
 through C parallel to AB. Then LM, 
 KR, and AB are II lines. Why? There- 
 fore CR = RB. Why ? 
 
 QUERY. If a line is parallel to one of two parallels, it is parallel 
 to the other. Contrast this statement with the statement of Theorem 34. 
 
 Theorem 36 
 
 131. The line which joins the mid-points of two sides of 
 a triangle is parallel to the third side and equal to one 
 half of it. c 
 
 ALB 
 
 Given the triangle ABC in which the line KR joins the mid- 
 points of AC and BC. 
 
 To prove that KR is II to AB and = 
 
 Proof 
 
 1. Draw KM II to AB. 1. 45. 
 
 2. M is the mid-point of CB. 2. 130. 
 
 3. KM and KR coincide. 3. Why ? 
 
 4. KR is II to AB. 4. Why? 
 
 5. Draw RL II to CA. 5. 45. 
 
 6. L is the mid-point of AB, and 6. 130. 
 
 AL=BL = AB. 
 
 7. ALRK is a parallelogram, 7. Why ? 
 
 8. KR = AL. 8. 85. 
 
 9. KH = AB. 9. 65. 
 
70 PLANE GEOMETRY 
 
 EXERCISES 
 
 133. Find the number of sides of a polygon if the stfm of its inte- 
 rior angles exceeds the sum of its exterior angles by 34 right angles. 
 
 134. Find the number of sides of a polygon if the sum of its 
 interior angles exceeds the sum. of its exterior angles by 1260. 
 
 135. Find the number of sides of a polygon if it is regular 
 and one exterior angle is one seventeenth of one interior angle. 
 
 136. ABCD is a parallelogram. AD and EC are divided into 
 five equal parts, and the corresponding points of division are 
 joined. Show that these lines divide the diagonals of the paral- 
 lelogram into five equal parts. 
 
 137. If K is any point on AB of the triangle ABC and R is 
 the mid-point of AK, L the mid-point of AC, and M the mid-point 
 of CK, then KRLM is a parallelogram. 
 
 138. The lines joining the mid-points of the sides of a triangle 
 divide it into four congruent triangles. 
 
 139. The line bisecting two sides of a triangle bisects all lines 
 drawn to the third side from the opposite vertex. 
 
 132. Bases of a trapezoid. The bases of a trapezoid are the 
 
 two parallel sides. 
 
 Theorem 37 
 
 133. The line joining the mid-points of the nonparallel 
 sides of a trapezoid is parallel to the bases and equal to half 
 their sum. 
 
 A B 
 
 Given the trapezoid ABCD with KR joining the mid-points of 
 the nonparallel sides AD and CB. 
 
 To prove that KR is II to AB and DC and = (AB + DC). 
 
BOOK I 71 
 
 Proof 
 
 1. Draw KM II to AB. 1. 45. 
 
 2. CM=MB. 2. 129. 
 
 3. M and R coincide. 3. Why ? 
 
 4. KR coincides with KM. 4. Why ? 
 
 5. ## is II to ^J3. 5. Why ? 
 
 6. # is H to DC. 6. Why ? 
 
 7. Draw Z>, cutting KR at Z. 7. 19. 
 
 8. L is the mid-point of BD. 8. 130. 
 
 9. KL = \AE, 9. 131. 
 
 10. LR = DC. 10. Why? 
 
 11. KL + LR = i ,45 + J- DC. 11. Why ? 
 
 12. # = | (.45 + - DC). 12. Why ? 
 
 EXEKCISES 
 
 140. If a line bisects one nonparallel side of a trapezoid 
 'and is parallel to one base, it bisects the other nonparallel 
 
 side also. 
 
 141. The line joining the mid-points of two adjacent sides of 
 a quadrilateral is equal to one half of one of the diagonals of the 
 quadrilateral. 
 
 134. Trisection. If a magnitude is divided into three equal 
 
 parts, it is said to be trisected. . . . . 
 
 A K R B 
 
 If AK = KR = RB, AB is tri- 
 sected, K and R being the points of 
 trisection. 
 
 Let AM be the median of the tri- 
 angle ABC, with K and R the points 
 of trisection. Then R, the point 
 nearest the side to which the me- B M C 
 
 dian is drawn, is called the basal 
 point of trisection, and K is called the vertical point of trisection. 
 
72 PLANE GEOMETRY 
 
 135. Inequalities. In a number of theorems now to be 
 proved, inequalities occur. To the handling of inequalities, 
 either alone or in connection with equations, certain laws of 
 .operation (axioms) apply. 
 
 The signs of inequality are >, which is read " is greater than," 
 and <, which is read "is less than." 
 
 136. Axiom VII. The whole is greater than any of its parts. 
 
 137. Axiom VIII. If the first of three magnitudes is greater 
 than the second and the second is greater than the third, the 
 first is greater than the third. 
 
 In algebraic symbols this axiom states that if a > I and I > r, 
 then a > c. 
 
 138. Order of inequalities. Two inequalities are said to 
 exist in the same order if the left member in each is the 
 greater member or the less member, as the case may be. 
 
 Thus 9 > 7 and 6 > 3 exist in the same order ; also x < m and 
 c < a exist in the same order. 
 
 Two inequalities are said to be in the reverse order if the 
 left member of one is its greater member and the right mem- 
 ber of the other is its greater member, or vice versa. 
 
 Thus the inequalities 7 > 3 and 8 < 10 are in reverse order. 
 
 139. Axiom IX. If the same number, positive or negative, 
 is added to or subtracted from each member of an inequality, 
 the results are unequal in the same order. 
 
 140. Axiom X. If both members of an inequality are multi- 
 plied or divided by the same positive number, tlie results are 
 unequal in the same order. 
 
 If a > 1) and x is positive, Axioms IX and X may be stated 
 in algebraic symbols thus : a -f- x > b + x, a x > b x, ax > bx, 
 
 x x 
 
BOOK I 73 
 
 141. Axiom XL If the corresponding members of two or more 
 inequalities which are in the same order are added, the sums are 
 unequal in the same order. 
 
 Thus, if a > x and b > y, then a -f b > x -f y. 
 
 142. Axiom XII. If unequals are subtracted from equals, the 
 results are unequal in the reverse order. 
 
 Thus, if a > b, then x a < x b, while if c < d, x c > x d. 
 
 EXERCISE 142. Write down one or more numerical inequalities 
 (such as 8>3) and test the truth of Axioms VIII to XII. 
 
 QUERY 1. If the members of an equation are subtracted from the 
 corresponding members of an inequality, what is the form of the result? 
 In what order is it? 
 
 QUERY 2. If the members of an inequality are subtracted from the 
 corresponding members of an equation, what is the result? In what 
 order is it ? 
 
 QUERY 3. In the inequality a + x > b, may k be substituted for x if 
 k = x ( t 
 
 QUERY 4. In an inequality may any term be replaced by its equal 
 without making any other change ? Illustrate. 
 
 QUERY 5. If a number is added to the greater member of an in- 
 equality, what is the result? If added to the less member? 
 
 QUERY 6. Change added to to subtracted from in Query 5 and answer. 
 
 QUERY 7. If in Axiom XII subtracted from is changed to added to, what 
 other change must be made to make the resulting statement true ? 
 
 QUERY 8. May a term be transposed from one member of an in- 
 equality to the other without destroying the inequality ? 
 
 QUERY 9. May all the terms in each member of an equation be 
 transposed to the opposite side of the sign of equality ? 
 
 QUERY 10. In 7 3 + 8>4 2+1 transpose every term and inspect 
 the result. 
 
 QUERY 11. What conclusion may be drawn from the result in 
 Query 10? 
 
 QUERY 12. If x + 8 >9, show that x>l. 
 
 QUERY 13. If 2 x 6 >4, show that x>5. 
 
 QUERY 14. If - + 5 <6, show that x<2. 
 
74 PLANE GEOMETRY 
 
 Theorem 38 
 
 143. The medians of a triangle are concurrent in the 
 basal point of trisection of each. 
 
 Given the triangle ABC with the medians AF and CH, and G 
 the mid-point of AC. 
 
 To prove that 
 
 (I) AF and CH intersect at some point 0. 
 (II) AF= 3 OF and CH=3 OH. 
 (Ill) The third median BG passes through O, making BG = & OG. 
 
 Proof 
 
 (I) 
 
 1. Z BA C +^BCA < 2 rt. A. 1. 66. 
 
 2. Z 1< Z 4 C and 2. 136. 
 
 3. Zl-fZ2<Z7L4C + Z.SC,4. 3. 141. 
 
 4. Z 1 + Z2 < 2 rt. Zs. 4. 137. 
 
 5. AF and C// are not II and 
 must meet at some point O. 
 
 (II) 
 
 1. Draw LA, GR, FK, and MB II 1. 45. 
 to CH, points R and K being 
 
 on ,4. 
 
 2. Since ^G = C, AR = ##. 2. 130. 
 In like manner, J5/C = KH. 
 
BOOK I 75 
 
 3. AH= HB. 3. Hypothesis. 
 
 4. AR = RH = HK = KB. 4. 39. 
 
 5. As AR = RH=HK, 5. 129. 
 
 AP = PO = OF. 
 Then AF = 30F. 
 
 6. In like manner it can be 6. "Why ? 
 proved that CH = 30H. 
 
 (Ill) 
 
 This proof shows that O is the basal point of trisection of 
 CH and AF. In the same way EG and AF can be proved to inter- 
 sect in their basal point of trisection. But each median of a triangle, 
 as AF, has only one basal point of trisection. Therefore the three 
 medians of a triangle intersect in that point. 
 
 EXERCISES ON MEDIANS AND MID-POINTS 
 
 143. A BCD is a parallelogram. The diagonals intersect in R. 
 The line DK bisects AB at K and cuts AC in L. Prove that DL 
 is twice LK and LR equals one sixth of A C. 
 
 144. If K is the middle point of the side BC of the triangle 
 ABC and BR and CL are perpendiculars from B and C respec- 
 tively to AK, produced if necessary, prove that BR equals (!L. 
 
 145. If a median of a triangle is perpendicular to one side, the 
 triangle is isosceles. 
 
 146. If two medians of a triangle are equal, the triangle is 
 isosceles. 
 
 147. The side BC of the triangle ABC is produced to the point 
 K. The angle ACS is bisected by the line CR, which meets AB 
 at R. A line is drawn through R parallel to BC, meeting AC at 
 L and the line bisecting the exterior angle ACK at G. Show 
 that R L equals LG. 
 
 148. In a right triangle the median from the vertex of the 
 right angle is one half the hypotenuse. 
 
 149. State the converse of the theorem of the preceding exercise 
 and, if it is true, prove it. 
 
76 PLANE GEOMETKY 
 
 150. BC is the base of the isosceles triangle ABC. The side 
 BA is produced its own length to K, and KC is drawn. Prove 
 that the triangle BCK is a right triangle. 
 
 HINT. Xote the equal angles in the figure and then consider the 
 sum of the angles of the triangle BCK. 
 
 151. Draw KR from K on AC of triangle ABC to R on BC so 
 that AK is one fourth of AC smdBR is one fourth ofBC. Prove 
 that KR is three fourths of AB. 
 
 152. In the triangle ABC the line KR parallel to AB joins K 
 on AC to R on BC and is equal to two fifths of AB. Prove AK 
 equal to three fifths of A C. 
 
 153. If lines be drawn from any vertex of a parallelogram to the 
 mid-points of the opposite sides, they will divide the diagonal which 
 they intersect into three equal parts. 
 
 154. ABC is a triangle. K is the 
 mid-point of BC and R the mid-point 
 of AK. If BR produced meets AC in 
 
 L, prove AL equal to one third of AC. A B 
 
 155. The lines joining the mid-points of the adjacent sides of 
 a quadrilateral form a parallelogram. 
 
 156. The lines joining the mid-points of the opposite sides of 
 a quadrilateral bisect each other. 
 
 157. K is taken on AB of the parallelogram ABCD so that AK 
 is one fifth of AB. The lines DK and AC intersect at R. Prove 
 that AR is one sixth of AC. 
 
 144. Corollary to 69. An exterior angle of a triangle is 
 greater than either opposite interior angle. 
 
 HINTS. Z 1 = Z A + Z C. Therefore Z 1 > /.A or 
 ZC, by Axiom VII, 136. 
 
 EXERCISE 158. In the triangle ABC the side AC equals the 
 side BC. AK is drawn to any point K on BC. Prove that the 
 angle AKC is greater than the angle CAB. 
 
BOOK I 
 Theorem 39 
 
 77 
 
 145. If two sides of a triangle are unequal, the angles 
 opposite them are unequal and the greater angle lies oppo- 
 site ike greater side. 
 
 A B 
 
 Given the triangle ABC with CB greater than CA. 
 To prove that Z CA 13 > Z.B. 
 
 Proof 
 
 1. On CB lay off CK equal to CA 1. 
 
 19. 
 
 and draw A K. 
 2. K lies between C and 7). 
 
 4. CK=CA. 
 
 5. Z1 = Z2. 
 
 7. Z1>Z7J. 
 
 8. Therefore Z.CAB>Z.B. 
 
 2. CB is given greater than CA. 
 
 3. 136. 
 
 4. Construction. 
 
 5. 29. 
 
 6. 144. 
 
 7. 65. 
 
 8. 137. 
 
 EXERCISES 
 
 159. In the triangle ABC the side A C equals BC. AK is drawn 
 to any point K on BC. Prove by use of 145 that the angle CKA 
 is greater than the angle KA C. 
 
 160. If the diagonals of a quadrilateral are unequal and bisect 
 each other at right angles, the figure is a rhombus. 
 
 HINTS. Let A BCD be the quadrilateral, K the point of intersection 
 of the diagonals, and A C greater than BD. Prove that ABCD is a 
 parallelogram with equal sides. Then, in the triangle A BK, show that 
 the angle KB A is greater than the angle KAB. 
 
78 PLANE GEOMETRY 
 
 146. Postulate VI. Any side of a triangle is less than the sum 
 of the other two sides. ', 
 
 A still broader and more usual statement is: A straight 
 line is the shortest line between two points. 
 
 Theorem 40 
 
 147. The sum of any two sides of a triangle is greater 
 than the sum of two lines drawn from a point within it 
 to the extremities of the third side. 
 
 A B 
 
 Given the triangle ABC with the lines KA and KB drawn from 
 the point K within the triangle to the extremities of AB. 
 
 To prove that AC+CB>KA + KB. 
 
 Proof 
 
 Extend A K to meet CB in some point L 
 
 1. AC + CL>AK+ KL. (1) 1. 146. 
 
 2. Also KL + LB> KB. (2) 2. 146. 
 S.AC+CL + LB + KL > AK 3. (1) + (2), 141. 
 
 + KL + KB. 
 
 4. AC+CL+LB>AK+KB. (3) 4. 139. 
 
 5. CL + LB = CB. (4) 5. Why ? 
 
 6. AC + CB>AK + KB. 6. From (4) and (3), by 65. 
 
 EXERCISES 
 
 161. In a quadrilateral any side is less than the sum of the 
 other three sides. 
 
 162. ABODE is a pentagon. Draw CE and prove that the 
 perimeter of ABCE is less than that of the pentagon. 
 
BOOK I 79 
 
 163. If the lines AB and CD intersect, then the sum of AB 
 and CD is greater than the sum of AC and DB. 
 
 164. If from a point within a triangle lines are drawn to the 
 vertices, their sum is greater than half the sum of the sides of 
 the triangle. 
 
 HINT. Study the figure to obtain three inequalities which, if their 
 corresponding members be added, will give a suggestive result. 
 
 Theorem 41 
 
 148. If two angles of a triangle are unequal, the sides 
 opposite them are unequal and the greater side lies oppo- 
 site the greater angle. 
 
 C 
 
 A B 
 
 Given the triangle ABC in which the angle CAB is greater than 
 the angle B. 
 
 To prove that BOA C. 
 
 Proof 
 
 Suppose All to be the line through A such that Z.RAB (or 
 
 1. Z/l>Z7J. 1. Hypothesis. 
 
 2. Z.4>Z1. 2. 65. 
 
 3. Therefore AR lies between 3. 136. 
 AB and AC and must inter- 
 
 sect BC in some point, K. 
 
 4. Now in AABK, 4. 99. 
 
 AK=BK. 
 
 5. But A K + KC>A C. 5. 146. 
 
 6. BK + KC(=BC)>AC. 6. -65. 
 
80 
 
 PLANE GEOMETRY 
 
 149. Corollary. The perpendicular from a point outside 
 a straight line is the shortest line from the point* to the line. 
 
 HINTS. Let A B be any line, K any point out- 
 side AB, KR a JL to AB, and KL any other 
 line from K to AB. Compare Zl with Z2. 
 Conclusion ? 
 
 A. 
 
 
 QUERY. How does the hypotenuse of a 
 right- triangle compare with each of the other two sides ? 
 
 R B 
 
 Explain. 
 
 EXERCISES 
 
 165. In the triangle ABC the side AC equals the side AB. If 
 AK is any line from A to BC, prove that AK is less than AB. 
 
 166. A rectangle and a parallelogram have the same base and 
 equal altitudes. Which has the greater perimeter ? Prove it. 
 
 Theorem 42 
 
 150. If two triangles have two sides of one equal 
 'respectively to two sides of the other and the included 
 angle of the first greater than the included angle of the 
 second, the third side of the first is greater than the third 
 side of the second. 
 
 Given the triangles ABC and FGH, in which AB equals FG, 
 BC equals GH, and the angle ABC is greater than the angle G. 
 To prove that AC> FH. 
 
BOOK I 
 
 81 
 
 Proof 
 
 2. 
 3. 
 
 4. 
 5. 
 
 6. 
 
 7. 
 8. 
 
 9. 
 10. 
 11. 
 12. 
 
 Superpose AFGH upon 
 A ABC so that the side GH 
 will fall on its equal BC, 
 the point G falling at B, the 
 point H at (7, and the point 
 F at some point K. 
 Line GF will fall within the 
 angle A BC in the position BK. 
 Let the line BR bisect /.ABK 
 and intersect A C at It. 
 Draw #A\ 
 In &ABR aui&KBR, 
 
 BK = BR. 
 
 Z1=Z2. 
 
 .4 = G'^, or its equal BK. 
 AABR and AKBR are con- 
 gruent. 
 
 1. 20. 
 
 C.R + /L4 > CYC 
 AOFH. 
 
 2. By hypothesis, 
 
 Z6' 
 
 3. 28. 
 
 4. 19, Postulate I. 
 
 5. Why ? 
 
 6. Construction. 
 
 7. Hypothesis. 
 
 8. 25. 
 
 9. 27. 
 
 10. 146. 
 
 11. 65. 
 
 12. Why ? 
 
 EXERCISES 
 
 167. If a parallelogram is not a rectangle, its diagonals are 
 unequal. 
 
 168. Referring to the figure of Ex. 80, p. 51, suppose that the 
 ray of light from A to the mirror was reflected to E from some 
 other point than L. Show that the distance the light travels would 
 be longer than ALE. 
 
 169. Two towns situated certain distances from a straight river 
 bank each wish to secure a water supply from a common pump- 
 ing station on the bank of the river, from which a pipe line is 
 to run to each town. Where should the station be placed so that 
 the length of the pipe lines may be the shortest possible ? 
 
82 PLANE GEOMETRY 
 
 170. If two oblique lines drawn from a point to a straight 
 line meet the line at unequal distances from the.- foot of the 
 perpendicular drawn from the point to the 
 
 line, they are unequal, and the more remote 
 is the greater. 
 
 HINTS. Extend A C to R, making CR = A C, 
 and draw R K and RB. Then AB + RB>AK 
 + RK. Why? Conclusion? 
 
 171. If from a point within a triangle \ 1 ^ 
 lines be drawn to the vertices, their sum 
 
 is less than the sum of the three sides of the triangle. 
 
 HINT. Use Theorem 40 and Axiom XI ; then use Axiom X. 
 
 172. In the triangle ABC the median AK is drawn forming an 
 acute angle A KB. Which is the greater, AB or AC? 
 
 Theorem 43 
 
 151. If two triangles have two sides of one equal respec- 
 tively to two sides of the other and the third side of the first 
 greater than the third side of the second, the included angle 
 of the first is greater than the included angle of the second. 
 
 G 
 
 Given the triangles ABC and FGH, in which AC equals FH, 
 EC equals GH, and AB is greater than FG. 
 
 To prove that 
 
BOOK I 
 
 83 
 
 Proof 
 
 Let us suppose 
 
 (1) 
 (2) 
 
 or ZC>Z//. (3) 
 
 1. If we suppose (1) is true, then 
 it follows that AB < FG. 
 
 2. But AB>FG. 
 
 3. Hence (1) is not true. 
 
 4. If we suppose (2) is true, 
 it follows that A ABC is con- 
 gruent to AFGH. 
 
 5. Then AB = FG. 
 
 6. But AB>FG. 
 
 1. Hence (2) is not true. 
 
 8. Now one of the statements 
 (1), (2), or (3) must be true. 
 
 9. Hence (3) is true. 
 
 No other possibilities exist. 
 
 1. 150. 
 
 2. Given. 
 
 3. A supposition which by cor- 
 rect reasoning leads to a 
 false conclusion is a false 
 supposition. 
 
 4. 25 
 
 5. Why ? 
 
 6. Given. 
 
 7. A supposition is false which 
 makes it possible to prove 
 that AB = FG. 
 
 8. There are no further alter- 
 natives. 
 
 9. (1) and (2) have been proved 
 untrue. 
 
 152. Proof by reductio ad absurdum. The proof of the pre- 
 ceding theorem is an example of a method of proof which has 
 long had the Latin name . reductio ad absurdum. Frequently 
 such a proof is called indirect. 
 
 The two names, however, are not interchangeable. For example, 
 in Theorem 36 line KR was proved parallel to AB by showing that 
 it coincided with KM. This proof is indirect but is not a proof 
 by reductio ad absurdum. An indirect proof consists in showing 
 that any figure which fulfills the requirements of the theorem must 
 necessarily be the one mentioned in the hypothesis. 
 
84 PLANE GEOMETEY 
 
 One essential feature of a proof reductio ad absurdum consists 
 in making all the assumptions possible under th<e hypothesis 
 (usually three) and then disproving the truth of all but one. 
 Another essential feature of the method is the manner of dis- 
 proving each of the false assumptions made. This consists in 
 reasoning from the assumption to a conclusion which is seen to be 
 false because it contradicts some statement of the hypothesis, some 
 obvious fact, or some previously established theorem. Of course 
 this result disproves the truth of the assumption, for if correct 
 reasoning leads from an assumption to a conclusion which is untrue, 
 then the assumption on which the reasoning is based is untrue. 
 
 Such a method is conclusive, if we make all the assumptions 
 both true and false which are possible under the given hypothesis. 
 Of these we must know beforehand that one mustjbe true and only 
 one can be true. If, then, we prove all of the assumptions false 
 but one, that one must be true. 
 
 An indirect proof has certain advantages. Often it is shorter 
 than a direct proof and occasionally it is easier to obtain. While 
 a direct proof for Theorem 43 is possible, it would be longer than 
 that given in the text, and probably the only way of solving 
 Ex. 102 by the theorems of Book I is a proof by reductio ad 
 absurdum. 
 
 QUERY 1. What theorems, if any, before Theorem 43 are proved by 
 reductio ad absurdum? 
 
 QUERY 2. What theorems of Book I have been proved by super- 
 position? In what other theorems has Postulate II been used? 
 
 QUERY 3 What is meant by reasoning in a circle ? Give an. example. 
 
 EXERCISES ON INEQUALITIES 
 
 173. If the diagonals of a parallelogram are unequal, it is not 
 a rectangle. 
 
 174. If two sides of a triangle are unequal and the median to the 
 third side is drawn, the angles formed with the base will be unequal. 
 
 175. Prove Theorem 41 by the indirect method. 
 
 HINT. AC equals BC, is less than BC, or is greater than BC. Then 
 use Theorem 39. 
 
BOOK I 85 
 
 176. In the triangle ABC, if K is any point in AB and R any 
 point in BC, prove that AB plus BC is greater than AK plus KR 
 plus EC. 
 
 177. The difference between two sides of a triangle is less than 
 the third side. 
 
 178. If one side of a triangle is divided into two parts by a 
 perpendicular from the opposite vertex, each part of that side is 
 less than the adjacent side of the triangle. 
 
 179. AB and BC are equal sides of the triangle ABC. If BC 
 is produced to the point K and KA is drawn, the angle BAK is 
 greater than the angle BKA . 
 
 180. Any point (except the vertex) in one of the equal sides 
 of an isosceles triangle is unequally distant from the extremities 
 of the base. 
 
 181. ABC is a triangle in which KB and KC bisect the angles 
 B and C respectively. Show that if AB is greater than AC, KB 
 is greater than KC. 
 
 182. In the quadrilateral ABCD, AD is the longest and BC the 
 shortest side. Prove that the angle B is greater than the angle D 
 and that the angle C is greater than the angle A. 
 
 183. If in the triangle ABC the angle A is bisected by a line 
 meeting BC at A", AB is greater than BK and .4 C is greater than CK. 
 
 HINT. Compare the angles at AT with the two at ^4. 
 
 184. If lines are drawn from any point within a triangle to 
 the extremities of one side, they will form an angle greater 
 than the angle formed by the other two 
 
 sides. 
 
 185. AK is a median of triangle ABC. 
 Prove that AB plus AC is greater than "\ \* /'' 
 
 twice A K. ;>r' 
 
 HINT. Extend A K to R, making KR equal 
 
 to AK, and draw BR and CR. Then study the triangle ABR, or the 
 triangle .4 CE. 
 
86 PLANE GEOMETRY 
 
 186. A BCD is a parallelogram. K, R, L, and M are points on 
 AB, EC, CD, and DA respectively such that AK equals CL and 
 BR equals DM. Prove that KRLM is a parallelogram. 
 
 153. Geometric fallacies. Many fallacious geometric proofs 
 exist in which the error is fairly well concealed. One of the 
 best known of these is the proof that 
 any triangle is isosceles. 
 
 Let A ABC be any triangle. Draw 
 CL the bisector of Z.ACB and let 
 KR be _L to AB at its mid-point R. 
 Call the intersection of these two 
 lines 0. Then draw Oil and OM _L C~~ ~~f 
 
 respectively to AC and B C. 
 
 Then A OCM is congruent to A OCIL 50 
 
 Therefore HO = OM, 
 
 and IIC^MC. (1) 27 
 
 Now AO = BO. 116 
 
 Therefore A A OH = ABOM. 97 
 
 Therefore All = BM. (2) 27 
 
 (1) + (2), A n -4- 7/C = ZM/ + Jl/C, vsAC = BC. 
 
 Therefore any triangle is isosceles. 
 QUERY. Where is the error in the "foregoing ? 
 
 HISTORICAL NOTE. The earliest Greek textbook on mathematics 
 was written by Hippocrates of Chios about 440 n.c. There were many 
 other writers of elements of geometry before Euclid, but his work 
 superseded theirs and held the field alone. Nearly seven centuries 
 after Euclid, Theon of Alexandria, who was the father of Hypatia, 
 the heroine of Kingsley's novel, prepared an edition of the "Elements " of 
 Euclid for use in his own classes. On the destruction of Alexandria 
 and its University, A.D. 640, some of the learning centered there was 
 preserved in various Syrian cities on the east coast of the Mediter- 
 ranean. There the Arabs of Bagdad became acquainted with many of 
 the products of Greek culture. The translation of Euclid's "Elements " 
 into Arabic was begun in the reign of Harun-al-Rashid (786-809) 
 
BOOK I 87 
 
 and completed under Al Mamun (813-833). After the Arabs had con- 
 quered Spain in 711, among other treasures they brought Euclid's 
 " Elements " with them, and here for over three centuries they guarded it 
 jealously. About the year 1120, however, Adelhard of Bath, an English 
 monk studying in Spain, succeeded in translating the " Elements " 
 from Arabic into Latin. Courses in Euclid's " Elements " soon found a 
 place in many universities of Europe, especially in those of Germany 
 and Italy, but it was not until after the invention of printing that 
 the >study of Euclid became common in the higher schools. Ten years 
 before Columbus discovered America a Latin version of the " Elements " 
 was published in Venice. By the year 1516 five different editions had 
 been printed. One of these was at Paris. When Constantinople fell 
 into the hands of the Turks in 1453, many educated Greeks fled to Italy, 
 taking with them numerous valuable manuscripts of Greek literature 
 Euclid's " Elements " among them. Thus it happened that the Paris 
 edition which has been mentioned contained the commentaries of Theoii 
 and Hypsicles. The first translation of the "Elements" into English 
 was made in 1570 by Sir Henry Billingsley. 
 
 The Germans modified the " Elements " considerably, as did also the 
 French. The English, however, have adhered to Euclid's " Elements " 
 very closely. In this respect the United States has followed the lead of 
 the English. 
 
 Euclid's " Elements " was first used at Harvard about 1726 and at 
 Yale about six years later. 
 
 The axiom given in 45 is essentially the famous parallel axiom of 
 Euclid. Many mathematicians have attempted to regard this axiom as 
 a theorem and obtain a proof of it based on Euclid's other axioms. 
 Some, Legendre among them, thought they had obtained valid proofs, 
 but all the attempts at proof, when carefully examined, were found to 
 assume at some point the truth of the axiom itself. Indeed, the mere 
 fact that Euclid was acute enough not to place this axiom with the 
 others is regarded by some as enough evidence in itself to stamp him 
 as a great mathematician. 
 
 At the present time no well-informed person attempts to prove this 
 axiom. Correct notions in regard to it were arrived at independently by 
 three men : Gauss (1777-1855), Bolyai (1802-1860), and Lobachevsky 
 (1793-1856). Lobachevsky set himself, not the problem of proving the 
 axiom, but of finding out what would result if instead of it he assumed 
 the following : Through a point outside a line more than one parallel to 
 
88 PLANE GEOMETRY 
 
 the line can be drawn. Leaving the other axioms of Euclid unchanged, 
 he was then able to develop a perfectly consistent system of geometry 
 very different from that of Euclid. His system is consistent because 
 it leads to no contradictory conclusions to no two theorems which 
 make precisely opposite assertions. It does, however, lead to theorems 
 which contradict theorems of Euclidean geometry. For example, one 
 theorem of non-Euclidean geometry asserts : The sum of the angles 
 of any triangle is less than two right angles (see 66). Moreover, in 
 Lobachevsky's geometry similar triangles do not exist. 
 
 Later Riemann replaced Euclid's axiom by the following : Through 
 a point outside a line no parallel to the line can be drawn. He then 
 developed a system of geometry different from that of both Euclid 
 and Lobachevsky. 
 
 All this may be somewhat confusing, but a proper conception of 
 what mathematics really is will help to clear up the difficulty which 
 exists. A distinguished American mathematician denned mathematics 
 as the science which draws necessary conclusions. This implies, of 
 course, a body of conclusions drawn from certain fundamental assump- 
 tions, which do not necessarily deal with numbers, as in arithmetic 
 and algebra, or with space, as in geometry. The simple fact is that 
 Lobachevsky changed one of Euclid's assumptions, and consequently 
 many of the necessary conclusions were different from those of Euclid. 
 
BOOK II 
 
 THE CIRCLE 
 
 154. Circle. A circle is a closed plane curve every point 
 of which is equally distant from a point in the plane of 
 the curve. 
 
 155. Center of a circle. The center of a circle is the point 
 in its plane which is equally distant from -every point of 
 the circle. 
 
 156. Radius. A radius of a circle is a straight line from 
 the center to the circle. 
 
 157. Diameter. A diameter is a straight line through the 
 center terminating in the circle at both ends. 
 
 158. Arc. An arc is any continuous portion of a circle. 
 
 In the adjacent figure the entire curved line 
 is a circle, C is the center, CK a radius, A B a 
 diameter, and KB an arc. ^i x >g 
 
 159. Circumference. The length of a circle 
 is the circumference. 
 
 Some of the more useful inferences from the preceding defini- 
 tions will now be stated. 
 
 160. A diameter is equal to the sum of two radii. 
 
 161. Equal circles are congruent. 
 
 162. All radii of the same circle or of equal circles are 
 equal. 
 
90 PLANE GEOMETRY 
 
 163. All diameters of the same circle or of equal circles 
 are equal. " 
 
 164. Circles having equal radii or equal diameters are 
 equal. 
 
 165. If the centers of two equal circles which lie in the 
 same plane be made to coincide, the circles will coincide. 
 
 166. Semicircle. A semicircle is an arc which is half the 
 circle. 
 
 167. Minor arc. An arc less than a semicircle is called a 
 minor arc. 
 
 168. Major arc. An arc greater than a semicircle is called 
 a major arc. 
 
 The word arc alone will be understood to 
 mean either a major or a minor arc. To avoid 
 all ambiguity it is customary to denote a 
 minor arc by two letters and a major arc by 
 three. Thus, in the adjacent figure the arc 
 AB denotes the minor arc and arc ACB denotes the major arc. 
 
 169. Central angle. A central angle is an angle whose 
 vertex is at the center of a circle and whose sides are two 
 of its radii. 
 
 170. Intercept and subtend. An angle is said to intercept 
 the arc cut off between its sides, and the arc is said to subtend 
 the angle. 
 
 Two parallel lines may also be spoken of as intercepting two 
 arcs of a circle. 
 
 NOTE. The use of the terms circle and circumference has long been 
 confusing. Both words have been used to designate the curve itself. 
 The word circle has been and still is used to denote both the curved 
 line itself and the portion of the plane bounded thereby. When the 
 latter is meant we shall use the phrase the area of the circle. 
 
BOOK II 91 
 
 Theorem l 
 
 171. In the same circle or in equal circles if two central 
 angles are equal, their intercepted arcs are equal. 
 
 Given the two equal circles B and F in which the central 
 angles B and F are equal and the intercepted arcs are AC and 
 EG respectively. 
 
 To prove that the arc A C equals the arc EG. 
 
 Proof 
 
 1. If Z.B and Z- F are equal and 1. By hypothesis Z.B = Z.F, and 
 in the same circle, let one 20. 
 
 angle rotate about the center 
 until the two coincide. 
 If Z.B and ZF are in differ- 
 ent circles, place the circle 
 with center F on the circle 
 with center B so that the 
 point F falls on the point B 
 and the sides of /.F fall on 
 the sides of /-B. 
 
 2. Then E falls on A and G falls 2. 162. 
 on C. 
 
 3. No point of arc GE can fall 3. 165. 
 outside or inside of arc CA. 
 
 4. Therefore arc EG coincides 4. Why ? 
 with arc A C and is equal to it. 
 
92 
 
 PLANE GEOMETBY 
 
 K t 
 
 172. Corollary 1. Any diameter of a circle bisects it. 
 
 Proof. Let B be the center and AC any 
 diameter. Then the straight angle ABC = the 
 straight angle ABC ; that is, Z 1 = Z 2. There- c 
 fore arc^lA'C = arc^A'C ( 171). 
 
 173. Corollary 2. In the same circle or 
 in equal circles, if two central angles are 
 unequal, the greater angle intercepts the 
 greater arc. 
 
 HINTS. Let Zl be greater than Z2. Lay off 
 ZABL equal to Zl. EL falls outside /.ABC. 
 Why? ArcAL = arc A72. Why? Conclusion? 
 
 EXERCISES 
 
 1. Two intersecting diameters divide a circle into two pairs of 
 equal arcs. 
 
 2. Two perpendicular diameters divide a circle into four 
 equal arcs. 
 
 Theorem 2 
 
 174. In the same circle or in equal circles, if two arcs 
 are equal., they subtend equal central angles. 
 
 Given two equal circles B and F in which the arc AC equals 
 the arc EG. 
 
 To prove that 
 
BOOK II 
 
 93 
 
 Proof 
 
 1. Place the circle with center F 1. 20 and 165. 
 on the circle with center B so 
 
 that FE will coincide with its 
 equal BA, the point F falling 
 on B, and the point E on the 
 point .4, and make the arc EG 
 fall on the arc A C. 
 
 2. The point G will fall on the 2. Hypothesis, 
 point C. 
 
 3. Therefore GF coincides with 3. 19. 
 ('II. 
 
 4. Therefore Z F = Z B. L Why ? 
 
 175. Corollary l. If a line bisects a circle, it is a diameter. 
 
 Proof. Let a straight line CA make arc 
 AKC = &YcARC. Now suppose the center B 
 is not on AC and draw BC and BA. Then 
 since arc CKA = arc CJIA, by 174 one angle 
 CBA = the other angle C5.4 ; that is, Z 1 = Z 2. 
 Therefore Z.CBA is a straight angle, and 
 hence the line CBA is a straight line; that 
 is, the center B must lie on AC. Therefore AC is a diameter. 
 
 176. Corollary 2. In the same circle or in equal circles, if two 
 arcs are unequal, the greater arc subtends the greater central angle. 
 
 HINT. Prove in a manner like that of 173. 
 
 177. Chord. A chord is a straight line whose 
 extremities are on the circle. 
 
 A chord is said to subtend an arc. Every chord, 
 except a diameter, subtends two unequal arcs, a 
 minor arc arid a major arc. If neither is specified, the minor arc 
 will be the one meant. 
 
 QUERY. Is a diameter a chord ? 
 
94 PLANE GEOMETRY 
 
 Theorem 3 
 
 178. In the same circle or in equal circles, if two arcs 
 are equal, their chords are equal. 
 
 Given, in the equal circles whose centers are respectively B 
 and G, the chords AC and FH, subtending the equal arcs AC and 
 FH respectively. 
 
 To prove that chord AC = chord FH. 
 
 Proof 
 
 1. Draw the radii AB, CB, FG, 1. 19. 
 and HG. 
 
 2. Arc A C = arc Fit. 2. Hypothesis. 
 
 3. In A ABC and FGH, 3. 174. 
 
 4. BA == GF and EC = GH. 4. 162. 
 
 5. A4CiscongruenttoAF//. 5. Why? 
 
 6. Therefore 6. Why ? 
 
 chords C = chord FH. 
 
 EXERCISES 
 
 3. If three diameters, AB, FG, and KR, divide a circle into six 
 equal arcs, prove that the six angles at the center contain 60 each. 
 
 4. A 9 K, B, and R are points in order on a circle with center C 
 such that arc AK equals arc BR. Prove that the angle KCR 
 equals the angle A CB. 
 
BOOK II 95 
 
 5. AB is a chord of a circle equal to its radius. Prove that 
 the arc AB is one sixth of the whole circumference. 
 
 6. If the arcs AB, BC, CD, and DE of the same circle are 
 equal, then the chords AC, BD, and CE are equal. 
 
 7. K, It, and L are three points on a circle such that arcs KR, 
 ILL, and LK are equal. The chords KR, RL, and LK are drawn. 
 Prove that the triangle KRL is equiangular. 
 
 Theorem 4 (Converse of Theorem 3) 
 
 179. In the same circle or in equal circles, if two chords 
 are equal, their subtended arcs are equal. 
 
 Given (using the figure of 1 78) the equal circles whose centers 
 are respectively B and G, and chord AC equal to chord FH. 
 
 To prove that arc AC = arc FH. 
 
 HINTS. A .1 B C is congruent to A FGH. Why ? 
 
 ^B = ZG. Why? 
 
 Arc A C = arc FH. 171 
 
 QUEJIY 1. If two circles are not equal, would equal chords subtend 
 arcs equal in length ? Explain. 
 
 QUERY 2. If two circles are not equal, would arcs equal in length 
 have equal chords? Explain. 
 
 EXERCISES 
 
 8. If chords AB, BC, CD, and DE are equal, then chords AC, 
 BD, and CE are equal. 
 
 9. A, K, B, and R are points in order on a circle such that the 
 chord AB equals the chord KR. Prove that the swcAK equals 
 the arc BR. 
 
 10. Two diameters of a circle are perpendicular to each other. 
 Prove that the chords joining their extremities form a square. 
 
96 PLANE GEOMETRY 
 
 Theorem 5 
 
 180. In the same circle or in equal circles, if two minor 
 arcs are unequal, the greater arc has the greater chord. 
 
 Given the equal circles whose centers are respectively C and 
 H y in which minor arc AB is greater than minor arc FG, and 
 chords AB and FG. 
 
 To prove that chord AB > chord FG. 
 
 Proof 
 
 1. Draw radii AC, BC, FJI, and 1. 19. 
 GH. 
 
 2. In A ABC and FGH, 2. Why ? 
 
 CA = HF, 
 and CB = IJG. 
 
 3. Z.O/.IL 3. 176. 
 
 4. Therefore 4. 150. 
 
 chord AB > chord FG. 
 
 EXERCISES 
 
 11. AK is a diameter and KB and KC are chords such that KB 
 lies between AK and KC. Prove that the chord K] is greater 
 than the chord KC. 
 
 12. A diameter of a circle is greater than any other chord of 
 the circle. 
 
BOOK II 
 
 9T 
 
 Theorem 6 (Converse of Theorem 5} 
 
 181. Li the same circle or in equal circles, if two chords 
 are unequal, the (jr eater chord has the greater minor arc. 
 
 Given the equal circles whose centers are respectively C and 
 H y with chord AB greater than chord FG. 
 
 To prove that 
 
 arcAB > arcFG. 
 
 Proof . 
 
 1. Draw radii AC, EC, FH, and 1. 19. 
 GH. 
 
 2. In A ABC and FGH, 2. Why ? 
 
 AC = FH. 
 
 3. EC = GH. 3. Why ? 
 
 4. Chord -4 B> chord -fC?. 4. Hypothesis. 
 
 5. Z.OZ.H. 5. 151. 
 
 6. Therefore arc . 4 > arc.ro. 6. 173. 
 
 EXERCISES 
 
 13. AB is a diameter and BR and EK are chords with BK the 
 greater. Prove chord AK is less than chord AR. 
 
 14. /? is the center and AB, BC, and CA are chords of a circle. 
 If ^ is greater than BC and C is greater than CA, prove 
 the angle ARC less than 120. 
 
98 PLANE GEOMETRY 
 
 Theorem 7 
 
 182. If a line passes through the center of a circle and is 
 perpendicular to a chord, it bisects the chord and the arcs 
 subtended ~by it. 
 
 Given LR passing through C, the center of the circle, and 
 perpendicular to chord AB at K. 
 
 To prove that AK=KB, arcAR=arcBR, and arcALarcBL. 
 
 Proof 
 
 1. Draw the radii AC and BC. 1. 19. 
 
 2. In the rt.&AKC and BK, 2. Why ? 
 CK = CK and CA = CB f 
 
 3. A^tfCiscongraenttoAB/tC'. 3. Why ? 
 
 4. A K = KB. 4. Why ? 
 
 5. Also Z1=Z2. 5. Why? 
 (5. arc A R = arc Rll. 6. 171. 
 
 7. Z ,4 CL = Z.BCL. 7. Why V 
 
 8. Therefore arc/li = arcZ. 8. 171. 
 
 EXERCISES 
 
 15. If a diameter bisects a chord, it is perpendicular to the 
 chord and bisects its arcs. 
 
 HINT. In a figure like that of 182, prove the angles at K equal. 
 
 16. If a diameter bisects an arc, it bisects the chord of the arc 
 at right angles. 
 
 17. If a line bisects a chord and one of its arcs, it is perpendicular 
 to the chord, passes through the center, and bisects the other arc. 
 
BOOK II 
 Theorem 8 
 
 99 
 
 133. In the same circle or in equal circles, if two chords 
 are equal, they are equally distant from the center. 
 
 Given the chord GF equal to the chord AB in circle C, and CH 
 and CK their respective distances from the center. 
 
 To prove that 
 
 CH=CK. 
 
 Proof 
 
 1. Draw radii CG and CA. 1. 19. 
 
 2. In .&AKC and GHC, 2. 120. 
 
 Z1 = Z2 = 1 rt.Z. 
 
 3. Chord AB = chord GF. 3. Hypothesis. 
 
 4. GH = i GF and AK = -J ,4 J3. 4. 152. 
 
 5. GH = AK. 5. Why? 
 
 6. CG = CA. 6. Why? 
 
 7. A ,4 ATMs congruent to A (77/C. 7. Why ? 
 
 8. Therefore C// = CK. 8. Why ? 
 
 EXERCISES 
 
 18. AB and ^4C are equal chords and AK is a diameter. Prove 
 that the angle BA K is equal to the angle CAK. 
 
 19. The arcs AB, BC, and CA of a circle are equal. Show that 
 the distance of the chords AB, BC, and CA from the center is 
 one half the radius. 
 
100 PLANE GEOMETRY 
 
 Theorem 9 
 
 184. In the same circle or in equal circles, if two chords 
 are equally distant from the center, they are equal. 
 
 Given (using the figure of 183) the chords AB and FG with 
 their respective distances from the center, CK and Cff, equal. 
 
 To prove that chord AB = chord FG. 
 HINT. Prove that A A CK is congruent to A GCH. 
 
 EXEECISES 
 
 20. Two chords from one extremity of a diameter making equal 
 angles with it are equal. 
 
 21. If the perpendiculars from the center upon two chords are 
 equal, the major arcs subtended by the chords are equal. 
 
 Theorem ID 
 
 185. In the same circle or in equal circles, if two chords are 
 unequal, the greater is at the less distance from the center. 
 
 Given in the circle C the chord AB greater than the chord 
 FG, and their respective distances from the center, CK and CH. 
 To prove that CK < CH. 
 
 Proof 
 
 1. Let EL be a chord equal to 1. 42. 
 the chord FG and let CR be 
 
 _L to EL. Join K and R. 
 
 2. KB = AB and BR = EL. 2. 182. 
 
BOOK II 101 
 
 3. Chord AB > chord FG or its 3. Hypothesis. . r , , 
 equal, chord EL. 
 
 4. KB>BR. 4. 140. .- . , - 
 
 5. Z 3 > Z 2. (1) 5. 145. 
 
 6. Z 3 H- Z 4 = Z 2 + Z 1. (2) 6. 40. 
 
 7. From (1) and (2), Z 4 < Z 1. 7. 142. 
 
 8. CK<CR. 8. 148. 
 
 9. But CJR = C7/. 9. 183. 
 10. CK<CH. 10. Why? 
 
 QUERY. Does Theorem 10 hold when one chord is a diameter? 
 
 EXERCISES 
 
 22. The arcs AB, BC, and CD of a circle are equal. K is the 
 mid-point of the arc AB, KR is perpendicular to the chord AB at 
 R, and CH is perpendicular to the chord BD at //. Prove that 
 CH is greater than KR. 
 
 23. AB is a, diameter, G the center of a circle, and AK and AR 
 chords such that K lies between B and 7?. The line GH is perpendic- 
 ular to AK at //, and GL is perpendicular to AR at Z. The line HL is 
 drawn. Prove that the angle LHG is greater than the angle GLH. 
 
 Theorem 11 
 
 186. In the same circle or- in equal circles, if two chords 
 are unequally distant from the center, the chord at the 
 greater distance is the less. 
 
 Given (using the figure of 185) chords FG and AB, with 
 distance CH greater than distance CK. 
 
 To prove that chord FG < chord AB. 
 
 HINTS. CR>CK. Why? 
 
 Z1>Z4. Why? 
 
 Z2<Z3. Why? 
 
 BR<RK,etc. Why? 
 
 QUERY. Does Theorem 11 hold when one chord passes through the 
 center ? 
 
102 PLANE GEOMETRY 
 
 EXERCISES 
 
 < 
 24.- AH is a:diameter, () the center, and BK and BR chords of 
 
 a- circle. 7, (> is perpendicular to BK at 7, and 077 is perpendicular 
 to 7272 at //. If OL is less than OH, prove that chord A K is less 
 than chord AR. 
 
 25. If in the same circle or in equal circles two chords are 
 unequally distant from the center, the chord at the greater dis- 
 tance subtends the less minor arc. 
 
 26. If two chords of a circle bisect each other, they are diameters. 
 
 EXERCISES ON CHORDS, ARCS, AND CENTRAL ANGLES 
 
 27. Two intersecting chords making equal angles with the 
 diameter passing through their point of intersection are equal. 
 
 28. If a diameter is drawn through a point within a circle 
 equidistant from two points upon it, the arc between the two 
 points is bisected. 
 
 29. If a circle is divided into six equal arcs, any chord joining 
 adjacent points of division is equal to the radius of the circle. 
 
 30. If a line bisects an arc and is perpendicular to the chord of 
 the arc, it will, if produced, pass through the center of the circle. 
 
 31. Two points, each equidistant from the ends of a chord, 
 determine a line passing through the center of the circle. 
 
 32. A is the center of a circle and K is a point outside. KRL 
 and KHG make equal angles with AK and cut the circle in R and 
 L and in 77 and O respectively. Prove chords 727, and HG equal. 
 
 33. B is the mid-point of the arc A C of a circle, and BK and BR 
 are drawn perpendicular to the radii OA and OC respectively. 
 Prove that BK equals BR. 
 
 34. If from the extremities of any diameter perpendiculars are 
 drawn upon any chord (produced if necessary), the feet of the 
 perpendiculars are equidistant from the center of the circle. 
 
 35. If two unequal chord* intersect in a circle, the greater 
 chord makes the less acute angle with the diameter through the 
 point of intersection. 
 
BOOK II 
 
 103 
 
 36. A and B are the centers of two circles which intersect in 
 K and R. KL is drawn to the mid-point of AB. A line through 
 K perpendicular to KL cuts one circle in H and the other in G. 
 Prove that the chord KH equals the chord KG. 
 
 187. Tangent. A tangent to a circle is a straight line 
 which, however far it may be produced, has only one point 
 in common with the circle. 
 
 If a straight line is tangent to a circle, 
 the circle is also tangent to the line. 
 
 The point common to the circle and to 
 the tangent is called the point of tangency K 
 or point of contact. 
 
 Thus, in the accompanying figure, KR is 
 the tangent and P the point of contact. 
 
 188. Circumscribed polygon. A polygon 
 whose sides are tangent to a circle is called 
 a circumscribed polygon. 
 
 189. Tangent from an outside point. A tangent to a circle 
 from an outside point means a line-segment whose extremities 
 are the outside point and the point of contact. 
 
 tangent p 
 
 In the above figure the line-segments AB and AK are tangents 
 to the circle from the outside point A . 
 
 190. Common tangent. A common tangent to two circles is 
 a straight line tangent to each. 
 
104 PLANE GEOMETRY 
 
 Theorem 12 
 
 191. If a line is perpendicular to a radius at its outer 
 extremity, it is tangent to the circle. 
 
 A NC _/ B 
 
 Given the circle C and AB perpendicular to the radius CP at 
 P, which is on the circle. 
 
 To prove that AB is tangent to the circle. 
 
 Proof 
 
 1. Draw CK, where K is any 1. 19. 
 point on AB except P. 
 
 2. CK>CP. 2. 149. 
 
 3. K is outside the circle. 3. Why ? 
 
 4. Hence any point of AB except 4. 187. 
 P is outside the circle, and 
 
 AB is tangent to the circle. 
 
 EXERCISES 
 
 37. ABCDE is a circumscribed equiangular pentagon. OK and 
 OR are radii to the points of contact of sides AB and AE. How 
 many degrees in the angle KOR ? 
 
 38. Perpendiculars to a diameter at its extremities are parallel 
 tangents. 
 
 QUERY. Can either of two tangents to a circle from an outside point 
 be perpendicular to the chord joining the points of contact? 
 
BOOK II 105 
 
 Theorem 13 (Converse of Theorem 12) 
 
 192. If a line is tangent to, a circle, it is perpendicular 
 to the radius drawn to the point of tangency. 
 
 Given the circle K y the line AB, tangent to it at 1?, and 
 radius KR. 
 
 To prove that KR is _L to AB. 
 
 Proof 
 
 1. Draw LK to L, any point on 1. 19. 
 AB except^. 
 
 2. . LK>KR. 2. 154 and 187. 
 
 3. Hence KR is the shortest line 3. 149. 
 from K to AB, and KR is _L 
 
 to AB. 
 
 QUERY. To how many radii of a circle can one tangent be 
 perpendicular ? 
 
 EXERCISES 
 
 39. Tangents to a circle at the extremities of a diameter are 
 parallel to each other. 
 
 40. Tangents to a circle at the extremities of two perpendicular 
 diameters meet in G, H, L, and M. Prove that GHLM is a circum- 
 scribed square. 
 
 41. If two tangents to a circle are parallel, the points of 
 contact and the center are in a straight line. 
 
106 PLANE GEOMETRY 
 
 Theorem 14 
 
 193. If two tangents are drawn to a circle from an 
 outside point, 
 
 (1) the tangents are equal; 
 
 (2) the line joining the outside point to the center bisects 
 the angle between the tangents and the angle between the 
 radii drawn to the points of contact. 
 
 R 
 
 Given AK and AR tangents from the point A outside the circle 
 whose center is C, the line AC, and the radii CK and CR to the 
 points of contact K and R. 
 
 To prove that (1) AK=AR. 
 
 (2) Z1 = Z2 awd 
 
 Proof 
 
 1. In &AKC and ARC, 1. Why ? 
 
 2. CA = CA. 2. Why ? 
 
 3. CK=CR. 3. Why? 
 
 4. A yl KC is congruent to A A RC. 4. Why ? 
 
 5. Therefore AK=AR,Z.l=Z.2, 5. Why ? 
 and Z 3 = Z 4. 
 
 EXERCISES 
 
 42. Two circles have two common tangents which do not 
 pass between them. The points of tangency are A and B for one 
 tangent and C and D for the other. Prove that A B equals CD, 
 
BOOK II 
 
 10T 
 
 43. The line A'B touches a circle at K, the line AL at L, and 
 the line BR at R. Prove that AB equals the sum of AL and BR. 
 
 44. In a circumscribed quadrilateral the sum of two opposite 
 sides equals the sum of the other two. 
 
 194. Line of centers. The line-segment joining the centers 
 of two circles is called 
 
 the line of centers. 
 
 195. Tangent circles. 
 If two circles have only 
 one point in common, each 
 is tangent to the other. 
 
 196. Externally tangent circles. If each of two tangent 
 circles is outside the other, they are tangent externally. 
 
 197. Internally tangent circles. If one of two tangent cir- 
 cles is within the other, they are tangent internally. 
 
 198. Internal common tangent. A common tangent of two 
 circles which passes between them is an internal common tangent. 
 
 199. External common tangent. A common tangent which 
 does not pass between two circles is an external common tangent. 
 
 QUERY 1. What is the greatest and the least number of common 
 tangents two circles can have ? 
 
 QUERY 2. What other possible numbers of common tangents to 
 two circles are there? 
 
 QUERY 3. What are the relative positions of the two circles in each 
 of the preceding cases ? 
 
108 
 
 PLANE GEOMETRY 
 
 QUERY 4. Can two circles have more than two, less than two, or no 
 common external tangents? < 
 
 QUERY 5. What are the relative positions of the two circles in each 
 of these cases ? 
 
 QUERY 6. Change external to internal and answer the two preceding 
 queries, 
 
 QUERY 7. Can equal circles be tangent externally? internally? 
 
 EXERCISES 
 
 45. The centers of two unequal circles and the point of 
 intersection of their two external common tangents are in the 
 same straight line. 
 
 46. The centers of two circles and the point of 
 intersection of their two internal common tangents 
 are in the same straight line. 
 
 200. Common chord. The common chord of 
 two intersecting circles is the chord joining 
 the points of intersection. 
 
 Theorem 15 
 
 201. If two circles intersect, the line of centers bisects 
 their common chord at right angles. 
 
 Given the circles K and R intersecting at A and B y the line of 
 centers KR, and the common chord AB. 
 
 To prove that KR bisects AB at right angles,. 
 
BOOK II 109 
 
 Proof 
 
 1. Draw the radii KA, KB, RA, 1. 19. 
 and RB. 
 
 2. Then K is equally distant 2. Why ? 
 from A and B. 
 
 3. R is also equidistant from 3. Why ? 
 A and . 
 
 4. Therefore KR is the mid- 4. 118. 
 perpendicular of AB. 
 
 QUERY 1. If one of three unequal circles is tangent to each of the 
 other two, how many points of contact must there be ? 
 
 QUERY 2. How many may there be? 
 
 QUERY 3. What are the relative positions of the circles in each case ? 
 
 QUERY 4. Has Theorem 15 more than one converse? If so, state 
 each. 
 
 QUERY 5. Of the fifteen theorems on circles, which have more than 
 one converse ? 
 
 EXERCISES 
 
 47. Three circles pass through two points K and R. Prove 
 that their centers are in a straight line. 
 
 48. A is the center of a circle, K is a point on the radius AB. 
 With K as the center and KB as the radius another circle is 
 described. Prove that the two circles are tangent. 
 
 HINT. Suppose the circles intersect at a point //, distinct from B. 
 Draw the mid-perpendicular of BH. Can this mid-perpendicular pass 
 through both centers? 
 
 49. A is the center of a circle. The radius AB is produced 
 to C. With C as the center and CB as the radius another circle 
 is described. Prove that the two circles are tangent. 
 
 HINT. Draw KB perpendicular to AB and show that all points of 
 circle A except point B are on one side of KB and all points of circle C 
 except point B are on the other side of it. 
 
 50. Prove the theorem of Ex. 49, using the hint of Ex. 48. 
 
110 PLANE GEOMETRY 
 
 Theorem 16 
 
 202. If two circles are tangent externally or internally, 
 the centers and the point of tangency are in a straight line. 
 
 Case I. When the circles are tangent externally. 
 
 Given the circles A and B tangent externally at K. 
 
 To prove that the points A, K, and B are in a straight line. 
 
 Proof 
 
 1. Suppose the line of centers 1. 19. 
 does not pass through K, but 
 
 cuts the circle A in R and the 
 circle B in L. Draw the radii 
 AK and BK and the line LR. 
 
 2. Now AK = AR. 2. Why ? 
 
 3. BK = BL. 3. Why? 
 
 4. A K -f KB < A R + II L + LB ; 4. Why ? 
 that is, A KB is shorter than 
 
 any other line from A to B 
 by the portion RL. 
 
 5. Therefore A , A", and B are in 5. 146. 
 the same straight line. 
 
 Case II. When the two circles are tangent internally. 
 
 Given the circles F and G tangent internally at H. 
 
 To prove that the points F, G, and 11 are in a straight line. 
 
 1. Suppose the circle M is tan- 1. Why ? 
 gent to the circle F at //. 
 Draw IIM. 
 
BOOK II 111 
 
 2. Circle M will be tangent to 2. 195. 
 circle G. 
 
 3. Then points G, H, and M are 3. Case I. 
 in the straight line HM. 
 
 4. Also points F, H, and M are 4. Case I. 
 in the straight line HM. 
 
 5. F, G, and H are all in the 5. Steps 3 and 4. 
 same straight line. 
 
 203. Corollary 1. The line of centers of two externally tangent 
 circles is equal to the sum of the two radii. 
 
 204. Corollary 2. The line of centers of two internally tangent 
 circles is equal to the difference of the two radii. 
 
 205. Corollary 3. A line perpendicular to the line through 
 the centers of two tangent circles at their point of contact is the 
 common tangent of the two circles. 
 
 EXERCISES 
 
 51. If two circles are tangent externally, two tangents drawn 
 to them from any point in their common internal tangent are equal. 
 
 52. If two circles are tangent externally, their common internal 
 tangent bisects the portion of their common external tangent 
 between the points of contact. 
 
 206. Concentric circles. Two or more circles which have 
 the same center are called concentric circles. 
 
 EXERCISES ON TANGENTS AND TANGENT CIRCLES 
 
 53. AB and AC are tangent to a circle at B and C respec- 
 tively. Prove that AK, a line from A to the center, is the mid- 
 perpendicular of BC. 
 
 54. AB and AC are two tangents to a circle at B and C respec- 
 tively. A third tangent touching the arc BC cuts AB at K and 
 AC at R. Prove that the perimeter of the triangle AKR is equal 
 ioAB plus vlC. 
 
112 
 
 PLANE GEOMETRY 
 
 55. The segments of two internal common tangents of two cir- 
 cles included between their respective points of contact are equal 
 
 56. Tangents CA and DB of a circle drawn at the extremities 
 of the diameter AB meet a third tangent CD at C and D respec- 
 tively. K is the center of the circle. Prove that the angle CKD 
 is a right angle. 
 
 57. Tangents to a circle at the extremities of any pair of diam- 
 eters which are not perpendicular to each other form a rhombus. 
 
 58. If the opposite sides of a circumscribed quadrilateral are 
 parallel, the four sides are equal. 
 
 59. The contact points of the exterior common tangents 
 of two circles are A and B for one circle and C and D for 
 the other. Prove AB parallel to CD. 
 
 60. Tangents from the same point to 
 two concentric circles cannot be equal. 
 
 HINT. Let A C and AR be the tangents, 
 B the center, and H the mid-point of AB. 
 Draw HC and HR. Then study triangles 
 AHC and AHR (see Ex. 148 and 150). 
 
 61. If a point lies on each of two circles and also on the line 
 through their centers, the circles are tangent to each other. 
 
 62. A and B are the points of contact of a common exterior tan- 
 gent to two circles. One common interior tangent cuts AB at K and 
 the other common exterior tangent at R. Prove that AB equals KR. 
 
 R 
 
 207. Secant. A secant is a line which cuts a circle in two 
 points. 
 
BOOK II 
 
 113 
 
 Theorem 17 
 
 208. If two arcs of a circle are intercepted by two 
 parallel lines, the arcs are equal. 
 
 A H B H A < 
 
 FIG. 1 
 
 FIG. 2 
 
 FIG. 3 
 
 Case I. When one of the parallels is a tangent. 
 
 Given (Fig. 1) the circle C and AB, the tangent at JT, parallel 
 to the secant KR, the two intercepting the arcs 772 and x. 
 
 To prove that 
 
 arc m = arc x. 
 
 1. Draw the diameter HL. 
 
 2. Then HL is _L to AB. 
 
 3. HL is _L to KR. 
 
 4. Therefore arc m = arc x. 
 
 Proof 
 
 1. 19. 
 
 2. 192. 
 
 3. 47. 
 
 4. 182. 
 
 Case II. When both parallels are tangents. 
 
 Given (Fig. 2) the circle C and two tangents parallel to each 
 other and touching the circle at H and L respectively. 
 
 To prove that arc HKL = arc HRL. 
 
 Proof 
 
 1. Let KR, II to one tangent, form 1. 45. 
 the arcs m, x, a, and c. 
 
 2. A^R is II to both tangents. 
 
 3. Then m = x and a = c. 
 
 4. Hence arc HKL = arc HRL. 
 
 2. If KR cuts one parallel, it 
 would cut both. 46. 
 
 3. Case I. 
 
 4. 31. 
 
114 PLANE GEOMETRY 
 
 Case III. When loth parallels are secants. 
 
 HINT. Draw AB II to KR, touching the circle and forming the arcs 
 a and c, as indicated in the figure. 
 
 QUERY 1. Is the statement which follows a converse of Theorem 17 ? 
 If two lines intercept equal arcs of a circle, the lines are parallel. 
 
 QUERY 2. Is this statement true? 
 
 QUERY 3. Has Theorem 17 more than one converse? Illustrate. 
 
 QUERY 4. Is the following modification of Theorem 17 true? If in 
 the same circle or in equal circles two arcs are intercepted by parallel 
 lines, the two arcs are equal. 
 
 209. Inscribed polygon. A polygon whose vertices lie on 
 a circle is called an inscribed polygon. 
 
 EXERCISES 
 
 63. AD and EC are the nonparallel Sides of the inscribed 
 trapezoid A BCD. Prove that the arc ABC equals the arc DAB, 
 and prove that the arc ADC equals the arc BCD. 
 
 64. The nonparallel sides of an inscribed trapezoid are 
 equal and the diagonals are equal. 
 
 65. Two circles intersect in A and B. 
 
 Through .4 and B parallels K 
 are drawn, points K and H being on one 
 circle and R and L on the other. Prove 
 that the chords A B, KH, and RL are equal 
 
 66. Two chords of a circle perpendicular to a third at its 
 extremities are equal. 
 
 210. Degree of arc. If an arc whose central angle is a right 
 angle is divided into ninety equal parts, one of these parts is 
 called a degree of arc. 
 
 From the preceding definition and 56 it follows that a 
 central angle of one degree intercepts on the circle an arc of 
 one degree. 
 
BOOK II 115 
 
 211. Relation of a central angle to its intercepted arc. The 
 
 number of angle degrees in a central angle is eqvial to the 
 number of arc degrees in its intercepted arc. 
 
 The usual statement of this relation is the theorem 
 
 A central angle is measured by its intercepted arc. 
 The truth of this theorem will be assumed. 
 
 The same number, therefore, expresses the measure of an arc of 
 a circle and of its subtended central angle, but this does not assert 
 that an angle is the same thing as an arc. It merely asserts that 
 they may be equal numerically, just as the number of acres in a 
 iield may be equal numerically to the number of rods of fence 
 around the field, without implying that the field and the fence or 
 an acre and a rod are the same thing. 
 
 212. Notation for numerically equal arcs and angles. If 
 
 the last sentence of 211 is clearly understood, we may 
 write, referring to the adjacent figure in which C is the 
 center, Z.C arc A B. Later we shall prove that Z F is 
 measured by one half the number of degrees 
 in the intercepted arc HG, and this state- 
 ment may be written /.F=. i- arc HG. This, 
 however, must be understood to mean : 
 The number of degrees of angle in angle F 
 equals one half the number of degrees of 
 arc in arc HG. A similar meaning must be assigned to 
 Z C = arc AB. In dealing with arcs and angles we shall 
 gain in brevity if the numerical relations which exist are 
 written as equations. This is perfectly correct, since the 
 two members of the equation are equal numbers. Of course 
 the usual laws of operation on equations may then be 
 applied to these statements. The equation form of state- 
 ment is used in the proofs .of Theorems 20, 21, and 23 
 which follow. 
 
116 
 
 PLANE GEOMETEY 
 
 The length of an arc in feet or inches must not be confused 
 with the number of degrees it contains. In the adjacent figure, the 
 angle C is a central angle in both circles. There- 
 fore arc AB and arc KR contain the same number 
 of degrees. But arc AB is shorter than arc KR. 
 In the same circle or in equal circles, however, 
 two arcs of the same number of degrees are of 
 equal length. Thus, if each of arcs AB and HL 
 of the adjacent figure contains ?^, they are also 
 of equal length in feet or inches. 
 
 213. Inscribed angle. An angle whose ver- 
 tex is on a circle and whose sides are chords 
 of the circle is called an inscribed angle. 
 
 214. Segment. A segment of a circle is a figure bounded 
 by an arc of the circle and its chord. 
 
 An angle is said to be inscribed in a segment or an arc if its 
 vertex is on the arc and its sides pass through the ends of the arc. 
 
 Thus the angle C may be spoken of as inscribed in the segment 
 A CB or in the arc A CB. 
 
 215. Sector. A sector of a circle is a figure bounded by 
 two radii and their intercepted arc. 
 
 QUERY 1. How many degrees of angle are there 
 about the center of a circle ? 
 
 QUERY 2. How many degrees of arc are there in 
 a circle? 
 
 QUERY 3. What figure may be called either a segment or a sector? 
 
 QUERY 4. A chord forms how many segments? 
 
 QUERY 5. Two radii form how many sectors ? 
 
BOOK II 117 
 
 EXERCISES 
 
 67. The circumference of a circle is 30 ft. How many degrees 
 in an arc of this circle 10 ft. long ? in one 4 ft. long ? 
 
 68. An arc of a circle containing 18 is 3 ft. long. Find the 
 circumference of the circle. 
 
 HISTORICAL NOTE. The division of a right angle into ninety de- 
 grees is equivalent to dividing a circle into three hundred and sixty 
 degrees. This part of our mathematical heritage, as well as the present 
 division of the day into hours, minutes, and seconds, comes from the 
 Babylonians. Why they adopted such a scheme of division we do not 
 know. Cantor, a great historian of mathematics, suggests that their 
 division of a circle came from their combining an astronomical belief 
 with a geometrical fact. At first, according to his supposition, they 
 supposed that the sun went round the earth in a circle once in three 
 hundred and sixty days, thus making its daily motion ^^ of the whole. 
 Moreover, they were familiar with the fact that a chord of a circle equal 
 to the radius has an arc which is exactly one sixth of the circle ( 348). 
 Such an arc seemed a natural division of the circle, since the sun would 
 take sixty days to pass over it. In some such way they were led to what 
 is called the sexagesimal division of the circle. 
 
 About the time the metric system was devised (1790) an attempt 
 was made in France to put the division of a right angle on a decimal 
 (or rather centesimal) basis. The right angle was divided into one 
 hundred grades, each grade into one hundred minutes, and each minute 
 into one hundred seconds. A large force of computers, after great ex- 
 penditure of time and effort, calculated the necessary trigonometric and 
 logarithmic tables to correspond. No attempt has been made to transfer 
 to the centesimal system the many standard reference tables, the countless 
 records of observation, and extensive data of various kinds which workers 
 in mathematics and science use. Until this is done the undoubted merits 
 of the. centesimal system will not be apparent. As a consequence, the 
 system has not been used outside of France, and only slightly there. 
 
 Although this system and the metric system of measurement, which 
 is closely associated with it, possess ^great practical advantages over our 
 present or English system of measurement, the difficulty and expense of 
 making the necessary changes in machinery and instruments prevent 
 its adoption. 
 
118 
 
 PLA^E GEOMETRY 
 
 Theorem 18 
 
 216. An inscribed angle is measured by one half the 
 number of degrees in its intercepted arc. 
 
 FIG. 1 
 
 Given the angle BAG, an inscribed angle. 
 
 To prove that /.BAG is measured by one half the arc BC. 
 
 Proof 
 
 Case I. When the center K is on one side of the angle (Fig. 
 
 1. Draw CK. 1. 19. 
 
 2. Then in AACK, 2. Why ? 
 
 3. Also Zl = Z.4 + Z C. 3. Why ? 
 
 4. Z1 = 2Z,1. 4. 65. 
 
 5. But Zl is measured by arc 5. 211. 
 EC. 
 
 6. /.A is measured by one half 6. Since Zl is measured by the 
 the arc EC. arc EC, and Z .1 = \ Z 1. 
 
 Cose 77. TTA^n ^e center K falls within the angle (Fig. 2). 
 
 HINTS. Draw the diameter AR. Z2 is measured by what? Z3 is 
 measured by what ? Z 2 + ^ 3, which equals Z. BA C, is measured by 
 what ? Conclusion ? 
 
BOOK II 
 
 119 
 
 Case III. When the center is outside the angle (l?ig. 3). 
 
 HINTS. Draw the diameter AR. ZBAR is measured by what? Z5 
 is measured by what? Z BAR Z5, which equals Z4, is measured by 
 what? Conclusion? 
 
 217. Corollary 1. An angle inscribed in a semicircle is a 
 right angle. 
 
 HINT. How many degrees in the arc intercepted 
 by the angle ? 
 
 218. Corollary 2. Two angles inscribed in 
 the same segment are equal. 
 
 HINTS. How is /.B measured ? Z A'? Conclusion? 
 
 219. Corollary 3. The opposite angles of an inscribed 
 quadrilateral are supplementary. 
 
 HINTS. If A BCD is an inscribed quadrilateral, what arc measures 
 Za? Zc? What relation exists between these two arcs? Conclusion? 
 
 QUERY 1. If a chord bisects an inscribed angle, does it bisect the 
 intercepted arc ? 
 
 QUERY 2. If a line is drawn from the mid-point of an arc to the 
 vertex of an angle inscribed in that arc, does it bisect the angle ? 
 
 EXERCISES 
 
 69. Two angles of an inscribed triangle are 40 and 70 
 respectively. Find the number of degrees in each of the arcs 
 subtended by the sides of the triangle. 
 
 70. AB, BC, and CA are the sides of an in- 
 scribed triangle. The arc AB contains 135, and 
 the arc .BC is twice the arc AC. Find each, angle 
 of the triangle. 
 
 71. Assign values to three angles of an in- 
 scribed quadrilateral and then determine the num- 
 
 ber of degrees in each of the four arcs subtended by its sides. 
 
 72. An inscribed angle ABC contains 30. Prove that the 
 chord AC equals the radius of the circle. 
 
120 PLANE GEOMETRY 
 
 Theorem 19 
 
 220. If two vertical angles are formed by tied inter- 
 secting chords, each angle is measured ~by one half the 
 number of degrees in the sum of their intercepted arcs. 
 
 K 
 
 Given the chords AB and KR intersecting in the point L. 
 
 To prove that /. 1 and Z. 2 are each measured by i (arc AK 
 -f- arc BR), and that Z. 3 and Z. ALR are each measured by 
 (arc AR + arc BK). 
 
 HINTS. Draw the chord BK. Then Zl = Z2. Why? 
 
 But ' Zl = ^K + ZB. Why? 
 
 How is Z K measured ? Z B ? Conclusion ? 
 
 QUERY. If a chord in the figure of 220 bisects Zl and Z2, does 
 it bisect the arc BR and the arc A A'? 
 
 EXERCISES 
 
 73. If arc AR in the figure of 220 is 27 and arc BK is 52, 
 find the number of degrees in Z. 3. 
 
 74. If arc AR in the figure of 220 is 30 and Z3 is 75, 
 find the number of degrees in the arc KB. 
 
 75. Two sides of an inscribed triangle subtend arcs which are 
 one fifth and one eighth of the circle respectively. Find the 
 angles of the triangle. 
 
 76. Three angles of an inscribed quadrilateral are 85, 95, and 
 80. Can these angles be consecutive angles of the quadrilateral 
 in the order given ? 
 
 77. If two equal chords intersect, the point of intersection 
 divides both in the same way. 
 
BOOK II 
 
 121 
 
 Theorem 20 
 
 221. If an angle is formed by a tangent and a chord 
 drawn from the point of contact, the angle is measured by 
 one half the number of degrees in its intercepted arc. 
 
 K 
 
 FIG. 1 
 
 Given AB, a tangent to a circle C at K, and the chord KR. 
 
 To prove that Z.BKR is measured ly \ arc KR, and Z.AKR 
 by 1 arc KHR. 
 
 Proof 
 
 Case /. When the chord KR is not a diameter (Fig. -?). 
 
 1. Draw chord RL II to AB. 
 
 2. Then Z.R = Z.BKR. 
 
 3. Now Z R = J arc KL. 
 
 4. Therefore 
 
 Z BKR = i arc KL. 
 
 5. But arc KR = arc KL. 
 
 6. Therefore BKR is measured 
 by J arc KR. 
 
 Now consider the angle AKR. 
 
 1. Draw ## so that Z.AKH is 
 less than 90. 
 
 2. Then </AKH= arc #7f. 
 
 3. Z ##72 = I arc 7/tf. 
 
 4. Therefore Z.AKH + Z HKR, 
 or Z.AKR, is measured by 
 
 1. 45. 
 
 2. Why ? 
 
 3. Why ? 
 
 4. Why ? 
 
 5. Why ? 
 
 6. Why ? 
 
 1. 
 
 is an obtuse angle. 
 
 2. Why ? 
 
 3. Why ? 
 
 4. Why ? 
 
 
 arc 
 
122 PLANE GEOMETRY 
 
 Case II. When the chord KR is a diameter (Fig. 2). 
 
 1. Arc KMR = 180. 1. Why ? 
 
 2. Z BKR = 90. 2. Why ? 
 
 3. ThereforeZ/O2 is measured 3. Why ? 
 by i arc KMR. 
 
 In like manner /.AKR is measured by J arc KGR. 
 
 QUERY 1. By what arcs is an exterior angle of an inscribed polygon 
 measured ? 
 
 QUERY 2. If arc KR in Fig. 1 of 221 is 50, how many degrees in 
 ZAKRt 
 
 QUERY 3. In Fig. 1 of 221, if arc LR equals are KH equals 70, 
 how many degrees in angle HKR ? 
 
 QUERY 4. In Fig. 1 of 221, if chord LR = chord RK, can you find 
 arc LK1 How? Z#? How? 
 
 EXERCISES 
 
 78. If a chord of a circle bisects the angle between a tangent 
 and a chord drawn from the point of contact, does it bisect 'the 
 intercepted arc ? Prove. 
 
 79. If chords A B and A C of a circle make equal angles with 
 the tangent at A, they are equal. 
 
 80. A tangent at the mid-point of an arc of a circle is parallel 
 to the chord of the arc. 
 
 81. A BCD is an inscribed polygon. The chords AB, BC, and 
 CD subtend arcs which are one third, one fourth, and one fifth of 
 the circle respectively. Find the number of degrees in each exterior 
 angle of ABC D. 
 
 82. A, B, and C, three points on a circle, and 7?, a point within, 
 are the vertices of quadrilateral ABCR. Prove that the angle A 
 is less than the exterior .angle at C. 
 
 83. Two angles of an inscribed triangle are 46 and 62 
 respectively. Find the arcs subtended by its sides. : 
 
BOOK II 123 
 
 Theorem 21 
 
 222. The angle between two secants, or two tangents, 
 or a tangent and a secant, which intersect outside the 
 circle, is measured by one half the number of degrees in 
 the difference of the intercepted arcs. 
 
 B 
 
 FIG. 1 FIG. 2 FIG. i) 
 
 Given (Fig. 1) the circle BKR, and AKB and ARC two 
 secants intersecting at -4, a point outside the circle. 
 
 To prove that ZA is measured by (arc BC arc KR). 
 
 Proof 
 
 1. Draw BE, making Zl with 1. 19. 
 CR. 
 
 2. Then Zl =A + Z#; 2, Why'? <: 
 
 3. Whence : 3. 51. 
 
 4. Zl-iarcUC. 4. 
 
 5. B = |arcA^ - , 5, Why? 
 
 6. Therefore Z ^4 is measured by 6. 65. 
 
 | (arc 5 C arc A'72). 
 
 HINTS. The proof when .4 B and ^4 C are tangents (Fig. 2) and when 
 one is a tangent and the^bllier a secSrri:t"(F1g.:;3) can be worked out in 
 like manner. 
 
 ""223; ^Corollary ~l. : '-The~ angie~ft&w~e r en~two tang'enU' fo~' d~cir- 
 d M measured by the- supplement i of . the smaller Lqf-.th&^two 
 intercepted arcs. 
 
124 PLANE GEOMETRY 
 
 Outline of proof. In Fig. 2 of 222 let the minor arc BC = x. 
 Then the major arc BKC = 360 - x. Then, by Theorem 21, 
 Z A = J- [(360 - aj) - x] = 1 (360 - 2 a: ) = 180 - x. 
 
 EXERCISES 
 
 84. If the arc/a? (Fig. 1, 222) is 28 and the arc BC is 86, 
 find the number of degrees in the angle A . 
 
 85. If the arc KR (Fig. 1, 222) is 28 and the angle A is 65, 
 find the number of degrees in the arc BC. 
 
 86. If the minor arc BC (Fig. 2, 222) is one fourth of the 
 major arc BC, find the number of degrees in the angle A. 
 
 87. If the angle A (Fig. 3, 222) contains 46 and the arc BC 
 is five times the arc BR, find the number of degrees in the arc BR 
 and in the arc BC. 
 
 88. If a secant bisects the angle between two tangents to a 
 circle, does it bisect each arc intercepted by the tangents ? 
 
 Theorem 22 
 
 224. If three points are not in the same straight line, 
 one circle and only one can pass through them. 
 
 H. L 
 
 _________ ... 
 
 A K B P 
 
 Given A, B, and C, three points not in the same straight line. 
 
 To prove that one circle and only one can pass through A, 
 ) and C. 
 
BOOK II 125 
 
 Proof 
 
 1. Join A to B and B to C. Let 1. 41, 42. 
 Z.\ be the angle at B, 
 
 which is less than a straight 
 angle. Let KL be the mid- 
 perpendicular of AB. Draw 
 MP J_ to A B produced, through 
 R, the mid-point of BC. 
 
 2. Then KL is II to MP. 2. 44. 
 
 3. Z MRS > 90. 3. 144. 
 
 4. Now draw HR, the mid- 4. 46. 
 perpendicular of BC. 
 
 Then HR lies within the angle 
 MRB and intersects KL at 
 some point O. 
 o. Then AO=OB = OC. 5. 116. 
 
 6. Hence a circle with radius OC 6. Previous step, 
 and center will pass through 
 
 A, B, and C. 
 
 7. The lines A'L and HR inter- 7. Why ? 
 sect in only one point. 
 
 Therefore but one circle can 
 pass through A, B, and C. 
 
 QUERY 1. How many circles can pass through two points? 
 QUERY 2. Where do all their centers lie? 
 
 QUERY 3. Answer the question in Ex. 88 if the bisector passes 
 through the center of the circle. 
 
 225. Concyclic points. Three or more points which lie on 
 a circle are called concyclic points. 
 
 EXERCISES 
 
 89. Prove that a circle will pass through the vertices of a right 
 triangle if the center is the mid-point of the hypotenuse and the 
 radius is half the hypotenuse. 
 
 90. Prove that the vertices of a rectangle are concyclic points. 
 
126 PLANE GEOMETRY 
 
 Theorem 23 
 
 226. If two opposite angles of a quadrilateral are 
 supplementary, a circle can be circumscribed about it. 
 
 D G 
 
 Given the quadrilateral ABCD, with the angles A and C 
 supplementary. 
 
 To prove that a circle can be circumscribed about ABCD. 
 
 Proof 
 
 1. Let ABD be a circle passing 1. 224. 
 through A , B, and D. 
 
 Now the fourth vertex may fall(i) outside the circle, as at K, 
 (2) inside the circle, as at 72, or (3) upon the circle, as at C. 
 
 (1) Let L be the point where one of the sides of Z. K cuts the 
 circle. Complete the quadrilateral BADL. 
 
 2. Z.A + ^BLD = 180. 2. 219. 
 
 3. Z.BLD>Z.K. 3. 144. 
 
 4. Hence Z.A +^K< 180. 4. From statements 2 and 3. 
 
 But this is contrary to the hypothesis. Therefore the fourth 
 vertex cannot fall outside the circle. 
 
 (2) Let G be the point where BR cuts the circle. Draw chord GD. 
 
 5. Then /. A + /L G = 180. 5. Why ? 
 
 6. RRD>/-G. 6. Why? 
 
 1. Hence Z. A +Z.BRD > 180. 7. From statements 5 and 6. 
 
 But this contradicts the hypothesis. Hence the fourth vertex 
 cannot fall inside the circle. Then it must fall on the circle, as at C. 
 
BOOK II 127 
 
 EXERCISES 
 
 91. Prove by Theorem 23 that a circle can be circumscribed 
 about a rectangle. 
 
 92. Are the vertices of a square coney clic ? of a rhombus ? 
 of a parallelogram ? Prove. 
 
 93. In the trapezoid ABCD the angle A equals the angle B 
 and the angle C equals the angle D. Are the points A, B, (7, 
 and D concyclic ? Prove. 
 
 94. Kis a point outside the triangle A B C. 
 KR, KL, and KM are the perpendiculars to 
 AB, CA, and BC respectively, produced if 
 necessary. Prove that the points A, R, L, 
 and K are concyclic, that the points C, L, 
 
 K, and M are concyclic, and that the points B, K, R, and M are 
 concyclic, and locate the center of each circle. 
 
 MISCELLANEOUS EXERCISES ox CIRCLES 
 
 95. The chords AB and CD intersect at 0. The angle AOC is 
 30 and the arc DB is 42. Find the arc A C. 
 
 96. The parallel sides of an inscribed trapezoid subtend arcs 
 of 100 and 110 respectively. Find its angles. 
 
 97. ABCD is an inscribed quadrilateral. The arc AB is 48 
 and the arc BC is 92. The axe DC is one fourth of the arc AD. 
 Find the number of degrees in each angle of the polygon. 
 
 98. An inscribed angle A intercepts an arc ?/, and a central angle 
 B in the same circle intercepts an arc x. If x plus y equals 98 and 
 the angled is 8 more than three times the angle A, find the angle A 
 and the angle B. 
 
 99. KR and ML, two sides of the inscribed quadrilateral 
 KRLM, when produced meet in A. The diagonals of the quadri- 
 lateral meet at 0. From H on A K a tangent HB is drawn touching 
 the circle at B so that the arc BK is three times the arc BR. - The 
 arc RL is 70, the arc ML is 80, and the arc MK is 130. Find the 
 number of degrees in the angles BHK, RAM, RLM, KOR, and ARL. 
 
128 PLANE GEOMETRY 
 
 100. Assign numerical values to the angles of an inscribed 
 quadrilateral such that its opposite sides produced will meet in 
 K and L respectively. Then bisect the 
 
 angle K and the angle L and determine 
 the angle between the bisectors. 
 
 101. A chord of a circle equals the 
 radius. Show that the minor arc of the 
 chord is 60. 
 
 102. AB is a diameter, and the chord 
 AK is equal to the radius. Draw BK 
 
 and prove that the angle A is 60 and that the angle B is 30. 
 
 103. Chords AB and AC are equal to the radius. Draw BC 
 and determine the number of degrees in each angle of the 
 triangle ABC. 
 
 104. If the two non parallel sides of a trapezoid are equal, 
 its vertices are concyclic. 
 
 105. AB, a chord of a circle, is the base of an isosceles triangle 
 whose vertex C is without the circle and whose equal sides inter- 
 sect the circle in D and E respectively. Prove that -CD is equal 
 to CE. 
 
 106. A is the center of a circle and K is any outside point. 
 R is the mid-point of AK. A circle with R as the center and R A 
 as the radius cuts the first circle in H and L. Prove that KH and 
 KL are tangent to the first circle. 
 
 107. The perimeter of an inscribed equilateral triangle is equal 
 to one half the perimeter of the circumscribed equilateral triangle. 
 
 108. The diagonals of an inscribed equilateral polygon of five 
 sides are equal. 
 
 109. The bisectors of two opposite angles of an inscribed 
 quadrilateral meet the circle in A and B. Prove that AB is a 
 diameter of the circle. 
 
 110. ABC is an inscribed equilateral triangle. DE joins the 
 middle points of arcs BC and AC. Prove that DE is trisected by 
 the sides BC and AC. 
 
BOOK II 129 
 
 111. Any secant through the point of contact of two tan- 
 gent circles subtends arcs of which the two in one circle 
 contain the same number of degrees 
 
 respectively as the two in the other 
 circle. 
 
 HINT. Draw the common tangent. . 
 
 112. A secant through the point of 
 contact, A, of two tangent circles cuts 
 
 them in K and R respectively. Prove that the radii drawn to K 
 and R are parallel. 
 
 113. AB, a diameter, is extended half its length to K. Tangents 
 from K cut the tangent at A in R and L respectively. Prove that 
 the triangle KRL is equilateral. 
 
 114. ABC is an inscribed triangle. The bisectors of angles A, 
 B, and C meet in D. A D produced meets the circumference in E. 
 Prove that DE is equal to the chord CE. 
 
 115. Two circles intersect at A and B, and a tangent to each 
 circle is drawn at A meeting the circles at R and S respectively. 
 Prove that the triangles ABR and ABS & g 
 
 are mutually equiangular. 
 
 116. A common external tangent is 
 drawn to two circles which are exter- 
 nally tangent. Show that the chords 
 drawn to the points of tangency from 
 
 the points where the line through the centers cuts the circles are 
 parallel in pairs. 
 
 CONSTRUCTIONS 
 
 227. Problems of construction. Thus far it has been assumed 
 that such figures as were needed could be drawn with a suffi- 
 cient degree of accuracy to serve as aids in the demonstration 
 of the exercises and theorems. Methods of constructing figures 
 composed of straight lines or of circles or of a combination of 
 the two will now be studied. These methods are theoretically 
 exact. We say theoretically because, however sharp our pencil, 
 
130 PLANE GEOMETEY 
 
 we cannot draw a geometrical line ; nor can we draw, even 
 with the best ruler, a line which is perfectly straight. While 
 the figures that result even with exact methods are not abso- 
 lutely accurate, they are precise enough for many practical 
 purposes ; and what is more important from a scientific point 
 of view, they enable us to demonstrate the existence of many 
 geometric relations which until now we have been obliged to 
 assume. For example, in 42 and other places the existence 
 of a perpendicular from a point to a line was assumed. In 
 235 we shall show how to construct such a line. 
 
 228. Postulates of construction. The possibility of three 
 simple operations, which underlie all the construction work 
 of elementary geometry, is assumed: 
 
 1. Through any two points a straight line can be drawn. 
 
 2. A given line-segment can be extended indefinitely beyond 
 either extremity. 
 
 3. A circle can be described with any given point as a center 
 and any line of a given length as a radius. 
 
 There are, besides the three assumptions here explicitly 
 stated, a very great number of others which are implicitly 
 taken for granted. For example : Every straight line is con- 
 tinuous ; Every angle is divisible into any number of equal 
 parts ; Two straight lines can intersect in only one point ; etc. 
 
 229. Instruments used in construction. From the postulates 
 just stated it follows that the instruments which we shall require 
 are only two : the ruler (straightedge) and the compasses. 
 
 The ruler is used for drawing a straight line between two points 
 and for extending a given straight line-segment. 
 
 The compasses are employed for drawing circles and arcs, and 
 are also used for transferring distances ; that is, for making one 
 line-segment equal to another one. 
 
 The ruler and the compasses are the two instruments which 
 from the earliest times have been associated with the problems of 
 
BOOK II 131 
 
 geometry. We still'limit ourselves to their use for several reasons. 
 First, they are easy to obtain and simple to use. Second, although 
 many of the constructions here performed can be carried out more 
 simply by a T-square or a parallel ruler, still the two instruments 
 most essential to a draughtsman are the ruler and the compasses. 
 Third, to include another instrument for example, one making 
 possible the drawing of ellipses would so enlarge the field of 
 construction work that even an efementary knowledge of it could 
 not be obtained in the time available. 
 
 230. Solution of a construction problem. The solution of a 
 construction problem may best be divided into four parts : 
 
 1. Given. The length of certain lines, the size of certain 
 angles, the position of certain points or lines, must be defi- 
 nitely stated, for they are the "given" elements. 
 
 2. Required. A precise statement of what is to be done 
 must be made. 
 
 3. Construction. The figure "required" may well be first 
 sketched in outline but must also be actually constructed by 
 the drawing of straight lines and circles, the whole properly 
 lettered, and the portion of it which is the required figure 
 specified. 
 
 4. Proof. Here it is necessary to assume that the required 
 figure has been constructed as stated in 3, and to prove that 
 if so constructed it is the required figure. 
 
 This means that 1 and 3 furnish the hypothesis on which the 
 proof is based. 
 
 The written solution of a construction should be accompanied 
 by a figure showing all the arcs whose intersections determine 
 the important points. 
 
 The student should note carefully how Constructions 1-9, 
 which follow, illustrate the various points of 230, and he should 
 observe them in the solutions he himself works out. He should 
 also note that the " given " and the " required " lines in the figures 
 are continuous and all other lines are dotted. 
 
132 PLANE GEOMETRY 
 
 Construction l 
 
 231. Bisect a given line-segment. 
 
 \/ 
 \K 
 
 A 
 
 
 B 
 
 1. Given the straight line-segment AB. 
 
 2. Required to find a point C on AB such that AC = CB. 
 
 3. Construction. Describe two arcs with A and B as centers and 
 with a radius great enough to give two points of intersection, 
 K and R. 
 
 Draw KR cutting AB at C. 
 
 Then AC = CB. 
 
 4. Proof. RA = RB and #4 = KB. Const. 
 Therefore A'72 is the mid-perpendicular of AB. 118 
 NOTE. In performing this construction a radius should be taken 
 
 large enough to make points R and K fall rather far apart. A line 
 determined by two points very near each other cannot be drawn as 
 accurately as one determined by points a considerable distance apart. 
 
 QUERY 1. How can the mid-perpendicular of a line be constructed? 
 QUERY 2. How can a circle "be drawn through two fixed points? 
 QUERY 3. How can a circle be drawn through three fixed points 
 not in a straight line ? (See 224.) 
 
 QUERY 4. In Construction 1 does AB bisect each arc KR? Why? 
 
 EXERCISES 
 
 117. Divide a line into four equal parts. 
 
 118. Construct the mid-perpendicular of a given line. 
 
 119. Bisect a given arc. 
 
 HINT. Construct the mid-perpendicular of the chord of the arc. 
 
 120. Circumscribe a circle about a given triangle. 
 
BOOK II 133 
 
 Construction 2 
 232. Construct a triangle, given the three sides. 
 
 B a C 
 
 1. Given the three lines a, b, and c. 
 
 2. Required to construct a triangle with a, b, arid c as sides. 
 
 3. Construction. Construct the line EC equal to a. 
 
 With B as the center and c as the radius, describe the arc xy. 
 With C as the center and b as the radius, describe an arc 
 cutting arc xy at A. 
 
 Then ABC is the required triangle. 
 
 4. Proof. BC = a, CA b, and AB = c. Const. 
 
 NOTE. In constructions and computations relating to triangles it is 
 a convenient and a fairly general custom to denote the angles of any 
 triangle ABC by the capital letters A, B, and C, and to denote the' sides 
 
 opposite these angles by the small letters a, b, and c respectively. 
 
 * 
 
 233. Altitude of a triangle. An altitude of a triangle is a 
 perpendicular from any vertex to the side opposite, produced 
 if necessary. 
 
 EXERCISES 
 
 121. Construct an equilateral triangle with a given line as a side. 
 
 122. Construct an isosceles triangle, given the base and the 
 altitude upon it. 
 
 123. Construct a parallelogram, given two adjacent sides and 
 one diagonal. 
 
 QUERY 1. How can an angle of 60 be constructed? an angle 
 of 120? 
 
 QUERY 2. How is the supplement of a given angle constructed ? 
 
134 PLANE GEOMETRY 
 
 Construction 3 
 
 234. At a given point in a line construct a perpendicular 
 to the line. ^ 
 
 4 
 
 V 
 
 
 / 
 
 ^ 
 
 
 / 
 
 \ 
 
 
 / 
 
 \ 
 
 
 A / , 
 
 2 \ 
 
 B 
 
 V K 7 
 
 ? 
 
 1. Given the line AB and the point K on AB. 
 
 2. Required to construct a A. to AB at K. 
 
 3. Construction. With K as the center and any convenient 
 radius, describe two arcs cutting AB at L and R respectively. 
 
 With L and R as centers and a greater radius than before, 
 describe two arcs intersecting at G. Draw line GK. 
 
 Then GK is the required perpendicular. 
 
 4. Proof. Draw GL and GR. 
 In A GLK and GRK, 
 
 KL = KR and GL = GR. Const. 
 
 GK = #. Why ? 
 
 A Z# = A tffl/r. Why ? 
 
 Z1=Z2. Why? 
 
 Therefore /T is J_ to ^4. Why ? 
 
 EXERCISES 
 
 124. Construct a right angle. 
 
 125. Given an acute angle, construct its complement, 
 
 126. Construct a right triangle, given the two shorter:, sjdes, . : .' .-; 
 
 127. .Construct, a tangent to a, circle at a given- point on ite; ; 
 
 128. Construct a right triangle, given the hypotenuse: an.&aJside;, 
 , 129. Construct a : recta-ngle'j given one side.. and a. Diagonal: 
 
 QUERY. How can an isosceles right triangle be constructed-? ah- 
 angle . -of .45r?.::.:. -J:::-. ."i-T;-; :: .:.,,' '-..'. 
 
BOOK II 135 
 
 Construction 4 
 
 235. From a given point outside a line construct a 
 perpendicular to the line. 
 
 K 
 
 1. Given the line AB and the point K outside the line. 
 
 2. Required to construct a A. to AB from K. 
 
 3. Construction. With K as the center and a radius great enough 
 to cut AS, describe arcs cutting AB at points R and L. 
 
 Then with R and L as centers and any radius greater than half 
 of 7?Z, describe two arcs intersecting at G. 
 Then draw KG. 
 Then KG is the required perpendicular. 
 
 4. Proof. KR = KL and GR = GL. Const. 
 Therefore KG is perpendicular to RL and AB. 118 
 
 EXERCISES 
 
 130. Given a triangle having one obtuse angle, construct its 
 three altitudes. 
 
 131. Through a given point outside a line construct a parallel 
 to. a given line. 
 
 HINT. Use Construction 4 and then Construction 3. 
 
 NOTE. In determining points by the intersections of arcs, it is desir- 
 able to make the arcs cut each other as in Construction 4. Then the 
 points of intersection can be definitely located. If the arcs cut each 
 other at a small angle, they appear to coincide for some distance near 
 the point of crossing, and the exact point of intersection is not clearly 
 determined. 
 
130 PLANE GEOMETRY 
 
 Construction 5 
 236. Bisect a given angle. 
 
 'K -C 
 
 1. Given the angle BAG. 
 
 2. Required to bisect the angle BAC. 
 
 3. Construction. With A as the center and any radius AR, de- 
 scribe two arcs cutting the sides of Z. RA C at R and A'. Then with 
 any radius greater than half R K and with R and A as centers, 
 describe two arcs intersecting at L. 
 
 Draw A L. 
 
 Then /.BAL Z.CAL. 
 
 4. Proof. Draw RL and KL. 
 
 Then AARL = AAKL. Why ? 
 
 Therefore Z BA L=Z.CAL. Why ? 
 
 QUERY 1. May the radius of the arcs which intersect at L equal the 
 radius of arc R A? 
 
 QUERY 2. What point is equidistant from the sides of a triangle? 
 
 QUERY 3. How can such a point be located? 
 
 QUERY 4. How can an angle of 30 be constructed ? an angle of 15 ? 
 
 QUERY 5. How can an angle of 45 be constructed? of 22^? of 
 135? of 150? 
 
 QUERY 6. How can a right angle be trisected? 
 
 QUERY 7. Can any given angle be trisected ? 
 
 QUERY 8. In Construction 5 does AL bisect the arc KR ? Why? Is 
 the chord R K perpendicular to AL ? Why ? 
 
 EXERCISES 
 
 132. Inscribe a circle in a given triangle. 
 
 133. Given a circular arc AB, find by construction the center 
 of the circle of which the arc is a part. 
 
BOOK II 137 
 
 Construction 6 
 
 237. Given one side and the vertex, construct an angle 
 equal to a given angle. 
 
 G 
 
 1. Given FG one side and F the vertex of an angle, and the 
 angle ABC. 
 
 2. Required to construct an angle equal to Z.ABC with F as 
 the vertex and FG as one side. 
 
 3. Construction. With B as the center and any radius, describe 
 an arc cutting AB at R and BC at A'. With BK as the radius and 
 F as the center, describe arc VL cutting FG at L . With chord KR 
 as the radius and L as the center, describe an arc cutting arc VL 
 at H. Draw FH. 
 
 Then Z/-'=Z#. 
 
 4. Proof. Draw RK and HL. 
 
 Then A BKR = A FLH. Why ? 
 
 Therefore Z F = Z . Why ? 
 
 EXERCISES 
 
 134. Construct a triangle, given two sides and their included 
 angle. 
 
 135. Construct a triangle, given one side and the two adjacent 
 angles. 
 
 136. Construct a triangle, given two of its angles. How many 
 such triangles are possible ? 
 
 137. Construct a triangle, given two sides and the angle opposite 
 one of them. 
 
138 PLANE GEOMETKY 
 
 138. Given a, b, and angle A. Determine the number of possible 
 triangles if angle A is acute and a is less than b. '< 
 
 139. Proceed as in Ex. 138 if a is equal to b and the angle A 
 is acute. 
 
 140. Proceed as in Ex. 138 if a is greater than b and (1) the 
 angle A is acute, (2) the angle A is right, (3) the angle A is obtuse. 
 
 141. Construct an isosceles triangle, given the base and the 
 vertical angle. 
 
 142. Through a given point outside a line draw a line parallel 
 to the given one. 
 
 HINT. Draw any line through the point cutting the given line at A". 
 Then at the given point construct an angle equal to one of the angles 
 at K so that the two will form a pair of corresponding angles. 
 
 143. Carry out the construction of Ex. 142 by using 245 only. 
 
 The adjacent figure gives the outlines of a Gothic arch. 
 The curves used in such arches are 
 circles. 
 
 QUERY 1. Are the centers of the arcs EC 
 and EF above or below the line DPI 
 
 QUERY 2. Is the center of arc BC on AB 
 or AB produced? 
 
 QUERY 3. Have the arcs BC and EF the 
 same center ? D~~A C F 
 
 EXERCISES 
 
 144. Construct a Gothic arch similar to the one above in which 
 AC is two inches and B is one and one- 
 
 half inches distant horn AC. 
 
 145. The adjacent form with added detail 
 is often the basis of a design for a window. 
 If the points A, B, and C are the centers of 
 the arcs opposite, show how to construct the 
 inscribed circle. 
 
BOOK II 139 
 
 Construction 7 
 
 238. Divide a given straight line into three or more 
 equal parts. 
 
 ,'S / 
 
 /\z*(_./b o/ v.< 
 
 " i ^sa 
 
 c 
 
 1. Given the line AB. 
 
 2. Required to divide the line AB into (say) jive equal parts. 
 
 3. Construction. Draw. 4 G, making Z.BAG about 30. 
 
 Take AK on AG as nearly equal to one fifth of AB as can be 
 estimated and lay off on AG in succession KR, RL, LM, and MP 
 each equal to A K. 
 
 Draw BP. 
 
 Then at K, R, L, and M respectively construct angles equal to 
 Z.BPG and in a corresponding position. 
 
 If necessary, extend the sides of these angles at K, R, L, and 
 M to cut AB in Z, H, 0, and V respectively. 
 
 Then AZ = ZH = HO = OV = VB. 
 
 4. Proof. Draw A S parallel to PB. 
 
 By construction, AK = KR = RL = LM = MP, and from the 
 construction of the angles at A, K, R, L, and M it follows that 
 AS, KZ, RH, LO, and MV are II to PB. Why ? 
 
 Therefore AS, KZ, RH, LO, MV, and PB are II to each 
 other. Why ? 
 
 AZ = ZH = HO == OV= VB. Why ? 
 
 QUERY 1. Why is it best to make the angle BA G in the above figure 
 acute rather than obtuse ? 
 
 QUERY 2. Why is AK made approximately one fifth of AB1 
 
140 
 
 PLANE GEOMETKY 
 
 EXERCISES 
 
 146. Construct an equilateral triangle, given a line equal to its 
 perimeter. 
 
 147. Construct a square, given a line equal to its perimeter. 
 
 148. Divide a given rectangle into five congruent rectangles. 
 
 149. Construct a rhombus, given its perimeter. How many such 
 figures may be constructed with a given perimeter ? 
 
 150. Construct a triangle, given one side and the two medians 
 from its extremities upon the other sides. 
 
 Construction 8 
 
 239. Construct a tangent to a circle from an outside 
 point. 
 
 1. Given a circle with center A y and a point K outside it. 
 
 2. Required to construct a tangent to the circle A from K. 
 
 3. Construction. Draw AK. Construct 7/6', the mid-perpen- 
 dicular of AK, cutting it at B. With B as the center and BA 
 as the radius, describe a semicircle, AK, cutting the circle A at R. 
 Draw KR. 
 
 Then KR is the required tangent. 
 
 4. Proof. Draw AR. Then Z.AEK is a right angle. Why ? 
 Therefore KR is tangent to the circle A at R. Why ? 
 
 QUERY. Ho-pr many tangents may be drawn to a circle from an 
 outside point? 
 
BOOK II 141 
 
 EXERCISES 
 
 151. Given the radii a and />, construct two circles tangent 
 internally and two others externally. (See Exs. 48 and 49.) 
 
 152. Given two tangent circles. From what points may two 
 equal tangents, one to each circle, be drawn ? 
 
 Construct two such tangents. 
 
 153. A is the center for the arcs BC and 
 KL. Construct a circle between the two arcs 
 tangent to each. 
 
 The figure below is the basis of many 
 designs for windows and doors in churches. This design 
 is often employed in the decorations about the entrance and 
 in those about the windows. 
 
 The point B is the center from which AD and KR are 
 described. 
 
 QUERY. In what position is the point C 
 with reference to the two arcs KR and ADI 
 
 154. If the arches of the adjacent figure 
 are equilateral, that is, if AB equals BD 
 equals AD, construct the design for a given 
 length of AB. 
 
 HINT. Observe that one can make the design by starting either 
 with the arc AD or with the shortest arcs of the figure. 
 
 155. Given a circle and a point A" outside it, construct a 
 triangle having K as one vertex and its sides tangent to the 
 circle. 
 
 156. Given a circle and a point K outside it, construct a 
 rhombus having K as one vertex and its sides tangent to the 
 circle. 
 
 157? In Ex. 153, if the arcs BC and KL are on opposite sides 
 of A, construct a circle tangent to each arc. 
 
142 PLANE GEOMETRY 
 
 Construction 9 
 
 240. With a line of given length as a chord con- 
 struct a segment such that every angle inscribed in it 
 shall equal a given angle. 
 
 1. Given line a and the angle BAG. 
 
 2. Required to construct a segment whose chord is equal to 
 a arid such that every angle inscribed in it shall equal /.BAG. 
 
 3. Construction. Select K, a point on A C, such that its distance 
 from AB is less than a. With K as the center and a as the 
 radius, describe an arc cutting AB, as at L. Draw KL. Construct 
 FG and HO, the mid-perpendiculars of KL and A L respectively. 
 Extend HO and GF until they intersect at R. With R as the 
 center and RK as the radius, construct a circle. This circle will 
 pass through A and L. Then the segment LQK is the required 
 segment. 
 
 4. Proof. KR = RL and RL = RA. Why ? 
 
 Therefore the circle whose center is R and whose radius is RK 
 passes through A and L. 
 
 By construction, KL = a. 
 
 Also, any angle inscribed in segment L QK equals Z BA C. Why? 
 
BOOK II 143 
 
 EXERCISES 
 
 158. Construct a triangle, given one side and the opposite angle. 
 How many such triangles are possible ? 
 
 159. Construct a triangle, given one angle, the side opposite, 
 and the median on that side. 
 
 160. Construct a triangle, given one angle, the side opposite, 
 and the altitude on that side. 
 
 LOCI 
 
 241. Examples of loci. 1. Where are all the points whose 
 distance from a given line AB is half an inch ? Answer : 
 
 They are on two lines, one on * 
 
 each side of AB, both par- i* 
 
 allel to AB and half an inch r-* 
 
 from it. |* 
 
 2. If two parallel lines AB 
 
 and CD are six inches apart, ^ g 
 
 where are all the points whose fy 
 
 distance from each line is the Ioctls 
 
 same ? Ansiver : On a line par- f , 
 
 allel to AB and CD, between _J 
 
 them, and three inches from ^ & 
 
 each. i 
 
 3. Where are all the points 3j 
 
 whose distance from one end of A ! B 
 
 a given line-segment AB is equal 
 to their distance from the other 
 end ? Answer : On the line per- 
 pendicular to AB at its mid-point. (See 116 and 117.) 
 
 QUERY. Consider any point, as C or _D, which is not on the per- 
 pendicular bisector of AB. Is it equidistant from A and 5? 
 
144 
 
 PLANE GEOMETRY 
 
 4. If two lines AS and CD intersect at K, where are all 
 the points equally distant from each line ? Answer : On the 
 bisectors of the angles formed by these two lines. (See 121 
 and 122.) The bisectors are per- 
 pendicular to each other. Why ? 
 
 Each of the four preceding 
 questions deals with what is 
 called a locus (plural loci, pro- 
 nounced 15 'si), a Latin word 
 meaning " place." , 
 
 242. Locus. A locus is a figure containing all the points, 
 and only those points, which fulfill a given requirement. 
 
 For example, the locus determined in 1 and 4 of 241 con- 
 sists of two lines. Although every point in either of these 
 lines satisfies the required condition, neither line alone can 
 properly be called the locus, because it does not contain all 
 the points which fulfill the given requirement. 
 
 When looked at in another way : 
 
 A locus is the path of a point which moves in such a way 
 that its every position fulfills a given requirement. 
 
 In plane geometry only those loci are considered which lie 
 wholly in one plane. 
 
 In the following queries determine the locus required. No 
 lengthy -detailed proof of the correctness of -the result is re- 
 quired, an accurate figure and a brief common-sense expla- 
 nation of why the proposed lines are the required locus is 
 all that is desired. Be sure to determine the entire locus. 
 Sometimes it consists of more than one line. 
 
 QUERY 1. What is the locus of a point two inches from a fixed points? 
 
 QUERY 2. Is a semicircle the locus of points equidistant from its 
 center? Explain. 
 
 QUERY 3. A fixed line AB, one side of a triangle, is three inches 
 long; the median drawn to it is two inches in length. How many 
 
BOOK II 145 
 
 triangles satisfying these conditions can be drawn ? What is the locus 
 of the vertices of all these triangles? 
 
 QUERY 4. Given two parallel lines, AB and KR, nine inches apart. 
 What is the locus of a point twice as far from AB as from KR ? 
 
 QUERY 5. One side of a triangle is two inches and the altitude upon 
 it is one inch. How many triangles satisfy these conditions? What is 
 the locus of the vertices of all these triangles? 
 
 QUERY 6. On one side of AB as a hypotenuse how many right tri- 
 angles can be constructed ? What is the locus of all the vertices of the 
 right angles of those triangles ? If right triangles be constructed on both 
 sides of AB as the hypotenuse, what is the locus ftf the vertices of the 
 right angles ? 
 
 QUERY 7. A fixed point A' is the point of intersection of the diag- 
 onals of a rectangle. One diagonal is four inches. How many rec- 
 tangles satisfy these two conditions? What is the locus of the vertices 
 of all the rectangles? 
 
 QUERY 8. A man walks across the floor so that he remains equi- 
 distant from the east and south walls of a room. What is his path ? 
 
 QUERY 0. One side of a square is six inches. A point moves so that 
 its distance from one side of the square is always one inch. What is 
 the locus of the poiiufc? Is the locus a continuous line? 
 
 QUERY 10. Lines are drawn parallel to one side of a triangle and 
 terminating in the other two. What is the locus of their mid-points? 
 
 QUERY 11. Lines parallel to one base of a trapezoid terminate in 
 the nonparallel sides. What is the locus of their mid-points? 
 
 QUERY 12. Draw KR, any line outside a given square. Inside the 
 square draw fifteen or twenty lines, terminated by its sides and parallel 
 to KR. Locate the mid-points of the lines and draw the line on which 
 these mid-points lie. What is the locus of the mid-points if instead of 
 twenty lines the number is infinite? 
 
 QUERY 13. In Query 12 replace the square by an equiangular hexa- 
 gon and determine the locus. 
 
 QUERY 14. A coin, diameter one and one-half inches, is moved 
 around another coin, diameter seven-eighths inches, the two coins 
 constantly touching each other and remaining in the same plane. 
 What is the path of the center of the larger coin ? 
 
 QUERY 15. K is the center of a fixed circle whose radius is five 
 inches. A circle whose diameter is one inch moves tangent to the first 
 and in its plane. What is the locus of the center of the smaller circle ? 
 
146 PLANE GEOMETKY 
 
 QUERY 16. K is any point outside the line AB. Lines are drawn 
 from K and terminate in AB. What is the locus of their mid-points? 
 
 QUERY 17. A hay barn, forty feet by eighty, stands in a meadow. A 
 horse is tied outside the barn to one corner by a rope one hundred feet long. 
 Draw a diagram showing the boundary of the area over which he can graze. 
 
 QUERY 18. With line AB as a side, triangles are described in which 
 the angle opposite AB is 45. What is the locus of their vertices? 
 
 QUERY 19. In. Query 18 change 45 to 60 and answer. 
 
 QUERY 20. K is any point outside a circle. Draw from K ten or 
 fifteen lines which terminate in the circle but do not cut it. Then 
 locate the mid-points*of these lines. What is their locus ? If the locus 
 is not apparent, change the position of K, or enlarge the circle, or do 
 both, and repeat the work until the form of the locus : is apparent. 
 
 QUERY 21. Triangles are constructed on one side of the same base 
 AB, all the angles opposite A B being equal. The bisectors of all the 
 equal angles formed will pass through a single point K. Given A B and 
 the point K, determine the locus of the third vertex of all such triangles. 
 
 243. Intersection of loci. Frequently the solution of a 
 construction problem depends on the location of one or more 
 points whose position fulfills two conditions. When this is 
 the case the locus for each condition should be constructed 
 separately. It should be noted that the points of intersec- 
 tion of the loci satisfy both conditions. The determination 
 of their position is often the first and sometimes the most 
 difficult step of the solution. 
 
 Example. Locate all points 
 which are equally distant from 
 two fixed points A and B and half 
 an inch from a third point C. \ /l \ 'C 
 
 Solution. Construct KR, the 
 mid-perpendicular of AB. With 
 C as the center and with a radius 
 of half an inch, describe the circle H. The circle cuts KR in 
 the points O and G, which are the points required in the example. 
 
 Proof. KR is the line which contains all the points, and only 
 those points, which are the same distance from A as from B. Why ? 
 
BOOK II 147 
 
 The circle C contains all those points, and only those points, whose 
 distance from C is half an inch. Now a point to be the same dis- 
 tance from A and B and a distance of half an inch from C must be 
 on both these loci. and G are two such points and the only two. 
 Discussion. The relative position of A, B, and C could be such 
 that the circle CH might cut KR, touch KR, or fail either to cut 
 or touch KR. The first case gives two points, as in the figure, 
 the second would give one, and the third would give none. 
 
 Observe that the following problem of construction requires 
 in its solution the location of a point by the intersection 
 of two loci. 
 
 EXERCISE 161. Construct a circle passing through a given point 
 and tangent to a fixed line at a given point of the line. 
 
 QUERY 1. A point is three inches from a fixed point A and two 
 inches from another fixed point B. When are there two solutions? 
 one? none? Explain. 
 
 QUERY 2. A point is one inch from a line AB and four inches from 
 a point K. May there be no solution for this ? two ? more than two ? 
 Explain. 
 
 QUERY 3. A point is equidistant from A and B and six inches from 
 the nearest point of a circle whose radius is two inches. Can there be 
 no solution? more than two? Explain. 
 
 QUERY 4. Locate a point which is equidistant from two intersect- 
 ing lines, and also equidistant from two given points. Can there be no 
 solution ? one ? more than one ? Explain. 
 
 QUERY 5. Locate a point which is two inches from a circle whose 
 radius is six inches, and four inches from a given straight line. Discuss 
 the possible solutions. 
 
 QUERY 6. AB and AC are two lines. KR, is a line which is not 
 parallel to AB or A C. A line HL two inches long and parallel to KR 
 terminates in AB and A C or these lines produced. Locate H and L. 
 Discuss. (See pp. 151-153 for a more exact treatment of loci.) 
 
 244. General directions for solving a construction problem. 
 If the solution is not apparent after a brief examination, 
 proceed as follows: 
 
148 PLANE GEOMETRY 
 
 1. Taking a common-sense view of the problem and its 
 data, draw (do not construct) as accurately as possible a 
 figure which approximately satisfies the conditions. Then, 
 using your knowledge of geometry, especially that of loci 
 (243) and of Constructions 1-9 (pp. 132-142), try to 
 devise a way of constructing the required figure. If the 
 desired method is still not apparent, ask yourself questions 
 like the following: Can I construct this part of the figure 
 (a triangle, perhaps) from the given data ? If so, can I then 
 construct another part, and finally the whole figure ? 
 
 2. If no solution results from 1, draw a new figure which 
 conforms to the data but has a different shape from the first. 
 Then repeat the other steps in 1 with the new figure. Some- 
 times success comes only after several figures have been drawn 
 and studied. 
 
 While conforming strictly to the conditions of a construction 
 problem we may select one of many possible positions for the 
 given points and lines and one of many possible sets of values 
 for the lengths for the given distances. The permissible varia- 
 tions here may result in no solution, in one solution, in a defi- 
 nite number of solutions, or in infinitely many solutions. In 
 the following exercises the word discuss (see the example of 
 243) calls for an analysis of the permissible changes in the 
 data and the consequent variation in the number of solutions. 
 
 MISCELLANEOUS EXERCISES IN CONSTRUCTION 
 
 162. Construct a parallelogram, given two adjacent sides. 
 
 163. Construct a circle of a given radius tangent to a given 
 circle at a given point thereon. 
 
 164. Given the three radii, construct three circles each tangent 
 externally to the other two. 
 
 HINT. Sketch three circles in the required position and study the 
 distances between their centers. 
 
BOOK II 
 
 149 
 
 165. Given two 'fixed circles, construct a third circle of a given 
 radius tangent internally to the other two. 
 
 166 . Given two fixed circles, construct a third circle of a given ra- 
 dius tangent externally to one and internally to the other. Discuss. 
 
 167. Construct a circle having its center in a fixed line and 
 passing through two fixed points. 
 
 HINT. What is the locus of the centers of all the circles passing 
 through two fixed points? 
 
 168. Construct a circle of a given radius touching a fixed circle 
 and a fixed line. 
 
 HINT. Where are the centers of all circles of given radius which 
 (1) touch the given line? (2) touch the given 
 circle? Where are the centers of the circles satis- 
 fying both conditions? 
 
 169. Construct four circles inside a square 
 each tangent to one side of the square and to 
 two of the circles. 
 
 170. Construct four circles inside a square 
 
 each tangent to two sides of the square and to two of the circles. 
 
 171. Construct the designs given below. 
 
 172. Construct a rectangle, given its perimeter and one side. 
 
 173. Given two points on a circle as the points of contact 
 of the sides of a circumscribed rhombus, construct the rhombus. 
 
 174. Inscribe a regular hexagon in a given circle. (See Ex. 5, 
 p. 95.) 
 
150 
 
 PLANE GEOMETRY 
 
 175. Inscribe an equilateral triangle in a given circle. 
 
 176. Construct an equilateral triangle with its sides touching 
 a given circle. 
 
 HINTS. Draw an equilateral triangle and inside it draw a circle 
 touching the sides of the triangle at K, It, and L respectively. Draw 
 radii OR, OK, and OL. How many degrees are there in Z/tO/t? 
 Conclusion ? 
 
 177. Construct a right triangle, given the hypotenuse and one 
 acute angle. 
 
 HINT. What is the locus of the vertex of the right angle of all right 
 triangles constructed on the given hypotenuse? 
 
 178. Construct the bisector of the angle between two given 
 nonparallel lines without extending them to their point of 
 intersection. 
 
 179. Given two fixed circles, construct their common external 
 tangent. 
 
 HINT. LetAC = R-r. How is B C drawn? AKt BL1 
 
 180. Through one vertex of a triangle draw a line equidistant 
 from the other two vertices. Discuss. 
 
 181. Given AB and A C, two intersecting lines, and K any point 
 within the angle BAC. Draw through K a line terminated by AB 
 and A C and bisected by K. 
 
 182. Given the triangle ABC. Find by construction one or 
 more points on its circumscribed circle without using its center 
 or its radius. 
 
BOOK II 151 
 
 245. General directions for solving a locus problem. The 
 
 complete solution of a locus problem which is not at once 
 obvious involves two distinct processes: 
 - 1. The drawing of a good figure and locating, as accurately 
 as possible, enough points which answer the requirements to 
 indicate the position and character of the lines composing the 
 desired locus. 
 
 Step 1 is really a matter of experiment and discovery, and 
 while it is of the utmost importance, the description of its details 
 is not required in any written or oral solution of a problem. 
 
 2. The drawing of the locus discovered in step 1, the 
 assertion that the line or the system of lines so drawn is the 
 locus, and the proof of the assertion. This geometrical proof 
 consists of two parts r 
 
 a. The proof that every point on the proposed locus 
 satisfies the given condition. 
 
 b. The proof that any point which satisfies the given 
 condition is on the proposed locus. 
 
 Frequently step 1 is unnecessary because of the simple char- 
 acter of the locus. As an illustration of step 2 note question 3 
 of 241. Then turn to 116 and 117. These two theorems 
 really deal with the locus of a point which is equally distant from 
 the extremities of a line. The first proves that every point of 
 that description is on the mid-perpendicular of the line. This is 
 a under step 2. The second proves that every point which is 
 equally distant from the extremities of the line is on the mid- 
 perpendicular of the line. This is b under step 2. 
 
 Question 4 of 241 has in a way been touched upon in 121 
 and 122. These two theorems constitute a and b respectively 
 under step 2. 
 
 The following example illustrates 245, and shows how to 
 attack the more difficult locus problems and indicates just 
 what should be stated in the solution itself. 
 
152 
 
 PLANE GEOMETRY 
 
 246. Example. Solution of a locus problem. 
 
 From point K outside a circle whose center is A secants are 
 drawn. What is the locus of the mid-points of the chords 
 so formed ? 
 
 Draw a number of secants, each making about the same angle 
 with the two adjacent to it, and locate very accurately the mid- 
 points of the corresponding chords, as in the figure below. 
 
 Now it can be seen that points 1, 2, 3, etc., seem to form an are 
 of a circle. Inspection indicates that the center of this circle is on 
 AK. Closer inspec- 
 tion shows that the 
 mid-perpendicular of 
 the chord from A to 
 point 1 will nearly 
 
 bisect line AK. It | -A 
 
 seems very probable, 
 therefore, that AK 
 is a diameter of the 
 circle. It thus ap- 
 pears reasonable to 
 infer that the required locus is that portion of the circle having 
 A K as the diameter which lies within the circle whose center is A. 
 
 Thus we discover the probable locus. If we had not done so, we 
 ought to have moved K away from or toward the circle, or used a 
 larger circle for A, or both, and to have inspected carefully the 
 resulting figure. 
 
 In every case, if we are to discover anything worth while, an 
 accurate figure is absolutely necessary. Moreover, we must not 
 always expect to discover the locus by just one drawing, however 
 carefully it may be done. Finally, however excellent the figure, 
 we must study it attentively, calling to our aid in the process 
 whatever intuition and geometrical imagination we possess. 
 
 The preceding work of discovery, while absolutely essential in 
 determining the nature of a locus, need never be written down 
 in the solution of a locus problem. All that need be included in 
 any solution is illustrated by what follows. 
 
BOOK II 153 
 
 Given a circle with center 4, and a point K outside the circle. 
 
 Required the locus of the mid-points of all the chords formed by 
 secants from K. 
 
 Solution. On AK as a diameter describe a circle cutting A at 
 B and C. Then arc BA C is the required locus. 
 
 Proof. (1) Let L be any point on arc BAC. Draw the secant 
 KG L R and chord AL. 
 
 Since AK is a diameter, Z.ALK = 90. Why ? 
 
 Then RL = LG. Why? 
 
 Since L is any point on BAC, the preceding proof holds for 
 every point on arc BA C. 
 
 (2) Let P be the mid-point of any chord, as HM, formed by the 
 secant KMH. Let HM cut the arc BAC at O. Draw AO. 
 
 90. Why? 
 
 HO == MO. Why ? 
 
 Hence P must coincide with 0. Why ? 
 
 Therefore arc BAC is the required locus. 
 
 EXERCISES 
 
 183. Given a point K on a fixed circle, find the locus of the 
 mid-points of all the chords drawn from K. 
 
 184. Given a point A" within a fixed circle, find the locus of the 
 mid-points of all the chords drawn through K. 
 
154 PLANE GEOMETRY 
 
 185. From a point A' outside a fixed circle a number of secants 
 are drawn each cutting the circle in two points and terminating 
 in it. What is the locus of their mid-points and of the mid- 
 points of the external part of each ? 
 
 HISTORICAL NOTE. In the field of mathematics there is scarcely any 
 name more noted than that of Euclid, the first professor of mathematics 
 in the first university in the world, the University of Alexandria. His 
 great reputation rests securely on his "Elements," a textbook on elemen- 
 tary mathematics (see page 86), devoted mainly to geometry but in part 
 to algebra and to the theory of numbers. The "Elements " summed up in 
 such a masterly way the work of his predecessors that though similar 
 works existed before it, none has come down to us. This last, of course, 
 may have been due in part to the standing which his connection with the 
 University of Alexandria gave him. The known facts regarding Euclid 
 are few. We know that he was a Greek and that he was born, probably 
 at Athens, about 330 B. c. and died about 275 B. c. His selection (300 B. c.) 
 for the University might indicate that he was eminent among the 
 mathematicians of his day. Yet such a conclusion is merely probable. 
 The " Elements " itself offers little evidence on this point, for it is known 
 to be largely a compilation, and whether any important part of it was 
 original with Euclid or whether he made any original discoveries in 
 mathematics is a matter of conjecture. 
 
BOOK III 
 
 RATIO, PROPORTION, AND SIMILAR FIGURES 
 
 247. Introduction. The results obtained in Book I cluster 
 about the theorems on congruent triangles and those in Book II 
 about circles. In both books equal lines and equal angles 
 were dealt with, but only two magnitudes were considered 
 at a time. It will be found that the work of Book III is 
 largely concerned with figures which have proportional lines 
 and especially with what are termed similar polygons. One 
 property of such polygons is that the lengths of any two 
 sides of one are proportional to the lengths of the corre- 
 sponding sides of the other. Hence in Book III we shall 
 deal very largely with four magnitudes at a time. As in 
 Books I and II, so also in Book III equal angles will play 
 an important part. But while Books I and II deal continu- 
 ally with two equal lines, Book III will be concerned almost 
 wholly with four proportional lines. Therefore, some clear 
 notions of ratio, measurement, and proportion, which are 
 really algebraic in character, must be grasped before the 
 strictly geometrical work of Book III can be advantageously 
 taken up. 
 
 248. Ratio. The ratio of one number, a, to a second num- 
 ber, 5, is the quotient obtained by dividing the first by the 
 
 second, or 
 o 
 
 The ratio of a to b is also written a : >, the colon meaning 
 " divided by." 
 
 155 
 
156 PLANE GEOMETRY 
 
 It follows from the above that all ratios are fractions and 
 all fractions may be regarded as ratios. 
 
 Thus 3:2, - > - > - > 6:2, or - are ratios and fractions. 
 
 249. The terms of a ratio. The first term, or the numerator, 
 in a ratio is called the antecedent:, and the second term, or 
 the denominator, is called the consequent. 
 
 EXERCISES 
 
 Write the following ratios as fractions and, if possible, reduce 
 the fractions to their lowest terms : 
 
 2. 27 inches : 10 yards. 
 
 3. (o-4):(-2). 6 
 
 7. Separate 80 into two parts in the ratio of 2 to 3. 
 
 8. Separate 154 into three parts in the ratio of 2 : 3 : 6. 
 
 9. The value of a ratio is -f and the antecedent is 12 less than 
 the consequent. Find the ratio. 
 
 250. Measurement. We may speak of the ratio of two con- 
 crete numbers if they have a common unit of measure. The 
 ratio of 5 yards to 4 feet is ^ 6 -, the common unit of measure 
 being 1 foot. Obviously, no ratio exists between 5 years and 
 3 feet ; that is, a ratio can exist only between magnitudes of 
 the same kind. 
 
 If we say a piece of paper contains 54 square inches, we are 
 expressing by the number 54 the ratio of the surface of the 
 paper to the surface of a square whose side is 1 inch. 
 
 Every measurement, then, is the determination of a ratio, 
 either exact or approximate. 
 
BOOK III 157 
 
 251. Magnitudes which have no common unit of measure. 
 Consider the right triangle ABC in which AB=BC = \. 
 
 The student has used in arithmetic the fact, which will be proved 
 in 284, that _, 
 
 Therefore ^4C 2 = 1 2 + 1 2 , 
 
 or A C = V2. 
 
 Now V2 cannot be exactly expressed by deci- 
 mals or by a common fraction. Its first seven figures 
 are 1.414213, and the decimal part is nonrepeating. 
 If we suppose AB divided into ten equal parts, one of these would 
 be contained 14 times and a little over in AC. Again, divide AB 
 into 100 equal parts. Then one of them would be contained 141 
 times and a little over in AC. Again, divide AB into 1000 equal 
 parts. Then one of these would be contained 1414 times and 
 a little over in A C. Here, no matter how minute we make the 
 equal divisions of AB, one of them is never exactly contained 
 in A C. This illustrates the fact that A B and A C have no common 
 unit of measure. 
 
 The circumference of a circle and its diameter is another 
 example of two magnitudes having no common unit of 
 measure. Here C -^ />= 3.14159 -f a never-ending, nonrepeat- 
 ing decimal. 
 
 252. Commensurable magnitudes. If two magnitudes have a 
 common unit of measure, they are said to be commensurable. 
 
 253. Incommensurable magnitudes. If two magnitudes 
 of the same kind have no common unit of measure, they are 
 said to be incommensurable. 
 
 The distinction between commensurable and incommensurable 
 magnitudes is never one which can be observed by our sense of 
 sight or of touch. For instance, there might be two commensurable 
 lines such that no measure greater than one millionth of an inch 
 would be contained exactly in both. Such a unit of measurement 
 is too small to see even with a microscope, so that our senses 
 
158 PLANE GEOMETRY 
 
 would make these lines appear incommensurable ; but according 
 to our definition these magnitudes are as truly commensurable as 
 if the common measure were an inch. It appears, therefore, that 
 the distinction between commensurable and incommensurable 
 numbers would never occur to the mechanic, the draftsman, or 
 the engineer, since he could never use his instruments with suffi- 
 cient accuracy to detect it. Mathematics, however, does not deal 
 with lines of chalk or of ink. Although we use drawings to sug- 
 gest to our minds the real geometric figures, the geometrical line 
 does not exist outside our thoughts. It is entirely ideal. Hence 
 properties or magnitudes too minute to appeal to our senses, but 
 not beyond the reach of exact thought, may be very real and 
 important in the study of the science of geometry, where one 
 seeks relations which are not merely approximately true but 
 absolutely so. 
 
 QUERY 1. The decimal .272727 + is never-ending. Is it incommen- 
 surable with 1 ? 
 
 QUERY 2. Reduce y\ to a decimal. What bearing has the result 
 on the answer to the previous query? 
 
 254. Proportion. Four numbers, a, 5, c, and c?, are in 
 proportion if the ratio of the first pair equals the ratio of 
 the second pair. 
 
 This proportion is written either in the form (1) a : b = c : d 
 
 The second form is the usual algebraic way of writing an 
 equation in which each member is a simple fraction. For this 
 reason it is preferable, as the axioms and rules of operations 
 already familiar to the student will seem to him more directly 
 applicable to it than to the first form. 
 
 Since a proportion is an equation of a special type, certain 
 special relations exist among the four numbers involved which do 
 not exist in every equation. As the need for them arises, -these 
 special relations will be stated as theorems of proportion and 
 proved. 
 
BOOK III 
 
 159 
 
 EXERCISES 
 
 4 7 
 10. Solve l = i - 
 
 5 x 
 
 11. Solve = 
 6 
 
 5 
 
 12. Solve 4:^-3=7:2 - x. 
 
 If, referring to the adjacent figure, we write = > 
 
 the length of AK the length of AC *. .? 
 we mean - j = - ~ T Whether the 
 
 the length of / the length of ^C 
 
 unit of measure is some common standard, as the inch or the 
 
 centimeter, or a wholly arbitrary length is not 
 
 stated and really makes no difference. But 
 
 whatever the unit, AK in the proportion means 
 
 the number of times the unit of length is con- 
 
 tained in AK. A like meaning is understood 
 
 for BK y AC, and EC. Similarly, in all propor- 
 
 tions involving the lines of a figure the terms 
 
 of the proportion are really numbers. A 1C ~B 
 
 255. Corresponding segments of two systems of intersecting 
 lines. If two or more concurrent lines are cut by another 
 group of two or more lines, as in the 
 adjacent figure, certain segments 
 taken four at a time are spoken of as 
 corresponding. Sometimes two seg- 
 ments are spoken of as correspond- 
 ing to two others. The segments of 
 the four following groups in the 
 order named are corresponding : 
 
 1. 5, c, e, f. 3. M, v, x, y. 
 
 2. v, w, y, z. 4. a, g, b, h. 
 
 QUERY. Are the following groups corresponding ? 
 
 1. a, b, g, h. 2. u, w, x, y. 3. a, b, u, v. 4. a, b, h, i. 
 
 Usually the term corresponding as used in Book III will apply to two 
 or more concurrent lines cut by two or more parallels. 
 
160 PLANE GEOMETKY 
 
 Theorem l 
 
 256. If a line parallel to one side of a triangle divides 
 a second side into two commensurable segments, it divides 
 the third side into corresponding segments which are 
 proportional. 
 
 
 Given the triangle ABC in which KR, parallel to AB, intersects 
 AC and BC at K and R respectively and forms the commensurable 
 segments CK and KA. 
 
 *i 4 CK CR 
 
 To prove that = _ - 
 
 Proof. Apply to A K and KC the common unit of measure CL. 
 Suppose it is contained h times in CK and ni times in K A . Through 
 C and the points of division draw parallels to AB, cutting CB. 
 
 Then CR will be divided into h equal parts and BR into m 
 equal parts. . 129 
 
 CK h ... 
 
 C* 7? 7i 
 
 Also - = - (2) Why? 
 
 Therefore from (1) and (2), 
 
 = Why? 
 
 KA RB 
 
 EXERCISE 13. In the figure above, if A K is 6 inches, KC is 
 10 inches, and .BC is 24 inches, find BR and CR. 
 
BOOK III 161 
 
 257. Alternation theorem of proportion. If four terms are in 
 proportion, they are in proportion by alternation; that is, the 
 first is to the third as the second is to the fourth. 
 
 Given f = . (1) 
 
 To prove that ~ c = d' (2) 
 
 Proof. Multiplying by bd in (1), 
 
 ad = be. (3) Why ? 
 
 Dividing by cd in (3), - = - Why ? 
 
 c a 
 
 258. Fourth proportional. A fourth proportional to three 
 
 numbers a. b, and c is the number a; if - = - 
 
 b x 
 
 EXERCISES 
 
 14. Find a fourth proportional to 7, 3, and 18. 
 
 15. Write by alternation f = 
 
 16. Solve the following proportions for x. Write each by 
 alternation and solve. Compare the two results in each case. 
 4 : 5 = 6 : x ; a : b = c : x. 
 
 17. Find a fourth proportional to 8, 9, and 36. 
 
 HISTORICAL NOTE. The Greeks never succeeded in uniting the ideas 
 of number and magnitude. To them number meant nothing but inte- 
 gers and common fractions. Irrationals were not numbers. In all their 
 arithmetical work they were hampered by a number system even more 
 cumbersome than that of the Romans. This made proportion a rather 
 formidable idea. In modern symbols Euclid's definition of a propor- 
 tion is as follows : Four numbers, a, b, c, and d, are in proportion if for 
 any integers, m and n, we have ma > = < nb, according as me > = < nd. 
 
162 PLANE GEOMETRY 
 
 Theorem 2 
 
 t, 
 259. If a line parallel to one .side of a triangle divides 
 
 a second side into two incommensurable segments, it 
 divides the third side into two corresponding segments 
 which are proportional to the first pair. 
 
 C 
 
 A B 
 
 Given the triangle ABC with KR parallel to AB, and CK and 
 KA incommensurable. 
 
 CK CR 
 
 To prove that 
 
 KA RB 
 
 T> x a CK CR ' u 4.1. 4- CK CR ' 
 
 Proof. Suppose - = is not true, but that - = is 
 
 YY J\. J t -O j V A. J t -/ 
 
 true, and that the point T is between R and B. (1) 
 
 Now take a point S between T and B so that RS is commen- 
 surable with CR and draw SP parallel to KR. 
 
 Then f = if' ( 2) 256 
 
 CK KA 
 
 From(l), ~c^ = ^' ^ 
 
 From (2), = (4) 
 
 From (3) and (4), f| = || - (5) Why ? 
 
 But since #P < KA and RS> RT, the numerator of the second 
 member of (5) is less than that of the first, while the denominator of 
 the second member is greater than that of the first. On both grounds 
 
BOOK III 163 
 
 the second member is less than the first and not equal to it. 
 Consequently the assumption (1), which leads to this false result 
 must be incorrect. In like manner we obtain a false result if T 
 is assumed to fall on RB produced. Hence RB must be the true 
 fourth proportional to CK, KA, and CR, and 
 
 CK _ CR 
 KA ~ RB ' 
 
 EXERCISES 
 
 18. Take the fraction J* and try to find an equal fraction 
 having a numerator smaller than 15 and a denominator greater 
 than 32. What point in the proof of Theorem 2 does this 
 illustrate ? 
 
 19. Prove that part of Theorem 2 in which the point T is 
 assumed to fall on RB produced. 
 
 260. Corollary. A line parallel to one side of a triangle and 
 cutting the other two divides them into four corresponding seg- 
 ments which are proportional. 
 
 Outline of proof. Using the figure of 259, from Theorems 1 
 and 2 we have 
 
 = 2. . m 
 
 KA RB 
 Therefore, from (1), f = ~ 257 
 
 EXERCISES 
 
 20. Using the figure of 259, if- CK is 9, KA is 6, and RB is 8, 
 find CB. 
 
 21. Using the figure of 259, if AK is 12, CK is 18, and CB 
 is 25, find CR and RB. 
 
 22. Using the figure of 259, if AC = 4, CR = 24, and CK is 
 twice KA, find CK, KA and RB. 
 
164 PLANE GEOMETRY 
 
 261. Inversion theorem of proportion. If four terms are in 
 proportion, they are in proportion by inversion ; \that is, the 
 second is to the first as the fourth is to the third. 
 
 Given ? = . (1) 
 
 To prove that - = - . (2) 
 
 a c 
 
 Proof. Multiplying in (1) by bd, 
 
 be, = ad. (3) 
 
 Dividing in (3) by ac, - = - Why ? 
 
 a c 
 
 262. Addition theorem of proportion. If four terms are in 
 proportion, they are in proportion by addition ; that is, the sum 
 of the first two is to either as the sum of the second two is to 
 the corresponding one. 
 
 Given - = - (T) 
 
 To prove that --^ = ^l* , (2) 
 
 b d 
 
 and L+* = f?. (3) 
 
 a c 
 
 Proof. Adding 1 to each member of (1), 
 
 From (4), ' = ' (5) 
 
 Write (1) by inversion. Then (3) can be obtained as was (5). 
 
 EXERCISES 
 
 23. Write ^ = \^ by inversion and by alternation and each 
 result by addition. Are the statements thus obtained proportions ? 
 
 24. If a : b = c : d, prove that a + b : a = c + d : c. 
 
BOOK III 165 
 
 Theorem 3 
 
 263. If a line parallel to one side of a triangle cuts 
 the other two sides, the two sides are in proportion to 
 their corresponding segments. 
 
 Given the triangle ABC with KR, parallel to AB, forming 
 segments CK and KA of CA and CR and RB of CB. 
 
 To prove that 
 
 1././X 
 
 Z? = 
 
 \SJL\- ji^n 
 
 ~CR = ~RB 
 
 
 
 
 CK 
 
 CR 
 
 
 
 Proof 
 
 ~KA = 
 
 
 (1) 
 
 260 
 
 CK -f- 
 
 H vnm il ^ 
 
 KA 
 
 CR -f- RB 
 
 (2) 
 
 262 
 
 1} > KA 
 
 
 RB 
 
 Thus (2) becomes 
 
 CA 
 KA ~ 
 
 CB 
 
 RB' 
 
 (3) 
 
 Why? 
 
 From (3), 
 
 CA 
 
 KA 
 
 RB' 
 
 ( 
 
 257 
 
 From (1), 
 
 CK 
 
 CR ~ 
 
 KA 
 
 ~RB' 
 
 (5) 
 
 257 
 
 Hence, from (4) and (5), 
 
 CA _ 
 CB ~ 
 
 KA CK 
 
 RB ~ CR ' 
 
 
 
 EXERCISES 
 
 25. In the figure above, if CK is 6, CR is 4, and A K is 10, find BR. 
 
 26. In the figure above, if A C is 16, CR is 8, and BR is 6, find A K. 
 
 27. In the figure above, if AC is 20, BC is 24, and BR is 6, 
 find CK. 
 
166 PLANE GEOMETRY 
 
 Theorem 4 
 
 264. If two nonparallel lines are cut by three or more 
 parallels, the corresponding segments of the two trans- 
 versals are proportional. 
 
 H 
 
 
 L X 
 
 M 
 
 / \ 
 
 Given two nonparallel lines AD and KM cut by the parallels 
 AK, BR, CL, and DM. 
 
 , AB BC CD 
 
 To prove that _ = __. 
 
 Proof. Let A D and KM intersect in H. 
 
 Then ff = ti' (1) 263 
 
 |f = ff- (2) , 2 60 
 
 A D 7?^ 
 
 From (1) and (2), ' = Why ? 
 
 A/I JKLi 
 
 7?C C D 
 
 In like manner it can be proved that > and so on for 
 
 ,, , /l^L> jL^U 
 
 any number of parallels. 
 
 j z? /^ r n 
 A> x>C C x/ 
 
 Therefore __ = _ = _. 
 
 QUERY. If the word nonparallel is omitted in the statement of 
 Theorem 4, is the resulting theorem true? 
 
 EXERCISE 28. In the figure above, AB is 8, BC is 7, CD is 10, 
 and KM is 30. Find KR, RL, and 
 
BOOK III 167 
 
 Theorem 5 (Converse of the Corollary-, 260) 
 
 265. If a line divides two sides of a triangle into pro- 
 portional corresponding segments, it is parallel to the 
 third side. 
 
 Given the triangle ABC and line KR dividing sides AC and 
 BC so that ^ = ? 
 
 KA RB 
 
 To prove that KR II to AB. 
 
 Proof. Suppose KR is not II to AB. Then draw KH II .to AB., ,. 
 
 Then fhif" (2); 260 
 
 /- r> C* J-f 
 
 From (1) and (2), = ' ( 3 ) 
 
 Thus (4) becomes ^ = J^' ( 5 ) 
 
 From (5), RB = J75. Why ? 
 
 Therefore .R and H coincide, Why? 
 
 and since KH is II to AB, KR is II to AB. 
 
 EXERCISE 29. ABC and ABK are two triangles on opposite 
 > sides of AB. Through L on BK a parallel to the line from K to C 
 cuts BC in M, and another line through L parallel to KA cuts AB 
 in //. Prove that BH : BA BM:BC. 
 
168 PLANE GEOMETRY 
 
 266. Corollary. (Converse of Theorem 3.) If a line outs two 
 sides of a triangle so that the two sides are proportional to two of 
 their corresponding segments, the line is parallel to the third side. 
 
 EXERCISES 
 
 30. A line intersects two sides of a triangle and the lengths 
 of the four corresponding segments formed are respectively 10, 
 6, 18, and 9. Is the line parallel to the third side? 
 
 31. Using the figure of 265, if AK is 10, KG is 15, and 
 CR is 18, what value must BC have so that KR will be parallel 
 to AB? 
 
 32. R is a point in the common side of the triangles ABC and 
 ABH. RL, parallel to A C, meets BC in L ; and RK, parallel to AH, 
 meets BH in K. Draw LK and CH and prove them parallel. 
 
 33. In the triangle ABC a straight line cuts AB&tL and A C at 
 R and the third side produced at K. If AR equals AL, prove that 
 BK:CK = BL: CR. 
 
 HINTS. Let CH, parallel to KL, cut AB at H. Then BK-.CK 
 -Bl'.LH. Lastly; prove LH equal to CR. 
 
 267. Similar polygons. Two polygons are szm7ar(symbol~) 
 if the angles of one are equal 
 
 respectively to the angles of the 
 other and the sides are propor- 
 tional each to each. 
 
 Thus the polygons ABCDEF 
 and A'B'C'D'E'F' are similar if 
 
 (1) 
 
 and 
 
 AJL- BC CD - DE EF =- FA 
 A'B'~ B'C' ~ C'D' D'E' E'F' F'A 1 
 
BOOK III 
 
 169 
 
 268. Corresponding lines. In similar (or congruent) poly- 
 gons two lines which are in like position with respect to the 
 equal angles are called corresponding or homologous lines. 
 
 In triangles the corresponding sides lie opposite the equal 
 angles. 
 
 269. Corollary. Corresponding sides of similar polygons are 
 in proportion by 267 and 268. 
 
 QUERY 1. Must similar polygons have the same number of sides? 
 of angles ? 
 
 QUERY 2. Are all squares similar? all rectangles? all equilateral 
 triangles ? 
 
 QUERY 3. Can two polygons be mutually equiangular without being 
 similar? Illustrate. 
 
 QUERY 4. Are the sides of the following polygons proportional each 
 to each : (a) two squares? (6) two rectangles ? (c) two parallelograms? 
 (') two regular polygons? (e) two regular polygons having the same 
 number of sides? 
 
 QUERY 5. Can the sides of two polygons be proportional each to 
 each and the two polygons not be similar? Illustrate. 
 
 QUERY G. If two triangles are similar to a third, are they similar to 
 each other? Why? 
 
 QUERY 7. Can two rectangles be drawn whose sides are not propor- 
 tional each to each ? Illustrate. 
 
 QUERY 8. Can two parallelograms be drawn whose sides are propor- 
 tional each to each and whose angles are not equal each to each? 
 Illustrate. 
 
 QUERY 9. Are there any two polygons which must have their sides 
 proportional each to each if the polygons are mutually equiangular? 
 
 QUERY 10. Are there any two poly- 
 gons which must be mutually equiangular 
 if their sides are proportional each to 
 each? 
 
 QUERY 11. Are equal triangles similar? 
 Why? 
 
 QUERY 12. In the three rectangles of 
 the adjacent figure, a: is a third proportional (see 283) to b and a. 
 Are any two of the three rectangles similar? Which? Prove. 
 
 * o 
 
 
 F 
 
 
 a 
 
 
 
 
 G 
 
 
 H-*-H 
 

 
 170 PLANE GEOMETRY 
 
 Theorem 6 
 
 270. If two triangles are mutually equiangular, they 
 are similar. 
 C 
 
 C' 
 
 AH B A B' 
 
 Given the triangles ABC and A'B'C', in which the angle A 
 equals the angle A' y the angle B equals the angle B', and the 
 angle C equals the angle C'. 
 
 To prove that A ABC AA'B'C f . 
 
 Proof. Since Z C = Z C', we can place AA'B'C 1 on A ABC so 
 that point C' falls on point C, side C'A' on CA, side C'B' on CB, 
 and yt',6' in the position KR. 
 
 By hypothesis, Zl = Z.t. 
 
 Hence A^ is II to AB. Why ? 
 
 (~* fC C* T* 
 
 Therefore cx = cF <*> Why ? 
 
 C 'A ' C'B 1 
 
 But (1) is the same as = , t (2) 
 
 C JT. C JJ 
 
 Since Z-B^LB 1 , we may now place AA'B'C' on AABC so 
 that point 5' falls on 5, side .B'C" on BC } side .SM' on BA } and 
 X'C' in the position HL. 
 
 By hypothesis, /.A = Z.2. 
 
 Hence 7/Z is II to AC. Why ? 
 
 - <*> *>' 
 
 Fro m(2 )and( 3 ) ( - = = . Why? 
 
 Therefore AABC AA'B'C'. 267 
 
BOOK III 
 
 171 
 
 271. Corollary i. A line cutting two sides of a triangle and 
 parallel to the third side forms a second triangle similar to the first. 
 
 272. Corollary 2. If two angles of one triangle are equal re- 
 spectively to two angles of another, the triangles are similar. 
 
 273. Gunther's scale. The adjacent figure shows a diag- 
 onal scale (enlarged) whose unit is one inch. It is used in con- 
 nection with a pair of dividers for determining the distance 
 between two points in inches and hundredths of an inch. 
 
 To understand the scale consider first the triangle ABC, in 
 which BC represents -^ of an inch. What are the lengths in 
 hundredths of an inch represented by the portions of the lines 
 1, 2, 4, etc. which are included between lines BA and CA ? 
 
 EXERCISES 
 
 34. What is the distance from the point marked on AB to the 
 point marked on the oblique line 6-7 ? 
 
 35. What is the distance from each of the three points marked 
 on the oblique line 2-3 to the line AB? 
 
 36. In the triangle ABC the side AB is 8, BC is 12, and CA 
 is 16. A line parallel to BC cuts AB at K so that AK is 6 and 
 cuts AC at R. Find the other sides of the triangle AKR, 
 
172 
 
 PLANE GEOMETRY 
 
 37. The sides of a trapezoid are 10, 15, 16, and 36 respectively. 
 The two longer sides are parallel. The other two sides t are extended 
 to meet. Find the perimeter of the smaller triangle so formed. 
 
 38. In the figure of 264, if AC is 18, AK is 16, and CL is 24, 
 find HA. 
 
 39. A line is drawn from the vertex C of triangle ABC to 
 K on the opposite side, produced if necessary, 
 
 making angle A CK equal to angle B. What two 
 triangles of the figure are similar ? Prove. 
 
 40. CM is a median of the triangle ABC. 
 KRL, parallel to AB, cuts AC in K, CM in K, 
 and EC in L. Prove that KR is equal to II L. 
 
 41. If an 8-inch square is cut as indicated 
 in ABCD and the parts arranged in g 
 EFGHj it may be made to appear that 
 
 the areas of the two figures are equal or 5 
 that 64 square inches equals 65 square 
 inches. Where is the error ? Prove. 
 
 B 
 
 Theorem 7 
 
 274. If an angle of one triangle equals an angle of 
 another and the sides including those angles are propor- 
 tional, the triangles are similar. 
 
 c' 
 
 B' 
 
 A^ 
 
 Given the triangles ABC and A'B'C' in which angle C equals 
 angle C' and 
 
 To prove that 
 
 C'A 1 ~~ C'B' 
 
 AABC ^ 
 
BOOK III 173 
 
 Proof. Since /. = /^C', we may place AA'B'C' so that point 
 
 C' falls on C, side C'A' on CA, and side C'B' on CB. Then 
 A 'B' will fall in the position KR. 
 
 CA CB CA CB 
 Now, by hypothesis, = , or = - 
 
 Hence ^ is II to AB. 266 
 
 Therefore AKRC ~ AABC, 271 
 
 or AABC^A.i'JJ'C 1 . Why? 
 
 EXERCISES 
 
 42. Proportional compasses are often used in enlarging or 
 reducing drawings. These instruments have a clamp, K, which 
 may be set at any point in the slots in the legs. 
 Where must K be placed with respect to A, B, 
 C, and D so that the dimensions of a copy of 
 a drawing will be 60% greater than- in the origi- 
 nal, AB being the distance between two points in 
 the latter? 
 
 43. The bases of a trapezoid are 12 and 18 
 respectively ; its altitude is 10. The non parallel 
 sides are produced to meet. Find the altitude of 
 each triangle thus formed. 
 
 44. The diagonals of quadrilateral AB CD inter- 
 sect in point K so that A K : KC = BK : KD. Prove that ABCD is 
 a trapezoid. 
 
 45. Two corresponding medians of similar triangles form two 
 pairs of similar triangles. 
 
 46. In the figure of 274, if CK equals one half A K and AB 
 is 24, find A'B'. 
 
 47. How would the points C and D of the proportional com- 
 passes be placed on the scale of 273 in order to set the compasses 
 so that the distance CD will be 1.43 inches ? 
 
174 PLANE GEOMETRY 
 
 Theorem 8 
 
 275. If the sides of two triangles are respectively pro- 
 portioned, the triangles are similar. 
 
 .C 
 
 .C' 
 
 Given the triangles ABC SLndA'B'C' in which 
 AB _ BC_ CA 
 A'B' ~ We ~~ C'A' ' 
 
 To prove that A ABC ~ A A'B' C'. 
 
 Proof. On CA lay off CK equal to C'A ' and on CB lay off CR 
 equal to C 'B'. Draw KR. 
 
 Then A CKR ~ACAB. 274 
 
 CA CB AB 
 
 But CK = C'A', 
 
 and/ CR=B'C'. 
 
 CA EC AB 
 Hence (2) becomes -^ = = 
 
 From (1) and (3), ^| = |f>- ( 4 ) 
 
 From (4), A 'B' = /a?. Why ? 
 
 Hence A .1 7>"C" is congruent to A KRC, Why ? 
 
 or A A'B'C' ^ A yl/3C. Why ? 
 
 NOTK. From Theorems 6 and 8 it follows that the triangle is unique 
 among polygons. If either one of the conditions of the definition of 
 267 holds for two triangles, the other follows by 270 and 275 
 as a necessary consequence. But for any polygons except triangles 
 one of the conditions of 267 may exist without the other. 
 
BOOK III 175 
 
 EXERCISES 
 
 48. A triangle is constructed with the longest sides respectively 
 of three similar unequal triangles. Is the triangle similar to 
 the other three ? Prove. 
 
 Can such a triangle be 
 constructed ? 
 
 49. A garage is to be 
 18 feet wide. The roof 
 
 is to rise 8 feet as shown in the adjacent figure. How shall a 
 carpenter's square be placed on the rafter to mark the proper 
 angle at which to cut the upper end ? the lower end ? 
 
 276. General addition theorem of proportion. 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. 
 
 Given = ' 1 
 
 a 4- c -\- e a c e 
 To prove that - - = - =/ (2) 
 
 Proof. Let = == ( 3 ) 
 
 Then a = br, (4) 
 
 c = dr, (5) 
 
 " e=fr. (6) 
 
 Adding (4), (5), and (6), 
 
 a + c + e = br + dr +fr = (b + d +f)r. . (7) 
 
 Therefore ^-- . W 
 
 From (3) and (8), 
 
 a -\-c-\-e_a_c_e 
 . b + d+f~~b~~d~J 
 
176 
 
 PLANE GEOMETRY 
 
 EXERCISES 
 
 50. Prove that the perimeters of two similar triangles are 
 in the same proportion as any two corresponding sides. 
 
 51. The sides of a triangle are 9, 10, and 
 17 respectively. The perimeter of a similar 
 triangle is 108. Find the sides of the latter. 
 
 52. Two triangles are similar if their 
 sides are perpendicular each to each. 
 
 HINT. Extend the perpendicular sides until 
 they meet. What relations exist between the 
 angles 1 and 2 ? C and 3 ? 3 and 4 ? etc. 
 
 Theorem 9 
 
 277. In two similar triangles any two homologous sides 
 are proportional to 
 
 (I) two corresponding altitudes ; 
 (II) two corresponding medians ; 
 (III) the bisectors of two corresponding angles. 
 
 KRL 
 
 A KRL 
 
 Given the similar triangles ABC and A'B'C', in which RC 
 bisects the angle ACB and R'C' bisects the angle A'C'B', also 
 CL and C'L', two corresponding medians, and CK and C'K', two 
 corresponding altitudes. 
 
 To prove that 
 
 AB 
 
 B'C' C'A' 
 
 m 
 
BOOK III 177 
 
 Proof. (I) In right triangles AKC and A'K'C', 
 
 Hence 
 and 
 
 By hypothesis, 
 From (2) and (3), 
 
 (II) By hypothesis, 
 
 and 
 
 ABCL~AB'C'L'. 274 
 
 Therefore ^, = 4^,' (5) 
 
 A 7\ Y~* 
 
 ~ A A'K'C 
 
 '', 
 
 
 272 
 
 h 
 h 1 
 
 AC 
 ~ A'C 1 ' 
 
 
 (2) 
 
 269 
 
 AB 
 
 BC 
 
 CA 
 
 /0\ 
 
 
 A'B' 
 
 B'C' 
 
 C'A' 
 
 ( 6 ) 
 
 
 AB 
 
 BC 
 
 CA 
 
 h 
 
 
 A'B 1 
 
 B'C' 
 
 C'A' 
 
 A' ' ' 
 
 
 BC 
 
 BA 
 
 %BA 
 
 BL 
 
 
 B'C 1 
 
 B'A' 
 
 %B'A' 
 
 ~B'L'' 
 
 
 (III) ABCR^AB'C'R'. 272 
 
 7. Kf^ 
 
 Hence b' = Wc'' (6 > Why? 
 
 AB BC CA h m b 
 1- i-om (4), (o), and (6), _ = _ = = - = _ = _. 
 
 278. Corresponding lines of similar polygons. Theorem 9 is a 
 special case of the much more general theorem which follows : 
 
 In two similar polygons any two lines similarly drawn are in 
 the same ratio as any two corresponding sides. 
 
 EXERCISES 
 
 53. Two triangles are similar if their sides are parallel each 
 to each. 
 
 54. Prove that the altitudes of two similar trapezoids are 
 proportional to any two corresponding sides. 
 
 55. Prove that two corresponding diagonals of two similar 
 hexagons are proportional to any two corresponding sides. 
 
178 PLANE GEOMETRY 
 
 279. Drawing to scale. Every map is drawn to scale and has 
 (or should have) its scale indicated ; that is, a statement should be 
 given saying that one inch (or some other unit of length) on the 
 drawing represents a certain number of inches, feet, or miles, or 
 some other distance, in the country of which the map is a repre- 
 sentation. Similarly, the plans of a house, the detailed drawings 
 for the pattern of a small casting, or those for the construction of 
 an ocean liner are all drawn to scale. Such drawings are based on 
 the principles of proportion and the relations of similar triangles. 
 By the use of scale drawings engineers often avoid long calcu- 
 lations and difficult field measurements. Certain lines or angles 
 of a field may be measured and from them an accurate drawing 
 made. Then the measurement of certain distances in the drawing 
 will give the distance between objects in the field, distances which 
 it would frequently be difficult if not impossible to measure or a 
 laborious task to compute by any other method. 
 
 The example cited indicates the value of what is called the 
 graphical method of solving problems. It is used widely in all 
 kinds of construction work and one of its great advantages is that 
 it presents the results of the whole computation to the eye so that 
 its various relations can be readily grasped. Its greatest advantage 
 for certain classes of problems, however, is its brevity compared 
 to trigonometric or other methods of calculation. 
 
 280. Instruments used in graphical constructions. Besides 
 the straightedge and compasses, three other instruments are 
 desirable for graphical solutions. These 
 
 are a triangle, a protractor, and a scale 
 marked in inches and tenths of inches. 
 
 The triangle has one right angle. Usu- 
 ally the other angles are either 60 and 
 30 respectively, or each is 45. The right 
 triangle can be used for constructing a 
 perpendicular at the point K of the line - 
 AB, as indicated in Fig. 1. A series of par- 
 allels can be accurately and quickly drawn 
 
 by means of the straightedge (or T-square) and the triangle, 
 used in the manner indicated in Fig. 2 on the following page. 
 
BOOK III 
 
 179 
 
 Fig. 3 shows a protractor, an instrument used for measuring 
 angles and for laying off angles of a required size. The point (' 
 is called the center of the protractor. The instrument is placed 
 on the angle A CB in the correct position for measuring it ; that 
 is, with the center of the protractor at the vertex and the zero 
 mark on one side of the angle. 
 The other side of the angle 
 coincides with the mark 52. 
 Hence the angle A CB is equal 
 to 52. To lay off an angle 
 
 of 65, first draw a line CA, Fio. 2 
 
 then place the protractor in 
 
 the position indicated and make a point K on the paper oppo- 
 site the 65 mark. Next remove the protractor and draw CK. 
 Then the angle ACK is equal to 65. 
 
 In the following exercises the student should have a scale 
 marked in inches divided into tenths, a protractor for laying off 
 and measuring angles, a pair of compasses, and a triangle for 
 drawing right angles. The direction " Solve graphically " means 
 "draw to scale the 
 polygons required 
 and draw the lines 
 whose lengths are 
 sought. Then meas- 
 ure these lines and 
 by use of the scale 
 on which the figure 
 has been drawn com- 
 pute their length." 
 
 With a scale marked in inches and tenths of inches, the inches 
 and tenths of an inch in the length of a line can be read directly, 
 and the hundredths of an inch, which are the fractions of a tenth, 
 can be estimated with considerable accuracy. 
 
 While graphical solutions are sufficiently accurate for many prob- 
 lems of the engineer, they are not exact. In order to secure good 
 practical results observe the following directions : Choose as large a 
 scale as possible; use good paper ; use a fairly hard drawing pencil 
 
 FIG. 3 
 
180 PLANE GEOMETRY 
 
 with a well-sharpened point and draw figures accurately, making 
 all measurements of lines and angles with great care. If still 
 closer approximations are required than can be obtained graphi- 
 cally, it is necessary to use trigonometry. 
 
 EXERCISES FOR GRAPHICAL SOLUTION 
 
 56. Measure the number of de- 
 grees in the angle of the adjacent 
 figure. 
 
 57. Construct an acute angle and 
 measure it. 
 
 58. Construct on a given line as 
 a side an angle of 38. 
 
 59. Measure the three lines on 
 
 the right, obtaining their length to 
 
 hundredths of an inch. Thus the 
 
 first line is 1.56 inches in length. _______ 
 
 60. Using the protractor and scale, draw an angle of 56 whose 
 sides are 2.3 inches and 3.7 inches respectively. 
 
 61. Draw to scale a triangle in which one angle is 70 and the 
 sides including it are 28 inches and 35 inches respectively. Then 
 draw the altitude on the side 35 and determine its length. 
 
 62. Draw to scale a triangle whose sides are 15, 20, and 25 
 respectively. 
 
 63. Find the altitude on side 18 of a triangle whose sides are 
 10, 15, and 18 respectively. 
 
 64. Find the median on the longest side of a triangle whose 
 sides are 10, 12, and 18 respectively. 
 
 65. Find the length of the bisector of the smallest angle of the 
 triangle whose sides are 9, 12, and 15 respectively. 
 
 66. One angle of a parallelogram is 40 and the two sides are 
 24 and 36 respectively. Find each diagonal. 
 
BOOK III 181 
 
 67. The longest diagonal and the shortest side of a parallelo- 
 gram are respectively 40 and 16. The angle between them is 28. 
 Find the other side and the angle it makes with the diagonal. 
 
 68. The sides of a triangle are 18, 20, and 34 respectively. Find 
 its three altitudes. 
 
 69. The sides of a triangle are 33, 47, and 55. Find the length 
 of the bisector of the greatest angle. 
 
 70. Find the shortest median of the triangle whose sides are 
 115, 252, and 277. 
 
 71. Find the least altitude of a triangle whose sides are 164, 
 225, and 349. 
 
 72. If one angle of a rhombus is 60, show that one diagonal is 
 about 1.73 times as long as the other. 
 
 73. In the polygon ABODE side AB = 14, EC = 12, CD = 15, 
 DE = 20, Z.B = 120, Z C = 105, and Z.D = 130. Draw the 
 polygon to scale and find AE and Z-A and ZE. 
 
 74. In the figure of Ex. 73 draw A C and A D and perpendiculars 
 from A to CD, from B to AC, and from E to AD. Then find the 
 length of .4 C, AD, and each of the three perpendiculars. 
 
 75. In the figure of 290 measure AK, KB, CK, and KD with 
 care. Then substitute in AK x KB = CK x KD and find each 
 product. This should give an approximate numerical verification 
 of the truth of Theorem 12. 
 
 76. In the figure of 297 measure A C, BC, KB, and KA. Then 
 substitute in KB : KA = CB: CA and obtain an approximate veri- 
 fication of the truth of this theorem. 
 
 77. In a circle of radius 2 inches draw a figure like that of 
 294 and measure AB, BC, BK, and BR. Then substitute in 
 AB x BC =' BK x BR and thus obtain a numerical verification of 
 the truth of the theorem. Is the agreement obtained here closer 
 than that secured in Exs. 75 .and 76 ? Explain. 
 
 281. Mean proportional. A mean proportional between two 
 numbers a and b is the number x if a:x = x:b, 
 
182 
 
 PLANE GEOMETRY 
 
 Theorem 10 
 
 282. If a perpendicular is drawn from the vertex of the 
 right angle to the ki/potenuse of a right triangle, 
 
 (1) the two triangles formed are similar to each other 
 and to the given triangle; 
 
 (2) the perpendicular is a mean proportional between 
 the segments of the hypotenuse; and 
 
 (3) the square of either side about the rigid angle 
 equals the producl of the whole hypotenuse and the seg- 
 ment adjacent to that side. 
 
 Given CK perpendicular to hypotenuse AB of right triangle ABC. 
 
 (1) To prove that AACX AJ3CK ^ A ABC. 
 
 Proof. &ACK, BCK, and ABC are right triangles. Why ? 
 
 And . Z.A + Zl = Z.A + Z.B. Why ? 
 
 Therefore Z 1 = Z B. Why ? 
 
 Now Z is in &ABC and in ABCK, and Zl is in A C.I A'. 
 Therefore A A CK ^ABCK^^ABC. 272 
 
 (2) To prove that AK: C7f = CKiKB. 
 
 Proof. By (1), AACK ^ ABCK, and in them Zl = Z.B and 
 
 J /T, opposite Z 1 _ CA', opposite Z .4 
 CA', opposite ZJ5 ~~ KB, opposite Z 2 
 
 (3) To prove that AC 2 = AB x AK, and BC 2 = AB x BK. 
 
 =Z2. 
 
 Then 
 
 269 
 
BOOK III 183 
 
 Proof. By (1), AACK^AABC, and in themZl = Z/? and 
 
 A K, opposite Zl _ _AC, opposite Z3 
 .4 C, opposite Z yt, opposite Z A CB 
 
 Then /1~C 2 = AB x AK. 
 
 In like manner BC 2 = AB x BK. 
 
 283. Third, proportional. A third proportional to two num- 
 bers a and b is the number x of a : b = b : x. 
 
 EXERCISES 
 
 78. A BCD is a parallelogram. DR cuts A B in K and CB produced 
 in R. How many similar triangles are thus formed ? Prove that 
 AK-.AD = AB:CR. 
 
 79. From any point R on a circle, RK is drawn perpendicular 
 to the diameter AB. Then KH is drawn perpendicular to the 
 radius RL. Show that RH is a third proportional to AL and KR. 
 
 80. If the two sides about the right angle of a right triangle 
 are a and 3 a respectively, the altitude upon the hypotenuse 
 divides it into two segments one of which is nine times the other. 
 
 Exs. 81-86 refer to the figure of Theorem 10. 
 
 81. If AKis 4 and AB is 20, find CK, AC, and EC. 
 
 82. If A C is 6 and AB is 18, find AK, CK, and EC. 
 
 83. If AB is 25 and CK is 12, find AK and BC. 
 
 84. If AC is 12 and BK is 18, find AK, AB, and BC. 
 
 85. If ,4C=10 and BK= 8, find AB. 
 HINTS. AC 2 = ABxAK. Let AK = x. 
 Then a; (a: + 8) = 100, 
 
 and x 2 + 8 x + 16 = 116. 
 
 Whence x + 4 = V116 = 10.77 etc. 
 
 86. If AK is 12 and JBC is 20, find AB. 
 
184 
 
 PLANE GEOMETRY 
 Theorem 11 
 
 284. In any right triangle the square of the hypotenuse 
 equals the sum of the squares of the other two sides. 
 
 Given the right triangle ABC in which AB is the hypotenuse. 
 To prove that AB 2 = AC 2 -f BC 2 . 
 
 Proof. Draw CK _L to A B. 
 
 Then (1) AC* = AB x AK, by (3) of 282, 
 
 and (2) ~BC* = ABx BK. 
 
 Adding (1) and (2), AC* + C 2 == (.4 B x A K) -f (AB x BK) 
 
 = AB(AK 
 
 EXERCISES 
 Exs. 87-89 refer to the figure of Theorem 1 1. 
 
 87. If A C is 12 and BC is 35, find AB. 
 
 88. If AB is 277 and AC is 115, find BC. 
 
 89. If AK is 16 and BC is 15, find BK and AB. 
 
 90. Find the diagonal of a rectangle whose sides are respec- 
 tively 616 and 663. 
 
 91. AB and AC are equal sides of the isosceles triangle ABC, 
 AK is the mid-perpendicular of BC, and KR is perpendicular to 
 A C, meeting it in R. Prove that AK* = AB x AK. 
 
EUCLIDE^ Elements, 
 
 PROP. XLVIL 
 
 7*f right-angled 
 triangles, BAG, 
 the fquarg BE, 
 which is made, of 
 the fde BC that 
 fubtendsthe light 
 angle BAG, is e- 
 qual to both the 
 fyuates BG, CH, 
 which are made 
 of the fdcs AB, 
 AC containing the 
 right angle. 
 
 Join AE, and 
 AD; and draw 
 AM parallel to CE, 
 
 Becaufe the angle DBG a^rFBA, add the an* a *z.Jx e 
 gle ABC common to them both ; then J. the an- 
 
 tie ABD-FBC. Moreover AB = FB, and b29.<fc/ 
 D *=rBC^ therefore is the triangle ABDrrFBC, c 4. i. 
 But the Pgr.BM^ ^ ABD.and the I J gf.<?BG= d 41.1. 
 2 FBC(for GAG rs one right line by Hypothefrs, 
 and 14. i.) e therefore is the Pgr.BM=BO. By e 6* <w, 
 the fame way of argument is the Pgr. CMnCH. 
 Therefore is the whole BE^EG+CH. Which 
 3VM to be (temffnftiated* 
 
 Schol. 
 
 This moi! excellent and ufeful theoreme hatfe 
 d^ferved the title of Pjtha&oiai his theoreme, be- 
 caufe he was the ijwentor of it. By the help of 
 which the addition and fubtt raft ion of fquares 
 are performed ; to which purpofe Urve the two 
 lollov/ing problejoiei. 
 
 Tin; DEMONSTRATION OF THE THEOREM ON THIS PAGE is FROM A COPY OF 
 BARROW'S EDITION OF EUCLID USED BY A HARVARD STUDENT IN 1735 
 
BOOK III 
 
 185 
 
 92. In the figure of Ex. 91 show that BC* is 4 CR x AB. 
 
 93. Ctfis an altitude of the triangle A B C. Prove that/C 2 -f BK* 
 
 285. Right triangles in which one angle is 45 or 30. An 
 
 important application of Theorem 11 is its use in connection with 
 right triangles in which one acute angle is 45 or in which one is 
 30. When one angle is 45 the two problems which arise in 
 practice are illustrated in Figs. 1 and 2, while the two problems 
 
 involving 30 (or 60) are illustrated in Figs. 3 and 4. Note how 
 the relations proved in Theorems 24 and 23 of Book I are used 
 in 3 and 4 below. 
 
 1. Given one angle 45 and one side about the right angle s, to 
 find the other angle and the other two sides. 
 
 2. Given one angle 45 and the hypotenuse h, to find the other 
 angle and the other two sides. 
 
 3. Given one angle 30 and the hypotenuse s, to find the other 
 angle and the two sides. 
 
 4. Given one angle 30 and one side about the right angle a, to 
 find the other angle and the other two sides. 
 
 The relation of 3 and 4 above to any equilateral triangle as 
 indicated in Fig. 4 should be carefully noted. 
 
 In. general, if we know one angle of any right triangle is 30, 
 45, or 60, and know also any one side, we can compute the other 
 angle and the other sides. 
 
 In Exs. 98-102, inclusive, give answers in simplest algebraic 
 form, merely indicating the roots involved,. 
 
186 PLA^E GEOMETRY 
 
 EXERCISES 
 
 94. If the acute angles of a right triangle are each 45 and one 
 side about the right angle is m, show that the hypotenuse is m V2. 
 
 95. If the acute angles of a right triangle are each 45 and the 
 
 hypotenuse is s, show that each leg is v2. 
 
 2i 
 
 96. If one acute angle of a right triangle is 30 and the hypot- 
 
 A-ii . 
 
 enuse is in, show that the longer of the other two sides is V3. 
 
 97. If one acute angle of a right triangle is 30 and the longer 
 
 side about the right angle is a, show that the hypotenuse is ^ V3. 
 
 o 
 
 98. The hypotenuse of an isosceles right triangle is 12. Find 
 the other sides* 
 
 99. The side of an equilateral triangle is 13. Find the altitude. 
 
 100. The altitude of an equilateral triangle is 7. Find the side. 
 
 101. In the triangle ABC, A is 30, B is 45, and b is 10. Find 
 a and c. 
 
 HINT. Draw a perpendicular to AB from C. 
 
 102. In the triangle ABC, A is 30, b is 12, and c is 16. Find a. 
 
 103. In the triangle ABC, A is 45, B is 120, and b is 10. 
 Show that c is 2.988+ and a is 8.164 -f. 
 
 HINT. Draw a perpendicular from C to AB produced. 
 
 104. In triangle ABC, A is 45, b is 20, and c is 40. Show that 
 a is 29.4+. 
 
 105. In triangle ABC, A is 30, B is 135, and b is 20. Show 
 that c is 7.320+ and a is 14.142+. 
 
 286. Introduction to trigonometry. Section 285 treats the 
 important though special problem of computing the parts of 
 a right triangle if one side is given and if it is known that 
 one acute angle is 30, 45, or 60. The general problem, in 
 which one angle has any value between and 90, requires 
 the methods of trigonometry. 
 
BOOK III 187 
 
 287. Trigonometric ratios. Let KAR be any angle. From any 
 point on either side, as^TT, draw BC a perpendicular to the other 
 side. Let BC= a, A C= b, and AB= c. 
 
 a side opposite 
 
 Ihe sineZ.A = - = - f , 
 
 c hypotenuse 
 
 b side adjacent 
 
 CvSine Z. A = = 
 
 c hypotenuse 
 
 a side opposite 
 
 and tangent Z 4 = - = -- p 
 
 6 side adjacent 
 
 The foregoing ratios are commonly written in an abbreviated 
 
 form as sin A = -i cos A = -i and tan A = - respectively. 
 c c o 
 
 It is important to observe that the value of each ratio 
 depends wholly on the size of the angle A, not on the size of the 
 triangle ABC. Thus 
 
 BC B C 
 
 These ratios are numbers. Their values to three decimals 
 for every angle from 1 to 89 is given on page 188. 
 
 The method of computing the ratios for 30, 45, and 
 60 follows easily from the work on 285. From the 
 adjacent figure, for example, 
 
 c 
 -------- , 
 
 cos 30 = - = - 
 
 C 
 
 e 
 
 and tan 30 = ~ = - = -i =.577 +. 
 
188 
 
 PLANE GEOMETRY 
 TRIGONOMETRIC TABLE 
 
 Angle 
 
 Sin 
 
 Cos 
 
 Tan 
 
 Angle 
 
 Sin 
 
 Cos 
 
 Tan 
 
 1 
 
 .017 
 
 .9998 
 
 .017 
 
 45 
 
 .707 
 
 .707 
 
 .000 
 
 2 
 
 .035 
 
 .9994 
 
 .035 
 
 46 
 
 .719 
 
 .695 
 
 .036 
 
 3 
 
 .052 
 
 .9986 
 
 .052 
 
 47 
 
 .731 
 
 .682 
 
 .072 
 
 4 
 
 .070 
 
 .9976 
 
 .070 
 
 48 
 
 .743 
 
 .669 
 
 .111 
 
 5 
 
 .087 
 
 .996 
 
 .087 
 
 49 
 
 .755 
 
 .656 
 
 .150 
 
 6 
 
 .105 
 
 .995 
 
 .105 
 
 50 
 
 .766 
 
 .643 
 
 .192 
 
 7 
 
 .122 
 
 .993 
 
 .123 
 
 51 
 
 .777 
 
 .629 
 
 .235 
 
 8 
 
 .139 
 
 .990 
 
 .141 
 
 52 
 
 .788 
 
 .616 
 
 .280 
 
 9 
 
 .156 
 
 .988 
 
 .158 
 
 53 
 
 .799 
 
 .602 
 
 .327 
 
 10 
 
 .174 
 
 .985 
 
 .176 
 
 54 
 
 .809 
 
 .588 
 
 .376 
 
 11 
 
 .191 
 
 .982 
 
 .194 
 
 55 
 
 .819 
 
 .574 
 
 .428 
 
 12 
 
 .208 
 
 .978 
 
 .213 
 
 56 
 
 .829 
 
 .559 
 
 .483 
 
 13 
 
 .225 
 
 .974 
 
 .231 
 
 57 
 
 .839 
 
 .545 
 
 .540 
 
 14 
 
 .242 
 
 .970 
 
 .249 
 
 58 
 
 .848 
 
 .530 
 
 .600 
 
 15 
 
 .259 
 
 .966 
 
 .268 
 
 59 
 
 .857 
 
 .515 
 
 .664 
 
 16 
 
 .276 
 
 .961 
 
 ' .287 
 
 60 
 
 .866 
 
 .500 
 
 .732 
 
 17 
 
 .292 
 
 .956 
 
 .306 
 
 61 
 
 .875 
 
 .485 
 
 .804 
 
 18 
 
 .309 
 
 .951 
 
 .325 
 
 62 
 
 .883 
 
 .469 
 
 .881 
 
 19 
 
 .326 
 
 .946 
 
 .344 
 
 63 
 
 .891 
 
 .454 
 
 .963 
 
 20 
 
 .342 
 
 .940 
 
 .364 
 
 64 
 
 .899 
 
 .438 
 
 2.050 
 
 21 
 
 .358 
 
 .934 
 
 .384 
 
 65 
 
 .906 
 
 .423 
 
 2.144 
 
 22 
 
 .375 
 
 .927 
 
 .404 
 
 66 
 
 .914 
 
 .407 
 
 2.246 
 
 23 
 
 .391 
 
 .921 
 
 .424 
 
 67 
 
 .921 
 
 .391 
 
 2.356 
 
 24 
 
 .407 
 
 .914 
 
 .445 
 
 68 
 
 .927 
 
 .375 
 
 2.475 
 
 25 
 
 .423 
 
 .906 
 
 .466 
 
 69 
 
 .934 
 
 .358 
 
 2.605 
 
 26 
 
 .438 
 
 .899 
 
 .488 
 
 70 
 
 .940 
 
 .342 
 
 2.747 
 
 27 
 
 .454 
 
 .891 
 
 .510 
 
 71 
 
 .946 
 
 .326 
 
 2.904 
 
 28 
 
 .469 
 
 .883 
 
 .532 
 
 72 
 
 .951 
 
 .309 
 
 3.078 
 
 29 
 
 .485 
 
 .875 
 
 .554 
 
 73 
 
 .956 
 
 .292 
 
 3.271 
 
 30 
 
 .500 
 
 .866 
 
 .577 
 
 74 
 
 .961 
 
 .276 
 
 3.487 
 
 31 
 
 .515 
 
 .857 
 
 .601 
 
 75 
 
 .966 
 
 .259 
 
 3.732 
 
 32 
 
 .530 
 
 .848 
 
 .625 
 
 76 
 
 .970 
 
 .242 
 
 4.011 
 
 33 
 
 .545 
 
 .839 
 
 .649 
 
 77 
 
 .974 
 
 .225 
 
 4.331 
 
 34 
 
 .559 
 
 .829 
 
 .675 
 
 78 
 
 .978 
 
 .208 
 
 4.705 
 
 35 
 
 .574 
 
 .819 
 
 .700 
 
 79 
 
 .982 
 
 .191 
 
 5.145 
 
 36 
 
 .588 
 
 .809 
 
 .727 
 
 80 
 
 .985 
 
 .174 
 
 5.671 
 
 37 
 
 .602 
 
 .799 
 
 .754 
 
 81 
 
 .988 
 
 .156 
 
 6.314 
 
 38 
 
 .616 
 
 .788 
 
 .781 
 
 82 
 
 .990 
 
 .139 
 
 7.115 
 
 39 
 
 .629 
 
 .777 
 
 .810 
 
 83 
 
 .993 
 
 .122 
 
 8.144 
 
 40 
 
 .643 
 
 .766 
 
 .839 
 
 84 
 
 .995 
 
 .105 
 
 9.514 
 
 41 
 
 .656 
 
 .756 
 
 .869 
 
 85 
 
 .996 
 
 .087 
 
 11.43 
 
 42 
 
 .669 
 
 .743 
 
 .900 
 
 86 
 
 .9976 
 
 .070 
 
 14.30 
 
 43 
 
 .682 
 
 .731 
 
 .933 
 
 87 
 
 .9986 
 
 .052 
 
 19.08 
 
 44 
 
 .695 
 
 .719 
 
 .966 
 
 88 
 
 .9994 
 
 .035 
 
 28.64 
 
 45 
 
 .707 
 
 .707 
 
 1.000 
 
 89 
 
 .9998 
 
 .017 
 
 57.29 
 
BOOK III 189 
 
 EXERCISES 
 
 106. Derive the ratios for 45 and for 60 by the use of 287 
 and compare the results with the values given in the table on 
 page 188. 
 
 107. Construct an angle of 20 with a protractor and complete 
 a triangle like ABC of 287. Measure a, b, and c, and then 
 compute approximate values for the ratios of 20. Compare the 
 results obtained with those given in the table. 
 
 108. Show that the value of the cosine and the tangent of the 
 angle A do not depend on the size of the right triangle ABC 
 of 287. 
 
 109. Read from the table (a) sin 30, (b) cos 30, (c) tan 30, 
 (d) sin 18, (e) sin 71, (/) cos 8, (g) cos 32, (h) tan 64, (i) tan 79. 
 
 110. From the table determine the value of A if 
 
 (a) sin .4 = .438 (d) cos A =.225 (g) tan A = 2.904 
 
 (b) sin .4 = .731 (e) cos.4 = .956 (A) ton^= .325 
 
 (c) sinyl = .966 (/) cosvl = .629 (i) tan .4 =1.428 
 
 111. About what value has the angle A if 
 
 sin A = .590 ? tan A = .362 ? cos A = .425 ? 
 
 288. Use of the trigonometric ratios. If one side and one 
 acute angle of a right triangle can be determined in any 
 way, the other sides can be computed by means of the 
 trigonometric ratios. 
 
 Example. In the triangle ABC the 
 angle A is 24 and A C is 42. Find EC. 
 
 Solution. tan A = - > 
 
 b 
 
 tan 24 = ^. 
 
 a - 42 x tan 24 
 
 = 42 (.445) = 18.69. 
 Therefore EC = 18.69. 
 
190 . PLANE GEOMETRY 
 
 From the triangle ABC of 287 we have the general results 
 
 sin A = -, therefore a = csinA and c = ; similarly, cosA = -, 
 
 sinA " c 
 
 therefore b = c cos A and c = . ; and tan A = , therefore a 
 
 coaA b 
 
 = b tanA and b= . 
 tan A 
 
 EXERCISES 
 
 It is customary to letter a right triangle as in 287. With that 
 understanding solve the following right triangles : 
 
 112. A = 25, c = 50. Find Z.B, a, and b. 
 
 113. A = 70, b = 40. Find Z, a, and c. 
 
 114. A = 48, a = 30. Find Z#, b, and c. 
 
 115. The shadow of a certain tree on level ground is 60 feet 
 long, and the sun's rays make an angle of 52 with the ground. 
 How high is the tree ? 
 
 116. A balloon is held captive by a rope 800 feet long. An 
 observer notes that the rope makes an angle of about 76 with 
 the ground, which is level. How high is the balloon ? 
 
 117. If c = 125 and a = 26, find Z A, Zfl, and b. 
 
 118. If c = 200 and b = 106, find Z.A, Z.B, and a. 
 
 119. If a = 117 and b = 130, find Z.A, Z.B, and c. 
 
 120. The base of an isosceles triangle is 20 and the angle oppo- 
 site is 46. Find the other sides. 
 
 121. The sides of a right triangle are 3, 4, and 5. Find to the 
 nearest integer the number of degrees in the acute angles. 
 
 122. A chord of a circle whose radius is 10 feet subtends at the 
 center an angle of 40. How long is the chord ? 
 
 HINT. Use the cosine of one of the angles to get the half chord. 
 
 123. The radius of a circle is 32 feet. What angle is subtended 
 at the center by a chord 32 feet long ? 
 
 124. One of the equal angles of an isosceles triangle is 56; 
 the base is 18 inches. Find the equal sides. 
 
 HINT. Use the cosine of one of the angles. 
 
BOOK III 191 
 
 125. Find the altitude on the base of an isosceles triangle in 
 which the vertical angle is 80 and the base is 32 feet. 
 
 126. From one station a balloon was observed to have an 
 angle of elevation of 11, and at the same instant from another 
 station an angle of 8. The two 
 
 stations were at the same level, two ^____^ .^..--- = ^^ : ^ \x 
 
 kilometers apart, to the south of, H 24^ y >l 
 and in the same vertical plane with, 
 
 the balloon. How high was the balloon and how far was it from 
 the nearer station ? 
 
 HINTS. - = tan 11, or x = y tan 11. 
 
 y 
 
 ^ = tan 8, or x = y tan 8 + 2 tan 8, etc. 
 
 127. Two observation stations at the same level are seven kilo- 
 meters apart. An airplane crossing the line connecting the two 
 stations has an elevation of 32 from one station and 40 from 
 the other. Find the height of the airplane at the time. 
 
 289. Theorem of proportion on two pairs of equal products. 
 
 If the product of tivo numbers equals the product of two others, 
 the numbers of one pair can be made the first and fourth terms 
 of a proportion of which the numbers of the other pair are the 
 second and third. 
 
 Given ax = by. (1) 
 
 (i ?y 
 
 To prove that -== (2) 
 
 b x 
 
 Proof. Dividing (1) by by, j = - (3) 
 
 OC' 
 
 EXERCISES 
 Write as proportions : 
 
 128. ab = kr. 131. x 2 - y z = x 2 - 4 x -f 4. 
 
 129. xy = 12. 132. AC AD = AB . CD. 
 
 130. a:y = l. 133. AB' 2 -- 
 
192 PLANE GEOMETRY 
 
 Theorem 12 
 
 290. If two chords of a circle intersect, the product of 
 the segments of one equals the product of the segments 
 of the other. 
 
 B 
 
 Given the chords AB and CD intersecting in point K within 
 the circle. 
 
 To prove that AK x KB = CK x KD. 
 
 HINTS. Draw A C and BD and prove that &AKC^~&BKD. 
 
 Then prove that AK : CK = KD : KB, etc. 
 
 EXERCISES 
 
 134. In the figure above, if AK is 8, BK is 12, and CK is 6, find 
 the length of KD. 
 
 135. In the figure above, if A K is 8, BK is 24, and CD is 28, find 
 the length of CK and KD, each to two decimals. 
 
 136. In the figure above, AB is 24, CK is 8, and DK is 12, find 
 AK and DK to two decimals. 
 
 137. On the smooth ice of a lake three vertical standards, 
 each four feet long, were set up in a straight line at intervals 
 of one mile. On looking through a telescope across the top 
 of the first standard to the top of the third it was found that 
 the line of sight fell 8 inches below the top of the second 
 standard. Why? Find from the preceding facts the diameter 
 of the earth. 
 
BOOK III 193 
 
 291. Solution of geometric identities; suggestions for prov- 
 ing the truth of a geometric identity. Theorems 12, 13, and 
 14 assert that the product of two lines equals the product of 
 two others. With such simple identities one should proceed as 
 follows : 
 
 1. Write the identity as a proportion by 289. 
 
 2. Seek two triangles of which the four lines are sides. If the 
 triangles are not complete, try to discover the lines necessary to 
 complete them. 
 
 3. Then prove the triangles similar and use 269. The required 
 result will easily follow. 
 
 4. When, as will sometimes happen, the preceding steps fail, 
 try to discover if the four lines of the proportion are corresponding 
 segments of two intersecting lines cutting two parallels. 
 
 5. After Theorems 15-18, inclusive, have been mastered, try 
 to use one or more of them if the four preceding steps fail. 
 
 6. Consider the possibility and the effect of substituting one 
 line for its equal. 
 
 When the identity involves sums and products, an intelligent 
 use of transposition and factoring in connection with some one or 
 more of the preceding steps will frequently result in a solution, 
 as in the following : 
 
 Example. In the triangle ABC sides AB and AC are equal. 
 R is a point on AC such that BR equals EC. 
 
 Prove that BC 2 - CR* = A R . CR. 
 Transposing, BC* = AR CR + C^ 2 . 
 
 Factoring, BC 2 = CR ( A R + CR) 
 
 = CR -AC. 
 Therefore A C : BC = BC : CR, etc. 
 
 In still more difficult problems one should consider in con- 
 nection with the preceding, or by itself, the possibility and the 
 effect of substituting the square of one line for the sum or the 
 difference of the squares of two others, or the reverse. 
 
194 PLA^E GEOMETRY 
 
 EXERCISES 
 
 138. EC is the common chord of two intersecting circles. 
 The chord BA is tangent to one circle, the chord BK to the other. 
 Prove that BC 2 = A C . CK. 
 
 139. In the parallelogram ABCD a line DE is drawn meeting 
 the diagonal AC in F, the side BC in G t and the side AB produced 
 in E. Prove that DF - AE = AB FE. 
 
 Theorem 13 
 
 292. If from a point without a circle a secant terminat- 
 hi(j in the circle and a tangent be drawn, the square of the 
 tangent equals the whole secant times its external segment. 
 
 Given the secant AR, cutting the circle in K and J?, and the 
 tangent AB. 
 
 To prove that AB 2 = AR x AK. 
 
 HINTS. Draw BK and BE, and prove that &AKB ^AABR. 
 
 Then prove that A R : A B = A B : A K, etc. 
 
 293. Corollary. If from a point without a circle two secants 
 terminating in it are drawn, the product of one secant and its 
 external segment equals the product of the other secant and its 
 external segment. 
 
 EXERCISES 
 
 140. InthengureofTheoreml3,if.45is8and.4 J Ris20,rind/lA'. 
 
 141. A tangent to a circle from A is 12 inches long. AKR is a 
 secant cutting the circle so that the chord K R is 18 inches. Find A K. 
 
BOOK III 
 
 195 
 
 142. Tangents to two intersecting circles from any point on 
 their common chord produced are equal. 
 
 143. AB and A C are tangents 
 to a circle whose center is K, at 
 B and C respectively. BR drawn 
 _L to CK produced meets it in Pi. 
 Prove that AC : CK = CR : BR. 
 
 144. If the centers of arcs A C 
 and BC are at B and A respec- 
 tively, the arch is equilateral. 
 
 Construct an equilateral arch 
 
 in which AB is two inches and 
 
 inscribe a circle in it. r IT . U 
 
 HINTS. Let the radius of the arc BC= R. TheuAH= - Let KL = 2 r. 
 By 292, AK-AL = AH*. Substituting, (R-2r)R = < Whence 
 
 7J '-17? 
 
 #-2r = -andr = ,etc. 
 4 8 
 
 Theorem 14 
 
 294. In any triangle the product of two sides equals the 
 altitude on the third side multiplied by the diameter of the 
 circumscribed circle. 
 
 Given the triangle ABC, the altitude BK, and BR the diameter 
 of the circumscribed circle. 
 
 To prove that ABxBC= BK X BK. 
 HIXTS. Draw CR and prove that &ABK &BCR. 
 Then prove that AB ; BK = BR : BC, etc. 
 
196 PLANE GEOMETEY 
 
 EXERCISES 
 
 145. Find the altitude on the side AB of a triangle ABC in 
 which AB is 21, BC is 32, and AC is 20. 
 
 HINTS. Draw the altitude CK. Let K/l equal x and C/f equal y. 
 Then z 2 + # 2 = 400 and (21 - x) 2 + f = 1024, etc. 
 
 146. The sides of a triangle are 4, 13, and 15 respectively. 
 Find the altitude on side 13 and the diameter of the circum- 
 scribed circle. 
 
 147. The sides of a triangle are 13, 14, and 15 respectively. 
 Find the altitude on the side 14, and the diameter of the circum- 
 scribed circle. 
 
 Theorem 15 
 
 295. The bisector of an angle of a triangle divides the 
 side opposite into segments proportional to the sides 
 adjacent to them. 
 
 Given the triangle ABC in which CK bisects the angle ACS 
 and meets AB at K. 
 
 To prove that AK : KB = AC:BC. 
 
 Proof. Through A draw a parallel to CK intersecting BC pro- 
 duced in R. 
 
 Then A K : KB = R C : CB. (1) 260 
 
 Now Z1 = Z2. Why? 
 
 Z1=Z4. Why? 
 
BOOK III 19T 
 
 = Z3. Why? 
 
 = Z4. Why? 
 
 RC = AC. (2) Why? 
 
 From (1) and (2), A K : KB = A C : BC. Why ? 
 QUERY. Is Theorem 15 true when AC is equal to CB1 
 
 EXERCISES 
 
 148. The sides of a triangle are 8, 12, and 15 respectively. 
 Find the segments of the side 15 made by the bisector of the 
 angle opposite. 
 
 149. The sides of a triangle are 4, 8, and 10 respectively. The 
 perimeter of a similar triangle is 132. Find the segments of the 
 greatest side of the latter made by the bisector of the angle 
 opposite. 
 
 150. CK is the median of the triangle ABC. The bisectors of 
 the two angles at K meet A C and BC in L and R respectively. 
 Prove that LR is parallel to A B. 
 
 151. State and prove the converse of Theorem 15. 
 
 296. Segments of a line made by a point. If in the adjacent 
 figure AB is a line of given length and K is a point on AB 
 or AB produced, KA and KB are called the segments of the 
 line AB. 
 
 When the point of division is on the line, the line is divided 
 internally ; when the point of division is on the line produced 
 
 in either direction, the line . 
 
 is divided externally. A B A B K 
 
 Note that in either case the segments are measured from 
 the point of division to the extremities of the line, and that 
 when the point is on the line produced, one segment is 
 longer than the line itself. 
 
PLANE GEOMETRY 
 Theorem 16 
 
 297. If a triangle is not isosceles, the bisector of an 
 exterior angle at any vertex divides the side opposite exter- 
 nally into segments proportional to the adjacent sides. 
 
 A B K 
 
 Given the triangle ABC in which CK, the bisector of the exte- 
 rior angle at C, meets the side AB produced at K. ^ t& 
 
 To prove that KB : KA = CB : CA. 
 
 HINTS. Draw BR parallel to CK, cutting AC in R. Is KB'.KA 
 equal to CR : CA1 Why? Compare CB and CR. 
 
 QUERY 1. Does the proof of Theorem 16 hold for an isosceles 
 triangle ? 
 
 QUERY 2. Is Theorem 16 true when the triangle is isosceles? 
 
 QUERY 3. What condition determines whether the bisector of the 
 exterior angle at C (using the figure of 297) meets the side opposite 
 produced to the right or meets it produced to the left? 
 
 EXERCISES 
 
 152. The sides of a triangle are 8, 12, and 16 respectively. 
 Find the segments of side 16 made by the bisector of the exterior 
 angle at the vertex opposite. 
 
 153. State and prove the converse of Theorem 16. 
 
 154. The bisectors of the interior and the exterior angle at C 
 of the triangle ABC cut the base in K and R respectively. Prove 
 that KA xRB=RA x KB. 
 
BOOK III 
 
 199 
 
 Theorem 17 
 
 298. If two convex polygons are similar, they can be 
 divided into the same number of triangles which are 
 similar each to each. 
 
 A 
 
 Given the convex polygon ABODE similar to the convex poly- 
 gon A'B'C'D'E'. 
 
 To prove that these polygons may be divided into the same 
 number of triangles similar each to each and similarly placed. 
 
 Proof. Draw from two corresponding vertices A and A' all 
 the diagonals possible. 
 
 At) JjC ,, _ , _. ^.^_ 
 
 J_1JCJ1 
 
 Therefore 
 
 A'B 1 
 
 ~ B'C' * 
 ~ AA'B'C'. 
 
 
 5 *vM 
 
 274 
 
 And 
 
 Z2 
 EC 
 
 = Z3. 
 
 = Z4. 
 AC 
 
 (1) 
 
 (2) 
 
 Why? 
 Why? 
 
 Why? 
 
 B'C' 
 
 A'C' 
 
 
 BC 
 
 CD 
 
 (3) 
 
 Why? 
 
 B'C' 
 
 C'D 1 
 
 From (2) and (3), 
 
 AC 
 
 CD 
 
 W 
 
 Why? 
 
 A'C' 
 
 C'D' 
 
 By (1) and (4), 
 
 AACD 
 
 ^ A A' C'D'. 
 
 
 Why? 
 
 In like manner any triangle in the first polygon can be proved 
 similar to the similarly placed triangle in the second. 
 
200 PLANE GEOMETRY 
 
 QUERY. If the perimeters of two polygons are in the same ratio as 
 the sides taken in pairs, are the polygons similar ? 
 
 EXERCISE 155. ABODE and A'B'C'D'E' are similar pentagons. 
 K is a point within the first and K' a point within the second. 
 Lines are drawn from K to the vertices of the first polygon, and 
 from A*' to the vertices of the second. If the triangle KAB is 
 similar to the triangle K'A 'B', prove that the triangle KBC is sim- 
 ilar to the triangle K'B'C'. 
 
 Theorem 18 (Converse of Theorem 17) 
 
 299. If two convex polygons can be divided into the 
 same number of triangles similar each to each and simi- 
 larly placed, the polygons are similar. 
 
 E' 
 
 D' 
 
 B A B' 
 
 Given the convex polygons ABCDEF and A'B'C'D'E'F' divided 
 into triangles so that the triangle ABC is similar to the triangle 
 -A'JS'C', the triangle ACD is similar to the triangle A'C'D', etc., 
 and in the same relative position. 
 
 To prove that 
 
 the polygon ABCDEF^ the polygon A'B' C'D'E'F'. 
 HINTS. Z.B = Z.B'. Why? 
 
 ZBCD = ZB'C'D',etc. Why? 
 
 *- = 4. Why? 
 
 B'C' A'C' 
 
 - = ^-. Why? 
 
 C'D' A'C' 
 
 Therefore - = --^-,> etc. 
 
BOOK III 201 
 
 EXERCISE 156. Can a line parallel to the bases of a trapezoid 
 divide it- into two similar trapezoids ? Prove. 
 
 QUERY 1. Are Theorems 17 and 18 true for reentrant polygons? 
 QUERY 2. Why is the word convex necessary in the statement of 
 Theorems 17 and 18? 
 
 300. The pantograph. The pantograph is an instrument for 
 enlarging or reducing drawings, being especially serviceable 
 in tracing curved or broken irregular lines. The instrument 
 is easily understood from the adjacent figure. 
 
 The points A and B move in circles about the fixed pivot F, 
 while the points C and D can move in circles about B and A 
 respectively. For these 
 reasons any of the points 
 C, D, or E can, within 
 certain limits, be made 
 to move about at will. 
 
 The instrument is ad- 
 justed for use by means 
 of the pins at A, B, C, 
 -and D. When the pins are properly placed they must be so 
 situated that the figure ABCD will be a parallelogram. If E 
 is a point such that AFAD~AFBE, the points F, D, and 
 E will remain in a straight line. Then if D is a tracer which 
 follows the drawing, a pencil at E will draw the enlargement. 
 By placing the tracer at E and the pencil at D, a reduced 
 drawing KDR . of K'ER 1 can be made. 
 
 EXERCISES 
 In the preceding figure 
 
 157. Prove that KR : K'R' = FA : FB. 
 
 158. Determine the boundaries of the area over which D can 
 move. 
 
202 PLANE GEOMETRY 
 
 Theorem 19 
 
 301. If two parallels are cut by three or more con- 
 current transversals, the corresponding segments of the 
 parallels are proportional. 
 
 \H 
 
 Given two parallel lines, KH and K'H', cut by four concur- 
 rent lines through A, in K, R, L, H, and K', R', L', and H' 
 respectively. 
 
 , KR EL LH 
 
 To proved _ = _ = _. 
 
 Proof. Triangles AKR and ARL are similar respectively to 
 
 angles A'K'R' and 
 
 Therefore 
 In like manner 
 From (3) and (4), 
 
 A'R'L'. 
 AR 
 
 KR 
 
 (1) 
 (2) 
 (3) 
 
 (4) 
 LH 
 
 Why ? 
 Why ?' 
 
 Why ? 
 Why ? 
 
 AR ( 
 AR 
 
 K'R' 
 RL 
 
 AR' 
 KR 
 
 R'L' 
 RL 
 
 K'R' 
 RL 
 
 R'L' 
 LH 
 
 R'L' 
 KR 
 
 L'H' 
 RL 
 
 K'R 1 
 
 R'L' 
 
 L'H' 
 
 QUERY. If KRL and H in the left-hand figure above are fixed and 
 A moves off to a great distance, what does the figure KRR'K' ultimately 
 become ? 
 
BOOK III 203 
 
 EXERCISES 
 
 159. Two lines from the vertex of a triangle divide a line which 
 is parallel to the base and which terminates in the other two sides 
 into three parts each equal to one ninth of the base. Prove that 
 the two lines are divided into segments in the ratio of two to one. 
 
 160. If n lines from the vertex of a triangle divide a line which 
 is parallel to the base and which terminates in the other two sides 
 into n -\- 1 parts each equal to one twentieth of the base, what is 
 the ratio of the two segments into which each of the n lines is 
 divided ? 
 
 302. The subtraction theorem of proportion. If four num- 
 bers are in proportion, they are in proportion by subtraction; 
 that is, the first minus the second is to either as the third 
 minus the fourth is to the corresponding one. 
 
 Given . H- (1) 
 
 a b c d 
 
 To prove that = > (2) 
 
 d 
 
 a bc d /ON 
 
 and = - - (3) 
 
 a c 
 
 Proof. Subtracting 1 in (1), 
 
 j 1 = - 1. (4) Why ? 
 
 Whence . 2_IL* = Lzil ( 5 ) 
 
 b d 
 
 Write (1) by inversion. Then (3) can be obtained as was (5). 
 
 EXERCISES 
 
 161. Give the proof of (3) as suggested above. 
 
 162. If the sides of a trapezoid taken in order are proportional 
 to the sides of another in the same order, are the trapezoids 
 similar ? Prove. 
 
204 PLANE GEOMETRY 
 
 Theorem 20 (Converse of Theorem 19) 
 
 < 
 
 303. If three nonparallel transversals intercept pro- 
 portional corresponding segments on two parallels, the 
 transversals are concurrent. 
 
 \R' 
 
 /K \fi >^ /K \R \L 
 
 Given the nonparallel transversals KK', RR', and LV cutting 
 the two parallels KL and K'V at K, R, L, and K', R', L', 
 respectively in such manner that 
 
 V"D -nj 
 
 J^L = J*. (i) . 
 
 K'R' R'L' 
 To prove that KK', RR', and LL' are concurrent. 
 
 Proof. Let us suppose that KK' and RR' intersect at B, and 
 that RR' and LL' intersect at 0. Consider the first figure. 
 
 K"*? BR 
 
 = ^ (2} 271 
 
 RL OR 
 and WT = OK'- < 3 > 
 
 From (1), (2), and (3), ||=||- (4) 
 
 OR + OK' BR + BR' 
 Then from (4), QR , ^7 - , (5) 
 
 RR 1 RR' 
 = - 
 
 From (6), OR' = BR', hence O and B coincide and the trans- 
 versals KK\ /!/!', and /,/>' are concurrent. 
 The proof for the second figure is similar. 
 
BOOK III 205 
 
 EXERCISES 
 
 163. Prove that the two nonparallel sides and the line joining 
 the mid-points of the bases of a trapezoid are concurrent. 
 
 164. Examine the proof of Theorem 20 and determine what 
 changes are necessary to make it apply^to the second figure. 
 
 165. If four nonparallel transversals intercept proportional 
 corresponding segments on two parallels, the transversals are 
 concurrent. 
 
 
 FIELD PROBLEMS WITH A TAPE LINE 
 
 In Exs. 166-173 the use of a fifty-foot tape line, a ball of 
 cord, and a sufficient number of long, straight stakes for marking 
 all important points is assumed. With this equipment three simple 
 field problems are possible : 
 
 1. Setting three stakes in a straight line at random distances 
 or at given distances apart. 
 
 2. Setting four stakes so that the line (cord) joining two of them 
 is perpendicular to the line (cord) joining the other two (Ex. 168). 
 
 3. Setting four stakes so that the line joining two of them is 
 parallel to the line joining the other two (Ex. 169). 
 
 166. The height of a flagpole cannot be measured directly, but the 
 length of its shadow can. What simple method with accompanying 
 measurements will enable one to compute 
 
 the height of the pole ? 
 
 167. Is a triangle whose sides are 3, 4, jj 
 and 5 respectively or any multiple of these 
 numbers a right triangle ? Prove. 
 
 168. How can a tape line and the fact, 
 just proved be used to locate stake D so 
 that the string stretched from it to C will 
 
 be perpendicular to the string stretched from A to B ? Could a mason 
 mark the outside line for a foundation in this way ? Is it done so 
 in practice ? Why not use a square ? 
 
206 
 
 PLANE GEOMETRY 
 
 169. How can a tape line be used to place four stakes so that 
 they will determine two parallel lines ? 
 
 170. In the left-hand figure below, the distance between A 
 and B, two points on opposite sides of a river, is desired but 
 cannot be measured directly. By using a tape what lines in the 
 neighborhood of A can be laid out and measured so that the dis- 
 tance AB can then be computed ? Assume values for the measured 
 lines and compute AB. 
 
 171. In the right-hand figure above, the points A and B are 
 accessible, but the line AB cannot be directly measured. What lines 
 can be laid out and measured that will enable one to compute AB? 
 Assume values for the measured lines and compute AB. 
 
 172. In surveying practice a line is prolonged through an 
 obstacle a given distance by the following .method : In the ad- 
 jacent figure suppose it is required to prolong KB through so 
 
 H 
 
 that BH shall be 90 feet. Lay off AB = 45 feet. Lay out AE, 
 making an acute angle with A B. Locate C and D so that A E 2 A C 
 = 4 CD. Lay out BCF so that BF = 2 BC. Lay out FDGH so 
 that FH = 4 FD = 2 FG. Lay out EGI so that El = 2 EG. Then 
 IH will satisfy the conditions set. Prove that this is. true. The 
 
BOOK III 
 
 20T 
 
 H 
 
 surveyor checks the accuracy of his work by measuring IH. If 
 it equals AB, he considers his work is correct. Prove that this 
 checks the work. 
 
 173. In the adjacent figure the 
 distance A B between two inacces- 
 sible points oft'shore is desired. 
 How must the points K, R, L, M, 
 and H (and other points if neces- 
 sary) be located, and what lines 
 must be measured in order to com- 
 pute AB? Assume numerical values for the measured lines and 
 compute AB. 
 
 MISCELLANEOUS NUMERICAL EXERCISES 
 
 174. Two straight stretches of railroad track, I A and HB, are 
 connected by a circular curve. The chord AB is 100 feet. The 
 distance from the center of the 
 
 chord to the nearest point of the 
 
 track is 8 feet. Find the radius 
 of the curve of the inner rail. 
 
 175. In triangle ABC, b is 8, 
 a is 12, and c is 10. Find the seg- 
 ments of a made by the bisector 
 of Z.I. 
 
 176. In the triangle of Ex. 175 find the segments of a made by 
 the bisector of the exterior angle at A . 
 
 177. CK is the altitude on the hypotenuse AB of the triangle 
 ABC. AB is 25 and CK is 12. Find AK and AC. 
 
 178. In the triangle of Ex. 177, if AB is 13 and AK Is 4, find 
 A C and CK. 
 
 179. In the triangle ABC, c is 18, I is 20, and a is 24. Find 
 the altitude on side 18. 
 
 180. For the triangle of Ex. 179 find the diameter of the 
 circumscribed circle. 
 
208 PLANE GEOMETKY 
 
 181. The altitude of an equilateral triangle is -8. Find to three 
 decimals the length of one side. < 
 
 182. ABC &nd.AKR are secants to the same circle. AB is 12, 
 chord EC is 8, and chord KR is 15. Find AK and the tangent to 
 the circle from A . 
 
 183. The radii of two circles are 8 and 20 respectively. The 
 distance between their centers is 40. How far from the center of 
 the smaller circle does (a) the external common tangent cut the 
 line of centers ? (b) the internal common tangent ? 
 
 184. AB and KR are two chords of a circle intersecting at 0. 
 AB is 24, OR is 18, and OK is 3. Find AO and 0%. 
 
 185. The hypotenuse of a right triangle is 37 and another 
 side is 35. The perimeter of a similar triangle is 308. Find the 
 shortest side of the second triangle. 
 
 186. In the triangle ABC, c is 8 inches, Z.A is 45, and /-B is 60. 
 Find b and a. 
 
 187. In the triangle ABC, c is 10, b is 16, and Z.A is 30. 
 Find a. 
 
 188. How far is a swimmer from a cliff 400 feet high if when 
 his eye is level with the surface of the water the top of the cliff 
 is just visible ? (Use 3960 miles for the earth's radius.) 
 
 HINT. Let h = the height of the mountain and d the distance 
 required, both in miles. Then d 2 = h(Ji + 2R). 
 
 189. How high is a mountain if a swimmer thirty miles away 
 can see its top from the surface of the ocean ? 
 
 HINT, h (h + 2R) = d 2 . Since h is small compared to 2R, this 
 
 d 2 
 becomes very "closely 2 Rh = d 2 . Whence h 
 
 / i 
 
 190. If h is one's height in feet above the surface of the ocean 
 and d the distance in miles to an object on its surface just visible 
 
 l3~k 
 on the edge of the horizon, show that d = \~Q~> approximately. 
 
BOOK III 209 
 
 191. A lookout 100 feet above the surface of the ocean observes 
 an object close to the surface as it comes into view on the edge of 
 the horizon. . How far away is the object ? 
 
 192. A' is any point on the semicircle AB. AKcuts the tangent 
 at B in R and BK cuts the tangent at A in L. Prove that AB is a 
 mean proportional between AL and BR. 
 
 193. From the vertex D of the parallelogram ABCD a straight 
 line is drawn cutting AB at R and CB produced at K. Prove that 
 CK is a fourth proportional to AR y AD, and AB. 
 
 194. AB is a diameter of a circle and KB is a tangent. AK cuts 
 the circle in R. Prove that AR AK = Z/T. 
 
 195. Two chords of a circle, AB and CD, are produced to meet 
 in K. Then KR is drawn parallel to A D, meeting CB produced 
 at R. Prove that KR is a mean proportional between BR and CR. 
 
 196. If two lines parallel respectively to two sides of a triangle 
 are drawn through the point of intersection of the medians, they 
 trisect the third side. 
 
 197. Prove Theorem 20 thus : Draw a line through B and L 
 cutting K'R' at H. Then show by the use of the theorems on 
 proportion that L' coincides with 7/; that is, that the line through 
 L and L' passes through B. 
 
 198. Through A', any point in the common base of the triangles 
 ABC and ABR, lines are drawn parallel to AC and AR, meeting 
 BC and BR in the points F and G 
 
 respectively. Prove that FG is par- 
 allel to CR. 
 
 199. Two nonintersecting circles 
 are cut by a third circle. The two 
 common chords intersect at L. Secant 
 LKR cuts the first circle in K and R, and secant LGH cuts the 
 second in G and //. Prove that LK x LR = LG x LH. 
 
 HINT. Draw the tangents LA, LB, and L< '. 
 
PLANE GEOMETRY 
 
 Construction 1 
 
 304. Construct a wean proportional between two given 
 lines. 
 
 Given the lines a and b. 
 
 Required to construct a line 2*, so that a : : x = x : b. 
 
 Construction. Draw the indefinite straight line KG and lay off KL 
 equal to a and LR equal to I. Bisect KR at 0. With as a center 
 and OK as a radius, construct semicircle KR. Construct a perpen- 
 dicular to KR at L, cutting the semicircle at //. 11 L is the required 
 line. 
 
 HINTS. Draw KH and HR and use 282. 
 
 EXERCISES 
 
 200. Construct x if x = V<7//, a and b being lines of given length. 
 
 201. Construct a line V(J inches long. 
 
 HINTS. Let :r l>e the line. Then a: 2 = 6. Therefore 2:x = x: 3. 
 
 202. Construct a line a V 8 inches long, a being a line of given 
 length. 
 
 HINTS. Let a; be a V3. Then x* = 3 a 2 = (3 a), a : x = x :3 a, etc. 
 
 203. Construct # if x is <7 V f , being a given line. 
 
 204. Construct a; if .1- is -- Vo, ee. being a given line. 
 
 2i 
 
BOOK III 211 
 
 Construction 2 
 305. Construct a fourth proportional to three given lines. 
 
 Given the lines a, &, and c. 
 
 Required to construct a line x, so that a : b = c : x. 
 
 Construction. Draw AH and AL, making angle A about 30. On 
 AH lay off AR equal to a, on RH lay off RK equal to c, and on 
 AL lay off AB equal to b. Draw BR, and through K construct KG 
 parallel to BR, cutting AL at C. Then EC is the required fourth 
 proportional. 
 
 The proof should be supplied by the student. 
 
 EXERCISES 
 
 285. Construct x if x is equal to > a, b, and c being lines of 
 given length. 
 
 206. Construct if a; is equal to ab, a and b being 'lines of 
 given length. 
 
 HINT. #! = a ft, etc. 
 
 207. Construct a third proportional to a and b, two given lines. 
 
 208. Construct x if a 2 = foe, a and 5 being two given lines. 
 
 209. Construct x if a : b = x : c where a, b, and c are lines of 
 given length. 
 
 210. Construct x if - - = - where a, b, and c are lines of 
 , , , a x c 
 
 given length. 
 
212 
 
 PLANE GEOMETRY 
 Construction 3 
 
 306. Divide a given line into parts proportional to two 
 or more given lines. 
 
 b- 
 
 Given the lines a, &, and c, and the line AB. 
 
 Required to divide AB into parts proportional to a, b, and c. 
 
 Construction. Draw AC, making Z.A about 30. From A on 
 AC lay off All, RL, and LH equal respectively to a, ft, and c. 
 Draw J3//, and through R and L construct parallels to BH, cutting 
 A B at K and G respectively. Then AK, KG, and GB are the 
 required parts of the line AB. 
 
 The proof is left to the student. 
 
 Construction 4 
 
 307. Upon a given side homologous to a side of a given 
 polygon construct a polygon similar to the given one. 
 
 Given the polygon ABODE and the line A'B'. 
 Required to construct a polygon similar to ABODE with AB 
 and A'B' corresponding sides. 
 
BOOK iii 
 
 Construction (outline of method only). Draw all the diagonals 
 from A. Then construct Z2 =Z1, /.B' =/.B, and produce the 
 sides of Z 2 and Z B' until they meet in C'. 
 
 Then construct Z 4 = Z 3, Z 6 = Z 5, etc. 
 
 The remainder of the construction and the proof are left to the 
 student. 
 
 CONSTEUCTIONS INVOLVING PROPORTION 
 
 211. Given the perimeter, construct a triangle similar to a 
 given triangle. 
 
 212. Given the perimeter, construct a rectangle similar to a 
 given rectangle. 
 
 213. Divide one side of a triangle into two parts proportional 
 to the other two sides. 
 
 214. Construct a triangle, given two sides, a and b, and the 
 ratio of a to the third side c, expressed by two other given lines, 
 h and k. 
 
 HINT. Find a fourth proportional to A, k, and a. 
 
 215. Construct a triangle, given one side, a, the ratio of a to b, 
 and the ratio of a to t. 
 
 216. Draw a line parallel to one side of a rectangle, cutting off 
 a rectangle similar to the given one. 
 
 217. Inscribe in a given circle a triangle similar to a given one. 
 
 HINT. Circumscribe a circle about the given triangle. Join its center 
 to the vertices and study the central angles so formed. 
 
 218. Circumscribe about a given circle a triangle similar to a 
 given one. 
 
 219. Given one altitude, construct a triangle similar to a given 
 one. 
 
 220. Construct a trapezoid similar to but not equal to a given 
 one. 
 
BOOK IV 
 
 SURFACE MEASUREMENT. AREAS 
 
 308. Congruence and equality. It follows from the defini- 
 tion of congruence ( 24) that congruent figures are equal. 
 
 Suppose two plane figures are composed of, or may be divided 
 into, the same number of parts. Then if for each part of the first 
 there is one part of the second congruent to 
 it, and only one, the two figures are equal. 
 
 Hence it follows that two equal figures 
 are not always congruent. 
 
 Thus the rectangle A and the right triangle 
 B may be put together to form the trapezoid 
 or the pentagon on the right, and these two 
 polygons are equal but not congruent. 
 
 309. Unit of surface. Comparison of* 
 the size of plane figures requires meas- 
 urement of length, since the unit of surface is a square whose 
 side is the unit of length. This square is called the unit square. 
 The unit of surface in practical use is the square inch or the 
 square centimeter, or some multiple of them, as the square 
 yard or the square meter respectively. 
 
 310. Area. The area of a plane figure is the number 
 which expresses the ratio between its surface and the surface 
 of the unit square. 
 
 It is evident that any closed plane surface, whether bounded 
 by straight lines or by curves or in part by both, has an area. To 
 find that area may be easy or it may be difficult. An important 
 
 214 
 
BOOK IV 
 
 A 
 
 part of the work of plane geometry is finding the area of the 
 simple plane figures. 
 
 311. Area of a rectangle. The plane figure the measurement 
 of whose surface most clearly illustrates the notion of area 
 is a rectangle whose sides are exact multiples of the unit of 
 length. Thus, rectangle ABCD 8 five units long and three 
 units wide. It can, by parallels to n r 
 the sides, be divided into fifteen 
 
 squares each equal to the unit 
 square K. We say the area of the 
 rectangle is fifteen square units. 
 
 If the length of the base, AB, or 
 that of the altitude, A D, of the rec- 
 tangle is a mixed number like 5^, 
 the area can still be stated in terms 
 of K and is as before the product 
 
 of the base by the altitude. This is still true even if the base 
 and the altitude are both mixed numbers. 
 
 When the base or the altitude of the rectangle, or both, and 
 the side of the square have no common unit of measure, the area 
 of the rectangle can still be expressed in terms of K and is equal 
 to the product of the base and altitude. 
 
 The proof of this last is difficult. It will be omitted, and Theo- 
 rem 1, of 312, will be assumed as the basis of all work on area, 
 
 Theorem .1 
 
 312. The area of a rectangle is the product of its base 
 and altitude. 
 
 It should be noted that the product of two lines always means 
 the product of their numerical measures. 
 
 313. Corollary 1. The area of a square is the square of one of 
 its sides. 
 
 314. Corollary 2. The area of a right triangle is one half the 
 product of the two sides about the right angle. 
 
216 
 
 PLANE GEOMETRY 
 
 315. Example. A report of the Commissioner of Education of 
 the United States shows the number of pupils doing work of 
 secondary-school grade to be as follows : 
 
 School year . . . 
 Pupils (in thousands) 
 
 1899-1900 
 
 719 % 
 
 1904-1905 
 
 876 
 
 1909-1910 
 1131 
 
 1914-1915 
 1587 
 
 The above data can 
 be represented graph- 
 ically by means of rec- 
 tangles (bars) which 
 have equal bases but 
 whose altitudes vary 
 with the number of 
 pupils enrolled. 
 
 Solution. Decide upon 
 a common unit for the 
 base of eacn rectangle 
 and let one unit of alti- 
 tude represent 100,000 
 pupils. We have then the 
 accompanying diagram : 
 
 Pupils doing work of 
 Secondary Grade in the 
 Schools of the United States 
 
 1,500,000 - 
 
 1,000,000- - 
 
 500,000 
 
 EXERCISES 
 1. The following data are taken from the above report : 
 
 Year 
 
 1890 
 
 1895 
 
 1900 
 
 1905 
 
 1910 
 
 1915 
 
 Public and private high-school pupils 
 
 
 
 
 
 
 
 studying geometry (in thousands) 
 
 59 
 
 114 
 
 168 
 
 219 
 
 252 
 
 346 
 
 Graph the above data as in the foregoing example. 
 2. The following data are taken from the same report : 
 
 Year . 
 
 1890 
 
 1895 
 
 1900 
 
 1905 
 
 1910 
 
 1915 
 
 Pupils studying algebra (in thousands) 
 
 127 
 
 245 
 
 347 
 
 444 
 
 465 
 
 636 
 
 Graph the above data. 
 
BOOK IV 
 
 217 
 
 3. The same report shows the following percentages of the 
 public and private high-school enrollment to be studying algebra : 
 
 Year 
 
 1890 
 
 1895 
 
 1900 
 
 1905 
 
 1910 
 
 1915 
 
 Percentages of enrollment studying 
 algebra 
 
 43% 
 
 52% 
 
 55% 
 
 56% 
 
 57% 
 
 49% 
 
 
 
 
 
 
 
 
 Graph the foregoing data. 
 
 4. Secure (if possible) and graph data showing for your school 
 the percentage of pupils in each of the grades 9-12, inclusive. 
 
 5. The number of guns organized into batteries by the four 
 great allies on November 11, 1918, was as follows : 
 
 Nation 
 
 
 
 France 
 
 Italy 
 
 E upland 
 
 America 
 
 Number of 
 
 guns (in 
 
 hundreds) . 
 
 116 
 
 77 
 
 70 
 
 30 
 
 Represent these data graphically. 
 
 6. The number of battle planes in each army at the date of the 
 armistice was as follows : 
 
 Nation .... 
 
 France 
 
 Germany 
 
 England 
 
 Italy 
 
 America 
 
 Austria 
 
 Belgium 
 
 Number of planes 
 
 
 
 
 
 
 
 
 (in hundreds) . 
 
 33 
 
 27 
 
 18 
 
 8 
 
 7 
 
 (i 
 
 1.5 
 
 Represent these data graphically. 
 
 316. Equivalence. If two plane figures have the same area, 
 they are said to be equivalent. Equivalence is denoted by the 
 symbol =c=. Congruent figures, since they can be made to 
 coincide, are equivalent. 
 
 The word equivalent when applied to areas conveys the 
 same meaning as the word equal. 
 
 The words triangle, rectangle, etc. are often used for the 
 area of a triangle, the area of a rectangle, etc. 
 
218 PLANE GEOMETRY 
 
 317. Base. Either one of two adjacent sides of a paral- 
 lelogram can be considered its base. 
 
 318. Altitude. An altitude of a parallelogram is a line drawn 
 from any point in one base perpendicular to the opposite side. 
 
 A parallelogram has two bases and two altitudes. By " the base 
 and the altitude of a parallelogram " we mean either base and the 
 corresponding altitude. Similarly, " the base and the altitude of a 
 triangle " means any side and the corresponding altitude. 
 
 Theorem 2 
 
 319. The area of a parallelogram is the product of its 
 base and altitude. 
 
 D C 
 
 A K 
 
 Given the parallelogram ABCD, and DK its altitude on the 
 base AB. 
 
 To prove that the area of ABCD = AB x DK. 
 
 Proof. Extend AB and draw the altitudes OR and DK. 
 
 A A DK = ABCR. Why ? 
 
 Taking ABCR from quadrilateral A ROD leaves the parallelo- 
 gram A B CD. Why? 
 Taking AADK from quadrilateral ARCD leaves the rectangle 
 KRCD. Why? 
 Then parallelogram ABCD =c= rectangle KRCD. 51 
 The area of KRCD = DK x DC. 312 
 But DC = AB. Why? 
 Therefore the area of the parallelogram ABCD = AB x DK. 
 
BOOK IV 219 
 
 EXERCISES 
 
 7. The area of a rectangle is 192 square feet and its base is 
 three times its altitude. Find the base and the altitude. 
 
 8. The sides of a parallelogram are 10' and 20' respectively 
 and one angle is 45. Show that the area is 141.42 + square feet. 
 
 NOTE. The notation ' and " has long stood for minutes and seconds 
 respectively of an arc. In building-plans and engineers' drawings, how- 
 ever, it is widely used to designate feet and inches respectively. In 
 many of the numerical exercises this notation will be used. 
 
 Theorem 3 
 
 320. T}ie area of a triangle is one half the product of 
 its lose and altitude. 
 
 Given the triangle ABC and its altitude CK. 
 To prove that the area of A ABC = - -^ 
 
 Proof. Through C draw a line parallel to AB, and through A 
 draw a line parallel to BC intersecting the first in R. 
 
 AABC = AARC. Why? 
 
 Therefore the area of A AEC=\ the area of parallelogram A BCR. 
 But the area of A BCR = AB x CK. 319 
 
 Therefore the area of AABC = : - 
 
220 PLANE GEOMETRY 
 
 EXERCISES 
 
 9. Two sides of a triangle are 12" and 15" respectively, and 
 their included angle is 30. Find the area, 
 
 10. Find the altitude and the area of an equilateral triangle 
 whose side is 11. 
 
 11. Find the altitude and the area of an equilateral triangle 
 whose altitude is 15. 
 
 12. The sides of a triangle are 40', 50', and 60' respectively. 
 Draw the triangle to scale and determine graphically its three 
 altitudes. Using each altitude, compute the area and compare 
 the three results. 
 
 HINT. Let 1 inch represent 10 feet. 
 
 13. The triangle ABC and the triangle ABK have a common 
 base, AB, and are on the same side of it. The line CK is parallel 
 to AB. Prove the two triangles equivalent. 
 
 14. KR is parallel to AB of triangle ABC and cuts A C in A" and 
 BC in R. AR and BK are drawn. Prove that the triangle ARC is 
 equivalent to the triangle BKC. 
 
 15. A BCD is a quadrilateral and K the mid-point of AD. KR, 
 parallel to AB, and KL, parallel to CD, cut BC in R and L respec- 
 tively. AR, BK, CK, and DL are drawn. Name the pairs of 
 equivalent triangles thus formed. Prove the triangles of each 
 pair equivalent. 
 
 16. The hypotenuse of a right triangle is 58. Another side 
 is 40. Find the third side and the area of the triangle. 
 
 321. Altitude of a trapezoid. The altitude of a trapezoid 
 is the perpendicular drawn from any point in one base to 
 the other, produced if necessary. 
 
 In the figure on the opposite page, a is the altitude of the trapezoid 
 A BCD. Again, BK is also the altitude of A BCD. 
 
 QUERY. Are all altitudes of a given trapezoid equal ? 
 
BOOK IV 
 Theorem 4 
 
 221 
 
 322. The area of a trapezoid is one half the product of 
 its altitude and the sum of its bases. 
 
 b 
 
 Given the trapezoid ABCD in which b and c are the bases and 
 a is the altitude. 
 
 To prove that the area of trapezoid ABCD = 
 
 Proof. Draw the diagonal BD and the altitude BK. 
 
 BK = a. Why ? 
 
 a x I 
 
 Now 
 
 and 
 
 AABD = 
 
 ADCB = 
 
 (1) Why ? 
 
 7>'/\ x c a x <' 
 
 (2) Why? 
 
 But the trapezoid A BCD = AA BD + ADCB. (3) 
 Therefore, from (1), (2), and (3), the trapezoid 
 
 ABCD _ 
 
 EXERCISES 
 
 17. Find the area of a trapezoid whose bases are 46 inches and 
 42 inches respectively, and whose altitude is 2 feet. 
 
 18. Find the area of the trapezoid ABCD in which the base AB 
 is 100", AD is 20", Z.A is 30, and ^B is 45. 
 
222 PLANE GEOMETRY 
 
 19. The bases of a trapezoid are 20' and 29' respectively, and 
 the nonparallel sides are 17' and 10' respectively. Find the area 
 of the trapezoid. 
 
 20. Find the area of a triangle in which two sides are 10" and 
 12" respectively, and their included angle is 45. 
 
 21. In triangle ABC, c is 16','* is 20', and Z^ is 120. Find the 
 area of the triangle. 
 
 22. Find the area of a parallelogram in which two adjacent 
 sides are 3' 4" and 2' 3" respectively, and their included, angle 
 is 150. 
 
 23. Find the area of a right triangle in which the hypotenuse 
 and one side are 53" and 28" respectively. 
 
 24. The area of an. equilateral triangle is 100 V3. Find its side. 
 
 323. Area of an irregular polygon. It is practically impossible 
 to express the area of an irregular polygon of four or more sides 
 in terms of any simply defined bases and altitudes. In such cases 
 the area really depends on the lengths of the sides and the angles 
 which the adjacent sides form. The solution of all such problems 
 is fully treated in trigonometry. Certain special cases, however, 
 can be solved by the principles of geometry ; and the most gen- 
 eral case itself can be solved graphically. If the polygon is a 
 large one, such as a field, it becomes an important matter to make 
 the measurements of sides and angles as few as possible. A prac- 
 tical method in such cases consists in dividing the polygon into 
 triangles and trapezoids and measuring the bases and the alti- 
 tudes. This method is illustrated in Exs. 25 and 26, where lines 
 only are measured. Another method of determining the area 
 by line measurement only is by measuring the sides and all the 
 diagonals from one vertex. This really divides the polygon into 
 triangles, in each of which the three sides are known. Their 
 areas can then be computed graphically or by Problem 1, 334. 
 Ex. 27 illustrates the computation of the area of an irregular 
 polygon from measurements of its sides and its angles only. 
 
BOOK IV 
 
 223 
 
 EXERCISES 
 
 25. In the first figure the dimensions are in rods, the angles at 
 A", A', and L being right angles. Find the area of ABCDE in acres. 
 
 26. In the second figure, 
 AK is 60 rods, BL is 50 
 rods, DM is 40 rods, ER 
 is 45 rods, FK is 42 rods, 
 KR is 60 rods, RL is 54 
 rods, LM is 10 rods, 
 and MC is 35 rods. The 
 angles at K, R, L, and M 
 are right angles. Find the 
 area of ABCDEF in acres. 
 
 27. In a field ABCDE, 
 AB is 50 rods, BC is 35 
 rods, CD is 85 rods. The 
 angles A, B, C, and D are 
 respectively 110, 120, 
 
 100, and 115. Draw the polygon to scale and compute AE, DE, 
 and the area graphically. 
 
 324. Trapezoidal rule. The area bounded by a curved line, a 
 straight line, and two perpendiculars to the latter can be found 
 approximately as follows : 
 
 Divide AB into a number of equal parts and measure the length 
 of the perpendiculars from 
 the points to the curve. 
 Then consider each part, 
 as AMLK, a trapezoid. 
 
 The distances a v 2 , etc. 
 are called offsets. AM B 
 
 EXERCISE 28. Show that the area of a figure like ABRLK can 
 be approximately obtained by the trapezoidal rule as follows : 
 
 A dd half the sum of the first and last offsets to the sum of the other 
 offsets and multiply the result by the distance between the offsets. 
 
224 
 
 PLANE GEOMETEY 
 
 Theorem 5 
 
 325. If two triangles have an angle of one equal to an 
 angle of the other, their areas are to each other as the 
 product of the sides including the angle of the first is to 
 the product of the sides including the angle of the second. 
 
 A 
 
 H KG B' 
 
 A B 
 
 Given the triangles ABC and A'B' C' in which the angle A equals 
 the angle A'. 
 
 area A ABC AB x AC 
 
 Proof. Lay off A 'R =AC and A'K = AB and draw RK. Draw 
 also altitudes RH and C'G. 
 
 Then AABC = AA'KR. 
 
 Now 
 
 AA'KR _|(yl7f x/Z//) 
 AA''C"~ 
 
 Substituting 
 
 From (1) and 
 
 A'tf 
 
 /~)1* vy 
 
 ^(A'^'xc'f;) A'' 
 
 AA'HR^AA'GC*. 
 
 RH _ A'R 
 C 7 G~ A'C 1 ' 
 
 for -^ in (2) gives 
 
 AA'KR __ A'K x A'R 
 AA'B'C' ~ 
 AABC 
 
 C'G 
 
 (1) Why ? 
 
 (2) 320 
 
 (3) Why? 
 
 A'B' xA'C' 
 ABxAC 
 
 AA'B'C' -A'B' xA'C' 
 
 EXERCISE 29. Two triangles are equivalent if two sides of one 
 are equal respectively to two sides of the other and the included 
 angles are supplementary. 
 
BOOK IV 225 
 
 EXERCISE 30. If two triangles have an angle of one supple- 
 mentary to an angle of the other, they are to each other as the 
 product of the sides including the angle of the first is to the 
 product of the sides including the angle of the second. 
 
 HINT. Place the triangles with the two supplementary angles adjacent. 
 
 Theorem 6 
 
 326. The areas of two similar triangles are to each 
 other as the squares of any two homologous sides. 
 
 .C 
 
 Jf 
 
 Given the similar triangles ABC and A'B'C'. 
 
 AABC AB 2 AC 2 BC 2 
 
 a ^^__ = _ = __ = _ 
 
 Proof. Since /.A = /.A', 
 
 AABC AB x AC Al 
 
 _^,. r\\ &<>* 
 
 x A'B' X A'C~'' ( 
 
 AB EC AC 
 
 But == 2 
 
 Substituting in (1) for . its equal r -, obtained . from (2) 
 
 yl O ./I x> 
 
 AABC AB AB AB* 
 
 AA'B'C 1 ~ A'B' X A'B' ~ A*W*' 
 
 AB 2 ~BC* AC* 
 
 "From (2), = = = =1' ( 4 ) Why? 
 
 A'B' 2 B'C'* A'C'* 
 
 From (3) and (4), 
 
 AABC AB* AC* BC* 
 
 AA'B'C 
 
226 PLANE GEOMETRY 
 
 EXERCISES 
 
 31. In triangle ABC side AB is 9", AC is 10", and EC is IT". 
 The area of ABC is 36 square inches. A line parallel to AB cuts 
 EC in K and AC in R so that the area of the triangle CKR is 
 9 square inches. Find the sides of the triangle CKR. 
 
 32. The sides of a triangle are 4", 5", and 6" respectively. 
 Find the sides of a similar triangle whose area is nine times 
 the area of the first. 
 
 33. The area of a triangle is 300 square feet and one side is 18'. 
 The corresponding side of a similar triangle is 24'. Find its area. 
 
 34. A line parallel to EC of the triangle ABC divides it into 
 two parts of equal area. If AB is 10", find the two parts into 
 which the parallel divides it. 
 
 35. K on AB and R on A C of the triangle ABC are points such 
 that the trapezoid BCRK is seven sixteenths of the triangle ABC. 
 If AB is 20 centimeters, find AK. 
 
 36. One side of a triangle is 20". Two parallels to another side 
 divide the triangle into three equivalent parts. Find the three 
 segments into which parallels divide the given side. 
 
 Theorem 7 
 
 327. The areas of two similar convex polygons are to 
 each other as the squares of any two homologous sides. 
 D 
 
 D' 
 
 E'< ^ 
 
 Given the similar convex polygons P and P', in which the sides 
 AB, BC, etc. correspond respectively to the sides A'B', 5'C', etc. 
 
 P IB 2 ~BC* 
 To prove that etc. 
 
 P' A'B' 2 B'C' 2 
 
BOOK IV 227 
 
 Proof. From A and A' draw all the diagonals possible. 
 
 Then AABC^AA'B'C", AACD^ AA'C'D', etc. Wny ? 
 
 AB BC CD 
 
 By hypothesis, jjj, = ^ = ^ > etc. (1) 
 
 J~r>2 T> x-Y 2 ~ T-\2 
 
 Therefore =o = = - 2 ' etc. (2) Why? 
 
 .!'' 5'C" C'Z)' 
 
 and 
 
 A.4C.P 
 
 AA'C'D' 
 
 AADE 
 
 From (2), (3), (4), (5), etc. 
 
 AABC AACD AADE AB* BC 
 
 (6) , 
 
 AA'B'C' AA'C'D 1 AA'D'E' ^B' 2 Wc~' 2 
 From (6), 
 
 AA'B*C> + AA'C^ AA'D'E' = jjjT* = |5 ' 6tC ' 276 
 
 rni_ r P AB BC 
 
 Therefore -, = rs = rs > etc. 
 
 B'C' 
 
 EXERCISES 
 
 37. The corresponding sides of two similar polygons are 10' and 
 24' respectively. The area of the first is 1352 square feet. Find 
 the area of the second. 
 
 38. The areas of two similar polygons are 100 and 256 respec- 
 tively. One side of the first is 8. Find the corresponding side of 
 the second. 
 
 39. The corresponding sides of two similar polygons are 14' 
 and 48' respectively. Find their area if the area of a similar 
 polygon equivalent to their sum is 2500 square feet. 
 
228 PLANE GEOMETEY 
 
 Theorem 8 
 
 * < 
 
 328. If on the three sides of a right triangle as the 
 homologous sides similar polygons of n sides be con- 
 structed, the area of the polygon on the hypotenuse 
 equals the sum of the areas of the polygons on the 
 other two sides. 
 
 Given the right triangle ABC and the similar polygons P 1? P 2 , 
 and P 3 described on AC, AB, and BC respectively as the homolo- 
 gous sides. 
 
 To prove that P^^P^P^. 
 
 p fr 2 
 Proof. = 327 
 
 p K 
 
 and 7T = -==r ( 2 ) Why? 
 
 3 -o'' 
 
 
 But . = 1. Why? 
 
 EC 
 
 Therefore P 3 =0= P l + P 2 . 
 
 QUERY. Is the theorem of 284 a special case under Theorem 8 
 above ? 
 
BOOK IV 229 
 
 EXERCISES 
 
 40. The corresponding sides of two similar polygons are 
 10 meters and 24 meters respectively. What is the corresponding 
 side of a polygon similar to the first two and equivalent to 
 their sum ? 
 
 41. The corresponding sides of two similar polygons are 
 70 centimeters and 2 meters respectively. Find the correspond- 
 ing side of a similar polygon equivalent to their difference. 
 
 42. The corresponding sides of two similar polygons are 1' 
 arid 1' 4" respectively. Find the corresponding side of a polygon 
 similar to them whose area is (1) four times their combined 
 areas ; (2) seven times their combined areas. 
 
 329. Projection. The projection of one line-segment upon 
 another is the segment of the second line included between 
 the perpendiculars drawn from the extremities of the first 
 line to the second, produced if necessary. 
 
 In Figs. 1, 2, and 3 are shown the projections of a line KR 
 on a second line AB. In Fig. 1 the projection is LM, in Fig. 2 
 
 AL ^*i B 
 
 A L M B AL B R 
 
 FIG. 1 FIG. 2 FIG. 3 
 
 it is LR, and in Fig. 3 it is L M. These correspond to the three 
 possibilities, (1) the first line-segment may not meet the sec- 
 ond, (2) it may meet the second, (3) it may cross the second. 
 
 QUERY 1. Can the projection of a line-segment be greater than the 
 line-segment itself? Explain. Can it be less? Can it be equal to it? 
 Explain. 
 
 QUERY 2. Can the projection of a line-segment be zero? Explain. 
 
230 
 
 PLANE GEOMETRY' 
 
 QUERY 3. In the triangle ABC the, line CK is drawn perpendicular 
 to AB. What is the projection of A C on AB ? of EC on AB ? 
 
 QUERY 4. In an obtuse-angled triangle ABC, altitudes AK, nil, and 
 CL are drawn. What is the projection of AB on AC? AB on BC1 
 BC on ACt BC on AB? AC on BC? AC on AB? 
 
 | 
 
 EXERCISES 
 
 43. A line 12 inches long makes an angle of 30 with another 
 line. Find the length of the projection of the first line on the 
 second. 
 
 44. A line KR is 1' 8" long and makes an angle of 45 with a 
 second line AB. Find the length of the projection of KR on AB. 
 
 45. A line a inches long makes an angle of x *withKll. Find 
 the length of its projection on KR if x is 30, 45, 60, 90, 120, 
 135, 150. 
 
 Theorem 9 
 
 330. In any triangle the square of the side opposite an 
 acute angle equals the sum of the squares of the other two 
 sides minus tivice the product of one of these sides ~by the 
 projection of the other side upon it. 
 
 FIG. 1 
 
 TIG. 2 
 
 Given the triangle ABC, in which AK is the perpendicular 
 to BC, segment BK is the projection of side AB on side BC, and 
 the angle B is acute. 
 
 To prove that b 2 = c 2 + a 2 2 ax. 
 
BOOK IV 231 
 
 Proof. In Fig. 1, 6 2 = 7r + (a - xf. (1) Why? 
 
 In Fig. 2, b* = It + (a; - a) 2 . (2) Why ? 
 
 Both (1) and (2) give b' 2 = tf + x 2 -2ax + a 2 . (3) 
 In Figs. 1 and 2, . 
 
 h* + x 2 = (?. (4) 
 
 From (3) and (4), b 2 = c 2 + 2 - 2 ax. (5) 
 
 NOTE. Theorem 9 is an important theorem because of its value in 
 trigonometry. In that subject it is used to compute the angles of any 
 triangle when the lengths of the sides are known. In the right triangle 
 
 .1 JjK of Fig. l,cos B = - or x = c cos B. Substituting this value in equa- 
 c 
 
 tion (5) of the proof gives 
 
 &2 = C 2 + a 2 _ 2 ac - cos B. 
 Similarly, c 2 = a 2 + Ir- -2ab- cos C. 
 
 QUERY. Does Theorem 9 enable one to compute the altitude of a 
 triangle given the three sides? Explain. 
 
 EXERCISES 
 
 46". In the figures of Theorem 9 draw CR perpendicular at R to 
 BA, produced if necessary, and prove that b 2 = c z -f- a 2 2 c x BR. 
 
 47. If in the figure of Theorem 9 the side ABis7",AC is 15", 
 and EC is 20", find BK. Then find AK. 
 
 48. The sides of a triangle are 10', 17', and 21' respectively. 
 Find by the use of Theorem 9 the altitude on the side 21' and the 
 -area of the triangle. 
 
 49. The sides of a triangle are a, = 6, b = 5, and c = 4. Using 
 the equation given in the. note above and the table on page 188, 
 find the approximate values of the angles of the triangle. 
 
 50. From an inspection of the two formulas of the note above 
 state a similar formula which would involve cos .4. 
 
232 
 
 PLANE GEOMETRY 
 Theorem 10 
 
 331. In an obtuse-angled triangle the square of the side 
 opposite the obtuse angle equals the sum of the squares of 
 the other two sides plus twice the product of one of these 
 sides l>y the projection of the other side upon it. 
 
 Given the triangle ABC, in which AK is the perpendicular to 
 BC, BK is the projection of AB on BC, and the angle B is obtuse. 
 
 To prove that b 2 = c 2 + a 2 + 2 ax. 
 
 Proof. I' 2 = tf -f (x + a)\ (1) 'Why ? 
 
 or // 2 =A 2 + .x 2 + 2^ + a 2 . (2) Why? 
 
 Now 7i 2 + a a = e*. (3) Why? 
 
 From (2) and (3), tf = c 2 + a 2 + 2 ax. 
 
 EXERCISES 
 
 51. In the figure of Theorem 10 draw CR perpendicular to 
 AB produced at R and prove that I? = a* + c 2 + 2 AB x .##. 
 
 52. The sides of a triangle are 18", 20", and 34" respectively. 
 Find by the use of Theorem 10 the altitude on the side 18" and 
 the area of the triangle. 
 
 53. The sides of a triangle are 12', 17', and 25' respectively. 
 Find the altitude on the side 25' and the area. 
 
 54. If the angle C of the triangle ABC is 120, prove that 
 AB* = J3C 2 + AC 2 -f A C x BC. 
 
BOOK IV 
 
 233 
 
 332. Peaucellier's linkage. The figure below gives a top view 
 of a linkage, sometimes called compound compasses, designed to 
 draw a straight line without a ruler. This form of the instru- 
 ment was invented by A. Peaucellier of Nice. He published a 
 description of his instrument in 1873. Some time afterward he 
 was awarded for his invention the " Prix Montyou," the great 
 mechanical prize of the Institute of France. In the figure, CKPR 
 
 is a rhombus, AR = AK, AB = BC, and points A and B are fixed. 
 The bars meeting at the six points A , B, C, R, P, and K are hinged. 
 Hence, if the rhombus be moved, R and K will describe a circle 
 about A, C will move in a circle about B, and P, carrying a pen- 
 cil, will move in a straight line. The dotted lines are used only 
 in the proof which follows. 
 
 Points A, C, and P lie in a straight line, the perpendicular 
 bisector of KR. 
 
 AR ' 2 = AC ' J + CR ' 2 + 2 C// .1 C. Why ? (1) 
 
 AR 2 - ~cJl 2 = A C (A C + 2 C//) (2) 
 
 = JC-.-lP. (3) 
 
 If P.Vis A.toAL, 
 
 T heref ore 
 
 From (5) and (3), 
 
 Whence 
 
 AP AN 
 AL . .-LV = .4C .4P. 
 .1L AN = (AR* - C 
 
 AN= (AR 2 - C 
 
 (6) 
 
234 
 
 PLANE GEOMETRY 
 
 Since AR, CR, and AL are fixed lengths, the length of AN is 
 constant. Therefore, from any position of P a _L drawn to A /> 
 will always meet it in the same point N ; that is,/* will move 
 in a straight line. 
 
 NOTE. The proofs of Theorems 9 and 10 illustrate the fact that a 
 demonstration is more easily followed if the lines involved are denoted 
 by one letter instead of two. It is often a great help in solving exercises 
 to simplify the notation in this way. 
 
 QUERY. How can one determine from the three sides of a triangle 
 whether it has a right angle ? an obtuse angle ? an acute angle ? 
 
 EXERCISE 55. The sides of a triangle in which C is an acute angle 
 are a, b, and c respectively. Show that the projection of the side 
 
 a on the side b is and that the altitude on the side b is 
 
 Theorem 11 
 
 333. In any triangle the sum of the squares of two 
 sides equals half the square of the third side plus twice 
 the square of the median drawn to the third side. 
 
 Given the triangle ABC, in which CK is the median to AB. 
 
 = L + 2 m\ 
 
 To prove that 
 
 Proof. Draw CR _ 
 
 If CR coincides with CK, A ABC is isosceles and the truth of 
 the theorem immediately follows. If CR and CK do not coincide, 
 then Zl and Z2 will be unequal, Let ^2 be obtuse, 
 
BOOK IV 235 
 
 Now A K = K B = C - - (1) Why ? 
 
 L 
 
 Then, in ABCK, a 2 = + 2 + 2 ar, (2) 331 
 and, in A .1 C X, fl a = * + a - 2 * ( 3 ) 33 
 
 EXEECISES 
 
 56. The sides of a triangle are 16 centimeters, 30 centimeters, 
 and 34 centimeters respectively. Find the median drawn to the 
 longest side. 
 
 57. The sides of a triangle are 1 meter 40 centimeters, 1 meter 
 60 centimeters, and 2 meters respectively. Find to two decimals 
 the length of the median to the side 1 meter 40 centimeters. 
 
 58. ABC is a right triangle and CK the median on the 
 hypotenuse. Prove that 4C/C = AB . 
 
 NOTE. Determining the exact position of the big guns of the enemy so 
 that they can be destroyed by counter fire is an important part of modern 
 warfare. One way of doing this is to have at each of three or more 
 suitably placed stations certain instruments by which the report of the 
 big gun is noted and also the instant at which the sound arrives at each 
 station. A comparison of results shows the precise time-difference in 
 the arrival of the sound at the three stations. From the thermometer 
 and barometer readings the velocity of sound can be computed. From 
 this data the position of the gun can 
 then be calculated. 
 
 59. Three stations A, B, and 
 C are in a straight line so that 
 AB = BC = 2 miles. The report 
 of a big gun reaches B one second after it reaches A and reaches C 
 two seconds after it reaches B. The velocity of sound is 1120 feet 
 per second. How far is the gun from A, B, and C? 
 
236 PLANE GEOMETRY 
 
 HINTS. Let t = the number of seconds sound requires to travel from 
 G to A and let v equal the velocity of sound in miles per second. Then 
 GA = vt, GB = v(t + 1), and GC = v (t + 3). , 
 
 By 333, GA 2 + GC 2 = 2GB 2 + 2.^B Z . (1) 
 
 Substituting in (1) and solving, 
 
 = 18.1147 miles = 18 miles 605 feet etc. 
 
 V 2 3^ 2 
 
 EXERCISE 60. In the preceding exercise draw the triangle ACG 
 to scale and determine by means of a protractor the angle GA C. 
 
 Problem 1 (Computation) 
 
 334. Compute the area of a triangle in terms of its 
 three sides. 
 
 Given the triangle ABC in which the angle A is acute. 
 To compute the area, T, of A ABC in terms of the sides a, 
 5, and c. 
 
 Solution. Draw altitude CR. 
 
 th 
 Then 2*== (1) 320 
 
 Z 
 
 But h=^/P-a?. (2) Why? 
 
 Therefore (1) becomes 
 
 r = vy-^ (3) why? 
 
 Now since A ABC must have two acute angles, if /-A is acute, 
 then a* = b* + c 2 -2ex. (4) 330 
 
BOOK IV 237 
 
 Solving (4) for x 3 x = * + C *~ * - (5) 
 
 Substituting this value for x in (3), 
 
 Equation (6) expresses the area of a triangle in terms of its 
 sides, but in a form extremely inconvenient to use for numerical 
 computation. It can be put in a form better adapted for such a 
 purpose as follows : 
 
 T ..... *\ 1 1 J> i 
 
 
 f & _j_ 6 2 _ ^2X 
 
 Why ? (7) 
 
 T 2\V 
 
 2c /' 
 
 ^ 2c / 
 
 c 1/2 be 
 
 + # H- c 51 - a 
 
 2 \ /2 bc-b 2 - c 2 + a' 2 \ 
 
 /O\ 
 
 2 Ml 
 
 2c 
 
 A 2c / 
 
 ( 8 ) 
 
 c J[(#> + 2^-f c 2 )- 
 
 a 2 ][a a -(^-2fic + c a )] 
 
 (9) 
 
 ~ 9 \ 
 
 
 4c 2 
 
 = h^ 
 
 -<> 2 -^ 2 - 
 
 -;<i- ? n 
 
 (10) 
 
 = i V(6 + c 
 
 i + a) (6 + c a 
 
 , )(a+ ^_ c)(a _^ +c) . 
 
 (11) 
 
 In order to 
 
 simplify (11) 
 
 still further, let 
 
 
 
 a -{-b -\- c - 
 
 = 2*. 
 
 (12) 
 
 
 2a = 
 
 = 2. 
 
 (13) 
 
 (12)-(13), 
 
 - -f b -f c = 
 
 = 2s 2a = 2(s a). 
 
 (14) 
 
 Iii like manner a -f- b 
 
 -c = 2(s-c), 
 
 (15) 
 
 and 
 
 a b 
 
 + c = 2(s-b). 
 
 (16) 
 
 Substituting from (12), (14), (15), and (16) in (11), the latter 
 becomes 
 
 T=l^2sx 2(s -a)2(8- c)2(8 -b) (17) 
 
 = J Vl6*(* - a)(s - b)(s - c) (18) 
 
 Finally, T= Vs(s- a)(s- b)(s- c). (19) 
 
238 PLANE GEOMETRY 
 
 Example. Compute the area of a triangle whose sides are 
 9", 10", and 17" respectively. 
 
 Solution. T = Vx(s a) (s b)(s c). 
 
 a + b + c = 2s = 9 +10 + 17 =36. Whence s = 18. 
 Then T= Vl8 (18 - 9) (18 - 10) (18 - 17) 
 
 = Vl8- 9-8.1 = V3 2 x 2 x 3 2 x 2 3 
 
 = V3 4 x 2 4 = 3 2 x 2 2 = 36. 
 Note that s is not the sum of the three sides, but half the sum. 
 
 QUERY. In the preceding example what is the easiest method of 
 finding one or all of the altitudes? 
 
 MISCELLANEOUS NUMERICAL EXERCISES 
 
 In the following exercises, whenever possible, use equation 19 
 of 334 : 
 
 61. Find the area of a triangle whose sides are respectively 7', 
 15', and 20'. 
 
 62. Find the area of a triangle whose sides are respectively 
 15", 28", and 41". 
 
 63. Find the altitudes of the triangle whose sides are 35, 52, 
 and 73 respectively. 
 
 64. Find the -area and the altitude on the side 200 of a triangle 
 whose sides are 87, 119, and 200 respectively. 
 
 65. Find the radius of the circumscribed circle of a triangle 
 whose sides are 116, 231, and 325 respectively. 
 
 HINT. See 294. 
 
 66. Find the area of a trapezoid whose sides are 10 meters, 
 17 meters, 30 meters, and 51 meters respectively, the last two 
 being the bases. 
 
 67. ABODE is a pentagon. The lengths in yards of AB, BC, 
 CD, DE, and EA are respectively 60, 111, 39, 75, and 117. 
 The diagonal AC is 153 yards arid AD is 120 yards. How many 
 acres are there in the pentagon ? 
 
BOOK IV 
 
 239 
 
 68. How many tiles 4 inches square will it take to cover 
 the floor of a halt 60' 4" long and 15' 8" wide ? t ,, 
 
 69. A floor is 14' 6" by 18' 3". How many tiles >8 ' . 
 
 in the shape of a regular hexagon, each side being 
 
 one half an inch, will be required to cover it if the 
 cement between the tiles covers one twelfth of the 
 floor space ? 
 
 70. Find the area of the cross section of an I-beam 
 
 with dimensions as given in the adjacent figure. ' ' ' 
 
 71. The figure below represents a map of Kansas. To what 
 scale is it drawn if a line from the southeast corner to the north- 
 west corner is 440 miles long ? Compute the distance from Wichita 
 to Kansas City. Compute approximately the area of the state. 
 
 Kansas 
 KANSAS 
 
 
 NEW 
 MEXICO 
 
 Wichita 
 
 
 k 300 miles ^ 
 
 72. Compute from the figure above by means of the scale 
 the perimeter and the area of New Mexico. 
 
 73. The area of a right triangle is 30 square feet. The hypote- 
 nuse is 13 feet. Find the other two sides. 
 
 HINTS. Let x and y represent the sides about the right angles 
 respectively. 
 
 Then x 2 + f = 169, (1) 
 
 and | = 30. (2) 
 
 From (2), 2xy = l20. (3) 
 
 From (1) and (3), x 2 + 2 xy + y 2 = 289, (4) 
 
 and x z - 2 xy + f = 49. (5) 
 
 From (4), x + y = 17. 
 
 From (5), x y = 7 etc. 
 
240 PLANE GEOMETRY 
 
 74. The area of a right triangle is 6| square yards and the 
 hypotenuse is 17'. Find the other two sides. 
 
 75. The homologous sides of two similar polygons' are 35' and 
 12' respectively. Find the homologous side of a similar polygon 
 whose area is sixteen times the sum of the areas of the first two. 
 
 HISTORICAL NOTE. The discovery of the proof of the formula of 
 334 is credited to Hero of Alexandria (about 100 B.C.). Hero's proof 
 of the formula for the area of a triangle is given in Ball's "History of 
 Mathematics," pp. 92-93. Though very different from the proof here 
 given, being more geometrical in character, it is easily understood. It 
 is remarkable that a mind so able could not 
 derive the correct formula for the area of a 
 quadrilateral. His formula is 
 
 in which o^, a 2 , and & are denned as in the 
 adjacent figure. 
 
 Hero's activities extended to astronomy, surveying, and physics as 
 well as to mathematics. Eighteen centuries before Watt, Hero devised 
 an engine with steam as the motive power, which ran on the same prin- 
 ciple as that of a rotary lawn-sprinkler. 
 
 MISCELLANEOUS EXERCISES 
 
 76. A median of a triangle divides it into two equivalent parts. 
 
 77. H is any point on the diagonal AC of the parallelogram 
 A BCD. Draw DH and BH and prove that the triangles ABH and 
 ADH are equivalent. 
 
 78. R is any point on the median A K of the triangle ABC. Draw 
 BR and CR and prove the triangles ABR and ACR equivalent. 
 
 79. The line joining the mid-points of the bases of a trapezoid 
 divides it into two equivalent parts. 
 
 80. K is any point on the side AD and R any point on the side 
 DC of the parallelogram A BCD. Draw lines AR, BR, KB, and KC 
 and prove that the triangles BKC and ARB are equivalent. 
 
BOOK IV 241 
 
 81. K is any point within the parallelogram A BCD. Draw 
 lines from K to the four vertices and prove that the sum of 
 the areas of the triangles KA B and KCD is one half the area 
 of the parallelogram. 
 
 82. The mid-points of two sides of a triangle are joined to any 
 point in the third side. Prove that the area of the quadrilateral 
 thus formed is one half the area of the triangle. 
 
 83. The area of a triangle is one half the product of its perim- 
 eter and the radius of its inscribed circle. 
 
 84. The radius of the inscribed circle of a triangle whose 
 sides are a, l>, and c respectively is 
 
 >f 
 
 85. The sum of the perpendiculars 
 
 from any point within an equilat- 
 eral triangle to the sides is equal to the altitude of the triangle. 
 
 86. A BCD is a quadrilateral divided into two equivalent parts 
 by A'/i, the point K being on AB and the point R on CD. The point 
 L is on KB. A parallel to LR through K cuts CD (not CD 
 produced) in //. Prove that LH also 
 
 bisects the quadrilateral. 
 
 87. ABCD is a quadrilateral such 
 that K, a point within the triangle 
 ABC, makes the area of the quadri- 
 lateral AKCD one half that of ABCD. 
 
 A parallel to AC through K cuts BC in the point R. Draw the 
 line AR and prove that the quadrilateral A R CD is equivalent to 
 the triangle ABR. 
 
 88. Parallels are drawn through the vertices of the triangle ABC 
 cutting the sides or the sides produced in K, R, and L respectively. 
 Draw KR, RL, and LK and prove that the area of the triangle KRL 
 is twice that of ABC. 
 
242 
 
 PLANE GEOMETRY 
 
 
 ^V' 
 
 V 
 nr 
 
 ET B 
 K 
 
 \ 
 
 \ 
 
 \ 
 \ 
 
 \ 
 
 89. R and K are any two points on the sides CD and EC re- 
 spectively of the parallelogram A BCD. A parallel to AK through 
 D cuts AR produced in L. A paral- 
 lel to AL through K cuts DL pro- 
 duced in M. Prove that AKML and 
 
 A BCD are equivalent. 
 
 90. Prove that the area of the 
 square on the hypotenuse of a right 
 triangle equals the sum of the areas 
 of the squares 011 the other two sides. 
 
 HINTS. In the adjacent figure the tri- 
 angle A CK is congruent to the triangle 
 BCM. Therefore the rectangle OMCG 
 is equivalent to the square BCKR etc. 
 
 91. Prove that, in the right triangle ABC, a 2 -f- b 2 = c 2 , using 
 the adjacent figure, which was devised by President Garfield in 
 an original proof of the Pythagorean Theorem. 
 
 HISTORICAL NOTE. One of the great teachers of the 
 times preceding Euclid was Pythagoras. After study 
 and travel in various Mediterranean countries he 
 returned to Samos, his birthplace, where his teach- 
 ing had little success. Later he went to Croton, a 
 prosperous Greek colony in southern Italy, and here 
 great numbers of enthusiastic disciples attended his 
 school. He was a moralist and a philosopher and tried 
 to construct a mathematical basis for his moral and philosophical teach- 
 ings. Unfortunately the political results of his teaching brought him 
 and his school under suspicion, and in a popular uprising many of his asso- 
 ciates were killed and he himself had to flee. A year later, in 500 B.C., 
 he met his death in a similar disturbance in the city of Metapontum. 
 
 In all his work the ideal of Pythagoras was high knowledge, not 
 wealth, being the object of his study. In geometry the discovery and 
 proofs of numerous theorems are due to him. The one of Ex. 90, though 
 known to the Hindus and the Egyptians for centuries before his time, 
 has long been called the Pythagorean Theorem. It seems certain that 
 he was the first to prove it. But which one of the fifty or more proofs 
 now known is his has not been determined. 
 
BOOK IV 243 
 
 Problem 2 (Construction) 
 
 335. Construct a square equivalent to the sum of two 
 given squares. 
 
 HINT. Study Theorem 8 ( 328) and devise a method of proof. 
 
 Problem 3 (Construction) 
 
 336. Construct a polygon equivalent to the sum of two 
 given similar polygons and similar to them. 
 
 HINT. Study Theorem 8 ( 328) and devise a method of proof. 
 
 EXERCISES 
 
 92. Construct a square equivalent to the difference of two 
 given squares. 
 
 93. Construct a square equivalent to the sum of three given 
 squares. 
 
 Problem 4 (Construction) 
 
 337. Construct a square equivalent to a given triangle. 
 
 B 
 
 Given the triangle ABC with base b and altitude a. 
 Required to construct a square equivalent to A ABC. 
 
 Analysis. Let x be the side of the required square. Then its 
 area is x*. 
 
 Now the area of A ABC = ^-- 
 
 2i 
 
 Hence x*=. 
 
 Then %:x = x:b. Why? 
 
 
 
 The construction and proof are left to the student. 
 
244 PLANE GEOMETRY 
 
 NOTE. It should be noted that heretofore any description of the 
 method of thinking out the solution of a construction problem has been 
 carefully excluded from the solution itself. In many problems involv- 
 ing equivalent areas, however, it seems best to include as one step of 
 the detailed explanation the analysis which led to the solution. It is 
 desirable especially with those problems in which a simple equation so 
 clearly and simply indicates what the subsequent steps should be. 
 
 EXERCISES 
 
 94. Construct a square equivalent to a given parallelogram. 
 
 95. Construct a square equivalent to a given trapezoid. 
 
 96. Construct a square which is three fifths of a given rectangle. 
 
 97. Construct a square which will be to a given square in the 
 ratio of 7 to 4. 
 
 Problem 5 (Construction) 
 
 338. On a given line as the base, construct a triangle 
 equivalent to a given parallelogram. 
 
 D J/ C 
 
 \ 
 
 Given c, the base of the required triangle, and the parallel- 
 ogram ABCDj whose altitude is a and whose base is &. 
 
 Required to construct a triangle equivalent to ABCD on c as 
 the base. 
 
 Analysis. Let x be the unknown altitude of the required 
 triangle. 
 
 Then its area is That of the parallelogram is ab. 
 ft 
 
 ("IT 
 
 Then -^ = ab. 
 
 
 
 Hence ^:a = b:x. Why? 
 
 The construction and proof are left to the student. 
 
BOOK IV 245 
 
 339. Use of an equation in construction problems. Most of 
 the problems of Book IV require the construction of a figure 
 equivalent to a given one or to a certain definite part of a given 
 one. Such problems can be solved most readily as follows : 
 
 First, express as an equation the stated or implied relation. 
 Then represent the unknown line by x and express the area of 
 the required figure in terms of x and the given parts. 
 
 Then express the area of the given figure in terms of its 
 dimensions. 
 
 Lastly, use the conditions stated in the problem and state an 
 equation, as was done in 337 and 338. An inspection of this 
 equation, when put into the form of a proportion, will show 
 whether a mean proportional or a fourth proportional is required 
 to solve the given problem. 
 
 It is worth noting that the solution of a construction problem 
 involving areas can very often be made to depend on either a 
 mean proportional or a fourth proportional, or on both. 
 
 EXERCISES 
 
 98. On a given base construct a triangle equivalent to a given 
 square. 
 
 99. On a given base construct a triangle equivalent to a given 
 triangle. 
 
 100. On a given base construct a triangle equivalent to the sum 
 of two given triangles. 
 
 QUERY 1. How many triangles may be constructed which satisfy 
 the conditions of 338 ? 
 
 QUERY 2. Is more than one solution possible for the problem of 
 335 ? of 336 ? of 337 V 
 
 QUERY 3. How many parallelograms may be constructed. equivalent 
 to a given square ? 
 
 QUERY 4. How many solutions are possible for the following : On 
 a given base construct a parallelogram equivalent to a given square. 
 
 QUERY 5. How many solutions are possible for the following : Given 
 the two bases and another side, construct a trapezoid equivalent to a 
 given square. 
 
246 PLANE GEOMETRY 
 
 Problem 6 (Construction) 
 
 340. Transform a polygon of n sides into an equivalent 
 polygon ofn1 sides. 
 
 T? 
 
 D 
 
 B 
 
 Given the polygon ABCDEF , having n sides. 
 
 Required to transform the polygon ABCDEF into an equiv- 
 alent polygon having n 1 sides. 
 
 Construction (outline of method). Draw any diagonal which 
 joins two alternate vertices, as A C. 
 
 Through B draw BH II to A C. 
 Extend DC (or LA) to meet BH at A'. 
 
 Draw AK. 
 
 Then AKDEF is equivalent to ABCDEF . and has n 1 
 sides. 
 
 HINT. &ACB**&ACK. Why? 
 
 EXERCISES 
 
 101. Using the method of Problem 6, transform a quadrilateral 
 into an equivalent triangle. 
 
 102. Transform a pentagon into an equivalent triangle. 
 
 103. Transform a hexagon into an equivalent triangle. 
 QUERY. How may a polygon of n sides be transformed into an 
 
 equivalent triangle? 
 
 104. Construct a triangle equivalent to the sum of two given 
 triangles. 
 
 HINT. Use Problem 6. 
 
BOOK IV 
 
 247 
 
 NOTE. Brevity, completeness, and precision, or, in one word, concise- 
 ness, are desirable qualities in all effective writing; and they are espe- 
 cially desirable in writing out the solution of a construction problem. 
 In the following example observe particularly that all necessary lines 
 are actually constructed and that the arcs for the construction of a per- 
 pendicular and those for constructing the needed parallels are drawn. 
 Note how the accompanying diagrams cut down lengthy explanations. 
 
 Example. Given the hypotenuse, construct a right triangle 
 equivalent to a given triangle. 
 
 Given the triangle ABC and the line EF. 
 
 Required to construct a right triangle having EF as the 
 hypotenuse and equivalent to &ABC. 
 
 Analysis. Let c and h be respectively the base and the alti- 
 tude of the given triangle ABC and let x be the unknown altitude 
 upon the hypotenuse m, which equals EF, of the required triangle. 
 
 Then f = ^- .. (1) . 320 
 
 Hence m : c = h : x. (2) 289 
 
 Construction. Construct x, a fourth proportional to m, c, and h. 
 
 Then construct KR J_ to EF at its mid-point and describe 
 semicircle EF. On OR lay off OL equal to x. Construct LH _L to 
 OL, cutting the semicircle in F. Draw EV and FV. Then EVF 
 is the required triangle. 
 
 Proof. Z.EVF= 90. Why ? 
 
 The altitude of A E VF on EF = L = x. 
 
 * AEVFo &ABC by (1) and (2). 
 
248 PLANE GEOMETRY 
 
 QUERY 1. Is a solution of the preceding example possible for all 
 values of m, c, and A? Explain. 
 
 QUERY 2. What is the relation between in, c, and h which makes a 
 solution possible ? 
 
 PROBLEMS OF CONSTRUCTION INVOLVING AREAS 
 
 105. Construct a triangle equivalent to a given trapezoid. 
 Discuss (see 244). 
 
 106. Divide a triangle into five equivalent parts by lines drawn 
 from one vertex. Discuss. 
 
 107. Construct a square five times as large as a given square. 
 Discuss. 
 
 108. Through a given point within a parallelogram draw a line 
 dividing it into two equivalent parts. Discuss. 
 
 109. Divide a parallelogram into three equivalent parts by lines 
 through one vertex. Discuss. 
 
 110. Given two sides, construct a triangle equivalent to a given 
 triangle. Discuss. 
 
 111. Divide a parallelogram into two equivalent parts by a 
 line perpendicular to one side. Discuss. ^ 
 
 112. Given one diagonal, construct a 
 rhombus equivalent to a given parallelo- 
 gram. Discuss. 
 
 113. R L in the adjacent figure divides 
 the quadrilateral ABCD into two equiva- 
 lent parts. Draw through K a line which will also divide ABCD 
 into two equivalent parts. Discuss. 
 
 114. Construct an equilateral triangle whose area is twice that 
 of a given equilateral triangle. Discuss. 
 
BOOK Y 
 
 REGULAR POLYGONS AND THE MEASUREMENT 
 OF THE CIRCLE 
 
 Problem 1 (Construction) 
 341. Inscribe a square m a yiven circle. 
 
 D 
 
 Given a circle whose center is K. 
 Required to inscribe a square in the circle. 
 
 Construction . Draw the diameter CD and construct the diam- 
 eter AB J_ to it. Draw the chords AC, CB, ED, and DA. 
 Then ])BCA is the required square. 
 The proof is left to the student. 
 
 EXERCISES 
 
 In Exs. 1-9 the result obtained from any one may be used, 
 where possible, in any of those following. In problems dealing 
 with regular polygons (see 79) R, unless otherwise stated, rep- 
 resents the radius of the circumscribed circle. 
 
 1. Circumscribe a square about a given circle. 
 
 2. Inscribe a regular octagon in a given circle. 
 
 249 
 
250 
 
 PLANE GEOMETRY 
 
 3. Inscribe a circle in a given square. 
 
 4. Describe a method by which regular polygons of 16 sides, 
 32 sides, 64 sides, etc. may be inscribed in a circle/ 
 
 5. If the alternate vertices of a regular polygon of an even 
 number of sides are joined, the polygon thus formed is regular. 
 
 NOTE. Problems 1-6 of this book, and the exercises which follow 
 each, deal with certain regular polygons which can be inscribed in a 
 circle and circumscribed about a circle. It will be assumed that for 
 these regular polygons the center of the inscribed circle coincides with the 
 center of the circumscribed circle. Later this assumption will be proved. 
 
 342. Apothem. The apotliem of a regular polygon is the 
 perpendicular from the center of its circumscribed (or in- 
 scribed) circle to any side of the polygon. 
 
 EXERCISES 
 
 6. Show that the side of an inscribed square is R V2, that 
 
 7? 
 
 its apothem is V2, and that its area is 2R 2 square units. 
 
 7. Show that the side of a regular inscribed octagon is 
 W2-V2. 
 
 HINTS. Let AB in the adjacent figure be the 
 side of an inscribed square and BL the side of 
 the regular inscribed octagon. Compute in terms 
 of R the value of each of the following lines in 
 the order given : AB, BK, KH, KL, and BL. 
 
 8. Show that the apothem of a regular in- 
 scribed octagon in terms of R is -^^ + V2. 
 
 HINTS. In the figure of Ex. 7 let OC equal BL. Then MC is how 
 much ? HM is how much ? 
 
 9. Show that the area of a regular inscribed octagon is 2 R 2 V2. 
 10. A regular octagon is inscribed in a circle whose radius is 
 
 10. Show that its side is 7.65 + ? its apothem 9.23 +, and its 
 area 282.84 + square units. 
 
 A 
 
BOOK V 251 
 
 Problem 2 
 343. Inscribe a regular hexagon in a given circle. 
 
 Given a circle whose center is K. 
 
 Required to inscribe a regular hexagon in the circle. 
 
 Construction. Draw any radius KA. Then with A as the center 
 and A K as the radius, describe an arc cutting the circle at B. 
 With B as the center and KA as the radius, describe an arc cut- 
 ting the circle at C. In like manner obtain points D, E, and F. 
 Draw the chords AB, EC, CD, DE, EF, and FA. They will form 
 the required regular hexagon. 
 
 Proof. Draw the radius KB. 
 
 Then A ABK is equilateral. Why ? 
 
 Hence Z.K= 60, Why ? 
 
 and arc AB = 60, or one sixth of the circumference. Why ? 
 
 Hence arc ABCDEF is five sixths of the circle, and arc FA 
 is one sixth. 
 
 That is, chord FA = chord AB, 178 
 
 and the hexagon ABCDEF is equilateral. 
 
 Now Z D = one half of 240 = 120. Why ? 
 
 In like manner Z.E, Z.F, etc. are each equal to 120, 
 and the hexagon is equiangular. 
 
 Therefore the polygon ABCDEF is a regular hexagon inscribed 
 in the circle. 79 
 
252 
 
 PLANE GEOMETRY 
 
 EXERCISES 
 
 In Exs. 11-18 the result of any one may be used, if needed, 
 in any one following. 
 
 11. Inscribe an equilateral triangle in a given circle. 
 
 12. Inscribe a regular dodecagon in a given circle. 
 
 13. Circumscribe a regular hexagon about a given circle. 
 
 14. Circumscribe an equilateral triangle about a circle. 
 
 15. Show that the apothem of a regular inscribed hexagon is 
 
 ^ V3 and that the area is - V3. 
 
 Z 2i 
 
 16. Show that the side of a regular inscribed dodecagon is 
 j?\/2 Vs. the apothem -^\/2 + V3, and the area 3J? 2 . 
 
 t 
 
 17. Show that the side, the apothem, and the area of a regular 
 inscribed dodecagon in a circle whose radius is 20 inches are re- 
 spectively 10.35 + inches, 19.31 -f- inches, and 1200 square inches. 
 
 18. Construct the above designs. 
 
 19. The radius of a circle is 18. Show that the area of its 
 regular circumscribed hexagon is 1122.36 -J- square units. 
 
 20. Show that the side, the apothem, and the area of an in- 
 
 n O p2-y/Q 
 
 scribed equilateral triangle are respectively R VS, -~ > and - 
 
 21. Show that the side, the apothem, and the area of a circum- 
 scribed equilateral triangle are respectively 2 R V3, R, and 3 R 2 V3. 
 
BOOK V 253 
 
 344. Mean and extreme ratio. A line is divided by a point 
 in mean and extreme ratio when one segment is a mean 
 proportional between the whole line and the other segment. 
 
 B 
 
 AB:AK=AK:KB AB:AR=AR:RB 
 
 FIG. 1 FIG. 2 
 
 There are two cases : (1) when the point K is on the given 
 line AB (Fig. 1) ; (2) when the point R is on the given line 
 BA produced (Fig. 2).- 
 
 As case (2) is not used in elementary geometry, it will 
 be given no further consideration here. 
 
 Problem 3 (Computation) 
 
 345. Express the numerical values of the two segments 
 of a line of length a divided internally in mean and extreme 
 ratio by a point on the line. 
 
 A x K a-x B 
 
 Given the line AB of length a, and the point K dividing the 
 line in extreme and mean ratio. 
 
 Required to express the values of the segments AK and KB in 
 terms of a. 
 
 Solution. Let K be the point which divides AB internally in 
 mean and extreme ratio. Then, by the definition of 344, 
 
 AB:AK = AK:KR (1) 
 
 Let AK=x. 
 
 Then KB = a x. 
 
 Substituting in (1), a : x = x : (a, x). (2) 
 
254 PLANE GEOMETRY 
 
 Then x 2 = a 2 - ax. (3) 
 
 x 2 + ax = a?. (4) 
 
 Completing the square, 
 
 ^ 
 
 Therefore x=^ + ^ V5, or ^ ( 1 -f VS), (7) 
 
 and x = -| - I V5, or |(-i- V5). (8) 
 
 Since (8) gives a negative value of 35, it is neither ^4/f nor BK 
 of the figure. Therefore (7) gives the length of AK. 
 
 Hence BK = a x = a {- > + "^ ) = o "^^ 
 
 \ ^a j& / .J ^ 
 
 or /f = -(3 V5). (9) 
 
 EXERCISE 22. Show that the segments of a line 12' long 
 divided internally in mean and extreme ratio are 7.42' and 
 4.57'+ respectively. 
 
 NOTE. The problem of 345 was of great interest to the early Greek 
 geometers. Plato (429-347 B.C.) was the first to discuss and solve the 
 problem. Later Eudoxus (408-355 B.C.), who is usually ranked as third 
 among the Greek mathematicians, added several other theorems con- 
 cerning it. Euclid in the thirteenth book of his " Elements " has five 
 theorems on the "golden section." This name was used by the Greeks 
 to designate what now is usually called mean and extreme ratio. The 
 Greeks knew little of modern algebra and nothing of the Hindu nota- 
 tion. Hence their proofs are very much more difficult in appearance 
 than the proof given of Problem 3. The actual geometrical construc- 
 tion, however, follows in 346. 
 
 The numerous applications of the " golden section " in architecture 
 and other forms of art indicate that this division of a line possesses 
 aesthetic as well as mathematical significance. 
 
BOOK V 255 
 
 Problem 4 (Construction) 
 
 346. Divide a given line internally in extreme and 
 mean ratio. 
 
 Given the line AB. 
 
 Required to locate a point K on AB such that AB : AKAK:KB. 
 
 Construction. Bisect AB at L. Construct EM _L to AB at B. Lay 
 off BG equal to BL. Construct a circle with G as the center and 
 BG as the radius. Draw AG, cutting the circle in F and H. Lay 
 off AK on AB equal to AF. 
 
 Then K is the required point. 
 
 Proof. AH:AB=AB:AF. (1) 292 
 
 From (1), (AHAB):AB = (ABAF):AF. (2) 302 
 
 Now AB = FK, (3) Why? 
 
 and AF = AK. (4) Why? 
 Substituting from (3) and (4) in (2) gives 
 
 (AH - FH):AB = (AB - AK):AK. (5) 
 
 But AH FH=AF or AK, (6) 
 
 and A B - A K == KB. (7) 
 
 Therefore, using (6) and (7), (5) becomes 
 
 AK:AB=KB:AK t (8) 
 
 or AB:AK=AK:KB. (9) Why? 
 
 Therefore K is the required point. 344 
 
 347. Decagon. A decagon is a polygon having ten sides. 
 
256 
 
 PLANE GEOMETRY 
 
 Problem 5 (Construction) 
 348. Inscribe a regular decagon in a (jweii circle. 
 
 L 
 
 Given a circle whose center is A. 
 
 Required to inscribe a regular decagon in the circle. 
 
 Construction. Draw any radius AB. Then divide AB internally 
 in mean and extreme ratio as indicated in Fig. 2. With AK, the 
 larger segment, as the radius and B of Fig. 1 as the center, de- 
 scribe an arc cutting the circle at C. With C as the center and AK 
 as the radius, describe an arc cutting the circle, at F. In like 
 manner obtain points G, H, L, M, O, R, and S. Draw chords BC, 
 CF, etc. They will form the required decagon. 
 
 Proof. In Fig. 1 draw AC and lay off AK on radius AB equal 
 to AK of Fig. 2 and draw KC. Then in Fig. 1, 
 
 AB:AK = AK:KB. 
 But BC AK. 
 
 Hence AB : BC = BC : KB. 
 
 Now Z.B = Z. 
 
 Hence A A B C ^ A CBK. 
 
 Then 
 But 
 
 .I/;:. 1C = CB'.CK. 
 AB = AC. 
 
 (1) 
 
 Why ? 
 
 (2) 
 
 Why ? 
 
 (3) 
 
 
 (5) 
 
 274 
 
 (6) 
 
 Why ? 
 
 (7) 
 
 Why ? 
 
BOOK V 
 
 Hence BC = CK. (8) 135 
 
 Then from (2) and (8), C K = A K. (9) 
 
 Now Z . I + Z /? + Z EC A = 180. (10) Why ? 
 
 From (7), B = Z.BCA. (11) 29 
 
 From (5), Z3 = Z.4. (12) Why? 
 
 From (<*), Z4 = Z.l. (13) Why? 
 
 Also Z/*r.M=Z3+Z4. (14) 
 
 By (12), (13), and (14), Z.BCA = 2Z.1. (15) 
 
 By (11) and (15), Z = 2Z.l. (16) 
 
 Substituting from (15) and (16) in (10) gives 
 
 Z.I + 2ZJ + 2Z.4 =180. 
 Therefore Z.4 = 36, ^ 
 
 and arc BC is one tenth of the circle. Why? 
 
 Then arc CHSB is nine tenths. Hence arc SB is one tenth of 
 
 the circle. 
 
 Therefore BCFGH - - - is a regular inscribed polygon. Why ? 
 
 QUERY. How can a regular pentagon be inscribed in a circle? a 
 regular polygon of twenty sides ? of forty ? of eighty ? a five-pointed star ? 
 
 EXERCISES 
 
 In the next four exercises the results of any one may be used, 
 if possible, in any one following. 
 
 23. Show that the side of a regular inscribed decagon is 
 
 f(VE-i). A, 
 
 HINT. Study equation (7) of Problem 3 and its 
 results. 
 
 24. Show that the apothem of a regular in- 
 
 p i 
 scribed decagon is 4 v 10 -f- 2 V5. 
 
 HINTS. In the adjacent figure what is the value of AB, the side of 
 a regular inscribed decagon, in terms of /?? of AK in terms of Rl 
 Then use the triangle ACK. 
 
258 PLANE GEOMETRY 
 
 25. Show that the area of a regular inscribed decagon is 
 
 26. If the radius of a 'circle is 8 units, show that the side 
 of its regular inscribed decagon is approximately 4.94 units. 
 
 27. Construct the two designs which are given above. 
 QUERY. How can a regular inscribed pentagon be constructed ? 
 
 Problem 6 (Construction) 
 
 349. Inscribe a regular pentadecagon (15 sides) in a 
 given circle. 
 
 Given a circle whose center is 0. 
 
 Required to inscribe a regular 15-sided polygon in the circle. 
 " The construction and proof are left to the student. 
 
 HINTS. Let EC be the side of a regular inscribed hexagon and BK 
 the side of a regular inscribed decagon. How many degrees are there 
 in Zl? What fraction of the circumference is arc KCt 
 
BOOK V 259 
 
 EXERCISES 
 
 28. How can a regular polygon of 30 sides be constructed? 
 of 60 sides ? of 120 sides ? 
 
 29. How many regular polygons of less than 1025 sides can be 
 constructed by the use of Problems 1-6, inclusive ? 
 
 30. Write in order, the smallest first, the series of numbers 
 involved in the answer to the preceding exercise. 
 
 HISTORICAL NOTE. In the latter part of the eighteenth century Gauss 
 (1777-1855) gave considerable attention to the subject of regular poly- 
 gons. When he was but nineteen years old he devised a method of 
 constructing a regular polygon of seventeen sides. Later he showed that 
 if the number of sides of a polygon is a prime number and expressed by 
 the formula 2 n + 1, the polygon can be constructed by ruler and com- 
 passes. By bisecting the arcs subtended by the sides of each regular in- 
 scribed polygon thus obtained other regular polygons having 2, 4, 8, 16, 
 etc. times as many sides can be constructed. Allowing for this, Gauss 
 expressed his results by the more general formula 2 m (2 n + 1), where m 
 is any positive integer or zero and n is zero or any integer which 
 makes 2 n + 1 a prime number. But m and n must not be zero at the 
 same time. 
 
 Besides Gauss's construction of the regular 17-sided polygon at least 
 three others have been devised : one by J. A. Serrett (1819-1885), one 
 by A. L. Crelle (1780-1855), and one in 1897 by L. Gerard. The first 
 five prime numbers given by the formula are 3, 5, 17, 257, and 65,537. 
 F. J. Richelot carried out the solution for the regular polygon of 257 
 sides, and Oswald Hernes (1826-1909), after ten years of labor, com- 
 pleted the solution for a regular polygon of 65,537 sides. Since 2 n + 1 
 in Gauss's formula must be a prime number, the number of polygons 
 possible is a difficult problem in the theory of numbers. Very recently 
 Professor L. E. Dickson has shown that in 2 H + 1 if n .^ 100, the num- 
 ber of regular polygons which may be inscribed in a circle is 24, for 
 it is 52, and for n =100,000 it is 206. 
 
 EXERCISE 31. Form a table showing that Gauss's formula 
 2 m (2 n -f 1) includes all the polygons which are given in the 
 answer to Ex. 30, except those having 15, 30, etc, sides. 
 
260 PLANE GEOMETRY 
 
 Theorem 1 
 
 350. A circle can be circumscribed about any regular 
 polygon. E 
 
 Given any regular polygon ABODE 
 
 To prove that a circle can be described passing through the 
 points A, B, C, D, E, etc. 
 
 Proof. Let a circle be described through A, B, and C, point A' 
 being its center. 
 
 Draw the radii AM , KB, KG, and the line A/A 224 
 
 Z1+Z2=Z3+Z4. Why? 
 
 Z2=Z3. Why? 
 
 Therefore . Z 1 = Z 4. Why ? 
 
 AB = CD. Why? 
 
 KB = KC. Why ? 
 
 Hence A A KB is congruent to ACKD, Why ? 
 and A K = KD. 
 
 Therefore the circle K passes through D'. In like manner it 
 can be proved that the circle K passes through the other vertices. 
 
 EXERCISE 32. An equiangular polygon inscribed in a circle is 
 regular if it has an odd number of sides. 
 
 351. Radius of polygon. The radius of the circumscribed 
 circle of a polygon is called the radius of the polygon. 
 
BOOK V 261 
 
 Theorem 2 
 352. A circle can be inscribed in any regular poly cj on. 
 
 Given any regular polygon ABCDE 
 
 To prove that a circle may be drawn tangent to AB, J>C, CD, etc. 
 
 Proof. Circumscribe a circle about ABCDE . Let A' be its 
 center. 
 
 Now A B = BC = CD = DE, etc. Why ? 
 
 These chords are the same distance from the center. Why ? 
 
 Draw the apothem KR. 
 
 With K as the center and KR as the radius, a circle can be 
 drawn which is tangent to AB, BC, CD, etc. 191 
 
 353. Corollary. The centers of the inscribed and circumscribed 
 circles of a regular polygon coincide. 
 
 354. Center of polygon. The center of a regular polygon is 
 the common center of its inscribed and circumscribed circles. 
 
 355. Angle at the center of polygon. The angle at the center 
 of a regular polygon is the angle between the two radii drawn 
 to the extremities of any side. 
 
 EXERCISES 
 
 33. The angle at the center of a regular polygon is the supple- 
 ment of an angle of the polygon. 
 
 34. The apothem of an equilateral triangle is one half the radius. 
 
262 PLANE GEOMETRY 
 
 Theorem 3 
 356. If a circle is divided into three or more ; equal arcs, 
 
 (1) the chords joining the adjacent points of division 
 form a regular inscribed polygon ; 
 
 (2) tangents at the points of division form a regular 
 circumscribed polygon. 
 
 Given the circle R-, points A, B, C, D, E, etc., dividing it into 
 equal arcs; chords AB, BC, etc.; and tangents at A, B, C, etc., 
 intersecting at G, H, K, L, etc. 
 
 (1) To prove that ABODE is a regular inscribed polygon. 
 Proof. Chord AB = chord BC = chord CD, etc. Why? 
 
 Hence the inscribed polygon AD is equilateral. Why? 
 
 Now Z2 is measured by one half arc AEC, or by half the whole 
 circle minus half the sum of the two equal arcs AB and BC. Why ? 
 
 Also Z3 is measured by one half of arc BFD-, that is, by half 
 the whole circle minus half the sum of two arcs each equal to arc 
 BCorsiYcAB. Why? 
 
 Hence Z2 = Z3. 
 
 In like manner Z2 = Zl, etc., 
 
 and the inscribed polygon BCD - is equiangular. 
 
 Therefore the inscribed polygon BCD - is regular. Why ? 
 
BOOK V 263 
 
 (2) To prove that GHKL is a regular circumscribed 
 polygon. 
 
 Proof. Arc DE = arc DC. Why ? 
 
 Z.LED = Z.LDE and Z.KDC = ^KCD, Why ? 
 
 ^LDE = Z.KDC, Why? 
 
 and chord DE = chord DC. Why ? 
 
 Therefore AEDL is congruent to ADCK and each triangle is 
 isosceles. Why ? 
 
 Hence L = Z.K. Why? 
 
 In like manner /.K =/.H /.G, etc., 
 and the polygon GHKL ... is equiangular. 
 
 Now EL = LD = DK = KC =CH= HB, etc. Why ? 
 
 Then ZA' = KH = HG, etc. 
 
 Therefore the polygon GHKL ... is equilateral. 
 
 Therefore the circumscribed polygon GHKL -is regular. Why ? 
 
 QUERY 1. At what point in the proof of Theorem 3 could 223 
 have been used? 
 
 QUERY 2. Was 193 used in the proof of Theorem 3 ? 
 
 QUERY 3. Can a proof that GHKL is regular be based on these two 
 sections ? 
 
 EXERCISES 
 
 35. The side of the circumscribed equilateral triangle of a 
 circle is twice that of the inscribed equilateral triangle. 
 
 36. An angle of a regular circumscribed polygon is the sup- 
 plement of the angle measured by the smaller arc intercepted 
 by two adjacent sides. 
 
 37. An equiangular polygon circumscribed about a circle is 
 regular. 
 
 38. An equilateral polygon circumscribed about a circle is 
 regular if it has an odd number of sides. 
 
264 PLANE GEOMETRY 
 
 Theorem 4 
 
 357. If each of two regular polygons has -n sides, the 
 polygons are similar. 
 
 E 
 
 Given the regular polygons AD and 4 'D f , each having n sides. 
 To prove that AD is similar to A'D'. 
 Proof. Since the polygons are regular, 
 
 . AB = BC = CD,eic., (1) 
 
 and A'B' = B'C' = C'D', etc. (2) 
 
 A P TiC* C 1 T) 
 
 From (1) and (2), - = , = , etc. (3) Why ? 
 
 By hypothesis each polygon is equiangular. 
 
 Henee Z A = < 2n ~ , (4) 125 
 
 and * ^ A , = . (5) Why? 
 
 From (4) and (5), Z A = Z .4 '. (6) 
 
 Similarly, Z = Z ', Z C = Z C', etc. (7) 
 
 Therefore from (3), (6), and (7), polygons AD and A'D' are 
 similar. Why ? 
 
 EXERCISE 39. The diagonals of a regular pentagon form another 
 regular pentagon. 
 
BOOK V 
 Theorem 5 
 
 265 
 
 358. The perimeters of two regular polygons of the 
 same number of sides have the same ratio as their radii 
 or their apothems. 
 
 Given the regular polygons ABC - and A'B'C' , each of 
 n sides, with perimeters p and p\ radii r and r\ and apothems 
 a and a 1 respectively. 
 
 p r a 
 
 To prove that - = = . 
 
 p' r' a' 
 
 Proof. Let A' and A'' be the respective centers of the inscribed 
 circles. Draw KB and K'B'. 
 
 Why ? 
 
 (1) Why? 
 Why ? 
 
 (2) Why ? 
 
 (3) Why? 
 
 Then 
 
 Then 
 
 From (1) and (2), 
 
 Now 
 
 AB+BC+CD + ete. 
 
 AABK^AA'B'K'. 
 r _ AB 
 r'~ A'B'' 
 
 AAKL^AA'K'L'. 
 } a 
 
 a AB 
 
 AB EC CD ... __ 7 , 
 
 ^=^=c^' eto - < 4 > wh ^ ? 
 
 AB 
 
 _ p 
 
 A'B'+B'C'+C'D'+ete. A'B' p' 
 From (3) and (5), 5 = 5 = 5' 
 
 (5) 276 
 
266 PLANE GEOMETRY 
 
 Theorem 6 
 
 359. The area of a regular polygon is one, half the 
 product of its perimeter and its apothem. 
 
 Given a regular polygon ABC < of n sides with apothem a 
 and 'perimeter p. 
 
 To prove that the area of ABC is 5-*-. 
 
 2k 
 
 Proof. Let K be the center of the inscribed circle and s one 
 side of the polygon. Draw KA and KB. 
 
 The area of AABK = ^^ - Why ? 
 
 The polygon can be divided into n triangles, each congruent 
 to AABK. 
 
 ft VV o >y 
 
 Hence the area of the n triangles = n , or - n x s. Why ? 
 
 2i 2i 
 
 But n x s = p, the perimeter of the polygon. 
 
 Therefore the area of the regular polygon ABC = - 
 
 J 
 
 EXERCISES 
 
 40. The radius of a circle is 10". Find the apothem, the side, 
 and the area of its inscribed and its circumscribed squares. 
 
 41. The radius of a circle is 1. Find the apothem, the side, 
 and the area of its regular inscribed and circumscribed hexagons. 
 
 360. Computation of the circumference of a circle. The 
 circumference of a circle has been defined as the length of the 
 circle. A clear notion of what is meant by the circumference 
 and the method of computing it can be obtained as follows : 
 
BOOK V 267 
 
 In the adjacent figure the perimeter of the square A B CD 
 is obviously less than the circumference. Now bisect the arcs 
 AB, BC, CD, and DA at K, R, L, and M respectively, and 
 draw chords A K, KB, BR, etc. Since AK+ KB > AB (Why ?) 
 it follows that the perimeter of the octagon AKBR CLDM is 
 greater than the perimeter of the square. Further, the perim- 
 eter of the octagon is less than the circumference. Now 
 bisect the arcs AK, KB, BR, etc. at 
 F, G, H, etc. respectively, and form the 
 regular sixteen-sided figure AFKGB . 
 NowAF+FK>AK. (Why?) There- 
 fore it follows that the perimeter of the 
 sixteen-sided polygon is greater than that 
 of the octagon, and it also is less than 
 the circumference of the circle. The 
 process just outlined can be continued indefinitely and regular 
 polygons having 32, 64, 128, etc. sides can be constructed, 
 in each of which the perimeter is greater than the perimeter 
 of the preceding polygons and less than the circumference. 
 
 If we assume that these polygons are constructed, it is 
 apparent that the perimeters will increase with the number of 
 gides of the polygons and approach as near as we please the 
 length of the circumference of the circle. 
 
 If regular circumscribed polygons of 4, 8, 16, 32, etc. 
 sides be constructed about the same circle, it can be shown in 
 like manner that the perimeters of the successive polygons 
 decrease and approach as near as we please the length of 
 the circumference. 
 
 Further, the apothem oW>vW> apothem sW. (Why?) 
 In like manner the apothem of any of the inscribed polygons 
 is greater than the apothem of the preceding one and in the 
 process we are here considering can be made to approach as 
 near as we please to the radius. 
 
268 PLANE GEOMETRY 
 
 The following theorem will' be assumed: 
 
 ^ 
 / 
 
 Theorem 7 
 
 361. As the number of sides of a regular inscribed or 
 circumscribed polygon is successively doubled, the perim- 
 eter approaches as near as we please to the length of the 
 circumference of the circle and the apothem approaches 
 the radius. 
 
 362. Corollary. The circumferences of two circles are in the 
 same ratio as their radii or their diameters. 
 
 363. Definition of TT. The number TT (pronounced pi), used 
 in calculations on the circle, is the number obtained by divid- 
 ing the circumference of a circle by its diameter; that is, 
 
 From the above, C irD or C = 2 7r.fi. 
 
 364. The ratio TT is the same for all circles. This follows 
 from 362 and 363. 
 
 EXERCISES 
 
 42. The radius of a circle is 10'. Show that the circumference 
 is 20 TT feet. 
 
 43. The circumference of a circle is 32 TT inches. Find the 
 radius and the diameter of the circle. 
 
 44. A circle is inscribed in a square whose side is V 5". Find 
 the circumference in terms of TT. 
 
 45. The circumference of a circle is 100 TT inches. Find the 
 radius of a circle whose circumference is four times as great. 
 
 In Problem 7, which follows, a formula is derived by means of 
 which an approximate value of IT can be computed. 
 
BOOK V 269 
 
 Problem 7 (Computation) 
 
 365. If s is the side of any regular inscribed polygon, 
 x the side of a regular inscribed polygon having twice as 
 many sides as the first, and R the radius of the circle 
 circumscribing both, express x in terms of s and R. 
 
 C 
 
 Given the circle JET; AB, or 5, and BC, or x, the sides of reg- 
 ular inscribed polygons of n sides and 2 n sides respectively ; 
 and the radius R. 
 
 Required to express x in terms of s and R. 
 
 Solution. Draw the diameter CK, cutting AB at L. Draw BK 
 and BIL 
 
 Then CBK is a rt. Z. Why ? 
 
 Also CK is _L to A B. Why ? 
 
 Therefore CB* = CK x CL, (1) 282 
 
 or x 2 = 2R x CZ,. (2) 
 
 Therefore x = ^2R x CL. (3) 
 
 Now CL = R LH. 
 
 And since JLB = | , LH = -Jtf 2 - ^ (5) Why ? 
 
 From (4) and (5), CL 
 From (3) and (6), x 
 
270 PLANE GEOMETRY 
 
 NOTE. The formula (7) is a general result useful in computing the 
 ratio of the circumference to the diameter of a circle; Note carefully 
 the meaning of the formula. Thus, if s is the side of 1 an inscribed 
 square, x is the side of a regular inscribed octagon ; if s is the side of 
 an inscribed equilateral triangle, x is the side of a regular inscribed 
 hexagon, etc. 
 
 EXERCISES 
 
 The word formula in the following exercises refers to (7), 365. 
 46. Bead and express wholly in words : 
 
 47. In the first part of the preceding exercise if s^ replaces ,<? 10 , 
 what other change is necessary ? Make the change and express 
 the resulting statement verbally. 
 
 48. In the second formula of Ex. 46 substitute 3 for n and 
 express the resulting statement in words. Do the same with 6 
 substituted for n. 
 
 49. Use the formula to show that the side of a regular inscribed 
 dodecagon is R V 2 V3. 
 
 HINT. Here s = R ( 343). 
 
 50. Explain how the formula can be used to compute the side 
 of a regular inscribed octagon in a circle of given radius. 
 
 366. Computation of TT. The perimeter of a regular hexagon 
 (or any regular polygon) inscribed in a circle is less than the 
 circumference. The perimeter of a regular dodecagon in- 
 scribed in the same circle is greater than that of the regular 
 inscribed hexagon, but less than the circumference; and if 
 in the same circle as before regular polygons of 24, 48, 96, 
 etc. sides are inscribed, the perimeter of any one is greater 
 than the perimeter of the preceding one, but each perimeter is 
 less than the circumference ( 360). If we start with a circle 
 of given radius and compute the length of one side of each 
 
BOOK V 
 
 271 
 
 of these regular polygons and the corresponding perimeter, we 
 shall obtain, if this process is continued, a closer and closer 
 approximation to the length of the circle; that is, to the 
 circumference. By dividing this perimeter by the diameter of 
 the circle an approximate value of TT can be obtained. 
 
 In the table below, n denotes the number of sides of a 
 regular inscribed polygon, R the radius of its circumscribed 
 circle, * the length of one side, P the perimeter, and D the 
 diameter. P -5- D will be an approximate value of C-*-D, or TT. 
 
 n 
 
 R 
 
 s 
 
 P 
 
 P ~- D or ir 
 
 6 
 12 
 
 1 
 
 1 
 
 6 
 
 6-2 = 3 
 
 1 
 
 
 
 
 24 
 
 1 
 
 
 
 
 48 
 
 1 
 
 
 
 
 96 
 
 1 
 
 .06543817 
 
 6.28206396 
 
 3.14103 + 
 
 192 
 
 1 
 
 .03272346 
 
 6.28290510 
 
 3.14145 + 
 
 384 
 
 1 
 
 .01636228 
 
 6.28311544 
 
 3.14155 + 
 
 768 
 
 1 
 
 .00818126 
 
 6.28316941 
 
 3.14158 .+ 
 
 jR, s, and P-+-D for the hexagon are easily filled out, since 
 for that polygon x = R ( 343). For the side of an inscribed 
 dodecagon the computation is as follows : 8 & = R = ]_. Substi- 
 tuting in (7) of 365, 
 
 x or s 12 = 
 
 2 - V3 = V2 - 1.7320508 = V.2679497. 
 
 The most nearly accurate result as indicated in the table gives 
 TT = 3.14158.. In the work as outlined above the computation of 
 every side after the first depends on the degree of accuracy of 
 
272 PLANE GEOMETRY 
 
 the computation of all those which precede. Thus, s for a polygon 
 of 768 sides depends on s } for a polygon of 384 sides, and this in 
 turn on one of 192 sides, and so on back to the computation of 
 the side of a dodecagon. Therefore, to insure the correctness of a 
 final result to a certain number of decimals, the first computation 
 must be carried farther than the later ones ; that is, if the side of 
 the polygon of 768 sides is desired correct to eight decimals, the 
 computation of s for a dodecagon must be carried to fourteen or 
 sixteen decimals or possibly more. Careful attention to these 
 matters gives the value TT = 3.14159 -f-. This is usually written 
 TT = 3.1416. In work which does not require great accuracy, 
 7r = - 2 y 2 - is a convenient value often used. 
 
 EXERCISE 51. Compute s, P, and TT in the table of 366 for 
 the dodecagon, obtaining results to five decimals. 
 
 HISTORICAL NOTE. No number is more important in mathematics 
 and its applications than the number TT. The significant role it has played 
 in the development of mathematics can merely be hinted at in a short 
 note, for the history of TT runs back over four thousand years. Early 
 
 geometers perceived that the area of a circle is - . They saw that 
 a square whose side was a mean proportional between C and - would 
 
 have the same area as the circle, and called the problem, the solution 
 of which required the finding of C y , "squaring the circle." This is 
 essentially the problem of determining the value of TT, and it was from 
 the considerations just stated that the attempts of mathematicians to 
 determine ir by the ruler and compasses began. 
 
 The earliest reference to TT is by an Egyptian priest, Ahmes (1700 B.C.), 
 
 (8 D\ 2 
 - 1 , where D is the diameter. We 
 
 2 2 
 
 64 D 2 
 
 use f or the area - , where ir = 3.14 16. Therefore - = Solving 
 4 4 ol 
 
 this, TT = 3.1604, the value used by the Egyptians. 
 
 The value 3 for ?r, probably derived from the Babylonians, is mentioned 
 in the Bible in two places : 1 Kings vii, 23 ; and 2 Chronicles iv, 2. 
 
 Archimedes (born 287 B.C.), perhaps the greatest of all mathematicians 
 except Newton, proved by a method similar to that of 366, that the value 
 of TT lies between 3} and 3}J; that is, between 3.1428 and 3.1408. 
 
BOOK V 273 
 
 Where great accuracy is not required, 3^ is still often used. The 
 astronomer Ptolemy (A.D. 150) gave TT = 3 T y^, which equals 3.14166, a 
 value more nearly correct than that given by Archimedes. 
 
 Accurate values for ir were known to the Chinese at a comparatively 
 early date. Chang Heng (A.D. 78-139) gave TT = VI6, or 3.1623. In the 
 fifth century A.D. Tan Chung Chin gave a value for TT between 3.1415927 
 and 3.1415926, each value correct to the sixth decimal ! By methods 
 which he did not explain he inferred that these values were equivalent 
 to 2j 2 - and f f g respectively; and it is a remarkable fact that TT = f f *{ 
 was obtained independently eleven centuries later by Adrian Anthonisz 
 (1527-1607), a European mathematician. 
 
 Viete (1540-1603), using polygons of 6 2 16 , or 393,216, sides, found a 
 value of TT correct to nine decimals. About the same time another com- 
 puter carried the work correctly to 17 decimals and another to 35 decimals. 
 
 With Newton's invention of the calculus, methods of computing the 
 value of TT were discovered far simpler than that requiring the use of 
 inscribed polygons. By the aid of the calculus it was shown that 
 
 Still other series more convenient for the calculation of TT were derived. 
 Using- modern methods, Shanks, in 1873, obtained the value of TT to 
 707 decimals. 
 
 It may be observed that for a long time mathematicians endeavored 
 to find a common fraction equal to ?r; this meant that they assumed 
 that it was a rational number. Then, looking deeper, they began to 
 suspect that TT was an irrational number like VlO, a value given more 
 than once as an approximation. In 1766 Lambert gave an imperfect 
 proof that TT was irrational. In 1794 Legendre completed Lambert's 
 proof; but the possibility of obtaining the value of TT by ruler and 
 compasses still remained, for x = VlO = V2 5 is easily constructed by 
 305. In 1882, however, Lindeman proved that the problem of "squar- 
 ing the circle," that is, of constructing with ruler and compasses a 
 square having the same area as a given circle, was impossible. 
 
 Theoretic considerations rendered the study of the nature of TT im- 
 portant, but the lengthy decimals with which we can now express it are 
 never needed. For, in the words of Newcomb, "Ten decimals are suf- 
 ficient to give the circumference of the earth to a fraction of an inch, 
 and thirty decimals would give the circumference of the whole visible 
 universe to a quantity imperceptible with the most powerful microscope." 
 
274 PLAHE GEOMETRY 
 
 Theorem 8 
 
 367. The length of an arc of a circle in linear units is 
 to the circumference as the number of degrees in the arc 
 is to 360. 
 
 Given the circle O whose radius is R, and the arc AB whose 
 length is x and which contains a. 
 
 To prove that x:2TrR=a: 360. 
 
 Proof. The length of the circle A KB is 2irR. The arc x sub- 
 tends a central angle of a and the circumference subtends a 
 central angle of 360. 
 
 Therefore x:27rR = a: 360. 211 
 
 QUERY. Why are linear units specified in Theorem 8? 
 
 EXERCISES 
 
 Unless otherwise directed the value 3.1416 is to be used for TT 
 in all computations on the circle. 
 
 52. Find the length of an arc of 20 in a circle whose radius is 4. 
 
 53. An arc of a circle is 8 TT inches long and contains 30. Find 
 the radius of the circle. 
 
 54. The radius of a circle is 10 and the length of an arc is 5 TT. 
 How many degrees are there in the arc ? 
 
 55. A wheel stands in water which reaches halfway to the center 
 of the hub. How many degrees are there in the arc bounding the 
 submerged portion ? 
 
BOOK V 
 
 275 
 
 56. How long does it take the minute hand of a clock to move 
 through 132 ? 
 
 57. How long does the hour hand of a clock require to move 
 through 87 ? 
 
 58. The hands of a clock are 6' and 7' long respectively. How 
 far does the outer extremity of the minute hand travel in the time 
 from 8.30 A.M. to 1.55 P.M. ? the outer extremity of the hour 
 hand ? (Use - 2 y 2 - for *) 
 
 59. The radius of a wheel on a freight cai: is 21 inches. If 
 the car runs 35 miles per hour, how many revolutions does the 
 wheel make per minute ? (Use 
 
 - 2 T 2 - for TT.) 
 
 60. The centers of two pul- 
 leys (wheels) are 3' apart. The 
 radii are respectively 6" and 2'. 
 What is the length of the belt 
 which runs around the two ? 
 
 HINTS. In the adjacent figure 
 the length of the belt is AB plus arc BC plus CD plus arc DEA . Draw 
 KA, KR, and RB. Then draw RL parallel to AB, cutting KA at L. 
 
 61. Through how many degrees does a point 011 the equator 
 revolve about the axis of the earth in 9 minutes ? Through how 
 many miles ? (Use 3960 miles as the 
 
 radius of the earth.) 
 
 B 
 
 62. The radii of two pulleys are 
 8" and 2" respectively and their line 
 of centers is 20". Find the length of 
 a belt passing around them and crossed as in the adjacent figure. 
 
 368. Area of a circle. The area of a parallelogram which 
 is not a rectangle, of a triangle, or of a trapezoid depends in 
 a simple manner on the area of a rectangle, though no one 
 of the three figures can be completely divided into rectangles. 
 
276 
 
 PLANE GEOMETRY 
 
 The area of a circle is very much more difficult to measure 
 than that of any figure bounded by straight lines. The part of 
 a plane bounded by a circle cannot be divided up into squares 
 or rectangles, however small they may be, nor can its area 
 be made to depend in any simple way on either. 
 
 Just what is meant by the area of a circle and how it may be 
 determined can be understood by reference to the adjacent figure. 
 The area of the inscribed square is less than 
 the area of the circle. The area of the regular 
 inscribed octagon is greater than that of the 
 square but less than the area of the circle. If 
 regular inscribed polygons of 16, 32, 64, 128, 
 etc. sides are constructed, the area of each 
 polygon is greater than that of the preceding 
 one and less than that of the circle. If this 
 process is continued, a polygon can be obtained whose area is as 
 nearly equal to the area of the circle as we please. A similar 
 illustration can be obtained from regular circumscribed polygons 
 of 4, 8, 16, 32, etc. sides. 
 
 The facts, together with 359-361, illustrate the truth of 
 Theorem 9, which will be assumed. 
 
 Theorem 9 
 
 369. The area of the sector of a circle equals one half 
 the 2^oduct of its radius and its arc. 
 
 370. Corollary 1. The area of a circle equals one half the 
 product of its radius and its circumference. 
 
 371. Corollary 2. The area of a circle is irfi?. 
 
 Proof. By 363, C = 7rZ>, or 2 irR. Substituting 2 TrR for C in 
 
 C x R 
 
 the formula for the area of a circle, - ( 370), gives 
 
 <0 
 
 area of circle 
 
 x R 
 
BOOK V 277 
 
 372. Corollary 3. Tlie areas of two circles are to each oth et- 
 as the squares of 'their radii. 
 
 QUERY 1. If a circle is divided into n equal arcs, is a sector bounded 
 by one arc and two radii approximately a triangle ? 
 
 QUERY 2. What is the altitude approximately of the triangle ? the 
 base ? the area ? 
 
 QUERY o. What,. then, is the approximate area of the entire circle? 
 
 QUERY 4. If //, the number of equal arcs, is doubled, are the answers 
 to the preceding queries changed? 
 
 QUERY 5. Suppose the number of arcs is indefinitely great, what 
 bearing have these conclusions on the truth of Theorem 9? 
 
 EXERCISES 
 
 63. If the radius of a circle is 10", find the circumference and 
 the area. 
 
 64. The area of a circle is 16 TT. Find the diameter. 
 
 65. The circumference of a circle is 10 TT. Find the diameter 
 and the area. 
 
 66. The radii of two circles are respectively 9 and 12. Find 
 the diameter of a circle equivalent to their sum. 
 
 67. Find the radius of a circle whose circumference equals the 
 sum of the circumferences of the two circles in Ex. 66. 
 
 68. The radii of two circles are respectively 3 and 18. Show, 
 without finding the area of either, that the area of the larger is 
 36 times the area of the smaller. 
 
 69. The radius of a circle is 20". Find the radius of a con- 
 centric circle dividing the portion of the plane bounded by the 
 first into two equivalent parts. 
 
 70. The area of a circle is divided by two concentric circles 
 into three equivalent parts. The radius of the smallest circle is 10. 
 Find the radii of the other circles. 
 
 71. Semicircles are described on the sides of a right triangle as 
 diameters. Prove that the area of the semicircle on the hypotenuse 
 equals the sum of the areas of the other two. 
 
278 PLANE GEOMETRY 
 
 Theorem 10 
 
 373. The area of a sector is to the area of the circle as 
 the angle of the sector is to four right angles. 
 
 jr- <B 
 
 Given the sector AOB in the circle whose center is 0. 
 
 To prove that the area of the sector AOB : TrR 2 = Z.O: 360. 
 
 Proof. The area of the circle = ^ , or TrR 2 . (1) 371 
 
 o -p/- A Z? vx Z> 
 
 The area of sector A OB = - 
 
 From (1) and (2), 
 
 * : i MGABxR - ^^ /ON 
 
 area of sector A OB : area of circle = - : - (3) 
 
 Or area of sector A OB : area of circle = arc AB : C. (4) 
 
 But arc AB : C = Z. : 360. (5) 367 
 
 From (4) and (5), area of sector AOB : irR 2 = ZO : 360. 
 
 NOTE. There is a close analogy between the area of a sector and 
 that of a triangle. In applying this formula care must be exercised to 
 avoid the error of multiplying the arc in degrees by the radius in 
 linear units. 
 
 EXERCISES 
 
 72. The 1910 Census Report of the United States shows each 
 of four parts which make up the entire population to be as follows : 
 (1) native born, of native parents, 54 % ; native born, with 
 one or both parents foreign born, 20.5%; foreign born, 14.7%; 
 and negroes, 10.7%. 
 
BOOK V 
 
 279 
 
 Display the above data by assigning to each part that sector 
 of a circle which is its correct portion of the entire population. 
 
 sectors are to each other as 
 
 20.5 % 
 
 Native born, 
 one or both 
 parents 
 foreign 
 born 
 
 Solution. Since the areas of the 
 their central angles, we must first 
 find the central angle which cor- 
 responds to each percentage given. 
 Now 54% of 360 = 194, 20.5% of 
 360 = 74, 14.7% of 360 = 53, and 
 10.7% of 360 = 39. 
 
 The central angles are then con 
 structed, as in the adjacent figure. 
 
 73. The same report gives 
 the following percentages for 
 the population of Dayton, Ohio : 
 native born, of native parents, 
 62 % ; native born, with one or 
 
 both parents foreign born, 21.9%; foreign born, 11.9% ; and 
 negroes, 4.2 % . 
 
 Display the above data as before. 
 
 74. For Des Moines, Iowa, the census report gives the follow- 
 ing facts : native born, of native parents, 62.3 % ; native born, 
 with one or both parents foreign born, 22.3 % ; foreign born, 
 12%; and negroes, 3.4%. 
 
 Display the above data. 
 
 75 . What fraction of the area of a circle is the area of a sector 
 in it whose angle is 18 ? 30 ? 45 ? 24 ? 108 ? 
 
 76. The area of a circle is 100 square feet. What is the area 
 of a sector of 36 in this circle ? of 60 ? of 135 ? of 40 ? 
 
 77. Find the area of a sector of 20 in a circle whose radius is 10'. 
 
 78. The area of a sector is 12 TT square inches in a circle whose 
 radius is 6". Find the angle of the sector. 
 
 79. Find the circumference of a circle in which a sector of 40 
 has an area of 16 TT square feet. 
 
280 PLANE GEOMETRY 
 
 80. The radius of a circle is 7'. An arc in it is 4' long. Find 
 the area of the corresponding sector. (Use - 2 y 2 - for TT.) 
 
 81. The hour hand of a clock is 8 feet long. Find, the area of 
 the sector over which it moves in the time from 12.18 P.M. 
 to 3.40 P.M. 
 
 Theorem 11 
 
 374. The area of a segment ivhose arc is less than a 
 semicircle equals the area of the corresponding sector minus 
 the area of the triangle bounded by two radii and the 
 chord of the segment. 
 
 The proof is left to the student. 
 
 QUERY. How is the area of a segment whose arc is greater than a 
 semicircle found? 
 
 EXERCISES 
 
 82. Find the area of a segment whose arc is 90, the radius of 
 the circle being 10'. 
 
 83. In the preceding exercise let the arc be 30 and solve. 
 
 84. Let the radius of a circle be 20" and x be the number of 
 degrees in the arc of a segment. Find the area of the segment 
 when x is 45, 60, 120, 150, 300. 
 
 85. Three equal circles, radius 10", are tangent each to the 
 other two. Find the area of the triangular portion of the plane 
 bounded by arcs of the three circles. 
 
 86. Show that the area of the segment bounded by a side of 
 
 -R 2 
 an inscribed square and its arc is -j- (TT 2). 
 
 87. Show that the area of the segment of any circle bounded 
 by a side of a regular inscribed hexagon and its subtended arc is 
 
BOOK V 281 
 
 375. Graphical determination of the area of a segment of a 
 circle. Note that by 373 the area of a sector can always be 
 found if its arc or central angle and the radius of the circle 
 are given. But given its arc or central angle 
 
 and the radius of the circle, the area of a seg- 
 
 ment cannot be found by means of 374 
 
 and the geometry so far presented unless the 
 
 angle is 30,' 36, 45, 60, 72, 90, ,120, 
 
 135, or 150. For example, if the radius is 
 
 10 and the arc of the segment is 80, we must have recourse 
 
 to graphical methods or to trigonometry. 
 
 To solve the problem just stated by the graphical method 
 we can draw the circle and the segment to scale, as in the 
 adjacent figure. Then measurement of the base AB and the 
 altitude OR of triangle ABO will enable one to determine its 
 area. Since the area of the sector AOB is | of 100 TT, the 
 area of the segment AKB easily follows. 
 
 376. Approximate formula for the area of a segment of a 
 circle. Frequently in practice it is much easier to measure AB, 
 the chord, or the base of the segment in the 
 
 adjacent figure, a'nd.RJf, the mid-perpendicular 
 of chord AB, than to measure the arc AKB or 
 its central angle. Let b denote AB and h denote 
 
 KK. Then the area of AKBR is -- + 
 
 253 
 
 This formula is approximate, but gives very accurate results. 
 
 EXERCISES 
 
 88. Find graphically the area of a segment whose arc is 112 
 in a circle whose radius is 10". 
 
 89. Find graphically the area of a segment whose arc is 280 
 in a circle whose radius is 20'. 
 
282 PLANE GEOMETRY 
 
 90. Test the accuracy of the formula of 376, thus : Find the 
 area of a semicircle in the usual manner, then by the use of the 
 formula, and compare the two results. 
 
 91. Test the accuracy of the formula of 376 on segments 
 whose arcs are 90 and 60 respectively. 
 
 92. Find by 374 and by 376 the area of the segment whose 
 chord is the side of a regular inscribed hexagon and compare the 
 two results. 
 
 93. The radius of a circle is 10'. The height of a segment in 
 it is 2'. Find the area of the segment. 
 
 MISCELLANEOUS EXERCISES 
 
 94. A central angle whose intercepted arc is equal to the 
 radius of the circle is called a radian. Show that one radian 
 equals 57.29 + degrees. 
 
 95. Show that TT radians equal 180. 
 
 96. How many degrees are there in -^ radians? ? -^-? 
 
 IT 3 ? 
 
 97. How many radians are there in 90 ? 135 ? 60 ? 240? 
 30 ? 45 ? (Give answers in terms of TT.) 
 
 98. The diameter of a cylinder is 22". Find the pressure 
 on the piston LR when the steam pressure is 
 
 90 pounds to the square inch. What is the thrust 
 along the piston rod AB? 
 
 99. The earth turns on its axis once in about 
 24 hours. Through how many degrees does a 
 
 point on the equator rotate in 5 hours? Through how many 
 miles does it rotate about the earth's axis ? (R = 4000 miles.) 
 
 100. If the earth rotates on its axis once in 23 hours and 
 56 minutes, how many miles does a point 30 north of the equator 
 move in 2 hours and 20 minutes ? (R = 3960 miles.) 
 
BOOK V 
 
 283 
 
 101. A continuous belt. runs around two pulleys, B and D, 
 which are mounted so as to turn 
 
 freely with the shafts AB and CD, 
 and so that if either wheel rotates the 
 other does also. If pulley D makes 
 1200 R. P. M. (revolutions per minute), 
 how many does B make ? 
 
 102. If point K on the belt travels 
 
 14 feet per second, how many R.P.M. does each wheel make ? 
 
 103. Belting runs around pulleys A and BC. Pulleys BC 
 and BD are ' firmly fastened to the same shaft. Pulley BD is 
 belted to E. If A makes 1400 R.P.M., how many does E make ? 
 
 104. How many times does cogwheel A turn while B turns 
 once ? -As A turns 30 times, how often does B turn ? 
 
284 PLANE GEOMETRY 
 
 105. In the figure below, the cogwheels B and C turn together, 
 being made of a single piece of iron. While A revolves 36 times 
 how often does the wheel B revolve ? the wheel C ? l!he wheel Z> ? 
 
 106. While D revolves 20 times how often does C revolve ? 
 B? A ? 
 
 107. While B revolves 60 times how often does A revolve ? 
 C? D? 
 
 HISTORICAL NOTE. In one year the student has obtained a fairly 
 complete notion of the fundamentals of plane geometry which were 
 known to the Greeks. During the last three hundred years, however, the 
 labors of various mathematicians have built up the vast structure of 
 modern geometry an extensive body of mathematics using principles 
 and methods of which Euclid never dreamed. 
 
 Euclidean geometry concerns itself largely with measurement ; in fact, 
 it is often called metric geometry. Only occasionally, as in Theorem Iti, 
 page 110, does it deal with a descriptive theorem one into which no 
 notion of measurement enters. On the other hand, modern geometry, 
 or, more accurately, modern projective geometry, deals largely with 
 descriptive theorems. 
 
 Regarded by some as the greatest geometrician since Euclid, Jacob 
 Steiner (1796-1863) was perhaps the foremost contributor to the develop- 
 ment of modern geometry. He was the son of a poor farmer and did 
 not learn to read until he was thirteen. For a time he studied with 
 Pestalozzi, the famous exponent of certain methods of primary teaching. 
 Later he was a student in the University of Heidelberg. The publication 
 of some 'of his original work began in 1826. Other able mathematicians 
 
BOOK V 285 
 
 recognized his genius and succeeded in obtaining his election in 1834 
 to a special chair of geometry in the" University of Berlin. Of other 
 lines of mathematics Steiner remained in practical ignorance all his 
 life. It is said that his knowledge of algebra went only as far as the 
 quadratic equation. 
 
 Some incidents of the life of Jean Victor Poncelet (1788-1867), 
 another eminent worker in the field of modern geometry, are both 
 interesting and instructive. Educated in the Lyceum at Metz and in 
 the Polytechnic School of Paris, he obtained a commission in the 
 French engineers. During Napoleon's retreat from Moscow in 1812 
 he was left for dead on the field of Krasnoi. Captured by the Russians, 
 he was imprisoned in Saratov. There, in poor quarters, with no books 
 and with only what mathematics he remembered from his work in 
 school, he began a very remarkable series of original researches, re- 
 cording his discoveries on bits of paper brought him surreptitiously by 
 his jailer. On his release from prison in 1814 he took with him the 
 brilliant results of his solitary work, and in 1822 they were published 
 under the title "Treatise on the Projective Properties of Figures," a 
 work which will always remain a classic. 
 
SUPPLEMENTARY EXERCISES 
 
 BOOK I 
 
 1. If the base of an isosceles triangle is divided into three equal 
 parts, lines drawn from the vertex to the points of division are equal. 
 
 2. ABC is an equilateral triangle and K, R, and L are points on 
 the sides AB, BC, and CA respectively such that AK = BR = CL. 
 Prove that the triangle KRL is equilateral. 
 
 3. If ABC is an equilateral triangle and AB, BC, and CA are pro- 
 duced the same distance to K',R, and L respectively, the triangle whose 
 vertices are A', 72, and L is equilateral. 
 
 4. If each side of an equilateral triangle is divided into three equal 
 parts and the points of division nearest to each vertex are joined respec- 
 tively, a hexagon is formed which is equilateral and equiangular. 
 
 5. ABCDEF is a regular hexagon. The lines AE, AD, and A C are 
 drawn. Prove that the triangle ADE is congruent to the triangle A DC 
 
 6. Two isosceles triangles are equal if the base and the vertex angle of 
 one are equal respectively to the base and the vertex angle of the other. 
 
 7. ABC and ABK are two equilateral triangles on opposite sides 
 of the same base AB. The lines BR and BL are the bisectors of the 
 angles ABC and ABK respectively, meeting AC and AK in R and L 
 respectively. Draw RL and prove that the triangle BRL is equilateral. 
 
 8. K is the middle point of the side A C of the equilateral triangle 
 ABC and KR is perpendicular to BC at R. Prove that R C equals one 
 fourth of BC. 
 
 9. If from the vertex of one of the equal angles of an isosceles tri- 
 angle a perpendicular is drawn to the opposite side, it makes with the 
 base an angle equal to one half the vertical angle of the triangle, 
 
 HINT. Draw an auxiliary line from the vertex, 
 
 280 
 
SUPPLEMENTARY EXERCISES BOOK I 287 
 
 10. Investigate the proof of the preceding exercise and determine 
 whether the theorem' is true when the vertical angle is acute and when 
 it is obtuse. 
 
 11. If from any point K in one of the equal sides AB of an isosceles 
 triangle ABC, KR drawn perpendicular to the base BC meets CA 
 produced at R, prove that the triangle AKR is isosceles. 
 
 12. AC is the base of an isosceles triangle ABC. Any distance AL 
 is laid off on AB. On BC produced CR is laid off equal to AL. If RL 
 cuts A C at K, prove that KL equals KR. 
 
 HINT. Draw one auxiliary line from L parallel to AC. 
 
 13. A B K, BCL, and CDR are equilateral triangles drawn so that 
 the sides AB, BC, and CD of the parallelogram A BCD are sides of the 
 respective triangles. The triangle BCL overlaps the parallelogram ; the 
 others are adjacent to it. Prove that RL and KL are equal respectively 
 to the diagonals of the parallelogram. Investigate the case when two 
 triangles overlap the parallelogram and one is adjacent, and determine 
 if the corresponding theorem is still true. 
 
 14. The line joining the mid-points of the equal sides of an isosceles 
 triangle is perpendicular to the median drawn to the base. 
 
 15. If two sides of a triangle and a median to one of those sides are 
 equal respectively to two sides and the corresponding median of another 
 triangle, the triangles are congruent. 
 
 16. Any straight line drawn from any vertex to the opposite side of 
 a triangle is bisected by the line which joins the middle points of the 
 other sides of the triangle. 
 
 17. In any triangle ABC the line AK is perpendicular to BC, meet- 
 ing it at K. The point R is the mid-point of AK and L of AC. Prove 
 that the triangle RKL is congruent to the triangle RAL. 
 
 18. In the preceding exercise if H is the mid-point of KC, prove 
 that the triangle KHL is congruent to the triangle CHL. 
 
 19. If two medians are drawn from two vertices of a triangle and 
 produced their own length beyond the opposite sides, and these extrem- 
 ities are joined to the third vertex, these two lines will be equal and in 
 the same straight line, 
 
288 PLANE GEOMETRY 
 
 20. In the triangle ABC a line from K on AC to R on EC makes 
 AK equal one third of AC, and BR equal one third of BC. Prove that 
 ## is parallel to ,15. 
 
 HINTS. Draw a parallel to AB through K and one through L, the mid- 
 point of KG, meeting BC in Jf and N respectively. Then prove that the 
 points M and R are identical. 
 
 21. The point K on CB of the triangle ARC is taken so that CK 
 equals one third of CB. Similarly, R is chosen on CA so that CR equals 
 one third of CA. Prove that KR is one third of AB. 
 
 22. State the converse of Theorem 35, Book I. Is it true? Would 
 this converse, if true, have been of any service in demonstrating the 
 three preceding exercises ? 
 
 23. The mid-points of two opposite sides of a quadrilateral which is 
 not a parallelogram and the mid-points of the diagonals are the vertices 
 of a parallelogram. 
 
 24. The lines joining the mid-points of two opposite sides of a quad- 
 rilateral which is not a parallelogram and the mid-points of the diagonals 
 bisect each other. 
 
 25. If K and R are respectively the feet of the perpendiculars from A 
 on the lines bisecting the angles B and C of the triangle ABC, show 
 that KR is parallel to BC. 
 
 26. If four consecutive points, A, B, C, and D, which are in a 
 straight line are joined to an outside point A", then (1) the angle ABK 
 is greater than the angle ACK; (2) the angle ABK is greater than 
 the angle CKD] (53) the angle ABK is greater than the angle ADK. 
 
 27. In the triangle ABC, if AC is equal to or less than AB, show 
 that any straight line drawn through the vertex A and terminated by 
 BCis less than AB. 
 
 HINT. Designate the foot of the line drawn as above by .If. First com- 
 pare the angle C with the angle B and then compare the angle AKB with 
 the angle B. 
 
 28. In the triangle ABC, having the side AB greater than the side 
 BC, the median BK is drawn. Is the angle AKB always an obtuse angle ? 
 
 29. If in the triangle ABC, having the side AB greater than the 
 side A C, the median A K is drawn, any point on AK except K is nearer 
 to Cthan to , 
 
SUPPLEMENTARY EXERCISES BOOK II 289 
 
 30. The sum of the medians of a triangle is less than the sum of 
 the three sides of the triangle. 
 
 31. In the triangle ABC, AB is greater than EC, and BK, the bisector 
 of the angle B, meets A C in K. Prove that AK is greater than KC. 
 
 HINT. On BA lay off BE equal to BC and draw KB. Then compare KG 
 with KR and the angle A with the angle ARK. 
 
 32. The sum of the diagonals of any quadrilateral is less than the 
 sum of the four sides' but greater than half that sum. 
 
 33. From the mid-point K of the base AB of the isosceles triangle 
 ABC a line is drawn to cut BC in R and AC produced in L. Prove 
 that CR is less than CL. 
 
 BOOK II 
 EXERCISES ON CIRCLES 
 
 34. A parallelogram inscribed in a circle is a rectangle. 
 
 35. If an isosceles triangle is inscribed in a circle, the tangent at its 
 vertex makes equal angles with two of its sides and is parallel to the 
 third side. 
 
 36. The mid-perpendiculars of the sides of an inscribed quadrilateral 
 pass through a common point. Is this true for inscribed polygons of 
 more than four sides ? Prove. 
 
 37. The base angles of an inscribed trapezoid are equal. 
 
 38. The radius of a circle inscribed in an equilateral triangle is 
 equal to one third the altitude of the triangle. 
 
 39. What is the shortest line that can be drawn from a given 
 external point to a circle ? the longest line ? Prove. 
 
 40. The perpendicular from one vertex to the opposite side of an 
 equilateral triangle is one and one-half times the radius of the 
 circumscribed circle. 
 
 41. If perpendiculars are drawn to a tangent from the ends of any 
 diameter, 
 
 (1) the point of tangency will bisect the line between the feet of the 
 perpendiculars ; 
 
 (2) the sum of the perpendiculars will equal the diameter ; 
 
 (3) the center will be equally distant from the feet of the perpendiculars. 
 
290 PLANE GEOMETEY 
 
 42. If an equilateral triangle is inscribed in a circle, the distance of 
 each side from the center is equal to half the radius of the circle. 
 
 43. OA is a radius of a circle whose center is 0, and B is a point on 
 a radius perpendicular to OA. Through B the chord AC is drawn and 
 at C a tangent is drawn meeting OB produced in D. Prove that CBD 
 is an isosceles triangle. 
 
 44. A diameter AB is extended to K, making BK equal to the 
 radius. KR touches the circle at R and cuts the tangent at A in L. 
 Radius CR extended cuts the tangent AL in D. Draw KD and prove 
 that the triangle KLD is equilateral. 
 
 45. AB is a diameter extended half it's length to K. KR is a tan- 
 gent, R its point of tangency. A tangent at B intersects KR in H and 
 AR produced in L. Prove that the triangle HLR is equilateral. 
 
 46. If each of the equal angles of an inscribed isosceles triangle is 
 double the vertical angle and its vertices the points of contact of three 
 tangents, these tangents form an isosceles triangle each of whose equal 
 angles is one third its vertical angle. 
 
 47. If a straight line drawn through the point of contact of two 
 tangent circles terminates in the circles, the tangents at its extremities 
 are parallel. 
 
 ,48. Two circles are tangent either externally or internally, and 
 through the point of tangency two lines are drawn meeting one circle 
 in B and D and the other in E and C respectively. Draw the chords 
 BD and EC and prove them parallel. 
 
 49. If the angle between two secants intersecting outside a circle is 
 bisected by a third secant, does the latter bisect the arcs intercepted by 
 the first two ? Prove. 
 
 HINTS. Let ABC and ALM be the secants and AKR the bisector. Let 
 AR be on the same side of the center as AC. Draw the chords RL and RB. 
 Now the arc BK equals the arc LK or it does not. Assume that it does. 
 Then the arc RGB is less than the arc RML. Hence the chord RL is 
 greater than the chord RB. Why ? Fold the triangle RLA about RA as 
 an axis upon the triangle RAB. Then L must fall on AC, but cannot fall at 
 point B. Why? etc. 
 
 50. The line joining the mid-points of two parallel chords passes 
 through the center of the circle. 
 
SUPPLEMENTARY EXERCISES BOOK II 291 
 
 51. Two circles intersect at A. Chords BA and KA of one circle, 
 extended if necessary, cut the other in C and R respectively. If BK is 
 a diameter, prove that CR is a diameter. 
 
 52. Two circles are tangent externally at A. BC is tangent to the 
 two circles at B and C respectively. Prove that the circle described on 
 BC as a diameter passes through A. 
 
 53. Three circles are tangent to each other at the points A, B, and C 
 respectively. From A lines are drawn through B and C, meeting the 
 circle which passes through B and C at the points D and E respec- 
 tively. Prove that DE is a diameter. 
 
 54. If from any point on a circle perpendiculars be dropped upon the 
 sides of an inscribed triangle (produced if necessary), the feet of the 
 perpendiculars are in the same straight line. 
 
 EXERCISES IN CONSTRUCTION 
 
 55. Construct a chord of a circle, given its mid-point. 
 
 56. Construct an isosceles right triangle, given its hypotenuse. 
 
 57. Construct an isosceles triangle, given its perimeter and its base. 
 
 58. Construct an equilateral triangle, given its altitude. 
 
 59. Through a fixed point draw a line making a given angle with a 
 fixed line. 
 
 60. Construct a rhombus, given one side and one angle. 
 
 61. Construct a rhombus, given its perimeter and one diagonal. 
 
 62. Construct a triangle, given the angles adjacent to one side and 
 the altitude upon it. 
 
 63. Inscribe a square in a given circle. 
 
 64. Construct an isosceles trapezoid, given the bases and the distance 
 between them. 
 
 65. Given three fixed points K, R, and L not in the same straight 
 line. Construct a triangle the mid-points of whose sides are K, R, and 
 L respectively. 
 
 66. Construct a trapezoid, given two adjacent sides, the angle be- 
 tween them, and the difference of the bases. How many solutions 
 are there? 
 
292 PLANE GEOMETRY 
 
 67. Draw a line terminating in the equal sides of an isosceles triangle 
 forming a quadrilateral three of whose sides are equal: 
 
 68. Construct a triangle, given two sides and the median on the 
 third side. 
 
 HINT. Suppose ABC is the completed triangle and CK the given median. 
 Draw KR parallel to BC to R on A C. Can the three sides of the triangle 
 CK R be determined ? 
 
 69. Construct a triangle, given the three medians. 
 
 HINTS. Let the medians A A", BL, and CR of the triangle ABC intersect 
 at H. Draw LM parallel to A A', cutting CR at M. Can the triangle HLM 
 be constructed ? 
 
 70. Given one base and the two nonparallel sides of a trapezoid, 
 construct the trapezoid. How many solutions are there ? 
 
 71. Construct a trapezoid, given its four sides, two of which are 
 known to be the bases. How many solutions are there? 
 
 72. From a given point on a circle draw a chord bisected by a given 
 chord of the circle. 
 
 HINTS. Let A be the point and FG the chord. Draw two chords parallel 
 to the chord FG so that one chord passes through A and both are equally 
 distant from FG. Discuss the number of solutions and the limiting condi- 
 tions of the problem. 
 
 73. Construct a rectangle, given the sum and the difference of two 
 adjacent sides. 
 
 HINT. Let the sides be a and b respectively. Then (a + b) + (a b) 2 cr, 
 (a + 6) (a b) = 2 ft, etc. 
 
 74. Construct a right triangle, given the hypotenuse and the difference 
 of the other two sides. 
 
 75. Inscribe in a given circle a triangle whose angles are equal respec- 
 tively to those of a given triangle. 
 
 HINTS. Circumscribe a circle about the given triangle and draw radii to 
 the vertices. Study the angles at the center. 
 
 76. Construct a common internal tangent to two fixed circles, 
 HINT, Sec Ex, 179, p. 150. 
 
SUPPLEMENTARY EXERCISES BOOK II 293 
 
 77. Construct a right triangle, given the hypotenuse and the sum of 
 the other two sides. 
 
 HINTS. Let AB equal the sum. At A construct an angle KAB equal to 
 45. With B as -a center and the hypotenuse as a radius, describe an arc 
 cutting AK at L. From L construct LH perpendicular to AB. 
 
 78. Construct a right triangle, given the perimeter and one acute 
 angle. 
 
 HINTS. Let AB be the perimeter and the angle H the given angle. At A 
 construct an angle of 45 and on the same side of AB at B an angle equal 
 to one half the angle H. Let the sides of these angles meet in L. Construct 
 LR perpendicular to AB. Construct LV meeting R B at F, so that the 
 angle BLV equals one half the angle H. 
 
 79. Construct a triangle, given two of its angles and its perimeter. 
 
 EXERCISES IN Loci 
 
 80. What is the locus of the center of a circle which touches a fixed 
 line at a given point ? 
 
 81. Find the locus of the center of a circle of given radius which 
 touches a fixed line. Discuss. 
 
 82. Find the locus of the center of a circle which touches two fixed 
 lines. Discuss. 
 
 83. Find the locus of the center of a circle of given radius which 
 touches a fixed circle. 
 
 84. In a fixed circle what is the locus of the mid-points of all the 
 chords of a given length? 
 
 85. Find the locus of the vertices of all the triangles having the 
 common base AB and equal angles opposite AB. 
 
 86. A line cuts two fixed intersecting lines, one in K and the other 
 in R. The four angles on one side of KR are bisected, the bisectors 
 meeting in H and L. If KR, takes every possible position, what is the 
 locus of H and L ? 
 
 87. AB and AC, the sides of a right anglq, are prolonged indefinitely. 
 A line KR, six inches long, moves so that K remains on AB and R on 
 AC. Find the locus of the mid-point of KR. 
 
294 PLANE GEOMETRY 
 
 88. A and B are fixed points, K a variable point, on a circle. Upon 
 a line drawn through K and A points are located on both sides of K 
 such that their distance from K equals KB. What is %he locus of these 
 points ? 
 
 HINTS. Discover the locus as was done in 246. Then let L be a point 
 on the locus. Draw KB and LB. Is the angle BKL constant ? Is the angle L 
 constant ? 
 
 BOOK III 
 
 89. AB and AC are equal sides of a triangle. CK is the altitude on 
 AB. Prove that BC 2 = 2 AB x BK. 
 
 HINT. Draw the altitude on BC. 
 
 90. K is the mid-point of the arc AB. The chords KR and KL 
 cut the chord AB in // and G respectively. Prove that KG Z KR" 
 = KH-HR-KG. GL. 
 
 91. ABC is a triangle and K any point outside it. Through A' on 
 KA produced A'B' is drawn parallel to AB to meet KB produced in B'. 
 Then B'C', parallel to BC, is drawn to meet K C produced in C'. Prove 
 that the line through C parallel to CA will pass through A' and that 
 the triangle ABC is similar to the triangle A'B'C'. 
 
 92. If any polygon replaces the triangle, will a construction like that 
 of Ex. 91 give a polygon similar to the first? 
 
 93. K is the mid-point of BC, one of the parallel sides of the trape- 
 zoid A BCD. The lines AK and DK produced meet DC and AB in F 
 and G respectively. Prove that FG is parallel to AD. 
 
 94. The two parallel lines AB and CD are joined by AD and BC, 
 which intersect at 0. On AB and CD points K and R are taken so 
 that AK : KB = DR : R C. Prove that K, 0, and R are in a straight line. 
 
 95. The distance of the point of intersection of the medians of a 
 triangle from any line is one third the sum of the perpendiculars from 
 the vertices of the triangle to the line. 
 
 96. K is any point on a circle and AB is a diameter. KR is drawn 
 to AB and KL is drawn to BA produced, each making equal angles 
 with AK. Prove that AR x BL = AL x BR. 
 
SUPPLEMENTARY EXERCISES BOOK III 295 
 
 97. If each of three circles intersects the other two, the three common 
 chords are concurrent. 
 
 HINTS. Let AB and CD intersect 
 at P. Let KP cut the circles at G 
 and H. Then prove that G and H 
 must coincide with L. 
 
 98. CK is the bisector of the 
 angle C of the triangle ABC. If the 
 angle A is equal to the angle ACK, 
 prove that A C : A K = A B : CB. 
 
 99. Two circles touch internally at K. Two chords, KL and KR, 
 of the greater circle cut the smaller circle in G and H respectively. 
 Complete the triangles KGH and KLR and prove them similar. 
 
 100. If two circles are tangent to each other, the chords formed by 
 a straight line through the point of contact have the same ratio as the 
 diameters of the circles. 
 
 101. A common tangent to two unequal circles divides their line of 
 centers, produced if necessary, into two segments which are proportional 
 to the diameters of the circles. 
 
 102. AB is a diameter of a circle. Through A any straight line is 
 drawn cutting the circle in C and the tangent at B in K. Prove that 
 AC is a third proportional to AK and AB. 
 
 103. A tangent to a circle at K cuts two parallel tangents in R and L 
 respectively. Prove that the radius of the circle is a mean proportional 
 between KR, and KL. 
 
 104. Tangents from A to a circle whose center is K touch at B and C 
 respectively. BR is drawn perpendicular to CK produced. Prove that 
 ACxBR = CR x CK. 
 
 105. AB and AC are any chords of a circle and AD is a diameter. 
 The tangent to the circle at D intersects AB and AC produced at E 
 and F respectively. Draw BC and show that the triangles ABC and 
 AEF are similar. 
 
 106. K, R, L, and M are points on AB, BC, CD, and DA respec- 
 tively of rectangle ABCD such that KR = LM. Prove that 
 
296 PLANE GEOMETRY 
 
 107. KR is drawn from the mid-point of AC perpendicular to the 
 hypotenuse AB of the right triangle A BC. Prove that Btl ' 2 =AR 2 +BC' Z - 
 
 HIXT. Draw BK. 
 
 108. In a right triangle h is the hypotenuse and a and b are the other 
 two sides. The corresponding sides of another right triangle are //, A, 
 and B. If k : H a : A, prove the two triangles similar. 
 
 109. Every line parallel to the bases of a trapezoid and terminating 
 in the nonparallel sides, but not passing through the intersection of 
 the diagonals, is divided by the diagonals into three segments, two of 
 which are equal. 
 
 110. In the triangle ABC a line cuts AB produced in L, BC in 7v, 
 and AC in R. Prove that AL x BK x CR = AR x BL x CK. 
 
 HINTS. Draw perpendiculars from ^4, U, and C to RL, meeting it in 
 H, Jlf, and G respectively. Then study the triangles thus formed. 
 
 111. The length of the perpendicular to a chord, produced if neces- 
 sary, from any point on the circle is a mean proportional between the 
 perpendiculars from the same point upon the tangents drawn at the 
 extremities of the chord. 
 
 112. From A four lines are drawn so that the angle between each 
 adjacent pair is 45. A straight line cuts these four in K, jR, L, 
 and H respectively, making the triangle ARL isosceles. Prove that 
 KR*= KHx RL. 
 
 113. A common external tangent drawn to two circles cuts the line 
 of centers produced. Prove that radii to the points of contact form a 
 pair of similar triangles. 
 
 114. A B CD is a trapezoid, AB and CD the bases. A C and ED meet 
 in K. Prove that KA : KB = KG : KD. 
 
 115. R is the mid-point of the median AK of the triangle ABC. BR 
 extended cuts CA in L. Prove that A L is one third of A C. 
 
 116. The product of any two sides of a triangle is equal to the square 
 of the bisector of their included angle plus the product of the segments 
 of the third side made by that bisector. 
 
SUPPLEMENTARY EXERCISES BOOK III 297 
 
 HINTS. Circumscribe a circle about &ABC. Extend the bisector CK 
 to meet the circle in" R and draw BE. 
 
 Therefore CK : CB = A C : CK. (1) 
 
 From (1), AC x CB = CK 2 + CK xKR. (2) 
 
 But CK x KB = AK x KB. (3) 
 
 From (2) and (3), AC x CB = CK' 2 + AK x KB. 
 
 NUMERICAL EXERCISES 
 
 117. Two sides of a triangle are 10 and 12 respectively. The bisector 
 of the angle between them divides the third side into segments 5 and (> 
 respectively. Find the length of the bisector. 
 
 118. The sides of a triangle are 20, 39, and 45 respectively. Find 
 the length of the bisector of the angle opposite the side 39. 
 
 119. In the triangle ABC, -b is 24, a is 36, and c is 30. Find to 
 one decimal the length of the bisector of B. 
 
 120. Two sides of a parallelogram are 16 and 30 respectively and one 
 angle is 120. Find both diagonals. 
 
 121. In the figure of Ex. 120, if one angle of the parallelogram 
 is 45, find both diagonals. 
 
 EXERCISES IN CONSTRUCTION 
 
 122. In a given triangle inscribe a parallelogram similar to a given 
 one, one of whose angles equals one of the angles of the triangle. (The 
 vertices of the parallelogram must be on the sides of the triangle.) 
 
 HINTS. Let ABC be the triangle. Construct AKRL similar to the given 
 parallelogram, K being on AC and L on AB. Let AR cut BC in G. Then 
 G and A are two vertices of the required parallelogram. 
 
 123. Inscribe a rhombus in a given triangle. 
 
 124. Inscribe a square in a given triangle. 
 
 125. From K on the arc AB draw KL to L on the chord AB so that 
 KL is a mean proportional between AL and LB. 
 
 126. From a given point draw a secant terminating in a, circle so 
 that the external segment is .one half the whole secant. 
 
 HINTS. AB 2 = 2z 2 , etc. 
 
298 PLANE GEOMETRY 
 
 127. Construct a circle touching two fixed lines and passing through 
 a given point. 
 
 HINT. Let KA and KB be the given lines and P the givten point. Draw 
 KH bisecting the angle AKB. Draw PM _L to K H, meeting it at M. Extend 
 PM to Z>, making MD = PM. Let MP cut AK in E. Then, if T is the 
 required point of tangency for KA, WT 2 EP - ED, etc. 
 
 128. Construct a circle passing through two fixed points and touching 
 a fixed line. 
 
 BOOK IV 
 
 129. The area of a rhombus is one half the product of its diagonals. 
 
 130. The diagonals of a parallelogram divide it into four equivalent 
 triangles. 
 
 131. Through the vertices of a quadrilateral straight lines are drawn 
 parallel to the diagonals. Prove that the area of the parallelogram which 
 they form is twice that of the quadrilateral. 
 
 132. R and K are the mid-points of CD and AB respectively of the 
 parallelogram A BCD. L is any point in KR. Draw AL and DL and 
 prove that the area of the triangle A DL is one fourth the area of A BCD. 
 
 133. Through K and R, the mid-points of two sides of a triangle, 
 any two parallels are drawn cutting the third side, produced if necessary, 
 in H and L respectively. Draw KR and prove that the area of KRLH 
 is one half the area of the triangle. 
 
 134. ABC and ABK are two equivalent triangles on opposite sides of 
 AB. Prove that CK is bisected by the common base, produced if necessary. 
 
 135. AB and CD are the two nonparallel sides of a trapezoid and A' 
 is the mid-point of CD. Draw AK and BK and prove that the area of 
 the triangle ABK is one half that of A BCD. 
 
 136. In the triangle ABC the medians AK and BR produced intersect 
 a parallel to AB through C in L and M respectively. Draw AM and BL 
 and prove the triangles BCL and ACM equivalent. 
 
 137. AKB and ADL are two straight lines such that the triangles 
 AKL and ABD are equivalent. The parallelogram AB CD is completed 
 and BL is drawn cutting CD in R. Prove AK equal to CR. 
 
 HINT. KD is parallel to BL. Therefore the triangle AKD is congruent to 
 the triangle 
 
SUPPLEMENTARY EXEBCISES BOOK IV 299 
 
 138. The angle A and the side A B of the triangle ABC equal 
 respectively the angle K and the side KL of the triangle KRL. Prove 
 that the areas of the two triangles are to each other as AC is to KR. 
 
 139. AB CD is a parallelogram. Line HKL cuts AD at H and BC at 
 L and bisects AC at K. A parallel to AD through K cuts AB in R. 
 Draw CR, RL, and HR and prove that the triangle CRL is equivalent 
 to the triangle AHR. 
 
 140. K is the mid-point of the diagonal BD of the quadrilateral 
 A BCD. Draw ^4^T and CK and prove that the area of the quadrilateral 
 'ABCK is one half that of A BCD. 
 
 141. A BCD is a quadrilateral. ^4^ is drawn to K on BC. Line A'72 
 parallel to AB cuts ^4Z) (not ylD produced) in R. Draw line BR 
 and prove the quadrilateral AKCD equivalent to the quadrilateral 
 BRDC. 
 
 142. ABCD is a quadrilateral such that K, the mid-point of the 
 diagonal BD, lies on the same side of the diagonal A C as the vertex D. 
 Through K a parallel to AC cuts DC in JR. Draw AR and prove that 
 it divides ABCD into equivalent parts. 
 
 143. Through K, any point on a diagonal of a parallelogram, two 
 lines are drawn parallel respectively to two adjacent sides. Prove that 
 two of the parallelograms having a vertex at K are equivalent. 
 
 144. BK and CR are the medians from the vertices of the acute 
 angles of the right triangle ABC. Prove that 4 BK 2 + 4 CR 2 = 5 BC 2 . 
 
 NUMERICAL EXERCISES 
 
 145. Find the diagonals of a rhombus whose side is 10 feet and 
 whose area is 96 square feet. 
 
 146. The area of a rhombus is 336 and one side is 25. Find the 
 
 diagonals. 
 
 147. The base of an isosceles triangle is 10, and each of its base 
 angles is 48. Find the altitude on the base 10. 
 
 Altitude 
 
 HINT. - = tan 48, etc. 
 base 
 
300 PLANE GEOMETRY 
 
 148. Two sides of a triangle are 7 and 12 respectively, and their in- 
 cluded angle is 56. Find (1) the altitude on the side 12, (2) the area 
 of the triangle. 
 
 149. The sides of a certain triangle are 9, 11, and 15 respectively. 
 Find (1) the area of the triangle, (2) the altitude on the side 15, and 
 (3) the two angles adjacent to the side 15 (using the sine formula). 
 
 150. Two adjacent sides of a parallelogram are 7 and 12 respec- 
 tively. The angle between these two sides is 36. Find the altitude on 
 the side 12 and the area of the parallelogram. 
 
 HINT. Use the definition of the sine formula to find the altitude of the, 
 parallelogram. 
 
 151. Two adjacent sides of a parallelogram are 8 and 11 respectively. 
 If the altitude on the base 11 is 5, find the number of degrees in each 
 angle of the parallelogram. 
 
 152. Each of the nonparallel sides of a certain trapezoid is 8 inches. 
 If the parallel bases are 17 and 21 respectively, find each angle of the 
 trapezoid. 
 
 153. The diagonal of a certain rectangle is 15 inches long and makes 
 an angle of 18 with one base. Find the dimensions of the rectangle. 
 
 154. The dimensions of a certain rectangle are 7 by 12. Find the 
 angles which the diagonal makes with the sides of the rectangle. 
 
 CONSTRUCTION EXERCISES 
 
 155. Construct a square equivalent to a given pentagon. 
 
 156. Construct an equilateral triangle equivalent to a given square. 
 
 157. Divide a quadrilateral into two equivalent parts by a line 
 through one vertex. 
 
 158. Divide a quadrilateral into two equivalent parts by a line 
 through a given point on one side. 
 
 159. Divide a trapezoid into two equivalent parts by a line through 
 any point in either base. 
 
 160. Divide a trapezoid into two equivalent parts by a line parallel 
 to the bases. 
 
 161. Divide a parallelogram into two equivalent parts by a line 
 through a given point in one of the bases. 
 
SUPPLEMENTARY EXEECISES BOOK V 301 
 
 162. Divide a trapezoid into two equivalent parts by a line perpen- 
 dicular to the bases. 
 
 163. Construct on AB as a base an isosceles triangle equivalent to 
 the triangle ABC. 
 
 164. Divide a triangle into two equivalent parts by a line parallel to 
 one side. 
 
 165. Construct a triangle iu which the altitude and the base are 
 equal and which is equivalent to a given triangle. 
 
 166. Through a given point in one side of a triangle draw a line 
 dividing the triangle into two equivalent parts. 
 
 HINT. Apply 325 or draw the median to the side on which the given 
 point lies. 
 
 BOOK V 
 NUMERICAL EXERCISES 
 
 167. In Query 17 on page 146 of Book II compute the area over 
 which the horse can gra/e. 
 
 168. The radius of a circle is 10". Show that the difference between 
 its area and that of the regular circumscribed hexagon is 32.25 square 
 inches. 
 
 169. The radius of a circle is 13' and the chord of a segment is 10'. 
 Find the area of the larger of the two segments. 
 
 170. The side of a square is 10. A circle has the same perimeter as 
 the square. Compare the areas of the square and the circle. 
 
 171. The hands of a clock are 10" and 14" respectively. How far 
 does the outer extremity of the minute hand move in the time from 
 12.40 P.M. to 7.20 P.M. (Use - 2 T 2 - for TT.) 
 
 172. Under the same pressure how many times as much water per 
 minute will be delivered by a pipe whose inside diameter is 6" as by 
 one whose inside diameter is 2"? 
 
 173. A circular tank 8 feet in diameter and 40 feet high con- 
 tains 250 tons of water. Find the pressure on one square inch of 
 the bottom 
 
302 
 
 PLANE GEOMETRY 
 
 174. The rim of a circular saw 4 feet in diameter runs 10,000 feet 
 per minute. How many revolutions per 
 
 second does the saw make ? * 
 
 175. Show that the length of the 
 side of a regular inscribed pentagon is 
 
 -\/10-2V5. 
 
 HINTS. In the adjacerit figure let AB 
 and BC be sides of a regular inscribed 
 decagon. Then A C is the side of a regular 
 inscribed pentagon. Let BK be a diam- 
 eter and H the center. From previous 
 work BC is known in terms of E, and 
 BK is 2R. Then in the triangle BCK 
 find EL. Having BL and BC find LC, etc. 
 
 176. Show that the apothein of a regular inscribed pentagon is 
 f(l+V5). 
 
 that the area of a regular inscribed pentagon is 
 
 177. Show 
 5R 2 i 
 
 Vi 
 
 178. The radius of a circle is 10. Show that the area of its regular 
 inscribed pentagon is 237.76 + . 
 
 179. In the adjacent figure AB 
 is a diameter and KH a radius per- 
 pendicular to AB. L is the mid- 
 point of A K. LG equals LIL Prove 
 that KG is equal to the side of a 
 regular inscribed decagon. 
 
 HINTS. Compute in terms of E lines 
 LK, LH, LG, and KG in the order 
 named. The last result should agree 
 with that obtained in Ex. 23. Show 
 similarly that HG is equal to the side 
 of a regular inscribed pentagon. 
 
 180. From a study of the preceding exercise state a method different 
 from that of Problem 5 for inscribing a regular decagon and a regular 
 pentagon in a circle, 
 
INDEX 
 
 Addition, 164 
 
 Alternation, 161 
 
 Altitude, of parallelogram, 218 ; of 
 trapezoid, 220 ; of a triangle, 133 
 
 Angle, 6 ; acute, 34 ; at center of 
 polygon, 261 ;'central, 90 ; exterior, 
 of a triangle, 33; inscribed, 116; 
 obtuse, 34; right, 17; straight, 16 
 
 Angles, adjacent, 16; alternate-exte- 
 rior, 26 ; alternate-interior, 26 ; 
 complementary, 34 ; conjugate, 
 34 ; corresponding, 26 ; exterior, 
 26 ; interior, 26 ; supplementary, 
 29 ; vertical, 25 
 
 Antecedent, 156 
 
 Apothem, 250 
 
 Arc, 89 ; major, 90 ; minor, 90 
 
 Arch, Gothic, 138 
 
 Area, 214 ; of a circle, 215 ; of irreg- 
 ular polygon, 222 ; of rectangle, 
 215 
 
 Axiom, 7 
 
 Bases, of parallelograms, 218 ; of 
 
 trapezoid, 70 
 Bisection, 12 
 Boundaries, 6 
 
 Center of circle, 89 ; of polygon, 261 
 Central angle, 90 
 Chord, 93 ; common, 108 
 Circle, 89 ; center of a, 89 
 Circles, concentric, 111; externally 
 
 tangent, 107 ; internally tangent, 
 
 107 ; tangent, 107 
 Circumference, 89 
 
 Circumscribed polygon, 103 
 
 Commensurable, 157 
 
 Common chord, 108 
 
 Common tangent, 103 
 
 Concurrent lines, 62 
 
 Concyclic points, 125 
 
 Congruence, 9, 214 
 
 Consequent, 156 
 
 Construction, instruments used in, 
 130 ; postulates of, 130 ; problems 
 of, 129 ; solution of a, 131, 147 
 
 Constructions, 129 
 
 Converse, 30 
 
 Convex polygons, 64 
 
 Corollary, 28 
 
 Corresponding angles, 26 
 
 Corresponding lines, 169 
 
 Corresponding parts, 10 
 
 Decagon, 255 
 Definition, 3 ; of TT, 268 
 Degree, 26 ; of arc, 114 
 Demonstration, 7 
 Diagonal, 35 
 Diameter, 89 
 Dimensions, 4 
 Distance to a line, 62 
 Drawing of figures, 45 
 
 Equality, 214 ; of magnitudes, 8 
 Equiangular polygon, 65 
 Equiangular triangle, 52 
 Equilateral polygon, 46 
 Equilateral triangle, 52 
 Equivalence, 217 
 Euclid's "Elements," 86 
 
 303 
 
304 
 
 PLANE GEOMETRY 
 
 Extending a line, 47 
 External common tangent, 107 
 Externally tangent circles, 107 
 
 Fallacies, geometric, 86 
 Figure, geometric, 6 ; plane, 6 ; rec- 
 tilinear, 6 
 
 Figures, drawing of, 45-46 
 Fourth proportional, 161 
 
 Geometric fallacies, 86 
 Geometric figures, 6 
 Geometric magnitudes, 5 
 Graphical construction, 178 
 Gunther's scale, 171 
 
 Hexagon, 35 
 Homologous, 10 
 Hypotenuse, 23 
 
 Incommensurable, 157 
 Indirect proof, 83 
 Inequalities, 72 ; order of, 72 
 Inscribed angle, 116 
 Inscribed polygon, 114 
 Instruments, 130 
 Intercept, 90 
 
 Internal common tangent, 107 
 Internally tangent circles, 107 
 Intersection of loci, 146 
 Inversion, 164 
 Isosceles triangle, 13 
 
 Line of centers, 107 
 Line-segment, 5 
 
 Lines, 4; concurrent, 62; corre- 
 sponding, 169 ; parallel, 20 
 Linkage, 233 
 Loci, 143 
 Locus, 143 
 
 Magnitudes, 4 ; commensurable, 157; 
 geometric, 5 ; incommensurable, 
 157 
 
 Major arc, 90 
 
 Mean and extreme ratio, 253 
 
 Mean proportional, 181 
 
 Measurement, 156 
 
 Median, 56 
 
 Method of proof, 42 
 
 Minor arc, 90 
 
 
 Notation for equal arcs, 115 
 
 Order, of inequalities, 72; of ver- 
 tices, 46 
 
 Pantograph, 201 
 
 Parallel lines, 20 
 
 Parallelogram, 39 
 
 Parallels, postulate of, 21 
 
 Peaucellier's linkage, 233 
 
 Pentadecagon, 258 
 
 Pentagon, 35 
 
 Perpendicular, 16 ; mid-, 60 
 
 TT, computation of, 270 ; definition of, 
 268 
 
 Polygon, 34 ; angle at center of, 261; 
 center of, 261 ; circumscribed, 103 ; 
 equiangular, 65; inscribed, 114; 
 radius of, 260; regular, 34; ver- 
 tices, 46 
 
 Polygons, convey, 64; equilateral, 
 46; mutually equiangular, 46; 
 mutually equilateral, 46; reen- 
 trant, 64; similar, 168 
 
 Postulate, 7 
 
 Postulates of construction, 130 
 
 Projection, 229 
 
 Proof, 7 ; method of, 42-44 ; by re- 
 ductio ad absurdum, 83 
 
 Proportion, 158 
 
 Quadrilateral, 33 
 
 Radius, 89 ; of a polygon, 260 
 Ratio, 155; mean and extreme, 253 
 
INDEX 
 
 305 
 
 Ray, 5 
 
 Rectangle, 39 
 Reductio ad absurdum, 83 
 Reentrant polygons, 64 
 Relation of central angle to inter- 
 cepted arc, 115 
 Rhombus, 58 
 Right angle, 17 
 Right triangle, 23, 185 
 
 Secant, 112 
 Sector, 116 
 
 Segment, 116; area of, 281 
 Semicircle, 90 
 Similar polygons, 168 
 Solids, 4 
 Square, 58 
 Straight angle, 16 
 Straight line, 5 
 Subtend, 90 
 Subtraction, 203 
 Superposition, 22 
 
 Supplementary angles, 29 
 Surface, unit of, 214 
 Surfaces, 4 
 
 Tangent, 103 ; common, 103 ; exter- 
 nal common, 107 ; internal com- 
 mon, 107 ; from outside point, 103 
 
 Tangent cirdes, 107 
 
 Theorem, 7 
 
 Third proportional, 183 
 
 Transversal, 26 
 
 Trapezoid, 55 ; bases of, 70 
 
 Trapezoidal rule, 223 
 
 Triangle, 7 ; equiangular, 52 ; equi- 
 lateral, 52 ; isosceles, 13 ; right, 23 
 
 Trigonometric ratios, 187 
 
 Trigonometric table, 188 
 
 Trisection, 71 
 
 Unit of surface, 214 
 
 Vertex, 6 
 Vertical angles, 25 
 
ANNOUNCEMENTS 
 

 a 
 
 
 ."p 
 
 i* 
 
 p}^\ 
 ^ 
 
 . 
 
 V 
 
 ^ tv. 
 
 '' 
 
 
 M 
 
 ; 
 
TEXTBOOKS IN MATHEMATICS 
 
 FOR HIGH SCHOOLS AND ACADEMIES 
 
 Baker : Elements of Solid Geometry 
 
 Barker : Computing Tables and Mathematical Formulas 
 
 Beman and Smith : Higher Arithmetic 
 
 Betz and Webb : Plane Geometry 
 
 Betz and Webb : Plane and Solid Geometry 
 
 Betz and Webb : Solid Geometry 
 
 Breckenridge, Mersereau, and Moore : Shop Problems in 
 
 Mathematics 
 
 Cobb : Elements of Applied Mathematics 
 Hawkes, Luby, and Teuton : Complete School Algebra 
 Hawkes, Luby, and Teuton : First Course in Algebra (Rev.) 
 Hawkes, Luby, and Teuton : Second Course in Algebra 
 
 (Revised Edition) 
 
 Hawkes, Luby, and Teuton : Plane Geometry 
 Moore and Miner : Practical Business Arithmetic (Revised 
 
 Edition) 
 
 Moore and Miner : Concise Business Arithmetic 
 Morrison: Geometry Notebook 
 
 Powers and Loker : Practical Exercises in Rapid Calculation 
 Robbins : Algebra Reviews 
 Schorling and Reeve : General Mathematics 
 Smith : Algebra for Beginners 
 Wentworth : Advanced Arithmetic 
 Wentworth : New School Algebra 
 Wentworth and Hill : First Steps in Geometry 
 Wentworth- Smith Mathematical Series 
 
 Junior High School Mathematics 
 
 Higher Arithmetic 
 
 Academic Algebra 
 
 School Algebra, Books I and II 
 
 Vocational Algebra 
 
 Commercial Algebra 
 
 Plane and Solid Geometry. Also in separate editions 
 
 Plane Trigonometry 
 
 Plane and Spherical Trigonometry 
 
 GINN AND COMPANY PUBLISHERS 
 
HAWKES, LUBY, AND TOUTON'S 
 ALGEBRAS 
 
 By HERBERT E. HAWKES, Professor of Mathematics in Columbia University, 
 
 WILLIAM A. LUBY, Head of the Department of Mathematics, Junior 
 
 College of Kansas City, and FRANK C. TOUTON, State 
 
 Supervisor of High Schools, Madison, Wis. 
 
 FIRST COURSE IN ALGEBRA (Revised Edition) i 2 ny>, cloth, 
 302 pages, illustrated 
 
 SECOND COURSE IN ALGEBRA (Revised Edition) i2mo, cloth, 
 viii + 277 pages, illustrated 
 
 COMPLETE SCHOOL ALGEBRA (Revised Edition) lamo, cloth, 
 ix + 507 pages, illustrated 
 
 THE Hawkes, Luby, and Teuton Algebras offer a fresh treat- 
 ment of the subject, combining the best in the old methods of 
 teaching algebra with what is most valuable in recent developments. 
 The authors' unhackneyed and vital manner of presenting the sub- 
 ject makes a sure appeal to the interest of the student, while their 
 genuine respect for mathematical thoroughness and accuracy gives 
 the teacher confidence in their work. 
 
 Among the distinctive features of these algebras are the correla- 
 tion of algebra with arithmetic, geometry, and physics ; the liberal 
 use of illustrative material, such as brief biographical sketches of the 
 mathematicians who have contributed materially to the science ; early 
 and extended work with graphs ; and the introduction of numerous 
 " thinkable " problems. Prominence is given the equation through- 
 out, and the habit of checking results is constantly encouraged. 
 Thoroughness is assured by frequent short reviews. 
 
 The aim has been to treat in a clear, practical, and attractive man- 
 ner those topics selected as necessary for the best secondary schools. 
 The authors have sought to prepare a text that will lead the student 
 to think clearly as well as to acquire the necessary facility on the 
 technical side of algebra. The books offer a course readily adapt- 
 able to the varying conditions in different schools the " Complete 
 School Algebra " comprising a one-book course with material suffi- 
 cient for at least one and one-half year's work, and the " First Course" 
 and " Second Course " providing the same material, but slightly 
 expanded, in a two-book course. 
 
 119 
 
 GINN AND COMPANY PUBLISHERS 
 
GENERAL LIBRARY 
 UNIVERSITY OF CALIFORNIA BERKELEY 
 
 RETURN TO DESK FROM WHICH BORROWED 
 
 This book is due on the last date stamped below, or on the 
 
 date to which renewed. 
 Renewed books are subject to immediate recall. 
 
 9Mar'54 C" 
 
 FEB25I954BI 
 
 APR 2 
 6oep'57CS 
 C'D OD 
 
 AUii2?19S7 
 
 REC'D I 
 
 FEB13 
 
 19Nov'5| 
 REC'D LD 
 
 I60tt'60' Ef 
 
 ^ 
 
 D 21-100m-l,'54(1887sl6)476 
 
 REC'D LD 
 
 i 1961 
 
YD I mvo 
 
 YB 17297 
 
 918333 
 
 THE UNIVERSITY OF CALIFORNIA UBRARY