^OF-CALIFO%,
 
 THE UNIVERSITY OF CHICAGO 
 MATHEMATICAL SERIES 
 
 ELIAKIM HASTINGS MOORE 
 
 GENERAL EDITOR 
 
 SCHOOL OF EDUCATION 
 
 TEXTS AND MANUALS 
 
 GEORGE WILLIAM MYERS 
 
 EDITOR
 
 SECOND-YEAR MATHEMATICS 
 FOR SECONDARY SCHOOLS
 
 THE UNIVERSITY OF CHICAGO PRESS 
 CHICAGO, ILLINOIS 
 
 Hgents 
 THE BAKER &. TAYLOR COMPANY 
 
 NEW YORK 
 
 CAMBRIDGE UNIVERSITY PRESS 
 
 LONDON AND EDINBURGH
 
 Second -Year Mathematics 
 
 For Secondary Schools 
 
 By 
 
 GEORGE WILLIAM MYERS 
 
 Professor of the Teaching of Mathematics and Astronomy, College of 
 Education of the University of Chicago 
 
 and 
 
 WILLIAM R. WICKES ERNEST A. WREIDT 
 
 ERNST R. BRESLICH ARNOLD DRESDEN 
 
 Assisted by 
 ERNEST L. CALDWELL and ROBERT M. MATHEWS 
 
 Instructors in Mathematics in the University High School 
 of the University of Chicago 
 
 2 / 2 8 3~ 
 
 SCHOOL OF EDUCATION MANUALS 
 SECONDARY TEXTS 
 
 THE UNIVERSITY OF CHICAGO PRESS 
 CHICAGO, ILLINOIS
 
 All Rights Reserved 
 
 Published June 1910 
 Second Impression October 1910 
 
 Composed and Printed By 
 
 The University of Chicago Press 
 
 Chicago, Illinois. U.S.A.
 
 PREFACE 
 
 This book carries forward through the second high-school 
 year the combined type of material and the plan of treatment 
 of First-Year Mathematics. The two texts together cover the 
 essentials of what is commonly required of all pupils in the 
 first two years of secondary schools in this country, and include, 
 in addition, the elementary notions of plane trigonometry 
 through the solution of right triangles, as well as an introduc- 
 tion to some topics of formal algebra not usually treated iifc. 
 secondary texts. Each book constitutes a well-balanced and 
 not over-heavy year of work. This material so arranged at 
 the same time opens to the pupil a broader, richer, a more 
 useful and therefore a more alluring field of ideas than do the 
 two subjects of algebra and plane geometry treated separately. 
 
 It is felt that the material and treatment of these books 
 lays for the beginner a more stable foundation for future work 
 than does the usual order of a year of formal algebra, followed 
 by a year of formal demonstrative geometry, or of these sub- 
 jects in the reverse order. This judgment, founded on our 
 own experience, is confirmed by a recent extended inspection 
 of schools abroad. In England as well as in Germany and 
 France, the best secondary schools were seen to be using the 
 methods of combined mathematics almost exclusively, with 
 seeming advantage to their pupils. 
 
 Second-Year Mathematics lays chief emphasis on geometry, 
 as did the First-Year Mathematics on algebra. To take up 
 the work of these texts then requires no abrupt departure from 
 the order of subjects now prevailing in secondary curricula. 
 The plan of the books enables the work of the first high-school 
 year to connect smoothly and strongly with eighth-grade work
 
 viii . Preface 
 
 through both mensuration and general number, rather than 
 with one of these subjects in the first year, and the other 
 subject in the second year. 
 
 First-Year Mathematics thus becomes a continuation and 
 an outgrowth of these two arithmetical topics; and, without 
 losing hold on geometrical notions already begun, it develops 
 the customary topics of first-year algebra well into quadratics. 
 Toward the close of the first year geometrical ideas are revived 
 and considerable preliminary geometrical work is done. 
 
 The second-year book then begins with some constructive 
 and inductive geometry, and passes rapidly to demonstrative 
 geometry, employing for a time the half-experimental method 
 of superposition. By the employment of algebraic notation 
 and by the continued application of the equation to geo- 
 metrical matters, the hold on algebra is kept firm until the 
 opportunity arises to develop with profit other algebraic topics, 
 such as a completion of methods of solution of the quadratic 
 equation, a discussion of the roots, and the use of inequalities 
 in the solution of indeterminate equations. In the second 
 year, therefore, the algebraic ground already gained is not 
 only held, but is extended at least as far as is customary with 
 the algebra before the third year. The quadratic equation is 
 used from time to time throughout the second year. All of 
 plane geometry is taught with sufficient fulness. Thus, while 
 the first book may be styled algebra with associated arith- 
 metic and geometry, the second may be styled geometry with 
 associated algebra and trigonometry. 
 
 The plan is continually to command a retrospect and 
 prospect from the field of view of algebra during the first year 
 and a retrospect and prospect from the field of view of geome- 
 try during the second year. Nowhere is the horizon to be 
 needlessly restricted, nowhere are familiar fields to be too 
 soon lost to view, and nowhere is an exhilarating outlook and 
 forecast to be denied. Oftentimes a high-school pupil fails
 
 x 
 
 rightly to esteem a school subject because he cannot discern 
 its bearings on what has preceded and on what is to follow, or 
 even on the demands of the time. These books attempt to 
 bring to the early mathematical subjects the benefits of some 
 knowledge of these bearings. 
 
 The early and rather full use of the principles of congruency 
 (chap, i), of proportionality (chaps, ii and iv), of measurement 
 (chap, iii), and of inequality (chap, v), is worthy of remark. 
 This sequence of principles will be appreciated by many as 
 giving to the pupil in the order of difficulty a good grasp of the 
 four chief agencies of geometrical reasoning. More than 
 90 per cent, of the theorems of school geometry are proved by 
 them. The early part of four of the first five chapters thus 
 adds a new geometrical tool to the equipment of the pupil. 
 Furthermore, these new tools are added in close enough 
 proximity to enable each one to be seen in the light of com- 
 parison and contrast with others. It is pedagogically a waste 
 if not an impossibility to try to teach well one and only one 
 idea or method at a time. These principles are not taught 
 explicitly as geometric methods, they are taught by making 
 each one in turn the predominant method in demonstrations 
 long enough to give the pupil a clear idea of its meaning, and 
 a practical control of its use. 
 
 The authors are convinced that a topical treatment of the 
 theory of limits does not belong to the early years of a high- 
 school course, and in this they are in harmony with the writers 
 of some of the best among recent French and German second- 
 ary texts. What has been attempted here is to pave the way 
 toward a later thorough understanding of this important sub- 
 ject, rather than to induce the pupil to think he has mastered 
 the subject by the memorizing of a theorem. The question of 
 the existence of incommensurable lines and numbers is raised; 
 examples of these are given; their approximations by means to 
 continued fractions as well as by decimal fractions are studied;
 
 x Preface 
 
 whether ratio-theorems can be proved to hold true when such 
 magnitudes are to be compared, is discussed; the error made 
 in using a rational number as an approximation to the common 
 measure of two incommensurable numbers is evaluated and 
 shown to grow less and less as the process of approximation 
 is continued, calling forth intentionally the conviction that 
 there is a definite number toward which these approximations 
 tend. Finally, in the treatment of the ratio of the diameter 
 of a circle to the circumference, the notion of the limit of a 
 sequence is more fully developed, and it is hoped that as a 
 result of the preliminary work done throughout the book, 
 the pupil may now gain some fairly clear notion of that 
 concept. 
 
 This simple and sane treatment of incommensurables; the 
 concrete or experimental approach to new topics and to the 
 subject as a whole; the careful treatment of similarity and the 
 extension of it into the trigonometry of right triangles; the 
 reduced number of major propositions carefully though not 
 too fully treated, and the basing of other theorems upon them; 
 the uses of the graph, and the chapter on geometric algebra, 
 are some features of this text that will be appreciated by 
 students of the mathematical needs of secondary schools. 
 
 The critical reader will find in this text many places where 
 rigor is lacking. No one can be more conscious of the exist- 
 ences of these places than are the authors. But this book is 
 for boys and girls, many of whom nowadays reach the second 
 high-school year at fourteen. They are yet only children, and 
 the degree of rigor only has been attempted that is appreciable 
 and attainable by the larger number of them. Scientific rigor 
 is the limit toward which the high-school pupil approaches 
 step by step, but which he cannot reach by means of ideals 
 so far beyond his immature standards that he cannot even 
 appreciate that they are ideals. Criticisms on the point of 
 lack of rigor can then be looked upon only as differences of
 
 Preface xi 
 
 judgment as to what is appreciable and attainable rigor by 
 fourteen-year-old boys and girls. . 
 
 This text is published primarily for use in the classes of 
 the University High School. This school, however, possesses 
 no features that call for class material in any way different 
 from that needed by all good public high schools. It would 
 seem then that the book must be of interest to all teachers of 
 secondary mathematics who are seeking to improve their 
 work. The authors wish to emphasize that these books are 
 an attempt at finding a form of secondary texts in accord with 
 the pedagogical and mathematical advance of recent years. 
 The application of this advance to secondary problems is as 
 yet in an experimental stage, therefore these texts should not 
 be looked upon as final in the field, but rather as a reflection 
 of the present evolutionary stage of secondary mathematics, 
 to be perfected as the study of the problems of secondary 
 mathematics advances. Criticisms and suggestions looking to 
 the better carrying out of the plan are therefore invited. The 
 problem of getting better texts in mathematics for high schools 
 is quite as much a practical as a theoretical one, and the 
 co-operation of teachers is earnestly sought. 
 
 In conclusion, the' authors' acknowledgments are due to 
 those in administrative relations with the University High 
 School, especially Charles H. Judd, director of the School of 
 Education, and Franklin W. Johnson, principal of the Univer- 
 sity High School, and also to Eliakim Hastings Moore, pro- 
 fessor and head of the Department of Mathematics of the 
 University of Chicago; for their sympathy and encouragement 
 in the experimental work out of which this book has grown, 
 and for their very substantial help in bringing this publication 
 to the light of day. 
 
 THE AUTHORS 
 
 June, 1910
 
 TABLE OF CONTENTS 
 
 . 
 
 CHAPTER PAGE 
 
 I. CONGRUENCY OF RECTILINEAR FIGURES AND CIRCLES I 
 
 Circles 46 
 
 II. RATIO, PROPORTION, SIMILAR TRIANGLES .... 63 
 
 Ratio of Numbers and Segments 63 
 
 Ratio of Other Magnitudes 64 
 
 Commensurable and Incommensurable Segments 
 
 and Magnitudes 66 
 
 Proportion 69 
 
 Problems of Construction 81 
 
 Method of Analysis 86 
 
 Algebraic Exercises 87 
 
 Similar Triangles 89 
 
 Similar Polygons 99 
 
 Similar Right Triangles, Trigonometry .... 102 
 
 Problems on Right Angles 108 
 
 Relations of Trigonometric Ratios 116 
 
 Problems and Exercises 118 
 
 III. THE MEASUREMENT OF ANGLES BY ARCS OF THE CIRCLE 124 
 
 Inscribed Angles 130 
 
 The Method of Common Factors 142 
 
 The Method of Successive Division 143 
 
 IV. SIMILARITY AND PROPORTIONALITY IN CIRCLES . . . 146 
 
 To Draw Common Tangents. Second Method . 157 
 
 Solution of Quadratic Equations by the Formula . 161 
 
 The General Quadratic Formula 163 
 
 Problems in Physics Leading to Complete Quad- 
 ratic Equations 166 
 
 V. INEQUALITIES IN TRIANGLES AND CIRCLES . . . 168 
 
 xiii
 
 xiv Table of Contents 
 
 CHAPTER PAGE 
 
 VI. AREAS OF POLYGONS 207 
 
 The Area of the Triangle 207 
 
 Areas of Polygons 229 
 
 Proof by the Method of Analysis 234 
 
 Quadratic Equations in Two Unknowns . . . 237 
 
 Quadratic Equations Solved by the Graph . . 239 
 
 VII. REGULAR POLYGONS INSCRIBED IN, AND CIRCUM- 
 SCRIBED ABOUT, A CIRCLE 246 
 
 The Side of a Regular Inscribed Decagon . . . 253 
 
 The Side of a Regular Inscribed Pentagon . . . 253 
 Circles Circumscribed about, and Inscribed in, a 
 
 Regular Polygon 257 
 
 To Find the Circumference of a Circle .... 258 
 
 Area of the Circle 267 
 
 VIII. PROBLEMS AND EXERCISES IN GRAPHIC AND GEOMETRIC 
 
 ALGEBRA 271
 
 CHAPTER 1 
 
 CONGRUENCY OF RECTILINEAR FIGURES AND CIRCLES 
 2- / 2. >O" 
 
 1. A theorem is a statement, the truth of which is to be 
 proved. 
 
 EXAMPLE: If two angles are equal, their supplements are equal. 
 
 2. A problem is a statement of a construction to be made, 
 and to be proved correct. 
 
 EXAMPLE: To draw a line perpendicular to a given line at a given 
 point upon it. 
 
 3. A proposition is either a theorem or a problem. 
 
 A theorem consists of two parts, the hypothesis and the 
 conclusion. 
 
 The hypothesis is what is assumed to be true; and the 
 conclusion is what is to be proved to be a consequence of the 
 hypothesis. 
 
 In the theorem given above, "two angles are equal" is the hypothesis; 
 and "their supplements are equal" is the conclusion. 
 
 4. A construction is an exercise to draw some geometrical 
 figure without reference to a proof. 
 
 CONSTRUCTION I 
 
 5. To draw a circle of given center 
 and radius. 
 
 Let O (Fig. i) be the given center 
 and let the line, r, be the given radius. 
 
 Spread the compasses until the 
 distance between the tips is r. 
 
 Place the pin-tip on O and draw 
 the pencil-tip entirely around from A, 
 through B, C, D, to A.
 
 Second- Year Mathematics 
 
 FIG. 2 
 
 6. A plane figure bounded by a curve, every point of which 
 is equally distant from a point within the curve, is a circle. 
 
 The circle is sometimes defined as the curve that bounds 
 the figure (see FYM* p. 200). 
 
 7. The point within the curve, as O (Fig. i), is the center. 
 
 EXERCISES IN CONSTRUCTION 
 
 i. Draw a circle with any convenient radius. Without 
 changing the distance between the compass- 
 points step around the circle, and mark the 
 steps. Connect the alternate marks and 
 form a trilobe as shown in Fig. 2. 
 
 2. Draw a six-lobed figure as shown in 
 Fig. 3- 
 
 3. Mark a point on paper and draw a 
 circle about this point as center with a 
 radius of i inch; of inch; of inch; of 
 ij inches. 
 
 4. Mark a point on the blackboard, and 
 with crayon and string draw a circle about 
 this point as center with a radius of i foot; 
 of 9 inches; of 6 inches; of 15 inches. 
 
 8. The circles of either of the Exercises 3 and 4 are called 
 concentric circles because they have a common center. 
 
 CONSTRUCTION II 
 
 9. Given the three sides of a triangle, to construct the triangle. 
 Let a, b, and c (Fig. 4), be the three sides of the triangle. 
 To draw the triangle. 
 
 Draw a straight line, as A X, longer than the side to be 
 first laid off, say c. 
 
 * The abbreviation, FYM, stands for First-Year Mathematics. 
 
 FIG. 3
 
 Congruency of Rectilinear Figures and Circles 3 
 
 With A as center and with c as radius, mark an arc across 
 AX,asatB. ThenAB=c. 
 
 With A as center and with radius a, mark an arc, as i, 
 so that some of its points will look farther from B than the 
 length b, while other points of arc i will look nearer to B 
 than the length of b. 
 
 IB 
 FIG. 4 
 
 With B as center and with b as radius mark an arc, as 2, 
 across arc i. 
 
 Connect the crossing-point, C, of arcs i and 2 with A and 
 with B by straight lines. Then A B C is the required tri- 
 angle. 
 
 10. A portion of a plane completely bounded by three 
 straight lines, is a triangle. 
 
 EXERCISES IN CONSTRUCTION 
 
 A triangle having two equal sides is an isosceles triangle; 
 the third side is the base. 
 
 1. Draw a triangle having each of its sides i foot long. 
 
 A triangle having all sides equal is an equilateral triangle. 
 
 2. Which side is the base of the triangle in Exercise i ? 
 
 A triangle, no two of whose sides are equal, is a scalene 
 triangle. 
 
 Any side of a scalene or of an equilateral triangle may be regarded 
 as the base.
 
 Second-Year Mathematics 
 
 3. Construct an equilateral triangle on a side 15 inches long. 
 
 4. Is an equilateral triangle isosceles? Give reason for 
 
 answer. 
 
 5. Draw a triangle having sides of inch, f inch, and 
 i inch, beginning by laying off the shortest side first. 
 
 60' 
 
 FIG. 5 
 
 6. With crayon and string draw on the blackboard a tri- 
 angle having sides 6 inches, 8 inches, and 10 inches. 
 
 7. Describe how with a 66-foot chain stakes may be set in 
 the ground at the corners of a triangle of sides 30', 50', and 
 60'* (see Fig. 5). 
 
 8. Draw a triangle having sides 12", 12", and 16". 
 
 The definite portion of a line included between two points 
 is a line -segment. For example, the sides of a triangle are 
 line-segments. 
 
 CONSTRUCTION III 
 
 ii. Given the base and one of the equal sides of an isos- 
 celes triangle, to construct the triangle. 
 
 Take a line-segment i inch long for the base and another 
 line-segment inch long for the given side, and construct the 
 triangle with compasses on paper, or instead of the "inch" use 
 "foot," and construct the triangle with crayon and string on 
 the blackboard. 
 
 * The single prime ('), when written beside an arithmetical number, 
 means foot, or ]eet; the double prime (") means inch or inches.
 
 Congruency of Rectilinear Figures and Circles 5 
 
 The word "line" here and from now on means straight 
 line, unless some other meaning is expressly stated. 
 
 A line is supposed to be indefinitely extended in both 
 directions. 
 
 EXERCISES 
 
 1. The span of a roof is 24' and one of the two equal rafters 
 is 15'. Make a drawing to a convenient scale of the end 
 rafters. 
 
 2. One of two equal lines is denoted by #+12, and the 
 other by i$x. Find x, and the length of the equal lines. 
 
 3. Find the unknown, x, and the length of the equal sides 
 of an isosceles triangle if one of the equal sides is x 3 $x and 
 the other is 5^ + 9. 
 
 4. How many isosceles triangles will answer the conditions 
 of Exercise 3 ? Sketch the triangles. 
 
 5. Find x, and the two equal sides of an isosceles triangle, 
 the equal sides being denoted by the following pairs of num- 
 bers: 
 
 (1) x 2 x and $x 8 (6) x a +$x and 4(15 #) 
 
 (2) x 2 x and x+i$ (7) x 2 7 and 3(273;) 
 
 (3) x a and 7 6x (8) ^x^+^x and 12(1 x) 
 
 (4) x(x+4) and 3(2^ + 5) (9) $x(x+8) and 2(20 + 2^) 
 
 (5) #(#3) and i2(x 3) (10) 3^(10^3) and 20(2* i). 
 
 CONSTRUCTION IV 
 
 12. Given a side, to construct an equilateral triangle, 
 The solution is left to the student. 
 
 EXERCISES 
 i. The sides of an equilateral triangle may be denoted by 
 
 2 (5 x )> 3y~ 2 > an d 2 y- Find x > y> an d tne length of a side 
 of the triangle.
 
 Second-Year Mathematics 
 
 2. Find x, y, and the length of a side of an equilateral 
 triangle, the sides of one being designated by each of the 
 following sets of numbers: 
 
 (1) i2X, (2+y), and 3(6^-5) (3) 4*, 2(^+3), and 5(4-?) 
 
 (2) 3*, 29-5^, and 19 + 5? 
 
 (5) nx, 
 
 A.y 
 ' 
 
 1C 
 
 V 
 
 (4) x, y~3, and 
 , and 11(2^+9). 
 
 CONSTRUCTION V 
 
 13. Given two sides and twice the third side of a triangle, 
 to construct the triangle. 
 
 _ a _ Let a, b, and 2C be the given 
 
 line-segments. 
 
 To construct a triangle having 
 a, b, and c as sides. 
 
 Let us first find one-half of 
 the line 2c, i. e., let us bisect 2C. 
 Make AB = 2c. 
 
 With A as a center and with 
 a radius longer than half of 
 A B draw arcs i and 2 (Fig. 6). 
 With B as a center and 
 with the same radius draw the 
 arcs 3 and 4 across the arcs i and 2, respectively. Call the 
 crossing-points C and D. 
 
 Join C and D and call E 
 the point where the lines C D 
 and A B cross. 
 
 A E, or E B, is c, for either 
 is one-half of 2c. 
 
 Now construct a triangle with 
 a, 6, and A E, or E B, as sides 
 as in Construction II. 
 
 
 FIG. 6 
 
 ,'tf 
 
 l 
 
 A 
 
 >* 
 
 
 FIG. 7
 
 Congruency of Rectilinear Figures and Circles 7 
 
 Is it necessary that arcs 2 and 4 of Fig. 6 have the same 
 radii as arcs i and 3 ? (See Fig. 7.) 
 
 Is it necessary that the intersections lie on opposite sides 
 of the line to be bisected ? 
 
 If the radii of the arcs i and 2 of Fig. 7 are equal, what is 
 the only essential for the arcs 3 and 4 ? 
 
 OBSTRUCTION VI 
 
 14. To construct on a line an angle equal to a given angle, 
 and having its vertex at a given point on the line. 
 See Fig. 8 and FYM, p. 148. 
 
 15. The figure formed by extending two lines from a point 
 is an angle. Thus O A and O B (Fig. 9) extending from the 
 
 point O, form the angle A O B, and O'A' and O'B', extending 
 from the point O', form the angle A'O'B'.
 
 8 
 
 Second- Year Mathematics 
 
 The lines that form the angle are the sides, or arms, and 
 the point from which the lines extend is the vertex of the angle. 
 
 The magnitude, or size, of an angle is the amount of turn- 
 ing necessary to rotate a line around the vertex from one arm 
 to the other. 
 
 sir 
 
 A straight angle is an angle whose arms lie in the same 
 straight line on opposite sides of the vertex (Fig rr> ) ' The 
 two straight angles about a point make a perigon. A perigon 
 corresponds to a complete turn of a rotating line (Fig. n). 
 
 A right angle is one-half of a straight angle (see Fig. 12, I 
 
 An acute angle is an angle that is less than a right angle 
 (see Fig. 12, angle a). 
 
 An obtuse angle is an angle that is greater than a right 
 angle and less than a straight angle (see Fig. 12, angle c). __ 
 
 CONSTRUCTION VII 
 
 1 6. Given two sides and the included angle of a triangle, to 
 construct the triangle. 
 
 Given the sides a and b, and the angle C (Fig. 13).
 
 Congruency of Rectilinear Figures and Circles 9 
 
 To construct the triangle. 
 
 Make C A = 6, and at one end, as at C, construct an angle 
 equal to the given angle C (see Construction VI). 
 
 Make C B equal to a. 
 
 Connect B and A. 
 
 A B C is the required triangle. 
 
 EXERCISES 
 
 1. With sides i inch and i| inches long, and with the 
 included angle 60, construct a triangle, using a protractor to 
 lay off the given angle, and compasses for the construction. 
 
 2. Taking any convenient length for x, as ^ inch, construct 
 a triangle having two sides 3^ and 4X, and an included angle 
 of 45. 
 
 3. Construct an isosceles triangle with one of the equal 
 sides i inches, and the vertex-angle 55 (see 27). 
 
 4. Construct an isosceles triangle with base 2 inches and 
 base angles 15.
 
 10 
 
 Second- Year Mathematics 
 
 5. Construct a triangle with base inch and base angles 
 35 and 50. 
 
 17* Geometrical figures that have the same shafie are 
 similar figures. 
 
 18. Geometrical figures that have the same size, but not 
 the same shape, are equal or equivalent figures. 
 
 19. Geometrical figures, that have both the same size and 
 the same shape are congruent figures. Congruent figures may 
 be made to coincide. 
 
 20. There are some geometrical figures, such as straight 
 lines, angles, circles, etc., that can differ only in size. When 
 such figures have the same size they will be said to be equal, 
 rather than congruent; since the question of shape does not 
 arise in connection with them. 
 
 21. The sign, or symbol, of similarity is ^. Thus, a ~~ b 
 means a is similar to b. 
 
 22. The symbol of equality, or equivalency, is =. Thus, 
 a =b means a is equal to b. 
 
 23. The symbol of congruency is ^. Thus, ABC^ 
 A'B'C' means ABC and A'B'C' may be made to coincide. 
 
 PROPOSITION I 
 
 24. Theorem: // two sides and the included angle of one I 
 triangle are equal, respectively, to two sides and the included] 
 angle of another triangle, the triangles are congruent. 
 
 e 
 
 FIG. 14 
 
 Hypothesis: If the triangles ABC and A'B'C' (Fig. 14) 
 have A B=A'B', A C=A'C', and angle A =angle A';
 
 Congruency of Rectilinear Figures and Circles n 
 
 Conclusion: Then the triangles ABC and A'B'C' are con- 
 gruent, i. e., they can be made to coincide. 
 
 Proof 
 
 Imagine the triangle ABC placed upon the triangle 
 A'B'C' so that angle A shall fit exactly upon its equal angle, 
 A', A B falling upon A'B' and A C upon A'C'. 
 
 Because A B=A'B' (why?), B will fall upon B'. 
 Because A C=A'C' (why ?), C will fall upon C'. 
 
 Then B C will fall along and coincide with B'C', else there 
 would be two different straight lines connecting the two points 
 B and C, and this is evidently impossible. 
 
 Hence, the two triangles are congruent, i. e., they coincide, 
 or fit. 
 
 Since ABC and A'B'C' are any two triangles having two 
 sides and the included angle of one equal to the corresponding 
 parts of the other, the theorem is true. 
 
 EXERCISES 
 
 1. To find the distance B C across the lake (Fig. 15), 
 C A was extended through 
 
 A, so that CA=AC' and 
 B A through A so that 
 BA=B / A. B'C' was meas- 
 ured and found to be of a 
 mile long. Assuming the two 
 angles, marked with arcs, to 
 be equal, how long was B C ? 
 Give the reasons for your answer. 
 
 2. Prove: If AC=A'C', B C = B'C', and Z* C=ZC' 
 (Fig. 1 6), the two triangles are congruent. 
 
 3. If in Fig. 16, a=a', b=b', ZC=ZC', c=8(2*-s), 
 
 FIG. 15 
 
 * Z is the symbol for angle; Z s, for angles.
 
 12 Second-Year Mathematics 
 
 and ^B f = i4s r + 5, what are the values of x, r, s, Z.A, 
 and Z.B'? 
 
 FIG. 1 6 
 
 4. With sides and angles designated as in Fig. 1 6, assuming 
 a=a', b=b', and Z. C=Z.C', find x, y, and z if .4=ioo y, 
 
 z(z 20). 
 
 CONSTRUCTION VIII 
 
 25. Given two sides and twice the included angle of a tri- 
 angle, to construct the triangle. 
 
 Given the sides b and c and the angle 2 A (Fig. 17). 
 
 b 
 
 FIG. 17 
 
 To construct the triangle. 
 
 The given angle being the double of the desired angle of 
 the triangle, we must first obtain the desired angle. This is 
 done by bisecting the given angle, Fig. 18 (see FYM, p. 147). 
 
 Now construct a triangle as in Construction VII, with the 
 sides b and c and the included angle A O E.
 
 Congruence of Rectilinear Figures and Circles 13 
 
 26. A line that divides an angle into two equal parts is 
 called the bisector of the angle. 
 
 27. The angle that lies opposite to the base of an isosceles 
 triangle is the vertex-angle of the triangle. 
 
 EXERCISES 
 
 28. Prove the following exercises: 
 
 1. The bisector of the vertex- angle of an isosceles triangle 
 divides the triangle into two congruent parts. 
 
 2. The bisector of the vertex-angle of an isosceles triangle 
 bisects the base. 
 
 3. In an isosceles triangle the angles opposite the equal 
 sides are equal. 
 
 4. All points of the perpendicular bisector of a line-segment 
 are equidistant from the ends of the line. 
 
 5. An equilateral triangle is equiangular, i. e., the angles 
 are equal. 
 
 29. It is well for the student to become aware that we 
 have in our work, thus far, assumed certain things to be possible 
 and true without having proved them to be so; for example, 
 it has been assumed that a triangle may be picked up and 
 moved about in space without changing its shape or size. It 
 has also been assumed that through any two points only one 
 straight line can be drawn. Such fundamental assumptions 
 form the basis on which geometry is built; they are the axioms 
 and postulates of geometry. 
 
 30. An axiom is a statement the truth of which is assumed, 
 i. e., accepted without proof. 
 
 31. A postulate is a statement of the possibility of a con- 
 struction. 
 
 An axiom, like a theorem, has a hypothesis and a con- 
 clusion. How does an axiom differ from a theorem? A
 
 14 Second- Year Mathematics 
 
 postulate differ from a problem? A theorem differ from a 
 problem ? 
 
 32. Since every statement in a proof must be based directly, 
 or indirectly, upon an axiom, a postulate, or a definition, some 
 of the most important axioms and postulates are now stated. 
 
 AXIOMS 
 
 1. Magnitudes that are equal to the same magnitude are 
 equal to each other (comparison axiom). 
 
 2. Equals added to equals give equal sums (addition axiom). 
 
 3. Equals subtracted from equals give equal differences (sub- 
 traction axiom). 
 
 4. Equals multiplied by equal numbers give equal products 
 (multiplication axiom). 
 
 5. Equals divided by equal numbers (excluding division by 6} 
 give equal quotients (division axiom). 
 
 6. A whole is equal to the sum of all its parts. 
 
 7. A whole is greater than any of its parts. 
 
 8. The shortest distance Between two points is measured on 
 the straight line joining the points. 
 
 9. A figure may be moved about, or turned over in space, 
 without changing its form or size. 
 
 10. One and only one straight line can join two points. 
 Other axioms will be stated as they are needed. 
 
 POSTULATES 
 
 1. A straight line can be drawn joining two points. 
 
 2. A line-segment can be prolonged. * 
 
 3. A circle can be drawn with any point as center and with 
 any radius. 
 
 33. The Greek philosopher Plato (429-384 B. c.) taught 
 geometry as a basis for the study of philosophy. He was accord-
 
 Congruency of Rectilinear Figures and Circles 15 
 
 ingly much concerned about its logical purity. He decided 
 to restrict the lines that could be used in plane geometry to 
 the straight line and the circle, and the instruments that could 
 be used to 'the unmarked straight edge and the compasses. 
 Geometers since Plato's time have conformed to these restric- 
 tions, as a convenient means of denning the field of plane 
 geometry. 
 
 In the logic (deductive) of this book Plato's restrictions 
 are adhered to. Figures and construction lines used in estab- 
 lishing geometrical propositions must be made up of circles 
 and straight lines only, i. e., of such lines as can be drawn 
 with the unmarked ruler and the ungraduated circle. For 
 the deductive proofs neither the markings on a foot-rule or a 
 yard-stick, nor on a protractor can be used. 
 
 But another important part of geometrical study is to learn 
 how to find out things, how to discover new truths, i. e., how 
 to find out what is likely to be true and to state it in form 
 for logical proof or disproof. In this work of geometrical 
 exploration and discovery any sort of line or instrument that 
 furnishes a clue, affords a hint, or gives any sort of real help, 
 rational or intuitional, will be freely admitted. The graduated 
 measuring-stick, the protractor, cross-lined paper, graph, 
 logarithmic tables, algebraic equations, etc., are all welcomed 
 here as helps to inductive study and research. 
 
 CONSTRUCTION IX 
 
 34. To draw a triangle having given two angles and the 
 side included by the vertices of these angles. 
 
 Given the angles x and y (Fig. 1 9) and the side a. 
 
 To construct a triangle having the angles x and y and the 
 side a, included between their vertices. 
 
 On the indefinite straight line, A X, lay off A B=a.
 
 1 6 Second-Year Mathematics 
 
 At A make an angle equal to x, and at B an angle equal to y 
 (see Construction VI). 
 
 Prolong the sides of the constructed angles till they meet, 
 as at C. 
 
 a 
 
 A B C is the required triangle. 
 
 Does it seem possible to construct a triangle of any other 
 shape or size, using the parts a, x, and y of Fig. 19? 
 
 PROPOSITION II 
 
 35. Theorem: // two triangles have two angles and the 
 side included by their vertices in the one, equal, respectively, to 
 the corresponding parts in the other, the triangles are congruent. 
 
 Hypothesis: If the two triangles ABC and A'B'C' 
 (Fig. 20) have Z=Z', ZC=ZC', and a = a'; 
 Conclusion: The two triangles are congruent. 
 
 Imagine the triangle A B C so placed upon the triangle 
 A'B'C' (turned over if necessary) that side a coincides with 
 side a', the vertices B and C falling on the vertices B' and C', 
 respectively.
 
 Congruency of Rectilinear Figures and Circles 17 
 
 Let A fall on the same side of a' as A'. 
 
 Because Z-B=Z#' (why?), c will fall along c', and the 
 vertex A will fall somewhere on c' or its extension. 
 
 Because ZC=ZC' (why?), b will fall along b', and the 
 
 vertex A will fall somewhere on b' or its extension. 
 
 Because A falls at the same time upon c' and upon b', it 
 must fall at the only point c' and b' have in common, viz., at 
 their point of intersection, A'. 
 
 Hence, the triangles ABC and A'B'C' coincide, and are 
 congruent. But ABC and A'B'C' are any two triangles 
 having two angles and the intervening side of the one equal to 
 the corresponding parts of the other. 
 
 The theorem is therefore true. 
 
 EXERCISES 
 
 1. The perpendicular bisector of the base of an isosceles 
 triangle passes through the opposite vertex (use Proposition II, 
 p. 16). 
 
 Assume for the present that all right angles are equal. 
 
 2. In Fig. 21 the distances S A and S B of a lighthouse, S, 
 from the points A and B are desired. 
 
 The angle A B S' is made equal to the angle A B S, and 
 B A S' is made equal to B A S. A S' was measured and found
 
 i8 
 
 Second-Year Mathematics 
 
 to be 2,680 ft. and B S' was found to be 3,420 ft. How far 
 is S from A and B ? 
 
 3. The bisector 
 of the vertex-angle 
 of - an isosceles 
 triangle is perpen- 
 dicular to the base 
 (see Exercise 2, 
 P- 13)- 
 
 4. Oblique lines 
 drawn to a line from 
 a point on a per- 
 pendicular to the 
 line and makin 
 
 FlG - 2I 
 
 equal angles with the perpendicular are equal. 
 
 5. Oblique lines, drawn from a point on a perpendicular to 
 a line and making equal angles with the line, are equal. 
 
 Assume that the sum of the angles of any triangle is 180 (see p. 31, 3). 
 
 6. An equiangular triangle is equilateral. 
 
 An equiangular triangle is a triangle all of whose angles are equal 
 (see Exercise 5, p. 13). 
 
 7. If two angles of a triangle are equal, the sides opposite 
 the equal angles are equal, and the triangle is isosceles (see 
 FYM, p. 344, 9)- 
 
 c c' 
 
 A c BA c/ 
 
 FIG. 22 
 
 8. In triangles marked as in Fig. 22, supposing b = b', 
 A' r" r' n 
 
 JT! , \^ \^ , (I i 
 
 J T M 2 
 
 and B' = . Find x, y, n, a, c, and B'.
 
 Congruency of Rectilinear Figures and Circles ig 
 
 PROPOSITION III 
 
 36. Theorem: If the three sides of one triangle are equal, 
 respectively, to the three sides of another triangle, the triangles 
 are congruent. 
 
 Given, in Fig. 23, A B=A'B', A C=A'C', and B C=B'C'. 
 
 A B A B' 
 
 FIG. 23 
 
 To prove: A B C *= A'B'C'. 
 
 Place A B on A'B' so that A falls upon A' and B upon B'. 
 
 Let C fall on the same side of A'B' as C'. 
 
 Then if C does not fall on C' suppose it to fall at C", some 
 other point. 
 
 But A'C'=A'C" (why?) and B'C'=B'C" (why?). Hence 
 both A'C'C" and B'C'C" are isosceles triangles. 
 
 A perpendicular to C'C" at its middle point passes through 
 both A' and B' (see Exercise i, p. 17). 
 
 This is impossible unless the middle point of C'C" lies 
 on A'B' (see Axiom 10). The middle point of C'C" cannot 
 lie on A'B' without destroying the triangles, since C and C' 
 were made to fall on the same side of A'B'. 
 
 Therefore the supposition that C does not fall on C' is 
 false, and the triangles ABC and A'B'C' are congruent. 
 
 But ABC and A'B'C' are any two triangles having the 
 three sides of the one equal, each to each, to the three sides of 
 the other, and the theorem is then true.
 
 2o Second-Year Mathematics 
 
 This is an example of indirect proof. It consists in assum- 
 ing the opposite or the negative of what is to be proved, and 
 showing the assumption to lead to an absurdity. Hence 
 it is called a "reductio ad absurdum." 
 
 EXERCISES 
 
 1. A line drawn from the middle point of the base to the 
 opposite vertex of an isosceles triangle is perpendicular to the 
 base. 
 
 2. Oblique line-segments drawn from a point in a perpen- 
 dicular to a line and cutting off equal distances from the foot 
 of the perpendicular are equal. 
 
 3. Equal oblique line-segments drawn from a point in a 
 perpendicular to a line, meet the line at points equally distant 
 from the foot of the perpendicular (see 28, 3) . 
 
 37. The point of intersection of a line with a perpendicular 
 is called the foot of the perpendicular. 
 
 38. In congruent triangles, corresponding sides are sides 
 that lie opposite equal angles. 
 
 Corresponding angles are angles that lie opposite equal 
 sides. 
 
 39. The sides and angles of a triangle, or other figure, are 
 called the parts of the triangle, or figure. 
 
 0' 
 
 i . With the parts of two triangles designated as in Fig. 24, 
 and with a=a r , b=b', c=c', find r, s, t, and all the other angles 
 under the following conditions:
 
 Congruency of Rectilinear Figures and Circles 21 
 
 Angle A Angle A' Angle B Angle B' Angle C Angle C' 
 
 (1) 5(r+s) 2(29-5) 2(5+0 86-/ 4^ + 2/ 2(9+*) 
 
 (2) lr+$s + t 75-5 ^+5+43 90-5 ^+5+64 195-* 
 
 2. Supposing the sizes of the three angles of an equilateral 
 triangle may be represented by the numbers, r + 2s, $t ior, 
 and s+t, respectively, and that the sum of all three is 180, 
 find r, s, and J, and each of the angles. 
 
 40. Propositions I, II, and III are the most fundamental, 
 if not the most important, theorems of plane geometry. The 
 proofs of all three as given here furnish examples of the method 
 of establishing congruency by so superposing one figure upon 
 another as to make the figures coincide. 
 
 The proof of Proposition III is of this superposition type 
 mixed with the "reductio ad absurdum" type. These three 
 proofs should be thoroughly mastered. 
 
 AXIOM ii 
 All straight angles are equal. 
 
 PROPOSITION IV 
 
 41. Theorem: All right angles are equal. 
 
 To prove this use Axiom u just given and Axiom 5, p. 14. 
 
 42. The supplement of an angle is the difference obtained 
 by subtracting it from a straight angle (180). 
 
 43. The complement of an angle is the difference obtained 
 by subtracting it from a right angle (90). 
 
 PROPOSITION V 
 
 44. Theorem : The supplements of equal angles are equal. 
 Use Axioms n and 3. 
 
 PROPOSITION VI 
 
 45. Theorem: The complements of equal angles are equal. 
 The proof is left to the student.
 
 22 
 
 Second- Year Mathematics 
 
 46. Two angles having a common vertex and a common side 
 between them are adjacent angles. 
 
 EXERCISES 
 
 i. The sum of the adjacent angles formed by one line 
 meeting another is a straight angle. 
 
 This is evident from Axiom 6, p. 14. 
 
 ^/2?) One of two complementary angles is x and the other 
 is x 2 . Find x and the two angles. 
 
 3. One of two complementary angles is 6y and the other 
 is 4(^ + 5)- Find y and the angles. 
 
 4. One of two supplementary angles may be designated y 2 
 and the other 8y. Find y and the angles. 
 
 (Q The angles x and y are complementary and their differ- 
 ence is 10. Find x and y. 
 
 6. The angles r and 35 are supplementary and r 5 = 20. 
 Find r, s, and 35-. 
 
 7. The angles that lie opposite the equal sides in an isos- 
 celes triangle are represented in magnitude by 7^+8 and 
 80 $x, respectively. Find x and the two angles. 
 
 8. If one of two adjacent angles formed by one line meeting 
 another is a right angle, the other is also a right angle. Prove. 
 
 9. If two straight lines intersect (cross) one another, and 
 one of the angles formed is a right angle, the other three are 
 also right angles. Prove. 
 
 10. The sum of all the angles formed about a point on one 
 side of a line is a straight angle. Prove. 
 
 11. The sum of all the angles formed about a point is two 
 straight angles, or four right angles. Prove. 
 
 PROPOSITION VII 
 
 47. Theorem: If the sum of two adjacent angles is a 
 straight angle, their exterior sides form a straight line. 
 
 This follows directly from the definition of straight angle, 15.
 
 Congruence of Rectilinear Figures and Circles 23 
 
 EXERCISES 
 
 1. The bisectors of two supplementary adjacent angles are 
 perpendicular to each other. Prove. 
 
 When two lines intersect each 
 other at right angles, either line is 
 said to be perpendicular to the other. 
 
 2. The angle ABC (Fig. 25) 
 is a right angle. D B E is a 
 straight line. To what is the 
 sum, x+y, equal? 
 
 The opposite angles, as a and c, 
 or b and d (Fig. 26) formed by two 
 intersecting lines are called vertical 
 angles. 
 
 3. Prove that if two lines 
 intersect, the vertical angles are 
 equal. 
 
 4. The bisectors of two op- 
 posite, or vertical, angles form a straight line. 
 
 5. The bisector of an angle bisects also its vertical angle. 
 
 PROPOSITION VIII 
 
 48. Problem: From a point without a line to draw a per- 
 pendicular to the line. 
 
 See Fig. 27, and FYM, p. 146. 
 
 FIG. 26 
 
 FIG. 27
 
 Second-Year Mathematics 
 
 PROPOSITION IX 
 
 49. Theorem: Only one perpendicular can be drawn from 
 a point without a line to the line. 
 
 Let P O be a perpendicular from P to A B (Fig. 28). 
 To prove P O the only perpendicular from P to A B. 
 
 Proof: If there can be another 
 
 T-\ 
 
 perpendicular, let it be P C. Pro- 
 long P O to P 7 , making O P 7 =P O. 
 
 Join C and P 7 and prove P C O 
 and P 7 C O congruent triangles. 
 ~ P C, being a perpendicular to A B 
 (by supposition), P C P 7 is a straight 
 line (Proposition VII). 
 
 If then P C could be any other 
 perpendicular than P O, there would 
 be two straight lines connecting P 
 and P'. This is impossible. Why? (See Axiom 10, p. 14.) 
 
 Hence, there can be no other perpendicular from P to A B 
 than P O. 
 
 50. The symbol for perpendicularity is 1. 
 
 PROPOSITION X 
 
 51. Problem: From a point on a line, to draw a perpendicu- 
 lar to the line. 
 
 See Fig. 29, and FYM, p. 145. 
 
 o 
 
 FIG. 28 
 
 P 
 
 FIG. 29 
 
 1T23
 
 Congruency of Rectilinear Figures and Circles 25 
 
 PROPOSITION XI 
 
 52. Theorem: If each of two points of one line is equally 
 distant from two points of another line, the lines are per- 
 pendicular. 
 
 Let P and Q of either Fig. 30 or Fig. 31 be each equally 
 distant from the ends A and B of the line A B. 
 
 V 
 
 A" 
 
 " 
 
 -v A 
 
 FIG. 30 
 
 \ \ 
 
 
 
 A' 
 
 ^ 
 
 N 
 
 K? 
 
 \ 
 
 FIG. 31 
 
 To prove the lines P Q and A B perpendicular. 
 Proof: Connect P with A and B, and Q with A and B. 
 Prove triangles P A Q and P B Q congruent. 
 Prove triangle A O Q congruent to triangle B O Q. 
 Then ZQ O A= ZQ O B, and the lines P Q and A B are 
 perpendicular. Why ? 
 
 PARALLELS AND PARALLELOGRAMS 
 
 53. Parallel lines are lines which lie in the same plane and 
 do not intersect however far prolonged in both directions. 
 The symbol for parallelism is || .
 
 26 
 
 Second-Year Mathematics 
 
 AXIOM 12 
 
 One and only one parallel to a line can be drawn through 
 a point outside of the line. 
 
 PROPOSITION XII 
 
 / 54. Theorem: // each of two lines is parallel to a third 
 \line,Jhcy are parallel to each other. 
 
 Given A B and C D each parallel to E F. 
 
 A- 
 C- 
 
 -F 
 
 FIG. 32 
 
 To prove A B | CD. 
 The lines AB and CD 
 (Fig. 32) are supposed to lie 
 in the plane of the page. 
 
 Suppose A B could intersec t 
 C D, through some point as P. 
 Then, through P there would be two parallels to E F. 
 Why? This is impossible. Why? (See Axiom 12.) 
 
 A B and C D are then lines which lie in the same plane 
 and do not intersect. Hence, A B and C D are parallel. 
 
 This furnishes another example of the "reductio ad absurdum" 
 type of proof. 
 
 PROPOSITION XIII 
 
 55. Theorem: Two lines that are perpendicular to the 
 .same line are parallel to each other. 
 
 Let AB and CD both 
 be J_ E F (Fig. 33). 
 
 To prove A B and A. 
 C D parallel. 
 
 A B and C D are 
 supposed to lie in the C 
 plane of the page. 
 
 Suppose A B and 
 C D could intersect at 
 some point as P. FIG. 33
 
 Congruency of Rectilinear Figures and Circles 27 
 
 Then through P there would be two perpendiculars to 
 E F, which is impossible. Why ? 
 
 /. A B and C D do not intersect. 
 
 Since A B and C D lie in the same plane and do not inter- 
 sect, they are parallel. Why? 
 
 56. A line cutting two or more lines is called a transversal. 
 
 Thus E F (Fig. 34), is a transversal. 
 
 The angles a, b, c', and d' are ex- 
 terior angles. 
 
 The angles c, d, a', and b' are in- 
 terior angles. 
 
 The angles a and c', b and d', a' 
 and c, d and b' are alternate angles. 
 
 The angles a and a', b and b', c 
 and c', d and d' are corresponding 
 angles. 
 
 r 
 
 FIG. 34 
 
 The angles c and a', b' and d are alternate-interior angles. 
 The angles b and d', a and c' are alternate -exterior angles. 
 
 PROPOSITION XI 1 
 
 / 57. Theorem: If two alternate-interior angles, formed by two 
 I Urns and a transversal, are equal, the two lines are parallel. 
 
 Let the angles a 
 and a' (Fig. 35) , made 
 by A B and C D with 
 the transversal E F, 
 be equal. 
 
 To prove A B and 
 C D parallel. 
 
 The lines are all 
 supposed to lie in the 
 plane of the page. 
 
 A K HX r> 
 
 f\ 
 
 
 
 L 
 
 FIG. 35
 
 28 
 
 Second- Year Mathematics 
 
 Through the middle point, M, of G H draw a perpendicular 
 to A B, and prolong it to meet C D at L. 
 
 What parts of triangles M G L and M H K are equal ? 
 
 Prove M G L and M H K congruent. 
 
 M K H is a right angle. Why ? 
 
 Therefore M L G is a right angle. Why ? 
 
 A B and C D are both perpendicular to K L. Why ? 
 
 Therefore A B || C D. Why? 
 
 EXERCISE 
 
 i. If one of two parallel lines is perpendicular to a third 
 line, the other is also. Prove. 
 
 A corollary is a theorem, the proof of which is contained 
 in the proof of other propositions. 
 
 58. Corollary: If two lines are cut by a transversal in such 
 a way that the sum of two interior angles on the same side of the 
 transversal is equal to a straight angle, the two lines are parallel. 
 
 The arrangement of the proof is left to the student. 
 
 PROPOSITION XV 
 
 Theorem : If two lines cut by a transversal are parallel, 
 alternate-interior angles are equal. 
 
 Let A B and 
 C D be two paral- 
 lel lines, cut by 
 the transversal 
 E F (Fig. 36). 
 
 To prove angle 
 A H G equal to 
 angle HGD. 
 
 Proof: At H 
 on the line E F, 
 ZHGD. 
 
 _----N 
 
 NT 
 r / 
 
 ~7H 
 
 & 
 
 T) 
 
 Y 
 
 an angle can 
 
 FIG. 36 
 be drawn 
 
 ' ./ 
 
 equal to
 
 Congruency of Rectilinear Figures and Circles 29 
 
 If A B does not make angle A H G equal to H G D, suppose 
 some other line, as M H N does. Then M H G =H G D. 
 
 Then M N and C D are parallel. Why ? 
 
 But A B is assumed parallel to C D. 
 
 Hence, through H there would be two parallels to C D, 
 which is impossible. Why? (Axiom 12, p. 26.) 
 
 Therefore no other line than A B through H can make 
 with E F and C D the alternate-interior angles equal. 
 
 Hence, AH G=HGD. 
 
 This is another example of the indirect form of proof, known as 
 " reductio ad absurdum." 
 
 60. One theorem is said to be the converse of another, 
 when the hypothesis and the conclusion of the one are, respec- 
 tively, the conclusion and the hypothesis of the other. 
 
 Give examples of theorems and their converse; of axioms 
 and their converse. 
 
 The converse of a theorem is not true, simply because the 
 theorem is true. Some converses are true and some are not. 
 A proof is necessary before they can be accepted as true. 
 . Give cases in which the converse of a theorem is not true. 
 
 ('EXERCISES 
 . Supposing 'A B || ClT(Fig. 37) prove the following: 
 
 (2) b = V (4) d=d' (6) b=d' (8) b'=d. 
 
 FIG. 37
 
 30 
 
 Second-Year Mathematics 
 
 2. Prove also the following: 
 
 (1) a and b' are supplementary 
 
 (2) a' and b are supplementary 
 
 (3) c' and d are supplementary 
 
 (4) c and d' are supplementary. 
 
 3. A perpendicular to one of two parallels is perpendicular 
 to the other also. Prove. 
 
 4. The rails of two railroad tracks cross so that one of the 
 1 6 angles formed is 36. Make a sketch and state the value 
 of each of the other 1 5 angles. 
 
 5. If the sum of the interior angles on one side of a trans- 
 versal to two lines is less than a straight angle the lines meet. 
 
 61. Suppose an observer standing at the vertex of an angle 
 as at O (Fig. 38), and looking off over the angular space in 
 the direction O D, one side of the angle lies on his right and 
 the other on his left. 
 
 J> 
 
 
 
 FIG. 38 
 
 FIG. 39 
 
 An angle may thus be regarded as having a right side and 
 a left side (see Figs. 38 and 39). 
 
 EXERCISES 
 
 i . If the sides of two angles are parallel, right to right and 
 left to left (Fig. 40), the angles are equal. Prove. 
 
 2. If the sides of two angles are parallel, right to left and 
 left to right (Fig. 41), the angles are supplementary. Prove.
 
 Congruency of Rectilinear Figures and Circles 31 
 
 3. The sum of the three angles of any triangle is equal to 
 a straight angle, or to two right angles. Prove. (See Fig. 42. 
 See also FYM, pp. 54, 55.) 
 
 62. A triangle that contains a right angle is a right tri- 
 angle. 
 
 FIG. 41 
 
 63. A triangle that contains an obtuse angle is an obtuse 
 triangle. 
 
 FIG. 42 
 
 64. A triangle, all of whose angles are acute, is an acute 
 triangle.
 
 3 2 
 
 Second-Year Mathematics 
 
 EXERCISES 
 
 1. How many obtuse angles, at most, may a triangle have ? 
 How many right angles, at most? How many acute angles; 
 at most? How many acute angles, at least? Give reasons 
 for your answers. 
 
 2. The sum of the acute angles of a right triangle is a right 
 
 angle, i. e., the acute angles of a right 
 triangle are complementary. 
 
 3. If the sides of two angles 
 
 I 
 
 I 
 
 FIG. 43 
 
 FIG. 44 
 
 are perpendicular, right to right and left to left (Fig. 43), 
 the angles are equal. 
 
 4. If the sides of two angles are perpendicular, right to 
 left and left to right (Fig. 44), the angles are supplementary. 
 
 CONSTRUCTION X 
 
 65. Problem : To draw a line parallel to a given line through 
 a given point. 
 
 pi ^ -p Iet P be the given 
 
 point and A B be the 
 given line (Fig. 45). 
 
 i \t To draw a parallel to 
 
 X J3 A B through P. 
 
 With P as a center 
 
 \ 
 
 \ 
 \ 
 
 FIG. 45 
 
 and with a radius longer than the distance from P to A B, draw 
 an arc as C F.
 
 Congruency of Rectilinear Figures and Circles 33 
 
 With C as a center and with the same radius, draw the 
 arcPE. 
 
 With a radius equal to the distance from E to P and with 
 C as a center, draw an arc as at F across arc C F. 
 
 Join F and P. The line F P is the required parallel. 
 
 F P will later be proved parallel to A B. 
 
 66. A plane figure bounded by four straight lines is a 
 quadrilateral. 
 
 67. A quadrilateral having two pairs of parallel sides is a 
 parallelogram. 
 
 EXERCISES 
 
 1. The sum of the four angles of a quadrilateral is two 
 straight angles (see Exercise 3, 61, 
 
 and Fig. 46). 
 
 The angles marked with arcs (Fig. 47) 
 are exterior angles (see FYM, p. 55, 
 39). 
 
 2. The sum of the exterior angles 
 of a triangle is four right angles 
 (see FYM, 40, Exercise 4). 
 
 FIG. 46 
 
 FIG. 47 
 
 3. The sum of the exterior angles of a quadrilateral is four 
 right angles (see Fig. 47).
 
 34 
 
 Second-Year Mathematics 
 
 PROPOSITION XVI 
 
 68. Problem: To construct a parallelogram, having given 
 two adjacent sides and the included angle. 
 
 Let a and b be the two adjacent sides, and let x be the 
 included angle (Fig. 48). 
 
 a To construct the parallelogram. 
 
 I Lay off from the vertex of the angle 
 
 upon the sides (prolonged if 
 /~ necessary) the sides a and b, 
 ~~f as A Band A B. 
 
 / Through D draw a parallel 
 
 to A B and through B draw 
 a parallel to A D. 
 
 Let C denote the inter- 
 secting point of the parallels. 
 Then A B C D is the required parallelogram. Why ? 
 
 EXERCISES 
 
 1. Prove, that if two adjacent sides and the included angle 
 of one parallelogram are equal, respectively, to the correspond- 
 ing parts of another, the parallelograms are congruent. 
 
 2. Prove, the adjacent angles of a parallelogram are supple- 
 mentary. 
 
 3. Prove, the opposite angles of a parallelogram are equal. 
 
 PROPOSITION XVII 
 
 b 
 
 FIG. 48 
 
 69. Theorem: The op- 
 
 posite sides of a parallelo- 
 gram are equal. 
 
 p. 334, Problem 7. 
 
 FYM, 
 
 FIG. 49
 
 Congruency of Rectilinear Figures and Circles 35 
 
 PROPOSITION XVIII 
 
 ^^*"~ . ~^ ' ^^ 
 
 70. Theorem: If the opposite sides of a quadrilateral are 
 equal, the quadrilateral is a parallelogram. , 
 
 Converse of Proposition XVII. See p. 19, 36 and Proposition 
 XIV. 
 
 Let ABCD be a quadrilateral having AB=DC, and 
 AD=BC(Fig. 50). 
 
 To prove A B C D a 7) /*r 
 
 ~" |/> 7 O 
 
 parallelogram. 
 
 Draw a diagonal, as A C. 
 
 Prove the triangles A 
 ABC and ADC con- 
 
 tlG. 50 
 
 gruent. 
 
 Then since y=y' (why?) and since A C is a transversal to 
 A Band DC, AB||DC. Why? 
 
 Also, since x=x', and A C is also a transversal to A D 
 andBC, A D || B C. Why? 
 
 The quadrilateral is then a parallelogram. Why ? 
 
 PROPOSITION XIX 
 
 71. Theorem: If the opposite angles of a quadrilateral are 
 equal, the quadrilateral is a parallelogram. 
 
 B 
 
 FIG. 51 
 
 Let ABCD be a quadrilateral having a=c, and bd 
 (Fig- 50- 
 
 To prove the quadrilateral a parallelogram.
 
 36 Second-Year Mathematics 
 
 Proof 
 
 a+b+c+d=4 right angles. Why? 
 
 But c=a and b=d. Why? 
 
 Therefore a+d = a straight angle, or two right angles. 
 Why? 
 
 Regarding A D as a transversal to A B and DC, A B 1 1 
 D C (see 58). 
 
 Also d+c=a. straight angle or two right angles. 
 
 Therefore AD || BC. Why? 
 
 And A B C D is_a parallelogram. 
 
 PROPOSITION XX 
 
 72 . Theorem : The diagonals of a parallelogram bisect I 
 each other: 
 
 Given the parallelogram with the diagonals A C and B D 
 (Fig- 52). 
 
 & 
 
 FIG. 52 
 
 To prove A C and B D bisect each other. 
 
 Proof 
 
 Call the point of intersection of the diagonals E. 
 What parts of the triangle A B E are equal to corresponding 
 parts of D C E ? 
 
 Prove the triangles ABE and D C E congruent. 
 D E then equals E B, and A E equals E C. Why ? 
 See FYM, p. 334, Problem 8.
 
 Congruency of Rectilinear Figures and Circles 37 
 
 EXERCISES 
 
 1. Prove, conversely, if the diagonals of a quadrilateral 
 bisect each other, the quadrilateral is a parallelogram (use 
 
 Fig- 5 2 )- 
 
 2. Prove, a diagonal divides a parallelogram into two con- 
 gruent triangles (see FYM, p. 345). 
 
 PROPOSITION XXI 
 
 73. Theorem: If two sides of a quadrilateral are equal and 
 parallel, the quadrilateral is a parallelogram. 
 
 Given the quadrilateral J) p 
 
 A B C D with A D equal 
 and parallel to B C (Fig. 53). 
 
 To prove A B C D a paral- 
 lelogram. 
 
 Draw the diagonals A C FIG. 53 
 
 and B D ; let E be their intersection. 
 
 Regarding D B as a transversal to A D and C B, x=x f . 
 
 Why? 
 
 Regarding A C as a transversal to A D and C B, y=y'. 
 Why? 
 
 Prove A E=E C and D E=E B. 
 
 A B C D is then a parallelogram (see Exercise i above), 
 i . 1 ... 
 
 74. A rectangle is a parallelogram one of 
 whose angles is a right angle. 
 
 EXERCISES 
 
 1. Prove, that if one of the angles of a parallelogram is a 
 right angle, all the angles are right angles. 
 
 2. Prove, the diagonals of a rectangle are equal. 
 
 75. A square is a rectangle two of. whose 
 adjacent sides are equal.
 
 38 Second-Year Mathematics 
 
 EXERCISES 
 
 1. Prove, that if two adjacent sides of a rectangle are equal, 
 all the sides are equal. 
 
 2. Prove, the diagonals of a square 
 bisect each other perpendicularly. 
 
 76. A rhombus is a parallelogram two 
 of whose adjacent sides are equal. 
 
 EXERCISES 
 
 1. Prove, that if two adjacent sides of a parallelogram are 
 equal, all the sides are equal. 
 
 2. Prove, the diagonals of a rhombus bisect each other per- 
 pendicularly (see FYM, p. 346, Exercise 6). 
 
 77. A trapezoid is a quadrilateral having / \ 
 two parallel sides. L A 
 
 The parallel sides of a trapezoid are the bases. 
 A trapezoid whose non-parallel sides are equal is an 
 isosceles trapezoid. 
 
 A , ., , , , . ( two pairs of parallel sides is a parallelogram 
 A quadrilateral having j Qne of J *~^ | s;des fa a apezoid 5 
 
 . r. ( one right angle is a rectangle 
 
 A parallelogram having J ^ J jacen f sides equal | a rhombus 
 
 A parallelogram that is both a rectangle and a rhombus is a square. 
 
 , ( parallelogram \ r ctan gle { square 
 Quadrilateral 
 
 1 
 
 EXERCISES 
 
 i. Prove, that if the two angles at the ends of one base of 
 a trapezoid are equal, the trape- 
 zoid is isosceles. 
 
 Draw C E <\ D A. Prove C B 
 = D A (Fig. 54). 
 
 2. Prove, conversely, if the 
 FIG. 54 non-parallel sides of a trapezoid,
 
 Congmency of Rectilinear Figures and Circles 39 
 
 that is not a parallelogram, are equal, the angles at the ends 
 of one base are equal. 
 
 PROBLEMS 
 
 1. Given the base and one base angle of an isosceles tri- 
 angle, to construct the triangle. 
 
 2. Given one angle, one-half of another angle, and the side 
 included by the vertices of these two angles, to construct the 
 triangle. 
 
 To solve this it is necessary to construct an angle equal to the 
 double of a given angle. 
 
 Let a be the given half-angle (Fig. 55). 
 
 To construct an angle that is the double 
 of a. 
 
 With a convenient radius draw the arcs 
 M N and B D, the former with O as center 
 and the latter with A as center. 
 
 With the distance M N as radius and 
 with B as center draw an arc at C. _^____ 
 
 Then with C as center and with the same *t & * 
 
 radius,' draw the arc at D. Join A D. FIG. 55 
 
 B A D is then the double of a. Why ? 
 
 Now solve Problem 2. 
 
 3. Construct an angle equal to three times a given angle. 
 
 4. Construct an angle equal to the sum of two given angles. 
 
 5. Construct an angle equal to the difference of two given 
 angles. 
 
 6. Given two adjacent sides of a rectangle, to construct 
 the rectangle. 
 
 7. Given a side of a square, to construct the square. 
 
 8. Given three consecutive sides and the two included 
 angles of a quadrilateral, to construct the quadrilateral. 
 
 9. Given a diagonal of a square to construct the square.
 
 Second-Year Mathematics 
 
 10. Given the diagonals of a rhombus, to construct the 
 rhombus. 
 
 11. To construct the complement of a given angle. 
 
 12. To construct the supplement of a given angle. 
 
 13. Given the two sides including the right angle of a right 
 triangle, to construct the triangle. 
 
 EXERCISES 
 78. Prove the following theorems: 
 
 1. If the bisector of any angle of a triangle is perpendicular 
 to the opposite side, the triangle is isosceles. 
 
 2. If the middle points of the equal sides of an isosceles 
 triangle are joined to the opposite vertices, the triangles formed 
 with the bases are congruent. 
 
 3. The diagonals of a square are equal. 
 
 4. If the diagonals of a parallelogram are equal, the figure 
 is a rectangle. 
 
 5. Parallels intercepted between parallels are equal. 
 
 6. Prove the correctness of the construction of Construc- 
 tion X, p. 32. 
 
 7. Suppose c in Fig. 56 to be a right angle. Prove Za = 
 Z. b = Z d = a right angle. 
 
 a 
 
 FIG. 56 
 
 FIG. 57 
 
 8. /f the two parallels A B and C D are cut by the trans- 
 versal E F (Fig. 57), then:
 
 J 
 
 ,\ Congruency of Rectilinear Figures and Circles 41 
 
 (i) Za=Zc' (3) Z6=Zrf' (5) Z6'+Zc = i8o 
 
 7(2) Zc=Za' (4) Z& + Zc' = i8o (6) Zd+Zc'=i8o . 
 
 g. Prove, that if any one of the six relations of Exercise 8 
 is not true the lines A B and C D are not parallel. 
 
 10. Two of the angles of a triangle have the values indi- 
 cated by the following pairs of numbers. Find the third angle 
 in each case. 
 
 (1) 40 and 60 (5) 50 and a (9) $x and 2X 
 
 (2) 65 and 65 (6) r and 70 (10) x and y 
 
 (3) 12 and 98 . (7) A and B (n) a and b+c 
 
 (4) 37|andi23i (8) A/$ and B/8 (12) and 2^. 
 
 o 
 
 11. Prove, if two angles of one triangle are equal to two 
 angles of another, the third angles are equal, and the triangles 
 are mutually equiangular. 
 
 12. Prove, that an exterior angle of a triangle is equal to 
 the sum of the two opposite interior angles of the triangle. 
 
 13. One base angle of an isosceles triangle is J of the 
 vertex-angle. Find the angles of the triangle. 
 
 14. Find the angle between the bisectors of the acute angles 
 of a right triangle. 
 
 15. The bisector of an exterior vertex-angle of an isosceles 
 triangle is parallel to the base. Prove. 
 
 1 6. How many diagonals can be drawn from one vertex 
 in a quadrilateral? In a pentagon? In a hexagon? In a 
 heptagon ? In an octagon ? In an w-gon ? In a 2-gon ? 
 
 1 7. Into how many triangles do the diagonals of Exercise 16 
 divide the quadrilateral ? The pentagon ? The hexagon ? 
 The heptagon ? The octagon ? The w-gon ? The 2-gon ? 
 
 1 8. Prove, that the sum of the interior angles of an w-gon 
 is 2n 4 right angles, or ( 2) straight angles.
 
 42 Second- Year Mathematics 
 
 ig. Prove, that any interior angle of a regular w-gon is 
 
 2n 4 . . A (n 2\ . 
 
 right angles, or I straight angles. 
 
 n \ n / 
 
 20. Prove, that the sum of the exterior angles of an w-gon 
 is 4 right angles. 
 
 The distance between two parallels is measured on a common 
 perpendicular between them. See 60, 3, and 99. 
 
 21. Prove, that parallels are everywhere equally distant. 
 
 22. Prove, if two points on the same side of a line are 
 equally distant from the line, the line joining the two points is 
 parallel to the given line. 
 
 79. Algebraic solutions of equations based on geometrical 
 relations. 
 
 (i7)The sides of an equiangular triangle are denoted by 
 , 2Xy, and 14. Find x and y. 
 
 2. The angles of an equiangular triangle are denoted by 
 
 , 3(3^ 2^), and 60. Find x and y. 
 
 3. The three sides of an equiangular triangle are denoted 
 by > jx+2y, $(x + 2y), and Sx ^y + 2. Find x and y. 
 
 Solve the quadratics in the following by factoring, being 
 careful to obtain all solutions. 
 
 4. One pair of opposite sides of a parallelogram is denoted 
 by x 2 -\-x and 6(3 x}, and the other pair by y 2 y and 3(5 y). 
 Find x and y and the lengths of the sides. 
 
 5. Two opposite angles of a parallelogram are denoted by 
 x 2 +6 and 7(^+2). Find x and all angles of the parallelo- 
 gram. 
 
 6. The diagonals of a rectangle are denoted by x 2 x and 
 2(2^+7). Find both values of x and the diagonals. 
 
 7. The diagonals of a parallelogram divide each other so 
 that the segments of one are s 2 +s and 2(55 7), and of the
 
 Congruency of Rectilinear Figures and Circles 43 
 
 other, / 2 + 2/ and 8(3 /). Find s, t, and the lengths of the 
 diagonals. 
 
 Is the parallelogram a rectangle ? Why ? 
 
 8. The diagonals of a rhombus divide each other so that 
 the parts of one diagonal are denoted by x 2 and 3(2^+9), 
 and of the other by y 2 and 2(^+4). Find x, y, and both of 
 the diagonals. 
 
 9. Two of the four angles that the diagonals of a rhombus 
 make with each other are given by x 2 10 and 10(2^11). 
 Find x and all the four angles. 
 
 10. The distances from a point on the perpendicular bisec- 
 tor of a line-segment to the ends of the segment are given by 
 z 2 + 3z and 18(23). Find z and the distances from the point 
 to the ends of the line-segment. 
 
 11. The two sides that include the vertex-angle of an 
 isosceles triangle are given by x 2 x and 6(x 2). Find x 
 and the lengths of the sides. 
 
 12. The angles opposite the equal sides of an isosceles 
 triangle are given by x 3 i and 3(^+9). Find x and the 
 lengths of the sides. 
 
 13. Two angles of a triangle are equal. The sides oppo- 
 site these two angles are given by x 2 + $ and 3(^ + 5). Find 
 x and the lengths of the sides. 
 
 Sides and angles of polygons may be denoted by either positive or 
 negative numbers, since the sign of a number only indicates the direction 
 in which the line or angle is conceived as measured. 
 
 14. The angles opposite the equal sides of an isosceles tri- 
 angle are given by x a i and 8(# + i). Find x and the values 
 of the angles. 
 
 15. The lengths of the bisectors of the base angles of an 
 isosceles triangle are given by m 3 and i6(w 4). Find m and 
 the lengths of the bisectors.
 
 44 Second-Year Mathematics 
 
 1 6. The lengths of the perpendiculars from the vertices of 
 the base angles of an isosceles triangle to the opposite sides 
 are given by r 2 +4r and 12(2^3). Find r and the lengths 
 of the perpendiculars. 
 
 17. The lengths of the two lines drawn from the vertices 
 of the base angles of an isosceles triangle to points of the 
 opposite sides equidistant from the vertex of the vertex-angle 
 are given by c(c + $) and 21 (c 3). Find c and the lengths of 
 the two lines. 
 
 1 8. The following pairs of numbers denote the lengths of 
 the sides lying opposite a pair of equal angles of a triangle. 
 Find x and the lengths of the sides in each case. 
 
 (1) x 2 and 5# 4 (10) 3(5^ + 1) and 14* 
 
 (2) x 2 + i and 2(#+i8) (n) 4(y^ 2 i) and gx 
 
 (3) x 2 2 and 3# 4 (12) $x 2 and 2 5^ 
 
 (4) x 2 4 and x + i6 (13) $x 2 and 57103: 
 
 (5) x 2 ~4 and 4(2 x) (14) 4X 2 and 3(^ + 9) 
 
 (6) x 2 i and 2(^ + 7) (15) 4gx 2 i and 5(13 73:) 
 
 (7) x 2 +x and 4(9 x) (16) iSx 2 and 3(^ + 12) 
 
 (8) x(x + i) and 2$x* (17) 40 and ^x 2 + ijx 
 
 (9) Sx 2 and 2^+3 (18) x*gx* and 2(12133;)* 
 The only difficult step in solving the equation of Example 
 
 (18) is in factoring the first member. We apply here the 
 Remainder or Factor Theorem (see FYM, p. 314, 261). 
 
 Let us rehearse the reasoning given in the reference. Fac- 
 toring is finding exact divisors. Exact divisors are divisors 
 that give o for the remainder after dividing. The quickest 
 way to discover exact divisors, or factors, is to examine the 
 remainders. But the Remainder Theorem states that the result 
 of substituting any number (say 8 or r) for x in an expression 
 
 * Example (18) leads to the equation xi gx 2 + 26x 24 = 0. In 
 factored form this is (x z)(x $)(x 4) =o. The values of x are 
 then x = 2, # = 3, and 00 4.
 
 Congruency of Rectilinear Figures and Circles 45 
 
 containing x, is the same as the remainder after dividing the 
 expression by x 8, or x r. Hence, to find factors we begin 
 with o and substitute i, 2, 3, etc., then i, 2, 3, etc., until 
 the expression we are trying to factor gives a result o. 
 
 x minus the number that substituted for x gives o, is then a 
 factor of the expression. 
 
 The graph of 
 x 3 <)x 2 +26x 24 (Fig. 
 58), shows that there 
 are 3, and only 3, values 
 of x that will make the 
 expression equal o. 
 These are the three dis- 
 tances from o to the 
 crossing-points of the 
 curve on the horizontal 
 axis. There are no other 
 crossing-points than 
 those shown because 
 above #=4 and below 
 x = i the curve leaves 
 the horizontal axis 
 farther and farther. 
 
 Similarly the graph 
 shows that there are 
 as many crossing-points, 
 
 FIG. 58 
 
 and hence real values of x where the expression equals o, 
 as there are units in the highest exponent of x in the expres- 
 sion plotted. 
 
 19. Solve the following equations by this method of fac- 
 toring: 
 
 (1) X3&X 3 + I()X 12=0 (4) #3 4X*+X + 6=0 
 
 (2) X3 l JX a + I4X-?S=0 (5) X3X' IOX S=0 
 
 (3) X3 ~ 3* a #+3=0 (6) x^ T >
 
 46 
 
 Second- Year Mathematics 
 
 (7) x 4 7 
 
 (8) x4-$ 
 
 (9) X 4 I 
 
 (lo) X 4 -\-2X 3 I^X 
 
 8x=o 
 
 14^ + 24=0. 
 
 ___ __ CIRCLES 
 
 80. The curve which bounds a circle is the circumference. 
 
 8 1. A line-segment drawn through the center and termi- 
 nated both ways by the circumference is a diameter, as B C 
 (Fig. 59)- 
 
 82. A line-segment from the center to the circumference 
 
 Tariff e-nt 
 
 is the radius, as O A. 
 
 83. Any line-segment 
 terminated both ways by 
 the circumference is a 
 chord. 
 
 84. A line cutting the 
 circumference in two points 
 is a secant. 
 
 85. A line that touches 
 the circumference in but 
 
 one point, however far the line is prolonged, is a tangent. 
 
 86. Any part of the circumference is an arc, as A C (Fig. 59). 
 
 87. The parts into which a chord or a secant divides the 
 circle, are circular segments, or simply segments. 
 
 88. The portion of the circle included between two radii 
 and the arc they intercept, is a sector. 
 
 Circles are designated by reading the letter at their centers, 
 as the circle O (Fig. 59). 
 
 FIG. 59 
 
 CONSTRUCTION XI 
 89. Draw two circles having equal radii.
 
 Congruency of Rectilinear Figures and Circles 
 PROPOSITION XXII 
 
 47 
 
 90. Theorem: Tico circles having equal radii are equal, 
 and conversely, equal circles have equaljradj4._ 
 
 Given the circles whose centers are O and O', and having 
 A = O'A' (Fig. 60). 
 
 FIG. 60 
 
 To prove the circles are equal. 
 
 Proof: Place the circle O upon the circle O' so that the 
 center O falls on the center O'. 
 
 The circles then coincide throughout, for if they did not, 
 there would be points of one circumference farther from O 
 and O' than some points of the other, and for such points 
 the distances from O and O' would not be equal. 
 
 Conversely, since the circles are equal they may be made 
 to coincide, and hence the radii are equal. 
 
 EXERCISE 
 i. The diameters of equal circles are equal. 
 
 PROPOSITION XXIII 
 
 91. Problem: Given the circumference of a circle, to find 
 the center.
 
 Second- Year Mathematics 
 
 I. Draw any two non-parallel chords in the circle (see 
 
 DF 
 
 Fig. 61). 
 
 Bisect each of the chords with 
 perpendiculars. 
 
 These perpendiculars will inter- 
 sect (why ?) at some point, as at O. 
 
 O is the required center. 
 
 Assume that the perpendicular bi- 
 sector of a line-segment contains every 
 point that is equally distant from the 
 end-points of the line-segment. (Proved 
 on p. 174.) 
 
 II. Draw a single chord as AB, and bisect it as with 
 E F _L to it. 
 
 The middle of E F, as O, is the center of the circle. Why ? 
 
 PROPOSITION XXIV 
 
 92. Theorem: A diameter divides the circumference of a 
 circle into two equal parts; and conversely, any chord that divides 
 a circumference into two equal parts, is a diameter. 
 
 I. Draw a circle with a diameter, and show that if the cir- 
 cle be folded over this diameter as an axis, or hinge, the one 
 part will coincide with the other. 
 
 II. The proof of the converse is left to the student. 
 
 PROPOSITION XXV 
 
 93. Theorem: In the same circle or in equal circles, radii 
 that form equal angles intercept equal arcs on the circumference, 
 and conversely. 
 
 I. Given in circle O, the angle A O B made by the radii 
 O A and O B, equal to angle COD made by the radii .O C 
 and O D (Fig. 62). 
 
 To prove arc A B =arc CD. 
 
 Proof : Imagine the plane of the circle cut along A O and
 
 Congruency of Rectilinear Figures and Circles 49 
 
 B O and the sector A O B turned round O as a pivot, until 
 O A falls on O C. 
 
 Then A will fall on C. Why ? 
 
 O B will fall along, or will take the direction of, O D. 
 Why? 
 
 B will fall on D. Why ? 
 
 j 
 
 FIG. 62 
 
 Arc A B will coincide with arc CD, or the points of the 
 given circumference would be unequally distant from O. 
 Hence arc A B=arc C D. 
 
 II. Given circle O=circle O', and angle A O B, made by 
 the radii O A and O B, equal to angle A'O'B', made by the 
 radii O'A' and O'B'. 
 
 To prove arc A B=arc A'B'. 
 
 Proof: Place the circle O upon the circle O' so that O falls 
 on O' and O A along O'A'. 
 
 Then A will fall on A 7 . Why ? 
 
 O B will take the direction O'B' (why ?) and B will fall 
 onB'. Why? 
 
 Hence, arc A B = arc A'B'. 
 
 III. Converse. Given circles O and O' equal and arc 
 AB = arc CD = arc A'B'. 
 
 To prove ZA O B = ZC O D = Z A'O'B'.
 
 5o Second-Year Mathematics 
 
 Proof: Suppose the arc A B to be turned around O as 
 a pivot until arc A B coincides with C D. 
 
 Prove that the sectors A O B and COD coincide. 
 
 Next place circle O upon circle O' so that arc A B coincides 
 with arc A'B'. Why can this be done ? 
 
 Then prove sector A O B to coincide with sector A'O'B'. 
 
 Hence, in both cases the angles at the centers are equal. 
 
 94. Angles whose vertices lie at the center of a circle are 
 called central angles. 
 
 EXERCISE 
 
 i. Prove, in the same or in equal circles radii that form 
 unequal angles intercept unequal arcs ori the circumference, 
 and conversely. 
 
 PROPOSITION XXVI 
 
 95. Theorem: In the same circle or in equal circles, equal 
 arcs are subtended by equal chords, and conversely. 
 
 I. Given circle O, with arc A B=arc CD (Fig. 63). 
 
 fi 
 
 C 
 
 FIG. 63 
 
 To prove chord A B=chord C D. 
 This is left as an exercise for the student. 
 
 II. Given circle O=circle O', and arc A B=arc A'B'. 
 To prove chord A B =chord A'B'. 
 
 III. Conversely, given circle O, with chord AB=chord 
 CD.
 
 Congruency of Rectilinear Figures and Circles 51 
 
 To prove arc A B = arc C D. 
 
 Join the ends of the equal arcs with the center, or centers, prove 
 the triangles thus formed congruent, the central angles equal, and apply 
 Proposition XXV. 
 
 Let the student arrange the proof. 
 
 EXERCISES 
 
 1. If a circumference is divided into three equal parts, 
 and the points of division are joined by chords, the polygon 
 formed is an equilateral triangle. Prove. 
 
 2. The ends of a pair of perpendicular diameters are joined. 
 What kind of a polygon is formed by the joining lines ? Prove 
 your answer. 
 
 3. Draw a circle O= circle O', and a chord A B= chord 
 A'B'. Prove arc A B =arc A'B'. 
 
 PROPOSITION XXVII 
 
 96. Theorem: A radius, drawn perpendicular to a chord, 
 bisects the chord and also the arc subtended by the chord. 
 
 Given circle O and radius O C drawn perpendicular to 
 the chord A B (Fig. 64), intersecting 
 the chord at D. 
 
 To prove that chord A B and arc 
 A B are bisected by O C. 
 
 Join O A and O B. What kind of 
 triangle is A O B ? 
 
 Prove triangles A O D and BOD 
 congruent. 
 
 Then apply Proposition XXV, to 
 prove arc A C=B C. 
 
 EXERCISES 
 
 1. The diameter E C_L A B bisects the arc A E B. Prove. 
 
 2. A diameter that bisects a chord is perpendicular to the 
 chord, and bisects the subtended arc. Prove.
 
 52 Second- Year Mathematics 
 
 3. A line bisecting a chord and the subtended arc passes 
 through the center, and is perpendicular to the chord. Prove. 
 
 4. Show how to bisect a given arc. 
 
 5. Through a given point within a circle draw a chord that 
 will be bisected by the point. 
 
 6. Prove the line-segment connecting the middle point of 
 the two arcs into which a chord divides a circumference, is a 
 diameter perpendicular to the chord. 
 
 7. Prove a diameter bisecting an arc is the perpendicular 
 bisector of the chord that subtends the arc. 
 
 8. The perpendicular bisector of a chord passes through 
 the center of the circle and bisects the subtended arc. Prove. 
 
 9. Draw a circle through two fixed points. How many 
 such circles may be drawn ? 
 
 10. Draw a circle passing through three fixed points not 
 situated on a straight line. How many such circles may be 
 drawn ? 
 
 n. A line perpendicular to a chord and bisecting the 
 subtended arc passes through the center of the circle and bisects 
 the chord. Prove. 
 
 12. From Proposition XXVII and Exercises 2, 3, 7, 8, and 
 ii write down four conditions for a line to fulfil about the 
 center of a circle, a chord and the subtended arc, and show 
 that if a line fulfils any two of the conditions, the remaining 
 two are also fulfilled. 
 
 PROPOSITION XXVIII 
 
 97. Theorem: In the same circle, or in equal circles, equal 
 chords are equally distant from the center, and conversely. 
 
 Assume that the distance from a point to a line is measured 
 on the perpendicular from the point to the line. (Proved in 99.) 
 
 Given in the circle O, or in the circles O and O', chord A B 
 equal to chord A'B' (Fig. 65).
 
 Congruence of Rectilinear Figures and Circles 53 
 
 To prove O C = O C', and O C = O'C'. 
 
 Draw O A, O A', and O'A'. 
 
 Prove triangles O A C, O A'C', and O'A'C' congruent. 
 
 Conversely, given O C = O C' = O'C'. 
 
 FIG. 65 
 
 To prove chord A B=chord A'B' in either circle. 
 Prove triangles O A C, O A'C', and O'A'C' congruent. 
 
 98. By the distance from a point to a line is meant the 
 shortest distance. 
 
 PROPOSITION XXIX 
 
 99. Theorem: The distance from a point to a line is meas- 
 ured on a perpendicular from the point to the line. 
 
 GivenPD_LAB (Fig. 66). 
 
 To prove P D shorter than any 
 other line drawn from P to A B. 
 
 Let P E be any other line than P D 
 from P to A B. 
 
 Prolong P D to P', making 
 P / D=PD. 
 
 Join E and P'. EP'=EP. Why? 
 
 P E P' is not a straight line. Why ? 
 
 P P' is then shorter than P E +E P'. Why ?
 
 54 
 
 Second- Year Mathematics 
 
 Hence, half of P P', or P D, is shorter than half of P E + 
 EP'. Why? 
 
 P E being any other line than the perpendicular P D, the 
 line P D is the shortest from P to A B. 
 
 NOTE. See proof of the converse of this proposition in the chapter 
 on Inequalities, 208. State the converse. 
 
 100. The point where a tangent to a circle touches the 
 circumference, is called the point of tangency, or the point of 
 contact. 
 
 PROPOSITION XXX 
 
 101. Theorem: A tangent to a circle is perpendicular to 
 the radius drawn to the point of contact; and conversely, a line 
 drawn perpendicular to a radius at its outer end is a tangent to 
 the circle. 
 
 I. Let A B be tangent to the circle at the point C (Fig. 67). 
 
 To prove the radius O C_LA B. 
 
 Let D be any other point of 
 A B than C. 
 
 Join D with O. Then O D is 
 longer than O C. Why ? 
 
 Hence, all lines drawn from 
 O to other points of A B than C 
 are longer than O C. Why ? 
 
 Therefore, O C_LA B. Why? 
 The truth of the converse is here 
 
 D 
 
 FIG. 67 
 
 See note under 99. 
 assumed. 
 
 II. Conversely, let A B be _LC O, a radius, at C. 
 
 To prove A B tangent to the circle. 
 
 C is a point common to the line A B and the cir- 
 cumference. Why ? 
 
 Let the student prove any other point of A B, as D, farther
 
 Congruency of Rectilinear Figures and Circles 55 
 
 from O than the circumference, and hence all points of A B 
 other than C to be outside the circle. 
 
 Therefore A B is tangent to the circle at C. 
 
 PROPOSITION XXXI 
 
 1 02. Theorem: The arcs included between two parallel 
 secants are equal; and conversely, if two secants include equal 
 arcs, and do not intersect within the circle, they are parallel. 
 
 I. Given circle O and A B || C D, both cutting the circum- 
 ference (Fig. 68) at these points. 
 
 To prove arcs A C and B D equal. 
 
 Draw OE through the center _L AB, 
 and prolong it. 
 
 This line is also _L C D ( 60, 
 Exercise 3, p. 30). 
 
 A~E = E~B* and E~C = E~D . Why ? 
 
 Complete the proof. 
 
 II. Conversely, given A C=B D 
 
 (Fig. 68), A B and C D not intersecting within the cir- 
 cumference. 
 
 To prove A B ||CD. 
 
 Draw O E _L A B and prolong it to F. 
 
 Prove E~C=ED and CfF=D>. 
 
 Then refer to p. 52, Exercise 6, and complete the proof. 
 
 III. Prove the theorem with one of the lines, as A B, 
 tangent to the circle, as in Fig. 69. 
 
 IV. Prove the theorem with both parallels tangent to the 
 circle, as in Fig. 70. 
 
 103. Two circles are said to be tangent to each other if 
 both are tangent to the same line at the same point. This 
 
 * A E and E B mean are A E and are E B.
 
 Second- Year Mathematics 
 
 point is the point of tangency, or the point of contact of the 
 circles. 
 
 If the tangent circles lie wholly without each other they 
 are tangent externally (Fig. 71). 
 
 If one of the tangent circles lies within the other they are 
 tangent internally (Fig. 72). 
 
 PROPOSITION XXXII 
 
 104. Theorem: If two circles are tangent to each other, 
 their centers and the point of tangency lie in a straight line. 
 
 I. Let O and O' be the centers of two circles tangent 
 externally, T being the point of tangency (Fig. 71). 
 
 FIG. 71 FIG. 72 
 
 To prove O, O', and T lie in a straight line.
 
 Congruency of Rectilinear Figures and Circles 57 
 
 Prove that O T O' is a straight angle. 
 
 Prove that O T and O'T are in a straight line, by Proposition VII, 
 p. 22. 
 
 II. Prove the proposition for the case shown in Fig. 72. 
 EXERCISES 
 
 1. If two circles are tangent to each other externally, the 
 distance between their centers is the sum of their radii. Prove. 
 
 2. If the distance between the centers of two circles is equal 
 to the sum of their radii the circles are tangent externally. Prove. 
 
 3. If two circles are tangent internally the distance between 
 their centers is equal to the difference between the radii. Prove. 
 
 4. Draw a circle of given radius tangent to a given circle 
 (a) externally, (b) internally. 
 
 How many such circles can be drawn tangent to the same 
 circle ? 
 
 In what line do the centers lie ? 
 
 5. Draw a circle tangent to a given circle at a given point 
 in the circumference. 
 
 How many such circles can be drawn ? 
 In what line do the centers lie ? 
 
 6. Draw a circle through a given point and tangent to a 
 given circle. 
 
 7. The distance between the centers of two tangent circles 
 is 2^ inches. The radius of one is f inch. Draw the two 
 circles. 
 
 8. Draw a common tangent to two tangent circles. 
 
 9. How many circles can be drawn through the same two 
 points ? 
 
 In what line do the centers lie ? 
 
 10. The radii of three circles are i inch, ij inches, and 
 f inch, respectively. Draw the circles tangent to each other 
 externally.
 
 58 Second-Year Mathem-atics 
 
 PROPOSITION XXXIII 
 
 105. Theorem: The line joining the centers of two inter- 
 secting circles bisects the common chord perpendicularly. 
 
 Let O and O' be the intersecting circles (Fig. 73). 
 
 The student 
 
 FIG. 73 
 
 To prove O O'_LA B. 
 
 For proof see Proposition XI, p. 25, for both positions, 
 will supply the proof. 
 
 EXERCISES 
 
 1. ABC is a right triangle, /_A being an acute angle. 
 Let o denote the length of the side opposite Z.A, a, the length 
 of the side adjacent to Z.A, and h, the hypotenuse. Express 
 in these numbers the following ratios: 
 
 (1) the opposite side to the hypotenuse 
 
 (2) the adjacent side to the hypotenuse 
 
 (3) the opposite side to the adjacent side 
 
 (4) the hypotenuse to the opposite side 
 
 (5) the hypotenuse to the adjacent side 
 
 (6) the adjacent side to the opposite side. 
 
 2. Denote these six ratios as follows: 
 
 so/h C=h/o 
 
 c=a/h S=h/a 
 
 t o/a T=a/o 
 
 and express C in terms of s; S in terms of c; T in terms of t.
 
 Congruency of Rectilinear Figures and Circles 59 
 
 3. It has been shown (see FYM, pp. 198, 199). that 
 o*+a 2 =h 2 . Divide both sides of o 2 +a 2 =h 2 by h 2 . and ex- 
 press the resulting equation in terms of .s and c. 
 
 4. Solve the equation, s 2 +c 2 = i, for s; for c. 
 
 5. Express C in terms of c; 5 in terms of s; T in terms 
 of 5 and c; T in terms of s; T in terms of c. 
 
 6. Express c, t, C, S, and T in terms of s. 
 
 7. Express 5, t, C, S, and T in terms of c. 
 
 8. Express s, c, C, S, and T in terms of /. 
 
 9. Express s, c, t, S, and T in terms of C. 
 
 10. Express s, c, t, C, and T in terms of S. 
 
 1 1 . Express s, c, t, C, and 5 in terms of T. 
 
 a 
 
 FIG. 74 
 
 12. In a right triangle = 10, and s = %. Find h, C, c, a, 
 S, t, and T. 
 
 13. In a right triangle a =8 and c = J. Find h, S, s, o, 
 C, t, and T. 
 
 14. In a right triangle ^ = 15 and s=^. Find 0, C, c, 
 S, a, t, and T. 
 
 15. Two circles described on the diagonals of a rectangle 
 have areas denoted by x 2 2X and 4(7^ 26). Find x and 
 the areas of the circles. 
 
 1 6. Describe a circle that shall pass through two given 
 points and have a given radius.
 
 6o 
 
 Second- Year Mathematics 
 
 17. A diameter that bisects a chord bisects the central 
 angle between radii drawn to the ends of the chord. Prove. 
 
 1 8. The perpendicular bisectors of the sides of an inscribed 
 quadrilateral meet at a common point. Prove. 
 
 19. If two intersecting chords make equal angles with a 
 line joining their common point to the center, the chords are 
 equal. Prove. 
 
 20. Prove Za (Fig. 75) is two times Z#. The vertex of 
 a is the center of the circle. 
 
 FIG. 75 
 
 FIG. 76 
 
 21. Draw a pan* of perpendicular diameters in a circle, 
 connect their consecutive ends and prove the figure formed 
 by the connecting lines a square. 
 
 22. Draw a circle, step around it using the radius for a 
 step, connect every alternate mark, and prove that the con- 
 necting lines form an equilateral triangle. 
 
 23. A B C is an equilateral triangle and C D is perpen- 
 dicular to A B from the vertex C (Fig. 76). Express the 
 six ratios s, c, t, C, S, and T in triangle ADC (see Problem 
 2, p. 58). 
 
 24. Express a and o in terms of h (see Problem 3, p. 58). 
 
 25. Calculate the numerical values of all six of the ratios, 
 s, c, t, C, S, and T for angle A, using triangle ADC (Fig. 76).
 
 Congruency of Rectilinear Figures and Circles 61 
 
 26. What is the numerical value in degrees of Z. A ? Of 
 ? Of one of the parts of ZC made by C D ? 
 
 27. Denoting the ratios for a 60 angle by s^o, C6o> *6o> 
 $60, Ceo, and Tt , calculate the six numerical values from 
 triangle A D C of Fig. 76. 
 
 28. Denoting the six ratios for a 30 angle similarly, cal- 
 culate the numerical values of s 30 , c 30 , t 30 , C 30 , S 30 , and T 30 , 
 using triangle ADC (Fig. 76). ^ 
 
 29. ABC (Fig. 77) is a right triangle 
 having Z.A = /.B=^. Compare a 
 and o. Using a 2 +0 2 =/i 2 express h in 
 terms of o; of a; o in terms of h; of a; 
 
 a in terms of h. /} Q. C 
 
 30. Calculate from triangle ABC FlG ' 77 
 
 (Fig. 77) the numerical values of s 4S , c 4S , t 4S , C 4S , S 4S . and T 4S . 
 
 31. Given 5 40 =.643. Calculate to three decimal places 
 the numerical values of c 40 , / 40 , C 40 , S 40 , T 40 , using triangle 
 ABC (Fig. 78). 
 
 Use 105, Exercise 4. 
 
 451 
 
 FIG. 78 
 
 FIG. 79 
 
 32. Given s so =.766. Calculate to three decimal places 
 the values of c so , t so , C so , S 50 , and T so , using triangle ABC 
 (Fig. 78). 
 
 33. In a regular hexagon (Fig. 79), given the side I = 12 
 and 5 30 =^. Calculate the radius of the inscribed circle.
 
 62 Second-Year Mathematics 
 
 34. In a regular octagon (Fig. 80), given the side I equal 
 to 1 6 and c 32 ^= .924. Calculate the radius of the inscribed 
 circle. 
 
 FIG. 80 
 
 35. In a regular nonagon, given the side = 24 and 
 . 364. Find the radius of the inscribed circle.
 
 CHAPTER II 
 
 RATIO, PROPORTION, SIMILAR TRIANGLES 
 Ratio of Numbers and Segments 
 
 1 06. To measure the number a by the number b, a is 
 divided by b. The quotient a/b is the ratio of a to b. Thus, 
 the ratio of a to b indicates how many times b is contained in 
 a, or what fractional part a is of b. 
 
 The numbers a and b are the terms of the ratio a to b; 
 the first number a is the antecedent, and b the consequent. 
 
 107. The portion of a line bounded by two points on the 
 line is a line- segment, or briefly, a segment. 
 
 1 08. To measure a segment m, is to lay off on m repeatedly 
 a smaller segment, as n (see Fig. 81). 
 
 109. The number of times a segment b is contained, without 
 remainder, in a segment a is called the numerical measure of 
 a in the unit b. b is called a unit of measure. 
 
 Thus the numerical measure of A B (Fig. 81) is 4. CD is 
 a unit of measure of A B. 
 
 \(.m) A- 'B 
 
 (n) -D 
 
 FIG. 81 
 
 no. The ratio of two segments is the ratio of the numerical 
 measures of the segments, both segments being measured with 
 the greatest common unit. 
 
 But it can be shown that if any other common unit of 
 measure be taken, it will be contained in the greatest common 
 unit an integral number of times. Say it is contained m times. 
 
 63
 
 64 Second-Year Mathematics 
 
 Then both numbers expressing the measures of the magni- 
 tudes will be m times as large as before, but the ratio will be 
 
 the same, e. g., - = '-^=2 
 '7 m.'j 7 
 
 Ratio of Other Magnitudes 
 
 ill. The ratio of two angles or of two arcs or areas or of 
 any other two magnitudes of the same kind is, as in the case 
 of segments, the ratio of the numerical measures, both magni- 
 tudes being measured with the same unit. 
 
 Hence, the ratio of magnitudes is a number. It is the ratio 
 of the numerical measures of the magnitudes which are to be 
 compared. Consequently the ratio of two arcs may be the 
 same as (equal to) the ratio of two angles or the ratio of two 
 segments, etc. 
 
 EXERCISES 
 
 i. In Fig. 82, what is the numerical measure of A B ? Of 
 / . p CD? What is the 
 
 FIG. 82 
 
 ratio of A B to C D ? 
 2. What is the 
 ratio of M N to P Q 
 (Fig. 83)? What is the 
 numerical measure of M N ? 
 Of PQ? 
 
 3. In Fig. 84, what is 
 the ratio of htoh'? Of R to R' ? 
 
 o 
 p. 
 
 M- 
 
 FIG. 83 
 
 7? 
 
 FIG. 84
 
 Ratio, Proportion, Similar Triangles 
 
 4. Draw, on squared paper, a figure like Fig. 85 with 
 D E || B C. 
 
 Measure AD, D B, A E, and 
 E C correctly to two decimal places. 
 
 Find the ratio of AD to D B, 
 of A E to EC. How do these 
 ratios compare? 
 
 Squared paper may be used in getting 
 the lengths of the segments. 
 
 5. Draw a figure like Fig. 85. 
 
 Using i centimeter as the unit, measure A D and D B and 
 find the ratio AD:DB. Compare this ratio with the ratio 
 of A E to E C, found by using i inch as the unit of measure. 
 
 6. What is the ratio of Z A B C to Z A'B'C' (Fig. 86) ? 
 
 FIG. 85 
 
 FIG. 86 
 
 7. Cut from cardboard a circle. By rolling this circle 
 along a straight line, determine the length of the circle (cir- 
 cumference). What is the ratio of the circumference to the 
 
 length of the diameter? 
 
 8. Determine the 
 ratio of the perimeter 
 of a polygon to one of 
 
 m- 
 
 ~b/_ 
 
 'XT 
 
 FIG. 87 
 
 its sides, by using a 
 method like that in 
 Problem 7.
 
 66 Second-Year Mathematics 
 
 9. Draw a figure like Fig. 87, making lines I, II, III, etc., 
 parallel. Compare a/a', b/b', c/c f (use squared paper). 
 
 State the result as a general theorem. 
 
 10. Draw a triangle, as A B C (Fig. 88). Draw the bisector 
 
 AD A B 
 
 of angle B. Compare the ratio - with =--^ (use squared 
 
 ^ _L/ V-' -D \^r 
 
 paper). 
 
 C 
 
 FIG. 88 
 
 Commensurable and Incommensurable Segments and 
 
 Magnitudes 
 
 112. In the preceding problems, the ratio of two magni- 
 tudes was found by measuring each and finding the ratio of 
 the numerical measures. 
 
 A common measure of two segments can be found as follows: 
 Let A B and C D be the segments to be measured (Fig. 89). 
 Lay off the smaller segment C D on the larger segment A B, 
 leaving a remainder, as E B, which is less than C D. 
 
 A & 
 
 c r D 
 
 FIG. 89 
 
 . Lay off EB on CD leaving a remainder FD (<EB). 
 Lay off F D on E B. Suppose there is now no remainder. 
 Then the last remainder F D is a common measure of A B 
 and C D.
 
 Ratio, Proportion, Similar Triangles 67 
 
 For AB = 3CD+EB=3(CF+FD)+EB. (Why?) 
 =3(2EB + FD)+3FD. (Why?) 
 = 3 ( 2 .3FD+FD)+3FD. (Why?) 
 = 3 (7FD)+3FD = 2 4 FD. (Why?) 
 
 Similarly, CD = 2 EB+FD. (Why ?) 
 
 = 2 . 3 FD + FD = 7FD. (Why?) 
 
 Thus F D is a common measure of A B and C D. 
 This process is based upon the following law: 
 If two magnitudes, a and a l} have a common measure, let 
 a l the smaller, be used as a measure of the larger a. If there 
 is a remainder, say a 2 , this is less than a l . Using a 2 as a 
 measure of o I; the former measure, if there is a remainder as 
 a 3 , then a 3 is less than a 2 . If this process is continued, a 
 remainder, a n , will be reached which is contained exactly in 
 the former measure. This exact measure is the greatest com- 
 mon measure of the two given magnitudes. 
 
 The relation of a to a n may be expressed thus: 
 a =ma I +a a 
 a l =m I a 2 -\-a 3 
 a 3 =m 3 a 3 +a 4 
 
 .'. The ratio of a toa x or m-\ -=m-\ =w+- 
 a l (/, a x 
 
 113. By a similar process it can be shown that two magni- 
 tudes may not have a common unit of measure, as in the case 
 of the side and the diagonal of a square. We will show that 
 
 * means and so on according to the same lau\ It is 
 
 called a symbol of continuation.
 
 68 
 
 Second- Year Mathematics 
 
 if A B (Fig. 90) is laid off on A C, the remainder BjC can be 
 
 laid off twice on B C leaving a 
 
 C 
 B, 
 
 A 
 
 J) r remainder B 2 C. This can be 
 
 laid off twice on BjC, leaving a 
 remainder B 3 C. 
 
 No matter" how long we con- 
 tinue this process, there will always 
 be a remainder, which, however, 
 can be made indefinitely small. 
 p IG Q This is shown in the following 
 
 outline of proof. 
 
 Lay off (Fig. 91) A B (=B C) on A C to B x , with remainder 
 B X C. Draw B I A I _LAC. Then BjC, the remainder, equals 
 BxA^A.B. (Why?) 
 
 Use BjC as a measure of B C. It 
 is contained in it twice to B 2 with a 
 remainder B 2 C. Draw B 2 A 2 J_BC. 
 Then B 2 A 2 = A 2 B T =B 2 C. (Why ?) 
 
 Then B 2 C, the last remainder, is 
 contained in B^, the former measure, 
 twice to B 3 , with remainder B 3 C. 
 
 It is clear that this process can be FIG. 91 
 
 continued indefinitely, each measure 
 
 after the first, going into the preceding measure twice with a 
 remainder. 
 
 A C 
 
 Thus the ratio - cannot be found exactly, because in 
 
 D C 
 
 every measurement the hypotenuse of an isosceles right 
 triangle is measured by a side, and there is always a 
 remainder. 
 
 The ratio can, however, be found approximately to any 
 degree of accuracy.
 
 Ratio, Proportion, Similar Triangles 69 
 
 Thus: Since A C=B C+B X C, then ^ = 1+^ or 
 
 Jo U Jt> ^ 
 
 =-^ = i + approximately. 
 Jo C 
 
 B C B C AC 
 Again, since B C = 2B I C+B 2 C .'. ^-^ = 2+ B -^ , or ^ 
 
 i AC 
 
 = i + v~F , and 15-^ = 1 +i or i .5 approximately. 
 
 > 2 ^ > U 
 
 2+ BTC 
 Again, since B I C = 2 B 2 C+B 3 C, then ^^ = 2+^-.'. 
 
 D 2 (^ J5 2 V^ 
 
 AC i i 
 
 =-^ = iH or iH r or 1.4, and so on to any 
 
 B C i 2 + 5 
 
 desired degree of accuracy not absolute. 
 
 1.4 is an approximate value of 1/2, and it will be seen 
 later that the ratio of A C to A B, or B C, is equal to j/2. 
 
 114. Two magnitudes which have a common unit of meas- 
 ure are said to be commensurable. Magnitudes which do not 
 have a common measure are incommensurable. 
 
 EXERCISES 
 
 1. Draw three segments p = 12 cm, q= 6 cm, r =4 cm. Find 
 a common unit of measure and check by applying the unit to 
 each segment. 
 
 2. Given three segments a, b, and c, such that a can be 
 applied to b m times, leaving a remainder c; c can be applied 
 to a n times leaving a remainder d; d can be applied to c 
 I times exactly. What is the ratio of a to 6, when w=4, n=2, 
 
 Proportion 
 
 115. An equation whose members are equal ratios is a
 
 70 Second-Year Mathematics 
 
 i. Are the following statements proportions: 
 f = f? f- = f? J=lf? Give reasons. 
 
 2. Form several proportions. 
 
 116. Four segments are in proportion, or are proportional, 
 if their numerical measures are in proportion. 
 
 117. In a proportion, as k/l=m/n, k, I, m, and n are the 
 terms of the proportion. The first and last terms, as k and 
 n, are the extremes. The second and third terms, as I and m, 
 are the means of the proportion. 
 
 1 1 8. From a given proportion new proportions can be 
 formed by certain rules. Some of these were studied in First- 
 Year Mathematics, pp. 125-30, which should be reviewed 
 before going on with the next problems. 
 
 Multiply both sides of the proportion a/b=c/d by bd. 
 Taking the result for the conclusion and the given proportion 
 for the hypothesis, the following theorem is obtained: 
 
 PROPOSITION I 
 
 In a proportion, the product of the extremes is equal to the 
 product of the means.^__ 
 
 EXERCISES 
 
 1. Find the value of x in x/ 2=12/3. 
 
 2. Let ad=bc. Prove that the following statements are 
 proportions : 
 
 (1) a/b=c/d (5) b/a=d/c 
 
 (2) a/c=b/d (6) b/d=a/c 
 
 (3) c/a=d/b (7) c/d=a/b 
 
 (4) d/b=c/a (8) d/c=b/a 
 
 To prove (i), divide both members of equation ad = bc by bd. 
 119. In Problem 2, the following theorem is proved:
 
 Ratio ^Proportion, Similar Triangles 71 
 
 PROPOSITION II 
 
 Theorem : // the product of two factors is equal to the product 
 of two others, a proportion is formed by taking as means the two 
 factors of either product, and as extremes the two factors of the 
 other product. 
 
 EXERCISES 
 
 1 . Form proportions from p 3 v 3 =x 2 y 2 . 
 
 2. Form proportions from pq=m 2 n 2 . 
 
 3. Form proportions from x 2 y 2 =m 3 +n 3 . 
 
 4. Taking (i) of Problem 2 (118) as hypothesis, prove (2). 
 
 5. Taking (3) of Problem 2 (118) as hypothesis, prove (7). 
 
 Notice that (2) in Problem 2 (118) may be obtained from 
 (i), and (7) from (3), by letting the means interchange places. 
 
 This process of interchanging the means is called alternation. 
 
 6. Point out other proportions in Problem 2 (118), which 
 may be obtained by alternation from a proportion in the same 
 problem. 
 
 7. Assuming (i) of Problem 2 (118), prove (5). 
 
 8. Assuming (2) of Problem 2 (118), prove (3). 
 
 Notice that (3) Problem 2 (118), may be obtained from 
 (2) by inverting the ratios of proportion (2). Therefore (3) 
 is said to be obtained from (2) by inversion. 
 
 9. What proportion is obtained by inversion from (6) ? 
 from (7) ? 
 
 PROPOSITION III 
 
 Theorem: From a given proportion, proportions may be 
 obtained: (i), by interchanging the means (alternation); (2), 
 by interchanging the extremes (alternation); (3) by inverting 
 both ratios (inversion).
 
 72 Second-Year Mathematics . 
 
 EXERCISES 
 
 1 20. Solve the following exercises: 
 
 i. In Figs. 92 and 93, the segments m, n, a, 6 are in pro- 
 portion. Write the proportion and then form other propor- 
 tions by alternation, and by 
 inversion. 
 
 a 
 
 y 
 
 FIG. 92 FIG. 93 
 
 2. Prove the theorem: A line that bisects one side of a tri- 
 angle, and is parallel to a second side, bisects the third side. 
 
 Hypothesis: A D=D B (Fig. 94). 
 
 DE||BC; 
 
 Conclusion : A E = E C . 
 Proof: Draw E F j| A B. 
 Prove that E F=D B =A D. 
 Prove triangle A D E^F E C. 
 ThenAE=EC. 
 
 3. Prove A D, D B, A E, and E C (Fig. 94) are proportional. 
 
 4. Since E F || A B 
 (Fig. 94), it follows that 
 BF=FC. (Why?) (See 
 Problem 2.) 
 
 Prove that DE=JBC. 
 
 jj p (j 5. Prove the theorem: 
 
 p If three or more parallel 
 
 lines cut two transversals, 
 
 and if the segments intercepted on one of the transversals are 
 equal, the segments intercepted on the other are equal also.
 
 Ratio, Proportion, Similar Triangles 73 
 
 x=y (Fig. 95). Why? 
 Then AI = AH. 
 Therefore A'B' = B'C'. 
 Similarly B'C' = C'D'. 
 
 t 
 
 7 
 
 t V 
 
 7 
 
 9- \ 
 
 FIG. 95 
 
 6. Prove that A B, A'B', B C, B'C', etc., are proportional. 
 
 7. Prove the following theorem: A line drawn through the 
 middle point of one of the sides of a trapezoid parallel to the 
 bases, bisects the other side. 
 
 PROPOSITION IV 
 
 Theorem: // two parallel lines cut two intersecting trans- 
 versals, the segments of one transversal are proportional to the 
 corresponding segments of the other. 
 
 Hypothesis: A B || AjB, (Fig. 96). 
 AT A and B t B in- 
 tersect at C; 
 
 bVBUUUUU . 
 
 W A A, 
 
 M CA ' 
 
 CB 
 
 A./ 
 
 \ 
 
 M f 
 
 \ p 
 
 A/ 
 
 vs 
 
 (2) CA, 
 
 ( \ 
 
 CB X 
 BB X 
 
 / 
 
 \ 
 
 CA CB lG -9 
 
 Proof of (i): Take A A x as a measure of C A x . It will be 
 contained an integral number of times, say twice, with prob- 
 ably a remainder, A 2 C.
 
 74 Second-Year Mathematics 
 
 But B B T will measure B Z C twice and leave a remainder, 
 as B 2 C (Problem 5, 120). 
 
 Use the remainder A 2 C as a measure of the former meas- 
 ure A A!. 
 
 It will be contained in it an integral number of times + a 
 remainder as A 3 A t . 
 
 But B 2 C will go into B B r the same number of times as did 
 A 2 C in A A! and leave a remainder. (Why ?) 
 
 If this process is continued and the successive divisions 
 come to an end because a remainder is found that is con- 
 tained exactly in the preceding measure, this remainder is a 
 common measure of A C and A A x ( 112). 
 
 But we have also seen that the units of measure of A X C 
 and A A x cut B C by lines || to A^ into units that measured 
 B^ and B B z the same number of times as did the units of 
 measure of A X C and A A z . 
 
 Consequently the numerical measures of A X C and A A x are 
 the same as those of B^ and B B!, and consequently their 
 
 ratios are equal. That is, -T-^T- =^-^- . 
 
 A Aj x> tj I 
 
 Prove conclusions (2) and (3) in like manner. 
 
 121. If the lines A A z and A X C are incommensurable, the 
 successive divisions never come to an end because there is 
 always a remainder. But we have seen ( 113) that the ratios 
 A X C : A X A and B T C : BjB can be found approximately to any 
 desired degree of accuracy, and at every approximation the 
 ratios will be found equal. 
 
 The two ratios thus formed are then equal, since if they 
 were said to be unequal, differing by d, we could prove them 
 to differ by less than d (see also "Approximation of Values of 
 Irrational Numbers," First-Year Mathematics, p. 295).
 
 Ratio, Proportion, Similar Triangles 75 
 
 COROLLARY I 
 
 A line parallel to one side of a triangle divides the other two 
 sides proportionally. 
 
 Prove Proposition IV for Fig. 97. 
 
 Bl 
 
 \ 
 
 FIG. 97 
 
 PROPOSITION V 
 
 Theorem: // any number of parallel lines cut two trans- 
 versals, the segments of one transversal are proportional to the 
 corresponding segments of the other. 
 
 ZtXV ' 
 
 &/ a/\Qf r 
 
 \C 
 - 
 
 FIG. 98 
 (See Fig. 98.) Prove a=a', b=b'. Use Proposition IV. 
 
 EXERCISES 
 
 1. Prove Proposition V for the case in which D B || E C. 
 
 2. Prove: Two lines that cut two given intersecting lines, 
 and make the corresponding segments of the given lines pro- 
 portional, are parallel. (Converse of Proposition IV, p. 73.)
 
 76 Second-Year Mathematics 
 
 It is assumed that the cutting lines do not intersect each other 
 between the given lines. 
 
 _ _ . AB AB t 
 Hypothesis: g-Q=g-Q~ ( Fl g- 99); 
 
 Conclusion: B B t || C C,. 
 
 YT 
 
 FIG. 99 
 
 Proof (indirect method): Suppose E^ x not parallel to 
 C C T . Then C C 2 can be drawn parallel to B B T . Therefore 
 
 M = ABi why? But AB = ABl Why? 
 BC B.C/ BC B^/ 
 
 ^liJ^-li. Why? /.B.C.-B^. Why? 
 
 rijL.a 5!^! 
 
 This is impossible (why?) and the assumption that B B, 
 is not parallel to C C t is wrong. Therefore B B T || C C x . 
 
 3. Prove (Fig. 99) that, if ^|=^ , then B B, || C C,. 
 
 4. Prove Problem 2 for Fig. 100. 
 
 X / 
 
 / 
 
 FIG. 100
 
 Ratio, Proportion, Similar Triangles 
 
 77 
 
 5. Prove: The line joining the middle points of two sides 
 of a triangle is parallel to the third side (the converse of Prob- 
 lem 2, 120). 
 
 7 
 
 7 
 
 \ 
 
 FIG. ioi 
 
 6. Prove that the line through the midpoints of the non- 
 parallel sides of a trapezoid is parallel to the bases (Fig. ioi). 
 
 PROPOSITION VI 
 
 Theorem: The medians of a triangle meet in a common 
 point. 
 
 Let two of the medians meet at F (Fig. 102) and prove that 
 the segment A G drawn from A through F is a median. 
 
 Draw B H || D F. Then A F=F H. Why? 
 
 Draw H C. Then E F || H C. Why ? 
 
 Figure B H C F is a parallelogram. Why ? 
 
 /. B G = G C. Why? Hence, A G is the median of tri- 
 angle A B C to the side B C. 
 
 EXERCISES 
 
 i. Prove in Fig. 102 that F 
 is a trisection point of the 
 medians, i.e., FG= JAG, FE = 
 JBE, FD=|DC. 
 
 The intersection point of the 
 medians of a triangle is its center 
 of gravity, for, if supported under 
 that point, the triangle will be 
 found to balance.
 
 Second- Year Mathematics 
 
 PROPOSITION VII 
 
 Theorem: The bisector of an interior angle of a triangle 
 divides the opposite side into segments whose ratio is equal to 
 the ratio of the adjoining sides of the triangle, 
 
 Hypothesis: Z# = Zy (Fig. 103); 
 
 D 
 FIG. 103 
 
 Conclusion : 
 
 AD AB 
 
 DC BC' 
 Proof: Draw A E || B D. 
 Prolong C B to meet A E in E. 
 AD EB 
 
 Then 
 
 DC BC' 
 
 (Why?) Prove Zz=Zw. 
 
 Then E B =A B (why ?) and 
 
 AD AB 
 
 DC BC' 
 
 (Why ?) 
 
 EXERCISES 
 
 1. In a triangle A B C, A B = 5 in., B C=4 in., A C=3 in., 
 A'B' is drawn parallel to A B making A'C= in. Find the 
 lengths of all segments in the figure. 
 
 2. Prove that the quadrilateral whose vertices are the mid- 
 points of the sides of a triangle and one vertex of the triangle, 
 is a parallelogram. 
 
 3. Prove that the midpoints of the sides of a quadrilateral
 
 Ratio, Proportion, Similar Triangles 79 
 
 may be taken as vertices of a parallelogram. Prove that this 
 parallelogram is equal to one-half the quadrilateral. 
 4. State and prove the converse of Proposition VII. 
 
 122. A point P on a segment A B, divides A B into the 
 parts A P and P B. 
 
 A B P 
 
 FIG. 104 
 
 The segment A B may be thought of as described by a 
 moving point, for example, by letting A move to B or by letting 
 B move to A. The segment thus has two directions. If one 
 of the two directions of a segment is considered positive, the 
 other direction is negative. Thus, if A B is positive (+), B A 
 is negative ( ). 
 
 Taking a point P on a segment A B, then (+AP) + 
 (+PB) = (+AB), expresses the fact that the whole is equal 
 to the sum of its parts. If P is on the extension of A B, as in 
 Fig. 104 (+AP) + (PB) = + (AB), since P B is negative, 
 being in direction opposite to that of A P. Thus the equa- 
 tion holds in both cases. Because the sum of A P and 
 P B is A B, A P and P B are called parts of A B, and A B 
 is said to be divided into two parts externally by P. Here 
 the two parts are measured one from A to P, and the other 
 from P to B, the same as when P is between A and B. 
 
 123. Theorem: The bisector of an exterior angle of a tri- 
 angle divides the opposite side externally into segments whose 
 ratio is equal to the ratio 
 
 of the adjoining sides of 
 the triangle. 
 
 Using Fig. 105, follow 
 the proof of Proposition VII. 
 
 State and prove the A 
 converse of this theorem.
 
 8o Second-Year Mathematics 
 
 PROPOSITION VIII 
 
 If two or more ratios are equal, the sum of the antecedents 
 is to the sum of the consequents as any antecedent is to its conse- 
 quent. 
 
 Hypothesis: a/b=c/d=e/f=g/h=. . . . ; 
 
 Conclusion: " ..... =a/b=c/d = . . 
 
 b+d+f+h+ ..... 
 
 Proof: a/b=a/b 
 
 c/d=a/b(?) 
 e/f=a/b(?) 
 g/h=a/b(?) ..... 
 Therefore, ab=ba(?) 
 cb=da(?) 
 <*-/*(?) 
 
 gb=ha(?) ..... 
 
 Adding (a-\-c+e+g+ . .)b = (b+d+f+h + . .)a. (?) 
 a+c+e+g+ . . . a 
 
 EXERCISES 
 
 1. If a/b=c/d, prove that r-r^ = ^/^- 
 
 ~j~(t 
 
 2. Prove that, if a/b=c/d, z=a/c=b/d. 
 
 3. If a/b=c/d, prove that 
 
 ab cd ab cd 
 
 or in words: In a proportion the sum (or difference) of the 
 terms of one ratio is to the antecedent or consequent as the sum 
 (or difference} of the terms of the other ratio is to its antecedent 
 or consequent.
 
 Ratio, Proportion, Similar Triangles 81 
 
 The resulting proportion is said to be obtained from the 
 given proportion by addition if the sum is taken, and by 
 subtraction if the difference is taken. 
 
 The names composition and division are sometimes given 
 to these processes. 
 
 T , ., , , ., . a + b c+d 
 
 4. If a b=c d, prove that r= -. . 
 
 ab ca 
 
 xa+b abx 
 
 5. Apply addition and subtraction to 
 
 x+a b a+b+x ' 
 
 ix 
 
 6. Apply addition and subtraction to - =3. 
 
 V i+x V i x 
 
 7. Solve for x: - ^== - - . (Ans. x= i.) 
 2 V x VX+S YX 
 
 FIG. 1 06 
 
 8. In triangle ABC, - (Fig- 106). 
 
 , BA BC , BA BC 
 Prove that = ; and 
 
 . 
 
 Problems of Construction 
 
 124. In a proportion, as a/b=c/d, d is the fourth propor- 
 tional to a, b, and c; and a is the fourth proportional to d, c, 
 and b. 
 
 i . To construct the fourth proportional to three given seg- 
 ments. 
 
 Given: the segments A B, CD, and E F.
 
 82 
 
 Second- Year Mathematics 
 
 Required: to construct the fourth proportional to A B, 
 C D, and E F. 
 
 a) Algebraic solution: Let x be the fourth proportional. 
 Measure A B, C D, E F, and substitute these numbers in the 
 
 AB EF 
 proportion . 
 
 Solve this equation for x. 
 
 Construct a segment whose measure is x. This is the re- 
 quired fourth proportional. 
 
 b) Geometric solution : On one of two intersecting lines, as 
 P Q, lay off P S=A B, S T = C D (Fig. 107). 
 
 o On the other, as 
 
 PR, lay off PU=EF. 
 Draw SU. 
 
 Draw T X || S U. 
 Then U X is the 
 required fourth propor- 
 
 U 
 
 FIG. 107 
 
 X 
 
 tional. 
 
 Prove by Proposition IV. 
 
 2. To construct the product of two numbers m and n. 
 Notice that i : m = n : mn. 
 
 3. To construct the ratio of two numbers m and n. 
 Observe that n : i = m : m/n. 
 
 4. Find the fourth proportional to i, 2, and 8; to pq, pr, 
 and p. 
 
 5. Solve for *: ^57=4/13. 
 
 /c c i c P~Q P+Q P~Q 
 
 6. Solve for x: x : - - =- * : * . 
 
 pr p-q p-q 
 
 125. In a proportion, as a/b=b/c, c is called the third 
 proportional to a and b, and a is the third proportional to c 
 and b.
 
 Ratio, Proportion, Similar Triangles 
 
 1. Construct the third proportional to two segments 
 
 (a) Algebraically (see 124). 
 (6) Geometrically. 
 
 2 . To divide a segment A B internally and externally in 
 the ratio of m/n, i. e., (i) to find a point, P, on A B so that 
 A P/P B=w/, and (2) to find a point, P', on the extension 
 
 A P' 
 
 of A B so that ^T^-^W/W (see 122). 
 
 I. To divide A B internally in the ratio m/n. 
 Let A B (Fig. 108) be the 
 
 given segment. Draw a line 
 A C through A and lay off 
 A D = m and D E =n. Draw 
 E B. Through D draw 
 D P || E B. Then P divides 
 A B internally in the ratio 
 of m/n. 
 
 II. To divide A B externally in the ratio m/n. 
 
 Draw as before A D =m (Fig. 109) but D E=w. Join E to 
 
 AP' m 
 
 B and draw D P' II E B. Then 
 
 P'B 
 
 Prove. 
 
 FIG. 109 
 
 3. A segment AB = i8 is divided internally or externally 
 
 A P 
 
 at a point P. What is the ratio =r= for A P = 2 ? 3? 6? 9? 
 
 r D 
 
 12? 2O? 24?
 
 84 Second-Year Mathematics 
 
 4. Given a segment AB = i8. To find points P on the 
 
 AP 2 AP . AP AP 3 
 
 line AB so that --; ^=6, ^4, 
 
 5. The sum of two unknown segments is s and the ratio 
 is the same as the ratio of two given segments m and n. Con- 
 struct the unknown segments. 
 
 126. If a segment is divided internally and externally in 
 the same ratio, it is said to be divided harmonically. 
 
 The endpoints of the segment and the two points of 
 division are called four harmonic points. 
 
 If four harmonic points, as A, B, C, D (Fig. no) are joined 
 to a point P outside of line A B, the lines P A, P B, P C, and 
 and P D form an harmonic pencil. 
 
 FIG. no 
 
 EXERCISES 
 
 1 . Prove that the bisector of an angle of a triangle and the 
 bisector of the external angle having the same vertex divide 
 the opposite side harmonically. 
 
 2. Prove that if the segment A B is divided harmonically 
 by points C and D, the segment C D is divided harmonically 
 by the points A and B. 
 
 3. To divide a given segment, A B, into segments propor- 
 tional to several given segments, x, y, and z. 
 
 On a ray,* as A C (Fig. in) lay off x, y, z successively and 
 * A line extending in one direction from a point, is a ray.
 
 Ratio, Proportion, Similar Triangles 85 
 
 join B to the last point of division D. Draw parallels to 
 B D at the points of division. Use Proposition V, p. 75. 
 
 B 
 
 FIG. in 
 
 4. To divide a segment into equal parts (Fig. 112). 
 
 Construction is the same as 
 in Problem 3, using equal seg- 
 ments instead of x, y, and z 
 in Problem 3. 
 
 5. A segment A B =30 cm. 
 A point P is 10 cm from A. 
 Determine a point P' so that A, 
 FIG. 112 P, B, P 7 , are harmonic points. 
 
 6. The segment XY is divided harmonically by the points 
 
 i /i i \ 
 A and B. Prove that =i ^FT + ^5 ( Fi S- IJ 3)- 
 
 Proof: 
 
 XA 
 
 XB 
 
 X 
 
 A 
 
 y 
 
 
 B 
 
 
 AY 
 
 BY 
 
 
 
 
 FIG. 113 
 
 
 AY 
 
 BY 
 
 (Why?) 
 
 XY-XA XB 
 
 -XY 
 
 the 
 
 "XA 
 
 XB' 
 
 XA 
 
 XB ^ 
 
 figure). 
 
 
 
 
 
 
 
 
 XY 
 
 
 XY 
 
 XY 
 
 XY 
 
 
 
 
 
 -1=1 
 
 
 '' 2= XA + 
 
 
 
 
 
 "XA 
 
 XB' 
 
 XB 
 
 = + ' dividin S both sides '
 
 86 Second-Year Mathematics 
 
 7. The segment XY is divided harmonically by the points 
 A and B. Prove that 
 
 i , / i i \ 
 
 BY/ ' 
 
 Method of Analysis 
 
 127. When no obvious solution of a problem presents 
 itself, the following method may be used to discover the solu- 
 tion: 
 
 (1) Assume that the problem is true, and consider results 
 which might follow. Continue deducing other results from 
 these results, until a known proposition or statement is reached 
 (analysis) . 
 
 (2) Then reverse the process, thus obtaining the solution 
 (proof). 
 
 This method, called the method of analysis, is often useful 
 in attacking problems and original propositions. 
 
 1. Prove that if a : b=c : d, then (a+b) : b = (c+d) : d. 
 
 Analysis: 
 
 a+b c+d 
 
 (1) Assume ~~r~ = ~j~ 
 
 CL 
 
 (2) Then ad+bd = bc + bd. 
 
 (3) ad=cb. 
 
 (4) a:b=c:d. 
 
 Proof: 
 
 (1) a:b=c:d. 
 
 (2) ad=cb. 
 
 (3) ad+bd=cb+bd. (?) 
 
 (4) d(a+V)=b(c+d), (?) 
 
 (5) (a+b):b = (c+d):d. (?) Q. E. D. 
 
 2. If a/b=c/d, prove the four following relations: 
 
 (i) c 2 /d 2 =ca/bd.
 
 (2) 
 
 Ratio, Proportion, Similar Triangles 87 
 
 a + c a 2 d 
 
 b+d b*c 
 
 a a +c a == ab+cd 
 w> a-c*~ab-cd' 
 
 . a+b+c+d b+d 
 
 W.^THT : -^~ 
 
 3. If a/b=b/c prove that 
 
 a 2 +6 2 6 2 +c 2 
 (2) 
 
 (3) 
 
 a c 
 
 a+c _a c 
 
 b a +c a ~~b a -c*' 
 
 a a +ab b a +bc 
 
 a c 
 
 4. If a/b=c/d=e/f, prove that 
 
 . a 2 b a 
 
 (2) 
 
 lame+nc_a 
 (3) lb-tnf+nd = b' 
 
 Algebraic Exercises 
 Solve the following exercises algebraically: 
 
 1. The segments of the lines of a pencil, or ray of lines, 
 from P made by a pair of parallels may be designated as 
 shown in Fig. 114. Find x, y, and the lengths of the segments 
 included by the parallels. 
 
 2. A pencil like that of Fig. 114 is cut by two parallels, 
 so that PA=4, PB = 3, PC=s, PD = 7, and AA' = i-;y, 
 B W=x i, C C'=2X 3, and D D / = 2^+i3. 
 
 Find x, y, and the last four segments named.
 
 88 
 
 Second- Year Mathematics 
 
 3. As in Fig. 114, if PA=3, PB=PC = i, PD = n, and 
 y-x, EW=x+y-2, CC'=3-y~3X, and D D' = 
 2y x, find x, y, and the last four segments. 
 
 FIG. 114 
 
 4. Two intersecting lines are cut by a pair of parallels so 
 that segments may be designated as shown in Fig. 115. Find 
 x and the designated segments. 
 
 Can you give any geometrical meaning to the negative 
 value of x ? 
 
 FIG. 115 
 
 5. If the parallels cut so that PA =9 4*, A A 7 =34^, 
 PB = 2o# 3, and BB / =3 4X, find x and the segments of 
 the non-parallels. Draw a figure to illustrate for both values 
 oi x. 
 
 6. With the segments in the same order as in Exercise 5 
 denoted by 3^1, Jx, 8 + 2X, and 2X + i, find x and the 
 four segments.
 
 Ratio, Proportion, Similar Triangles 89 
 
 Similar Triangles 
 
 128. Similar triangles have been denned (see 82-84, 
 First-Year Mathematics) as follows: 
 
 Two triangles having the corresponding angles equal and 
 the ratios of the corresponding sides equal, are similar. 
 
 129. Similar polygons are denned in the same way. Then 
 to prove that two polygons are similar, we must show that the 
 corresponding angles are equal and that the corresponding sides 
 are in proportion. Corresponding angles, sides, or other lines 
 are often called homologous angles, sides, or lines. 
 
 To prove that two triangles are similar, however, it is 
 sufficient to prove that only some of these conditions are 
 satisfied. 
 
 It will be shown that two triangles are similar under any 
 one of the following three conditions: 
 
 (1) If two angles of one triangle are equal each to each 
 to two angles of the other. 
 
 (2) If the ratio of two sides of one triangle equals the ratio 
 of two sides of the other and the angles included by these 
 sides are equal. 
 
 (3) If the corresponding sides are in proportion. 
 
 130. Many problems can be solved by the aid of similar 
 triangles. 
 
 PROBLEMS 
 
 i. Surveyors use similar triangles to lay off parallel lines. 
 Let A C be a given line. /^^ rC 
 
 To construct a line parallel Zh ~I^u 
 
 to AC. 
 
 Lay off a triangle, as A B C 
 (Fig. 1 1 6). FIG. 116
 
 9 o 
 
 Second- Year Mathematics 
 
 On A B mark a point, as D. Determine a point E on 
 
 A B C B 
 
 C B, so that =-rjr =:prw Draw D E. This is the required 
 D B Jb r> 
 
 parallel. 
 
 Triangles D B E and A B C are then similar to each other 
 by case (2), 129. 
 
 Therefore Z.x= Z.y by definition of similar triangles and 
 D E || A C. 
 
 2. A perpendicular to a line from a remote point may be 
 located by similar triangles. 
 
 Let A be the point and B C 
 the line to which the perpen- 
 dicular from A is to be drawn 
 (Fig. 117). Draw AB. From 
 a convenient point D on A B 
 draw D EJ_B C. 
 
 If A F represents the required 
 perpendicular it is seen that triangle B E D^ triangle B F A 
 case (i), 129. Therefore F can be located by the fourth 
 
 DB EB 
 
 proportional in ^~g=p-g . 
 
 PROPOSITION IX 
 
 Theorem: Two triangles are similar if two angles of one 
 are equal to two angles 
 of the other. 
 
 Hypothesis: x=x'. 
 y=/(Fig. 118); 
 
 Conclusion: A ABC 
 w~ AA'B'C'. 
 
 Proof: Prove C=C'. 
 
 Lay off CD=C'A', 
 andCE = C'B'. FIG. 118
 
 Ratio, Proportion, Similar Triangles 
 
 Draw D E. Prove AD E C 
 Prove x"=x. 
 
 Then D E || AB,, . 
 
 = and 
 
 C/A ' C/B/ 
 
 Similarly prove 
 
 CA CB 
 C'B' B'A 
 
 j 
 - and = 
 
 CA"CB 
 C'A' 
 
 (Why?) 
 
 i . Problem : To find the height of a chimney. 
 
 Let A'B' be a vertical stick and A'C' its shadow (Fig. 119). 
 
 C A' 
 
 FIG. 119 
 Let A C be the shadow of the chimney. 
 
 Prove that A B = A'B' ^ . 
 A (^ 
 
 What is the height of the chimney if A'B' =4 ft., A'C' = 
 9ft., AC = io8ft.? 
 
 EXERCISES 
 
 1. Triangles having the corresponding sides parallel, or 
 perpendicular, are similar. Prove. 
 
 2. If lines are drawn joining the midpoints of the sides 
 of a triangle, another triangle is formed which is similar to 
 the first triangle. 
 
 AB 
 .AB 
 
 3. In triangle ABC (Fig. 120) j ^ Jjj 
 
 .V . CF 
 
 P rovethat ED = '
 
 Second- Year Mathematics 
 
 4. The bases of a trapezoid are 18 and 24, and the altitude 
 is 12. Find the area. 
 
 Produce the non-parallel sides 
 until they meet and compare the 
 altitudes of the similar triangles. 
 
 5. The shadow cast by a 4-ft. 
 vertical rod is 5^ ft., and the 
 shadow cast by a spire is 220 ft. 
 How high is the spire ? 
 
 FIG. 120 
 
 6. A man at a window sees a point on the ground in line 
 with the top of a post 2 ft. 8 in. from the foot of the post and 
 finds that the post is 3 ft. high, and 24^ ft. from a point just 
 under the window. How high is the window from the ground ? 
 
 7. A boy wishes to know how far it is from the shore of 
 a lake to an island, and makes the measurements shown in 
 Fig. 121, also D B = io rd. Find the required distance. 
 
 8. The boy now wishes 
 to lay off a half-mile course 
 having the island A as one 
 end. How long must D E 
 be made that the line of 
 vision from B to E shall 
 cut a line parallel to D E 
 through A at a point J mile 
 from A ? 
 
 9. Make a drawing 
 which shall represent two 
 
 stakes in a lake or river, and lines measured on the land by 
 which to find the distance between the stakes. 
 
 10. Make a similar drawing and show how to locate the 
 stakes a given distance apart. 
 
 11. The non-parallel sides of a trapezoid of bases 60 and 
 1 8 and of altitude 6, are produced until they meet. What 
 
 B 
 
 FIG. 121
 
 Ratio, Proportion, Similar Triangles 
 
 93 
 
 are the altitudes of the triangles formed on the bases of the 
 trapezoid ? 
 
 12. The base of a triangle is 72 in., and its altitude is 
 12 in. Find the upper base of the trapezoid cut off by a line 
 parallel to the base of the triangle and 8 in. from it. 
 
 13. The bases of a trapezoid are nj and 5, and the non- 
 parallel sides are 5^ and 3. Find the sides of the smaller tri- 
 angle formed by producing the sides of the trapezoid. 
 
 PROPOSITION X 
 
 Theorem: Two triangles are similar if the ratio of two 
 sides of one equals the ratio of two sides of the other, and the 
 angles included between these sides are equal. 
 
 AC C B 
 Hypothesis: x=x' and = '' 
 
 FIG. 122 
 
 Conclusion: A A B C^ AA'B'C'. 
 
 Proof: Lay off C'D-C A, C'E = C B. Draw D E. 
 
 I C'. (Why ?) 
 
 DC' EC' 
 AfC'^Cm' (Why?) 
 .'. D E || A'B'. 
 
 Prove AD E C^ AA'B'C'. 
 C^ AA'B'C'.
 
 94 Second-Year Mathematics 
 
 EXERCISES 
 
 i. To determine the distance across a river. 
 
 Sighting across the river with telescope A (Fig. 123), place 
 rods at B and C in the line of sight. Take readings of rods at 
 E and D. Depress the telescope sighting at C and take a 
 reading at F. From the rod readings find D F and E C. 
 
 FIG. 123 
 
 Since EC || DF, /^DAF-AEAC, and 
 F P 
 
 ,.AE-AD.|f. 
 
 E D can be found by subtracting A D from A E. 
 
 2. Two isosceles triangles are similar, if an angle in one 
 is equal to the corresponding angle in the other. Prove. 
 
 3. Two right triangles are similar if the ratio of the sides 
 including the right angle in one, is equal to the ratio of the 
 corresponding sides in the other. Prove. 
 
 4. Two sides of a triangle are 14 and 3.5 in. and the in- 
 cluded angle is 75. Two sides of another triangle are 20 and 
 5 and the included angle is 75. Are the triangles similar? 
 Give reason for your answer. 
 
 5. Prove that two parallelograms are similar if they have 
 two corresponding angles equal, and the including sides in 
 proportion. 
 
 6. Two rectangles are similar if the ratio of two adjacent 
 sides of one is equal to the ratio of the corresponding sides 
 of the other.
 
 Ratio, Proportion, Similar Triangles 95 
 
 PROPOSITION XI 
 
 Theorem: Two triangles are similar if the corresponding 
 sides are in proportion. 
 
 AB BC CA 
 
 Hypothesis: _-_ = _ (Fig. 124); 
 
 FIG. 124 
 
 Conclusion: A A B C^ AA'B'C 7 . 
 
 Proof: Layoff CD = C'A' 
 andCE = C'B'. 
 
 DrawDE. Prove AD C E^AA C B, 129. (i) 
 
 A B B C 
 
 : =7=TF (Similar triangles have their sides in proportion.) 
 
 AB BC 
 
 A 7 !"' = B 7 c > * ( y h yp thesls -) 
 
 , f A B _ A B (Magnitudes equal to equal things 
 A'B' D E ' are equal to each other.) 
 
 Prove AD C E^ AA'B'C'. (2) 
 
 Therefore AA B C^ AA'B'C' (from (i) and (2)). 
 
 PROPOSITION XII 
 
 Theorem: The perimeters of similar triangles are to each 
 other as any two homologous sides. 
 
 Denoting the sides of two similar triangles by a, b, c and
 
 Second-Year Mathematics 
 
 a', b', c', respectively; we have a/a'=b/b'=c/c f . Then by 
 
 a+b+c a b c 
 
 Proposition VIII, p. 80, , = = 7- = -. 
 a'+b'+c' a' b' c' 
 
 EXERCISES 
 
 1. The perimeter of a triangle is 15 cm., and the sides of 
 a similar triangle are 4. 5 cm., 6. 4 cm., and 7.1 cm. Find 
 the lengths of the sides of the first triangle. 
 
 2. A race course in the shape of an equilateral triangle is 
 \ mile long. Find the sides of a similar course 80 ft. long; 
 1,760 ft. long; i mile long. 
 
 3. The perimeters of two similar triangles are # 2 +3# + a 
 and 1 6, and a pair of homologous sides are respectively yx 
 and 8. Find the value of x. 
 
 4. The perimeters P and P' of two similar triangles, and a 
 pair of homologous sides a and a' may be expressed in the fol- 
 lowing table. Find the values of x, y, and k. 
 
 0=35 
 a=x 2 
 
 a=x+i 
 
 a' = 2 
 a' =4 
 a' =6 
 ' 
 
 0'=3 
 
 FIG. 125 
 
 = 5 
 
 a =40 a' = 2X b 
 
 5. The homologous alti- 
 tudes of two similar tri- 
 angles are to each other as 
 their homologous sides, or 
 as their perimeters. 
 
 Show that the triangles 
 T and T' (Fig. 125) are 
 similar.
 
 Ratio, Proportion, Similar Triangles 
 
 97 
 
 6. The homologous medians of two similar triangles are to 
 each other as any two homologous sides, and as the perim- 
 eters. See Fig. 126. (amx^a'm'x'.} Prove. 
 
 c' 
 
 FIG. 126 
 
 PROPOSITION XIII 
 
 The perpendicular to the hypotenuse from the vertex of 
 a right triangle divides 
 
 the triangle into triangles 
 similar to each other, and 
 to the given triangle. 
 
 x=x' (?) (Fig. 127) 
 
 z=z' 
 
 y % 
 
 z' 
 
 FIG. 127 
 
 .'. AA D C^ AB D C^ AA B C. 
 
 131. In the proportion a/b=b/c, b is called a mean pro- 
 portional between a and c. 
 
 EXERCISES 
 
 1 . What is a mean proportional between 4 and 9 ? 
 Denoting a mean proportional by x, we have ^/x=x/g. (?) 
 
 x 1 = 4 9 ( ?), x= 6. Prove both values correct. 
 
 2. What are mean proportionals between 2 and 18? 10 
 and 90? 8 and 200? 20 and 180? 
 
 3. Find a mean proportional between a 3 and b 2 ; between 
 c 2 and d 2 . 
 
 4. Find a mean proportional between x 2 + 2xy-\-y 2 and 
 x 2 2xy+y 2 .
 
 9 8 
 
 Second- Year Mathematics 
 
 132. Since triangle ADC (Fig. 127), is similar to triangle 
 
 DEC, it follows that ? ^=^- E . That is 
 JJ U D r> 
 
 In a right triangle, the perpendicular from the vertex of the 
 right angle to the hypotenuse is a mean proportional between 
 the segments of the hypotenuse. 
 
 This affords a way of finding geometrically a mean pro- 
 portional between two numbers. 
 
 PROBLEMS 
 
 i. Construct a mean proportional between any two segments. 
 Let a and b be the given segments. Required to construct 
 a mean proportional, x (Fig. 128). 
 
 X* ' 
 
 *" v. 
 
 x 
 
 
 / 
 
 \ 
 
 
 I/ 
 
 \ 
 \ 
 
 
 / a 
 
 b \ 
 
 
 A C 
 
 L 
 
 FIG. 128 
 
 
 On any line, as A L, lay off A E=a and B C=b. 
 
 At B draw B DA C. 
 
 Draw a circle on A C as a diameter, meeting B D at D. 
 Then B D is the required mean proportional. 
 
 Proof: Assume any angle inscribed in a semicircle to be a R. A. 
 (proved in 163, 2) .'. Z A D C = i R. A., then see 132. 
 
 2. Construct a mean proportional between 25 and 16. 
 
 3. The mean proportional between a and b is the square 
 root of the product of a and b. Prove. 
 
 Problems 3 and i suggest a way of finding the square root 
 of a number geometrically. 
 
 4. Construct the square root of 6 to two decimal places. 
 
 On squared paper lay off in the same way as a and b
 
 Ratio, Proportion, Similar Triangles 
 
 99 
 
 (Problem i) two factors of 6, as 3 and 2, to the scale of i =sum 
 of the sides of 10 small squares. Erect a perpendicular as x in 
 Problem i, and measure this perpendicular. 
 
 5. Find geometrically to two decimal places the square 
 root of 8. 
 
 6. Find geometrically to two decimal places the square 
 root of 5. 
 
 7. In a right-angled triangle either side including the right 
 angle is a mean proportional between the hypotenuse and the 
 adjacent segment of the hypotenuse, made by the perpendicular 
 from the vertex of the right angle to the hypotenuse. 
 
 Proof follows from similarity of the triangles ABC and A C D and 
 of triangles A B C and D B C (Proposition XIII). 
 
 8. In triangle ABC ZC = i R. A., C DA B. A D = 2, 
 D B =30. Find the lengths of A C and C B (Fig. 129). 
 
 The segment of the hypotenuse adjacent to either side is 
 called the projection of that side. 
 
 9. The radius of circle O is 12.5 and a chord, as AC, 
 equals 15. Find the projection of A C upon the diameter 
 through A (Fig. 130). 
 
 C 
 
 30 
 
 AD -B 
 
 FIG. 129 
 
 Similar Polygons 
 PROPOSITION XIV 
 
 Theorem: Homologous diagonals divide similar polygons 
 into similar triangles. 
 
 Hypothesis: AB C D . . . . ^A'B'C'D' .... (Fig. 131); 
 Conclusion: \I^AI
 
 100 
 
 Second- Year Mathematics 
 CD DE 
 
 '.?=/(?) 
 
 CE EF , DE 
 
 ^Er,=^r^f as both ratios equal . 
 
 Therefore AH A IF, etc 
 
 EXERCISES 
 
 133. Solve the following exercises: 
 
 i. The perimeters of similar polygons are to each other as 
 any two homologous sides. 
 
 Denoting the sides of the polygons by a, b, c, . . . . and a', b', c' 
 
 a b c 
 
 . . . . then , = = = Why? 
 
 a b c 
 
 a+b+c .... a b c 
 
 a' + b' + c' a' b' c'' 
 
 2. State and prove the converse of Proposition XIV. 
 The ZBCA (Fig. 132) being a right angle, show the 
 
 following : 
 
 3. c : a=a : m or that a 2 =m c.
 
 Ratio, Proportion, Similar Triangles 
 
 101 
 
 4. c : b=b : n and b*=n c. 
 
 5. a a -\-b a = (m+n)cc a (Add Ax. applied to Exercises 
 3 and 4). 
 
 6. And a 2 /b 2 =m/n. (Div. Ax.). 
 
 7. Also a a =c a -b 3 and b 2 =c*-a 2 . (?) 
 
 8. It has been shown that m:h=h:n or h = -\/mn. (?) 
 
 9. From Exercise 5 and Exercise 7 it follows that in a 
 right triangle the square of the hypotenuse is equal to the sum 
 of the squares of the other two sides, and the square of either 
 side about the right angle is equal to the square of the hypote- 
 nuse less the square of the other side. 
 
 10. a = i2and& = 5. Find C 
 c, m, n, and h (Fig. 132). 
 
 11. a=8andc = io. Find 
 b, m, n, and h. 
 
 12. w=9| and = 5l- 
 Find a, b, c, and h. 
 
 Solve Exercise 12 first for 
 h by Exercise 8, then for a by Exercise 5, a a =m a +h a , and 
 check results in Exercises 3, 4, and 6. 
 
 13. Find geometrically a line-segment whose length squared 
 
 shall be f of the 
 square of the length 
 of a given segment 
 (i.e., the ratio of their 
 squares shall be 5 : 3). 
 Draw a line as A B, 
 and divide it into 
 
 A segments = 5 to 3 
 (Fig. 133). 
 
 8 
 
 FIG. 133 
 
 Draw semicircle, 
 and perpendicular. Lay off C D the length of the given line.
 
 102 
 
 Second- Year Mathematics 
 
 Draw D E || A B. C E is the required line. 
 
 Proof: CE : CD = CB : C A. (?) 
 
 CE 2 : CD 2 =CB 2 : C A 2 . (?) 
 
 But C B 2 : C A 2 =w : n = $ : 3 (see Exercise 6). 
 
 CE 2 :CD 2 = 5:s. Hence C E 2 =f C D 2 . 
 
 14. Solve Exercise 13 when C D is longer than C A. 
 
 15. Solve Exercise 13 when the ratio is 7 : 4. 
 
 Similar Right Triangles, Trigonometry 
 134. Solve the following problems: 
 
 1. All right triangles having an angle of 30 are similar. 
 Why? 
 
 2. On squared paper draw a right triangle having an angle 
 of 30. Measure the sides, and find the ratio of the side 
 opposite the angle 30 to the hypotenuse. 
 
 Prove that this ratio is the same for all right triangles 
 having an angle of 30. 
 
 3. In the triangle of Problem 2, find the ratio of the side 
 opposite the angle 60 to the hypotenuse. 
 
 Prove that this ratio is constant for all right triangles 
 having an angle of 60. 
 
 4. In a right triangle having an angle of 45, find the ratio 
 of the side opposite the angle 45 to the hypotenuse. 
 
 Prove that this 
 
 Mr ratio is constant for all 
 
 right triangles having 
 an angle of 45. 
 
 5. Draw with a 
 protractor an angle 
 of 40. From any 
 points, as A x , A 2 , A 3 , 
 A 4 . . . . (Fig. 134)
 
 Ratio, Proportion, Similar Triangles 103 
 
 on either side of the angle, draw perpendiculars to the other 
 side. Measure A^, and A t O and find their ratio. 
 
 6. Prove that the ratio of the side opposite the angle 40 
 to the hypotenuse is the same for all triangles of Fig. 1 34. 
 
 135. The constant ratio of the opposite side to the hypote- 
 nuse as in Fig. 1 34, is called the sine of angle 40. 
 
 1. Find the sine of 70; of 50; of 20. 
 
 2. On square-ruled paper, draw an acute angle whose 
 sine is . Find the number of degrees in the angle. 
 
 3. Find the number of degrees in an acute angle whose 
 sine is |; .2; .75. 
 
 136. Many problems in science, and in surveying and other 
 branches of engineering, are solved by the aid of the ratios of 
 the sides of right triangles. 
 
 Let AOD (Fig. 135) 
 represent any acute angle. 
 Construct a right triangle 
 containing the angle AOD, 
 by drawing from any point 
 
 P, on either side, a line ^ 
 
 PM perpendicular to the ft P D 
 
 other side. FlG - J 35 
 
 The ratio of M P to O P is called the sine of angle O 
 (written sin O). 
 
 The ratio of M O to O P is called the cosine of angle O 
 (written cos O). 
 
 The ratio of M P to O M is called the tangent of angle O 
 (written tan O). 
 
 EXERCISES 
 
 i . Prove that the sine of a given acute angle O is the same 
 for all right triangles that have the angle O, i. e., that the 
 sine of O does not depend upon the size of the right triangle 
 that contains the angle O.
 
 IO4 
 
 Second-Year Mathematics 
 
 2. Prove that the cosine of a given acute angle is constant. 
 
 3. Prove that the tangent of a given acute angle is constant. 
 
 137. It is customary to denote the angles of a triangle by 
 capital letters, and the sides opposite by the corresponding 
 small letters. Let ABC (Fig. 1 36) be any right triangle 
 (C being the right angle), and let a, b, c denote the lengths of 
 the sides opposite the angles A, B, C, respectively. 
 
 The definitions of 136 may then be stated briefly as follows: 
 
 a opposite side 
 sin A=-=-p- 
 
 c hypotenuse 
 
 b adjacent side 
 cos A =- = -T^- . 
 
 c hypotenuse 
 
 a opposite side 
 
 tan A=T=-^T 7-r- . 
 
 b adjacent side 
 
 138. It is important to note that the sine, cosine, and tan- 
 gent of an angle are numbers, the ratios or quotients of certain 
 lengths. 
 
 The ratios have definite, fixed values for any given angle, 
 i. e., they are constant numbers for a given angle. 
 
 The sine, cosine, and tangent of an angle, as defined in 
 137, for acute angles cannot be negative, since no one of the 
 numbers a, b, c (Fig. 136) is negative. 
 
 i. Give the sine, cosine 
 (Fig. 136). 
 
 PROBLEMS 
 :, and tangent of the acute angle B
 
 Ratio, Proportion, Similar Triangles 
 
 2. In Fig. 136, show that 
 
 (1) sin A=cos B 
 
 (2) cos A=sin B 
 
 (3) tanA = 
 
 i 
 
 tan B 
 
 sm A 
 
 (4) tan A = . 
 
 cos A 
 
 3. Show that the sine of an angle is less than (or at most 
 equal to) i. 
 
 In Fig. 136, c*=a*+b 2 (133, Problem 9, and 142, 
 First-Year Mathematics). 
 Hence, a <c. Why? 
 Therefore, sin A =a/c< i . Why ? 
 
 4. Show that the cosine of an angle is less than i . 
 
 5. On square-ruled paper, with a protractor, draw angles 
 of 40, 50, 25, 65. Find by measuring the lines, the sine, 
 cosine, and tangent of each angle and compare the results 
 with the values in the following table : 
 
 
 25 
 
 s 
 
 40 
 
 65 
 
 Sine 
 
 .4226 
 
 .7660 
 
 .6428 
 
 .0063 
 
 Cosine 
 
 .0063 
 
 .6428 
 
 . 7660 
 
 .4226 
 
 Tangent 
 
 .466? 
 
 i . 1918 
 
 8^01 
 
 2 . I44 1 } 
 
 
 
 
 
 
 6. The lengths of the sides of a right triangle are repre- 
 sented by a, b, and c, the hypotenuse being c. Find the sine, 
 cosine, and tangent of angle A and of angle B under each of 
 the following conditions: 
 
 (1) a=3, 6=4 (3) 6 = 15, c = i7 
 
 (2) a = s, = 13 (4) a = 7, = 25. 
 
 7. On square-ruled paper draw the angle whose sine is . 
 Find the cosine and tangent of the angle, and the number of 
 degrees in the angle. 
 
 8. Draw the angle whose cosine is . Find the sine and 
 tangent of the angle, and the number of degrees in the angle.
 
 io6 
 
 Second- Year Mathematics 
 
 9. Draw the angle whose tangent is i. Find the sine and 
 cosine of the angle, and the number of degrees in the angle. 
 
 10. Draw an equilateral triangle 
 (ABC, Fig. 137). 
 
 Prove that the altitude B H bisects 
 A C and the Z. 
 
 Letting # represent the length of AH, 
 show that A B = 2#, andB H=#j/3. 
 
 1 1 . In the right triangle A H B 
 
 A * H C 
 
 (Fig. 137), show that the sine, cosine, 
 and tangent of 30, and of 60, have 
 the values indicated in the following table : 
 
 H 
 
 FIG. 137 
 
 
 
 Sine 
 
 Cosine 
 
 Tangent 
 
 
 
 
 * 
 
 i / 
 
 i / * 
 
 
 
 _ 
 
 vv 5 
 
 3V 
 
 
 60 
 
 4i/3~ 
 
 i 
 
 / 
 
 
 
 
 
 V 6 
 
 * Prov 
 
 e that -7= = $V 
 
 '3- 
 
 
 
 Compare these values with the results found in Problems 7 
 and 8. A B 
 
 12. Draw an isosceles right triangle 
 (ABC, Fig. 138). 
 
 Prove that Z A = /_ B =45. 
 
 Letting x represent the length of each of 
 the equal sides, show that A E=x\/2. 
 
 13. From Fig. 138, show that the sine, cosine, and tangent 
 of 45 have the values indicated in the following table: 
 
 x 
 
 
 Sine 
 
 Cosine 
 
 Tangent 
 
 4<; . 
 
 4 i/a* 
 
 4 1/2* 
 
 I 
 
 
 
 
 
 * Prove that = 4. 1/2. Then since 1/2 = i .4143 + , ^ or J 
 
 1/2 |/ 2 
 
 .7071 +. Compare these values with results found in Problem 9.
 
 TABLE OF SINES, COSINES, AND TANGENTS OF 
 ANGLES FROM i-8g 
 
 Angle 
 
 Sine 
 
 Cosine 
 
 Tangent 
 
 Angle 
 
 Sine 
 
 Cosine 
 
 Tangent 
 
 l 
 
 0175 
 
 .9998 
 
 OI7S 
 
 46 
 
 7193 
 
 .6947 
 
 1-0355 
 
 2 
 
 349 
 
 9994 
 
 0349 
 
 47 
 
 7314 
 
 .6820 
 
 1.0724 
 
 3 
 
 0523 
 
 .9986 
 
 .0524 
 
 48 
 
 743 1 
 
 .6691 
 
 I .1106 
 
 4 
 
 .0698 
 
 .9976 
 
 .0699 
 
 49 
 
 7547 
 
 .6561 
 
 1.1504 
 
 5 
 
 .0872 
 
 .9962 
 
 .0875 
 
 50 
 
 .7660 
 
 .6428 
 
 1 .1918 
 
 6 
 
 .1045 
 
 9945 
 
 .1051 
 
 5i 
 
 .7771 
 
 .6293 
 
 1-2349 
 
 7 
 
 .1219 
 
 9925 
 
 .1228 
 
 52 
 
 .7880 
 
 6l57 
 
 1.2799 
 
 8 
 
 .1392 
 
 993 
 
 .1405 
 
 53 
 
 .7986 
 
 .6018 
 
 1.3270 
 
 9 
 
 .1564 
 
 9877 
 
 .1584 
 
 54 
 
 .8090 
 
 5878 
 
 I-37 6 4 
 
 10 
 
 1736 
 
 .9848 
 
 I 7 6 3 
 
 55 
 
 .8192 
 
 5736 
 
 I .4281 
 
 ii 
 
 .1908 
 
 .9816 
 
 .1944 
 
 56 
 
 .8290 
 
 5592 
 
 1.4826 
 
 12 
 
 .2079 
 
 .9781 
 
 .2126 
 
 57 
 
 8387 
 
 5446 
 
 1-5399 
 
 13 
 
 .2250 
 
 9744 
 
 2309 
 
 58. 
 
 .8480 
 
 5299 
 
 i . 6003 
 
 14 
 
 .2419 
 
 973 
 
 2493 
 
 59 
 
 8572 
 
 5150 
 
 1.6643 
 
 15 
 
 .2588 
 
 9 6 59 
 
 .2679 
 
 60 
 
 .8660 
 
 .5000 
 
 1.7321 
 
 16 
 
 .2756 
 
 .9613 
 
 .2867 
 
 61 
 
 .8746 
 
 .4848 
 
 i . 8040 
 
 i? 
 
 .2924 
 
 9563 
 
 357 
 
 62 
 
 .8829 
 
 4695 
 
 1.8807 
 
 18 
 
 .3090 
 
 95 11 
 
 3249 
 
 63 
 
 ..8910 
 
 4540 
 
 i .9626 
 
 iQ 
 
 3 2 56 
 
 945-5 
 
 3443 
 
 64 
 
 .8988 
 
 4384 
 
 2.0503 
 
 20 
 
 .3420 
 
 9397 
 
 .3640 
 
 65 
 
 .9063 
 
 .4226 
 
 2 1445 
 
 21 
 
 .3584 
 
 9336 
 
 3839 
 
 66 
 
 9i35 
 
 .4067 
 
 2 . 2460 
 
 22 
 
 3746 
 
 .9272 
 
 .4040 
 
 67 
 
 .9205 
 
 397 
 
 2-3559 
 
 2 3 
 
 397 
 
 .9205 
 
 4245 
 
 68 
 
 .9272 
 
 3746 
 
 2-4751 
 
 24 
 
 .4067 
 
 9i35 
 
 4452 
 
 69 
 
 9336 
 
 3584 
 
 2 . 605 I 
 
 25 
 
 .4226 
 
 .9063 
 
 .4663 
 
 70 
 
 9397 
 
 .3420 
 
 2-7475 
 
 26 
 
 4384 
 
 .8988 
 
 .4877 
 
 7i 
 
 9455 
 
 3256 
 
 2 .9042 
 
 27 
 
 4540 
 
 .8910 
 
 595 
 
 72 
 
 95 11 
 
 .3090 
 
 3-0777 
 
 28 
 
 4695 
 
 .8829 
 
 53i7 
 
 73 
 
 9563 
 
 .2924 
 
 3.2709 
 
 29 
 
 .4848 
 
 .8746 
 
 5543 
 
 74 
 
 .9613 
 
 .2756 
 
 3-4874 
 
 30 
 
 .5000 
 
 .8660 
 
 5774 
 
 75 
 
 9 6 59 
 
 12588 
 
 3-7321 
 
 31 
 
 515 
 
 8572 
 
 .6009 
 
 76 
 
 9703 
 
 .2419 
 
 4.OIO8 
 
 32 
 
 5 2 99 
 
 .8480 
 
 .6249 
 
 77 
 
 9744 
 
 .2250 
 
 4-33 I 5 
 
 33 
 
 5446 
 
 .8387 
 
 .6494 
 
 78 
 
 .9781 
 
 .2079 
 
 4 7046 
 
 34 
 
 5592 
 
 .8290 
 
 6745 
 
 79 
 
 .9816 
 
 .1908 
 
 5 J 446 
 
 35 
 
 573 6 
 
 .8192 
 
 .7002 
 
 80 
 
 .9848 
 
 1736 
 
 5-6713 
 
 36 
 
 .5878 
 
 .8090 
 
 .7265 
 
 81 
 
 .9877 
 
 .1564 
 
 6-3138 
 
 37 
 
 .6018 
 
 .7986 
 
 7536 
 
 82 
 
 9903 
 
 .1392 
 
 7-"54 
 
 38 
 
 6i57 
 
 .7880 
 
 7813 
 
 83 
 
 9925 
 
 .1219 
 
 8.1443 
 
 39 
 
 6293 
 
 .7771 
 
 .8098 
 
 84 
 
 9945 
 
 .1045 
 
 9-5 I 44 
 
 40 
 
 .6428 
 
 .7660 
 
 .8391 
 
 85 
 
 .9962 
 
 .0872 
 
 11.4301 
 
 4i 
 
 .6561 
 
 7547 
 
 .8693 
 
 86 
 
 .9976 
 
 .0698 
 
 14.3006 
 
 42 
 
 .6691 
 
 743 1 
 
 .9004 
 
 87 
 
 .9986 
 
 523 
 
 19.0811 
 
 43 
 
 .6820 
 
 73 J 4 
 
 93 2 5 
 
 88 
 
 9994 
 
 0349 
 
 28.6363 
 
 44 
 
 .6947 
 
 7 r 93 
 
 9657 
 
 89 
 
 .9998 
 
 0175 
 
 57.2900 
 
 45 
 
 7071 
 
 .7071 
 
 I . OOOO 
 
 
 

 
 io8 Second-Year Mathematics 
 
 139. The preceding table gives the values, correct to four 
 decimal places, of the sine, cosine, and tangent of angles i 
 to 89, inclusive. 
 
 Notice how the sine, cosine, and tangent change as the 
 angle increases from i to 89. 
 
 Problems on Right Angles 
 
 In measuring angles needed for the solution of problems 
 in science and engineering, the surveyor's compass (Fig. 139) 
 and the engineer's transit (Fig. 140) are used. 
 
 FIG. 139 
 
 140. Imagine E H (Fig. 141) to be rotated (raised or 
 elevated) in the plane about E to the position E A. The angle 
 x, thus formed, is the angle of elevation of A from E. If E H 
 is rotated (lowered or depressed) to the position E B, the angle 
 y, thus formed, is the angle of depression of B from E.
 
 Ratio, Proportion, Similar Triangles 109 
 
 FIG. 140 
 
 *^,'y 
 
 H 
 
 FIG. 141
 
 Second- Year Mathematics 
 
 141. In the following problems find the unknown parts of 
 the triangles, first from a scale drawing, and then by the aid 
 of the sine, cosine, and tangent of the angles. 
 
 i . The rope of a flag-pole is stretched out so that it touches 
 the ground at a point 20 ft. from the foot of the pole, and 
 makes an angle of 73 with the ground. Find the height of 
 the flag-pole. 
 
 With a ruler and protractor, draw the 
 right triangle to scale (Fig. 142). By measure- 
 ment, x is found to represent 66 ft. nearly. 
 
 More accurate results can be obtained by 
 using the tangent of Z.A, thus: 
 
 x . 
 
 = tan 7V 
 20 
 
 = 3. 2 709, from the table on p. 107. 
 
 FIG. 142 
 
 Therefore # = 20X3.2709=65.418. 
 
 The angle A (Fig. 142) is the angle of eleva- 
 tion of the top of the flag-pole from the point 
 A (see 140). 
 
 2. A balloon is anchored to the ground by a rope 260 ft. 
 long, making an angle A of 67 with the ground. If the rope- 
 line is straight, what is the height of the balloon? 
 
 Use the sine of angle A ( 137). 
 
 3. A kite-string 300 ft. long is fastened to a 
 stake. The distance from the stake to a point 
 directly under the kite is 102.6 ft. Find the 
 height of the kite, supposing the kite-string to 
 be straight. 
 
 What is the angle of elevation of the kite from 
 the stake ? 
 
 /J 102.6 C 
 
 FIG. 143 
 
 Draw the right triangle to scale (Fig. 143) and measure 
 a and /.A.
 
 Ratio, Proportion, Similar Triangles in 
 
 102.6 
 
 cos A= =.3420. 
 300 
 
 But cos 70= .3420, from the table, p. 107. 
 .*. A = 70, the angle of elevation of the kite. 
 
 =sin 70= .9397, from p. 107. 
 300 
 
 Therefore a = 300 X . 9397 = 281.91. 
 
 The value of a can also be obtained from the equation 
 
 a = l/3oo 2 102. 6 2 (See 133, Problem 9.) 
 
 4. A vertical pole, 8 ft. long, casts on level ground a shadow 
 9 ft. long. Find the angle of elevation of the sun. 
 
 Use the tangent ratio. 
 
 5. The angle of elevation of an aeroplane at a point A on 
 level ground, is 60. The point C on the ground directly 
 under the aeroplane is 300 yd. from A. Find the height of 
 the aeroplane. 
 
 6. What is the angle of elevation of the top of a hill 
 5oo]/3 ft. high, at a point in the plain 1,000 ft. from the top 
 of the hill ? 
 
 7. What is the angle of elevation of a road that rises i ft. 
 in a distance on the road, of 50 feet ? 
 
 8. A road makes an angle of 6 with the horizontal. How 
 much does the road rise in a distance of 100 ft. along the 
 horizontal ? 
 
 9. On a tower is a search-light 140 ft. above sea-level. The 
 beam of light is depressed (lowered) from the horizontal, 
 through an angle of 20, revealing a passing boat. How far 
 is the boat from the base of the tower ? 
 
 142. We know that all right triangles in which the following 
 parts are equal, each to each, are congruent: 
 I. The two sides including the right angle.
 
 ii2 Second-Year Mathematics 
 
 II. A side and one acute angle. 
 
 III. The hypotenuse and one of the other sides. 
 
 143. In other words, if in a right triangle two parts in 
 addition to the right angle are given (at least one being a side), 
 the triangle is completely determined, and may be constructed 
 from these given parts. 
 
 144. The problems of 141 show that when two parts of 
 a right triangle are given (as indicated in 142), the remaining 
 parts may be computed by the methods of scale drawing or by 
 using the sine, cosine, and tangent of the angles. 
 
 Which method gives the more accurate results ? 
 
 145. The branch of mathematics which shows how to 
 solve triangles by the aid of the sine, cosine, and tangent of 
 the angles is called trigonometry. The ratios, sine, cosine, and 
 tangent are called trigonometric ratios. An angle has three 
 other ratios, but they will not be used in this book. 
 
 PROBLEMS 
 
 146. Solve the following problems: 
 
 1 . A boat passes a tower on which is a search-light 1 20 ft. 
 above sea-level. Find the angle through which the beam of light 
 must be depressed, from the horizontal, so that it may shine 
 directly on the boat when it is 400 ft. from the base of the tower. 
 
 2. From the top of a cliff 150 ft. high, the angle of depres- 
 sion of a boat is 25. How far is the boat from the top of 
 the cliff? 
 
 3. When an aeroplane is directly over a town C the angle 
 of depression of town B, 2\ miles from C, is observed to be 
 10. Find the height of the aeroplane. 
 
 4. From an aeroplane, at a height of 600 ft., the angle of 
 depression of another aeroplane, at a height of 150 ft., is 39. 
 How far apart are the two aeroplanes ?
 
 Ratio, Proportion, Similar Triangles 113 
 
 5. Two persons, 1,200 ft. apart, observe an aeroplane 
 directly over the straight line from one to the other. One 
 person finds the angle of elevation of the aeroplane to be 35; 
 the other, at the same time, from his position, finds it to be 
 55. Find the height of the aeroplane. 
 
 6. On the top of a 'tower stands a flagstaff. At a point A 
 on level ground, 50 ft. from the base of the tower, the angle 
 of elevation of the top of the flagstaff is 35. At the same 
 point A the angle of elevation of the top of the tower is 20. 
 Find the length of the flagstaff. 
 
 7. A balloon rises vertically from a point C, on level ground. 
 At a point A, 200 yd. from C, the balloon is observed when 
 the angular elevation at A is 40. Six minutes later the angular 
 elevation at A is 60. Find the altitude of the balloon at the 
 time of the second observation, and the number of yards it 
 rises per second, supposing the motion to be uniform. 
 
 8. From a point A on the bank of a river flowing due east 
 the angle of elevation of the top of a tree on the other side 
 is 45. At a point B, 70 yd. south of A, the angular eleva- 
 tion is 30. Find the width of the river. 
 
 9. At a window 20 ft. from the ground, the angle of de- 
 pression of the base of a tower is 15, and the angle of eleva- 
 tion of the top of the tower is 37. What is the height of the 
 tower ? 
 
 10. A boy wishes to determine the height H K of a fac- 
 tory chimney. He places a transit first at B and then at A 
 and measures the angles x and y. The transit is on a tripod, 
 3^ ft. from the ground. A and B are two points in line with 
 the chimney and 50 ft. apart. What is the height of the chim- 
 ney if the ground is level and if #=63 and ;y =33^ ? (Fig. 144.) 
 
 Show how the height of the chimney could be readily found by 
 measuring shadow-lengths, without using angles. One method would 
 thus furnish a check on the other.
 
 Second-Year Mathematics 
 
 In Fig. 145 
 
 w/z=tan 63 = i .9626 
 
 w 
 
 =tan 33^ = .6620 
 
 (i) 
 
 (2) 
 
 K 
 
 7 63 C 
 
 FIG. 144 
 
 50 3 
 
 FIG. 145 
 
 I 
 
 (i) and (2) are simultaneous equations in which w and z 
 are the unknown numbers. To eliminate w, by substitution, 
 we have, from (i) 
 
 7^ = 1.96262 (from (i)) (3) 
 
 By substituting (3) in (2), 
 i .96262 
 
 = .6620. (from (2)) (4) 
 
 Substitute this value of z in (3), 
 
 50+2 
 
 Find the value of z in (4) . 
 thus obtaining the value of w. 
 
 11. A balloon is directly over a straight road. The angles 
 of depression of two buildings on the road are 34 and 64. 
 If the buildings are 65 yards apart, how high is the balloon? 
 
 12. From a lighthouse, situated on a rock, the angle of 
 depression <of a ship is 12, and from the top of the rock, it is 8. 
 
 'The height of the lighthouse above the rock is 45 ft. Find 
 the distance of the ship from the rock. 
 
 13. From an aeroplane the angles of depression of the top 
 and bottom of a flag-pole 55 ft. high, are 45 and 157, respec- 
 tively. Find the height of the aeroplane. 
 
 14. Village B (Fig. 146) is due north of village C. An 
 army outpost is located at a point A, 8 miles due west of C.
 
 Ratio, Proportion, Similar Triangles 
 
 B bears 60 east of north from A. An aeroplane is observed 
 to fly from C to B in a quarter of 
 an hour. Find the average hori- 
 zontal speed of the aeroplane. 
 
 15. To measure the width of a 
 river flowing due east, a man selects 
 a point A from which a tree at C, 
 on the other side bears 60 east of 
 north. He then walks east from A 
 
 A 
 
 C 
 
 FIG. 146 
 
 until he finds a point B from which C bears 30 west of north. 
 A B is found to be 300 yards. Find the width of the river, 
 
 CH (Fig. 147). 
 
 Show that x = 150, and 
 
 A 
 
 H B 
 
 FIG. 147 
 
 1 6. Two aeroplanes start from 
 city C at the same time. Aero- 
 plane A flies south at the average 
 rate of 15 miles an hour. Aero- 
 
 plane B flies west. At the end of of an hour, aeroplane B 
 is observed to bear 51^ west of north from aeroplane A. How 
 far apart are the aeroplanes at the time of observation? 
 What is the average speed of aeroplane B ? 
 
 147. Problems on isosceles triangles may be solved by 
 using the two right triangles into which an altitude line from 
 the vertex- angle divides it. 
 
 PROBLEMS 
 
 1 . The distance from a cannon to a straight road is 7 miles. 
 If the range of the cannon is 10 miles, what length of the road 
 is commanded by the cannon ? 
 
 Show that ABC (Fig. 148) is an isosceles triangle, and that A H 
 bisects B C. In the right triangle A B H, find the length of B H. 
 
 2. The arms of a pair of compasses are opened to a dis-
 
 n6 
 
 Second- Year Mathematics 
 
 tance of 6.25 cm. between the points. If the arms are 
 11.5 cm. long, what angle do they form ? 
 
 In the isosceles triangle ABC (Fig. 149) draw the altitude A H. 
 Find the number of degrees in x. 
 
 10 
 
 B H C 
 
 FIG. 148 
 
 3. A pair of compasses is opened to an angle of 50. What 
 is the distance between the points if the arms are 12. 5 cm. 
 long? 
 
 Draw the altitude of the isosceles triangle. 
 
 4. A cannon with a range of 1 1 miles can shell a length of 
 13 miles of a straight road. How far is the cannon from the 
 road? 
 
 5. A clock-pendulum, 20 inches long, swings through an 
 angle of 6. Find the length of the straight line between the 
 farthest points which the lower end reaches. 
 
 6. A clock pendulum is 25 in. long. Through what angle 
 does the pendulum swing if the distance between the farthest 
 points which the lower end reaches is 6 in. ? 
 
 7. Two firemen are playing a stream of water on the wall 
 of a burning building from a fire-hose which throws water 
 1 20 ft. The distance on the ground from the firemen to the 
 wall is 100 ft. What is the greatest distance on the wall 
 which can be reached by the water ? 
 
 Relations of Trigonometric Ratios 
 
 148. Important relations between the sine, cosine, and 
 tangent of an angle can be shown by simple formulas.
 
 Ratio, Proportion, Similar Triangles 
 
 117 
 
 i. Prove that if A is any acute angle 
 (sin A) 2 + (cos A) 2 = i. 
 
 In Fig. 150 sin A=a/c 
 
 cos A=b/c 
 
 Squaring (i) and (2), 
 
 (sin A) 2 =a 2 /c 2 
 (cos A) 2 =& 2 /c 2 
 
 Adding (3) and (4), 
 
 (sin A) 2 + (cos A) 2 = 
 
 B 
 
 a 
 
 (i) 
 
 (3) 
 (4) 
 
 (5) 
 
 A b C 
 
 FIG. 150 
 
 In Fig. 150, show that a 2 +6 2 =c 2 . 
 
 Show that (5) may be written (sin A) 3 + (cos A) 2 = i. (6) 
 (sin A) 2 is usually written sin 2 A: similarly (cos A) 2 and 
 (tan A) 2 are written cos 2 A and tan 2 A. 
 
 2. In Fig. 150, prove that sin 2 B+cos 2 B = i. 
 
 3. Using the formula sin 2 #+cos 2 x=i, show that 
 
 sin xV-i cos 2 x* (i) 
 
 and cos x = l/i sin 2 x (2) 
 
 4. The sine of an angle x is f . Find the cosine of x. 
 Use formula (2), Problem 3. 
 
 5. The cosine of an angle A is . Find the sine of A. 
 
 * The double sign is not used before the root sign because a negative 
 sine or cosine have no meaning under the foregoing definitions for acute 
 angles (138).
 
 1 1 8 Second- Year Mathematics 
 
 6. From Fig. 150, show that 
 
 sin A 
 
 tan A= , and 
 
 cos A 
 
 sin B 
 
 tanB= 
 
 cos B 
 
 7. From the formulas of Problems 3 and 6, find cos y and 
 tan y, if sin y = %. 
 
 Problems and Exercises 
 
 149. The sine, cosine, and tangent of an angle are called 
 functions of the angle. 
 
 Find the values of the other two functions here used, when 
 
 1. sin B =0.5 4. cosB = j/3 7. sin B=o.6 
 
 2. sinB=J]/3 5. sin B = Jj/2 8. cos B=o.6 
 
 3. cosB=J 6. cosB = |]/2 9. cos B =0.8 
 10. Find the values of the other functions, when tan B=f. 
 From Problem 6, 
 
 sin B , 
 
 C -oiB =f (I) 
 
 From Problem 2, 
 
 sin 2 B+cos 2 B = i (2) 
 
 Equations (i) and (2) are to be solved as simultaneous 
 
 equations in the two unknowns, sin B and cos B. Sin B is 
 eliminated by substitution, as follows: 
 
 From (i) sin B=f cos B (3) 
 
 Substituting (3) in (2) T 9 T cos 2 B+cos 2 B N = i (4) 
 
 Clearing of fractions 9 cos 2 B + i6 cos 2 B = i6 (5) 
 
 25 cos 2 B = i6 (6) 
 
 cos 2 B =|| (7) 
 
 cos B=f (8) 
 From equation (3) sin B =f and cos B =f .
 
 Ratio, Proportion, Similar Triangles 119 
 
 Find the values of the other functions, when: 
 
 11. tanB=f 14. tanB=Jj/3 17. cosB=J 
 
 12. tanB = i/3 15. tanB = 2 18. sinB=i 
 
 13. tanB = i 16. tan 6=0.5 T 9- cosB=w 
 150. The two fundamental relations 
 
 sin 2 A+cos 2 A = i 
 
 sin A 
 
 tanA = . 
 cos A 
 
 are true for any angle. They are therefore called identities, 
 and are sometimes written 
 
 sin 2 A+cos 2 A=i (i) 
 
 _sin A 
 
 tan A= (2) 
 
 cos A 
 
 The symbol, = , is read is, or is identical with. 
 These fundamental trigonometric identities may be used 
 in proving other relations between the sine, cosine, and tan- 
 gent of any angle. 
 
 EXERCISES 
 i 
 
 i. Prove: i+tan 2 A= 
 From formula (2) tan 2 A = 
 
 Therefore i + tan 2 A= i + 
 
 COS 2 A 
 sin 2 A 
 cos 2 A 
 sin 2 A 
 
 cos 2 A 
 _cos 2 A+sin 2 A 
 cos 2 A 
 
 EE , from formula (i). 
 
 cos 3 A ' 
 
 Prove the following identities: 
 sin A 
 
 2. cos A 
 
 tan A ' 
 
 cos A 
 
 3. tan A -: -=i. 
 
 sin A
 
 i2o Second-Year Mathematics 
 
 cos A tan A sin A ' 
 
 5. cos A tan A - r= 
 
 sm A 
 
 i 
 
 6. sm A - r^cos A. 
 
 tan A 
 
 i 
 
 '' tan 2 A sin 2 A 
 
 8. sin A+cos A= (i +tan A) cos A. 
 
 i . i 
 
 sm A cos A 
 
 j. . T sui n. zzn_ua n . 
 
 sm A tan A 
 
 1 -EEtanA+- I 
 
 i\j. T- - . . 10,11. 'i-T^T IT 
 
 cos A sin A tan A 
 
 n. 1/1 sin 2 A^sin A . 
 
 tan A 
 
 12. tan A cos AEEI/I cos 2 A. 
 
 i i 
 
 tan 2 A~sin 2 A ' 
 
 15. (i+tan 2 A)(i sin 2 A) = i. 
 
 i cos A 
 
 16. 
 
 cos A i tan A cos A ' 
 
 i 
 
 17. -- r sm A tan A^cos A. 
 cos A 
 
 18. tanAliH -- r ) 
 
 \ tan A/ 
 
 IQ ___ I __ _ 
 
 LCOS A sin AJL tan A 
 
 r i __ * ir , i i 
 
 LCOS A sin AJL tan Aj ' 
 
 151. Equations involving the trigonometric ratios of an 
 unknown angle are called trigonometric equations. The value 
 of the unknown angle in such equations may be found by
 
 Ratio, Proportion, Similar Triangles 121 
 
 algebraic methods, aided by the fundamental identities in 
 150. 
 
 PROBLEMS 
 
 1. Find the value of x in cos 2 # = J. 
 Taking the square root 
 
 cos x = %. 
 But cos 60 =J, by Problem n, p. 106. 
 
 Therefore, the original equation is satisfied if 60 be put 
 in place of x; that is, x=6o is a solution of the equation. 
 
 2. Find the value of xm4cos 2 x=^. 
 Dividing by 4, cos 2 #=f. 
 Taking the square root cos #= 
 But cos 3o = 
 Therefore, #=30 is a solution of the equation. 
 
 3. sin x : . Find x. 
 
 2 sin x 
 
 Multiplying by sin x, sin 2 ^=^ 
 
 sin x = V / \ = \V 2. 
 .'. #=45, by Problem 13, p. 106. 
 
 4. sin x-\ : =4. Find re. 
 
 sin x 
 
 Clearing of fractions, sin 2 #+i =| sin x 
 
 sin 2 # f sin x= i . 
 
 This is a complete quadratic equation in which sin x is 
 the unknown number. 
 
 Solving by the method of completing the square, 
 sm a x f sin x + H = -fa 
 sin x f=| 
 sin #=f f 
 = 2 or .
 
 122 Second-Year Mathematics 
 
 Sin x cannot be equal to 2 because the sine of an angle is 
 not greater than i (138, Problem 3). 
 Therefore sin x = %. 
 Therefore #=30. 
 
 5. tan 2 * 1= - . Find x. 
 
 cos x 
 
 It is helpful in solving equations to express all the functions 
 in terms of one of them. 
 
 sin 2 ^ i 
 
 By 150, (2), _-!= (i) 
 
 cos 2 * cos x 
 
 Clearing of fractions sin 2 # cos 2 #=cos x (2) 
 
 By 1 50, sin 2 * = i cos 2 x 
 
 .'. (2) may be written 
 
 1 cos 2 # cos 2 #=cos x (3) 
 2 cos 2 #+cos x i =o (4) 
 
 This is a complete quadratic equation in which cos x is the 
 unknown number. 
 
 Solving by the method of factoring, 
 
 (2 cos x i)(cos # + i)=o (5) 
 
 2 cos x i =o, or cos x=$ (6) 
 and cos x + i =o, or cos x= i (7) 
 
 Cos x cannot be equal to i (138). 
 
 cos x=% 
 And x=6o. 
 
 Find x in each of the following: 
 2 
 
 6. tan x= ^ . 
 
 tan x 
 
 T. 
 
 
 
 7. 4 sm x=- 
 
 sm x 
 
 8. 2 tan x cos x = i. 
 
 g. 3 cos x = 2 sin 2 x. 
 
 Express sin 2 * in terms of the cosine.
 
 Ratio, Proportion, Similar Triangles 123 
 
 10. 2 sin x = y"2 tan x. 
 
 11. tan 2 * 4 tan #+3=0. 
 
 12. tan* 2-| -- =o . 
 
 tan x 
 
 13. 7 sn 
 
 14. sin x cos 2 x= -]-. 
 
 1 6. sin ;-|-sin 2 ^=cos 2 ^. 
 
 17. 
 
 tan* 
 
 18. 2 sin 2 #+4 cos 2 * = 5 cos x.
 
 CHAPTER III 
 THE MEASUREMENT OF ANGLES BY ARCS OF THE CIRCLE 
 
 152. Suppose the angular space of 4R. A. about the center 
 of a circle to be divided into 360 equal angles. This may be 
 done by using the divisions on the edge of a protractor. One 
 of the angles so obtained may then be taken as a unit of meas- 
 ure for angles. It is called a degree of angle. How do the 
 sides of these angles divide the circumference? State the 
 theorem that applies. 
 
 Into how many equal arcs is the circumference then 
 divided ? One of these 360 equal arcs may be used as a unit 
 of measure for arcs. It is called a degree of arc. Thus a 
 degree of arc is an arc intercepted by the sides of a degree of 
 angle. 
 
 If now any central angle be drawn, 
 as ZAB C (Fig. 151), it will contain 
 a certain number, integral or fractional, 
 of these angular units, or degrees of 
 angle. 
 
 How many degrees of arc do you 
 think there are in the arc intercepted 
 by the sides of the angle of Fig. 151 ? 
 
 If an angle contains 15 angular 
 
 units, how many arc-units are there in the arc intercepted by 
 its sides ? 
 
 If a central angle contains 35^ degrees of angle, how many 
 degrees of arc are there in the intercepted arc ? 
 
 If the central angle contains m units or degrees of angle, 
 how many degrees are there in the intercepted arc ? Why ? 
 
 124
 
 Measurement of Angles by Arcs of the Circle 125 
 
 If the central angle is of 4 R. A. (72 degrees), what part 
 of the circle is the arc intercepted by its sides ? Why ? 
 
 153. We will now compare the ratio of two central angles 
 with the ratio of their intercepted arcs. Let angles ABC 
 and D E F, in the same circle or in two equal circles (Fig. 152), 
 be any two central angles such that an angle-unit x can be 
 found that is contained in each an exact number of times. 
 
 FIG. 152 
 
 Assume it to be contained in Z A B C m times, and in 
 Z D E F n times. 
 
 Then ZA B C=m Z* and ZD E F=n Z*. 
 w#_w 
 ~~ ' 
 
 i. e., the ratio of ZA B C to ZD E F is equal to the ratio 
 of their numerical measures, m and n. 
 
 The units of angle-measure into which Z A B C and 
 Z D E F are divided will intercept m equal arcs, y, on A C,* 
 and n equal arcs, y, on D F. Why ? 
 
 Then A~C=w y, and D~F=w y. 
 arc AC m y m 
 
 _ . -- _ ^. . _ I ry \ 
 
 arc D F n y n ' 
 
 ZA B C arc A C 
 
 Hence = =, . Why ? State the axiom used. 
 
 Z D E F arc D F 
 
 * A C and D F mean arc A C and arc D F.
 
 126 Second-Year Mathematics 
 
 154. We have proved that the ratio of two commensurable 
 central angles is equal to the ratio of the arcs they intercept 
 on the circumference. 
 
 But the question arises, do angles or other magnitudes of 
 the same kind, other than straight lines ( 113), exist, for 
 which there is no common unit of measure ? 
 
 1. Give a common measure or divisor of: 
 
 n^ ft. and 4^ ft. 
 mi. and 6J mi. 
 3 yd. and 7$ yd. 
 i T V and 3.25 
 6.23 and i .6342. 
 
 If these number pairs are reduced to the form of fractions having 
 a common denominator, the unit fraction having that same denominator 
 will be a common divisor. 
 
 2. Is there a rational* common divisor or measure for: 
 
 3 and 1 /^g ? 
 5 and 1/50 ? 
 2 and 1/9 ? 
 
 A simpler form for 1/^8 is 31/2, but j/2 (the square root 
 of 2) cannot be expressed exactly as an integer, or a rational 
 fraction. It is found approximately to be i .4142 . . . . , and 
 the number of decimal figures may be continued as far as we 
 please, but the exact value can never be thus expressed. So 
 there is no rational number that will exactly divide it. 
 
 Again, 3 and j/^g may be written 1/9 and i/^g. Of this 
 pair, 1/9 seems to be a common measure, as it is contained in 
 1/9 once, and in ]/i8> 1/2 times. But 1/2 times as we have 
 just seen is not an exact integral or fractional number of times. 
 Hence 3 and j/ig have no rational common measure. Such 
 numbers are said to be incommensurable. 
 
 * A rational number is a number that can be expressed exactly as 
 an integer or a fraction without the use of the radical sign or the equiva- 
 lent of the radical sign.
 
 Measurement of Angles by Arcs of the Circle 127 
 
 155. We shall see how incommensurable numbers may arise. 
 
 1 . What is the length of the diagonal A B of a rectangle 
 5 by 7 (Fig. 153)? (See 133, 9.) 
 
 Is there a common divisor for 5 and 1/74 ? for 7 and 1/74 ? 
 
 2. Find the length of the diagonal of a square whose side 
 is 6; 7; 8; 9; 10, a; b; c. 
 
 3. Is there a common unit of measure for the side and 
 diagonal of any square ? What num- 
 ber, incommensurable with unity, or i, 
 
 does the diagonal always seem to con- 
 tain as a factor (see -113) ? 
 
 We have seen that two numbers 
 and two line-segments may be incom- 
 mensurable. FlG 
 
 It is shown by higher mathematics 
 
 that the angles of a scalene triangle are cut by the medians 
 into angles that are incommensurable, i. e., angles for which 
 there is no common angle-unit of measure. 
 
 If then two central angles are drawn equal to two such 
 parts of an angle of a scalene triangle, would they have the 
 same ratio as their intercepted arcs? If so, how can it be 
 proved, since they have no common unit of measure by which 
 to express the ratio ? 
 
 Does the theorem proved for commensurable angles ( 153) 
 hold good for incommensurable angles ? 
 
 156. Suppose ZC A D and ZE B H (Fig. 154) to be two 
 incommensurable angles. If Z C A D, the smaller of the two 
 angles, be used as a unit of measure of ZE B H, it will measure 
 it an integral number of times, say n, with a remainder, say 
 Z F B H. It will intercept the same number n of equal arcs in 
 E H with remainder H F (see 1 53) .
 
 128 
 
 Second- Year Mathematics 
 
 Use the remainder, Z. F B H, as a measure of the first 
 measure, Z C A D. It is contained in it an integral number 
 of times, say m, with a remainder, if the angles are incom- 
 mensurable, but the remainder must be smaller than ZF B H; 
 but the number of angles measured off in Z C A D is the 
 same as the number of equal arcs of arc C D intercepted by the 
 sides of ZFBH. 
 
 FIG. 154 
 
 Continue this process indefinitely, and the remainders be- 
 come less each time the process is repeated, and the numerical 
 measures of the angles CAD and E B H are found to equal 
 the numerical measures of the arcs C D and E H to an error 
 as small as may be desired. 
 
 The ratio then of the angles is equal to the ratio of the 
 arcs. For if it be said the ratios differ by any amount, d, we 
 could by carrying the process sufficiently far prove they differ 
 by less than d (see 121). 
 
 157. A rigorous treatment of incommensurable cases is 
 beyond the bounds of a secondary text. Hereafter we will 
 give proofs for commensurable cases only, assuming the 
 theorems to be true for incommensurable magnitudes also. 
 
 We are now prepared by proof for commensurable angles, 
 and discussion for incommensurable angles, to recognize the 
 truth of the following theorem:
 
 Measurement of Angles by Arcs of the Circle 129 
 
 PROPOSITION I 
 
 158. Theorem: In the same circle or in equal circles, two 
 central angles have the same ratio as the arcs intercepted by 
 their sides. 
 
 Let ABC (Fig. 155) be any central angle. By the pre- 
 ceding theorem it may be compared with the perigon ABA. 
 
 ZABC arc AC .... 
 
 Then = ...... Why? 
 
 ZABA arcAMA 
 
 Expressing ZA B A and arc 
 A M A by the number of units 
 (degrees of angle and degrees of 
 arc) they contain, respectively, 
 we have 
 
 ZABC = arc AC 
 360 360 
 
 Multiplying through by 360, 
 
 ZABC=arc AC, in the 
 sense that the number of angle degrees in Z A B C is the same 
 as the number of arc degrees in the arc A C. 
 
 As A B C is any central angle, the number of degrees in 
 any central angle is the same as the number of degrees in the 
 arc intercepted by its sides. This is more technically stated 
 thus: 
 
 PROPOSITION II 
 
 159. Theorem: A central angle is measured by the arc 
 intercepted by its sides. 
 
 EXERCISES 
 
 1 60. Solve the following exercises: 
 
 i. By Proposition II, using rule and compasses only, 
 divide a circle (a) into 4 equal arcs; 
 (b) into 2 equal arcs. 
 
 FIG. 155
 
 1 3 o 
 
 Second-Year Mathematics 
 
 2. Divide a circle into three arcs in the ratio of i : 2 : 3. 
 
 3. Find an arc of 45; 75; 105; 165; 15. 
 
 4. If a circle is divided into 4 arcs in the ratio of i : 4 : 6 : 7 
 what is the number of degrees in each arc ? 
 
 5. The area of a circle is 616 sq. in. How many degrees 
 are there in an angle at the center that intercepts an arc 1 1 in. 
 long? 
 
 Assume area of circle = ?rR 2 and circumference = 27rR. Use ir=-f. 
 
 6. In Fig. 1 56, A B is a diameter. The number of degrees 
 in angle A O C and B O C are represented by x 2 -\-$x and 
 3# 2 + 1 2X. Find the values of x and the number of degrees in 
 arcs A C and B C. 
 
 A 
 
 
 
 FIG. 156 FIG. 157 
 
 7. In Fig. 157, ZABC is a right angle. ZABD = 
 2X 2 3, and ZD B C = iox 2 -15. 
 
 Find the values of x and the number of degrees in arcs 
 A D and D C. 
 
 Inscribed Angles 
 
 161. Let us next consider what relation a central angle 
 bears to an angle intercepting the same arc but whose vertex 
 is on the circumference. 
 
 An angle whose sides are chords and whose vertex is on 
 the circumference is called an inscribed angle.
 
 Measurement of Angles by Arcs of the Circle 131 
 
 CASE I. One side of the inscribed angle is a diameter. 
 
 In Fig. 158, ABC is a central angle. Extend A B to 
 meet the circumference as at D, and join D and C. 
 
 Then Z A D C is an inscribed angle 
 intercepting the same arc A C as does 
 the central angle ABC. 
 
 Compare angle ABC with the sum 
 of Z> and ZC of the triangle B D C 
 ( 78, 12). How do Z> and ZC 
 compare in size ? Why ( 28, 3) ? 
 
 Then how does Z.D compare with 
 ZABC? 
 
 What arc measures ZA B C? 
 Then what measures the inscribed angle ZA D C? 
 
 CASE II. The center of the circle lies between the sides of 
 the inscribed angle. 
 
 FIG. 159 
 
 As in Fig. 159, draw an inscribed angle A B C so that 
 the center of the circle lies between the sides of the angle. 
 Draw the diameter B D. By Case I what arcs measure 
 ZA B D and ZD B C? What then measures their sum, or 
 ZABC?
 
 132 Second- Year Mathematics 
 
 CASE III. The center of the circle lies outside of the in- 
 scribed angle. 
 
 As in Fig. 160, draw an inscribed angle A B C so that the 
 center of the circle lies outside of the angle. Draw the diameter 
 B D. By Case I, the half of what arc measures ZA B D ? 
 ZC B D ? What arc then measures the angle ABC, which 
 is the difference between Z s A B D and C B D ? 
 
 Are there any positions for an inscribed angle not included 
 by the three cases just considered ? We have now proved: 
 
 PROPOSITION III 
 
 162. Theorem: An inscribed angle is measured by one- 
 half the arc intercepted by its sides. 
 
 163. An angle is inscribed in a segment of a circle when 
 the vertex is on the arc of the segment and its sides touch the 
 end-points of the arc. 
 
 EXERCISES 
 
 1 . Prove that all angles inscribed in the same segment of a 
 circle are equal. 
 
 2. Prove that all angles inscribed in a semicircle are right 
 angles (see Fig. 161). A 
 
 3 
 
 C 
 
 FIG. 161 FIG. 162 
 
 3. Prove that all angles inscribed in a segment smaller 
 than a semicircle are greater than a right angle. 
 
 4. Prove that all angles inscribed in a segment larger than 
 a semicircle are less than a right angle.
 
 Measurement of Angles by Arcs of the Circle 133 
 
 5. Prove: If two chords intersect within a circle and the 
 end-points of the sides of two vertical angles be joined by two 
 straight lines, the triangles formed are mutually equiangular 
 (see Fig. 162). 
 
 Two triangles are mutually equiangular when each angle of one 
 triangle is equal to the corresponding angle in the other. 
 
 6. If two secants 
 meet without a circle, 
 the angle they form 
 is measured by one- 
 half the difference of 
 the arcs included be- 
 tween them. 
 
 Suggestion: Show 
 (Fig. 163) that ZB= FIG. 163 
 
 ZACD-ZB AC. 
 
 7. Prove: If two chords intersect within a circle, either 
 angle they form is measured by one-half the sum of the inter- 
 cepted arcs (see Fig. 164). 
 
 Prove Z A B C is measured by i (/fC+D~E). 
 
 Draw EC. A B C is an exterior angle 
 of triangle EEC. To the sum of what 
 two inscribed angles is it equal ? By what 
 arc then is it measured? Give complete 
 proof. 
 
 In similar manner show by what arc 
 Z D B A is measured. 
 
 8. Prove: An angle formed by a 
 tangent and a chord, is measured 
 by one-half the arc included by its 
 sides. 
 
 Suggestion: Prove by Fig. 165 Z A B C is measured by i B*C. 
 Draw the diameter B D. Z A B D being a right angle (why ?) is 
 measured by one-half the semicircle BCD. 
 
 Z C B D is measured by ? 
 
 Then what measures Z A B "C ? 
 
 FIG. 164
 
 134 
 
 Second- Year Mathematics 
 
 9. Similarly show by what arc the angle C B E is measured. 
 
 A_ B 
 
 io. Prove that an angle formed by a tangent and a secant 
 meeting without the circle, is measured by one-half the differ- 
 ence of the included arcs. 
 
 Suggestion: In Fig. 166 prove angle B is measured by one-half 
 (AC-A^D). ADC is an exterior angle of triangle A B D. ZB = 
 ZADC-ZBAD. Why? 
 
 FIG. i 66 
 
 1 1 . Prove that the angle formed by two tangents to a circle 
 is measured by one-half the difference of the arcs included by 
 them. 
 
 In Fig. 167, join the two points of tangency and use angle ADC 
 as in Fig. 166.
 
 Measurement of Angles by Arcs of the Circle 135 
 
 12. In laying a switch on a railway track a "frog" is used 
 at the intersection of two rails to allow the flanges of the 
 wheels moving on one rail to cross the other rail. Prove that 
 the angle of the frog, a (Fig. 168), made by the tangent to 
 
 3 
 
 FIG. 1 68 
 
 the curve and the straight rail D E, is equal to the central 
 angle O, of tfc arc B F. 
 
 (From "Real Applied Problems," School Science and Mathematics.) 
 
 PROPOSITION IV 
 
 164. Problem: Upon a given line-sect* to construct a seg- 
 ment of a circle in which an angle may be inscribed equal to a 
 given angle. 
 
 Given a line-sect b; and an angle A (Fig. 169). 
 
 To construct upon b as a chord, a segment of a circle in 
 which an angle equal to angle A may be inscribed. 
 
 What lines form an angle measured by the same arc as an 
 inscribed angle (see Exercise 8, 163) ? 
 
 Draw a line-sect equal to b (see Fig. 169). 
 * Line-sect is the same as line-segment.
 
 136 
 
 Second- Year Mathematics 
 
 FIG. 169 
 
 At one end-point A construct 
 an angle equal to the given angle 
 A, as E A C. 
 
 Then EA is to be made a 
 chord, and A C a tangent of a 
 circle. 
 
 It is required to find the center 
 of the circle. 
 
 On what line from point A 
 does it lie ? 
 
 On what other line from the 
 chord E A does it lie ? 
 
 Draw the circle and the seg- 
 ment required. 
 
 Give proof of the proposition 
 in full. 
 
 EXERCISES 
 165. Solve the following exercises: 
 
 1. Draw any acute angle and a line-sect. Construct the 
 segment required by the method of Proposition IV. Inscribe 
 angles in this segment, and compare them with the given 
 angle by using a protractor. 
 
 2. Draw a right angle and a line-sect. Construct the seg- 
 ment according to Proposition IV. Inscribe an angle in it 
 and test its size. 
 
 3. Draw an obtuse angle and a line-sect. Construct the 
 segment of a circle according to Proposition IV. Test the 
 result. 
 
 4. What is the size of the segment compared with the 
 semicircle in each of Exercises i, 2, and 3 preceding ?
 
 Measurement of Angles by Arcs of the Circle 137 
 
 What previous truths about inscribed angles do these 
 exercises confirm (see Exercises 2, 3, and 4, 163) ? 
 
 5. What name may be given to the chord formed by the 
 given line-sect in Exercise 2 ? State more than one reason 
 for your answer. 
 
 6. On a given line-sect, construct a segment of a circle 
 that shall contain an inscribed angle of 45; of 22; of 135. 
 Construct the angles with rule and compasses only. 
 
 7. On a given line-sect, construct a segment of a circle 
 that shall contain an angle of 60; of 30; of 120; using 
 rule and compasses only. Test your results with a protractor. 
 
 8. To draw a tangent to a given circle from a point with- 
 out the circle. 
 
 Let C be the center of the given circle, and E any given point 
 outside the circle. Connect C and E with a straight line. 
 
 Draw a semicircle on the line-sect C E as a diameter. 
 
 At what point will a line from E be tangent to C ? Why ? Are 
 there two such points? Why? See 163, 2. 
 
 ALGEBRAIC EXERCISES 
 1 66. Solve the following exercises algebraically: 
 
 1. Tf the arc AB (Fig. 170) is 4^8 degrees, how many 
 degrees are there in ZADB? 
 
 In ZACB? (C is the center 
 of the circle.) 
 
 2. If the arc AB (Fig. 170) 
 is 48, find x and the number of 
 degrees in Z A C B and Z A D B. 
 
 3. If the arc A D C (Fig. 171) 
 is 190, find the number of de- 
 grees in ZT A C and Z N A C. 
 
 .r IG. 
 
 4. If arc AMC (Fig. 171) is 
 
 represented by $x 2 +x, find the value of x, arc A D C being 1 90.
 
 138 
 
 Second-Year Mathematics 
 
 5. The arcs and angle being as marked in Fig. 172, find 
 x and y. 
 
 FIG. 173 
 
 6. Find x and y (Fig. 173), the arcs and angle between the 
 secants being as indicated in the figure. 
 
 7. When two tangents to a 
 circle make an angle of 60, 
 into what arcs do they divide the 
 circle ? 
 
 8. Into what arcs do two 
 tangents at right angles to each 
 other divide the circle ? 
 
 9. Two tangents include two 
 arcs of a circle, one of which 
 
 is four times the other. How many degrees are in the angle 
 they form ? 
 
 10. Two angles of an inscribed triangle are 82 and 76. 
 How many degrees are in each of the three arcs subtended 
 by its sides ? 
 
 11. A triangle A B C is inscribed in a circle, Z4=57, 
 Z.B=66. At A, B, and C tangents are drawn forming a cir- 
 cumscribed triangle A'B'C'. Find the three angles, A', B', 
 and C' (Fig. 174). 
 
 12. The angle between twb secants intersecting without a
 
 Measurement of Angles by Arcs of the Circle 139 
 
 circle is 76. One of the intercepted arcs is 243. Find the 
 other. 
 
 13. Two angles of a circumscribed triangle are 70 and 80. 
 Find the number of degrees in each of the three angles of the 
 inscribed triangle whose vertices are at the points of tangency. 
 
 14. The points of tan- 
 gency of a circumscribed 
 quadrilateral divide the 
 circle into arcs in the ratio 
 of 7 : 8 : 9 : 12. Find the 
 angles of the quadrilateral. 
 
 15. The vertices of an 
 inscribed quadrilateral 
 divide the circle into arcs 
 in the ratio of 3 : 4 : 5 : 6. 
 Find the angles of the 
 quadrilateral. 
 
 1 6. Two tangents to a circle from an outside point form an 
 angle of 70. What part of the circle is the larger arc included 
 by the points of tangency ? 
 
 17. If a race track is constructed on the plan of Exercise 
 1 6, the course being laid out on two tangents at an angle of 
 70 and the arc of the circle included by them, how much 
 farther than his competitor must a rider go whose path is 
 4 ft. farther from the rail ? 
 
 Assume circumference = 2irR. 
 
 1 8. The angle between two secants is 30 (Fig. 175). The 
 
 6x 2 + 2ox + 30 
 -- 
 
 number of degrees in arc D L is represented by 
 
 ' \) 
 
 O-v2 _ fj sv* _ T 
 
 in the arc B C. by - - - - . Find x and the number 
 
 *-S 
 of degrees in each of the two arcs. 
 
 Reduce the fractions to lowest terms.
 
 140 Second-Year Mathematics 
 
 19. In Fig. 176, ZA E D is 60, arc B C is represented by 
 
 x 2 + i2x-4$ 
 ; arc A D, by - . Find the number 
 
 of degrees in each of the two arcs. 
 
 FIG. 175 
 
 167. A method of proof, often valuable, for an exercise or 
 a theorem that involves the comparison or the measurement 
 of angles is that of bringing the given figure into some con- 
 nection with the circle. 
 
 For example : 
 
 Prove: The sum of the three angles of a triangle is two 
 right angles. 
 
 In Fig. 177, let ABC be any triangle; circumscribe a 
 circle about it (see 96, 10). The three 
 inscribed angles are measured by one- 
 half the sum of the three arcs A B, B C, 
 and C A. But the sum of the three arcs 
 A B, B C, and C A is the entire circum- 
 ference. Therefore one-half the circum- 
 ference or 180 (the measure of two right 
 angles) is the measure of the sum of the 
 three angles of the triangle, or AA + Z.B+ /^C = 2 R. A. 
 Q.E.D. 
 
 EXERCISES 
 
 1 68. Verify the following exercises by relating the given 
 figure to the circle :
 
 Measurement of Angles by Arcs of the Circle 141 
 
 1. At a given point in a line one and only one perpen- 
 dicular to the line can be drawn. 
 
 Let the two lines in Fig. 1 78 be J_ at point A. Draw a circle with 
 A as the center, and prove any other J_ at A impossible. 
 
 2. All right angles are 
 equal. 
 
 3. The sum of the ad- 
 jacent angles formed by one 
 straight line meeting another 
 is two right angles. 
 
 4. The sum of all the 
 angles formed at a point in 
 
 a line on the same side of the FlG 1?8 
 
 line is a straight angle. 
 
 5. If the sum of two adjacent angles is a straight angle, 
 their exterior sides form a straight line. 
 
 6. In an isosceles triangle the angles opposite the equal 
 sides are equal. 
 
 7. If two angles of a triangle are equal, the sides opposite 
 them are equal. 
 
 8. The sum of the acute angles of a right triangle is a 
 right angle. 
 
 9. An equilateral triangle is equiangular. 
 
 10. An equiangular triangle is equilateral. 
 
 1 1 . If two sides of a triangle are unequal the angles opposite 
 them are unequal, the greater angle lying opposite the greater side. 
 
 12. If two angles of a triangle are unequal the sides oppo- 
 site them . Complete the statement of the proposition, 
 
 and prove it. 
 
 13. That by drawing perpendiculars from the mid-points 
 of two adjacent sides, a circle can be circumscribed about any 
 rectangle.
 
 142 Second-Year Mathematics 
 
 14. Using the result of Exercise 13, prove that the diagonals 
 of a square: 
 
 (1) bisect each other; 
 
 (2) are perpendicular to each other; 
 
 (3) are equal; 
 
 (4) bisect the angles of the square. 
 
 15. Prove that the diagonals of a rectangle that is not a 
 square : 
 
 (1) are equal; 
 
 (2) bisect each other; 
 
 (3) are not perpendicular to each other; 
 
 (4) do not bisect the angles of the rectangle. 
 
 HIGHEST COMMON FACTOR 
 
 169. In no it was seen that to find the value of the ratio 
 of two magnitudes, a unit of measure, or a divisor, common 
 to the terms of the ratio, was needed. The largest integral 
 divisor, or the greatest common divisor (G.C.D.), of arithmetical 
 or algebraic numbers is used in reducing ratios, or fractions, 
 to lowest terms, and also to obtain the lowest common mul- 
 tiple (L.C.M.) of the denominators of fractions to be added 
 or subtracted, and to rid equations of fractions. 
 
 A method employed with arithmetical numbers is as follows: 
 
 To find the G.C.D. of 348 and 870. 
 
 The Method of Common Factors 
 
 Divide each of the given numbers by the 
 
 2)348, 870 smallest number that will divide both, and 
 
 3)174, 435 repeat the process with the successive quo- 
 
 29)58, 145 dents as long as they have a common factor, 
 
 2, 5 or divisor. The product of the divisors is 
 
 2X3X29 = 174 the G.C.D. In this example, 174 is found 
 
 to be the G.C.D. of 348 and 870.
 
 Measurement of Angles by Arcs of the Circle 143 
 
 170. Another method applied to line-segments in 112 
 is here shown with arithmetical numbers. 
 
 The Method of Successive Division 
 
 Divide the larger by the smaller, and then divide the divisor 
 
 by the remainder, and the last divisor by 
 
 348)870(2 the last remainder, repeating the process 
 
 696 until the division is exact. The last divisor 
 
 774)348(2 is the G.C.D. 
 
 348 If the last divisor is i, unity is the only 
 
 common divisor, and it is of no value in 
 reducing fractions, because the quotients obtained by dividing 
 by i are the same as the original numbers. When the last 
 divisor is unity we say the numbers are prime to each other, 
 or that they have no common divisor except unity. 
 For example, find the G.C.D. of 73 and 16. 
 
 The last divisor being i, we are sure 
 no number larger than i will exactly 
 divide 16 and 73. This can readily be 
 9) l6 (* tested. 
 
 - Note the important distinction be- 
 
 tween prime numbers and incommensur- 
 - . able numbers (see 113). 
 
 6 171. The principle on which the pro- 
 
 7)2(2 cess is based is: the common divisor of 
 2 any two numbers is also a common divi- 
 
 sor of one of them and of the number 
 obtained by adding to, or subtracting from, that one a multiple 
 of the other. 
 
 Examples 
 
 Find the G.C.D. of: 
 (1)174 and 273 (4) 350 and 425 
 
 (2) 174 and 275 (5) 3542 and 5016 
 
 (3) 263 and 765 (6) 3795 and 2865.
 
 144 Second-Year Mathematics 
 
 172. This method can be applied to algebraic polynomials. 
 The result is usually called the Highest Common Factor. 
 
 For example : 
 
 2 + 95^+30 
 4ox 2 + 136^ + 11 2 
 41) 41^82 
 
 The first term of the first remainder, Sx 2 , does not contain 
 $x 2 exactly. So we multiply it by 5, for as 5 is not a factor 
 of either of the given numbers, multiplying a remainder by 5 
 will not introduce a common factor. 
 
 Again, 41 is not a factor of either of the original num- 
 bers, so removing it from a remainder will not remove or de- 
 stroy a common factor. With these modifications the process 
 is the same as with arithmetical numbers. In this example 
 the last divisor, # + 2, is the H.C.F. 
 
 EXERCISES 
 
 173. Find the H.C.F. of the following polynomials by the 
 method of successive division: 
 
 As in ordinary division it is convenient to arrange the given poly- 
 nomials in the ascending or the descending powers of the same letter, 
 if they are not already so arranged. 
 
 1. 3a 2 +a 4 + i2a 16 and 
 
 2. c 2 + 2c 3 yc 20 and 
 
 3. 4c 2 +3c-io and 7c 2 +4c3 -3^-15
 
 Measurement of Angles by Arcs of the Circle 145 
 
 4. X 2X*+X 4 and 2X* 2X 3 
 
 5. <;* 13<; 2 +36 and c4 c3 
 
 6. 4a 3 4a 2 50+3 and ioa 2 190+6. 
 
 Find the H.C.F. of the following by either method: 
 
 7. x 2 2ix+2o and x 2 +6x 7 
 
 8. i5(w ri) 3 and iow 2 
 
 9. # 2 13^+30 and 5t; 2 
 
 10. m*m2 andw 2 +m 6 
 
 11. x^ * and x a 
 
 When three or more numbers are involved, find H.C.F. of two, then 
 the H.C.F. of this result and the next number, and so on. 
 
 12. a^ab 2 , 2a4 2a 2 b 2 , and a* 2a 2 b+ab 2 
 
 13. c4-^4 j c s+ds, cs+d*, and c 2 + 2cd+d 2 
 
 14. I2(m 6 n 6 ), i8(m 4 n 4 ), and 24 
 
 (m 3 mn 2 m 2 n+n 3 ) 
 
 15. 5^+4063, 7a 3 + 28a 2 6 + 28fl6 2 , and 3 (a 4 4a 2 6 2 ) 
 
 16. x 4 6+x 2 , x 4 io^c 2 + i6, and 3(x* + 2X 2 8).
 
 CHAPTER IV 
 
 SIMILARITY AND PROPORTIONALITY IN CIRCLES 
 PROPOSITION I 
 
 174. Problem: To construct a square equivalent to a given 
 rectangle. 
 
 Let a and b be given sides of a rectangle. 
 To construct a square equivalent to the rectangle a by b. 
 Upon a line-segment EF equal to the sum, a+b, as a 
 diameter, draw a circle (Fig. 179). At the point of division 
 C between a and b on the diameter, 
 erect a perpendicular and extend it 
 to the circle as at D. 
 
 Prove C D to be the side of the 
 required square. 
 
 Draw ED and F D. ZEDF 
 is a right angle. (Why ?) 
 FlG D C is a mean proportional 
 
 between the sects* of the hypotenuse 
 of the right triangle A D B. Why? See Proposition XIII, 
 
 p. 97. 
 
 COROLLARY 
 
 175. Theorem: A perpendicular to a diameter of a circle 
 at any point, extended to the circumference, is a mean propor- 
 tional between the sects of the diameter. 
 
 EXERCISE 
 
 Construct a square equal in area to the rectangle with 
 given sides c and d . 
 
 * The sects are the parts a and b into which the line-segment A B 
 is cut by the foot C of the perpendicular. 
 
 146
 
 Similarity and Proportionality in Circles 147 
 
 Measure to some scale the sides of the given rectangle and of the 
 constructed square, and note whether the product of the numbers of 
 the first two measures is equal to the square of the third. If there is a 
 difference, to what is it probably due ? 
 
 Measure each several times and take the average of the products of 
 the numbers expressing the measures of the sides of the rectangle, and 
 the average of the squares of the numbers expressing the measures of 
 the side of the square. How do the two averages compare ? 
 
 176. Draw a circle and a chord A B intersecting a chord 
 C D as at E (Fig. 180). Measure the four sects from the 
 point E. See whether E B E A=E C E D. 
 
 Draw on squared paper several 
 such figures with intersecting chords 
 in various positions. Measure care- 
 fully, and note the approximate 
 equality of the products of the sects 
 of each of the two chords. To what 
 is the difference, if any, probably due ? 
 
 177. If in Fig. i8o,AE.EB = 
 E C E D, then A E and E B may 
 
 be made the means of a proportion of which E C and E D 
 are the extremes (see 119). Draw A C and B D, and prove 
 the triangles ACE and B D E similar. 
 
 178. Show the same triangles similar by considering the 
 angles only (Fig. 180). If the triangles are similar, the corre- 
 sponding sides are proportional: 
 
 CE : AE=EB : ED. 
 
 Then, AE-EB = CE-ED. Why? Note carefully what 
 sects make the two equal products. These steps give a method 
 of proof for the following: 
 
 PROPOSITION II 
 
 1 79. Theorem : If two chords of a circle intersect, the product 
 of the sects of one is equal to the product of the sects of the other.
 
 148 Second-Year Mathematics 
 
 Prove the theorem. 
 
 Problem: Given the sides of a rectangle, to find the side 
 of an equivalent square by using Proposition II.' 
 
 In any circle large enough, draw a chord equal to the sum of two ad- 
 jacent sides of the rectangle. Draw a radius to the point of division. 
 What chord through this point is bisected at this point ? 
 
 EXERCISES 
 
 i. A chord of a circle D C (Fig. 181) cuts the mid-point 
 of chord A B at E, E D is 4 inches longer than E C, and 
 AB = i6 inches. Find the lengths 
 of E D and E C approximately to 
 
 2. The sects of two intersecting 
 chords are x+$ and x 6 in the 
 one, x + 2 and # 5 in the other, 
 find x and the length of each 
 chord. 
 
 3. The sects of intersecting 
 chords are as given below. Find x and the length of each 
 chord. 
 
 First Chord Second Chord 
 
 (1) X-4, X + S # + 3, X-4 
 
 (2) x+z, x+6 #4, #+18 
 
 (3) x-s, x-2 x->j,x+i4. 
 
 PROPOSITION III 
 
 1 80. Theorem: If from a point without a circle two secants 
 be drawn to the concave arc, the product of one secant and its 
 external sect is equal to the product of the other secant and its 
 external sect. 
 
 In Fig. 182, what two triangles are similar because mutually equi- 
 angular ?
 
 Similarity and Proportionality in Circles 
 
 149 
 
 What proportion can then be formed of the secants and their exter- 
 nal sects ? 
 
 Give complete proof of the theorem. 
 
 FIG. 182 
 
 EXERCISES 
 
 1 . If two adjacent sides of a rectangle are given, show how 
 other equivalent rectangles may be constructed by Proposi- 
 tion III without finding the numerical measures of the sides 
 of the given rectangle. 
 
 2. Given a and b , the sides of a rectangle. 
 
 By use of Proposition III construct three different rectangles 
 equivalent in area to that with sides a and b. 
 
 3. By use of Proposition III can a square be constructed 
 equivalent to a given rectangle ? 
 
 What must be the relative 
 size of the whole secant and 
 its external sect? If it were 
 drawn so as to make its 
 external sect equal to the 
 whole secant by what name 
 could the "secant" then be 
 called ? 
 
 181. Draw from a point P 
 
 FIG. 183 
 
 without a circle, a tangent PD (Fig. 183), and a secant P F, 
 cutting the circle at some point, as C. Draw D C and D F.
 
 150 . Second-Year Mathematics 
 
 What triangles contain as sides, the tangent, the secant, and 
 its external sect? 
 
 Show the triangles to be mutually equiangular and there- 
 fore similar. 
 
 Form a proportion making P D the mean proportional. 
 
 PROPOSITION IV 
 
 182. Theorem: If from a point without a circle a tangent 
 and a secant be drawn, the tangent is a mean proportional be- 
 tween the entire secant (to the concave arc) and its external 
 sect. Prove. 
 
 EXERCISES 
 
 1. A tangent and a secant are drawn from the same point 
 without a circle. The secant measured to the concave arc is 
 three times as long as the tangent, and the length of its external 
 sect is 10 feet. Find the length of the tangent and secant. 
 
 2. Two secants to the same circle from an outside point 
 are cut by the circle into chords of the circle and external 
 sects in the ratio of 5 to 3, and 5 to i respectively. The first 
 secant is 8 ft. long. Find the length of the second secant. 
 
 The following problems relate to two secants from an out- 
 side point as in Exercise 2. Find the length of the second 
 secant, and test by Proposition III. 
 
 Length of 
 First Secant 
 28ft. 
 28ft. 
 625 ft. 
 25 ft. 
 36 ft. 
 
 8. How do the tests for the five preceding problems show 
 whether Proposition III is true when the sects of one secant 
 are incommensurable with those of the other ? 
 
 Ratio of Sects of 
 
 Ratio of Sects of 
 
 First Secant 
 
 Second Secant 
 
 3- 
 
 5 : 
 
 : 2 
 
 3 
 
 : i 
 
 4- 
 
 3 : 
 
 : i 
 
 5 
 
 : 2 
 
 5- 
 
 4 : 
 
 i 
 
 5 
 
 : 4 
 
 6. 
 
 4 : 
 
 i 
 
 4 : 
 
 : 3 
 
 7- 
 
 7 = 
 
 2 
 
 7 : 
 
 : 3
 
 Similarity and Proportionality in Circles 
 
 9. In Fig. 184, A R = 
 
 , EB = 
 
 \/2, Find the value of x, and test by Proposition III. 
 
 c 
 
 FIG. 184 
 Solution of the equation of Exercise 9 : 
 
 V / I4 1/2 = 1/28 
 
 1/ 
 
 Squaring each member: 
 
 Hence : 
 Test: 
 
 x = 2 
 
 21/2 + 5 = 114 12 
 21/7 = 1/2 1/7 1/2 
 21/7 = 21/7 
 
 Remember that the square of the square root of a number, is the 
 number itself, by definition of square root. Thus the square of 1/8, 
 [or 0/8) a ], is 8; (Va + b) 2 is a + b; (l/c'+d*) 2 is c* + d*. 
 
 10. In Fig. 185, A 6 = 1/9-3;, 
 
 BD = 
 
 I/ac 3. Find x, and test by 
 Proposition IV. 
 
 Note that the square of VgX 
 
 is 9 x; that V5C + 3 Vac 3 = 
 
 V x 3 9; and that the square of 
 
 is 81 
 
 Are the tangents and 
 secants of Exercises 9 and 
 10 commensurable? 
 
 FIG. 185
 
 152 Second-Year Mathematics 
 
 11. In Fig. 185, AB = 2j/# 2 9; B C=4, DB = 
 
 Find x, and the tangent and secant. Test by Proposition IV. 
 
 12. In Fig. 184, A B = ]/; 3, D B= = , CB = 
 
 [, E B= . Find x and test by Proposition III. 
 
 l/X 2 
 Note that the product of V x 3 by V x2 is x ^Vx + 6. Also 
 
 13. In Fig. 184, AB = 2 + |/^+4, D B = 2 
 B E = 2 + j/#+5, B = 3 1/^+5. Find x and each line- 
 sect. Interpret the result. 
 
 Solution of Exercise 13: 
 
 (2 + V *+4~) (2 - 1/ ^+4) = (2 + 1/ ^c 
 
 Multiplying out 4 # 4=6+ I/ # + 5 # 5 
 
 -V 
 Squaring 
 
 Test: (2 + / 
 
 The principal square roots do not satisfy the equation. 
 Using the negative roots, 2 2 = i 4, 
 
 4 = 4- 
 
 The "secant" A B is thus shown to be a tangent. 
 The symbol -\/ denotes the positive square root. When the positive 
 and negative roots are desired the symbol i/ is used. 
 
 14. In Fig. 184, A~B = s + 
 
 B C=8 1/^+3 , B =3 + 1/^+3. Find y and each line- 
 sect. 
 
 15. In Fig. 184, AB = s-v/5, BC=8+ 3 i/2, EB = 
 2/2. FindDB. 
 
 1 6. In Fig. 184, DB = !/3 : B C = |/6 + 2i/2 : B E = 
 2]/6~ 31/2. Find A B. 
 
 17. In Fig. 184, the secant A B is the square of the external
 
 Similarity and Proportionality in Circles 153 
 
 sect. B C = 3l/f-2l/f, BE = 2l/J+8l/f. Find the se- 
 cant A B. 
 
 Let x be the external sect, then x 2 is the secant. 
 
 #3=6 '-4 ^ + 24 i-i6 
 
 1 8. In Fig. 185, the tangent A B is 1/3 + 51/7. The 
 secant B C is 3 + 51/7. Find the external sect B D. 
 Let * = B D. 
 
 19. In Fig. 185, A B = 1/21/3 , 6=41/3. Find the 
 sect B D. 
 
 20. In Fig. 185, A B = i, B 0=1/2 + 1/3. Show that D B 
 
 21. In Fig. 185, AB=\/l/a+l/6, B C = Vab, show that 
 
 22. InFig. 185, AE = Vx+y. The secant B C = Vx 2 -y 2 . 
 
 Find the external sect D B to be equal to l/x 2 y a . 
 
 xy 
 
 PROPOSITION V 
 
 183. Problem: To draw a common tangent to two circles 
 exterior to each other. 
 
 Let A and A' (Fig. 186) be the centers of two circles exterior 
 to each other. Connect A and A'. 
 
 Divide this line into sects proportional to the radii of the 
 circles as at C (see 125, 2). 
 
 Such a point of division is called a center of similitude of the two 
 circles. 
 
 From C draw a tangent to circle A'. 
 
 How can the points of tangency be found ? (See 165, 8.)
 
 154 
 
 Second-Year Mathematics 
 
 Let D' be one point of tangency. Draw D'A'. Draw 
 D'C, and extend it beyond circle A. 
 
 FIG. 186 
 
 Draw A D to line D'C and parallel to D'A'. 
 
 If A D can be proved to be a radius of circle A, by using 
 the constructions made and the fact of the similarity of the 
 triangles, then D'D is proved to be a tangent to both circles- 
 Why? 
 
 Similarly prove E F a tangent to both circles. 
 
 D D' and E F (Fig. 186) are called transverse tangents. 
 
 184. Prove C D'D and C E F common tangents to circles 
 ' and A (Fig. 187). 
 
 C 
 
 FIG. 187
 
 Similarity and Proportionality in Circles 
 
 155 
 
 If A A' be divided externally into sects proportional to the radii of 
 the circles, as A D and A'D' (Fig. 187), the constructions and proofs 
 are similar to those in Proposition V. Give full proof. 
 
 C D'D and C E F (Fig. 187) are called direct tangents. 
 The point C' is called an external center of similitude. 
 
 185. As the distance between the centers of the circles is 
 made to decrease, certain interesting changes are produced in 
 the common transverse and direct tangents, as shown in Figs. 
 188 to 192. 
 
 FIG. 188 
 
 Show how the points C, D, D', E, and F of Figs. 1 86 and 187 
 change position in Figs. 188 and 189. 
 
 G 
 
 When do two or more of such points coincide ? 
 In which figures do they disappear ?
 
 156 Second- Year Mathematics 
 
 What lines in Figs. 186 and 187 do the lines H G in Figs. 189 
 and 191 represent? 
 
 FIG. 190 
 
 FIG. 191 
 
 FIG. 192 
 
 FIG. 193
 
 Similarity and Proportionality in Circles 
 
 157 
 
 1 86. Prove that all lines from a center of similitude of two 
 circles, are cut proportionally by the circle. 
 
 Prove by Proposition V that in Figs. 193 and 194. 
 CB CD CE CF CH 
 
 CB' 
 
 because each is equal to 
 
 CA' 
 
 "CF CH' 
 
 F' 
 
 FIG. 194 
 
 To Draw Common Tangents. Second Method 
 187. A second method for drawing common tangents, is 
 
 here suggested, not based on similar figures. 
 
 To find by analysis the points of tangency, B and A, of the 
 
 common tangent to circles C and C' (see 127). 
 
 -8 A 
 
 FIG. 195 
 
 First, assume a common direct tangent B A to be drawn as 
 in Fig. 195, and C B and C'A, to be radii to the points of con- 
 tact. They are both _]_ to B A (why?), and therefore || to
 
 158 Second-Year Mathematics 
 
 each other (why?). If C'D is drawn || to A B, it completes 
 a rectangle A C'D B (why?). Then D B=A C', and D C = 
 B C A C'. If a circle be drawn with C as center and radius 
 C D (=B C-A C'), C'D is a tangent to it at D (why?), and 
 the radius of circle C to point of tangency of the common tan- 
 gent passes through D. 
 
 Second, reverse this process. 
 
 In the larger given circle draw a concentric circle D m 
 with radius =Rr. From C', center of smaller circle, draw 
 a tangent to O D m, locating point D (see 165, 8). 
 
 Extend line C D till it meets larger circle as at B. 
 
 Draw C'A || C B. 
 
 Join B and A. 
 
 B A is the common direct tangent required. Tell why. 
 
 i. In similar manner draw B'A', a second direct tangent. 
 
 1 88. Assume a transverse tangent B A to be drawn as in 
 Fig. 196, and the radii C'A and C B, extended, drawn to the 
 
 FIG. 196 
 
 points of tangency. Then draw C'D || A B, meeting C B as 
 at D. D B must equal C'A and be parallel to it, forming a 
 rectangle as in the preceding case ( 187).
 
 Similarity and Proportionality in Circles 
 
 159 
 
 Hence C D = C B + C'A (or R+r), and C'D is tangent to 
 the circle with center C and radius R+r (why?), and CD 
 passes through B, the point of tangency of the common tan- 
 gent on the larger given circle. The radius C'A drawn parallel 
 to C D determines the point of tangency A on the smaller 
 given circle. 
 
 Then to draw a common transverse tangent to two given 
 circle? C and C', reverse the process just given. 
 
 Make the construction, and give the proof. 
 
 i. In similar manner construct a second transverse tan- 
 gent B'A' (Fig. 196). 
 
 189. To express the area of a triangle in terms of the 
 sides and the diameter of the circumscribed circle. 
 
 In Fig. 197 let A B C be any triangle, with sides a, b, and c of known 
 length. Circumscribe a circle about it (see 96, 10). 
 
 FIG. 197 
 
 Draw a diameter from B, as B F. 
 Draw a perpendicular from B to A C, as B E. 
 
 The area of a triangle A B C = 
 
 , assuming the area of a tri- 
 
 angle equals one-half the base times the altitude. (i) 
 
 To express this area in terms of the sides and the diameter, B E 
 
 must be expressed in these terms. 
 
 Show AS B E C and A B F similar.
 
 160 Second-Year Mathematics 
 
 .-. = , (D = diameter B F.) 
 a D 
 
 BE-* (,) 
 
 Hence from (i) and (2) A B C=^=r , and we have thus proved 
 
 PROPOSITION VI 
 
 190. Theorem: The area of any triangle is equal to the 
 product of the three sides divided by twice the diameter of the 
 circumscribed circle. 
 
 Give a complete proof. 
 
 EXERCISES 
 
 1. The three sides of a triangle are 14, 8, and 12. The 
 diameter of the circumscribed circle is 14.1. Find the area 
 of the triangle. 
 
 2. The three sides of a triangle are 12, 10, and 8. The area 
 of the triangle is 39 . 7. Find the diameter of the circumscribed 
 circle. 
 
 3. The area of a triangle by Proposition VI is expressed 
 
 thus: A=-T=T. Express the diameter of the circumscribed 
 
 2D 
 
 circle in terms of the area and the three sides of the triangle. 
 
 4. The three sides of a triangle are 8, 12, and 14 and the 
 area is 47 . 66. Find the diameter of the circumscribed circle. 
 
 5. Using the facts that the area of a triangle is |& times the 
 altitude, and the area of the triangle is , find a formula 
 
 2 \J 
 
 for the altitude to the side b in terms of the other sides and the 
 diameter of the circumscribed circle. 
 
 6. Express in a similar way the altitude to side a; to side c. 
 
 7. Test the formulas for the three altitudes of a triangle 
 whose sides are 8, 12, and 14, the area 47.66; the diameter of
 
 Similarity and Proportionality in Circles 
 
 161 
 
 the circumscribed circle 14.1. Multiply each altitude by one- 
 half its base to find the area. 
 
 8. Test the formulas for the three altitudes of a triangle 
 with sides 10, 8, 12 and the area 39.7, and the diameter of the 
 circumscribed circle 12.09. 
 
 Solution of Quadratic Equations by the Formula 
 191. Two lights, L t and L 2 (Fig. 198), are placed 6 ft. 
 apart and the candlepower of L 2 is twice that of L x . A screen, 
 
 FIG. 198 
 
 S, is slipped along a rod until it is equally illuminated by both 
 lights. Find the distance, x, of the fainter light from the 
 
 screen. 
 
 It is 'a law of physics that intensities of illumination from 
 two lights, are directly proportional to the candlepowers of the
 
 162 Second-Year Mathematics 
 
 lights and inversely proportional to the squares of the dis- 
 tances from the lights to the screen that is illuminated. 
 
 Calling Ij and I 2 the intensities of illumination of the 
 screen by L x and L 2 , we have 
 
 I 2 x 2 
 
 But as the intensities are equal =^ = i 
 
 Hence, i = 
 
 2X 2 
 
 Or, x 2 i2X 36=0 (A) 
 
 Adding 72 to both sides 
 
 x 2 1 2^+36 = 72 
 Or, #-6 = 6i/2 
 
 And # = 6 61/2 = 14. 48 + , or 2.48 + 
 
 The results mean there are two places, one 14.48+ ft. to 
 the right of L T as shown in the figure and the other, between 
 L! and L 2 and 2.48+ ft. to the left of L,. 
 
 The fundamental equation of the solution just given is 
 equation (A). This is a particular example of the general 
 type 
 
 x 2 +px+q=o (B) 
 
 The solution of (B) solves at once all equations of this form. 
 
 Add --q to both sides of (B), 
 4 
 
 * 2 * 2 
 
 x*+px+^=^-q 
 144 
 
 Or, (*+>)' '-*!= 
 
 \ 2/ 4 
 
 Whence, *=--$l / /> 2 -4? (i)
 
 Similarity and Proportionality in Circles 163 
 
 Designating one root by # t and the other root by x 2 , 
 
 P 
 *i = --+il / /> 2 -4? (2) 
 
 2 
 
 and x, = --iV-4g (3) 
 
 Calling p the coefficient of the first power of x in the 
 quadratic x*+px+q=o and q the absolute term of the 
 quadratic, give descriptive phrases for the following: 
 
 ~ 
 
 Calling a;, the positive root of x 3 +px-{-q=o, and x, the 
 negative root, translate formulas (2) and (3) into words. 
 
 Calculate the value of x I +x 2 and state the result in words. 
 
 Calculate the value of x t x 2 and state the result in words. 
 
 Calculate the value of x t x 2 and state the result in words. 
 
 Equation (B) is itself a particular case of ax 3 +bx+c=o 
 in which a = i. Hence a more general form of the complete 
 quadratic is ax 3 +bx+c=o. Let us now solve this more 
 general form. 
 
 The General Quadratic Formula 
 
 192. Solve the equation ax 3 +bx+c=o, by completing the 
 square. 
 
 ax*+bx= c 
 
 , bx c 
 
 x*+= 
 
 a a 
 
 , b 1 , bx b 3 b 3 c b 3 -4ac 
 
 Add , x 3 -\ --- -- = --- = 
 
 4a 3 a 4<z 2 $a 3 a 4a 3 
 
 , b 
 Extracting sq. root, x-\ = 
 
 20
 
 164 Second-Year Mathematics 
 
 This statement of the value of x is called the General 
 Quadratic Formula, and is used in solving complete quadratic 
 equations. For every quadratic equation can be made to 
 have the form ax 2 +bx+c=o, in which: 
 
 a represents the coefficient of x 2 , 
 b represents the coefficient of x, 
 and c is the constant term when standing on the left 
 side of the equation. 
 
 Consequently by solving ax 2 +bx+c=o, we have solved every 
 quadratic equation in one unknown. 
 
 By this formula solve 3X 2 + $x 2=0. 
 
 Here a = 3, 6 = 5, andc = 2. 
 
 . _ 
 
 6 6 
 
 EXERCISES 
 193. Use the Quadratic Formula in solving: 
 
 1. 2:V 2 +5# + 2=0 5. 2X 2 +X = I$ 
 
 2. 6x 2 nx+$=o 6. ax 2 + (b a)x b=o 
 
 3. 2r 2 r 6=0 7. 2y a + (4a+b)y=2ab 
 
 4. 1 2m 2 i6am 30* =o Solve for y. 
 
 Solve for m, also for a. 8. 2Z 2 (2a + b)z+ab=o. 
 
 The sects of intersecting chords in a circle are as follows. 
 Solve for x by the Formula. 
 
 First Chord Second Chord 
 
 9- 3*-t 3^-3 2X + i, Sx-4 
 
 10. 2# + 5, X + I X + 2, 
 
 11. #5, 2#+3 
 
 12. 2X + 2, 3^ 3 
 
 13. ax+a, 3^3 ax+^a, x+i
 
 Similarity and Proportionality in Circles 
 
 165 
 
 14. In a circle of radius 26 in., what is the length of a 
 chord perpendicular to a diameter at a point 10 in. from the 
 center of the circle ? 
 
 15. In a circle of radius 10 in. the shortest distance from a 
 given point on the circumference to a given diameter is 8 in. 
 Find the segments into which the perpendicular from the point 
 divides the diameter. 
 
 1 6. Given a fragment of 
 a waterwheel, to construct a 
 new wheel of the same size. 
 D C (Fig. 1 99) is by measure- 
 ment 4', A B i'. Compute 
 the diameter. 
 
 FIG. 199 
 
 17. WW is a wall with a round corner of dimensions as 
 in Fig. 200, A B, 4', N E, 9". It is desired to find the radius 
 of the circle of which A N B is a part, and the location of 
 the center. 
 
 FIG. 201 
 
 1 8. To draw an arc of a circle as an "easement" of a 
 cornice, tangent to the cornice at B (Fig. 201) and passing 
 through A. 
 
 Problems 16, 17, and 18 are taken from "Real Applied Problems" 
 in School Science and Mathematics.
 
 1 66 Second-Year Mathematics 
 
 Problems in Physics Leading to Complete Quadratic Equations 
 
 The intensity of light on a screen varies directly as the 
 candlepower, and inversely as the square of the distance of 
 the source of the light to the screen. In the language of pro- 
 portion, the intensity is in direct ratio to the candlepower, and in 
 inverse ratio to the square of the distance (see again 191). 
 
 194. Solve the following problems by the Quadratic 
 Formula. 
 
 1. An electric lamp placed before a screen lights it as in- 
 tensely as two such lamps, each of the same candlepower as 
 the first, but placed three feet farther from the screen. How 
 far are the two lamps from the screen ? 
 
 2. Two lights 1,000 ft. apart, one of twice the candlepower 
 oi the other, appear equally bright when brought nearly into 
 line by an approaching vessel. What is the distance of the 
 vessel from the nearer light ? 
 
 3. As a boat approaches a harbor two range lights are 
 brought nearly into line. It is known that the nearer one is 
 of 500 candlepower, and the other 1,500 candlepower, and 
 that they are one-half mile apart. How far is the boat from 
 the nearer light when they are of exactly the same intensity ? 
 
 4. A candle lights a screen as brightly as do 7 equal candles 
 when placed 10 ft. farther from the screen. How far are the 
 lights from the screen ? 
 
 5. Assuming the law of heat to be essentially the same as 
 that of light, find how far from a stove a person is if by moving 
 4 ft. farther away, the intensity of heat is one-half as great as 
 it was before. 
 
 6. How far from a fire was a man if after moving 7 rods 
 nearer the fire the intensity of heat was 3 times as great as 
 before ?
 
 Similarity and Proportionality in Circles 167 
 
 7. How far is an object from a light if when moved toward 
 the light 3 feet, the intensity is 4^ times as great as before ? 
 
 8. An electric lamp lights a screen as much as b such lamps 
 of the same candlepower as the first, but placed c ft. farther 
 from the screen. How far are the b lamps from the screen ? 
 
 9. Use the result of Problem 8 to verify the results of 
 Problems i and 4. 
 
 10. How far from a fire was a man if after moving n rods 
 nearer the fire the intensity of heat was b times as great as 
 before ? 
 
 11. Use the result of Problem 10 to verify the result of 
 Problem 5. 
 
 12. As a vessel approaches a harbor two range lights are 
 brought nearly into line. It is known that the nearer one is 
 of a candlepower, and the farther one is of b candlepower, 
 and that they are c miles apart. How far is the vessel from 
 the nearer light when the two are of the same intensity ? 
 
 13. Use the result of Problem 12 to verify the results of 
 Problems 2 and 3.
 
 CHAPTER V 
 INEQUALITIES IN TRIANGLES AND CIRCLES 
 
 195. Experimental and Theoretical Treatment of the 
 Axioms of Inequality. 
 
 PROBLEMS AND EXERCISES 
 
 1. Construct triangles with sides as follows: 
 
 (1) a =8 6 = 8 c=3 
 
 (2) a=8 6 = 9 c=3 
 
 ( 3 ) a=8 6 = 10 c=3 
 
 (4) a=8 6 = 11 c=3 
 
 (5) 0=8 6 = 12 = 3 
 
 What do you notice in constructing parts (4) and (5) ? 
 Explain. 
 
 2. Is it possible to construct a triangle having any three 
 line-segments as sides ? If the lengths of two sides are given, 
 what condition must the third side fulfil? Calling the three 
 sides of a triangle a, b, and c, express by a formula a relation 
 which must hold between them. 
 
 Use the sign > or <, meaning greater than, and less than (see 198). 
 
 3. Construct triangles with sides as follows: 
 
 (1) a=8 6=8 c = 3 
 
 (2) a=8 6 = 7 c=3 
 
 (3) 0=8 6=6 c=3 
 
 (4) a = 8 6 = 5 c=3 
 
 (5) =8 6=4 c = 3 
 
 What do you notice in constructing parts (4) and (5) ? Is 
 the condition asked for in Exercise 2 fulfilled by the above sets 
 of sides ? Explain. 
 
 168
 
 Inequalities in Triangles and Circles 
 
 169 
 
 4. Was the condition arrived at in Exercise 2 sufficient to 
 make sure that a triangle can be constructed ? What further 
 condition must be fulfilled ? 
 
 Two of the fundamental truths on which geometry is 
 commonly based are expressed as Axioms 8 and 10, p. 14. 
 Re-read them. 
 
 The truths arrived at in Problems 1-4 will now be shown 
 to be immediate consequences of Axioms 8 and 10. 
 
 PROPOSITION I 
 
 196. Theorem: The sum of any two sides of a triangle is 
 greater than the third side, and their difference is less than the 
 third side. 
 
 Given any triangle ABC. 
 To prove: a+b>c; b+c>a; c+a>b. 
 ab<c; bc<a; ca<b. 
 ac<.b. 
 
 Proof: Consider the two points 
 A and B (Fig. 202). In what two 
 ways are these points connected? 
 We have then a+b>c. Why? 
 
 In a similar way prove b+c>a 
 and c+a>b. 
 
 Having proved a + b > c 
 and knowing b=b 
 
 FIG. 202 
 
 subtracting obtain a >cb, or cb<a. 
 
 In a similar way, prove the remaining five inequalities. 
 
 PROBLEMS AND EXERCISES 
 
 i. What obvious truth is used in obtaining the last six 
 inequalities from the first three in 196 ?
 
 170 Second-Year Mathematics 
 
 To keep sharply in mind on what fundamental truths our 
 statements are based and for convenient reference, this truth 
 is here stated as 
 
 AXIOM 13 
 
 The differences obtained by subtracting equal numbers from 
 unequal numbers are unequal in the same order as the minuends. 
 
 2. Three line-segments, a, b, and c, being given, it will be 
 possible to construct a triangle having these line-segments as 
 sides, if the sum of every two is greater than the third line- 
 segment; because, as was seen in the proof of Proposition I, 
 if these conditions are fulfilled, the difference of every two 
 sides is less than the remaining side. 
 
 3. Mention some sets of three out of the nine inequalities 
 in 196, from which the remaining six would follow. 
 
 4. State by inequalities the relations 
 which must hold between the sides of 
 the triangle in Fig. 203. 
 
 5. For what values of x do these rela- 
 F tions hold as expressed in the following 
 
 inequalities ? 
 
 Solution: x+4 + g>x+i2: q + x+i2>x+4; x+i2+x+4>g; 
 I3>i2; 2i>4; 2x+i6>g; 
 
 2x>-^ 
 *>-3* 
 
 6. For what value of x can the following expressions repre- 
 sent the lengths of the sides of a triangle ? 
 
 (1) a=x 5 b=x-\-'j c = i6 
 
 (2) a = 2X+T ) b = 2X + 2 C = 2I 
 
 (3) a=x+$ b=Sx c=i 
 
 (4) a = 7 b=x-$ c = g 
 
 (5) a = 2x 6 = 5 c=4*-7
 
 Inequalities in Triangles and Circles 171 
 
 197. Besides Axiom 13, p. 170, the following underlie the 
 solution of inequalities: 
 
 AXIOMS 
 
 14. The sums obtained by adding equal numbers to unequal 
 numbers are unequal in the same order as the unequal addends: 
 e.g., 5 <8 
 
 7<io 
 
 15. The products obtained by multiplying unequal numbers 
 by positive equal numbers are unequal in the same order as the 
 multiplicands: 
 
 e.g., io<i5 
 
 2=2 
 
 - M 
 
 20<30 
 
 1 6. The products obtained by multiplying unequal numbers 
 by negative equal numbers are unequal in the order opposite to 
 that of the multiplicands: 
 
 e.g., i2< 15 
 
 - 3 6>-45 
 
 17. The quotients obtained by dividing unequal numbers by 
 positive equal numbers are unequal in the same order as the 
 dividends: 
 
 e.g., 2o<3o 
 
 2=2 
 
 -D 
 
 io<i5 
 
 1 8. The quotients obtained by dividing unequal numbers 
 by negative equals are unequal in the order opposite to that of 
 the dividends: 
 
 e- g-, 5>4o 
 
 2 - 2 
 
 - D 
 
 25 < 20
 
 172 Second-Year Mathematics 
 
 19. The differences obtained by subtracting unequal numbers 
 from equal numbers are unequal in the order opposite to that of 
 the subtrahends: 
 
 e.g., 20 = 20 
 5> 3 
 
 198. The symbols for inequality are >, <, and =t=. 
 > means is greater than, < means is less than, and 4= means 
 is not equal to. 
 
 EXERCISES 
 
 1. Give some examples for each of the above axioms. 
 Which of these axioms were used in the solutions of Prob- 
 lem 6, 196 ? 
 
 2. Two sides of a triangle are 9 and 24 inches. Between 
 what limits must the third side be ? 
 
 3. There is $50 in the treasury of a club. The club wants 
 to buy furniture costing between $80 and $90. How much 
 should be raised at least ? 
 
 4. To pass in a certain study the final grade must be above 
 60. The final grade is obtained by taking the average of the 
 quarterly grade and the examination grade. The quarterly 
 grade being 80, find how much at least the examination grade 
 should be for passing ? 
 
 5. Other conditions being as in Exercise 4, how much at 
 least should the examination grade be if the quarterly grade 
 counts for in the final mark ? For ? For ? For \ ? 
 
 6. I live at a distance of 2\ miles from school. Leaving 
 home at 7 : 30 o'clock, at what rate must I walk to get to school 
 on time, if school begins at 8 : 20 ? At 8:30? At 8 145? At 
 9 o'clock ? 
 
 7. Solve Problem 6 for a pupil who lives i miles from 
 school.
 
 Inequalities ^n Triangles and Circles 
 
 173 
 
 8. A twentieth-century limited train wants to make the 
 distance between New York and Chicago (1,000 miles) in less 
 than 20 hours. During the first five hours it goes at the rate 
 of 45 miles per hour. During the next 7 hours it goes at the 
 rate of 57 miles per hour. How fast should it go thereafter 
 to cover the distance within the desired time ? 
 
 9. A's record average speed on a 2-mile run is 6 miles per 
 hour, and B's is 5! miles. How many feet can A afford to 
 give B as a handicap in a 2-mile race ? 
 
 199. By means of the truths discussed in the foregoing 
 articles of this chapter, prove 
 the following geometric exer- 
 cises and propositions: 
 
 1. Prove that a diameter 
 of a circle is longer than any 
 other chord of that circle. 
 
 See Fig. 204; notice that 
 KO + OL = MN. (Why?) 
 
 2. Prove that the distance F IG . 204 
 between the centers of two 
 
 intersecting circles is less than the sum of the radii, but 
 greater than the difference. 
 
 PROPOSITION II 
 
 200. Theorem: Any point not 
 on the perpendicular bisector of a 
 given line-segment is unequally 
 distant from the extremities of the 
 given line-segment. 
 
 Given: AC=CB; XY.LAB, 
 passing through C (Fig. 205). 
 
 Also, P lies outside X Y. 
 
 To prove P A=(=P B.
 
 174 Second- Ye,ar Mathematics 
 
 Proof: Join the point D, where PA and XY intersect, 
 with point B. 
 
 PB<PD+DB (?) 
 DB=DA (?) 
 /. PB<PD+DA (?) 
 orPB<PA (?) 
 orPB=^PA. Q.E.D. 
 
 i. Repeat the proof, 
 
 (a) taking P on the left of X Y; 
 
 (6) below A B. 
 
 In 28, 4, p. 13, we proved: 
 
 Any point on the perpendicular bisector of a given line- 
 segment is equally distant from the extremities of that line- 
 segment. 
 
 Using this proposition and Proposition II, make a state- 
 ment concerning the location of a point P, which is known to 
 be equally distant from the extremities of a given line-segment 
 as A B. This statement may be made as follows: 
 
 20 1. Any point equally distant from the end-points of a 
 line-segment must lie on the perpendicular bisector of the line- 
 segment, and any point on the perpendicular bisector must be 
 equally distant from the end-points of the line-segment. Or thus, 
 the locus of the points that are equally distant from the end-points 
 of a line-segment, is the perpendicular bisector of that line- 
 segment. 
 
 PROPOSITION III 
 
 202. Theorem: If a line contains two points that are 
 equally distant from the end-points of a given line-segment, it 
 is the perpendicular bisector of the given line-segment. 
 
 Given: (i) A line-segment A B and an indefinite line X Y 
 (Fig. 206);
 
 Inequalities in Triangles and Circles 175 
 
 (2) The points P and Q are on X Y; 
 
 (3) PA=PB; andQA = QB. 
 
 To prove the line X Y is the IX 
 
 perpendicular bisector of A B. ' 
 
 Proof: The perpendicular bi- 
 sector of A B passes through Q. (?) 
 See 201. 
 
 And also through P. (?) /\ 
 
 The perpendicular bisector of \Q 
 
 A B then passes through P and Q. 
 
 But the straight line X Y passes I 
 
 through P and Q. (?) ll 
 
 FIG. 206 
 Therefore, the straight line X Y 
 
 must be the perpendicular bisector of A B. (?) 
 
 203. In Axiom 8, we recognized that the shortest distance 
 between two points is measured along the straight line joining 
 
 them, i. e., all broken line-segments 
 like A P B and A Q B (Fig. 207), 
 joining two points like A and B 
 are longer than a straight line- 
 segment, like A B, joining these 
 two points. 
 
 * D It is now to be shown that of 
 
 F IG . 207 two such broken line-segments, 
 
 one of which lies entirely outside 
 
 of the other, the one that is nearer to the straight line is the 
 shorter. 
 
 PROPOSITION IV 
 
 204. Theorem: If from a point inside a triangle, line- 
 segments are drawn to the end-points of one side, the sum of 
 these line-segments is less than the sum of the other two sides.
 
 176 Second-Year Matliematics 
 
 Given: AAB C, a point P inside the triangle (Fig. 208). 
 To prove AP+PC<AB+B C. 
 
 A C 
 
 FIG. 208 
 
 Proof: Prolong A P until it intersects B C at some point, 
 asD. 
 
 We have now: 
 
 AP+PD<AB+BD (?) 
 PC<PD+DC (?) 
 
 adding: A P+P D+P C< A B+B D+P D+D C (?) 
 subtracting: A P+P C< A B+B D+D C (?) 
 
 .'. AP+PCXAB+BC. (?) 
 
 i . Prove the same proposition by prolonging the line C P. 
 
 EXERCISES 
 
 1. Two billiard balls are at A and B, respectively (Fig. 209). 
 To what point on the cushion X Y should the ball at B 
 
 be directed that it may, after reflection, hit the ball at A ? 
 
 Draw A A' perpendicular to X Y and so as to make C A'= 
 A C. Draw A'B meeting X Y at some point as P. Prove 
 that P is the required point, assuming that the reflection is 
 such as to make the angle of reflection equal to the angle of 
 incidence (see Figs. 209 and 210). 
 
 2. Prove that the distance from B to A by way of P is 
 shorter than by way of any other point on X Y. 
 
 Use 201 and Axiom 8, p. 14.
 
 Inequalities in Triangles and Circles 
 
 177 
 
 3. A ray of light reaching Q from P (Fig. 211), by way 
 of the reflecting surface X Y, travels along the shortest dis- 
 tance. Determine at what point the ray of light should strike 
 X Y to be reflected to Q. 
 
 FIG. 209 
 
 I 
 
 X 
 
 FIG. 210 
 
 Q 
 
 x- 
 
 FIG. 211 
 
 -Y 
 
 4. Prove that light is reflected so as to make the angle 
 of reflection equal to the angle of incidence. See Exer- 
 cise 3. 
 
 5. Prove that the sum of the 
 three line - segments joining a 
 point inside of a triangle with 
 the vertices, is less than the 
 perimeter of the triangle but 
 greater than its semf-perimeter 
 
 (see Fig. 212). FIG. 212 
 
 B
 
 i 7 8 
 
 Second-Year Mathematics 
 
 6. Determine between what limits x must lie in Fig. 213. 
 What values could x have if it were required to be an integer ? 
 
 7. Prove that the line joining one vertex of a triangle with 
 the middle of the opposite side is less than one-half of the sum 
 of the remaining two sides. 
 
 In Fig. 214 make DE = BD; -O 
 
 prove BE<BC + CE and prove 
 CE = BA. 
 
 6 
 
 FIG. 213 
 
 8. Prove that the sum of the diagonals of a quadrilateral is 
 less than the perimeter, but greater than the semi-perimeter. 
 
 9. Determine what integral numbers x may represent in 
 Fig. 215. 
 
 9 
 
 FIG. 215 
 
 3X+86 
 FIG. 216 
 
 10. Determine the integral values of x from conditions 
 shown in Fig. 216, 5#+4 and 4^31 being diagonals. 
 
 1 1 . Determine for what values of x the following inequali- 
 ties hold. Give a reason for each step. 
 
 (1) $X + 2X/2 
 
 (2) ^^<6 
 ' x-4 
 
 ( ?\ nv- ^i_ 
 
 \3) 2X
 
 Inequalities in Triangles and Circles 
 
 179 
 
 205. The properties of inequalities may be used to solve 
 problems in more than one unknown, when the unknowns are 
 restricted to positive integers. 
 
 PROBLEMS AND EXERCISES 
 i. Find what positive integers will satisfy the equation 
 
 x=i'j 2y (Why?) 
 
 Since x and y must be positive integers, we have 
 17 2y>o (?) 
 
 , and;y>o. (?) 
 Hence, the answer is: 
 
 y= i> 2 > 3. 4, 5> 6 > 7, 8. 
 * = i5 i3 " 9> 7 5. 3, I- ( ? ) 
 
 2. Determine what positive integers will satisfy the follow- 
 ing equations: 
 
 (1) 3*+;y-i5=o 
 
 (2) y=42-gx 
 
 (3) h-S=- 3 k. 
 
 17 
 
 FIG. 217 
 
 3. Drawing the graph of the equation # + 2^ = 17, solved 
 in Problem i, it becomes clear (Fig. 217) that in order to have
 
 180 Second-Year Mathematics 
 
 both x and y positive integers, we must have y<8% and y>o. 
 This agrees with the algebraic solution. 
 
 Verify graphically the solutions of Problem 2. 
 
 4. Determine, both algebraically and graphically, the posi- 
 tive integral roots of the following equations: 
 
 (1) zx-y = i8 
 
 (2) x 5^+21=0 
 
 (3) x + >jy=-i5. 
 
 5. A pipe-system, 100 ft. long, is to be built from pipes 
 5 ft. and 15 ft. long. How many pipes of each kind could 
 be used ? 
 
 6. If 20 yards of cloth were bought at one price, and 60 
 yards at another price, and if altogether $42.00 were spent, 
 what was the price of each kind, neither kind costing less 
 than 50 cents a yard ? 
 
 7. If the numerator and denominator of a fraction are both 
 increased by 5, the value becomes J. Of what fractions is this 
 possible ? 
 
 8. When some boys are arranged four in a row, one boy is 
 left out; if they are arranged 8 in a row, 5 boys are left out. 
 Determine how many boys there may have been, if there can 
 be at most 8 rows. 
 
 9. Find what numbers leave a remainder 2 when divided 
 by 3, and also leave a remainder 5 when divided by 9. 
 
 10. A sum of $45 is to be paid in five-dollar bills and one- 
 dollar bills. Determine how many bills of either kind should 
 be used, so that the total number of bills used should be least. 
 
 11. Determine the positive integral values of x and y 
 which will satisfy the equation
 
 Inequalities in Triangles and Circles 181 
 
 Solving (i) for x in terms of y and reducing the improper 
 fraction to a mixed number: 
 
 ~~ 
 x and y will be integers if - - is an integer. 
 
 i ~\~y 
 Let -=u, u being an integer. 
 
 2 
 
 Then i+y=2U or 
 
 y = 2U-i (3) 
 
 Substituting this in (2) 
 
 X = $U + 2 (4) 
 
 We verify by substitution that (3) and (4) satisfy (i) for all values 
 of u. 
 
 Now, to find the positive integral values of x and y which 
 satisfy (i), we must find what integral values of u will satisfy 
 the inequalities: 
 
 2u i >o and3W+2>o 
 we find 2u > i and 3^ > 2 
 or w>i and > . 
 These values are: w=i, 2, 3, 4 ..... 
 Substituting in (4): x=$, 8, n, 14 ..... 
 Substituting in (5) : y = 1,3, 5, 7 ..... 
 
 12. Determine the positive integers that will satisfy the 
 following equations: 
 
 (1) 4*+5;y = 2i (4) 50 + 246 = 291 
 
 (2) 3*+8;y = 25 (5) 7^-13^ = 12 
 
 (3) 9^ 2^=17 (6) 2/ 3W = 2O. 
 
 13. The amount of $75 is to be paid out in five-dollar and 
 two-dollar bills. How many bills of each kind may be used ? 
 
 14. Determine the numbers which will leave a remainder 
 3 when divided by 5, and also when divided by 7.
 
 1 82 Second-Year Mathematics 
 
 15. Two cogwheels with 170 and 130 cogs, respectively, en- 
 gage each other. At a certain moment two marked cogs are 
 opposite each other; after how many revolutions oT each 
 wheel will the same cogs again be opposite each other ? 
 
 1 6. At 12 o'clock the cars of two street-car companies leave 
 the same point. Cars of company A leave the same point 
 regularly every 7 minutes thereafter; those of company B 
 leave the same point every 9 minutes thereafter. At what 
 time is the car of company A, 5 minutes ahead of the, car of 
 company B ? 
 
 17. Under the same conditions as in Problem 16 find at 
 what time the car of company A is 5 minutes behind that of 
 company B ? At what times then are they 5 minutes apart ? 
 
 1 8. Other conditions being unchanged, find at what time 
 the cars will be i minute apart ? When will they leave at the 
 same moment? When will two cars of company A come in 
 succession ? 
 
 19. Solve a problem like Problem 16 if the cars start at 
 8 A. M., those of company A leaving every 13 minutes and those 
 of company B every 8 minutes. When will they be 2 minutes 
 apart ? 
 
 20. Ask and answer questions concerning Problem 19 
 similar to those asked in Problem 18. 
 
 21. A group of boys, numbering not more than 50, is ar- 
 ranged in rows of 4, leaving one over. If arranged in rows 
 of 5, there will be 2 over. How many boys are there ? 
 
 22. Determine the positive numbers which leave a re- 
 mainder i if divided by 3, and leave a remainder 7 if divided 
 by 8. 
 
 23. Into what two parts should 71 be divided so that one 
 part may be divisible by 3 and the other part by 4 ?
 
 Inequalities in Triangles and Circles 183 
 
 24. Gold alloy of 93 per cent, is to be made from 90 per 
 cent, and 95 per cent, alloy. How many pounds of each kind 
 may be used ? 
 
 25. How many pounds of the above kinds of alloy may be 
 used to make a new alloy of 92 percent.? Of 92^ percent.? 
 
 26. John says to William: "I divide my age by 5 and 
 have a remainder 2. I divide it by 4 and have a remainder i." 
 What is John's age ? 
 
 27. "Tell me the remainder you get when dividing your 
 age by 5; also when dividing it by 4," says a boy to his father, 
 "and I will tell your age." The answers being 3 and 2, re- 
 spectively, find the father's age. 
 
 28. Says A to B: "If I had 7 times as many eggs as I have 
 now and if you had 8 times as many eggs as you have now, 
 and you gave me one of your eggs, we would have the same 
 number of eggs." How many has each ? 
 
 206. On p. 14 we had Axiom 6, which expresses that a 
 whole is equal to the sum of all its parts. Connected with this 
 is the following: 
 
 AXIOM 20 
 
 A whole is greater than any of its parts* 
 
 PROBLEMS AND EXERCISES 
 
 1. Prove that an exterior angle of a triangle is greater than 
 either of the opposite interior angles. 
 
 2. Prove that the distance between the centers of two cir- 
 cles which lie entirely outside of each other, is greater than 
 the sum of the radii (see Fig. 218). 
 
 3. Prove that the distance between the centers of two cir- 
 cles, one of which lies entirely within the other, is less than the 
 difference of the radii (see Fig. 219). 
 
 * This axiom does not hold if we consider also negative numbers.
 
 1 84 
 
 Second- Year Mathematics 
 
 4. Draw a diagram representing a triangular field with 
 angles 36, 72, and 72. Measure and compare the sides. 
 Which side is smallest? 
 
 FIG. 218 
 
 5. Draw triangles with angles as follows: 
 
 (x) 4=35 5=65 C=8o 
 
 (2) ,4=20 5 = 130 C=3o 
 
 (3) 4=ioo 5 = 70 C = io. 
 
 Measure the sides of the triangles and 
 state in each case where the longest 
 side is situated with reference to the 
 angles, and where the shortest side is 
 situated. 
 
 6. What connection appears to 
 exist between the relative magnitude 
 of the angles of a triangle and the 
 relative magnitude of the sides ? 
 We shall now show that the truth which can be inferred 
 
 from the constructions in Problems 4 and 5 is a consequence 
 
 of the axioms and of propositions previously proved from the 
 
 axioms. 
 
 PROPOSITION V 
 207. Theorem: If two angles of a triangle are unequal, the 
 
 sides opposite them are unequal, the greater side lying opposite 
 
 the greater angle. 
 
 FIG. 219
 
 Inequalities in Triangles and Circles 185 
 
 Given: A A B C (Fig. 220). 
 
 To prove A C>A B. 
 
 Proof: Through the point B draw a line B D, so that 
 ZDBC=ZC. 
 
 (How may this construction be made ?) 
 
 B' 
 
 FlG. 220 
 
 Then DB=DC. Why? 
 Further, A D +D B > A B. Why ? ' 
 AOAB. Q.E.D. 
 
 EXERCISES 
 
 1. Prove that point D (Fig. 220) lies on the perpendicular 
 bisector of B C. 
 
 2. Prove the proposition (used in proving the above propo- 
 sition), which says that if a triangle has two equal angles the 
 sides opposite them must be equal. 
 
 Using these last two propositions, we now prove the con- 
 verse of Proposition V, stated thus.: 
 
 PROPOSITION VI 
 
 208. Theorem: // two sides of a triangle are unequal, the 
 angles opposite them are unequal, the greater angle lying oppo- 
 site the greater side. 
 
 Given: AA B C. A B< A C (Fig. 221). 
 
 To prove ZC< Z.B.
 
 1 86 
 
 Second- Year Mathematics 
 
 Proof: Can ZC=Z-6? Give a reason for your answer. 
 Can Z C > Z. B ? Give a reason for your answer. 
 How then must the angles C and B compare in size ? 
 
 A 
 
 C 
 
 FIG. 221 
 
 EXERCISES 
 
 1. Two sides of a triangle are 15 and 35; the angles oppo- 
 site them are #+5 and 5^ + 2, respectively. How large can 
 these angles be ? 
 
 2. Determine x from the data given in Fig. 222. Test to 
 see whether the conditions of Proposition VI are fulfilled. 
 
 3. The sides of a triangle are 
 28, 25, and 23; the angles opposite 
 them are 125 23;, 73 2*, and 
 4# 18, respectively. Determine 
 between what limits x must lie, 
 and what values the angles may 
 have. 
 
 4. A ship S is observed simultaneously from two forts at 
 A and B, respectively. As observed from A, S bears 50 N of 
 E and B bears 10 S of E; as observed from B, S bears 40 
 N of W, and A bears 10 N of W. From which fort is the 
 ship at the greater distance ? Find the distance from either 
 fort, the distance from A to B being 600 feet. 
 
 5. The angle of elevation of a tower from a point on a 
 horizontal plane 185 feet from the base, is 47; from a point 
 
 FIG. 222
 
 Inequalities in Triangles and Circles 187 
 
 215 ft. from the base the angle of elevation is 43. Between 
 what two numbers does the height of the tower lie ? 
 
 6. Prove that in a right-angled triangle, the hypotenuse is 
 the longest side. 
 
 PROPOSITION VII 
 
 209. Theorem: The perpendicular from a point to a line 
 is shorter than any other line connecting the point with the line. 
 
 Given the point A (Fig. 223) and the line X Y of indefinite 
 length ; 
 
 A 
 
 ' / 
 
 K 
 \ N> 
 \ 
 
 > 
 
 v\ 
 
 \ \ 
 
 c 
 
 D 
 
 FIG. 223 
 
 E F \ 
 
 x~ 
 
 ABiXY. 
 
 A C, A D, A E, A F, oblique lines. 
 
 To prove: A B< A C, A D, A E, A F, etc. 
 
 Suggestion: A B is a side in each of the right triangles A B C, A B D, 
 A B E, A B F, etc. Show how the theorem follows from this. 
 
 EXERCISES 
 
 1. Using Fig. 223, prove that A C=A F, if B C=B F, and 
 conversely. 
 
 2. Using Fig. 223, prove that A F >A E, if B F >B E. 
 Show also that angle A E F > angle A F E. 
 
 3. Using the same figure, prove that AF>AD, if 
 BF>BD. 
 
 Apply Exercises i and 2.
 
 1 88 Second- Year Mathematics 
 
 PROPOSITION VIII 
 
 210. Theorem: If two sides of one triangle are equal to 
 two sides of another triangle, but the angle included between the 
 two sides in the first is greater than the angle included by the 
 corresponding sides in the second; then the third side in the first 
 triangle is greater than the third side in the second. 
 
 Given A's A B C and D E F (Fig. 224) : 
 AB=DE; BC=EF; 
 To prove A C>D F. 
 
 \ >^T D ~F 
 
 \ X 
 
 FIG. 224 
 Proof: There are three possibilities: 
 
 II. 
 III. 
 
 I. Z.D> Z_A. In this case place AD E F on AA B C 
 so that D E falls on A B, D on A, E on B, and E F on the 
 same side of A B as B C. 
 
 The point F therefore falls below A C, as at F'. 
 
 Then draw F'C. 
 
 We have then: B F'=B C (?) 
 therefore ZBF'C = ZBCF' (?)
 
 Inequalities in Triangles and Circles 
 
 189 
 
 Further 
 and 
 
 therefore 
 therefore 
 
 ZAF'C >ZBF'C (?) 
 
 ZBCF'>ZACF' (?) 
 
 ZAF'C >ZACF' (?) 
 
 A C> A F' (?) 
 
 AC>DF (?) 
 
 . Q.E.D. 
 
 Another important truth has here been used for the first 
 time, namely: 
 
 AXIOM 21 
 If a>b and b>c, then a>c. 
 
 II. ZD= Z4. Again place AD E F on A A B C (Fig. 
 225), so that D E falls on A B, D on A, E on B, and E F on 
 the same side of A B as B C. 
 
 Now, the point F falls on A C, as at F". 
 
 Why cannot F fall on A C produced ( 36; Exercise 3) ? 
 
 We have consequently, 
 
 AF"<AC (?) 
 
 AC>DF (?) Q.E.D. 
 
 III. Z>< Z4 (Fig. 226). In this case, ZF>ZC. (?) 
 Place AD E F on AA B C so that E F falls on B C, E 
 on B, and F on C, and E D on the same side of B C as B A. 
 Repeat the form of proof given under I.
 
 190 
 
 Second-Year Mathematics 
 
 i. Prove Case III of the foregoing proposition by placing 
 ADEF on AABC, so that DE falls on A B. (See 
 Fig. 227. Use Proposition IV.) 
 
 F 
 
 
 FIG. 226 
 
 PROPOSITION IX 
 
 211. Theorem: If two sides of one triangle are equal to 
 two sides of another triangle, the third side of the -first triangle 
 being greater than the third side of the second; then the angle 
 opposite the third side of the first triangle is greater than the 
 angle opposite the third side of the second triangle.
 
 Inequalities in Triangles and Circles 
 
 191 
 
 Given A's P Q R and X Y Z (Fig. 228): 
 PQ=XY; QR=YZ; PR>XZ. 
 To prove Z<2>ZF. 
 
 P 7? X z 
 
 FIG. 228 
 
 Suggestions : What do we know about the triangle if Q = Y ? 
 
 Then, can Q = Y if P R>X Z, as here given ? 
 
 What do we know about P R and X Z if Q< Y ? Why ? 
 
 Then, is Q< Y, if P R>X Z, as here given ? 
 
 How, then, must angles Q and Y compare, if P R>X Z ? 
 
 Give full proof. 
 
 PROPOSITION X 
 
 212. Theorem: Any point not on the bisector of an angle 
 is unequally distant*- from the 
 sides of the angle. 
 
 Given ZAB C (Fig. 229); 
 B D bisects ZAB C; 
 a point P not on B D ; 
 PFJLBA; PExBC. 
 
 To prove P F^P E.f 
 
 Proof: As P F or P E must 
 intersect B D, assume that PF 
 intersects it at some point, as G. Fie. 229 
 
 * For distance of a point to a line, see 99, p. 53. 
 fFor symbol ^ see 198.
 
 192 Second-Year Mathematics 
 
 From G draw a line G H perpendicular to B C. Draw H P. 
 Then PE<PH. Why? 
 
 and P H< P G + G H. Why ? 
 /. PE<PG + GH. Why? 
 AlsoGH = GF. Why? 
 
 /. PE<PG + GF. Why? 
 or PE<PF, i.e., PE=PF. Q.E.D. 
 i. Repeat the proof taking P on the left of B D. 
 
 PROPOSITION XI 
 
 213. Theorem: Any point on the bisector of an angle is 
 equally distant from the sides of the angle. 
 
 Prove. (See FYM, 294.) 
 
 Using this conclusion and Proposition X, what can be 
 said about the location of a point known to be at equal dis- 
 tances from the sides of an angle? 
 
 The statement may be made thus: 
 
 PROPOSITION XII 
 
 214. Theorem: Any point equally distant from the sides of 
 a given angle must lie on the bisector of the angle; and any 
 point on the bisector of a given angle must be equally distant 
 from the sides of the angle; or thus: 
 
 The locus of all points equally distant from the sides of a 
 given angle is the bisector of that angle. 
 
 This is the second example of a locus met in this chapter. 
 
 EXERCISES 
 
 1. What is the location of all points in a plane which are 
 at a distance of 10 ft. from a given point P ? At a distance 
 of 4^ ft. from a point A ? At a distance a from a given point ? 
 
 2. What is the locus of all points in the plane at equal dis- 
 tances from two given parallel lines ?
 
 Inequalities in Triangles and Circles 193 
 
 What is the locus of all points in space equally distant from 
 two parallel lines ? 
 
 3. What is the locus of all points having a given distance 
 from a given line ? 
 
 4. What is the locus of all points in space known to have 
 a given distance from a given point ? Of all points in space 
 known to be at equal distances from two given points ? From 
 three given points ? 
 
 5. What is the locus of all points in space having a given 
 distance from a given line ? 
 
 6. Given a triangle ABC; it is required to determine a 
 point P that shall be equally distant from the three vertices 
 of the triangle. 
 
 The point P must be at equal distances from A and B (Fig. 230), 
 from B and C, and from C and A. 
 
 Where are all the points gathered 
 which are at equal distances from 
 A and B ? 
 
 On what line then must P lie ? 
 
 Where are all the points collected 
 which are at equal distances from 
 B and C ? A /- 
 
 On what second line then must p IG 2 , o 
 
 Plie? 
 
 How then may the point P be determined ? 
 
 7. Draw a circle passing through the vertices of an obtuse- 
 angle triangle. 
 
 How is the center of the circle found ? 
 
 8. Draw a circle passing through the vertices of an acute- 
 angle triangle. 
 
 9. Prove that the perpendicular bisectors of the three 
 sides of a triangle pass through a common point.
 
 194 
 
 Second- Year Mathematics 
 
 10. Given a triangle ABC: Determine a point that has 
 equal distances from the three sides of the triangle (Fig. 231). 
 
 1 1 . Inscribe a circle in an 
 acute-angle triangle. 
 
 If a circle is inscribed in a 
 triangle, the sides of the triangle 
 are tangent to the circle. 
 
 12. Inscribe a circle in an 
 obtuse-angle triangle. 
 
 13. Prove that thebisectors 
 
 FIG. 231 
 
 C 
 
 of the three angles of a triangle 
 pass through a common point. 
 14. Prove that the three altitude lines of a triangle drawn 
 from the vertices pass through a common point. 
 
 Through each vertex of triangle ABC draw a line parallel to the 
 opposite side forming a triangle, as D E F (Fig. 232). 
 
 Prove that the altitude lines of A B C are then perpendicular 
 bisectors of the sides of A^ E F. Then use Exercise 9. 
 
 FIG. 232 
 
 The propositions and exercises which have now been 
 proved enable us to prove some new propositions concerning 
 inequality of lines and angles in circles.
 
 Inequalities in Triangles and Circles 
 
 195 
 
 PROPOSITION XIII 
 
 215. Theorem: In the same circle or in equal circles, the 
 arcs subtended by unequal chords are unequal in the same order 
 as the chords; and, conversely, chords subtending unequal arcs 
 are unequal in the same order as the arcs. 
 
 Given circle A=circle B (Fig. 233): 
 Chord C D >chord E F. 
 To prove arc C D > arc E F. 
 
 FIG. 233 
 
 Proof: Draw radii A C, A D, B E, and B F. 
 
 Using Proposition IX, show that ZCAD>ZEBF. 
 
 Place Q B* on A, so that E B falls on A C, E on C, and 
 B on A, and F on the same side of C as D. 
 
 Then B F must come between A D and A C, as in position 
 A F'. ( ?) 
 
 Hence E F comes in the position C F', and F' falls on the 
 circumference between C and D. 
 
 Then arc C F'< arc C D (?) 
 also arc C F' = arc E F (?) 
 .'. arc E F< arc C D ( ?) Q.E.D. 
 
 * The symbol OB means the circle whose center is at B.
 
 196 Second-Year Mathematics 
 
 Converse. Given O A = B (Fig. 233) : 
 Arc CD>arcEF. 
 
 To prove chord C D > chord E F. 
 
 Proof: Draw radii AC, AD, B F, and B E, and place 
 O B on O A so that E B falls on C A, as above. 
 
 Since arc C D>arc E F, the point F will fall between C 
 and D, as at F', and the line B F will come on the same side 
 of A D as A C, as in position A F'. 
 
 Then we have: ZCAF'<ZCAD (?) 
 
 also ZCAF'=ZEBF (?) 
 
 .-. ZCAD >ZEBF (?) 
 
 Finally, use Proposition VIII to show that chord C D > 
 chord E F. Q.E.D. 
 
 PROPOSITION XIV 
 
 216. Theorem: In the same circle or in equal circles, un- 
 equal chords are unequally distant from the center of the circle, 
 the shorter chord lying at the greater distance; and, conversely, 
 chords unequally distant from the center are unequal, the chord 
 at the greater distance being the shorter chord. 
 
 Given P = oQ (Fig. 234): 
 Chord A B > chord D E. 
 To prove P P'< Q Q'. 
 
 Proof: 
 
 PP'JLAB, 
 
 QQ'_LDE. 
 
 Place Q on P, so that Q falls on P, D on B, and 
 chord D E in the position B C; then Q' will fall on Q", and 
 PQ"_LBC. Why? 
 
 DrawP'Q".
 
 Inequalities in Triangles and Circles 
 
 Then: P'B>Q"B. (?) 
 
 Hence Z B Q"P' > Z B P'Q" (?) 
 
 197 
 
 Observe that P P'J_A B, P Q"J_B C, and use Axiom 19 (p. 172). 
 
 FIG. 234 
 
 Hence P P' < P Q" ( ?) 
 also since PQ"=QQ' (?) 
 
 PP' <QQ' (?) Q.E.D. 
 
 Converse. Given (Fig. 234) 
 
 P = Q; PP'J_AB; QQ'_LDE; 
 P P'< Q Q' 
 
 To prove AB>DE. 
 
 Suggestion: Proceed with the steps of the foregoing demonstration 
 in the opposite order. 
 
 EXERCISES 
 
 1. Prove that a side of a regular inscribed decagon is less 
 than a side of a regular inscribed pentagon, inscribed in the 
 same circle, but that the side of the decagon is greater than 
 half the side of the regular pentagon. 
 
 2. Show that the greater the number of sides of a regular 
 inscribed polygon, the less is the length of one of its sides.
 
 198 Second-Year Mathematics 
 
 3. Prove that the distance from the center of a circle to 
 a side of a regular inscribed polygon is greater, the greater the 
 number of sides of the polygon. 
 
 4. The length of the chords AB and B C being 6# 14 
 and 4^ + 20, respectively (Fig. 235), and the lines P P' and 
 P P' 7 being 16 and 10, determine x and the chords. 
 
 We have: P'B = 3^-7 (?) 
 P"B = 
 
 Then: (3#-7)2 + i6* = Pl (?) (see Problem 9, p. 101). 
 
 and (2# + io) 2 + io 2 = PlB 2 (?) 
 
 = (2* + 10)2 
 
 or : gx 2 42X + 49 + 256 = ^x 2 + 403; +100+100 
 
 5^ 82^+105 =o. 
 
 82T/82*-4 5 105 . 
 ' * = Ip - (see 192 (C), p. 163). 
 
 82 68 
 
 x = - = i 1 5) or if 
 10 
 
 Then AB = y6, or -5! 
 CB=8o, or 25! 
 
 How is the truth of Proposition XIV illustrated by these 
 answers ? 
 
 5. The length of the lines A B and B C, P P', and P P" 
 (Fig. 235) being denoted by 1 I} 1 2 , d 1} and d 2 , respectively,
 
 Inequalities in Triangles and Circles 
 
 199 
 
 determine the unknown number in each of the following cases. 
 In every case test by Proposition XIV. 
 
 
 /, 
 
 /, 
 
 d, 
 
 rf, 
 
 (l).. 
 
 2a 7 
 
 43 14 
 
 2 
 
 i 
 
 (2).., 
 
 6 
 
 12 
 
 U-\- II 
 
 3M + 4 
 
 (3).. 
 
 x+* 
 
 *+5 
 
 6 
 
 4 
 
 (4).. 
 
 4/+ 14 
 
 IO/ 2 
 
 6 
 
 7 
 
 
 
 
 
 
 217. The formula for the solution of the general quadratic 
 equation: 
 
 ax 2 +bx+c=o, 
 
 30 
 
 (Compare 192 (C).) 
 
 Calling the first root x^ and the second root x 2 , we have: 
 
 By addition, we find: 
 
 - 2 b -b b 
 
 x I +x,= = or 
 
 2<z a a 
 
 By multiplication: 
 
 X-.X-, 
 
 b 2 (b 3 
 
 (2) 
 
 (3) 
 
 (4) 
 
 By means of Formulas (3) and (4) the sum of the roots and 
 the' product of the roots of a quadratic equation may be deter- 
 mined without solving the equation.
 
 2OO Second-Year Mathematics 
 
 Having given the equation 
 
 5# 2 
 we determine 
 
 \ 12 
 ) = > 
 / 5 
 
 b / i 2\ 
 
 the sum of the roots. 
 a 
 
 x t x a =-= -, the product of the roots. 
 
 v) 
 
 Knowing the sum and the product of the roots, we can 
 determine something about the sign of each of the roots, viz.: 
 
 I. If x j x 2 is positive, the roots must have like signs; i. e., 
 they are both positive, or both negative. Why ? 
 
 And if. Xi+x a also is positive, each of the roots is positive; 
 but if X-L +x 2 is negative, each of the roots is negative. Why ? 
 
 II. If x^Xz is negative, the roots must have unlike signs; 
 i. e., one is positive and the other one is negative. Why? 
 
 And if Xi+x a is positive, the positive root is greater in 
 numerical value; but if x I +x a is negative, the negative root 
 is greater in numerical value. Why ? 
 
 EXERCISES 
 
 218. By means of the above discussion, determine in each 
 of the following equations the sum and the product of the 
 roots, without solving the equations (but clearing of fractions 
 when necessary). Discuss also the signs of the roots: 
 
 i. sx z + i2X 18=0 
 
 3. 4# 2 I2X 8=O 
 
 4- 
 
 T.X 2 4 2X 
 
 5. ox = 
 
 V * X
 
 Inequalities in Triangles and Circles 201 
 
 219. From the formula for the solution of the general quad- 
 ratic equation, may be derived a way of testing whether the 
 answers of a given quadratic equation have a meaning or not 
 in our ordinary number-system, which is called the system of 
 real numbers. 
 
 Thus, if b 2 4ac is negative, the expressions for the roots 
 of the equation contain the square root of a negative number. 
 But the square root of a negative number does not exist among 
 the numbers which we have thus far dealt with; i. e., it is 
 not a real number. Why not ? 
 
 So that we have the following conclusions: 
 
 If b 2 4oc<o, the roots are not real. 
 
 If b 3 4ac>o, the equation has two unequal roots. 
 
 What can be said about the roots of the equation in case 
 b 2 40c=o? 
 
 EXERCISES 
 
 1. State without solving the equations, whether the roots 
 of the equations 1-5, 218, are real. 
 
 2. Determine the value of m for which the chords in Fig. 236 
 actually exist. 
 
 FIG. 236 
 
 We have: (w*) 2 +4 2 = (3*-4) 2 + 5 2 . Why? 
 i. e., w 2 # 2 + i6=9# 2 24^+16 + 25 
 (m* <j)x 2 + 24* 25 =o.
 
 Second-Year Mathematics 
 
 In this case a = m 2 9 
 6 = 24 
 c=-25 
 
 That the chords may actually exist x must be a real num- 
 ber. This will be true if 
 
 (24) 2 -4(w 2 - 9 )(-25)>o 
 i.e., 576 + ioofw 2 9oo>o 
 or if w 2 >if 
 i. e., if w>f or w< -. Why? 
 
 3. The meaning of 1 I} 1 2 , d x , and d 2 being as in Exercise 5, 
 p. 198, determine the value of m for which the chords have 
 real existence: 
 
 
 /i 
 
 1, 
 
 d, 
 
 d, 
 
 (l).. 
 
 2 1/ x m 
 
 6x 4 
 
 e 
 
 4 
 
 (2).. 
 
 8 
 
 i2m 
 
 * 
 
 5* 
 
 (3) 
 
 6x-S 
 
 2X+6 
 
 2m 
 
 i 
 
 (4).. 
 
 2X8 
 
 IO#+ 2 
 
 j/ 2x-\- yn 
 
 -j/ m x 
 
 
 
 
 
 
 * In this equation b is o. 
 
 PROBLEMS AND EXERCISES 
 220. Solve the following problems and exercises: 
 i . Construct a triangle ABC, the sides a and b and the 
 angle A, opposite one of them, being given (Fig. 237). 
 
 A' 
 
 C 
 
 FIG. 237 
 
 FIG. 238 
 
 Draw a line A C equal in length to b (Fig. 238). At the 
 extremity A of A C construct an angle equal to the given
 
 Inequalities in Triangles and Circles 
 
 203 
 
 angle A . Now, with C as a center and a as a radius, strike an 
 arc meeting the second side of Z.A at two points B and B'. 
 Both of the triangles A C B and A C B' have the required parts. 
 
 2. Draw the perpendicular C B" from C to A B (Fig. 238). 
 
 How many triangles would have been formed in case 
 a<CB"? Why? 
 
 How many if a = C B" ? Why ? 
 
 FIG. 240 
 
 3. Prove, that Z.B in /^A B C must be acute. 
 
 4. Prove, that Z.B' in /\A B'C must be obtuse. 
 
 5. Prove, that whenever the construction in Problem i 
 furnishes two triangles, one of those must be an acute-angle 
 triangle, and the other one an obtuse-angle triangle.
 
 2O4 
 
 Second- Year Mathematics 
 
 FIG. 241 
 
 6. Given two triangles ABC and D E F, having A B =D E, 
 B C=E F and Z.C opposite A B = Z.F, opposite D E. 
 
 7. Prove by the method of 
 superposition and by the use of 
 Exercises i to 6, that A A B C = 
 AD E F, in case Z_A and /_D 
 are both acute, both obtuse, or 
 both right angles. 
 
 221. The distance from a 
 point P to a circle whose center 
 is O, is the length of PA, that 
 portion of the line P O which 
 lies between P and the point 
 where P O meets the circle (see Fig. 239). 
 
 Hence the center O 2 of circle C 2 (Figs. 240 and 241) is at 
 a distance r 2 from circle C t ; the center O x of circle C z is at 
 a distance r t from circle C 2 . 
 
 EXERCISES 
 
 1. Find the locus of the points whose distance from a 
 given circle C is equal to a given length. 
 
 2. Find the locus of the center of the circles of given radius 
 a, which are externally or internally tangent to a given circle 
 (see Fig. 242). 
 
 3. Determine the points which shall lie at a given distance 
 from both circle A and circle B (Fig. 242). 
 
 Where are all the points collected that have the required distance 
 from circle A ? 
 
 Where are all the points collected that have the required distance 
 from circle B ? 
 
 Where are then found the points which have the required distance 
 from both circle A and circle B ? 
 
 4. Construct as many circles as possible with radius of 
 i inch, which shall be tangent at the same time to circle d
 
 Inequalities in Triangles and Circles 
 
 205 
 
 of radius 5 inches and to circle C 2 of radius 4 inches, when the 
 center O z and O 2 of the circles C r and C 2 are 6 inches apart. 
 
 5. Determine what relation must hold between the radii 
 fj and r 2 of circles C t and C 2 the distance, d, between their 
 centers and a line p, in order that there may be 8 circles of 
 radius p tangent to both C r and C 2 . 
 
 FIG. 242 
 
 6. Determine under what conditions there will be 6 circles; 
 2 circles; none, of the nature described in Exercise 5. 
 
 7. Can there ever be 4 circles of that kind? 
 
 Exercises for Algebraic Solution 
 222. Solve the following quadratics: 
 
 (1) r 2 gr 36=0 (6) 
 
 (2) / 2 + i5*=-44 (7) 
 
 (3) s 2 -72=6s (8) 
 
 (4) 3W 2 =6 7w (9) 
 
 = 28^ io;y 2 + 5 
 
 (5) 
 
 = 1# 2 20 
 
 (10)
 
 2o6 Second-Year Mathematics 
 
 223. Solve the following quadratics for y: 
 
 (1) i4^ = 3-8j 2 (5) 8^ 2 +8^ + 2c 2 = -i9C 2 -6>- 2 
 
 (2) a-y 2 = (ia)y (6) y 3 +py+q=o 
 
 (3) 2i6 2 = 23&j 6y 2 (7) aj 2 +&^+c=o 
 
 (4) 28m 2 = -i7&j+3>' 2 (8) xy*+zy=s. 
 
 224. Determine by inspection the nature of the roots of the 
 following quadratics: 
 
 (1) * 2 +6* + 5=o (6) 2$p 2 + i=o 
 
 (2) 4y 2 -gy + 2=o (7) / 2 ~ 
 
 (3) 4W 2 8w+4=o (8) s 2 
 
 (4) 3y 2 + 7? + 2=o (9) ^ 
 
 (5) r 2 -i=o (10) 8/ 2 -
 
 CHAPTER VI 
 AREAS OF POLYGONS 
 
 225. To measure the size of a plane surface is to compare 
 that surface with a standard surface, a unit of measurement. 
 A square is used as the unit of measurement for plane sur- 
 faces. The number of such units contained in a given sur- 
 face expresses the area of the surface in terms of that unit. 
 
 Thus, if a rectangle has sides of 4 in. and 6 in., it is easily 
 seen that a square inch may be laid off on it 24 times. We 
 say that 24 expresses the area of the rectangle, in terms of 
 square inches. 
 
 226. The areas of polygons may be found by dividing the 
 polygons into triangles, as in Figs. 243 and 244, and then 
 adding the areas of the triangles. 
 
 FIG. 243 
 
 FIG. 244 
 
 The Area of the Triangle 
 
 227. Thus it is seen that it is important to know how to 
 compute the area of a triangle. Several formulas will be worked 
 out by which the area of a triangle can be found. The parts 
 of the triangle that are known will enable the pupil to decide 
 which formula can be used. 
 
 207
 
 208 Second-Year Mathematics 
 
 Fig. 245 shows two triangles each one-half of a parallelo- 
 gram. Thus if we are able to find the area of a parallelogram, 
 the area of the triangle is found. 
 
 FIG. 245 FIG. 246 
 
 Fig. 246 shows how the area of a parallelogram can easily 
 be compared with the area of a rectangle. The question 
 arises, whether the rectangle and the parallelogram of Fig. 246 
 are equal to each other, and, more generally, under what con- 
 ditions such figures are equal to each other. This question 
 is answered by Exercise i, 229. 
 
 If we know how to compare parallelograms, we can com- 
 pare parallelograms with rectangles because a rectangle is a 
 special case of a parallelogram. 
 
 228. We have thus the following chain of problems leading 
 to the problem of computing the area of the triangle: 
 
 (1) to compare parallelograms 
 
 (2) to compare a parallelogram and a rectangle 
 
 (3) to compute the area of a rectangle 
 
 (4) to compute the area of a parallelogram 
 
 (5) to compare a parallelogram and a triangle 
 
 (6) to compute the area of a triangle. 
 
 PROPOSITION I 
 
 229. Theorem: Parallelograms having equal bases and 
 equal altitudes are equal. 
 
 Hypothesis: In parallelograms ABCD and EFGH 
 (Fig. 247, (i)), h=h'. AB=HG;
 
 Areas of Polygons 209 
 
 Conclusion: ABCD=EFGH. 
 
 Proof: Place ABCDonEFGH making A B coincide 
 with H G. 
 
 Then D C must fall in line E F, in position D'C', for the 
 altitudes of the parallelograms are equal. 
 
 XL 
 
 (2) 
 
 Prove AED'H^AFC'G. If triangle E D'H is sub- 
 tracted from the.quadrilateral H E C'G, H D'C'G ( = A B C D) 
 remains. If triangle F C'G is subtracted from the quadri- 
 lateral H E C'G, E F G H remains. 
 
 .'. Parallelogram A B CD=parallelogram E F G H. 
 
 Prove the theorem for Fig. 247 (2). 
 
 Hence, parallelograms having equal bases and equal alti- 
 tudes are equal. Q.E.D. 
 
 EXERCISES 
 
 1. Prove that the area of a parallelogram equals that of a 
 rectangle having the same base and altitude (see Fig. 248). 
 
 2. To find the number of surface units (area) in a rectangle, 
 when the number of linear units in the base and in the altitude 
 are known.
 
 Second- Year Mathematics 
 
 On squared paper draw a rectangle with corners at the points 
 
 (*> !) (5> !). (5, 3). C 1 , 3) ( see FJ g- 2 49)- 
 
 By counting the squares in A B C D (Fig. 249) find the area in 
 terms of the squares of the paper. 
 
 FIG. 248 
 
 3. On squared paper draw rectangles having the following 
 dimensions: 
 
 Base 
 
 i 
 
 7 
 
 I? 
 
 10 
 
 "LA. 
 
 
 
 
 
 
 
 Altitude 
 
 c 
 
 4 
 
 } 
 
 2 
 
 7 
 
 
 
 
 
 
 
 4. Using as surface unit a square whose side is the linear 
 unit, find the area of the rectangles in Exercise 3. 
 
 230. How can the number of surface units in a rectangle 
 be obtained from the number of linear units in the base and 
 altitude ? 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 3) 
 
 
 
 
 '5,3) 
 
 
 A 
 
 
 
 
 
 2 
 
 
 D 
 
 
 
 
 
 C 
 
 
 
 | 
 
 
 
 I 
 
 ^/^ 
 
 
 i 
 
 4 
 
 l i 4^ i 
 
 > 6 
 
 
 FIG. 249 
 
 b-. 
 
 FIG. 250 
 
 Let & denote the length of the base and h the length of the 
 altitude of a rectangle A B C D (Fig. 250). 
 
 Then the linear unit can be laid off b times on A B and 
 h times on A D.
 
 Areas of Polygons 
 
 211 
 
 Through the points of division of A B draw lines parallel 
 to A D, as E E', F F', etc. (Fig. 251) 
 
 Then rectangle A B C D is 
 divided into b congruent rec- 
 tangles R. Through the points 
 of division of 'A D, draw lines 
 parallel to A B, as K K', L L', 
 etc 
 
 Then each rectangle R is 
 divided into h congruent 
 squares. Denoting the num- 
 
 *-* 
 
 .' / 
 
 m 1 
 
 C 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 A, 
 
 R 
 
 
 
 /' 
 
 1 L 
 
 !* 
 
 
 
 
 K' 
 
 
 
 
 i 
 
 
 
 
 
 A E F 
 
 _'_ t 
 
 FIG. 251 
 
 her of squares in A B C D by S, it follows that 
 
 S = b-h. (a) 
 
 This equation may be stated in words thus: 
 
 PROPOSITION II 
 
 231. Theorem: The area of a rectangle equals the product 
 of the base by the altitude. 
 
 EXERCISES 
 
 i. Verify formula (a) for 6=4, h = i .4. 
 Lay off A 6=4 cm. and A D = i .4 cm. (Fig. 252). 
 
 M f 
 
 i 
 
 % 
 
 
 
 . 
 
 
 
 
 
 j 
 
 
 
 i 
 
 
 / A - 
 
 4 4 
 
 FIG. 252 
 
 Then A B C D = A B F E +E F C D. (What axiom ?) 
 (i) Counting squares, we find 
 
 A B F =4 sq. cm., E F C D=4X 1 4 (r sq. cm. 
 
 ABC D=4 sq. cm. +4X1*0 sq. cm. 
 
 =4(1 + T 4 ff ) sq. cm. = 5. 6 sq. cm.
 
 212 
 
 Second- Year Mathematics 
 
 (2) By formula (a) we find S=b ^=4X1 .4 = 5.6. 
 
 2. Verify formula (a) for 6=3.8, /z = 5; for 6=6, h = 2.6; 
 for 6=3, A =4. 2. 
 
 3. Verify formula (a) for 6=4.6, ^ = 2.4 (see Fig. 253). 
 
 FIG. 253 
 
 . cm. 
 = (8 + 1.2+1.6+. 24) sq. cm., etc. 
 
 4. Verify formula (a) for 6 = 1 .4, ^=0.32. 
 
 232. Exercises i, 2, 3, and 4, 229, show that formula (a) 
 holds when 6 and h are integers (whole numbers).. 
 
 Exercises i, 2, 3, and 4, 231, show that formula (a) 
 holds also when 6 and & are decimal fractions. 
 
 233. Now let 6 and h be irrational numbers, such as 
 
 6 = 1/12=3.464101 . . . . , 
 and ^ = 1/27 = 5.196152 ..... 
 
 If the theorem holds in this case, we should find: 
 
 lr2Xl27 = l2 2 3 3 3 =i8. (?) 
 
 Taking for 6 and h the values given in I, II, III, and IV 
 below, we find the corresponding values of 5 by multiplying.
 
 Areas of Polygons 
 
 213 
 
 
 6 
 
 h 
 
 5-6XA 
 
 I 
 
 ? .464 
 
 <. . 106 
 
 17.008044 
 
 II 
 
 3 .4641 
 
 ? .1061 
 
 17.00081001 
 
 Ill 
 
 3 .46410 
 
 <; . 10611; 
 
 1 7. 00008 32 1 <; 
 
 rv 
 
 ? .464101 
 
 s . 1061^2 
 
 17 QQQQQS^Q? 1 ^ 
 
 
 
 
 
 We see that by taking: 
 
 I, b and h accurate to within , 5 differs from 18 by less 
 
 I,OOO 7 
 
 , -S 1 differs from 18 by 
 
 S differs from 18 
 
 differs from 18 
 
 II, b and h accurate to within 
 
 less than i 
 
 III, & and /& accurate to within 
 
 1,0 
 
 by less than . 
 10,000 
 
 IV, b and h accurate to within ^ , 
 
 I ,OOO,OOO 
 
 by less than - . 
 t 100,000 
 
 By taking b and h to a sufficiently large number of decimal 
 places, the difference between S and 18 may be made as small 
 as may be wished. 
 
 To prove from these facts that our theorem holds true 
 when b and h are irrational numbers would require a fuller 
 discussion of such numbers than is desirable here. Sufficient 
 has been shown, however, to make us feel that this conclusion 
 is correct, for the inexactness of the result seems to come from 
 our inability to express exactly, without the root sign, the 
 dimensions of the rectangle, and not because the theorem is 
 untrue. 
 
 i. Make a similar discussion for 6 = 1/5 an( ^ ^ = 1/7; f r 
 & = j/2 and fc = i/3. 
 
 EXERCISES 
 
 i. Prove that any two parallelograms are to each other as 
 the product of their bases and altitudes.
 
 214 Second-Year Mathematics 
 
 2. Prove that the areas of parallelograms having equal 
 bases are to each other as their altitudes. 
 
 3. Prove that the areas of parallelograms having equal 
 altitudes, are to each other as their bases. 
 
 4. Prove that the area of a parallelogram is equal to the 
 product of the base and altitude. (Use Exercise 2.) 
 
 5. Prove that the area of a triangle is one-half the product 
 of the base and altitude. (Use Exercise 4 and Fig. 254.) 
 
 Thus the area of a triangle can be computed if the base and the 
 altitude are known. 
 
 6. Show that the area of a triangle can be expressed in 
 terms of any two sides, and the sine of the included angle. 
 
 In Fig. 254 let a = side D B, 6 = side D A; then ( 137, p. 104) 
 show that h = a sin D. .'. Area A B D = %b h = %b a sin D. 
 
 7. Prove that triangles having equal bases and equal alti- 
 tudes are equal. 
 
 8. Prove that the areas of triangles are to each other as the 
 products of the bases and altitudes. 
 
 9. Prove that the areas of triangles having equal bases are 
 to each other as the altitudes. 
 
 10. Prove that the areas of triangles having equal altitudes 
 are to each other as the bases. 
 
 u. To bisect a triangle by a line drawn from any vertex 
 to the opposite side. 
 See Exercise 7.
 
 Areas of Polygons 215 
 
 12. To divide a triangle into three equal triangles (trisect) 
 by lines from any vertex to the opposite side. 
 
 234. The area of a triangle can be expressed in terms of 
 the perimeter of the triangle and the radius of the inscribed 
 circle, by dividing it into three triangles whose common alti- 
 tude is the radius, and whose bases are the sides. 
 
 i. Prove that the area of a triangle is equal to one-half the 
 perimeter times the radius of the inscribed circle (see Fig. 255). 
 
 c 
 
 FIG. 255 
 
 The area of a triangle may i>e expressed in terms of the sides of the 
 triangle and the radius of the circumscribed circle. 
 
 2. Prove that the area of a triangle is equal to the product 
 of the three sides divided by four times the radius of the cir- 
 cumscribed circle. (See 189.) 
 
 The area of a triangle can be expressed in terms of the sides 
 alone, but in the proof of the formula, the following theorem 
 is needed: 
 
 PROPOSITION III 
 
 235. Theorem: The square on the hypotenuse of a right 
 triangle is equal to the sum of the squares on the sides including 
 the right angle. 
 
 Hypothesis: ABC (Fig. 256) is a triangle having a right 
 angle C; S I} S 3 , and 5" are the squares on the sides a, b, and 
 c, respectively;
 
 2l6 
 
 Second-Year Mathematics 
 
 Conclusion: S=S I +S 2 . 
 
 Proof: Draw CD_l_AB, dividing 5 into rectangles R T 
 and R 2 . Draw AE and C F. Show that triangle EBA 
 and Sj have equal bases and altitudes. 
 
 r D 
 
 FIG. 256 
 
 Then triangle E B A = J5,. 
 Similarly, prove that triangle FB C = J 
 But ABE^FB C. 
 For EB=BC (?) 
 AB=BF (?) 
 ZABE=ZFBC (?) 
 From (i), (2), (3) it follows that ^ = 1 
 
 Similarly, draw B G and C H, and prove S a =R 3 
 Therefore S=St+S a . Q.E.D. 
 
 (i) 
 (2) 
 (3) 
 
 (4)
 
 Areas of Polygons 217 
 
 This theorem is called the "theorem of Pythagoras" after the 
 Greek mathematirian Pythagoras, 569-500 B. c., who first proved it. 
 It is one of the most famous propositions of geometry, and numerous 
 proofs are on record. The proof which precedes is due to Euclid. A 
 few other proofs are given later (see 238). 
 
 EXERCISES 
 
 1. The area of a rectangle is i6y 3 4y, and the altitude is 
 4jy. The diagonal is 10, find y, and the sides of the rectangle. 
 
 2. The diagonal of a rectangle is 20 and one side is 16. 
 If x 3 +4* denotes the area, what does x equal ? 
 
 3. The area of a rectangle is 300 and one side is 15. Find 
 the perimeter. 
 
 4. The perimeter of a rectangle is 84, and the sides are to 
 each other as 3 to 4. Find the diagonal. 
 
 5. The area of a rectangle is 48, and the base is 8. Find 
 the diagonal. 
 
 6. The altitude of a rectangle is 24 and the area is 768. 
 Find the diagonal. 
 
 7. The area of a rectangle is 2C 3 2C, and the base is 20. 
 Find the diagonal. 
 
 8. The diagonal of a rectangle is 15 and the altitude is 9. 
 Find the area. 
 
 9. One side of a rectangle is 2X. The diagonal is x 2 + 1 , and 
 the area is 12. Find x, and the sides, and the diagonal. 
 
 Use the factor theorem. 
 
 PROPOSITION IV 
 
 236. Problem: To find the area of a triangle, in terms of 
 its sides, to be 1/5(5 a) (sb) (s c). 
 
 Hypothesis: In triangle ABC (Fig. 257) the sides a, b, 
 and c are known.
 
 2l8 
 
 Second- Year Mathematics 
 
 To express the area in terms of a, b, and c. 
 Proof: Area A B C = J6 h (?). 
 
 (i) 
 
 C 
 
 This gives the area in 
 terms of one side and the 
 altitude h, which is not 
 known. Therefore we will 
 express h in terms of the 
 sides, to substitute for h 
 
 < " in equation (i). 
 
 FIG. 257 
 
 h 2 =c 2 -(b-a') 2 (why?). (2) 
 
 h*=a 2 -a' 2 (why?). (3) 
 
 We must now eliminate a', which is not one of the three sides. 
 Subtracting, c 2 -a 2 +a' 2 -(b-a') 2 =o. (4) 
 
 Therefore c 2 a 2 b 2 + 2ba'=o. (5) 
 
 b 2 -c 2 +a 2 
 
 Solving for a', we find a'=- 
 
 (6) 
 
 Substituting in (3) the value of a' found in (6), we get 
 
 (7) 
 
 Equation (7) expresses A 2 in terms of the sides a, b, and c. 
 
 We could now substitute the value of h in equation (i) 
 and have a formula for the area of A B C in terms of a, b, 
 and c. But in order to get a more symmetrical result, the value 
 of h in (7) will be changed in form before substituting in (i). 
 b 2 -c 2 +a 2 \/ b 2 -c 2 + 
 
 From (7) h 2 = 
 
 \ 2b 
 
 ;r a* /' 
 
 +a 2 2ab b 2 +c 2 a 2 
 
 2b 
 
 (a+b) 2 -c 2 
 
 2b 
 c 2 -(a-b) 2 
 
 2 b 
 
 2b 
 
 , . 
 , (why?) 
 
 2b 
 
 (c+a-b)(c-a+b) 
 
 2b
 
 _ 
 Why? 
 
 Areas of Polygons 219 
 
 Denote (a + 6+c) by 2s. 
 
 Then a+bc = 2s 20 = 2(5 c) \ 
 
 a = 25 2a = 2(i- a) ' (9) 
 
 b = 2s 2b = 2(sb') / 
 Substituting (9) in (8) 
 
 2(sc)- 2s 2(56) 2(3 a)_45 (5 a)(s b)(s c) 
 ~ ~~ 
 
 Therefore h=jVs(s -a) (s-b) (s -c). Why? (10) 
 
 Substituting (10) in (i), 
 
 ABC = $6 \ 
 b 
 
 Therefore A B C = \s(s-a)(s-b)(s-c). (i i ) Q.E.D. 
 
 Since a + b + c = 2s, then s = % (a + b + c). 
 
 Formula (n) expresses the square root of the product of the half- 
 sum of the three sides, and the half-sum diminished by each side in 
 succession. 
 
 Problems 
 
 1. The sides of a triangle are 3, 5, and 6. Find the area. 
 Using the formula (n) of 236, 
 
 the area = 1/7 (73) (75) (7 6) = 1/7 -4 2 i =21/14, or 7.482 
 approximately. 
 
 2. The sides of a triangle are 34, 20, and 18. Find the 
 area. 
 
 3. The sides of a triangle are 10, 6, and 8. Find the area. 
 
 4. The sides of a triangle are 90, 80, and 26. Find the area. 
 
 5. The sides of a triangle are 70, 58, andi6. Find the area. 
 
 AREAS AND ALTITUDES 
 
 237. The altitudes of the triangle A B C to the sides a, b, 
 and c are h a , hf,, and h c , respectively (Fig. 258).
 
 Second- Year Mathematics 
 
 =^V s(sa}(sb}(sc) (i) (See 236, Formula (10).) 
 
 h a =-Vs(s-a)(s-b)(s-c) (2) 
 a 
 
 h c =-Vs(s-a)(s-b}(s-c). (3) 
 
 G 
 
 How can (2) and (3) be obtained from (i) by inspection? 
 
 Problems 
 
 i. In the triangle ABC, a = io, 6 = 17, c = 2i. Find h a . 
 
 2 
 
 2 / - 
 
 h a =-v s(s a)(s b)(s c) 
 
 1) =24 
 
 s a = i4, sb j, sc=3. 
 Substitute these values in the formula, and 
 
 24 14 7 ' 3=4 -3-2. 2- 7- 7- 3 = 
 -9.4- 49=i(2 3 ' 2 7)= = i6f 
 
 2. Find the area and the altitudes of each of the following 
 triangles: 
 
 (0 = 35. 6 = 29, c= 8 (4) a = i5, 6 = 20, ^ = 25 
 
 (2) = 70, 6=65, c= 9 (5) a= 8, 6= 8, c= 8 
 
 (3) a=45> ^=40, ^ = 13 (6) a = i7, 6 = 10, c= 9
 
 Areas of Polygons 221 
 
 3. The sides of a quadrilateral are as follows: A 6=29, 
 B C=8, C D=-28, D A = 2i, and the diagonal A C=3o. Find 
 its area, and the distance from D to A C. 
 
 4. The area of a triangle in terms of its sides and the 
 radius of the circumscribed circle is - (see 234, 2). Let 
 
 T denote the area of the triangle, then T . Solving this 
 
 equation for r, we get r==.. Using this formula for r, 
 
 4* 
 
 find the radius of the circles circumscribed about the tri- 
 angles of Problem 2. 
 
 5. To find the altitude and area of an equilateral triangle in 
 terms of its side. 
 
 Let each side be a. The altitude h, bisects the base. Why ? 
 & 2 =a 2 -($a) 2 =a 2 -a 2 = a 2 The area S=$a*h 
 
 -* VI 
 
 5=io 
 
 A =-1/3 or iai/3". S=la a i/3- 
 
 6. Find the areas of the following equilateral triangles 
 (a=side): (i) a = i2, (2) a = io, (3) a=4, (4) a=8, (5) a=c+d, 
 (6) a = 2mn. 
 
 Compare your answer of (4) with 2 (5). 
 
 7. Find the altitudes of the triangles of Problem 6. 
 
 8. Find the side of an equilateral triangle whose area is 
 (i) ^1/3 (2) 25!/3 (3) i 2 (4) 101/3. 
 
 9. Find the altitude of each of the triangles in Problem 8. 
 
 10. Find the area of a regular hexagon whose side is 
 (!) 12 (2) 8 (3) 10 (4) 7 (5) -6. 
 
 See FYM, p. 360, 6.
 
 222 
 
 Second-Year Mathematics 
 
 238. Prove the Pythagorean proposition by each of the 
 following figures (see footnote, p. 216). (In each figure C is 
 the vertex of the right angle of the right triangle ABC.) 
 
 Figs. 259 and 260. From the dimensions given in the 
 figures, find the areas of the triangles, and the square on c 
 by subtraction or addition. Use the algebraic method. 
 
 b K a f 
 
 FIG. 260 
 
 FIG. 261 
 
 Fig. 261. G C is a square on a+&. GE- 
 A E, ED, B D. Prove A B D E a square, 
 two rectangles G O 
 
 =D F=a Draw 
 Show that the 
 and O C are 
 
 equivalent to the four right triangles. 
 Subtract each in turn from G C. 
 
 Fig. 262. Complete the rectangle 
 CJKT. DrawCK. DrawCG||BE. 
 Show that K C G is a straight line. 
 Prolong D A to HJ, and EB to KT. 
 Compare A I K C with the square 
 AHJC, and KB 
 with C S and F E. 
 Fig. 263. Con- 
 struct A B D E a square. Through E 
 draw aline || to AC. Draw JLS as indi- 
 cated. Show that BC=HG, and that 
 D G and C K are squares, and that the p IG 
 
 right triangles are equal. 
 
 Fig. 264. Construct E H || B C, DK || BF. Draw C K L 
 and prove it to be || to B E. Compare A C K D with A L, etc.
 
 Areas of Polygons 
 
 223 
 
 Fig. 265. Construct DEF=ABC. Draw CG and CJ. 
 Prove G C J a straight line, and the quadrilaterals G H I J, 
 GABJ, CADF, CBEF equal, etc. 
 
 FIG. 264 
 
 Fig. 266. Find b in terms of a and c applying the law of 
 the secant and tangent. Use the algebraic method. 
 
 239. It has now been proved in 
 many ways that the square on the 
 hypotenuse of a right triangle equals 
 the sum of the squares on the sides 
 including the right angle. 
 
 FIG. 266 
 
 FIG. 267 
 
 Imagine the angle ABC 
 (Fig. 267) to increase, leaving 
 the lengths of the sides A B 
 and B C unchanged. Then 
 the squares on A B and B C 
 are not changed in size, but 
 as the end-points A and C 
 of A B and B C, are farther 
 apart, the square on AC 
 increases. 
 
 Therefore in an obtuse- 
 angled triangle, as Fig. 268,
 
 224 
 
 Second-Year Matliematics 
 
 the square on the side opposite the obtuse angle is greater 
 than the sum of the squares on the other two sides. 
 
 FIG. 268 
 
 FIG. 269 
 
 240. In a similar way, by decreasing angle ABC, until 
 it becomes acute, as in Fig. 269, we find that the square on 
 the side opposite the acute angle is less than the sum of the 
 squares on the other two sides. 
 
 EXERCISES 
 i. What is the relation between a 2 , b 2 , and c 2 (Fig. 270) 
 
 whenZ<9o? 
 when Z-B>9o? 
 
 2. How does the Z-6 com- 
 pare with a right angle when 
 
 c 
 
 FIG. 270 
 
 (2) b 2 a 2 +c 2 ? 
 
 (3) 6 2 <a 2 +c 2 ? 
 
 3. Is the Z.B > or < 90 when a, &, and c are, respectively,
 
 Areas of Polygons 225 
 
 (1) 3> 5^4 (5) 9. 14, 12 
 
 (2) 5, 13, 12 (6) z 2 -y 2 , z 2 +y 2 , 2zy 
 
 (3) 5,8,6 (7) x 2 -i,x 2 , 2X 
 
 (4) 6, 12, 8 (8) 2k*- 3 h 2 , 2 (k 2 +h 2 ), 2^hk? 
 
 241. The following two theorems will show by how much 
 the square on one side of a triangle differs from the sum of 
 the squares on the other two sides. 
 
 Problem : Let /_ B be an acute angle of triangle ABC 
 (Fig. 271). 
 
 c c 
 
 Let C D be perpendicular to A B. 
 To prove: b 2 =a 2 +c 2 2ca f . 
 Proof: b 2 =h 2 + (c-a') 2 (why?) 
 
 a 2 =h 2 +a' 2 . 
 
 Subtracting b 2 -a 2 = (c-a') 2 -a' 2 =c 2 -2ca'+a' 2 -a f2 . 
 Therefore b 2 a 2 =c 2 2ca'. 
 Solving for b 2 , b 2 =a 2 +c 2 - 2ca'. Q.E.D. 
 This shows that the product 2ca' is the amount by which 
 a a +c* exceeds b 2 . 
 
 a' (Fig. 271) is called the projection of a upon c. 
 
 242. The projection of a given line-segment upon a straight 
 line, is the segment of the line included between the perpen- 
 diculars to the line which pass through the extremities of the 
 given line-segment.
 
 226 
 
 Second- Year Mathematics 
 
 In Fig. 272 A'B' (or A B') is the projection of A B. 
 
 
 
 /!' 
 
 C 
 
 A 
 
 FIG. 272 
 
 The problem of 241 may now be stated thus: 
 
 PROPOSITION V 
 
 243. Theorem: In a triangle the square on the side oppo- 
 site an acute angle is equal to the sum of the squares of the 
 other two sides, diminished by two limes the product of one of 
 these two sides and the projection of the other upon it. 
 
 EXERCISES 
 
 i. b is opposite an acute angle, and c' is the projection of 
 c on a (Fig. 273). Test Proposition V with the figures below, 
 i. e., show that b 2 =a 2 +c 2 iac'. 
 
 h, 
 
 /oN r , 
 ' <Z ^.*- 
 
 C C 
 
 FIG. 273 
 
 2. a, b, and c are, respectively, 5, 10, and 9. (i) Is 
 
 90 ?* (2) Find a' the projection of a on c, and c' the pro- 
 jection of c on a. (3) a 2 +c 2 exceeds b 2 by how much? 
 (4) Compare 2a'c and 2ac r . 
 
 3. a, b, and c are, respectively, 5, 13, 12. (i) Find a' andc' 
 as in Exercise 2. (2) Is Z^trgo ? (3) Does Proposition 
 V hold true when Z_B is a right angle ? 
 
 * ^ means less than or greater than.
 
 Areas of Polygons 227 
 
 4. Find a' and c', when a, b, and c are respectively 
 
 (1) 7, 9, 6 (3) * 2 -;y, * a ~3y, 2*i/y 
 
 (2) II, 15, 12 (4) Z 2 -I, 2 2 -3, 2Z. 
 Check each in the equation 2a'c = 2ac'. 
 
 5. Show from Fig. 273 that c'=c cos B ( 137, p. 104), and 
 that the formula of Exercise i may be written 
 
 b 2 =a 2 +c 2 -2accos B* 
 
 6. Similarly write the equation for a 2 ; for b 2 . 
 
 PROPOSITION VI 
 
 244. Theorem: In an obtuse-angled triangle the square on 
 the side opposite the obtuse angle is equal to the sum of the 
 squares on the other two sides, increased by two times the prod- 
 uct of one of them and the projection of the other upon it. 
 
 Hypothesis: Let ZA B C be obtuse (Fig. 274); 
 
 C 
 
 Proof: b 2 =h 3 + (c+a') 2 and a 2 =h*+a' 2 . Why? 
 Therefore b 2 -a 2 =c 2 + 2ca'+a' 2 -a' 2 . 
 
 b*=a 2 +c 2 -\-2ca'. Q.E.D. 
 
 i . Show that the formula of Proposition VI may be written 
 b 3 =a'+c'-\-2ac cos B. 
 
 * The last term of this formula may be read : two times the product o) 
 the two sides into the cosine o) the included angle.
 
 228 
 
 Second-Year Mathematics 
 
 PROPOSITION VII 
 
 245. Theorem: In a triangle the sum of the squares of two 
 sides is equal to twice the square of one-half of the third side 
 increased by twice the square of the median to the third side. 
 
 Hypothesis: Let A B C be any triangle. In Fig. 275 let 
 m be the length of the median from C to A B ; 
 
 FIG. 275 
 
 Conclusion: a 2 +& 2 = 2(c/2 
 
 Proof: By Proposition VI, a 2 = 
 
 By the same, b 2 = (c/2) 3 +m 2 2(c/2)m'. 
 
 Addkig, a 2 +6 2 =2(c/2) 2 + 2w 2 . Q.E.D. 
 
 EXERCISES 
 
 1. The side opposite an obtuse angle is b, and c' is the 
 projection of c upon a (Fig. 276). Test Proposition VI from 
 the figure, i. e., show that b 2 =a 2 +c 2 + 2ac f . 
 
 2. (i) The number b 2 exceeds a 2 +c 2 by how much? 
 
 (2) Compare 2a'c and 2ac f (Fig. 276). 
 3. Find a' and c' when a, b, and c 
 are, respectively: 
 
 (1) 5, 15,12 
 
 (2) 6,12,8 
 
 (3) 7,ii,8 
 
 2S. 
 
 Check each in the equation 2a'c = 2ac'.
 
 Areas of Polygons 
 
 229 
 
 4. Find w^ (the median to the side b) when a, b, and c 
 are, respectively: 
 
 (i) 6, 10, 8 (2) 5, 13, 12 (3) 9, 15, 12. 
 
 5. Find m a under the conditions expressed in 4. 
 
 6. Show that w c = iV / 2(a 2 +6 2 )-c 2 in any triangle ABC. 
 
 PROPOSITION VIII 
 
 246. Theorem: 77te areas of similar triangles are to each 
 other as the squares of the homologous sides. 
 
 Denoting the areas of the given triangles by T and T' 
 (Fig. 277), we have 
 
 ri 7 7 7_ 7 
 
 \b h b h 
 
 Y' = W^' = b' ' h' 
 i-i (Why?) See 128. 
 
 Therefore 
 
 T' b' b' b' 
 
 Q.E.D. 
 
 FIG. 277 
 Areas of Polygons 
 
 PROPOSITION IX 
 
 247. Theorem: The areas of similar polygons are to each 
 other as the squares of the homologous sides. 
 
 Hypothesis: Polygon ABC.... ^polygon A'B'C'. . . . 
 (Fig. 278). Let P denote the area of A B C .... and P' 
 denote the area of A'B'C' . . . . ; 
 
 P d 3 
 Conclusion: =-.
 
 230 
 
 Second- Year Mathematics 
 
 Proof: Divide A B C and A'B'C' into triangles I, II, III, 
 and I', II', III', respectively, by drawing diagonals from 
 homologous vertices as B and B'. Then I^F, 11^ IF, etc. 
 (Why?) 
 
 C' 
 
 FIG. 278 
 Therefore: =, jp=J^, etc. (Why?) 
 
 But f^W 2 etc - (why?) 
 
 Therefore ^=^=^ .... etc. (Why?) 
 
 Therefore 
 
 P d 
 Therefore -= 
 
 I+II+IH+ .... ^11 
 F+IF+IIF+ . . .~~IF 
 
 Q.E.D. 
 
 'd' 2 ' 
 
 (Why ?) 
 
 EXERCISES 
 
 1. The side of .a triangle is 10 in. Find the corresponding 
 side of a similar triangle having twice the area. 
 
 2. Two similar triangles have two homologous sides 5 and 
 15, respectively. What is the ratio of their areas? 
 
 3. Two homologous sides of two similar polygons are 5 
 and 8. The area of the first is 150. Find the area of the 
 second.
 
 231 
 
 4. If one square is nine times as large as another, what is 
 the relative length of their homologous sides ? 
 
 5. The area of a polygon is 6| times the area of a similar 
 polygon. A side of the smaller is 4 ft. Find the length of 
 the homologous side of the larger. 
 
 6. Construct a pentagon similar to a given pentagon and 
 equivalent to J the area. 
 
 7. Construct a square equivalent to ^ of a given square. 
 
 8. The homologous sides of two similar hexagons are 9 in. 
 and 12 in., respectively. Find the homologous side of a similar 
 hexagon equivalent to their sum; equivalent to their difference. 
 
 PROPOSITION X 
 
 248. Theorem: The areas of two triangles that have an 
 angle in one equal to an angle in the other, are in the same 
 ratio as the product of the sides including the equal angles. 
 
 Given: AA B C and AA'B'C', Z4 = Z4' (Fig. 279). 
 
 C 
 
 FIG. 279 
 
 . 
 o prove: 
 
 ACB = AB AC 
 A / B / C /- A / B / . A , c ,
 
 232 Second-Year Mathematics 
 
 Proof: Lay off on A B, A B"=A'B', and on A C, A C" = 
 A'C'. Draw B"C, and B"C". 
 
 .'. A A B"C"^ A A'B'C'. ( ?) 
 ABC A B 
 AIJ"C == AB" '* ( See * 233 ' Exercise I0 ' P- 2I V 
 
 , AB"C" AC" 
 
 and AB^C = XC (?) 
 
 ABC AB-AC 
 
 By the division axiom 
 
 B" . A C" ' 
 
 But A B"=A'B', and A C"=A'C' 
 and AA B"C"^ A A'B'C'. 
 
 . ABC^AB.AC _ 
 
 ' ' A'B'C' A'B' - A'C' ' 
 
 Second method, using 233, Exercise 6, p. 214: 
 ABC =JAB AC sin ,4 (?) 
 A'B'C' = |A'B' . A'C' sin A' (?) 
 
 Since sin ^4=sin A' (see p. 103, 136, i) 
 ABC^AB AC 
 A'B'C' A'B' . A 7 C' ' 
 
 EXERCISES 
 
 1. Two triangles have an angle in each equal. The in- 
 cluding sides of one are 48 and 75, those of the other triangle 
 are 45 and 70. Find the relative areas of the triangles. 
 
 2. Two sides of a triangular building are 150 ft. and 130 ft. 
 What part of the whole building is included by 50 ft. on the 
 first side and 30 ft. on the second ? 
 
 3. A triangular lot extends 60 ft. and 80 ft. on two sides 
 from a corner. If a building is to front 50 ft. on the first side, 
 how many feet on the second side should it occupy to cover 
 
 of the lot ?
 
 Areas of Polygons 
 
 233 
 
 PROBLEMS 
 
 i . There are two similar polygons whose homologous sides 
 are 6 and 8. Find the side of a third similar polygon equiva- 
 lent to their sum. 
 
 Construct a right triangle having a = 6 and & = 8. Then c = io. (?) 
 Construct the similar polygons P r , P 2 , and P 3 (Fig. 280). 
 
 2 
 
 P, 10*' 
 
 i 2 + 8 2 100 
 io 2 100 
 
 P. + Ps 
 
 P 3 
 
 Hence P I +P 2 =P 3 . (?) 
 Therefore 10 is the required side. 
 
 FIG. 280 
 
 2. There are two similar polygons whose homologous sides 
 are 5 and 12. Find the side of a third similar polygon that 
 is equivalent to their sum. 
 
 3. There are two similar polygons whose homologous sides 
 are 21 and 35. Find the side of a third similar polygon 
 equivalent to their difference. 
 
 Construct a right triangle having = 21, and = 35. 
 
 4. To construct a figure similar to two given similar figures 
 and equivalent to their sum or their difference.
 
 234 
 
 Second-Year Mathematics 
 
 Proof by the Method of Analysis 
 
 In Fig. 281 the four sets of figures S T , S 2 , and S 3 , taken in 
 each set, are constructed similar figures, of which a, b, and c 
 
 (I) 
 
 a 
 
 (2) 
 
 b 
 
 
 
 S 3 
 
 5, 
 
 4 
 
 are homologous sides, equal to the sides and the hypotenuse 
 of a right triangle, respectively; see Fig. 281 (5).
 
 Areas of Polygons 
 
 235 
 
 Then, (i) |i= (?) 
 o 3 c 
 
 W J- (?) . 
 
 (3) ~~J5~~ ~^7~~ = I (?) 
 
 / \ C 1 i O C* / ")\ 
 
 (4) . . 6 I +o 2 =o 3 (?) 
 
 (5) and S 3 S I =S. 2 . (?) 
 
 Hence Fig. 281 (5) suggests the method of construction to 
 find the required homologous side for the required figure. 
 
 The figure can then be constructed similar to the given 
 figure. 
 
 The second part (2) of Problem 4 is a special case of the problem. 
 Its proof follows directly from the Pythagorean proposition, that is: the 
 square on the hypotenuse of a right triangle is equivalent to the sum of 
 the squares on the other two sides. 
 
 JfL. 
 
 5. To construct a figure equivalent to (i) the sum of any 
 number of given similar figures, (2) to any number of times 
 a given figure. 
 
 The special case of the squares is given here because it is 
 most easily comprehended. 
 
 Given: x, y, z, and w, the sides of given squares (Fig. 282). 
 
 Required: to find the side of a square equivalent to the 
 sum of the given squares.
 
 236 Second-Year Mathematics 
 
 (1) Show that C 2 =x 2 +y 2 +z 2 +w 2 . 
 
 (2) What relation between x, y, z, and w would make 
 
 6. To find a line equal to the square root of any number, 
 2, 3> 4, 5 
 
 Let x, y, z, w, . . . . in Figure 282 each = i, then a, 
 b, c . . . . are equal, respectively, to the 1/2, 1/3, 1/4, 
 1/5 See Fig. 282 (2). 
 
 Is it possible to solve Problems i and 2 when the given 
 figures are not similar? For example, to construct a figure 
 equivalent to the sum of a square and a pentagon. Obviously 
 if the pentagon can be transformed into an equivalent square 
 the problem can be solved. 
 
 7. To transform a pentagon into (i) an equivalent triangle, 
 (2) an equivalent square. 
 
 -A J< ->*r<5 * HE- 
 
 FIG. 283 
 
 Draw A F || E B and D G || E C (Fig. 283). 
 
 Then AEFB = AEAB, AEGC = AEDC. (?) 
 
 Hence AE F G=pentagon ABODE. (?) 
 
 J R L is a semicircle and R K _L J L. The area of Q 
 1(FG)(HE). (?) 
 
 The square Q = the pentagon A B C D E. (?)
 
 Areas of Polygons 237 
 
 From the above it will be seen that a polygon of n sides can 
 be transformed into an equivalent square. Hence, Problem i 
 is possible also when the given figures are not similar. In this 
 case, however, the required polygon can be drawn similar to 
 only one of the given polygons. 
 
 Quadratic Equations in Two Unknowns 
 
 249. In the right triangle ABC (Fig. 284) with sides 3 
 and 4, to construct a line through C so that the perimeters of 
 the two new triangles formed may be 
 equal. 
 
 Analysis: Consider the problem 
 solved and let C D be the required 
 line through C. The position of D 
 evidently is determined by determin- 
 ing A D. 
 
 Solution : Denoting the length of A D by x, and the length 
 of DBby;y, 3 +* + C D= 4 +;y+C D (?) 
 
 xy = i. (i) 
 
 2$. (2) 
 
 The values of x and y are the solutions of the system of 
 equations (i) and (2). 
 
 Solving (i) for x and substituting in (2), 
 
 (i+;y) 2 + 2(i+y)y+y* = 25 (3) 
 
 y*+y 6=0 
 (^1 = 2, (y,= -s 
 
 I #i=3 ( X 2 =-2 
 
 x=2, ;y= 3 satisfy equations (i) and (2) but do not 
 satisfy the conditions of the problem. Therefore this solu- 
 tion is disregarded and 3 and 2 are the required values of 
 x and y, respectively.
 
 238 
 
 Second-Year Mathematics 
 
 250. From the preceding solution it is seen that a system 
 of equations in two unknowns, when one of the equations is 
 of the first and the other of the second degree, may be solved 
 as follows: 
 
 Solve the linear -equation for one of the unknowns, x or y, 
 and substitute that value in the second-degree equation. This 
 will lead to a second-degree equation in the other unknown, 
 y or x, as (5), which is then to be solved. The values ofy or x, 
 thus found, may then be substituted in the first-degree equation, 
 as (i), to determine the corresponding values of the other 
 unknown. 
 
 This method of solving a system of equations in two 
 unknowns is called "elimination by substitution." 
 
 EXERCISES 
 
 1. Construct a right triangle whose perimeter is 30 and 
 whose hypotenuse is 13. 
 
 2. In the right triangle ABC (Fig. 285), the perimeters 
 of A C D and BCD are equal. C B =4, and D B = 2. Find 
 A C and A D. 
 
 D 
 
 FIG. 285 
 
 3. In the right triangle of Fig. 286, with sides 53;, 3^, and 
 x +y = 5- Construct the triangle. 
 
 4. Solve x*+y 3 = 2$ 
 
 yx=i. 
 
 5. Solve
 
 Areas of Polygons 
 
 239 
 
 6. Solve 
 
 Quadratic Equations Solved by the Graph 
 
 251. Problems which lead to quadratic equations in two 
 unknowns may be solved by means of the graph. 
 
 i. Solve x 3 +y' = 2$ and y x=i by the graph. By as- 
 suming values for x, and solving x 3 +y 3 = 2$ for y, we have 
 the following solutions of the equation x 2 +y 3 = 2$: 
 
 y=5 y= 
 
 y=4 y= 
 
 '3.4 
 
 FIG. 287 
 
 Plotting these solutions (Fig. 287) we find that the graph 
 of x 2 +y x = 2$ is a circle whose center is at the origin and 
 whose radius is 1/25, or 5. 
 
 The equation x a +y* = 2$ expresses the fact that the sum 
 of the squares of the co-ordinates of any point on the graph
 
 240 
 
 Second- Year Mathematics 
 
 of the equation is 25, e.g., O ~P 1 *=x 2 +y 2 = 2$. Hence 
 O P = 5- Moreover, a line every point of which has the same 
 distance from a given point is a circle. Therefore the graph 
 of x 2 +y 2 = 2$ is a circle whose radius is 5. 
 
 The points of intersection of this circle and the straight 
 line graph of equation y x = i, are Pj(3, 4), and P 2 ( 4, 3). 
 
 ; =3 
 
 j*=; 
 / y=' 
 
 and 
 
 Thus 
 and y. 
 
 2. In triangle ABC (Fig. 2$ 
 C=9o. Construct the triangle. 
 
 are the required values of x 
 18), AB= 5 , CDV-, angle 
 
 D 
 
 J3 
 
 FIG. 288 
 
 FIG. 289 
 
 3. The perimeter of the rectangle (Fig. 289) is 34. Find 
 the dimensions. 
 
 4. Solve by eliminating by substitution, and verify by 
 
 , r graphing: 
 
 (1) x 2 =s8-y 2 
 y ioy 
 
 (2) x a +y a =4o 
 x 3;y=o. 
 
 5. In triangle ABC (Fig. 
 290) draw D E parallel to A B 
 
 FIG. 290 
 
 so that D E is the mean pro- 
 
 portional between A C and D C. 
 
 6. Rectangles R, and R 2 have equal perimeters (Fig. 291), 
 /?! and R 3 have equal areas. Find the dimensions of R t .
 
 Areas of Polygons 
 
 241 
 
 7. In the right triangle ABC (Fig. 292), find the lengths 
 of the sides including the right angle. Area=2. 
 
 y 
 
 FIG. 291 
 
 w 2 + 2 = i7 Why? 
 m n=4 Why? 
 Adding 2 times (2) to (i) 
 
 Subtracting 2 times (2) from (i): 
 m 2 2 mn + n 2 = 9 
 
 (i) 
 
 (2) 
 
 (3) 
 
 (4) 
 
 FIG. 292 
 
 Thus the problem of solving (i) and (2) is reduced to 
 solving (3) and (4), i. e., solving the following system of 
 equations: 
 
 =~s (5) 
 
 w = 
 
 w w= 
 
 m n= 3 
 
 w n= 3 
 
 8. Solve Problem 7 by the graph. 
 
 Equation (i) of Problem 7 is a circle whose radius is -\/rj. 
 Equation (2) expresses the fact that the rectangle formed by 
 the co-ordinates of any point on its graph is constant and 
 equal to 4. This curve is called the rectangular hyperbola 
 (see Fig. 293). Find the points of intersection of the two curves 
 and determine their co-ordinates.
 
 242 
 
 Second- Year Mathematics 
 
 9. Plot on the same figure equations (3) and (4). Show 
 from the graph why the solutions of the systems (5) are the 
 same as the solution of (i) and (2). 
 
 -8 
 
 
 
 -6 
 
 I FIG. 293 
 
 10. Solve for m and n the following equations: 
 w 2 +w 2 13=0 
 mn 6=0 
 
 u. Verify by graphing the solutions of Problem 10. 
 
 12. Solve x 2 -\-y 2 =a and xy = b. 
 
 13. In the right triangle ABC (Fig. 294), x and y have
 
 243 
 
 the same values as in the right triangle D E F. Find the 
 values of x and y. 
 
 Suggestion: Eliminate x 1 or y 3 as in the case of a linear equation in 
 x and y. 
 
 D 
 
 FIG. 294 
 
 14. Verify, by graphing, the solutions of Problem 13. 
 
 15. Solve x 2 +y 2 = a, and x 2 y 2 =b. 
 
 PROBLEMS AND EXERCISES FOR REVIEW 
 
 1. Draw through a vertex of a triangle lines dividing it 
 (i) into two parts one of which shall be (a) , (6) , (c) ;} of 
 the other; (2) into parts in the ratio of 2 : 3 : 4. 
 
 2. Draw a parallel to one side of a triangle cutting off a 
 triangle which shall be (i) , (2) ?, (3) of the given triangle. 
 
 M 
 
 FIG. 295 
 
 3. To bisect a triangle by a line through a given point P 
 on the perimeter not the vertex of an angle (see Fig. 295). 
 
 Draw median B M, also P M, B D || P M, and P D. A? M D 
 - A? M B ( ?). . '. A A D P = quadrilateral D P B C. (?)
 
 244 Second-Year Mathematics 
 
 4. To trisect a triangle by lines through a point P on one 
 side not at the vertex. 
 
 Use the method of Exercise 3. 
 
 5. Construct (i) a right triangle, (2) an isosceles triangle, 
 (3) an obtuse-angled triangle each equal to a given triangle. 
 
 6. Construct an equilateral triangle equivalent to a given 
 triangle. 
 
 Transform the given triangle into an equal triangle having one 
 angle 60. Apply the theorem two triangles having an angle in each 
 equal are to each other as the products of the sides including the equal 
 angles (see 248, p. 231). 
 
 7. On a given base, construct a triangle equal to a given 
 triangle not having a side equal to the given base. 
 
 8. Find a line equivalent to the ]/6 (see p. 236, Problem 6). 
 
 9. Construct a triangle equivalent to a given hexagon. 
 
 10. Construct a circle equivalent to 6 times a given circle 
 (see Exercise 8). 
 
 11. Construct a pentagon equal to -f- of a given pentagon. 
 
 12. A line parallel to the base of a triangle cuts off a tri- 
 angle equal to f of it. If one side of the triangle is 12, how 
 far from the vertex does the line cut it ? 
 
 13. The diagonals of a rhombus are 2^14 and 2X, and 
 a side is x+i. Find x. 
 
 14. The radii of two circles are 25 and 24. Find the 
 radius of a circle equivalent to their difference. Use the law of 
 similar figures. 
 
 15. The area of one of three circles is equal to the sum 
 of the other two, and their radii are x, x 7, x+i. Find x. 
 
 1 6. The difference of two circles whose diameters are 
 x+2 and x is equivalent to a circle whose diameter is xj. 
 Find x. 
 
 17. The area of a rectangle is 60 and its diagonal is 13. 
 Find its dimensions.
 
 Areas of Polygons 245 
 
 1 8. The perimeter of a rectangle is 46 and its area is 120. 
 Find its dimensions. 
 
 19. The perimeter of a rectangle is 62 and its diagonal is 
 25. Find its area. 
 
 20. The sides of a triangle are 17, 10, and 9. The altitude 
 of a similar triangle upon the side homologous to the side 10 
 in the given triangle is 14^. Find all the sides of the second 
 triangle. 
 
 21. Find the side of a square equivalent to an equilateral 
 triangle whose side is 4^/3. 
 
 22. The ratio of two similar polygons is (i) 9 : 4, (2) 2 : 5, 
 and the shortest side of the first is 12. Find the shortest side 
 of the second. 
 
 23. The base of a triangle is 18 feet. Find the length of 
 a line parallel to the base which bisects the triangle. 
 
 24. The side of a square (or of any polygon, or the radius 
 of a circle) is a. Find the side (or radius) of a similar figure 
 K times as large. 
 
 25. Compute the altitude upon the hypotenuse of the 
 right triangle A B C in terms of its legs a and b. 
 
 26. The altitude and base of a rectangle are in the ratio 
 of 8 to 15 and the diagonal is 34 ft. Find the area. 
 
 27. Bisect any parallelogram by a line drawn through any 
 point on its perimeter. 
 
 28. To transform any triangle, ABC, into an equivalent 
 right triangle containing one of the acute angles. 
 
 29. The dimensions of a rectangle are in the ratio of 
 2ab to a 2 b 2 , and the diagonal is a 2 c 2 +& 2 c 2 . Find the area. 
 
 2d 
 
 30. The ratio of the sides of a rectangle is , and the 
 
 a 2 -i 
 
 diagonal is 6a 2 + 6. Find the sides.
 
 CHAPTER VII 
 
 REGULAR POLYGONS INSCRIBED IN, AND CIRCUMSCRIBED 
 ABOUT, A CIRCLE 
 
 252. A polygon that is equilateral and equiangular is a 
 regular polygon. Show that an equilateral triangle is a regular 
 polygon; that a square is a regular polygon. 
 
 A polygon whose vertices lie on a circle and whose sides 
 are chords is an inscribed polygon. The circle is said to be 
 circumscribed about the polygon. 
 
 A polygon whose sides are tangent to a circle is a circum- 
 scribed polygon. The circle is said to be inscribed in the 
 polygon. 
 
 To determine whether inscribed or circumscribed polygons 
 are regular, the following two theorems may be used: 
 
 PROPOSITION I 
 
 253. Theorem: If a circle is divided into equal parts and 
 if the successive points of division are connected by line-segments, 
 the polygon so formed is a regular inscribed polygon. 
 
 Hypothesis: A~B=B~C = CfD, etc. (Fig. 296); 
 
 C 
 
 Conclusion : 
 
 ( (a) AB=BC = CD, etc. 
 
 inn / 
 
 246
 
 Regular Polygons in and about a Circle 
 
 247 
 
 Prove (&) by the method of congruent triangles, using 
 AsB AF, ABC, BCD, etc. 
 
 PROPOSITION II 
 
 254. Theorem: If a circle is divided into equal parts, and 
 if tangents are drawn at the points of division, a regular cir- 
 cumscribed polygon is formed. 
 
 Prove AM A L^ ALB N^ANCO, etc. (Fig. 297). 
 
 FIG. 297 
 
 Then Z.A = /_B = Z.C, etc., the first condition for a 
 regular polygon. 
 
 MA=AL=LB=BN=NC, etc. Why? 
 
 AL=BN and LB=NC. Why? 
 
 .-. AB=BC. 
 
 In a similar way, prove B C = C D, etc., the second condi- 
 tion for a regular polygon. 
 
 EXERCISES 
 
 1. Prove that an equiangular circumscribed polygon is 
 regular. 
 
 2. Prove that an equilateral polygon inscribed in a circle 
 is a regular polygon. 
 
 Use 253.
 
 248 
 
 Second-Year Mathematics 
 
 PROPOSITION III 
 
 255. Problem: To inscribe a square in a given circle. 
 Construction : Draw the diameter A B _l_ diameter C D 
 
 (Fig. 298). Join A to D, D to B, etc. To prove that A D B C 
 
 is a square, Proposition I may be used. 
 
 A 
 
 PROPOSITION IV 
 
 256. Problem: To circumscribe a square about a circle. 
 Construction: In Fig. 298, draw tangents at A, C, B, D. 
 The proof follows from Proposition II. 
 
 EXERCISES 
 
 1. Denoting the side of the inscribed square by x, and the 
 radius of the circle by r, prove that x=r~\/2. 
 
 2. Express the side of the circumscribed square in terms 
 of the radius of the circle. 
 
 3. Express in terms of the radius, the areas of the inscribed 
 and circumscribed squares. 
 
 4. Express the perimeters of the inscribed and circum- 
 scribed squares in terms of the radius; in terms of the diameter. 
 
 5. The radii of two circles are 6 cm. and i cm., respectively. 
 Find the ratio of the areas of the inscribed squares; of the 
 circumscribed squares. How does the ratio of the areas 
 compare with the ratic of the squares of the radii ?
 
 Regular Polygons in and about a Circle 
 
 249 
 
 6. The area of a square is 16 sq. cm. Find the diameters 
 of the inscribed and circumscribed circles. 
 
 7. Prove that the point of intersection of the diagonals of 
 a square is the center of the inscribed and circumscribed 
 circles. 
 
 8. Show how to construct a regular 8-side, i6-side, 32-side. 
 
 PROPOSITION V 
 
 257. Problem: To inscribe a regiilar hexagon in a circle. 
 Construction: With A as center 
 
 and radius O A (Fig. 299), strike an 
 arc meeting the circle at B. 
 
 With B as center and the same 
 radius, strike an arc at C, D, etc. 
 Draw the chords A B, B C, C D, etc. 
 
 Then A B C D E F is the required 
 hexagon. 
 
 Proof: Draw O B. 
 
 Prove ZO=6o=of 360. 
 
 A~B = of the circle. Why ? 
 B~C = of the circle. Why ? 
 
 ABCDEFisa regular hexagon. Why ? 
 
 PROPOSITION VI 
 
 258. Problem: To inscribe an equilateral triangle in a 
 circle. 
 
 Solve the problem and give proof. 
 
 EXERCISES 
 
 1. Inscribe a regular i2-side, 24-side, 48-side, etc. 
 
 2. Given the radius r of a circle. Express in terms of r 
 the side of the regular inscribed hexagon (Fig. 300). 
 
 FIG. 299
 
 250 
 
 Second-Year Mathematics 
 
 3. Given the radius r of a circle, express in terms of r the 
 side of the regular inscribed triangle. 
 
 Bisect angle ACB (Fig. 301). .'. x = \ 36o = i2o, /. y = 6o. 
 
 AE I CD (why?), .'. A E 2 =r*--(why ?), .'. AB=r 1 /3 (why?). 
 
 4 
 
 4. Prove that the area of the equilateral inscribed triangle 
 
 s 
 
 5. To circumscribe about a circle a regular hexagon, an 
 equilateral triangle, a regular i2-side, 24-side, etc ..... 
 
 6. Prove that the side of the circumscribed equilateral 
 triangle is 2r]/3 and the area 3^ 2 v/3- 
 
 7. Prove that the side of the circumscribed regular hexagon 
 is n/3~, an d the area 2/- 2 j/3. 
 
 8. Express in terms of the radius: the perimeters (a) of the 
 inscribed and circumscribed regular hexagons; (b) of the 
 equilateral inscribed and circumscribed triangles. 
 
 9. Find the ratio of the inscribed and circumscribed equi- 
 lateral triangles. 
 
 10. Find the distance from the center of a circle to the 
 side of the equilateral inscribed triangle (the apothem). 
 
 11. Find the area of a regular hexagon whose side is 
 6 inches. 
 
 12. The radius of a circle is 10. Find the area of the 
 inscribed regular hexagon.
 
 Regular Polygons in and about a Circle 
 
 251 
 
 13. The diameter of a circle is 8. Find the area of the 
 regular inscribed hexagon. 
 
 14. Prove that in the same circle the area of a regular 
 inscribed hexagon is twice as large as that of the inscribed 
 equilateral triangle. 
 
 259. In order to construct a regular lo-side (decagon) 
 in a circle, it is necessary to divide a line-segment in extreme 
 and mean ratio. 
 
 A line-segment is divided in extreme and mean ratio when 
 the longer part is a mean proportional between the whole 
 segment and the shorter part. Thus a line-segment A B is 
 divided in extreme and mean ratio at C, when A B : A C = 
 A C : B C. 
 
 260. Problem: To divide a line-segment in extreme and 
 mean ratio. 
 
 Let A B be the given line- 
 segment (Fig. 302). 
 
 To find the point C, such 
 that A B : AC=AC : B C. 
 
 Construction: Draw B F 
 J.AB, and = A B. 
 
 With F as center and 
 radius F B draw F. 
 
 FIG. 302 
 
 Draw A F cutting the circle at E and D. 
 Lay off on A B, A C=A E. C is the required point. 
 AD AB 
 
 Proof: 
 
 AB AE 
 AD-AB 
 
 (?) (182, p. 150) 
 AB-AE 
 
 (i) 
 
 AB AE (?) (Exercise 3, p. 80). (2) 
 
 Substituting equal values in (2), noting that A B=E D (?) 
 
 and A E=A C (?), we obtain - ^=-7-7;, then -r-= 
 
 A D A \^ A. (^ 
 
 Q.E.D. 
 
 AC 
 : BC
 
 252 Second-Year Mathematics 
 
 EXERCISES 
 
 1. Divide a line-segment 5-in. long in extreme and mean 
 ratio. 
 
 Use compasses. 
 
 2. Calculate algebraically the length of the longer segment 
 of the line-segment in Exercise i; of one 6 in. long; of one 
 a in. long; of one b in. long. 
 
 PROPOSITION VII 
 
 261. Problem: To inscribe a regular lo-side (decagon) in 
 a circle. 
 
 Given O. 
 
 Required to inscribe a regular decagon (lo-side). 
 
 Construction: Divide the radius O A (Fig. 303) in extreme 
 
 and mean ratio at B (see 260), making 77^=^-7- 
 
 (J x> x> A 
 
 With A as center and O B as 
 radius, strike an arc at C. 
 
 With C as center and the same 
 radius, strike an arc at D, E, 
 etc. 
 
 Draw AC, CD, D E, etc. Then 
 A C D E . . . . is the required 
 decagon. 
 
 Proof: Draw O C and B C. Since 77^=^-7- and O B = 
 
 (J D D A 
 
 OA AC 
 
 AC ' then AC = BA' 
 
 and as Z^ = Z^, then AAOC^ABCA (why?). 
 OA CB OA i CB i 
 
 Then OC = CA (wh y ?) ' OC = ? ' = 
 
 andCB=CA.
 
 Regular Polygons in and about a Circle 253 
 
 Therefore m=n. Why? 
 
 Prove: p=o, and n = 2 o, m = 2 o, q+p = 2>o 
 
 o+w+<7+/> = i8o (Why?) 
 
 + 20 + 20 = 180 (Why?) 
 
 .'. 0=36. 
 
 Therefore A C= T ^ of circle. 
 .'. A C D E . . . . is a regular inscribed lo-side. Q.E.D. 
 
 PROBLEMS 
 
 1. To inscribe a regular 5-side (pentagon) in a given circle. 
 
 2. To circumscribe a regular ic-side about a given circle. 
 
 3. To circumscribe a regular 5-side, 2o-side, 4o-side, 
 etc about a given circle. 
 
 The Side of a Regular Inscribed Decagon 
 262. Given the radius r of a circle, and s the side of the 
 decagon. Compute the side of the regular inscribed decagon 
 (see Fig. 304). 
 
 (Why?) /. r*-rs=s' 
 
 (?) 
 
 Show that only the positive sign before FIG. 304 
 
 the radical can be used. 
 
 The Side of a Regular Inscribed Pentagon 
 263. Show that the side of a regular inscribed pentagon is 
 
 equal to ^rl/io 21/5. 
 
 In Fig. 305 let A C and C B be sides of a regular decagon, 
 
 then A B is the side of a regular pentagon. Why ? 
 
 In right triangle AFC, A~F 2 =AC 2 -FC 2 (i)
 
 254 
 
 Second-Year Mathematics 
 
 Use s for A F, 
 for since OF 2 =r 2 -s 2 (from AAFO), 
 
 " 1 ) f r A C, and r-vV-s 2 for F C, 
 = l/r 2 -i 2 . 
 
 Substituting these values in (i) 
 
 Whence $=-l/io 21/5. 
 4 
 
 (3) 
 
 But A B is 2s, hence A B, the side of the regular pentagon, 
 
 . r , 
 
 is -V 10 21/5. 
 
 2 
 
 EXERCISES 
 
 1. Find an approximate value of (1/5 1) to be 1.236 + . 
 
 2. Find an approximate value of 1/10 21/5 to be 2 . 351 + 
 
 3. Using the approximate value in Exercise 2, find the side 
 of a pentagon inscribed in a circle of radius 8; 10; 15; a. 
 
 4. Using the approximate value in Exercise i, find the side 
 of a decagon inscribed in a circle of 'radius 8; 10; 15; a. 
 
 5. The side of an inscribed pentagon is 18.8 in. Find the 
 radius of the circumscribed circle. 
 
 6. The side of an inscribed decagon is 14.83 in. Find the 
 radius of the circumscribed circle. 
 
 7. A man has a round table top which he wishes to change 
 into a pentagon. The diameter of the top is 2^ ft. What is 
 the length of the cut required ?
 
 Regular Polygons in and about a Circle 
 
 255 
 
 PROPOSITION VIII 
 
 264. Problem: To construct a triangle whose sides shall 
 be, respectively, the side of a regular hexagon, the side of a regular 
 decagon, and the side of a regular pentagon, inscribed in a 
 given circle. 
 
 Given O (Fig. 306). 
 
 Draw diameter A B, and radius 
 OC_LAB. 
 
 Bisect O B at M. 
 
 With M C as radius and M as 
 center, draw arc C E, cutting A B 
 at E. Join C and E. 
 
 Prove E O C to be the required 
 triangle : 
 
 (a) C O to be the side of a hexagon (radius). (a) 
 
 (6) E O to be the side of a decagon [$r(j/5 *)] See 262. 
 
 (c) EC to be the side of a pentagon [r I/ 10 
 See 263. 
 
 FIG. 306 
 
 Proof: In right triangle C O M, C M =-+r 2 , .'. C M = 
 
 4 
 
 p/5. Why? 
 
 EO = CM-OM = -i/5--=-(l/5-i), the side of a 
 decagon. (b) 
 
 In right triangle E O C, EC*-r*+|-(>/s-i)T- 
 
 -l/io 2 j/ 5, the side of a pentagon. (c) 
 
 EXERCISES 
 i. With radius equal to line-sect 
 
 _o , construct a 
 
 triangle with sides equal to the regular hexagon, decagon, and 
 pentagon by Proposition VIII.
 
 256 
 
 Second-Year Mathematics 
 
 With compasses test whether the line E O, used as a 
 chord, cuts the circle into ten equal arcs, and E C, into five 
 equal arcs. 
 
 2. The radius of a circle is 8 in. Find the side of a regular 
 inscribed decagon. 
 
 3. The radius of a circle is 10 in. Find the side of a 
 regular inscribed pentagon. 
 
 4. With radius = 7 in., find the side of the regular inscribed 
 decagon. 
 
 5. With radius = 7 in., find the side of the regular inscribed 
 pentagon. 
 
 6. The side of a regular pentagon is 5 in. Find the radius 
 of the circumscribed circle. 
 
 7. The side of a regular decagon is 10 in. Find the radius 
 of the circumscribed circle. 
 
 8. What part of the area of a circle is the area of the tri- 
 angle drawn in it by Proposition VIII, the area of the circle 
 being -n-R 3 ? 
 
 Use ir= . 
 
 265. Problem: 
 
 Since 15 is 
 
 PROPOSITION IX 
 To inscribe a regular ij-side. 
 of 360, the required polygon may be 
 obtained by constructing an angle of 
 24 with the vertex at O. Notice that 
 24 =60 -36. 
 
 Construct A B, the side of a regular 
 hexagon (Fig. 307), making angle 
 AOB=6o. 
 
 Then construct A C, the side of a 
 FIG. 307 
 
 regular decagon, making A O = 36. 
 Then B O C = 2 4 , and B~C is T V of the circle.
 
 Regular Polygons in and about a Circle 
 
 257 
 
 EXERCISES 
 
 1. Inscribe a regular 3o-side, 6o-side, etc 
 
 2. To circumscribe a regular polygon of i5-sides. 
 
 3. Show that regular polygons can be inscribed in a circle, 
 having the following number of sides: 2", 3 2", 5 2", 
 15 2". 
 
 Exercise 3 shows that a circle can be divided into 2, 3 2, 
 5 2, 15 2 equal parts. 
 
 Gauss proved that by the use of a compass and unmarked 
 straight edge, a circle could be divided into 2 >f + i equal parts 
 where n must be a number making 2 n + i a prime number. 
 Explain. 
 
 Circles Circumscribed about, and Inscribed in, a Regular Polygon 
 PROPOSITION X 
 
 266. Theorem: A circle can be circumscribed about any 
 regular polygon. 
 
 Let A B C D . . . . be a regular 
 polygon (Fig. 308). 
 
 To construct a circle circum- 
 scribed about A B C D 
 
 Construction: Draw a circle 
 through three vertices, as A, B, 
 and C (?). This is the required 
 circle. 
 
 Proof: We are to prove that FIG. 308 
 
 circle ABC passes through D, E, etc. 
 
 Prove A A O B^ AB O C. 
 
 .'. y=u, z=u (?), y+z = u+p (?). 
 
 Therefore, u+uu+p, and u=p. 
 
 Prove AGO D^ AC OB. Then O D = O B (?). 
 
 Circle ABC passes through D. ( ?)
 
 2 5 8 
 
 Second- Year Mathematics 
 
 In a similar way circle ABC may be proved to pass through 
 E, etc. Q.E.D. 
 
 PROPOSITION XI 
 
 267. Theorem: A circle can be inscribed in any regular 
 polygon. 
 
 Let A B C .... be a regular polygon (Fig. 309). 
 
 Required to draw the inscribed 
 circle. 
 
 Construction : Construct as in Pro- 
 position X the center O of the circum- 
 scribed circle. 
 
 From O draw OKiAB. With 
 O as center and radius O K draw 
 circle H K I. This is the required 
 circle. 
 
 Proof: Draw the circle circumscribed about ABC.... 
 (266). DrawOL^AE, OM_LB C. O K = O L = O M, 
 etc. Why ? 
 
 Since O K JL A B, .'. A B is tangent to H K I. (?) 
 Similarly prove that B C, C D, etc., are tangents to H K I. 
 
 Therefore H K I is inscribed in polygon ABC (?) 
 
 Q.E.D. 
 
 To Find the Circumference of a Circle 
 
 268. A regular polygon may be inscribed in a circle, by 
 dividing the circle into equal parts and joining the successive 
 points of division by segments. (See 
 
 Proposition I, p. 246.) 
 
 269. A regular polygon may be cir- 
 cumscribed about a circle by drawing 
 tangents at these points of division of 
 the circle. (See Proposition II, p. 247.) 
 
 270. A regular polygon may also FIG. 310
 
 Regular Polygons in and about a Circle 
 
 259 
 
 be circumscribed about a circle by drawing tangents parallel to 
 the sides of the inscribed polygon, as in Fig. 310. 
 
 To prove this, we will first prove the following theorem: 
 
 PROPOSITION XII 
 
 271. Theorem: If at the midpoint of the arcs subtended by 
 the sides of a given regular inscribed polygon, tangents are drawn 
 to the circle, they are parallel to the sides of the given polygon 
 and form a regular circumscribed polygon. 
 
 To prove that A B || A'B' (Fig. 311), show that both A B 
 and A'B' are perpendicular to O P. 
 
 To prove that C'A'B'D' .... is regular, show that arcs 
 Q P, P R, R S, etc., are equal. 
 
 EXERCISES 
 
 1. In Fig. 311, prove that line OB passes through B', 
 and state this exercise as a theorem. 
 
 Prove that B O bisects angle P O R, and therefore contains point B'. 
 
 2. Denoting the side of the circumscribed regular polygon 
 of w-sides, by S n and of the regular inscribed polygon of the 
 
 same number of sides by s n , prove that S n =^= " .
 
 260 
 
 Second- Year Mathematics 
 
 CD OF 
 
 Suggestion: In Fig. 312 
 
 Therefore ^ = V r*-(ls )> t wh y ? ) 
 
 Solve this equation for S n - 
 
 3. Taking the value of s 3 from 258, Exercise 3, find by 
 means of the formula of Exercise 2 the value of S 3 . 
 
 C F D 
 
 FIG. 312 
 
 4. In a similar way find the value of S 4 , using s 4 =r}/2, 
 as proved in 256, Exercise i. 
 
 5. Find Sf,, using 5 6 =r. 
 
 PROPOSITION XIII 
 
 272. Theorem: If a regular polygon of n-sides is inscribed 
 in a circle and the arcs subtended by the sides are bisected, a 
 regular polygon of 2n-sides may be drawn by joining each point 
 of division to the adjacent corners of the given polygon. Prove. 
 
 PROPOSITION XIV 
 
 273. Theorem: The perimeter of a regular inscribed 2n- 
 side is greater than the perimeter of the regular n-side inscribed 
 in the same circle. Prove. 
 
 PROPOSITION XV 
 
 274. Theorem: The perimeter of a regular circumscribed 
 2ii-gon is less than the perimeter of the regular n-gon circum- 
 scribed about the same circle. Prove.
 
 Regular Polygons in and about a Circle 261 
 
 275. From Propositions XIV and XV it follows that the 
 perimeter of the regular inscribed polygon increases as the 
 number of sides increases: 
 
 While the perimeter of the regular circumscribed polygon 
 decreases as the number of sides increases. 
 
 In the following discussion it will be shown that by in- 
 creasing the number of sides of the inscribed and circumscribed 
 regular polygons, the perimeters approach each other more 
 and more, and that the decimal fractions expressing these two 
 perimeters can be made to agree to a greater and greater 
 number of decimal places. It will be assumed that the cir- 
 cumference of the circle is less than the perimeter of any 
 polygon circumscribed about the circle, and it is easily proved 
 that the circumference is greater than the perimeter of any 
 polygon inscribed in the circle. Hence, the length of the cir- 
 cumference lies between the lengths of the perimeters of any 
 two inscribed and circumscribed polygons. 
 
 To calculate these lengths, the two following propositions 
 give formulas for determining the perimeters of circumscribed 
 and of inscribed polygons of a greater and greater number of 
 sides. 
 
 PROPOSITION XVI 
 
 276. Problem: To compute the side of the regular circum- 
 scribed 2n-gon in terms of the sides of the regular inscribed and 
 circumscribed n-gon. 
 
 Let A B be the side of the regular inscribed w-gon, denoted 
 by s n (Fig. 313). 
 
 Let C D be the side of the regular circumscribed 2-gon, 
 denoted by S 2n . E G the side of the regular circumscribed 
 w-gon denoted by S n . 
 
 To find S 2n in terms of s n and S n . 
 CD CG
 
 262 
 
 Second-Year Mathematics 
 S S 
 
 n A G S_n S n 
 
 2 
 
 S S S 
 Therefore - = " " , and S 2n S n =s n (S n S 2n ). 
 
 Or, 5 2W = 
 
 FIG. 313 
 
 EXERCISE. Denoting the perimeters of the inscribed and 
 circumscribed polygons by P 2n , pn, Pn, prove that 
 
 j. n 
 
 (2) 
 
 PROPOSITION XVII 
 
 277. Problem: Find the side of the regular inscribed poly- 
 gon of 2n-sides in terms of the side of the regular circumscribed 
 polygon of 2n-sides, and of the side 0} the regular inscribed 
 polygon of n-sides. 
 
 Let A B be denoted by s n (Fig. 314). 
 A H be denoted by s 2n . 
 C D be denoted by S 2n . 
 To find s 2n in terms of s n and S 2n . 
 
 Proof: ACHI^AHAK. Hence, 
 
 
 
 ~ - 
 AH AK 
 
 (Why?) 
 '
 
 Regular Polygons in and about a Circle 263 
 
 Whence, ~ = . (Why ?) 
 
 Therefore, S 2n =s 2n s 2n . (Why?) 
 
 and s 2n =- 
 
 V--7- 
 
 FIG. 314 
 
 (i) 
 
 EXERCISES 
 
 1. Prove that p 2n =vP 2n p n . (2) 
 
 2. It was proved in 256, Exercise i, that s 4 =/-j/2, and 
 in 256, Exercise 2, that S 4 d, the diameter of the circle. 
 
 Prove that P 4 ^d t and p 4 =dX 2. 828427 
 
 3. By means of the formula (2) of 276, prove that 
 P&=dX 3. 313780 . . . . , and by formula (2), Exercise i, of 
 this list, that p& =dX 3. 061467 
 
 4. It has been proved that P6=dX3-464ioi .... and 
 p 6 = $d. (P. 250, 8.) 
 
 Prove that/*, 2=^X3. 2 1 5390 . . . . and p l3 =dX$. 105828.
 
 264 
 
 Second- Year Mathematics 
 
 5. Prove the following: 
 
 Pi =2-598 . . 
 
 . . d 
 
 P 3 =5.196 . . 
 
 . . d 
 
 P 4 =2.828 . . 
 
 . . d 
 
 P 4 =4.000 . . 
 
 . . d 
 
 p 6 =3.000 . . 
 
 . . d 
 
 P 6 =3.464 - - 
 
 . . d 
 
 ps =3.061 . . 
 
 . . d 
 
 PS =3-3i4 - 
 
 . . d 
 
 I2 = 3.106 . . 
 
 . . d 
 
 P I2 =3.2i5 . . 
 
 . . d 
 
 i6=3- 121 
 
 . d 
 
 P l6 = 3.i83 . 
 
 . d 
 
 P. = 
 
 The table above shows how the decimal fractions expressing 
 the perimeters agree more and more closely as the number of 
 sides of the polygon is increased 
 
 278. The following table, which gives the decimal frac- 
 tions to six places, shows the approach of the perimeters still 
 better: 
 
 = 2.828427 
 =3.000000 
 =3.061467 
 =3.105828 
 P l6 =3.182598 . 
 
 P*4 =3-159659 ' 
 
 P 32 =3-I5l725 
 
 P 4 8 =3.146086 . 
 
 P 64 =3.144118 . 
 
 Pp 6 =3.142714 . 
 
 = 4.000000 
 = 3.464121 
 = 3.313708 
 
 d 
 
 P I92 =3. 141873 
 
 p ia 
 
 p l6 =3.121445 
 p 24 =3.132623 
 p 32 =3.136548 
 
 d 
 
 64 =3.140331 . 
 96 =3-I4I032 . 
 
 I28 = 3-i4i277 . 
 192=3-141452 - 
 2S 6 = 3- I 4i5 I 4 - 
 348 = 3 I 4i557 
 The last two perimeters agree to three decimal places. 
 
 Thus the circumference of the circle of diameter d, toward 
 
 which these perimeters approach, is found correct to three 
 
 decimal places. It equals dX 3. 141 
 
 Thus as the perimeters of the inscribed and circumscribed 
 
 polygons with increasing numbers of sides, approach each
 
 Regular Polygons in and about a Circle 265 
 
 other in length, both of them approach more and more closely 
 the length of the circumference of the circle. But however 
 close the length of the perimeter of any polygon may come to 
 the length of the circumference, there is always another poly- 
 gon the perimeter of which comes still closer to the length of 
 the circumference; and for every number given as expressing 
 the difference between any perimeter and the circumference, 
 we can find a polygon whose perimeter differs from the cir- 
 cumference by less than that number. This is expressed by 
 saying that the perimeters of the inscribed and circumscribed 
 polygons approach the circumference as a limit. 
 
 As is seen by the table above, the value of this limit can 
 be expressed more and more exactly by taking polygons of a 
 greater and greater number of sides. It cannot, however, be 
 determined exactly. 
 
 Continuing to increase the number of sides, we find in 
 the table above ^8192=^X3. 1415928 .... and p$ I92 = 
 
 From this it is seen that the circumference, being between 
 P 8lg2 and pBi 92 , can be expressed by 0=^X3.141592 .... 
 approximately, with an error less than i millionth. 
 
 The circumference of the circle is therefore a multiple of 
 the diameter, which may not be exactly expressed in figures, 
 for the number 3.141592 .... by which d is multiplied, is 
 an irrational number, and is commonly denoted by -T (the 
 first letter of -irepufttpeia, meaning circumference). 
 
 Thus C=tr d, or 2irr, or ir=C/d. TT then expresses the 
 ratio of the circumference to the diameter of the circle. 
 
 Archimedes (212 B.C.) found the value of it to be such that 
 3H<T<3l8 *>y finding the value of P 9 6 and p 9 &. 
 
 Ptolemaeus (150 A.D.) calculated TT =3. 14166. 
 
 At the end of the sixteenth century Vieta (1579 A.D.) found the 
 value of IT to 10 decimal places, and Ludolf van Eulen (1540-1610) to 
 20, 32, and 35 places. The value of ir has since been carried out to 
 more than 700 decimal places.
 
 266 
 
 Second- Year Mathematics 
 
 EXERCISES 
 
 1. The circumference of a circle is loocm. Find the 
 radius. 
 
 2. The radius of a circle is 10. Find the radius of a 
 circle (i) 4, (2) 9, (3) 25, (4) 3, (5) 17, times as large in area. 
 
 PROPOSITION XVIII 
 
 279. Theorem: The area of a regular inscribed polygon 
 is equal to the product of one-half of the perimeter and the per- 
 pendicular from the center to the side* 
 
 Proof: Area of A O ~B = 
 Area of B O C = 
 
 A B (Fig. 315). 
 B C. 
 
 FIG. 315 
 
 Therefore area of A B C D . . . .=$h (A B+B C + C D 
 . . . .) or area of A B C D . . . .=\h - p. 
 
 PROPOSITION XIX 
 
 280. Theorem: The area of a regular circumscribed polygon 
 is the product of one-half the perimeter and the radius. 
 
 The proof is similar to that of Proposition XVIII, that 
 area ABC.... =\r p. See Fig. 316. 
 
 *This perpendicular is called the apothem.
 
 Regular Polygons in and about a Circle 267 
 
 Area of the Circle 
 
 281. The irrational number TT has now been denned as 
 the limit approached by the series of numbers, 
 
 4.000000 2.828427 .... 
 
 3.464121 .... 3.000000 
 
 3.141662 .... 3.141557 .... 
 
 They approach each other, and approach TT, more and 
 more closely, the value of TT always lying between any two 
 corresponding numbers. 
 
 In preceding chapters (pp. 127-212) other irrational num- 
 
 AC 
 
 bers, 1/74, l/Tss, 1/27, -~^ = V 2 (p. 126), have been met. 
 j> C 
 
 Their value also was approximated by rational numbers. 
 But each is, as in the case of TT, the limit approached by its 
 successive approximations. 
 
 Theorems like "The area of a rectangle equals the product 
 of the base by the altitude" ( 231), can now be more clearly 
 seen to be true, even if the dimensions be irrational numbers. 
 
 The dimensions can be expressed by rational numbers 
 only to as close a degree of approximation as may be wished, 
 but they are expressed exactly by IT, V 2, 1/12, etc., and the 
 theorem concerned is absolutely true when they are thus used. 
 The logical proof of this, however, would involve a complete 
 study of limits, which is considered beyond the province of 
 secondary-school work, as is plain from 278. 
 
 The area of a circle lies between that of the circumscribed 
 polygon and that of the inscribed polygon. Why? As the 
 number of sides is increased 5 and s approach the area of the 
 circle and h equals or approaches r, so that S and 5. approach 
 \r c to as great a degree of accuracy as may be desired; or 
 S and s approach \r c as a limit. 
 
 This leads to the following:
 
 268 Second-Year Mathematics 
 
 PROPOSITION XIX 
 
 282. Theorem: The area of a circle is one-half the product 
 of the circumference and the radius, i. e., area of circle =\c r. 
 
 n-d 2 
 
 Show that the area of a circle =irr 2 , or . 
 
 4 
 
 EXERCISES 
 
 1. Prove that the circumferences of two circles are to each 
 other as the radii, or as the diameters, 
 
 2. Prove that the areas of two circles are to each other as 
 the squares of the radii, or as the squares of the diameters. 
 
 3. What is the ratio of the areas of two circles whose radii 
 are 5 in. and 10 in. ? 
 
 4. The area of a circle is 64. Find the diameter and 
 circumference. 
 
 5. What is the area of the ring formed by two concentric 
 circles whose radii are 5 in. and 6 in., respectively ? a in. and 
 & in., respectively ? 
 
 6. The areas of two circles are in the ratio of 2 to 4. What 
 is the ratio of the diameters ? 
 
 7. The circumference of a circle is 50 in. What is the 
 area? 
 
 8. The diameter of a circular table is 3 ft. What is the 
 circumference of the table ? the area ? 
 
 9. The radii of two circles are to each other as 3 : 5, and 
 their combined area is 3850. Find the radii of the circles. 
 
 TT 22 
 
 Use ir = . 
 
 7 
 
 10. The radii of two circles are to each other as 7 : 24, 
 and the radius of a circle equivalent to their sum is 50. Find 
 the radii of the other two circles.
 
 Regular Polygons in and about a Circle 269 
 
 283. Problem: To find the area of a sector of a circle. 
 
 Suggestion: The area of AA O C = iA~C O B (Fig. 317). If 
 AC isjbisected at D, the sum. of the areas of the two AsAOD + DOC 
 is}(Al) + DC) OB'. Why? 
 
 FIG. 317 
 
 If the number of divisions of A C be indefinitely increased, 
 and lines be drawn to the points of division, the sum of the 
 bases of the As will approach AC as a limit (?), and the 
 common altitude of the As will approach the radius as a 
 limit (?). Hence the sum of the As will approach the area 
 of the sector as a limit, and will approach AC r (?). 
 
 Hence the area of the sector A O C=^A C r. 
 
 This may be stated as a theorem: 
 
 284. The area of a sector of a circle is equal to one-half the 
 product of the arc of the sector, and the radius of the circle. 
 
 EXERCISES 
 
 1. The radius of a circle is 100 ft. The length of the arc 
 of a sector is 25 ft. Find the area of the sector. 
 
 2. The radius of a sector is 9 in., its area is 72 sq. in. 
 Find the length of the arc. 
 
 3. The area of a sector is 96 sq. ft., and the radius is 
 1 2 ft. How long is the arc ? 
 
 4. The area of a sector is a sq. ft., and the radius is r ft. 
 Find the length of the arc.
 
 270 Second-Year Mathematics 
 
 5. The radius of a circle is 8 in. Find the area of a sector 
 with arc 36. 
 
 6. Find the area of the segment whose arc is 36, in a 
 circle of radius 12 in. 
 
 Notice that 36 is ^ of the circumference. Thus the base of the 
 triangle is the side of a regular lo-side. 
 
 7. Find the area of a segment of arc 72, in a circle of 
 radius 20. 
 
 8. The area of a circle is 15,400 sq. in. Find the area 
 of a segment whose arc is 60.
 
 CHAPTER VIII 
 
 PROBLEMS AND EXERCISES IN GRAPHIC AND 
 GEOMETRIC ALGEBRA 
 
 285. In the preceding chapters algebraical processes have 
 been frequently used in the solution of geometrical problems. 
 This has been done by working with the numbers representing 
 the lengths of line-segments, the areas of surfaces, and other 
 magnitudes in the geometrical problem. 
 
 On the other hand, geometrical processes are of value in 
 the treatment of algebraical problems, as has been shown, 
 for instance, in the graphical solution of equations. 
 
 In a further application of geometry to algebra, the different 
 number expressions occurring in algebraical problems may be 
 represented by straight line-segments. To enable the student 
 to do this will be one of the purposes of this chapter. 
 
 286. Graphical problems and exercises. 
 
 1. The lengths of the sides of a triangular piece of land 
 are 125 rd., 54 rd., and 112 rd. A drawing is made of it on 
 a sheet of paper, the longest side of which is a little over 3 ft. 
 How long will the sides of the triangle in the drawing be ? 
 
 2. The dimensions of a rectangle are 12 and 17. Without 
 computing the length of the shorter side, construct a rectangle 
 similar to the given one and having a length 8 for its longer 
 side. 
 
 3. Construct a rectangle as in Problem 2 having a length 
 8 for its shorter side. 
 
 4. The sides of a triangle are 14 in., 10 in., and 19 in., 
 respectively, and are to be reduced to the ratio 1:7. Make 
 an accurate construction with the use of a ruler. In what 
 ratio will the area of the triangle then be reduced ? 
 
 271
 
 272 Second-Year Mathematics 
 
 5. Of triangle A B C we have given the sides A B, C B, 
 and the segment A D that the bisector of angle B cuts off 
 on the side A C. Construct the triangle. 
 
 6. Construct a triangle similar to a given triangle, and 
 having a given line / as one of its sides. How many different 
 triangles are possible, if the given triangle is scalene ? If it is 
 isosceles ? If it is equilateral ? 
 
 7. The lines a, b, and c being given, construct a line I such 
 
 a 
 
 287. To construct line-segments of lengths of fractions of 
 an inch not marked on the ruler. 
 
 1. To construct accurately a line of ^ of an inch. 
 Calling the required line x, we have x=%, or J= -. 
 
 OC 
 
 Hence, we see that the required line is obtained by constructing a 
 fourth proportional to 3 lines of, respectively, 7 inches, 4 inches, and 
 i inch. 
 
 2. Construct the lines whose lengths are given by the 
 following expressions, the letters denoting given lines and the 
 unit being chosen arbitrarily: 
 
 ^ (3)* <5)f (7) (9)? 
 
 S2J 
 
 (\ ^p / N 5 i f\ 7 /o\ <; / \ 7^ 
 
 2 ) (4) (6) - (8) A ( I0 ) 
 
 288. The graphic representation of statistics. 
 
 i. Make an accurate graph of the following population- 
 data for Chicago, representing a population of one million by 
 a line of 3 inches: 
 
 1872 .35 million 1882 .55 million 1892 1.45 million 
 1874 -4 " 1884 .65 " 1894 1.55 " 
 
 1876 .4 " 1886 .7 1896 1.6 
 
 1878 .45 " 1888 .8 " 1898 1.85 " 
 
 1880 .5 " 1890 1.2 " 1900 2. "
 
 Problems in Graphic and Geometric Algebra 273 
 
 289. Constructions. 
 
 1. To construct a rectangle whose area shall be 24 sq. in., 
 and whose base shall be 7 in. 
 
 2. To construct a rectangle whose altitude is given and 
 whose area is equal to that of a given square of known sides. 
 
 3. The area of a triangle is 15 sq. in. Construct a line 
 equal to its altitude when the base is 4 in. 
 
 4. For a triangle with an area of 17 sq. in. construct lines 
 equal to the altitudes for bases of length 2, 3, 4, 5 .... 13 in., 
 and graph the altitude as a function of the base. How does 
 the altitude change as the base increases? As it decreases? 
 
 290. The graphing of literal equations. 
 
 i. Graph a line whose equation is ax+by=c, where a, 
 b, and c represent given lines. 
 
 We find the following pairs of values from the given 
 equation: 
 
 if x=o, then y = - 
 
 and if yo, then x=- 
 
 a 
 c , c 
 
 (0 
 
 (2) 
 
 We construct lines equal to - and -, respectively, and plot 
 
 & 
 
 the points (i) and (2) (Fig. 318). 
 
 FIG. 318 
 
 Make graphs of the following equations: 
 
 2. 2#+4;y = 7 4. 
 
 3. 3^-2^ = 5 5. - 
 
 6. 2X ry=s
 
 274 Second-Year Mathematics 
 
 291. To find points coney clic with given points. 
 
 i. Having given 3 points, A, B, and C, to find points which 
 will be concyclic* with A, B, and C (see Fig. 319). 
 
 Draw the line AB and any line through C meeting A B, as 
 at P. Now, if we determine a point D on C P, such that 
 AP PB = CP PD, the point D will be concyclic with 
 A, B, andC. Why? 
 
 X 
 
 / 
 C' 
 
 FIG. 319 
 
 For the construction, draw AC; lay off a line P E, equal 
 to P B, on P X. Then, draw E F || A C, and lay off P D 
 equal to P F on P X. The point D is the required point. f 
 Why? 
 
 2. Having given 3 points P, Q, and R, determine a num- 
 ber of points concyclic with those points, without first drawing 
 the circle. 
 
 292. To represent areas and volumes by line-segments. 
 
 i . Represent by a line the area of a rectangle whose dimen- 
 sions are p and q. 
 
 Denoting the area by A, we have A=p q or ~ = . Why ? 
 
 P A 
 From this the method of construction is evident. 
 
 Construct lines whose lengths are represented by the 
 following expressions, taking a convenient unit of length in 
 each problem. 
 
 * Lying on the same circle. 
 
 t Admitting the use of a pair of triangles to draw parallel lines, the 
 construction is very rapid.
 
 Problems in Graphic and Geometric Algebra 275 
 
 2. ab 5. pqr 8. a*bc 
 
 3- 3^ 6. ^ 9. *;y 
 
 2#Z 4VX 
 
 4. - 7. - 10. *abc 
 
 5 <* 
 
 1 1 . Represent by a straight line the volume of a rectangular 
 block whose dimensions are given lines a, b, and c inches, 
 representing i cubic in. by a line of i in. 
 
 12. Represent by a straight line the volume of a cube 
 whose edge is equal to a given line x. 
 
 13. Make an accurate graph of the areas of squares whose 
 sides are equal to "; J"; $"; "; i"; i"; i$"; if"; 
 and 2". 
 
 14. Make an accurate graph of the volumes of cubes 
 whose edges are equal to \"\ \"\ \"; "; i"; ij"; \\"\ 
 if"; and 2". 
 
 15. Compare the two graphs of Problems 13 and 14. Does 
 the volume of a cube increase more or less rapidly than the 
 area of a square, when the sides of each are increased simul- 
 taneously ? 
 
 16. Given two triangles ABC and A'B'C', having Z_A = 
 Z.A'. Construct lines representing their areas and the pro- 
 ducts A B - A C and A'B' A'C'. 
 
 Then verify the proposition that two triangles which have 
 one angle equal are to each other as the products of the sides 
 including that angle. 
 
 293. Different methods for constructing a square equal to a 
 given rectangle. 
 
 i. Construct a square equal in area to a rectangle whose 
 dimensions are a and b. 
 
 By what relation are the side x of the square and the lines a and b 
 connected ?
 
 276 
 
 Second-Year Mathematics 
 
 2. Construct a square equal in area to a rectangle of 
 dimensions 5 and 7. 
 
 3. Construct a square equal in area to a triangle of area 21. 
 
 J) 
 
 FIG. 320 
 
 4. To construct a line, x, a mean proportional to two given 
 lines a and b, we may proceed in one of the three following 
 ways (see p. 98, Problem i): 
 
 (a) On a line X Y, lay off line-segments A B and B C 
 equal, respectively, to a- and b (Fig. 320). Describe a circle 
 on A C as a diameter, and erect line B D perpendicular to 
 AC at B, cutting the circle at D. The line B D is the 
 required line. Prove, using 132, p. 98. 
 
 FIG. 321 
 
 (b) On a line X Y lay off line-segments A B and A C 
 equal to a and b, respectively (Fig. 321). Describe a circle 
 on A B as a diameter and erect a line C D _l_ A B at C meet-
 
 Problems in Graphic and Geometric Algebra 277 
 
 ing the circle at D. The line A B is the required line. 
 Prove, using Problem 7, p. 99. 
 
 (c) On a line X Y (Fig. 322) lay off line-segments A B and 
 A C equal to a and b, respectively. Draw any circle passing 
 through C and B, such as circle O. Draw a tangent from 
 A to the circle. The line A D will be the required line. 
 Prove, using Proposition IV, p. 150. 
 
 FIG. 322 
 
 5. Under what conditions would method (6) or (c) have 
 advantage over method (a) ? 
 
 294. Construct lines that are mean proportionals to the 
 following pairs of lines: 
 
 1. 1 2" and 1 1" 5. 5" and 13" 
 
 2. 5" and 15" 6. 9" and 8" 
 
 3. 7" and 14" 7. i" and 6" 
 
 4. 4" and 9" 8. 10" and 7" 
 Use in .every case the method best suited to the problem.
 
 278 Second-Year Mathematics 
 
 9. Using an arbitrary unit of length, construct a line equal 
 to 1/3. 
 
 Representing the required line by x, we have ^ = 1/3, or x 2 =$; 
 i.e., i : x = x : 3, from which the method of construction becomes 
 evident. 
 
 10. Construct lines equal to: 
 
 (1) 1/7 (4) l/8_ (7) i/H 
 
 (2) i/6_ (5) 1/iS (8) I/a 
 
 (3) 1/12 (6) !/ a & (9) !/ 9 . 
 
 11. Two lines a and & being given, construct lines equal to 
 ab, I/a, 1/6, and I/ 06. 
 
 12. Prove graphically the relation: l/a& = l/a 1/6, i.e., 
 I/a : l/a6 = i : 1/6. 
 
 13. Three lines a, 6, and c being given, construct a line x 
 
 equal in length to: */ . 
 
 By the methods of 292 and 293, we construct a line y = 
 
 b* 
 
 , then construct a line x = yay. 
 
 14. Construct lines represented by the following expressions, 
 in which the letters and numbers represent lines that are given 
 in some arbitrary unit of length: 
 
 (i) 1/| (4) 2 fll/*a (7) 
 
 (5) 
 
 15. Having given the three sides of a triangle, construct a 
 line representing its area. 
 295. Miscellaneous constructions. 
 
 Construct lines equal to the roots of the following equa- 
 tions: 
 
 1. 4# 2 7=0 4. ax 2 c=o 
 
 2. 3# 2 =a 5. 4# 2 33; 1=0 
 
 3. 3# 2 2^ 5=0 6. 5
 
 Problems in Graphic and Geometric Algebra 279 
 
 7. The hypotenuse of a right triangle is 7; the sum of the 
 other two sides is 9. Construct the triangle. 
 
 8. Given a triangle ABC whose sides are 4, 7, and 8, it 
 is required to construct a right triangle by decreasing each of 
 the sides of triangle A B C by some fixed amount. 
 
 First compute the lengths of the sides of the new triangle, then 
 construct it. 
 
 9. Construct a line, x, equal to ^L+^L\/^. Then 
 prove, both geometrically and algebraically, that the line x is 
 the greater part of the line L divided in extreme and mean 
 ratio. 
 
 10. Construct a circle which shall pass through two given 
 points and be tangent to a given line, not parallel to the line 
 joining the two given points. (Make use of 293, 
 Problem 4 (c). 
 
 11. In the right triangle ABC (Fig. 323) the altitude line 
 AD is drawn; then D D' is drawn J_AC; D'D"xCD. 
 Express the lines A D, D D', D'D", etc., in terms of a, b, 
 and c. Similarly, the lines D E', E'E", etc. 
 
 12. Three lines, p, q, and r being given (of which r is the 
 greatest), construct lines equal to ^^ , - , * ^ , etc 
 
 13. The sides of an isosceles triangle being given, it is 
 required to construct lines equal to the radius of the circum-
 
 280 Second-Year Mathematics 
 
 scribed and inscribed circles, without constructing the triangle 
 itself. 
 
 296. Construction of more complicated irrational literal 
 expressions. 
 
 1. Two sides of a right triangle are a and b. Construct 
 the hypotenuse, x, and express the relation which exists be- 
 tween the lines x, a, and b. 
 
 2. The hypotenuse and one side of a right triangle being 
 given equal to c and a, respectively, construct the triangle and 
 express the relation, which exists between c, a, and the third 
 side b. 
 
 3. Construct a line equal to (i) ~\/p 2 +q 2 , (2) Vp 2 q 2 , 
 p and q being given lines. What relation must hold between 
 p and q in order that the last construction may be possible ? 
 
 4. Construct a line, y, equal to I/a 2 x 2 . 
 
 Give to x different values, and graph the corresponding 
 values of y. 
 
 5. Construct a line, y, equal to V / a 2 +x 2 . 
 
 Give to x different values, and graph the corresponding 
 values of y. 
 
 6. Using the constructions of 293 to 296 combined, con- 
 struct a line, x, equal to I/a 2 2bc. 
 
 Construct a line y equal to V zbc; then x = V a 3 y 3 . 
 
 7. Determine lines which may represent the following 
 expressions : 
 
 (i) I/a 2 -25 (6) x 
 
 (2) Vt*-rs (7) 
 
 (3) Va*+bc (8) ia 
 
 (4) 1/6" 4ac 
 
 (5) p + 1/pq + lab-pq 

 
 Problems in Graphic and Geometric Algebra 281 
 
 8. Construct lines equal to the solutions of the following 
 equations: 
 
 (i) x 2 -4X + 2=o (6) x 2 + (a-b)x+c=o 
 
 (2) # 2 + 3#+a=o (7) x 2 tx+s=o 
 
 (3) T>x 2 6x + 2=o (8) ax 2 + 2bx+c=o 
 
 (4) ax 2 +bx+c=o (9) x 2 -(a+b)x+(a-b)=o 
 
 (5) 2X 2 +6x+3=o (10) 2X 2 + (c-d)x + (c+d)=o. 
 9. Given a triangle ABC. To determine a line #, such 
 
 that the sides of the triangle ABC, decreased or increased 
 by x, may form the sides of a right triangle. 
 
 B 
 
 10. On the hypotenuse of the right triangle ABC (Fig. 324), 
 a segment A D is laid off equal to A B. D E is drawn _l_ A C, 
 and A E is drawn. Then A D' is made equal to A E, the per- 
 pendicular D'E' is erected and A E' drawn. It is required to 
 express A E, A E', etc., in terms of the sides of the triangle. 
 
 11. Construct a line equal to l x aV ab. 
 
 12. Construct a line equal to i/a. 
 
 Notice that \f a = V i V i a. 
 
 297. Construction of various geometrical figures. 
 
 i. Construct a triangle similar to a given triangle and 
 having an area equal to half the area of the given triangle; 
 having an area equal to J, , of the given triangle.
 
 282 Second-Year Mathematics 
 
 2. Construct a square equal in area to a regular triangle 
 inscribed in a circle of radius 2; 3; 4; a. 
 
 3. Construct an equilateral triangle equal in area to a 
 regular hexagon inscribed in a circle of radius 2 ; 3; 5; p. 
 
 4. A regular triangle is inscribed in a circle of radius 3. 
 It is required to construct a rectangle having base 5 and equal 
 in area to the triangle. 
 
 5. It is required to construct a circle of such radius that 
 the regular triangle inscribed in this circle shall be equal in 
 area to a given rectangle. 
 
 6. Construct a circle of such radius that the square in- 
 scribed in this circle shall be equal in area to the equilateral 
 triangle inscribed in a given circle. 
 
 7. Construct a circle of such radius that the area of the 
 inscribed regular hexagon shall be equal to the area of a 
 square inscribed in a given circle. 
 
 8. The hypotenuse and one side of a right triangle are 
 given lines. Construct a square equal in area to the triangle. 
 
 9. Construct a circle of such radius that the inscribed 
 square shall be equal in area to the area of a given triangle. 
 
 Let o= altitude and 5 = base of the given triangle. Construct a 
 circle with radius = Jy ab. Why will this be the required circle? 
 
 10. Construct a circle of such radius that the inscribed 
 regular triangle shall be equal in area to a given triangle. 
 
 11. In a given circle, using a diameter as one of the bases, 
 inscribe a trapezoid such that the altitude shall be one-half 
 of the upper base.
 

 
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