LIBRARY OF THE UNIVERSITY OF CALIFORNIA. GIFT - Class PLANE GEOMETRY BY C. A. HART i \ INSTRUCTOR OF MATHEMATICS, WADLEIGH HIGH SCHOOL, NEW YORK CITY AND DANIEL D. FELDMAN HEAD OF DEPARTMENT OF MATHEMATICS, ERASMUS HALL HIGH SCHOOL, BROOKLYN WITH THE EDITORIAL COOPERATION OF J. H. TANNER AND VIRGIL SNYDER PROFESSORS OF MATHEMATICS IN CORNELL UNIVERSITY , NEW YORK : CINCINNATI : CHICAGO AMERICAN BOOK COMPANY oo i 7 I*" GIFT COPYRIGHT, 1911, BY AMERICAN BOOK COMPANY. ENTERED AT STATIONERS' HALL, LONDON. 1I.-F. PLANE GEOMETRY. W. P. I PREFACE THIS book is the outgrowth of an experience of many years in the teaching of mathematics in secondary schools. The text has been used by many different teachers, in classes of all stages of development, and under varying conditions of sec- ondary school teaching. The proofs have had the benefit of the criticisms of hundreds of experienced teachers of mathe- matics throughout the country. The book in its present form is therefore the combined product of experience, classroom test, and severe criticism. The following are some of the leading features of the book : TJie student is rapidly initiated into the subject. Definitions are given only as needed. The selection and arrangement of theorems is such as to meet the general demand of teachers, as expressed through the Mathe- matical Associations of the country. Most of the proofs have been given in full. Proofs of some of the easier theorems and constructions are left as exercises for the student, or are given in an incomplete form ; but in every case in which the proof is not complete, the incompleteness is specifically stated. The authors believe that the proofs of most of the propositions should be complete, first, in order to serve as models for the handling of exercises ; second, to pre- vent the serious error of making the student feel contented with loose and slipshod reasoning which defeats the main pur- pose of instruction in geometry ; and third, as an excellent means of reviewing the previous theorems on which they depend. Tlie indirect method of proof is consistently applied. The usual method of proving such propositions as Arts. 189 and iii 223707 IV PREFACE 415, e.g., is confusing to the student. The method used here is convincing and clear. The exercises are carefully selected. In choosing the exercises, each of the following groups has been given due impor- tance : (a) Concrete exercises, including numerical problems and problems of construction. (6) So-called practical problems, such as indirect measure- ments of heights and distances by means of equal and similar triangles, drawing to scale as an application of similar figures, problems from physics, from design, etc. (c) The traditional exercises given in a more or less abstract setting. The arrangement of the exercises is pedagogical. Exercises of a rather easy nature are placed immediately after the theo- rems of which they are applications, instead of being grouped together without regard to the principles involved in them. In many instances the exercises are so arranged as to consti- tute a careful line of development, leading gradually from a very simple construction or exercise to others that are more difficult. For the benefit of the brighter pupils, however, and for review classes, large lists of more or less difficult exercises are grouped at the end of each book. TJie definitions of plane closed figures are unique. The stu- dent's natural conception of a plane closed figure is not the boundary line only, nor the plane only, but the whole figure composed of the boundary line and the plane bounded. All definitions of closed figures involve this idea, which is entirely consistent with the higher mathematics. The numerical treatment of magnitudes is explicit, the funda- mental principles being definitely assumed (Art. 336, proof in Appendix, Art. 595). This procedure is novel and is believed to be the only logical, and at the same time teachable, method of dealing with incommensurables. Teachers who find these subjects too difficult, however, can easily omit them without interruption of sequence, PREFACE v The area of a rectangle is introduced by actually measuring it, thereby obtaining its measure-number. This method permits the same order of theorems and corollaries as is used in the parallelogram and triangle. The correlation with arithmetic in this connection is valuable. The number concepts already found so useful and practical in the modern treatment of ratio and proportion have been developed in connection with areas, as well as in other portions of the book. Proofs of the superposition theorems and the concurrent line theorems will be found exceptionally accurate and complete. The many historical notes are such as will add life and interest to the work. The carefully arranged summaries throughout the book, and the collection of formulas of plane geometry at the end of the book, it is hoped, will be found helpful to teacher and student alike. Argument and reasons are arranged in parallel form. This arrangement gives a definite model for proving exercises, ren- ders the careless omission of the reasons in a demonstration impossible, leads to accurate thinking, and greatly lightens the labor of reading papers. Every construction figure contains all necessary construction lines. This method keeps constantly before the student a model for his construction work, and distinguishes between a figure for a construction and a figure for a theorem. The mechanical arrangement is such as to give the student every possible aid in comprehending the subject matter. The grateful acknowledgment of the authors is due to many friends for helpful suggestions ; especially to Miss Grace A. Bruce of the Wadleigh High School, New York City; to Mr. Edward B. Parsons of the Boys' High School, Brooklyn; and to Professor McMahon of Cornell University. CONTENTS PLANE GEOMETRY PAGE SYMBOLS AND ABBREVIATIONS . . . . . . viii INTRODUCTION ... 1 The Subject Matter of Geometry 1 Four Fundamental Geometric Concepts 2 The Four Concepts in Reverse Order 2 Definitions and Assumptions 3 Lines 4 Use of Instruments 6 Surfaces 7 Angles 8 . Assumptions .......... 11 Demonstrations . . . 13 BOOK I. RECTILINEAR FIGURES 19 Necessity for Proof ......... 21 Polygons. Triangles 22 Superposition .25 Measurement of Distances by Means of Triangles ... 32 Summary of Equal Triangle Theorems ..... 35 Loci 42 Summary of Unequal Angle Theorems . . . .64 Summary of Unequal Line Theorems . . . . .64 Parallel Lines 65 Summary of Parallel Line Theorems 73 Quadrilaterals. Parallelograms 84 Concurrent Line Theorems ....... 100 Construction of Triangles 106 Directions for the Solution of Exercises . . . ' . .110 Miscellaneous Exercises . . . . . .111 vi CONTENTS vii PAGE BOOK II. THE CIRCLE . . . . . . .113 Two Circles 131 Measurement 133 Constants and Variables. Limits ...... 137 Miscellaneous Exercises . . . . . . . . 156 BOOK III. PROPORTION AND SIMILAR FIGURES . . 1G1 Similar Polygons 176 Summary of Similar Triangle Theorems 183 \S Drawing to Scale 188 Miscellaneous Exercises . . ... . . . . 206 BOOK IV. AREAS OF POLYGONS 209 Transformation of Figures ........ 223 Miscellaneous Exercises . 239 Exercises of Greater Difficulty . . . . . . .241 BOOK V. REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 245 Measurement of the Circumference and of the Circle . . 258 Miscellaneous Exercises 274 Miscellaneous Exercises on Plane Geometry .... 276 Formulas of Plane Geometry ....... 282 APPENDIX TO PLANE GEOMETRY 284 Maxima and Minima 284 Variables and Limits. Theorems . . . . . .291 To Every Straight Line there belongs a Measure-number . . 294 Discussion of Common Measure of Lines ..... 295 An Angle can be bisected by only One Line .... 296 Note on Axioms ... . . . . . . . 296 INDEX 297 SYMBOLS AND ABBREVIATIONS = equals, equal to, is equal to. = does not equal. > greater than, is greater than. < less than, is less than. equivalent, equivalent to, is equiva- lent to. ~ similar, similar to, is similar to. ^ is measured by. _L perpendicular, perpendicular to, is perpendicular to. Js perpendiculars. || parallel, parallel to, is parallel to. || s parallels. . . . and so on (sign of continuation). v since. .-. therefore. ^ arc ; AD, arc AB. O, OJ parallelogram, parallelograms. O, circle, circles. Z, A angle, angles. A, A triangle, triangles. Q.E.D. Quod erat demonstrandum, lohich was to be proved. Q.E.F. Quod erat faciendum, which was to be done. The signs +, , x , * have the same meanings as in algebra. rt. right. str. straight. ext. exterior. int. interior. alt. alternate. def. definition. ax. axiom. post. postulate. hyp. hypothesis. prop. proposition. prob. problem. th. theorem. cor. corollary. cons. construction. ex. exercise. %. figure. iden. identity. com p. complementary. sup. supplementary. adj. adjacent. homol. homologous. viii PLANE GEOMETRY INTRODUCTION 1. The Subject Matter of Geometry. In geometry, although we shall continue the use of arithmetic and algebra, our main work will be a study of what will later be denned ( 13) as geometric figures. The student is already familiar with the physical objects about him, such as a ball or a block of wood. By a careful study of the following exercise, he may be led to see the relation of such physical solids to the geometric figures with which he must become familiar. Exercise. Look at a block of wood (or a chalk box). Has it weight ? color ? taste ? shape ? size ? These are called properties of the solid. What do we call such a solid ? A physical solid. Can you think of the properties of this" solid apart from the block of wood ? Imagine the block removed. Can you imagine the space which it occupied ? What name would you give to this space ? A geometric solid. What properties has it that the block possessed ? Shape and size. is it that separates this geometric solid from surrounding space ? thick is this surface? Ho'w many surfaces has the block? Where do they intersect ? How many intersections are there ? How wide are the intersections ? how long ? What is their name ? They are lines. Do these lines intersect ? where ? How wide are these intersections ? how thick ? how long ? Can you say where this one is and so distinguish from where that one is ? What is its name ? It is a point. If you move the block through space, what will it generate as it moves ? What will the surfaces of the block generate ? all of them ? Can you move a surface so that it will not generate a solid ? Yes, by moving it along itselji What will the edges of the block generate ? Can you move an edge so that it will not generate a surface ? What will the corners generate ? Can you move a point so that it will not generate a line ? What How 2 . PLANE GEOMETRY tfOUR FUNDAMENTAL GEOMETRIC CONCEPTS 2. The space in which we live, although boundless and unlimited in extent, may be thought of as divided h>to parts. A physical solid occupies a limited portion of space. The portion of space occupied by a physical solid is called a geometric solid. 3. A geometric solid has length, breadth, and thickness. It may also be divided into parts. The boundary of a solid is called a surface. 4. A surface is no part of a solid. It has length and breadth, but no thickness. It may also be divided into parts. The boundary of a surface is called a line. 5. A line is no part of a surface. It has length only. It may also be divided into parts. The boundary or extremity of a line is called a point. A point is no part of a line. It has neither length, nor breadth, nor thickness. It cannot be divided into parts- It is position only. THE FOUR CONCEPTS IN REVERSE ORDER 6. As we have considered geometric solid independently of surface, line, and point, so we may consider point indepen- dently, and from it build up to the solid. A small dot made with a sharp pencil on a sheet of paper represents approximately a geometric point. 7. If a point is allowed to move in space, the path in which it moves will be a line. A piece of fine wire, or a line drawn on paper with a sharp pencil, represents approximately a geometric line. This, how- ever fine it may be, has some thickness and is not therefore an ideal t or geometric, line. .8. If a line is allowed tp move in space, its path in general will be a surface. INTRODUCTION 3 9. If a surface is allowed* to move in space, its path in general will be a geometric solid. 10. A solid has threefold extent and so is said to have three dimensions ; a surface has twofold extent and is said to have two dimensions ; a line has onefold extent or one dimension ; a point has no extent and has therefore no dimensions. 11. The following may be used as working definitions of these four fundamental concepts : A geometric solid is a limited portion of space. A surface is that which bounds a solid or separates it from an adjoining solid or from the surrounding space. A line is that which has length only. A point is position only. DEFINITIONS AND ASSUMPTIONS 12. The primary object of elementary geometry is to deter- mine, by a definite process of reasoning that will be introduced and developed later, the properties of geometric figures. In all logical arguments of this kind, just as in a debate, certain fundamental principles are agreed upon at the outset, and upon these as a foundation the argument is built. In ele- mentary geometry these fundamental principles are called definitions and assumptions. The assumptions here mentioned are divided into two classes, axioms and postulates. These, as well as the definitions, will be given throughout the book as occasion for them arises. 13. Def. A geometric figure is a point, line, surface, or solid, or a combination of any or all of these. 14. Def. Geometry is the science which treats of the properties of geometric figures. 15. Def. A postulate may be defined as the assumption of the possibility of performing a certain geometric operation. Before giving the next definition, it will be necessary to introduce a postulate. 4 PLANE GEOMETRY 16. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. 17. Def. Two geometric figures are said to coincide if, when either is placed upon the other, each point of one lies upon some point of the other. 18. Def. Two geometric figures are equal if they can be made to coincide. 19. Def. The process of placing one figure upon another so that the two shall coincide is called superposition. This is an imaginary operation, no actual movement taking place. LINES 20. A line is usually designated by two capital letters, as line AB. It may be designated also by a small letter placed somewhere 011 the line, as line a. 21. Straight Lines. In 7 we learned that a piece of fine wire or a line drawn on a sheet of paper represented approximately a geometric line. So also a geometric straight line may be represented approximately by a string stretched taut between two points, or by the line made by placing a ruler (also called a straightedge) on a flat surface and drawing a sharp pencil along its edge. 22. Questions. How does a gardener test the strai^htness of the edge of a flower bed ? How does he get his plants set out in straight rows ? How could you test the straightness of a wire ? Can you think of a wire not stra ; ght, but of such shape that you could cut out a piece of it and slip it along the wire so that it would always fit ? If you re- versed this piece, so that its ends changed places, would it still fit along the entire length of the wire ? If you turned it over, would it fit ? Would the piece cut out fit under these various conditions if the wire were straight ? INTRODUCTION 5 23. JDef. A straight line is a line such that, if any portion of it is placed with its ends in the line, the entire portion so placed will lie in the line, however it may be applied. Thus, if AB is a straight line, and if any portion of AB, as CD, is placed on any other part of AS, with its ends in AB, 4 2 B every point of CD will lie in AB. FlG - 2 - A straight line is called also a right line. The word line, unqualified, is understood to mean straight line. 24. Straight line postulate. A straight line may be drawn from any one point to any other. 25. Draw a straight line AB. Can you draw a second straight line from A to B ? If so, where will every point of the second line lie ( 23) ? It then follows that : Only one straight line can be drawn between two points; i.e. a straight line is determined by two points. 26. Draw two straight lines AB and CD intersecting in point P. Show that AB and CD cannot have a second point in com- mon ( 2-3). It then follows that: Two intersecting straight lines can have only one point in com- mon; i.e. two intersecting straight lines determine a point. 27. Def. A limited portion of a straight line is called a line segment, or simply a line, or a segment. Thus, in Fig. 2, AC, CD, and DB are line segments. 28. Def. Two line segments which lie in the same straight line are said to be collinear segments. 29. Def. A curved line (or curve) is a line no portion of which is straight, as GH. 30. Def. A broken line is a line made up of different succes- sive straight lines, as Kl. FIG, 3, 6 PLANE GEOMETRY 31. Use of Instruments. Only two instruments are permitted in the constructions of plane geometry : the ruler or straight- edge for drawing a straight line, already spoken of in 21 ; and the compasses, for constructing circles or arcs of circles, and for transferring line segments from one position to another. Thus to add two lines, as AB and CD, draw, with a ruler, a straight line OX. Place one leg of the compasses at A and the other at B. Next place one leg at and cut off segment OM equal to AB. In a similar manner lay off MN equal to CD. Then M NX AB+CD = OM+ MN = ON. FIG. 4. Show how to subtract AB from ON. What is the remainder? Ex. 1. Can a straight line move so that its path will not be a surface ? If so, how ? Ex. 2. Can a curved line move in space so that its path will not be a surface ? If so, how ? Ex. 3. Can a broken line move in space so that its path will not be a surface ? Ex. 4. Draw three lines as AB, CD, and ^ p EF. Construct the sum of AB and CD ; of AB and EF-, of AB, CD, and EF. C D Ex. 5. Construct : () the difference be- E F tween AB and CD ; (6) the difference between FlG 5 CD and EF; (c) the difference between AB and EF. Add the results obtained for (a) and (6) and see whether the sum is the result obtained for (c). Ex. 6. Draw a line twrce as long as AB (the sum of AB and AB); three times as long as AB. INTRODUCTION 7 SURFACES 32. Plane Surface. It is well known that the carpenter's straightedge is applied to surfaces to test whether they are flat and even. If, no matter where the straightedge is placed on the surface, it always fits, the surface is called a plane. Now if we should use a powerful magnifier, we should doubt- less discover that in certain places the straightedge did not exactly fit the surface on which it was placed. A sheet of fine plate glass more nearly approaches the ideal. 33. Questions. Test the surface of the blackboard with a ruler to see whether it is a plane. How many times must you apply the ruler ? Can you think of a surface such that the ruler would fit in some positions (a great many) but not in all ? Can you think of a surface not plane but such that a piece of it could be cut out and slipped along the rest so that it would fit ? Would it fit if turned over (inside out) ? 34. Def. A plane surface (or plane) is a surface of unlimited F extent such that whatever two of its points are taken, a straight line joining them will lie wholly in the surface. 35. Def. A curved surface is a surface no portion of which is plane. 36. Def. A plane figure is a geometric figure all of whose points lie in one plane. Plane Geometry treats of plane figures. 37. Def. A rectilinear figure is a plane figure all the lines of which are straight lines. Ex. 7. How can a plane move in space so that its path will not t a solid ? / Ex. 8. Can a curved surface move in space so that its path will not be a solid ? If so, bow ? 8 PLANE GEOMETRY ANGLES 38. Def. An angle is the figure formed by two straight lines which diverge from a point. The point is the vertex of the angle and the lines are its sides. 39. An angle may be designated by a number placed within it, as angle 1 and angle 2 in Fig. 7, and angle 3 in Fig. 8. Or L M H FIG. 9. three letters may be used, one on each side and one at the vertex, the last being read between the other two; thus in Fig. 7, angle 1 may be read angle ABC, and angle 2, angle CBD. An angle is often designated also by the single letter at its vertex, when no other angle has the same vertex, as angle F in Fig. 8. 40. Revolution postulate. A straight line may revolve in a, plane, about a point as a pivot, and when it does revolve contin- uously from one position to another, it passes once and only once through every intermediate position. 41. A clear notion of the magnitude of an angle may be obtained by imagining that its two sides were at first collinear, and that one of them has revolved about a point common to the two. Thus in Fig. 8. we may imagine FG first to have been in the position FE and then to have revolved about F as a pivot to the position FG. 42. Def. Two angles are adjacent if they have a common vertex and a common side which lies between them ; thus in Fig. 7, angle 1 and angle 2 are adjacent; also in Fig. 9, angle HMK and angle KML are adjacent. INTRODUCTION 9 43. Two angles are added by placing them so that they are adjacent. Their sum is the angle formed by the two sides that are not common ; thus in Fig. 10, the sum of angle 1 and angle 2 is angle ABC. C B FIG. 10. 44. The difference between two angles is found by placing them so that they have a vertex and a side in' common but with the common side not between the other two. If the other two sides then happen to be collineaci*, the difference between the angles is zero and the angles are equal. If the other sides are not collinear, the angle which they form is the difference between the two angles compared ; thus in Fig. 11, the difference between angle 1 and angle 2 is angle ABC. A FIG. 11. 45. Def. If one straight line meets another so as to make two adjacent angles equal, each of these angles is a right angle, and the lines are said to be perpendicular to each other. Thus, if DC meets AB so that angle BCD and angle DCA are equal angles, each is a right angle, and lines An and CD are said to be 4 ' C perpendicular to each other. FIG. 12. 10 PLANE GEOMETRY 46. Def. If two lines meet, but are not perpendicular to each other, they are said to be oblique to each other. 47. Def. An acute angle is an angle that is less than a right angle ; as angle 1, Fig. 13. FIG. 13. FIG. 14. 48. Def. An obtuse angle is an angle that is greater than a right angle and less than two right angles ; as angle 2, Fig. 14. 49. Def. A reflex angle is an angle that is greater than two right angles and less than four right angles; as angle 2, Fig. 15. 50. Note. Two lines diverging from the same point, as BA and BC, Fig. 15, always form two positive angles, as the acute angle 1 and the reflex angle 2. Angle 1 may be thought of as formed by the revolution of a line counter- clockwise from the position BA to the position BC, and should be read angle ABC. Angle 2 may be thought of as 1 formed by the revolution of a line counter-clockwise from the position BG to the position BA, and should be read' angle CBA. 51. Def. Acute, obtuse, and reflex angles are sometimes called oblique angles. Ex. 9. (a) In Fig. 16, if angle 1 equals angle 2, what kind, of angles are they ? (6) Make a statement with regard to the lines AB and CD. (c) If angle 3 does not equal angle 4, what kind of angles are they ? (d) Make a statement with regard to the lines AB and CE. FIG. Ex. 10. A plumb line is suspended from the top of the blackboard. What kind of angles does it make with a horizontal line drawn on the blackboard ? with a line on the blackboard neither horizontal nor vertical ? INTRODUCTION 11 Ex. 11. Suppose the minute hand of a clock is at twelve. Where may the hour hand be so that the two hands make with each other : (a) an acute angle ? (6) a right angle ? (c) an obtuse angle, ? Ex. 12. Draw : (a) a pair of adjacent angles ; (6) a pair of non- adjacent angles. Ex. 13. Draw two adjacent angles such that : (a) each is an acute angle ; (6) each is a right angle ; (c) each is an obtuse angle ; (d) one is acute and the other right ; (e) one is acute and the other obtuse. Ex. 14. In Fig. 17, angle 1 + angle 2 = ? angle 3 + angle 4 = ? angle BAD + angle DAF = ? angle DAF - angle 3 = ? angle 2 + angle DAF- angle 4 = ? angle 4 + angle BAE angle 1 = ? Ex. 15. Name six pairs of adjacent angles in Fig. 17. Ex. 16. Draw two non-adjacent angles that have: (a) a common vertex; (6) a common side; (c) a common vertex and a common side. 52. Def. A line is said to be bisected if it is divided into two equal parts. 53. Def. The bisector of an angle is the line which divides the angle into two equal angles.* Ex. 17. Draw a line AB, neither horizontal nor vertical, (a) Draw freehand a line perpendicular to AB and not bisecting it :_ (6) a line bisecting AB and not perpendicular to it. ASSUMPTIONS 54. 1. TJiings equal to the same thing, or to equal things, are equal to each other. 2. If equals are added to equals, the sums are. equal. 3. If equals are subtracted from equals, the remainders are 4. If equals are added to unequals, the sums are unequal in the same order. 5. If equals are subtracted from unequals, the remainders are unequal in the same order. * The proof that every angle has but one bisector will be found in the Appendix, 599. 12 PLANE GEOMETRY 6. If unequals are subtracted from equals, the remainders are unequal in the reverse order. 7. (a) If equals are multiplied by equals, the products are equal; (b) if unequals are multiplied by equals, the products are unequal in the same order. 8. (a) If equals are divided by equals, the quotients are equal; (b) if unequals are divided by equals, the quotients are unequal in the same order. 9. If unequals are added to unequals, the less to the less and the greater to the greater, the sums are unequal in the same order. 10. If three magnitudes of the same kind are so related that the first is greater than the second and the second greater than the third, then the first is greater than 'the. third. 11. Tlie whole is equal to the sum of all its parts. 12. TJie whole is greater than any of its parts. 13. Like powers of equal numbers are equal, and like roots of equal numbers are equal. 14. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. (See 16.) 15. Straight line postulate I. A straight line may be drawn from any one point to any other. (See 24.) 16. Straight line postulate II. A line segment may be pro- longed indefinitely at either end. 17. Revolution postulate. A straight line may revolve in a plane, about a point as a pivot, and when it does revolve continu- ously from one position to another, it passes once and only once through every intermediate position. (See 40.) 55. Assumptions 1-13 are usually called axioms. That is, an axiom may be defined as a statement whose truth is as- sumed.* Ex. 18. Illustrate the first five assumptions above by using arithmeti- cal numbers only. Ex. 19. Illustrate the next five by using general numbers (letters) pnly. INTRODUCTION 13 DEMONSTRATIONS 56. It has been stated ( 12) that the fundamental princi- ples agreed upon at the outset as forming the basis of the logi- cal arguments in geometry are called definitions, axioms, and postulates. Every new proposition advanced, whether it is a statement of a truth or a statement of something to be per- formed, must by a process of reasoning be shown to depend upon these fundamental principles. This process of reasoning is called a proof or demonstration. After the truth of a state- ment has thus been established, it in turn may be used to establish new truths. The propositions here mentioned are divided into two classes. theorems and problems. 57. Def. A theorem is a statement whose truth is required to be proved or demonstrated. For example, " If two angles of a triangle are equal, the sides opposite are equal" is a theorem. There are two parts to every theorem : the hypothesis, or the conditional part ; and the conclusion, or the part to be proved. In the theorem just quoted, " If two angles of a tri- angle are equal" is the hypothesis; and "the sides opposite are equal " is the conclusion. Ex. 20. Write out carefully the hypothesis and the conclusion of each of the following : (a) If you do your duty at all times, you will be rewarded. (&) If you try to memorize your proofs, you will never learn geometry. (c) You must suffer if you disobey a law of nature. (d) Things equal to the same thing are equal to each other. (e) All right angles are equal. 58. Def. A corollary is a statement of a truth easily de- duced from another truth. Its correctness, like that of a theorem, must be proved. 59. Def. A problem, in general, is a question to be solved. As applied to geometry, problems are of two kinds, namely, problems of construction and problems of computation. 14 PLANE GEOMETRY 60. From the revolution postulate ( 40) and from the definition of a perpendicular ( 45) we may deduce the follow- ing corollaries : 61. Cor. I. At every point in a straight line there exists a perpendicular to the line. Given line AB and point C in j ' To prove that there exists a J_ to ,2-" AB at C, as CD. ' -E AC B Let CE (Fig. 18) meet AB at C, so that FIG. 18. Z 1 < Z 2. Then if CE is revolved about C as a pivot toward position CM, Z 1 will continuously increase and Z 2 will continuously decrease ( 40)- .-. there must be one position of CE, as (7Z>, in which the two A formed with AB are equal. In this position CD A.AB ( 45). 62. Cor. II. At every point in a straight line there exists only one perpendicular to the line. Given CD AB at C } i.e. Z. BCD To prove CD the only _L to AB \ at C. \ Let CE (Fig 19) be any line from C A C B other than CD. Let CE fall between OD FIG. 19. and CA* Then /.BCE>Z. BCD ; i.e. > ZD(L4. And Z.ECA<^DCA ( 54, 12). .-. ^.BOTand^CM are not equal and CE is not _L to AB ( 45). 63. 61 and 62 may be combined in one statement as follows : At every point in a straight line there exists one and only one perpendicular to the line. * A similar proof may be given if we suppose CE to fall between CE and CD, INTRODUCTION 64. Cor. m. All right angles are equal. B D 15 FIG. 20. Given Z.AOB and Z CPD, any two rt. A. To prove Z AOB = Z CPD. Place Z CPD upon Z A OB so that point P shall fall upon point 0, and so that PC shall be collinear with OA. Then PD and OB, both being _L to 0^4 at 0, must be collinear ( 62). .-. the two A coincide and are equal ( 18). 65. Cor. IV. If one straight line meets another straight line, the sum of the two adjacent angles is two right angles. Given str. line CD meeting str. line AP. at c t forming A BCD and DC A. To prove Z BCD + Z DC A = 2 rt. A. Let CE be J_ to AB at C ( 63). Then Z .BCD + Z DCE = 1 rt. Z ; .-. Z BCD = 1 rt. Z -/.DCE. r Again Z D(L4 = 1 rt. Z + Z D C7.E. .-. Z BCD + Z DC A = 2 rt. 4. c FIG. 21. 66. Cor. V. The sum of all the angles about a point on one side of a straight line passing through that point equals two right angles. Given A 1, 2, 3, 4, 5, and 6, all the A about point on one side of str. line CA. To prove Z1+Z2 + Z3+ The sum of the six A (Fig.22) is equal to /.AOB + Z BOC. 16 PLANE GEOMETRY 67. Cor. VI. The sum of all tKe angles about a point equals four right angles. Prolong one of the lines through the vertex and apply Cor. V. kl \ 68. Def. Two angles whose sum is one right FIG. 23. angle are called complementary angles, as angles 1 and 2, Fig. 25. Either of two such angles is said to be the complement of the other. 69. Def. Two angles whose sum is two right angles are called supplementary angles, as angles 1 and 2, Fig. 26. Either of two such angles is said to be the supplement of the other. An angle that is equal to the sum of two right angles is sometimes called a straight angle, as angle Fig. 26. 70. Def. Two angles are said to be vertical if the sides of each are the pro- longations of the sides of the other; thus, in Fig. 24, angles 1 and 2 are ver- tical angles, and angles 3 and 4 are like- wise vertical angles. 71. Def. An angle of one degree is one nine- tieth of a right angle. A right angle, therefore, contains 90 angle degrees. FIG. 24. C Ex. 21. In Fig. 25, angle ABC is a right angle. If angle 1 = 40, how many degrees are there in angle 2 ? How many degrees, then, are there in the comple- ment of an angle of 40 ? FIG. 25. Ex. 22. How many degrees are there in the complement of 20? 35 ? of a ? of | right angles ? of k right angles ? Ex. 23. In Fig. 26, angle 1 + angle 2, or angle ABC, equals 2 right angles. If angle 1 = 40, < how many degrees are there in angle 2 ? How of D B FlG - .many degrees, then, are there in .the supplement of .an angle of 40 ? INTRODUCTION 17 Ex. 24. How many degrees are there in the supplement of 20 ? of 140-? of o ? of n right angles ? Give these answers in right angles. Ex. 25. In the accompanying diagram : () If angle 1 = 65, how many degrees are there in each of the other three angles ? (6) If angle 2 = ra, how many right angles are there in each of the other three angles ? (c) If angle 3 = n right angles, how many degrees are there in each of the other three angles ? Ex. 26. Compare the supplement of an angle of 50 with its complement. *Draw a diagram showing the complement and the supplement of an acute angle, ABC. Ex. 27. Criticize the following exercise : How many degrees are there in an angle whose complement is f of its supplement ? in an angle whose supplement is f of its complement ? Ex. 28. If one straight line meets another straight line so that one angle is double its adjacent angle, how many degrees are there in each of the two adjacent angles ? Ex. 29. How many degrees are there in an angle whose complement and supplement together equal 194 ? Ex. 30. How many degrees are there in an angle which is f of its complement? Ex. 31. How many degrees are Inhere in an angle whose supplement is nineteen times its complement ? Ex. 32. If there are only five angles about a point, and each differs , by 15 from an angle adjacent, how many degrees are there in each \J^ angle ? Ex. 33. If there are only six angles about a point and they are all equal, how large is each angle ? 72. Def. Angles that are -supple- _ 2 -s mentary and adjacent are called, supple- mentary-adjacent angles, as A l~and 2. Ex. 34. If two angles are supplementary-adjacent and one of them is one half a right angle, how large is the other ? Ex. 35. Suppose that only three angles are formed about a point on one side of a straight line passing through the point. The greatest is four times the least, and the remaining one is 12 more than the least. How many degrees are there in each ? /) 18 PLANE GEOMETRY Ex. 36. (a) Can two angles be supplementary and not adjacent? Illustrate. (6) Can two angles be adjacent and not supplementary ? Illustrate. (c) Can two angles be both supplementary and adjacent ? Illustrate. (d) Can two angles be neither supplementary nor adjacent ? Illustrate. (e) Can two angles be both complementary and adjacent ? Illustrate. (/) Can two angles be both complementary and supplementary ? Illustrate. 73. From the definitions of complementary and supple- mentary angles may be deduced three additional corollaries as follows : 74. Cor. I. Complements of the same angle or of equal angles are equal. 75. Cor. II. Supplements of the same angle or of equal angles are equal. 76. Cor. III. If two adjacent angles are supplemen- tary, their exterior sides are collinear. Given Z A EG + Z CBD = 2 rt. A To prove BD the prolongation of AB. If BD is not the prolongation of AB, then some other line from B, as BE, must D ~B A be its prolongation. FIG. 29. Then Z ABC + Z CBD = 2 rt. A (By hyp.). Also Z ABC + Z CBE = 2 rt. A ( 65). This is impossible ( 54, 12) ; i.e. the supposition that BE is the pro- longation of AB leads, by correct reasoning, to the impossible conclusion that Z CBD = Z CBE. Hence this supposition itself is false. /. BD is the prolongation of AB. BOOK I RECTILINEAR FIGURES PROPOSITION I. THEOREM 77. If two straight lines intersect, the vertical angles are equal. Given two intersecting str. lines, forming the vertical A, and 3, also 2 and 4. To prove Z 1 = Z3 and Z2 = Z4. ARGUMENT = 2rt. A 4. 5. Likewise Z 2 = Z 4. Q.E.D. REASONS 1. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. Zs. 65. 2. Same reason as 1. 3. Things equal to the same thing are equal to each other. 54, 1. 4. If equals are subtracted from equals, the remain- ders are equal. 54, 3. 5. By steps similar to 1, 2, 3, and 4. 78. Cor. If two straight lines intersect so that any two adjacent angles thus formed are equal, all the angles are equal, and each angle is a right angle. 19 20 PLANE GEOMETRY Ex. 37. If three straight lines intersect at a common point, the sum of any three angles, no two of which are adjacent, as 1, 3, and 5, in the accompanying diagram, is equal to two right angles. Ex. 38. In the same diagram, show that : (a) Angle 1 + angle 3 = angle 7. (&) Angle 7 angle 4 = angle 6. (c) Angle 7 + angle 4 angle 3 = twice angle 1. (d) Angle 2 -f angle 4 + angle 6 = 2 right angles. Ex. 39. The line which bisects one of two vertical angles bisects the other also. Ex. 40. The bisectors of two adjacent complementary angles form an angle of 45. Ex. 41. Show the relation of the bisectors of two adjacent supple- mentary angles to each other. Ex. 42. The bisectors of two pairs of vertical angles are perpendicu- lar to each other. 79. The student will observe from Prop. I that the complete solution of a theorem consists of four distinct steps : (1) The statement of the theorem, including a general state- ment of the hypothesis and conclusion. (2) The drawing of a figure (as general as possible), to satisfy the conditions set forth in the hypothesis. (3) Tlie application, i.e. the restatement of the hypothesis and conclusion as applied to the particular figure just drawn, under the headings of " Given " and " To prove." (4) The proof, or demonstration, the argument by which the truth or falsity of a statement is established, with the reason for each step in the argument. 80. The student should be careful to quote the reason for every step in the argument which he makes in proving a theorem. The only reasons admissible are of two kinds : 1. Something assumed, i.e. definitions, axioms, postulates, and the hypothesis. 2. Something previously proved, i.e. theorems, corollaries, and problems. BOOK I 21 81. The necessity for proof. Some theorems seem evident by merely looking at the figure, and the student will doubtless think a proof unnecessary. The eye, however, cannot always detect error, and reasoning enables us to be sure of our conclu- sions. The danger of trusting the eye is illustrated in the following exercises.* Ex. 43. In the diagrams given below, tell which line of each pair is the longer, a or &, and verify your answer by careful measurement. a r n E6 I 1 Ex. 44. In the figures below, are the lines everywhere the same dis- tance apart ? Verify your answer by using a ruler or a slip of paper. \ \ \ \ \ \ \ v \\\\\\\\ Ex. 45. In the figures below, tell which lines are prolongations of other lines. Verify your answers. * These diagrams are taken by permission from the Report of the Committee on Geom- etry, Proceedings of the Eighth Meeting of the Central Association of Science an4 Mathematics Teachers, 22 PLANE GEOMETRY POLYGONS. TRIANGLES 82. Def. A line on a plane is said to be closed if it sepa- rates a finite portion of the plane from the remaining portion. 83. Def. A plane closed figure is a plane figure composed of a closed line and the finite portion of the plane bounded by it. 84. Def. A polygon is a plane closed figure whose boundary is composed of straight lines only. The points of intersection of the lines are the vertices of the polygon, and the segments of the boundary lines included between adjacent vertices are the sides of the polygon. 85. Def. The sum of the sides of a polygon is its perimeter. 86. Def. Any angle formed by two consecutive sides and found on the right in passing clockwise around the perimeter of a polygon is called an interior angle of the polygon, or, for brevity, an angle of the polygon. In Fig. 1, A ABC, BCD, CDE, DEA, and EAB are interior angles of the polygon. 87. Def. If any side of a polygon is prolonged through a vertex, the angle formed by the prolonga- tion and the adjacent side is called an exterior angle of the polygon. In Fig. 1, A 1, 2, 3, 4, and 5 are exterior angles. 88. Def. A line joining any two non-adjacent vertices of a polygon is called a diagonal; as AC, Fig. 1. 89. Def. A polygon which has all of its sides equal is an equilateral polygon. 90. Def. A polygon which has all of its angles equal is an equiangular polygon. BOOK I 23 91. Def. A polygon 'which is both equilateral and equi- angular is a regular polygon. 92. Def. A polygon of three sides is called a triangle ; one of four sides, a quadrilateral ; one of five sides, a pentagon ; one of six sides, a hexagon ; and so on. TRIANGLES CLASSIFIED WITH RESPECT TO SIDES 93. Def. A triangle having no two sides equal is a scalene triangle. 94. Def. A triangle having two sides equal is an isosceles triangle. The equal sides are spoken of as the sides * of the triangle. The angle between the equal sides is the vertex angle, and the side opposite the vertex angle is called the base. 95. Def. A triangle having its three sides equal is an equilateral triangle. Scalene Isosceles Equilateral TRIANGLES CLASSIFIED WITH RESPECT TO ANGLES 96. Def. A right triangle is a triangle which has a right angle. The side opposite the right angle is called the hypote- nuse. The other two sides are spoken of as the sides f of the triangle. 97. Def. An obtuse triangle is a triangle which has an obtuse angle. 98. Def. An acute triangle is a triangle in which all the angles are acute. * The equal sides are sometimes called the arms of the isosceles triangle. This term will be used occasionally in the exercises. t Sometimes the sides of a right triangle including the right angle are called the arms of the triangle. This term will be found in the exercises. 24 PLANE GEOMETRY Ex 46. Draw a scalene triangle freehand : () with all its angles acute and with its shortest side horizontal ; (&) with one right angle. Ex. 47. Draw an isosceles triangle : (a) with one of its arms hori- zontal and one of its angles a right angle ; (6) with one angle obtuse. 99. Def. The side upon which a polygon is supposed to stand is usually called its base ; however, since a polygon may be supposed to stand upon any one of its sides, any side may be considered as its base. The angle opposite the base of a triangle is the vertex angle, and the vertex of the angle is called the vertex of the triangle. 100. Def. The altitude of a triangle is the perpendicular to its base from the opposite vertex. In general any side of a triangle may be considered as its base. Thus in triangle EFG, if FG is taken as base, EH is the altitude; if GE is taken as base, ^Jfwill be the altitude; if EF is taken as base, the third alti- e tude can be drawn. Thus every FlG A triangle has three altitudes. It will be proved later that one perpendicular, and only one, can be drawn from a point to a line. 101. The sides of a triangle are often designated by the small letters corresponding to the capitals at the opposite vertices ; as, sides e, /, and g, Fig. 1. Ex. 48. Draw an acute triangle ; draw its three altitudes freehand. Do they seem to meet in a point ? Where is this point located ? Ex. 49. Draw an obtuse triangle ; draw its three altitudes freehand. Do they meet in a point ? Where is this point located ? Ex. 50. Where do the three altitudes of a right triangle meet ? 102. Def. The medians of a triangle are the lines from the vertices of the triangle to the mid-points of the opposite sides. BOOK I 25 103. Superposition. When certain parts of two figures are given equal, we can determine by a process of pure reason whether the two figures may be made to coincide. This process is far more accurate than the actual transfer- ence of figures, for we are free from physical errors such as have been referred to in 81. Problem. Given line AB less than CD. Apply AB to CD. Solution. Place point A upon point C. Make AB collinear with CD and A B let B fall toward D. Then B will fall , ^ between C and Z>, because AB < CD. Question. Under what hypothesis would B fall on D ? beyond D ? Problem. Given angle ABC less than angle BST. Apply angle ABC to angle BST. Solution. Place point B ~ upon point S arid make BA collinear with SB. Then BC will fall between SB and ST, because Z ABC , angle C = angle F. Apply triangle ABC to triangle DEF and. draw a diagram to illustrate each case. 26 PLANE GEOMETRY PROPOSITION II. THEOREM 105. Two triangles are equal if a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other. B Given A ABC and DEF, AG DF, Z. A Z D, and Z (7= Z F. To prove A ABC = A DEF. ARGUMENT 1. Place A ABC upon A DEF so that A G shall fall upon its equal DF, A upon D, G upon F. 2. Then .47? will become col- linear with DEj and B will fall somewhere on DE, or on its prolonga- tion. 3. Also CB will become col- linear with FE, and B will fall somewhere on FE, or on its prolonga- tion. 4. .'. point B must fall on point E. 5. .'. A ABC = A DEF. Q.E.D. REASONS 1. Any geometric figure may be moved from one posi- tion to another without change of size or shape. 54, 14. 2. Z A = Z D, by hyp. 3. Z C=ZF, by hyp. 4. Two intersecting str. lines determire a point. 26. 5. Two geometric figures are equal if they can be made to coincide. 18. BOOK I 27 Ex. 53. In the figure for Prop. II, assume AB = DE, angle A angle D, and angle B = angle E\ repeat the proof by superposition, marking the lines with colored crayon as soon as their positions are determined. B H D 6 93 K E Ex. 54. Place polygon I upon polygon II so that some part of I shall fall upon its equal in II. Discuss the resulting positions of the remaining parts of the figure. Ex. 55. If two quadrilaterals have three sides and the included angles of one equal respectively to three sides and the included angles of the other, and arranged in the same order, are the quadrilaterals equal? Prove. 106. Note. The method of superposition should be used for proving fundamental propositions only. In proving other propositions it is neces- sary to show merely that certain conditions are present and to quote theorems, previously proved, which state conclusions regarding such conditions. Ex. 56. If at any point in the bisector of an angle a perpendicular to the bisector is drawn meeting the sides of the angle, the two triangles thus formed will be equal. Ex. 57. If equal segments, measured from the point of intersection of two lines, are laid off on one of the lines, and if perpendiculars to this line are drawn at the ends of these segments, two equal triangles will be formed. Ex. 58. If at the ends of a straight line per-- pendiculars to it are drawn, these perpendiculars will cut off equal segments upon any line which bisects the given line and is not perpendicular to it. FIG. 1. FIG. 2. Ex. 59. If two angles of a triangle are equal, the bisectors of these angles are equal (Fig. 1). Ex. 60. If two triangles are equal, the bisector of any angle of one is equal to the bisector of the corresponding angle of the other (Fig. 2). 28 PLANE GEOMETRY PROPOSITION III. THEOREM 107. Two triangles are equal if two sides and the in- cluded angle of one are equal respectively to two sides and the included angle of the other. Given A ABC and MNO, AB == MN, AC = MO, and Z A = To prove A ABC = A MNO. M. ARGUMENT 1. Place A ABC upon A MNO so that AC shall fall upon its equal MO, A upon 3/> C upon 0. 2. Then AB will become col- linear with MN. 3. Point B will fall on point N.- 4. . ' . BC will coincide with NO. 5. .'. A ABC = A MNO. Q.E.D. 2. REASONS Any geometric figure may be moved from one posi- tion to another without change of size or shape. 54, 14. /.A = ZM, by hyp. 3. AB = MN, by hyp. 4. Only one str. line can be drawn between two points. 25. 5. Two geometric figures are equal if they can be made to coincide. 18. 108. Cor. Two right triangles are equal if the two sides including the right angle of one are equal respectively to the two sides including the right angle of the other. E*. 61. Prove Prop. Ill by placing 4B upou MN. BOOK I 29 109. Del. In equal figures, the points, lines, and angles in one which, when superposed, coincide respectively with points, lines, and angles in the other, are called homologous parts. Hence : 110. Homologous parts of equal figures are equal. Ex. 62. If two straight lines bisect each other, the lines joining their extremities are equal in pairs. HINT. To prove two lines or two angles equal, try to find two triangles, each containing one of the lines or one of the angles. If the triangles can be proved equal, and the two lines or two angles are homologous parts of the triangles, then the lines or angles are equal. The parts given equal may be more easily remembered by marking them with the same symbol, or with colored crayon. Ex. 63. In case the lines in Ex. 62 are perpendicular to each other, what additional statement can you make ? Prove its correctness. Ex. 64. If equal segments measured from the vertex are laid off on the sides of an angle, and if their extremities are joined to any point in the bisector of the angle, two equal triangles will be formed. Ex. 65. If two medians of a triangle are perpendicular to the sides to which they are drawn, the triangle is equilateral. Ex. 66. If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines joining the ends of these segments to the opposite ends of the base will be equal (Fig. 1). FIG. 1. FIG. 2. Ex. 67. Extend Ex. 66 to the case in which the equal segments are laid off on the arms prolonged through the vertex (Fig. 2). 30 PLANE GEOMETRY PROPOSITION IV. THEOREM 111. The base angles of an isosceles triangle are equal. B Given isosceles A ABC, with AB and BC its equal sides. To prove Z A = Z C. ARGUMENT 1. Let BD bisect Z ABC. 2. In A ABD and DBC, AB = BC. BD = BD. A ABD = A DBC. 6. Q.E.D. REASONS 1. Every Z has but one bi- sector. 53. 2. By hyp. 3. By iden. 4. By cons. 5. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. 107. 6. Homol. parts of equal fig- ures are equal. 110. 112. Cor. I. The bisector of the angle at the vertex of an isosceles triangle is perpendicular to the base and bisects it. 113. Cor. n. An equilateral ftiangle is also equi- angular. Ex. 68. The bisectors of the base angles of an isosceles triangle are equal. BOOK I 31 114. Historical Note. Exercise 69 is known as the pons asi- norum, or bridge of asses, since it has proved difficult to many beginners in geometry. This proposition and the proof here suggested are due to Euclid, a great mathematician who wrote the first systematic text-book on geometry. In this work, known as Euclid's Elements, the exer- cise here given is the fifth prop- osition in Book I. Of the life of Euclid there is but little known except that he was gentle and modest and " was a Greek who lived and taught in Alexandria about 300 B.C." To him is attributed the saying, " There is no royal road to geometry. " His appreciation of the culture value of geometry is shown in a story related by Stobaeus (which is probably authentic). "A lad who had just begun geometry asked, * What do I gain by learning all this stuff ? ' Euclid called his slave and said, ' Give this boy some coppers, since he must make a profit out of what he learns.' " EUCLID Ex. 69. By using the accompanying diagram prove that the base angles of an isosceles triangle are equal. HINT. Prove A ABE = A DEC. Then prove &ACE = &DAC. Ex. 70. (a) If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the vertex. Ex. 71. (a) If equal segjpients measured from the ends of the base are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base. 32 PLANE GEOMETRY Ex. 72. (a) If equal segments measured from the ends of the are laid off on the base of an isosceles triangle, the lines joining the vertex of the triangle to the ends of the segments will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the base prolonged (Fig. 1) . FIG. 1. FIG. 2. Ex. 73. (a) If equal segments measured from the ends of the base are laid off on the arms of an isoscles triangle, the lines drawn from the ends of the segments to the opposite ends of the base will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base (Fig. 2). Ex. 74. Triangle ABC is equilateral, and AE = BF = CD. Prove triangle EFD equi- lateral . A C 115. Measurement of Distances by Means of Triangles. The theorems which prove triangles equal are applied practically in measuring distances on the surface of the earth. Thus, if it is desired to find the distance between two places, A and B, which are separated by a pond or other obstruction, place a stake at some point accessible to both A and B, as F. Measure the distances FA and FB ; then, keeping in line with F and B, measure CF equal to FB, find, in line with F and A, measure FE. equal to FA. Lastly measure CE, and the distance from A to B is thus obtained, since AB is equal to CE. Can this method be used when A and B are on opposite sides of a hill and each is invisible from the other ? BOOK I 33 Ex. 75. Show how to find the distance across a river by taking the following measurements. Measure a convenient dis- tance along the bank, as J?T, and fix a stake at its mid-point, F. Proceed at right angles to BT from T to the point P, where F, S, and P are in line ; measure FT. Ex. 76. An army engineer wished to obtain quickly the approximate distance across a river, and had no instruments with which to make measure- ments. He stood on the bank of the river, as at A, and sighted the opposite bank, or B. Then without raising or lowering his eyes, he faced about, and his line of sight struck the ground at C. He paced the distance, AC, and gave this as the distance across the river. Explain his method. Ex. 77. Tell what measurements to make to obtain the distance between two inaccessible points, R and S (Fig. 1). Ex. 78. The fact that a triangle is deter- mined if its base and its base angles are given was used as early as the time of Thales (640 B.C.) to find the distance of a ship at sea; the base of the triangle was usually a lighthouse tower and the base angles were found by ob- servation. Draw a figure and explain. FIG. 1. Ex. 79. Explain the following method of finding Sfi (Fig. 2). Place a stake at $, and another at a convenient p point, Q, in line with S and E. From a con- venient point, as T, measure TS and TQ. Pro- long QT, and make TF equal to QT. Prolong ST, and make TB equal to ST. Then keep in line with F and B, until a point is reached, as G, where T and ft come into line. Then BG is equal to the required distance, US. Ex. 80. In an equilateral triangle, if two lines are drawn from the ends of the base, making equal angles with the base, the lines are equal. Is this true of every isosceles triangle? FIG. 2. 34 PLANE GEOMETRY PROPOSITION V. 'THEOREM 116. Two triangles are equal if the three sides of one are equal respectively to the three sides of the other. Given A ABC and RST, AB = RS, BC=ST, and CA = TR. To prove A ABC = A RST. 2. ARGUMENT Place A RST so that the longest side RT shall fall upon its equal AC, R upon A, T upon (7, and so that S shall fall oppo- site B. D.raw BS. 3. A AB sis isosceles. 5. A BCS is isosceles. 7. Zl- REASONS 1. Any geometric figure may be moved from one posi- tion to another without change of size or shape. 54, 14. 2. A str. line may be drawn from any one point to any other. 54, 15. 3. AB = RS, by hyp. 4. The base A of an isosceles A are equal. 111. 5. BC=ST, by hyp. 6. Same reason as 4. 7. If equals are added to equals, the sums are equal. 54, 2. BOOK I 35 ARGUMENT 8. .'. Z AC=Z. CSA. 9. .'. A ABC= A i.e. Q.E.D. REASONS The whole = the sum of all its parts. 54, 11. Two A are equal if two sides and the included /- of one are equal respec- tively to two sides and the included Z of the other. 107. Ex. 81. () Prove Prop. V, using two obtuse triangles and applying the shortest side of one to the shortest side of the other. (6) Prove Prop. V, using two right triangles and applying the shortest side of one to the shortest side of the other. 117. Question. Why is not Prop. V proved by superposition? SUMMARY OF CONDITIONS FOB EQUALITY OF TRIANGLES 118. Two triangles are equal if of one are equal respectively to of the other. a side and the two adjacent angles two sides arid the included angle three sides a side and the two adjacent angles two sides and the included angle three sides Ex. 82. The median to the base of an isosceles triangle bisects the angle at the vertex and is perpendicular to the base. Ex. 83. In a certain quadrilateral two adjacent sides are equal ; the other two sides are also equal. Find a pair of triangles which you can prove equal. 36 PLANE GEOMETRY Ex. 84. If the opposite sides of a quadrilateral are equal, the opposite angles also are equal. Ex. 85. If two isosceles triangles have the same base, the line joining their vertices bisects each vertex angle and is perpendicular to the com- mon base. (Two cases.) Ex. 86. In what triangles are the three medians equal ? Ex. 87. In what triangles are two medians equal ? Ex. 88. If three rods of different lengths are put together to form a triangle, can a different triangle be formed by arranging the rods in a different order ? Will the angles opposite the same rods always be the same? Ex. 89. If two sides of one triangle are equal respectively to two sides of another, and the median drawn to one of these sides in the first is equal to the median drawn to the corresponding side in the second, the triangles are equal. 119. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. The curve which forms the boundary of a circle is called the circumference. The fixed point within is called the center, and a line joining the center to any point on the circumference is a radius. 120. It follows from the definition of a circle that : All radii of the same circle are equal. 121. Def. Any portion of a circumference is called an arc. 122. Assumption 18. Circle postulate. A circle may be con- structed having any point as center, and having a radius equal to any finite line. 123. The solution of a problem of construction consists of three distinct steps : (1) The construction, i.e. the process of drawing the required figure with ruler and compasses. (2) TJie proof, a demonstration that the figure constructed fulfills the given conditions. (3) TJie discussion, i.e. a statement of the conditions under which there may be no solution, one solution, or more than one. BOOK I 37 PROPOSITION VI. PROBLEM 124. To construct an equilateral triangle, with a given line as side. \D Given line EC. To construct an equilateral triangle on EG. I. Construction 1. With E as center and EC as radius, construct circle CAD. 2. With C as center and EC as radius, construct circle EM A. 3. Connect point J,-at which the circumferences intersect, with E and C. 4. A ABC is the required triangle. II. Proof ARGUMENT 3. /. AB=BC = CA. A ABC is equilateral. Q.E.D. REASONS 1. All radii of the same circle are equal. 120. 2. Things equal to the same thing are equal to each other. 54, 1. 3 A A having its three sides equal is equilateral. 95. III. Discussion This construction is always possible, and there is only one solution, (See 116.) 38 PLANE GEOMETRY PROPOSITION VII. PROBLEM 125. With a given vertex and a given side, to construct an angle equal to a given angle. Given vertex A, side AB, and Z CDE. To construct an Z equal to Z CDE and having A as vertex and AB as side. / I. Construction 1. With D as center, and with any convenient radius, de- scribe an arc intersecting the sides of Z D at F and G. respec- tively. 2. With A as center, and with the same radius, describe the indefinite arc IH, cutting AB at /. 3. With / as center, and with a radius equal to str. line FG, describe an arc intersecting the arc IH at K. 4. Draw AK. 5. Z BAK = Z CDE, and is the Z required. II. Proof ARGUMENT 1. Draw FG and IK. 2. In A FDG and I A K, DF = AL 3. DG = AK. 4. FG = IK. REASONS 1. A str. line may be drawn from any one point to any other. 54, 15. 2. By cons. 3. By cons. 4. By cons. BOOK I 39 ARGUMENT 6. .'. /.EAK Q.E.D. REASONS 5. Two A are equal if the three sides of one are equal respectively to the three sides of the other. 116. 6. Homol. parts of equal figures are equal. 110. III. Discussion This construction is always possible, and there is only one solution. Ex. 90. Construct a triangle, given two sides and the included angle. Ex. 91. Construct an isosceles triangle, given the vertex angle and an arm. -Ex. 92. Construct a triangle, given a side and the two adjacent angles. Ex. 93. Construct an isosceles triangle, given an arm and one of the equal angles. Ex. 94. How many parts determine a triangle ? Do three angles determine it ? Explain. Ex. 95. Construct an isosceles triangle, given the base and an arm. " Ex. 96. Construct a scalene triangle, given the three sides. 126. Def. The bisector of an angle of a triangle is the line from the vertex of the angle bisecting the angle and limited by the opposite side of the triangle. A Ex. 97. In what triangles are the three bisectors equal ? Ex. 98. In what triangles are two bisectors, and only two, equal ? Ex. 99. In what triangles are the medians, the bisectors, and the altitudes identical ? B 40 PLANE GEOMETRY PROPOSITION VIII. PROBLEM 127. To construct the bisector of a given angle. JC V Given Z AB C. To construct the bisector of Z ABC. I. Construction 1. With B as center, and with any convenient radius, describe an arc intersecting BA at E and BC at D. 2. With D and E as centers, and with equal radii, describe arcs intersecting at F. 3. Draw BF. 4. BF is the bisector of Z. ABC. II. Proof ARGUMENT 1. In A EBF and 5^ = BD. 2. ^ = Z)* 7 . 3. BF = BF. 4. .-. AEBF= AFBD 5. .'. /. EBF = Z.FBD. REASONS 1. By cons. 2. By cons. 3. By iden. 4. Two A are equal if the three sides of one are equal respectively to the three sides of the other. 116. 5. Homol. parts of equal figures are equal. 110. BOOK I 41 ARGUMENT 6. .'. BF is the bisector of Z ABC. Q.E.D. REASONS 6. The bisector of an Z is the line which divides the Z into two equal Zs. 53. III. Discussion This construction is always possible, and there is only one solution. " Ex. 100. Draw an obtuse angle and divide it into : (a) four equal angles ; (&) eight equal angles. Ex. 101. Construct the bisector of the vertex angle of an isosceles triangle. Ex. 102. Draw two intersecting lines and construct the bisectors of the four angles formed. Ex. 103. Bisect an angle between two bisectors in Ex. 102, and find the number of degrees in each angle. Ex. 104. Construct the bisector of an ex^rior angle at the base of an isosceles triangle. Ex. 105. Construct the bisectors of the three angles of any triangle. What can you infer about them ? Can the correctness of this inference be proved by making a careful construction ? 128. Def. The distance between two points is the length of the straight line joining them. Thus if three points, A, B, and (7, are so located that AB = AG, A is said to be equidistant from B and G. Ex. 106. Find all the points on the blackboard which are one foot from a h'xed point, P, on the blackboard. Ex. 107. Draw a line, AB, on the blackboard and mark some point near the line, as P. Find all the points in AB that are a foot from P. Ex. 108. Mark a point, Q, on the blackboard. Find all the points on the blackboard which are : (a) ten inches from Q ; (6) four inches from Q. How far are the points of (6) from the points of (a), if the distance is measured on a line through Q ? 42 PLANE GEOMETRY LOCI 129. In many geometric problems it is necessary to locate all points which satisfy certain prescribed conditions, or to determine the path traced by a point which moves according to certain fixed laws. Thus, the points in a plane two inches from a given point are in the circumference of a circle whose center is the given point and whose radius is two inches. Again, let it be required to find all points in a plane two inches from one fixed point and three inches from another. All points two inches from the fixed point P are in the cir- cumference of the circle LMS, having P for center and having a radius equal to two inches. All points three inches from the fixed point Q are in the cir- cumference of the circle LRT, having Q for center and having a radius equal to three inches. If the two circles are wholly outside of each other; there will be no points satisfying the two prescribed conditions ; if the two circumferences touch, but do not intersect, there will be one point; if the two circum- ferences intersect, there will be two points. It will be proved later ( 324) that there cannot be more than two points which satisfy both of the given conditions. 130. Def. A figure is the locus of all points which satisfy one or more given conditions, if all points in the figure satisfy the given conditions and if these conditions are satisfied by no other points. A locus, then, is an assemblage of points which obey one or more definite laws. It is often convenient to locate these points by thinking of them as the path traced by a moving point the motion of which is controlled by certain fixed laws. BOOK I 43 131. In plane geometry a locus may be composed of one or more points or of one or more lines, or of any combination of points and lines. 132. Questions.* What is the locus of all points in space two inches from a given point ? What is the locus of all points in space two inches from a given plane ? What is the locus of all points in space such that perpendiculars from them to a given plane shall be equal to a given line ? What is the locus of all points on the surface of the earth midway between the north and south poles ? 23 from the equator ? 23 from the north pole ? 90 from the equator ? What is the locus of a gas jet four feet from the ceiling of this room ? four feet from the ceiling and five feet from a side wall ? four feet from the ceiling, five feet from a side wall, and six feet from an end wall ? Ex. 109. Given an unlimited line AB and a point P. Find all points in AB which are also : (a) three inches from P; (6) at a given distance, a, from P. Ex. 110. Given a circle with center O and radius six inches. State, without proof, the locus : (a) of all points four inches from ; (&) of all points five inches from the circumference of the circle, measured on the radius or radius prolonged. Ex. 111. Given the base and one adjacent angle of a triangle, what is the locus of the vertex of the angle opposite the base ? (State without proof.) Ex. 112. Given the base and one other side of a triangle, what is the locus of the vertex of the angle opposite the base ? (State without proof.) Ex. 113. Given the base and the other two sides of a triangle, what is the locus of the vertex of the angle opposite the base ? Ex. 114. Given the base of a triangle and the median to the base, what is the locus of the end of the median which is remote from the base ? Ex. 115. Given the base of a triangle, one other side, and the median to the base, what is the locus of the vertex of the angle opposite the base ? 133. Question. In which of the exercises above was a triangle determined 9 * In order to develop the imagination of the student the authors deem it advisable in this article to introduce questions involving loci in space. It should be noted that no proofs of answers to these questions are demanded. V 44 PLANE GEOMETRY PROPOSITION IX. THEOREM 134. Every point in the perpendicular bisector of a line is equidistant from the ends of that line. A C B Given line AB, its _L bisector CD, and P any point in CD. To prove PA = PB. ARGUMENT REASONS 1. In A APC and CPB, PC=PC. Z PGA = Z BCP. A APC = A CPB. PA = Q.E. D. 1. By hyp. 2. By iden. 3. All rt. A are equal. 64. 4. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. 107. 5. Hoinol. parts of equal fig- ures are equal. 110. Ex. 116. Four villages are so located that B is 25 miles east of A, C 20 miles north of A, and D 20 miles south of A. Prove that B is as far from C as it is from D. Ex. 117. In a given circumference, find the points equidistant from two given points, 4 and JS, BOOK I 45 135. Def. One theorem is the converse of another when the conclusion of the first is the hypothesis of the second, and the hypothesis of the first is the conclusion of the second. The converse of a truth is not always true ; thus, " All men are bipeds " is true, but the converse, " All bipeds.are men," is false. "All right angles are equal" is true, but "All equal angles are right angles" is false. 136. Def. One theorem is the opposite of another when the hypothesis of the first is the contradiction of the hypothesis of the second, and the conclusion of the first is the contradic- tion of the conclusion of the second. The opposite of a truth is not always true ; thus, " If a man lives in the city of New York, he lives in New York State" is true, but the opposite, " If a man does not live in the city of New York, he does not live in New York State," is false. 137. Note. If the converse of a proposition is true, the opposite also is true ; so, too, if the opposite of a proposition is true, the converse also is true. This may be evident to the student after a consideration of the fol- lowing type forms : (1) DIRECT (2) CONVERSE (3) OPPOSITE If A is B, If C is D, If A is not B, Then C is D. Then A is B. Then C is not D. If (2) is true, then (3) must be true. Again, if (3) is true, then (2) must be true. 138. A necessary and sufficient test of the completeness of a definition is that its converse shall also be true. Hence a definition may be quoted as the reason for a converse or for an opposite as well as for a direct statement in an argument. Ex. 118. State the converse of the definition for equal figures ; straight line ; plane surface. Ex. 119. State the converse of: If one straight line meets another straight line, the sum of the two adjacent angles is two right angles. Ex. 120. State the converse and opposite of Prop. IX. Ex. 121. State the converse of Prop. I. Is it true? 46 PLANE GEOMETRY PROPOSITION X. THEOREM (Converse of Prop. IX) 139. Every point equidistant from the ends of a line lies in the perpendicular bisector of that line. Given line US, and point Q such that QR = QS. To prove that Q lies in the _L bisector of RS. ARGUMENT 1. Let QT bisect Z RQS. 2. QR = QS. 3. .'.A RQS is isosceles. 4. .'. QT is the _L bisector of RS. 5. .'. Q lies in the _1_ bisector Of RS. Q.E.D. REASONS 1. Every Z. has but one bi- sector. 53. 2. By hyp. 3. A A having two sides equal is an isosceles A. 94. 4. The bisector of the Z at the vertex of an isosceles A is to the base and bisects it. 112. 5. By proof. 140. Cor. I. Every point not in the perpendicular bi- sector of a line is not equidistant from the ends of the line. HINT. Use 137, or contradict the conclusion and tell why the con- tradiction is false. BOOK I 47 141. Cor. II. The locus of all points equidistant from the ends of a given line is the perpendicular bisector of that line. HINT. See 143 and 144. 142. Cor. III. Two points each equidistant from the ends of a line determine the perpendicular bisector of tlie line. HINT. Use 139 and 25. 143. In order to prove that a locus problem is solved it is necessary and sufficient to show two things: (1) That every point in the proposed locus satisfies the prescribed conditions. (2) That every point outside of the proposed locus does not satisfy the prescribed conditions. Instead of proving (2), it may frequently be more convenient to prove : (2') That very point which satisfies the prescribed con- ditions lies in the proposed locus. 144. Note. In exercises in which the student is asked to "Find a locus," it must be understood that he has not found a locus until he has given a proof with regard to it as outlined above. The proof must be based upon a direct proposition and its opposite ; or, upon a direct propo- sition and its converse, Ex. 122. Find the locus of all points equidistant from two given points A and B. . Ex. 123. In a given unlimited line AB, find a point equidistant from two given points C and D not on this line. Ex. 124. Given a circle with center O, also a point P. Find all points which lie in the circumference of circle 0, and which are also (a) two inches from P ; (6) a distance of d from P. Ex. 125. Find all points at a distance of d from a given point P, and at the same time at a distance of m from a given point Q. Ex. 126. Given a circle with radius r. Find the locus of the mid- ^ points of the radii of the circle. Ex. 127. Given two circles having the same center. State, without proof, the locus of a point equidistant from their circumferences. JEx. 128. The perpendicular bisector of the base of an isosceles tri- mgle passes through the vertex. 48 PLANE GEOMETRY PROPOSITION XL PROBLEM 145. To construct the perpendicular bisector of a given straight line. \ /* \ / \ 1 \ A \ B A B i K\ > \ (C \ i t \ / \ \ / \ V Fi a. 1. FIG . 2. Given line AB (Fig. 1). To construct the perpendicular bisector of AB. I. Construction 1. With A as center, and with a convenient radius greater than half AB, describe the arc EF. 2. With B as center, and with the same radius, describe the arc HK. 3. Let C and D be the points of intersection of these two arcs. 4. Connect points C and D. 5. CD is the J_ bisector of AB. II. The proof and discussion are left as an exercise for the student. HINT. Apply 142. 146. Note. The construction given in Fig. 2 may be used when the position of the given line makes it more convenient. 147. Question. Is it necessary that GA shall equal CB? that CA shall equal DA (Fig. 1) ? Give the equations that must hold. BOOK I 40 PROPOSITION XII. PROBLEM 148. To construct a perpendicular to a given straight line at a given point in the line. > < 1 \ A D\ C IE B Given line AB and point C in the line. To construct a _L to AB at C- I. Construction 1. With C as center, and with any convenient radius, draw arcs cutting AB on each side of C7, as at D and E. 2. Then with D and E as centers, and with a longer radius, draw two arcs intersecting each other at F. 3. Draw Fa 4. FC is _L to AB at C. II. The proof and discussion are left as an exercise for the student. Ex. 129. Construct the perpendicular bisector of a line given at the bottom of a page or of a blackboard. Ex. 130. Divide a given line into four equal parts. Ex. 131. Construct the three medians of a triangle. Ex. 132. Construct a perpendicular to a line at a given point when the given point is one end of the line. (HINT. Prolong the line.) Ex. 133. Construct a right triangle, given the two arms a and />. ^*Ex. 134. Construct a right triangle, given the hypotenuse and an arm. Ex. 135. Construct the complement of a given angle. Ex. 136. Construct an angle of 45; of 135. Ex. 137. Construct a quadrilateral, given four sides and the angle , between two of them. PLANE GEOMETRY PROPOSITION XIII. PROBLEM 149. From, a point outside a line to construct a per- pendicular to the line. \C di Given line AB and point P outside of AB. To construct a _L from P to AB. I. Construction 1. With P as center, and with a radius of sufficient length, describe an arc cutting AB at points C and D. 2. With C and D as centers, and with any convenient radius, describe arcs intersecting at Q. 3. Draw PQ. 4. PQ is a _L from P to AB. II. The proof is left as an exercise for the student. The discussion will be given in 154. 150. Question. Must P and Q be on opposite sides of AB ? Is it, necessary that PC=QC? Ex. 138. Construct the three altitudes of an acute triangle. Do they seem to meet ? where ? Ex. 139. Construct the three altitudes of a right triangle. Where do they seem to meet ? Ex. 140. Construct the three altitudes of an obtuse triangle. Where do they seem to meet ? Ex. 141. Construct a triangle ABO, given two sides and the median drawn to one of them. Abbreviate thus : given a, />, and m a . BOOK I 51 151. Analysis of a problem of construction. In the more difficult problems of construction a course of reasoning is some- times necessary to enable the student to discover the process of drawing the required figure. This course of reasoning is called the analysis of the problem. It is illustrated in 152 and is more fully treated in the exercises following 274. 152. Note. In such problems as Ex. 141, it is well first to imagine the problem solved and to sketch a figure to represent the desired con- struction. Then mark (with colored crayon, if convenient) the parts supposed to be given. By studying carefully the relation of the given parts to the whole figure^ try to find some part of the figure that you can construct. This will generally be a triangle. After this part is constructed it is usually an easy matter to complete the required figure. Thus : Problem. Let it be required to con- struct an isosceles triangle, given an arm and the altitude upon it. By studying the figure with the given parts marked (heavy or with colored crayon), it will be seen that the solution of the ____^___ J problem depends in this case upon the construe- A C tion of a right triangle, given the hypotenuse and one arm. The right triangle ABD may now be constructed, and it will be readily seen that to complete the construction it is only necessary to prolong BD to (7, making BC = AB, and to connect A and C. Ex. 142. Construct a triangle .ABC, given two sides and an altitude to one of the given sides. Abbreviate thus : given a, b, and h b . Ex. 143. Construct a triangle ABC, given a side, an adjacent angle, and the altitude to the side opposite the given angle. Abbreviate thus : given a, B, and h(>. Ex. 144. Construct an isosceles triangle, given an arm and the me- dian to it. Ex. 145. Construct an isosceles triangle, given an arm and the angle which the median to it makes with it. Ex. 146. ' Construct a triangle, given two sides and the angle which a median to one side makes : (a) with that side ; (b) with the other side. Ex. 147. Construct a triangle, given a side, an adjacent angle, and its bisector. 52 PLANE GEOMETRY PROPOSITION XIV. THEOREM 153. If one side of a triangle is prolonged, the exterior angle formed is greater than either of the remote in- terior angles. B F \ D Given A ABC with AC prolonged to 7), making exterior Z DCB. To prove Z DCB > Z ABC or Z CAB. ARGUMENT 1. Let E be the mid-point of BC; draw AE, and pro- long it to F, making EF = AE. Draw CF. 2. In A ^ and EFC, BE = EC. 3. AE = EF. 4. Z ASM ='Z 5. .' . A ABE = A EFC. 6. REASONS 1. A str. line may be drawn from any one point to any other. 54, 15. 2. mid- By cons. E is the point of BC. 3. By cons. 4. If two str. lines inter- sect, the vertical A are equal. 77. 5. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. 107. 6. Homol. parts of equal fig- ures are equal. 110. BOOK I 53 ARGUMENT Z DOB > Z FCE. 9. Likewise, if BG is pro- longed to G, /iCAB. 10. But Z Z><7 = Z 11. .'. Z ZX75 > Z 12. REASONS 7. The whole > any of its parts. 54, 12. 8. Substituting Z B fqr its equal Z ,F<7J. 9. By bisecting line AC, and by steps similar to 1-8. 10. Same reason as 4. 11. Substituting its equal Z ACG. 12. By proof. for > /.ABC or Z CAB. Q.E.D 154. Cor. From a point outside a line there exists only one perpendicular to the line. HINT. If there exists a second JL to AB from A c K " P, as PC, then Z PGA and /. PKA are both rt. A and are therefore equal. But this is impossible by 153 155. 149 and 154 may be combined in one statement as follows : From a point outside a line there exists one and only one per- pendicular to the line. Ex. 148. A triangle cannot contain two right angles. Ex. 149. In the figure of Prop. XIV, is angle DGB necessarily greater than angle BCA ? than angle B ? 4 Ex. 150. In Fig. 1, prove that : (1) Angle 1 is greater than angle GAE or angle AEG ; (2) Angle 5 is greater than angle CBA or angle BA E ; (3) Angle EDA is greater than angle 3 ; (4) Angle 4 is greater than angle DAE. Ex. 151. In Fig. 2, show that angle 6 is greater than angle 7 ; also that angle 9 is greater than angle A, FIG. 2. C D E FIG. 1. A D B 54 PLANE GEOMETRY PROPOSITION XV. THEOREM 156. If two sides of a triangle are unequal, the anglv opposite tJie greater side is greater than the angle opposite the less side. Given A ABC with BG > BA To prove Z CAB > Z C. ARGUMENT 1. On BC lay off BD = AB. 2. Draw AD. 3. ThenZl = Z2. 4. Now Z2 > Z<7. REASONS 1. Circle post. 122, 157. 2. Str. line post. I. 54, 15. 3. The base A of an isosceles A are equal. 111. 4. If one side of a A is pro- longed, the ext. Z formed > either of the remote int. Zs. 153. 5. Substituting Z 1 for its equal Z 2. 6. The whole > any of its parts. 54, 12. 7. If three magnitudes of the same kind are so related that the first > the sec- ond and the second > the third, then the first Q.E.D. > the third. 54, 10. 157. Note. Hereafter the student will not be required to state postulates and definitions in full unless requested to do so by the teacher. 5. .-.Zl > Z(7. 6. But Z CAB > Z 1. 7. /. Zc^ > Z (7. BOOK I 55 158. Note. When two magnitudes are given unequal, the laying off of the less upon the greater will often serve as the initial step in de- veloping a proof. >^ Ex. 152. Given the isosceles triangle RST, with 8T the base and RT ^prolonged any length, as to K. Prove angle KSll greater than angle K. ^ Ex. 153. If two adjacent sides of a quadrilateral are greater respec- ^tively than the other two sides, the angle included between the two shorter sides is greater than the angle between the two greater sides. ^ Ex. 154. If from a point within a triangle lines are drawn to the ends 'of one of its sides, the angle between these lines is greater than the angle between the other two sides of the triangle. (See Ex. 151.) 159. The indirect method, or proof by exclusion, consists in contradicting the conclusion of a proposition, then showing the contradiction to be false. The conclusion of the proposition is thus established. This process requires an examination of every, possible contradiction of the conclusion. For example, to prove indirectly that A equals B it would be necessary to consider the only three suppositions that are admissible in this case, viz. : (1) ^ > B, (2) A < B, (3) A = B. By proving (1) and (2) false, the truth of (3) is established, i.e. A = 3. This method of reasoning is called reductio ad absurdum. It enables us to establish a conclusion by showing that every contradiction of it leads to an absurdity. Props. XVI and XA r II will be proved by the indirect method. 160. Question. Would it be possible to base a proof upon a contra- diction of the hypothesis ? 161. (a) In the use of the indirect method the student should give, as argument 1, all the suppositions of which the case he is considering admits, including the conclusion. As reason 1 the number of such possi- ble suppositions should be cited. (&) As a reason for the last step in the argument he should state which of these suppositions have been proved false. 56 PLANE GEOMETRY PROPOSITION XVI. THEOREM (Converse of Prop. IV) 162. If two angles of a triangle are equal, the sides opposite are equal. Given A RST with Z R = Z T. To prove r = t. ARGUMENT 1. r >t, r t ; then /.R> Z.T. 3. This is impossible. 4. Next suppose r < t] then Z.R the Z opposite the less side. 156. 3. By hyp., ^R = Z.T. 4. Same reason as 2. 5. Same reason as 3. 6. The two suppositions, r>t and r < t, have been proved false. 163. Cor. An equiangular triangle is also equilateral. Ex. 155. The bisectors of the base angles of an isosceles triangle form an isosceles triangle. BOOK I PROPOSITION XVII. THEOREM 57 (Converse of Prop. XV) 164. If two angles of a triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. Given A AB C with Z A > Z C. To prove a > c. ARGUMENT 1. a < c, a = c, or a > c. 2. First suppose a < c ; then Z.A c. Q.E.D. REASONS 1. In this case only three sup- positions are admissible. 2. If two sides of a A are un- equal, the Z. opposite the greater side > the Z opposite the les>-side. 156. 3. By hyp., /.A > Zc. 4. The base A of an isosceles A are equal. 111. 5. Same reason as 3. 6. The two suppositions, a BC. Let bisectors of angles A and B meet at O. Prove AO > BO. Ex. 159. A line drawn from the vertex of an isosceles triangle to any point in the base is less than one of the equal sides of the triangle. '' Ex. 160. If ABC and ABD are two triangles on the same base and on the same side of it such that AC = BD and AD = BO, and if AC and BD intersect at 0, prove triangle AOB isosceles. Ex. 161. Prove Prop. XVI by using the figure and method of Ex. 69. ' Ex. 162. Upon a given base is constructed a triangle one of the base angles of which is double the other. The bisector of the larger base angle meets the opposite side at the point P. Find the locus of P. Ex. 163. If the four sides of a quadrilateral are equal, its diagonals bisect each other. Ex. 164. The diagonals of an equilateral quadrilateral are perpen- dicular to each other, and they bisect the angles of the quadrilateral. Ex. 165. If two adjacent sides of a quadrilateral are equal and the other two sides are equal, one diagonal is the perpendicular bisector of the other. Tell which one is the bisector and prove the correctness of your answer. Ex. 166. If, from a point in a perpendicular to a line, oblique lines are drawn cutting off equal segments from the foot of the perpendicular, the oblique lines are equal. Ex. 167. State and prove the converse of Ex. 166. Ex. 168. If, from a point in a perpendicular to a line, oblique lines are drawn cutting off unequal segments from the foot of the perpendicular, the oblique lines are unequal. Prove by laying off the less segment upon the greater. Then use Ex. 166, 153, and 164. Ex. 169. If, from any point in a perpendicular to a line, two unequal oblique lines are drawn to the line, the oblique lines will cut off unequal segments from the foot of the perpendicular. Prove by the indirect method. Ex. 170. By means of Prop. XIV, prove that the sum of any' two angles of a triangle is less than two right angles. Ex. 171. Construct a triangle ABC, given two sides, a and 6, and the altitude to the third side, c. (See 152.) BOOK I 59 PROPOSITION XVIII. THEOREM 167. The sum of any two sides of a triangle is greater than the third side. A b Given A ABC. To prove a -f c > b- B (. ARGUMENT 1. Prolong c through B until prolongation BD = a. 2. Draw CD. 3. In isosceles A BDC, /.D = Z/)C'#. 4. But ZZ)CJ > ^DCB. 5. .-. ZDCM > ZD. 6. /. in A ADC, AD > 6. 7. .-. a -f c > b. Q.E.D. REASONS 1. Str. line post. II. 54, 16. 2. Str. line post. I. 54, 15. 3. The base A of an isosceles A are equal. 111. 4.' The whole > any of its parts. 54, 12. 5. Substituting /-D for its equal, /.DCS. 6. If two A of a A are unequal, the side opposite the greater Z is > the side opposite the less angle. 164. 7. Substituting a -f- c for AD. 168. Cor. I. Any side of a triangle is less than the sum (did greater than the difference of the other two. 169. Cor. II. Any straight line is less than the sum, of the parts of a broken line having the same extremities. 60 PLANE GEOMETRY 170. Note to Teacher. Teachers who prefer to assume that " a straight line is the shortest line between two points " may omit Prop. XVIII entirely. Then Prop. XVII may be proved by a method similar to that used in Prop. XV. (See Ex. 172.) Ex. 172. Prove Prop. XVII by using the hint contained in 158. Ex. 173. If two sides of a triangle are 14 and 9, between what limit- ing values must the third side be ? Ex. 174. If the opposite ends of any two non-intersecting line seg- ments are joined, the sum of the joining lines is greater than the sum of the other two lines. Ex. 175. Given two points, P and J?, and a line AB not passing through either. To find a point O, on AB, such that PO + OH shall be as small as possible. This exercise illustrates the law by which light is reflected from a mirror. The light from the object, P, is reflected and appears to come L from L, as far behind the mirror as P is in front of it. Ex. 176. If from any point within a triangle lines are drawn to the extremities of any side of the triangle, the sum of these lines is less than the sum of the other two sides of the triangle. HINT. Let ABC be the given triangle, D the point within. Prolong AD until it intersects BC at E. Apply Prop. XVIII. 171. Note to Teacher. Up to this point all proofs given have been complete, including argument and reasons. In written .work it is frequently convenient, however, to have students give the argument only. These two forms will be distinguished by calling the former a complete demonstration and the latter, which is illustrated in Prop. XIX, argument only. It is often a sufficient test of a student's understanding of a theorem to have him state merely the main points involved in a proof. This may be given in enumerated steps, as in Prop. XXXV, or in the form of a paragraph, as in Prop. XLIV. This form will be called outline of proof. BOOK I 61 PROPOSITION XIX. THEOREM 172. If two triangles have two sides of one equal respec- tively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. F /* Given two A ABC and DEF with AB = DE, BC = .RF, but ARGUMENT ONLY To prove AC > DF. 1. Place A D^F on A ABC so that DE shall fall upon its equal AB, D upon ^4, E upon 5. 2. J:F will then fall between AB and BC. Denote A DEF in its new position by ABF. 3. Draw BK bisecting /. FBC and meeting AC at JT. 4. Draw .O*. 5. In A J^JT and -fiTtfC, 5(7 = BF. 6. BKBK. 7. Z.FBK/.KBC. 8. .-. AFBK=AKBC. 9. .'.KF=KC. 10. ^JT+ KF>AF. 11. .-. ^^4-^(7>^^. 12. That is, ^C>^^. 13. .-. AC>DF. Q.E.D. 62 PLANE GEOMETRY Ex. 177. (a) Draw a figure and discuss the case for Prop. XIX when F falls on AC; when F falls within triangle ABC. (6) Discuss Prop. XIX, taking AC (the base) = DF, AB = DE, and angle CAB greater than angle D. Ex. 178. Prove Prop. XIX by using Fig. 1. HINT . In A A CE, Z CEA > Z BE A. :. Z CEA > Z EAB > Z EA C. FIG. 1. Ex. 179. Given triangle ABC with AB greater than BC, and let point P be taken on A B and point Q on DF. To prove /-B > ZlE. BOOK I 63 ARGUMENT 1. /-B Z#. 2. First suppose Z B < Z tf ; then ACZ.E. Q.E.D. REASONS 1. In this case only three sup- positions are admissible. 2. If two A have two sides of one equal respectively to two sides of the other, but the included Z of the first > the included Z of the second, then the third side of the first > the third side of the second. 172. 3. By hyp., AC>DF. 4. Two A are equal if twa sides and the included Z of one are equal respec- tively to two sides and the included Z of the other. 107. 5. Homol. parts of equal fig- ures are equal. 110. 6. Same reason as 3. 7. The two suppositions, Z,B < Z.E and Z = ZJE:, have been proved false. Ex. 182. If two opposite sides of a quadrilateral are equal, but its diagonals are unequal, then one angle opposite the greater diagonal is greater than one angle opposite the less diagonal. Ex. 183. If two sides of a triangle are unequal, the median drawn to the third side makes unequal angles with the third side. Ex. 184. If from the vertex 8 of an isosceles triangle RST a line is drawn to point P in the base BT so that HP is greater than P7 1 , then angle BSP is greater than angle PST. Ex. 185. If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles triangles. 64 PLANE GEOMETRY SUMMARY OF THEOREMS FOR PROVING ANGLES UNEQUAL 174. (a) When the angles are in the same triangle : If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. (&) When the angles are in different triangles: If two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. (c) An exterior angle of a triangle is greater than either re- mote interior angle. SUMMARY OF THEOREMS FOR PROVING LINES UNEQUAL 175. (a) When the lines are in the same triangle : The sum of any two sides of a triangle is greater than the third side. Any side of a triangle is less than the sum and greater than the difference of the other two. If two angles of a triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. (b) When the lines are in different triangles : If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. (c) Every point not in the perpendicular bisector of a line is not equidistant from the ends of the line. The perpendicular is the shortest straight line from a point to a line. Any straight line is less than the sum of the parts of a broken line having the same extremities. In some texts Exs. 168 and 169 are given as theorems. BOOK I 65 176. Note. The student should note that the proof of Prop. XIX depends upon the device of substituting a broken line for an equal straight line. This method is further illustrated by the following exercises. B E Ex. 186. Prove by the method discussed above that lines drawn to the ends of a line from any point not in its perpendicular bisector are unequal. Ex. 187. Construct an isosceles triangle, given the base and the sum of the altitude and a side. SOLUTION. Let b be the required base and let a 4- s be the sum of the altitude and a side. Imagine the problem solved, giving A ABC. By marking the given lines and studying the figure, the following procedure will be found to be possible : On an unlimited line CE lay off CB = b ; next draw the _L bisector of CB and upon it lay off FD = a + s. It is now necessary to break off a part of a + s to form AB. The fact that AB must equal AD should suggest an isosceles A of which BD will be the base and of which it is required to find the vertex. But the _L bisector of the base of an isosceles A passes through the vertex. Therefore, the solution is completed by drawing the JL bisector of BD, which determines A, and by drawing AB and AC. Ex. 188. Construct a triangle, given the base, an adjacent angle, and the sum of the other two sides. Ex. 189. Construct a right triangle, given one arm and the sum of the hypotenuse and the other arm. a + s PARALLEL LINES 177. Def. Two lines are parallel if they lie in the same plane and do not meet however far they are prolonged either way. 178. Assumption 19. Parallel line postulate. Two intersect- ing straight lines cannot both be parallel to the same straight line. 179. The following form of this postulate is sometimes more convenient to quote : Tlirough a given point there exists only one line parallel to a given line. 66 PLANE GEOMETRY PROPOSITION XXI. THEOREM 180. If two straight lines are parallel to a third straight line, they are parallel to each other. Given lines a and b, each II c. To prove a II b. ARGUMENT 1. a and b are either II or not II. 2. Suppose that a is not II b ; then they will meet at some point as D. 3 This is impossible. 4. .-. a II b. Q.E.D. REASONS 1. In this case only two sup- positions are admissible. 2. By clef, of II lines. 177. 3. Parallel line post. 178. 4. The supposition that a and b are not II has been proved false. 181. Def. A transversal is a line that intersects two or more other lines. 182. Defs. If two straight lines are cut by a transversal, of the eight angles formed, 3, 4, 5, 6 are interior angles ; 1, 2, 7, 8 are exterior angles ; 4 and 5, 3 and 6, are alternate interior angles ; 1 and 8, 2 and 7, are alternate exterior angles ; 1 and 5, 3 and 7, 2 and 6, 4 and 8, are corresponding angles (called also exterior interior angles). 5\6 V BOOK I 67 PROPOSITION XXII. THEOREM 183. If two straight lines are cut by a transversal making a pair of alternate interior angles equal, the lines are parallel. Given two str. lines MN and OQ cut by the transversal AB in points C and D } making Z. MCD = Z.QDC. To prove MN II OQ. ARGUMENT 1. or MN and OQ are either not II. 2. Suppose that MN is not II OQ ; then they will meet at some point as P, forming, with line DC, A PDC. 3. Then /.MCD > /_ QDC. 4. This is impossible. 5. .-. MN II OQ. Q.E.D. REASONS 1. In this case only two sup- positions are admissible. 2. By def. of II lines. 177. 3. If one side of a A is pro- longed, the ext. /- formed > either of the remote int. A. 153. 4. Z.MCD = /.QDC, by hyp. 5. The supposition that MN and OQ are not II has been proved false. 184. Cor. I. If two straight lines are cut by a trans- versal making a pair of corresponding angles equal, the lines are parallel. HINT. Prove a pair of alt. int. A equal, and apply the theorem. 68 PLANE GEOMETRY 185. Cor. n. If two straight lines are cut by a trans- versal making a pair of alternate exterior angles equal, the lines are parallel. (HINT. Prove a pair of alt. int. A equal.) 186. Cor. HI. If two straight lines are cut by a trans- versal making the suin of the two interior angles on the same side of the transversal equal to two right angles, the lines are parallel. Given lines m and n cut by the transversal t making Zl + Z2 = 2rt.Zs. To prove m || n. m L I i 2. 3. 4. 5. ARGUMENT = 2rt, A = 2rt.A Z2 = Z3. m n. Q.E.D. REASONS 1. By hyp. 2. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. Zs. 65. 3. Things equal to the same thing are equal to each other. 54, 1. 4. If equals are subtracted from equals, the remain- ders are equal. 54, 3. 5. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are II. 183. 187. Cor. IV. If two straight lines are perpendicular to a third straight line, they are parallel to each other. Ex. 190. If two straight lines are cut by a transversal making the sum of the two exterior angles on the same side of the transversal equal to two right angles, the lines are parallel. BOOK I 69 Ex. 191. In the annexed diagram, if angle Prove. = angle CHF, are AS and CD parallel? V * Ex. 192. In the same diagram, if angle HGB = angle GHC, prove that the bisectors of these ^\r angles are parallel. Ex. 193. If two straight lines bisect each other, the lines joining their extremities are parallel in pairs. Ex. 194. In the diagram for Ex. 191, if angle BGE and angle FHD are supplementary, prove AB parallel to CD. Ex. 195. If two adjacent angles of any quadrilateral are supplemen- tary, two sides of the quadrilateral will be parallel. PROPOSITION XXIII. PROBLEM 188. Through a given point to construct a line parallel to a given line. 8 S , */v r . ./ J /R K Given line AB and point P. To construct, through point P, a line II AB. I. Construction 1. Draw a line through P cutting AB at some point, as R. 2. With P as vertex and PS as side, construct Z YPS = Z BRP. 125. 3. XY will be II AB. II. The proof and discussion are left as an exercise for the student. Ex. 196. Through a given point construct a parallel to a given line by using : (a) 183 ; (b) 185 ; (c) 187, 70 PLANE GEOMETRY PROPOSITION XXIV. THEOREM (Converse of Prop. XXII) 189. If two parallel lines are cut ~by a transversal, the alternate interior angles are equal. tE -Y -B X c Given || lines AB and CD cut by the transversal EF at points and H. To prove Z A GH = Z DHG. ARGUMENT 1. Either Z A Gil = Z DHG, or Z.AGH=f= /.DHG. 2. Suppose Z. 4 GH=^ZDHG, but that line XF, through G, makes ZXGH = Z.DHG. 3. Then .YF II C7Z>. 4. But A B || CD. 5. It is impossible that AB and ^r both are || CD. 6. .'. Z AGH= /.DHG. Q.K.D. REASONS 1. In this case only two sup- positions are admissible. 2. With a given vertex and a given side, an Z may be constructed equal to a given Z. 125. 3. If two str. lines are cut by a transversal making a pair of alt. int. Z equal, the lines are || . 183. 4. By hyp. 5. Parallel line post. 178. 6. The supposition that Z A GH = Z DHG has been proved false. BOOK I 71 190. Cor. I. (Converse of Cor. I of Prop. XXII). // two parallel lines are cut by a transversal, the corresponding angles are equal. Given two || lines XY and MN % cut by the transversal AB, form- ing corresponding A 1 and 2. ^ \a N To prove Z 1 = Z2. \ MINT. /. 3 = Z. 2 by 189. \fl 191. Cor. II. (Converse of Cor. II of Prop. XXII). If two parallel lines are cut by a transversal, the alter- nate exterior angles are equal. 192. Cor. m. (Converse of Cor. Ill of Prop. XXII). If two parallel lines are cut by a transversal, the sum of the two interior angles on the same side of the trans- versal is two right angles. * Given two II lines XY and MN X cut by the transversal AB, form- ing int. A 1 and 2. Af ^__ To prove Zl+Z2 = 2rt. A. / B/ ARGUMENT ONLY Z3 = Z2. : 1 -f Z 2 = 2 rt. A. Q.E.D. 193. Cor. IV. A straight line perpendicular to one of two parallels is perpendicular to the other also. 194. Cor. V. (Opposite of Cor. Ill of Prop. XXII) ; If two straight lines are cut by a transversal making the sum of the two interior angles on the same side of the transversal not equal to two right angles, the lines are not parallel. (HINT. Apply 137, or use the indirect method.) 72 PLANE GEOMETRY 195. Cor. VI. Two lines perpendicular respectively to two intersecting lines also intersect. Given two intersecting lines AB and CD, and BE J, AB, GF _L CD. To prove that BE and CF also intersect. ARGUMENT ONLY 1. Draw CB. 2. /.FCD is a rt. Z. 3. .-. Zl < a rt. Z. 4. Likewise Z2 < a rt. Z. 5. .-. Z1 + Z2 < 2rt. A. 6. .-. BE and CF also intersect. Q.E.D. 196. Def. Two or more lines are said to be concurrent if they intersect at a common point. Ex. 197. If two parallels are cut by a transversal so that one of the angles formed is 45, how many degrees are there in each of the other seven angles ? Ex. 198. . If a quadrilateral has two of its sides parallel, two pairs of its angles will be supplementary. .-, Ex. 199. In the annexed diagram AB is \ \ parallel to CD, and EF is parallel to GH. A V A^- B Prove c \2 \ ft (a) angle 1 = angle 2 ; \ >^ (&) angle 1 + angle 3 = 2 right angles. F H Ex. 200. If a line is drawn through any point in the bisector of an angle, parallel to one of the sides of the angle, an isosceles triangle will be formed. Ex. 201. Draw a line parallel to the base of a triangle, cutting the sides so that the sum of the segments adjacent to the base shall equal the parallel line. Ex. 202. If two parallel lines are cut by a transversal, the bisectors of a pair of corresponding angles are parallel. BOOK I 73 Ex. 203. State and prove the converse of Ex. 190. Ex. 204. State and prove the converse of Ex. 202. Ex. 205. If a line is drawn through the vertex of an isosceles triangle parallel to the base, this Hue bisects the exterior angle at the vertex. Ex. 206. State and prove the converse of Ex. 205. Ex. 207. If a line joining two parallels is bisected, any other line through the point of bisection and limited by the parallels is bisected. Ex. 208. If through the vertex of one of the acute angles of a right triangle a line is drawn parallel to the opposite side, the line forms with the hypotenuse an angle equal to the other acute angle of the triangle. Ex. 209. The acute angles of a right triangle are complementary. Ex. 210. If two sides of a triangle are prolonged their own lengths through the common vertex, the line joining their ends is parallel to the third side of the triangle. Ex. 211. In an isosceles triangle, if equal segments measured from the vertex are laid off on the arms, the line joining the ends of the seg- ments is parallel to the base of the triangle. (HINT. Draw the bisector of the vertex Z of the A.) Ex. 212. Extend Ex. 211 to the case where the segments are external to the triangle. SUMMARY OF THEOREMS FOR PROVING LINES PARALLEL alternate interior angles equal, 197. (1) If two straight lines are cut by a trans- versal making the alternate exterior angles equal, corresponding angles equal, interior angles on one side of the transversal supplementary, the lines are parallel. (2) Two straight lines perpendicular to a third straight line are parallel to each other. (3) Two straight lines parallel to a third straight line are parallel to each other. (4) After parallelograms have been studied, the fact that the opposite sides of a parallelogram are parallel may be used ( 220). 74 PLANE GEOMETRY PROPOSITION XXV. THEOREM 198. Two angles whose sides are parallel, each to each f are either equal or supplementary. Given Zl and the A at E, with AB \\ EF and AC || DE. To prove Z 1 == Z 2 = Z3, and Z 1 + Z 4 = 2 rt. 4 Zl 5 = 2rt.Zs. ..ARGUMENT 1. Prolong .4 and DE until they intersect at some point as H. 3. Z2 = Z6. 4. .-. Z1 = Z2. 5. Z3=Z2. 6 .-. Z1='Z3. 7. Z2 Z4 = 8. .-. Z 1 + Z 4 = 2 rt. A REASONS 1. Str.linepost.il. 54,16. 2. Corresponding A of || lines are equal. 190. 3. Same reason as 2. 4. Things equal to the same thing are equal to each other. 54, 1. 5. If two str. lines intersect, the vertical A are equal. 77. 6. Same reason as 4. 7. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. A. 65. 8. Substituting Z 1 for its equal Z 2. BOOK 1 75 ARGUMENT 9. Z5=Z4. 10. .-. Zl + Z5= 2rt. A Q.E.D. REASONS 9. Same reason as 5. 10. Substituting Z5 for equal Z4. its 199. Note. Every angle viewed from its vertex has a right and a left side ; thus, in the annexed diagram, the right side of Z.A is r and the left side, I ; the right side of ZB is r and the left, I. FIG. 3. In the diagram of Prop. XXV, Zl and 22, whose sides are || right to right (ABto EF) and left to left (AC to EK) are equal ; while 2 1 and 24, whose sides are || right to left (AB to EF) and left to right (AC to ED) are supplementary. Hence : 200. (a) If two angles have their sides parallel right to right and left to left, they are equal. (b) If two angles have their sides parallel right to left and left to right, they are supplementary. Ex. 213. In Fig. 3, above, show which is the right side of each of the four angles about O. Ex. 214. If a quadrilateral has its opposite sides parallel, its opposite angles are equal. Ex. 215. Given two equal angles having a side of one parallel to a side of the other, are the other sides necessarily parallel ? Prove. Ex. 216. If two sides of a triangle are parallel respectively to two homologous sides of an equal triangle, the third side of the first is parallel to the third side of the second. Ex. 217. Construct a triangle, given an angle and its bisector and the altitude drawn from the vertex of the given angle. 76 PLANE GEOMETRY PROPOSITION XXVI. THEOREM 201. Two angles whose sides are perpendicular, each to each, are either equal or supplementary. H K E C D Given Zl and the A at P, with OAKC and OBj_HD. To prove Z1=Z2 = Z3, Zl+Z4=2rt. Zs,Zl+Z5 = 2 rt. A. HINT. Draw OE II CK and OF II DH. Prove O# J_ OA and OF J_ OB. Prove Z 7 = Z 1, and prove Z 7 = Z 2. 202. Note. It will be seen that Zl and Z2, whose sides are _L right to right (OA to PK} and left to left (OS to P#), are equal ; while Zl and Z4, whose sides are _L right to left (OA to PC) and left to right (OS to PT), are supplementary. Hence; 203. (a) If two angles have their sides perpendicular right to right and left to left, they are equal. (b) If two angles have their sides perpendicular right to left and left to right, they are supplementary. Ex. 218. If from a point outside of an angle perpendiculars are drawn to the sides of the angle, an angle is formed which is equal to the given angle. Ex. 219. In a right triangle if a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the right angle is divided into two angles which are equal respectively to the acute angles of the triangle. Ex. 220. If from the end of the bisector of the vertex angle of an isosceles triangle a perpendicular is dropped upon one of the arms, the perpendicular forms, witti the base an angle equal to half the vertex angle, BOOK I 77 PROPOSITION XXVII. THEOREM 204. The sum of the angles of any triangle is two right angles. Given A ABC. To prove /.A 4- Z ABC + Z C = 2 rt. A ARGUMENT REASONS 1. Through B draw D# II .4C. . 1. Parallel line post. 179. 2. Z 1 4- Z ABC 4- Z 2 2. The sum of all the A about = 2 rt. Z. a point on one side of a str. line passing through that point = 2 rt. A. 66. 3. Z1=Z^. 3. Alt. int. A of II lines are equal. 189. 4. Z 2 = Z C. 4. Same reason as 3. 5. .-. /-A 4- Z^.BC' -f ZC7 5. Substituting for A 1 and 2 their equals, Z A and (7, respectively. 205. Cor. I. In a right triangle the two acute angles are complementary. 206. Cor. II. In a triangle there can be but one right angle or one obtuse angle. 207. Cor. III. If two angles of one triangle are equal respectively to two angles of another, tlien the third angle of the first is equal to the third angle of the second. 208. Cor. IV. // two right triangles have an acute angle of one equal to an acute angle of the other, the other acute angles arc equal. Z^ -f /-ABC 4- ZC7 = 2 rt. A. Q.E.D. 78 PLANE GEOMETRY 209. Cor. V. Two right triangles are equal if the hy- potenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other. 210. Cor. VI. Two right triangles are equal if a side and an acute angle of one are equal respectively to a side and the homologous acute angle of the other. 211. Cor. VII. Two right triangles are equal if the hypotenuse and a side of one are equal respectively to tlie hypotenuse and a side of the other. B K A Given rt. A ABC and DEF, with AB = DE and BC=EF. To prove A ABC = A DEF. HINT. Prove DFKa, str. line ; then A DEK is isosceles. 212. Cor. VIII. The altitude upon the base of an isosceles triangle bisects the base and also the vertex angle. 213. Cor. IX. Each angle of an equilateral triangle is one third of two right angles, or 60. 214. Question. Why is the word homologous used in Cor. VI but not in Cor. V ? Ex. 221. If any angle of an isosceles triangle is 60, what is the value of each of the two remaining angles ? Ex. 222. If the vertex angle of an isosceles triangle is 20, find the angle included by the bisectors of the .base angles. Ex. 223. Find each angle of a triangle if the second angle equals twice the first and the third equals three times the second. Ex. 224. If one angle of a triangle is w and another angle Z, write an expression for the third angle. Ex. 225. If the vertex angle of an isosceles triangle is , write an expression for each base angle. BOOK I 79 Ex. 226. Construct an angle of 60 ; 120 ; 30 ; 15. Ex. 227. Construct a right triangle having one of its acute angles 60. How large is the other acute angle ? Ex. 228. Construct an angle of 150. Ex. 229. above. Ex. 230. Ex. 231. Ex. 232. Trove Prop. XXVII by using each of the diagrams given Given two angles of a triangle, construct the third angle. Find the sum of the angles of a quadrilateral. The angle between the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the two remaining angles. Ex. 233. The angle between the bisectors of the base angles of an isos- celes triangle is equal to the exterior angle formed by prolonging the base. Ex. 234. If two straight lines are cut by a transversal and the bisectors of two interior angles on the same side of the transversal are perpendicular to each other, the lines are parallel. Ex. 235. If in an isosceles triangle each of the base angles is one fourth the angle at the vertex, a line drawn perpendicular to the base at one of its ends and meeting the opposite side prolonged will form with the adjacent side and the exterior portion of the opposite side an equilateral triangle. A^ Ex. 236. Two angles whose sides are perpendicular each to each are either equal or supple- mentary. (Prove by using the annexed diagram.) Ex. 237. If at the ends of the hypotenuse of a right triangle per- pendiculars to the hypotenuse are drawn meeting the other two sides of the triangle prolonged, then the figure contains five triangles which are mutually equiangular. Ex. 238. If one angle of a triangle is 50 and another angle is 70, find the other interior angle of the triangle ; also the exterior angles of the triangle. What relation is there between an exterior angle and the two remote interior angles of the triangle ? B' B B' B 80 PLANE GEOMETRY PROPOSITION XXVIII. THEOREM 215. *dn exterior angle of a triangle is equal to the sum of the two remote interior angles. C Given AAEC with Zz><7 an exterior Z. To prove ZDCB=Z.A + ^B. The proof is left as an exercise for the student. HINT. Draw CE II AB. Ex. 239. The bisector of an exterior angle at the vertex of an isosceles triangle is parallel to the base. Ex. 240. If the sum of two exterior angles of a triangle is equal to three right angles, the triangle is a right triangle. Ex. 241. The sum of the three exterior angles of a triangle is four right angles. Ex. 242. What is the sum of the exterior angles of a quadrilateral ? Ex. 243. If the two exterior angles at the base of any triangle are bisected, the angle between these bisectors is equal to half the sum of the interior base angles of the triangle. Ex. 244. If BE bisects angle B of triangle ABC, and AE bisects the exterior angle DA (7, angle E is equal to one half angle C. Ex. 245. D is any point in the base BC of isosceles triangle ABC. The side B ~A D AC is prolonged through C to E so that CE = CD, and DE is drawn meeting AB at F. Prove angle EFA equal to three times angle AEF. Ex. 246. The mid-point of the hypotenuse of a right triangle is equidistant from the three vertices. Prove by laying off on the right angle either acute angle. BOOK I 81 PROPOSITION XXIX. THEOREM 216. The sum of all the angles of any polygon is twice as many right angles as the polygon has sides, less four right angles. Given polygon ABODE - , any polygon having n sides. To prove the sum of its A = 2 n rt. A 4 rt. A. ARGUMENT 1. From any point within the polygon such as 0, draw lines to the vertices. 2. There will be formed n A. 3. The sum of the A of each A thus formed = 2 rt. A. 4. .% the sum of the A of the n A thus formed = 2 n rt. A. 5. The sum of all the A about = 4 rt. A. 6. .-. the sum of all the A of the polygon = 2 n rt. A - 4 rt. Zs. Q.E.D. REASONS 1. Straight line post I. 15. 54, 2. Each side of the polygon will become the base of a A. 3. The sum of the A of any A is 2 rt, Zs. 204. 4. If equals are multiplied by equals, the products are equal. 54, 7 a. 5. The sum of all the A about a point = 4 rt. A. 67. 6. The sum of all the A of the polygon = the sum of the A of the n A - the sum of all the A about O. 217. Cor. Each angle of an equiangular polygon of n sides is equal to 2(n - 2) right angles. 82 PLANE GEOMETRY PROPOSITION XXX. THEOREM 218. // the sides of any polygon are prolonged in succes- sion one way, no two adjacent sides being prolonged through the same vertex, the sum of the exterior angles thus formed is four right angles. 4/7) Given polygon P with Z 1, Z 2, Z 3, Z4, its successive ex- terior angles. To prove Z 1 + Z 2 -f Z3 + ^4 H = 4 rt. A. ARGUMENT 1. Z1 + Z6 = 2 rt. Zs, Z2 + Z7 = 2 rt. A, and so on ; i.e. the sum of the int. Z and th'e ext. Z at one vertex = 2 rt. A. 2. .*. the sum of the int. and ext. zS at the n vertices = 2 n rt. Z. 3. Denote the sum of all the interior A by / and the sum of all the ext. A by E-, then #-)-/= 2 nrt. A. 4. But l=2wrt. Z-4rt. A 5. .-. J = 4 rt. Z; i.e. Zl + Z2 + Z3+Z4+- ... = 4 rt. A Q.E.D. REASONS 1. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. A. 65. 2. If equals are multiplied by equals, the products are equal. 54, 7 a. 3. Arg. 2. 4. The sum of all the A of any polygon = 2 n rt. A -4rt. Zs. 216. 5. If equals are subtracted from equals, the remain- ders are equal. 54, 3. BOOK I 88 219. Note. The formula 2 n rt. A - 4 rt. A ( 216) is sometimes more useful in the form (;i 2) 2 rt. A. Ex. 247. Find the sum of the angles of a polygon of 7 sides ; of 8 sides ; of 10 sides. Ex. 248. Prove Prop. XXIX by drawing as many diagonals as possible from one vertex. Ex. 249. How many diagonals can be drawn from one vertex in a polygon of 8 sides ? of 50 sides ? of n sides ? Show that the greatest num- ber of diagonals possible in a polygon of n sides (using all vertices) is n(n - 3) 2 Ex. 250. How many degrees are there in each angle of an equi- angular quadrilateral ? in each angle of an equiangular pentagon ? Ex. 251. How many sides has a polygon the sum of whose angles is 11 right angles ? 20 right angles ? 540 ? Ex. 252. How many sides has a polygon the sum of whose interior angles is double the sum of its exterior angles ? Ex. 253. Is it possible for an exterior angle of an equiangular'poly- gon to be 70 ? 72 ? 140 ? 144 ? Ex. 254. How many sides has a polygon each of whose exterior angles equals 12 ? Ex. 255. How many sides has a polygon each of whose exterior angles is one eleventh of its adjacent interior angle ? Ex. 256. How many sides has a polygon the sum of whose interior angles is six times the sum of its exterior angles ? Ex. 257. How many sides has an equiangular polygon if the sum of three of its exterior angles is 180 ? Ex. 258. Tell what equiangular polygons can be put together to make a pavement. How many equiangular triangles must be placed with a common vertex to fill the angular magnitude around a point ? TJL7 84 PLANE GEOMETRY QUADRILATERALS. ' PARALLELOGRAMS QUADRILATERALS CLASSIFIED WITH RESPECT TO PARALLELISM 220. Def. A parallelogram is a quadrilateral whose oppo- site sides are parallel. 221. Def. A trapezoid is a quadrilateral having two of its opposite sides parallel and the other two not parallel. 222. Def. A trapezium is a quadrilateral having no two of its sides parallel. PARALLELOGRAMS CLASSIFIED WITH RESPECT TO ANGLES 223. Def. A rectangle is a parallelogram having one right angle. It is shown later that all the angles of a rectangle are right angles. 224. Def. A rhomboid is a parallelogram having an oblique angte. It is shown later that all the angles of a rhomboid are oblique. 225. Def. A rectangle having two adjacent sides equal is a square. It is shown later that all the sides of a square are equal. 226. Def. A rhomboid having two adjacent sides equal is a rhombus. It is shown later that all the sides of a rhombus are equal. 227. Def. A trapezoid having its two non-parallel sides equal is an isosceles trapezoid. 228. Def. Any side of a parallelogram may be regarded as its base, and the line drawn perpendicular to the base from any point in the opposite side is then the altitude. 229. Def. The bases of a trapezoid are its parallel sides, and its altitude is a line drawn from any point in one base per- pendicular to the other. BOOK I PROPOSITION XXXI. THEOREM 85 230. Any two opposite angles of a parallelogram, are equal, and any two consecutive angles are supplementary. Given O ABCD. To prove : (a) /.A = Z (7, and /.B = /.D ; (b) any two consecutive A, as A and B, sup. ARGUMENT = /. Cand Z.B=Z,D. A A and B are sup. Q.E.D. REASONS 1. If two A have their sides II right to right and left to left, they are equal. 200, a. 2. If two II lines- are cut by a transversal, the sum of the two int. A on the same side of the transversal is two rt. A. 192. 231. Cor. All the angles of a rectangle are right angles, and all the angles of a rhomboid are oblique angles. Ex. 259. . If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. Ex. 260. If an angle of one parallelogram is equal to an angle of another, the remaining angles are equal each to each. Ex. 261. The bisectors of the angles of a parallelogram (not a rhom- bus or a square) inclose a rectangle, 86 PLANE GEOMETRY PROPOSITION XXXII. THEOREM 232. The opposite sides of a parallelogram are equal. Given O ABCD. To prove AB = CD and BC = AD. The proof is left as an exercise for the student. 233. Cor. I. All the sides of a square are equal, and all the sides of a rhombus are equal. 234. Cor. II. Parallel lines intercepted between the same parallel lines are equal. 235. Cor. III. The perpendiculars drawn to one of two parallel lines from any two points in the other are equal. 236. Cor. IV. A diagonal of a parallelogram divides it into two equal triangles. Ex. 262. The perpendiculars drawn to a diagonal of a parallelogram from the opposite vertices are equal. Ex. 263. The diagonals of a rhombus are perpendicular to each other and so are the diagonals of a square. Ex. 264. The diagonals of a rectangle are equal. Ex. 265. The diagonals of a rhomboid are unequal. Ex. 266. If the diagonals of a parallelogram are equal, the figure is a rectangle. Ex. 267. If the diagonals of a parallelogram are not equal, the figure is a rhomboid. Ex. 268. Draw a line parallel to the base of a triangle so that the portion intercepted between the sides may be equal to a given line. Ex. 269. Explain the statement : Parallel lines are everywhere equi- distant. Has this been proved ? BOOK I 87 Ex. 270. Find the locus of a point that is equidistant from two given parallel lines. Ex. 271. Find the locus of a point : (a) one inch above a given horizontal line ; (&) two inches below the given line. Ex. 272. Find the locus of a point: (a) one inch to the right of a given vertical line ; (6) one inch to the left of the given line. Ex. 273. Given a horizontal line OX and a line O Y perpendicular to OX. Find the locus of a point three inches above OJTand two inches to the right of O Y. 237. Historical Note. Rend Descartes (1596-1650) was the first to observe the importance of the fact that the position of a point in a plane is determined if its distances, say x and y, from two fixed lines in the plane, perpendic- ular to each other, are known, lie showed that geometric fig- ures can be represented by algebraic equations, and de- veloped the subject of analytic geometry, which is known by his name as Cartesian geome- try. Descartes was born near Tours in France, and was sent at eight years of age to the famous Jesuit school at La Fleche. He was of good fam- ily, and since, at that time, most men of position entered either the church or the army, he chose the latter, and joined the army of the Prince of Orange. One day, while walking in a street in a Holland town, he saw a placard which challenged every one who read it to solve a certain geometric problem. Descartes solved it with little difficulty and is said to have realized then that he had no taste for military life. He soon resigned his commission and spent five years in travel and study. After this he lived a short time in Paris, but soon retired to Holland, where he lived for twenty years, devoting his time to mathematics, philosophy, astronomy, and physics. His work in philosophy was of such importance as to give him the name of the Father of Modern Philosophy. DESCARTES 88 PLANE GEOMETRY PROPOSITION XXXIII. THEOREM 238. The diagonals of a parallelogram bisect each other. A D Given O ABCD with its diagonals AC and BD intersecting at 0. To prove AO = OC and BO = OD. HINT. Prove A OBC = A ODA. .-. AO = OC and BO = OD. Ex. 274. If through the vertices of a triangle lines are drawn parallel to the opposite sides of the triangle, the lines which join the vertices of the triangle thus formed to the opposite vertices of the given triangle are bisected by the sides of the given triangle. Ex. 275. A line terminated by the sides of a parallelogram and passing through the point of intersection of its diagonals is bisected at that point. Ex. 276. How many parallelograms can be constructed having a given base and altitude ? What is the locus of the point of intersection of the diagonals of all these parallelograms ? Ex. 277. If the diagonals of a parallelogram are perpendicular to each other, the figure is a rhombus or a square. Ex. 278. If the diagonals of a parallelogram bisect the angles of the parallelogram, the figure is a rhombus or a square. Ex. 279. Find on one side of a triangle the point from which straight lines drawn parallel to the other two sides, and terminated by those sides, are equal. (See 232.) Ex. 280. Find the locus of a point at a given distance from a given finite line AB. Ex. 281. Find the locus' of a point at a given distance from a given line and also equidistant from the ends of another given line. Ex. 282. Construct a parallelogram, given a side, a diagonal, and the altitude upon the pven side. BOOK I 89 PROPOSITION XXXI V. THEOREM (Converse of Prop. XXXII) 239. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. D 6. 7. .'. AB II CD. 8. Likewise /.EDA 9. /. BC II AD. 10. .-. ABCD is a O. BD. 1. S 2. B 3. B 4. B X 5. T 6. H 7. IJ 8. S 9. S Q.E.D. 10. B REASONS 1. Str. line post. I. 54, 15. By hyp. Given quadrilateral ABCD, with AB = CD, and BC=AD To prove ^LB CD a O. ARGUMENT 1. Draw the diagonal BD. 2. In A ABD and AB = CD. 3. BC = ^D. 4. D = BD. .-. A ABD = A BCD. 5. Two A are equal if the three sides of one are equal respectively to the three sides of the other. 116. Homol. parts of equal fig- ures are equal. 110. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are II. 183. Same reason as 6. Same reason as 7. By def. of a O. 220. Ex. 283. Construct a parallelogram, given two adjacent sides and the included anirle. 90 PLANE GEOMETRY Ex. 284. Construct a rectangle, given the base and the altitude. Ex. 285. Construct a square, given a side. Ex. 286. Through a given point construct a parallel to a given line by means of Prop. XXXIV. Ex. 287. Construct a median of a triangle by means of a parallelo- gram, (1) using 239 and 238 ; (2) using 220 and 238. Ex. 288. An angle of a triangle is right, acute, or obtuse according as the median drawn from i.ts vertex is equal to, greater than, or less than half the side it bisects. PROPOSITION XXXY. THEOREM 240. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. A D Given quadrilateral ABCD, with BC both equal and II to AD. To prove ABCD a O. OUTLINE OP PROOF 1. Draw diagonal BD. 2. Prove A BCD = A ABD. 3. Then ZCDB = /.ABD and AB II CD. 4. .'. ABCD is a O. Ex. 289. If the mid-points of two opposite sides of a parallelogram are joined to a pair of opposite vertices, a parallelogram will be formed. Ex. 290. Construct a parallelogram, having given a base, an adjacent angle, and the altitude, making your construction depend upon 240. Ex. 291. If the perpendiculars to a line from any two points in an- other line are equal, then the lines are parallel. Ex. 292. If two parallelograms have two vertices and a diagonal in common, the lines joining the other four vertices form a parallelogram. BOOK I 91 PROPOSITION XXXVI. THEOREM (Converse of Prop. XXXIII) 241. If the diagonals of a quadrilateral bisect earh other, the figure is a parallelogram. B C A D Given quadrilateral ABCD with its diagonals AC and BD in- tersecting at so that A0= CO and BO = DO. To prove ABCD a O. ARGUMENT ONLY 1. In A OBC and ODA, BO = DO. 2. GO AO. 3. Z3 = Z4. 4. .-.A OBC= A OD^. 5. .'. BC = AD. 6. Also Zl = Z2. 7. .'. 5(71! AD. 8. .-. ABOD is a O. Q.E.D. Ex. 293. In parallelogram ABCD, let diagonal AC be prolonged through ^4 and C to X and T, respectively, making ^1X= OF. Prove XB YD a parallelogram. Ex. 294. If each half of each diagonal of a parallelogram is bisected, the lines joining the points of bisection form a parallelogram. Ex. 295. Lines drawn from the vertices of two angles of a triangle and terminating in the opposite sides cannot bisect each other. Ex. 296. State four independent hypotheses which would lead to the conclusion, "the quadrilateral is a parallelogram." Ex. 297. Construct a parallelogram, given its diagonals and an angle between them. 92 PLANE GEOMETRY PROPOSITIOX XXXVII. THEOREM 242. Two parallelograms are equal if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II Given UJ I and II with AB EF, AD = EH, and /.A= /.E. To prove O.I = OIL ARGUMENT 1. Place O I upon Oil so that AB shall fall upon its equal EF, A upon E, B upon F. 2. Then AD will become col- linear with EH. 3. Point D will fall on H. 4. Now DC II AB, and #G II EF. 5. .-. DC7 and # are both II AB. 6. .*. DC will become collinear with #(?, and (7 will fall somewhere on HG or its prolongation. 7. Likewise ^(7 will become collinear with FG, and C will fall somewhere on FG or its prolongation. 8. .-. point C must fall on point G. 9. ,-, 71 =.OII, Q.E.J>. REASONS 1. Transference post. 54,14. 2. Z^l = Z.E, by hyp. 3. ^4Z> = EH, by hyp. 4. By def . of a O. 220. 5. AB and EF coincide, Arg. 1. 6. Parallel line post. 179. 7. By steps similar to 4, 5, and 6. 8. Two intersecting str. lines can have only one point in common. 26. 9. By def. of equal figures, 18. BOOK I 93 243. Quadrilaterals QUADRILATERALS CLASSIFIED 1. Opposite sides II : Parallelogram. - (a) Right-angled: Rectangle. Two adj. sides equal: Square. (6) Oblique-angled: Rhomboid. Two adj. sides equal : Rhombus. 2. Two sides II, other two non-11 : Trap- ezoid. (a) Two non-11 sides equal : Isosceles trapezoid. 3. No two sides II : Trapezium. Ex. 298. If it is required to prove a given quadrilateral a rectangle, show by reference to 243 that the logical steps are to prove first that it is a parallelogram ; then that it has one right angle. Ex. 299. If a given quadrilateral is to be proved a square, show that the only additional step after those in Ex. 298 is to prove two adjacent sides equal. Ex. 300. If a given quadrilateral is to be proved a rhombus, what are the three steps corresponding to those given in Ex. 298 and Ex. 299 ? Ex. 301. Since rectangles and rhomboids are parallelograms, they possess all the general properties of parallelograms. What property dif- ferentiates rectangles from rhomboids (1) by definition ? (2) by proof ? (See Ex. 266 and Ex. 267.) Ex. 302. (a) What two properties that have been proved distinguish squares from other rectangles ? (See Ex. 277 and Ex. 278.) (6) What two properties that have been proved distinguish rhombuses from other rhomboids? (See Ex. 277 and Ex. 278.) (c) Show that the two properties which distinguish squares and rhom- buses from the other members of their class are due to the common prop- erty possessed by squares and rhombuses by definition. Ex. 303. The mid-point of the hypotenuse of a right triangle is equi- distant from the three vertices. Ex. 304. If a line AB of given length is moved so that its ends always touch the sides of a given right angle, what is the locus of the mid-point of AB? 94 PLANE GEOMETRY PROPOSITION XXXVIII. THEOREM 244. If three or more parallel lines intercept equal seg- ments on one transversal, they intercept equal segments on any other transversal. Given II lines AG, BH, CJ, DK, etc., which intercept the equal segments AB, BC, CD, etc., on transversal AF, and which intercept segments GH, HJ, JK, etc., on transversal GL. To prove GH = HJ = JK, etc. ARGUMENT 1. Draw GM, HN, JR, etc. II A F. 2. Now AGMB, BHNC, CJRD, etc., are UJ. 3. .'. GM = AB, HN = BC, JR = CD, etc. 4. And AB = BC = CD, etc. 5. .% GM = HN = JR, etc. 6. Again GM, HN, JR, etc., are II to each other. 7. ,-. Zl = Z2 = Z3, etc. 8. And Z4 = Z5 = Z6, etc. REASONS 1. Parallel line post. 179. 2. By def. of a O. 220. 3. The opposite sides of a O are equal. 232. 4. By hyp. 5. Ax. 1. 54, 1. 6. If two str. lines are II to a third str. line, they are II to each other. 180. 7. Corresponding A of II lines are equal. 190. 8. If two A have their sides II right to right and left to left, they are equal. 200, a. BOOK I 9. .-. A GHM A HJN = AJKR, etc. 10. .'. GH = 77,7 = JK, etc. Q.E.D. 9. Two A are equal if a side arid the two adj. A of one are equal respectively to a side and the two adj. A of the other. 105. 10. Homol. parts of equal figures are equal. 110. 245. Question. Are the segments that the parallels intercept on one transversal equal to the segments that they intercept on another trans- versal ? Illustrate. 246. Cor. I. The line bisecting one of the non-paral- lel sides of a trapezoid and parallel to the bases bisects the other of the non-parallel sides also. 247. Cor. II. The line joining the mid-points of the non-parallel sides of a Ex trapezoid is (a} parallel to the bases; and (b} equal to A H D one half their sum. HINT, (a) Prove EF II 7? O and AD by the indirect method. (6) Draw GH II CD. Prove AH = GB ; then prove EF = GC =\(GC + HD} = 248. Cor. m. The line bisect- ing one side of a triangle and parallel to another side bisects tlie third side. 249. Cor. IV. The line joining the F^- mid-points of two Xx \ / sides of a triangle ^c~ ~^/? is parallel to the third side and equal to one half the third side. HINT. Draw CF II BA. Prove CF=AE = EB. 96 PLANE GEOMETRY PROPOSITION XXXIX. PROBLEM 250. To divide a given straight line into any number of equal parts. ,'X K >< P'-S R -'A''' >x , ! \ x x x \ .''' \ I H B A Given straight line AB. To divide AB into n equal parts. I. Construction 1. Draw the unlimited line AX. 2. Take any convenient segment, as AR, and, beginning at A, lay it off n times on AX. 3. Connect the nth point of division, as JT, with B. 4. Through the preceding point of division, as P, draw a line PH II KB. 188. 5. Then HB is one ?ith of AB. 6. .*. HB, if laid off successively on AB, will divide AB into n equal parts. II. The proof and discussion are left as an exercise for the student. Ex. 305. Divide a straight line into 7 equal parts. Ex. 306. Construct an equilateral triangle, having given the perim- eter. Ex. 307. Construct a square, having given the perimeter. 251. Def. The line joining the mid-points of the non- parallel sides of a trapezoid is called the median of the trape- zoid. Ex. 308. Show, by generalizing, that Cor. Ill, Prop. XXXVIII, may be obtained from Cor. I and Cor. IV from Cor. II. BOOK I 97 Ex. 309. The lines joining the mid-points of the sides of a quadri- latefal taken in order form a parallelogram. Ex. 310. What additional statement can you make if the quadri- lateral in Ex. 309 is an isosceles trapezoid ? a rectangle ? a rhombus ? a square ? Ex. 311. Lines drawn from the mid-point of the base of an isosceles triangle to the mid-points of its equal sides form a rhombus or a square. When is the figure a rhombus ? when a square ? Ex. 312. The mid-points of the sides of a quadrilateral and the mid- points of its two diagonals are the vertices of three parallelograms whose diagonals are concurrent. Ex. 313. What is the perimeter of each parallelogram in Ex. 312 ? Ex. 314. Construct a triangle, given the mid-points of its sides. Ex. 315. Through a given point within an angle construct a line, limited by the two sides of the angle, and bisected at the given point. Ex. 316. Every diagonal of a parallelogram is trisected by the lines joining the other two vertices with the mid-points of the opposite sides. Ex. 317. If a triangle inscribed in another triangle has its sides parallel respectively to the sides of the latter, its vertices are the mid- points of the sides of the latter. Ex. 318. If the lower base KT of trapezoid ESTK is double the upper base RS, and the diagonals intersect at 0, prove OK double OS, and Or double OH. Ex. 319. Construct a trapezoid, given the two bases, one diagonal, and one of the non-parallel sides. In the two following exercises prove the properties which require proof, state those which follow by definition, and those which have been proved in the text : Ex. 320. Properties possessed by all trapezoids : (a) Two sides of a trapezoid are parallel. (6) The two angles adjacent to either of the non-parallel sides are supplementary. (c) The median of a trapezoid is parallel to the bases and equal to one half their sum. Ex. 321. In an isosceles trapezoid : (a) The two non-parallel sides are equal. (6) The angles at each base are equal and the opposite angles are supplementary. (c) The diagonals are equal. 98 PLANE GEOMETRY PROPOSITION XL. THEOREM 252. The two perpendiculars to the sides of an angle from any point in its bisector are equal. ,C B DA Given /. ABC; P any point in BR, the bisector of Z.ABG\ PD and PE, the Js from P to BA and BC respectively. To prove PD = PE. ARGUMENT ONLY 1. In rt. A DBP and PBE, PB = PB. 2. Z DBP = /. PBE. 3. .-. A DBP = A PBE. 4. .'. PD PE. Q.E.D. 253. Prop. XL may be stated as follows : Every point in the bisector of an angle is equidistant from the sides of the angle. Ex. 322. Find a point in one side of a triangle which is equidistant from the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting lines and also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting lines and also equidistant from two given parallel lines. Ex. 325. Find a point equidistant from the four sides of a rhombus. Ex. 326. The two altitudes of a rhombus are equal. Prove. Ex. 327. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a given angle. Do not prove. BOOK I 99 PROPOSITION XLI. THEOREM (Opposite of Prop. XL) 254. The two perpendiculars to the sides of an angle from any point not in its bisector are unequal. C B G Given Z.ABG\ P any point not in BR, the bisector of Z.ABC\ PD and PE, Js from P to BA and BG respectively. To prove PD = PE. OUTLINE or PROOF Draw FG J_ BA ; draw PG. Then FE = FG. Now PF + FG > PG. .'.PE>PG. But PG > PD. .'. PE > PD. 255. Prop. XLI may be stated as follows : Every point not in the bisector of an angle is not equidistant from the sides of the angle. 256. Cor. I. (Converse of Prop. XL). Every point equi- distant from the sides of an angle li^es in tJie bisector of the angle. HINT. Prove directly, using the figure for 252, or apply 137. 257. Cor. II. The bisector of an angle is the locus of all points equidistant from, the sides of the angle. Ex. 328. What is the locus of all points that are equidistant from a pair of intersecting lines ? 100 PLANE GEOMETRY CONCURRENT LINE THEOREMS PROPOSITION XLIT. THEOREM 258. The bisectors of the angles of a triangle are con- current in a point which is equidistant from the three sides of the triangle. Given A ABC with AF, BE, CD the bisectors of A A, B, and G respectively. To prove: (a) AF, BE, CD concurrent in some point as O; (b) the point equidistant from AB, BC, and CA. ARGUMENT 1. BE and CD will intersect at some point as 0. 2. Draw OL, OH, and OG, J from O to AB, BC, and CA respectively. 3. v is in BE, OL *= OH. 4. v is in CD, OG = OH. 5. .'. OL = OG. REASONS 1. If two str. lines are cut by a transversal making the sum of the two int. A on the same side of the transversal not equal to 2 rt. A, the lines are not II. 194. 2. From a point outside a line there exists one and only one J_ to the line. 155. 3. The two J to the sides of an /- from any point in its bisector are equal. 252. 4. Same reason as 3. 5. Things equal to the same thing are equal to each other. 54, 1. BOOK I ARGUMENT 6. /. AF, the bisector of Z.CAB, passes through 0. 7. .. AF, BE, and CD are con- current in 0. 8. Also is equidistant from AB, BC, and CA. Q.E.D. REASONS 6. Every point equidistant from the sides of an Z. lies in the bisector of the Z. 256. 7. By def . of concurrent lines. 196. 8. By proof, OL = OH = OG. 259. Cor. The point of intersection of the bisectors of the three angles of a triangle is the locus of all points equidistant from the three sides of the triangle. Ex. 329. Is it always possible to find a point equidistant from three given straight lines ? from four given straight lines ? Ex. 330. Find a point such that the perpendiculars from it to three sides of a quadrilateral shall be equal. (Give geometric construction.) Ex. 331. Prove that if the sides AB and AC of a triangle ABC are prolonged to E and F, respectively, the bisectors of the three angles BAC, EBC, and BCF all pass through a point which is equally distant from the three lines AE, AF, and BC. Is any other point in the bisector of the angle BAC equally distant from these three lines? Give reason for your answer. Ex. 332. Through a given point P draw a straight line such that perpendiculars to it from two fixed points Q and E shall cut off on it equal segments from P. (HINT. See 246.) Ex. 333. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a given triangle. Do not prove. Ex. 334. Find the locus of a point in one side of a parallelogram and equidistant from two other sides. In what parallelograms is this locus a vertex of the parallelogram ? Ex. 335. Find the locus of a point in one side of a parallelogram and equidistant from two of the vertices of the parallelogram. In what class of parallelograms is this locus a vertex of the parallelogram ? Ex. 336. Construct the locus of the center of a circle of given radius which rolls so that it constantly touches a given circumference. Do not prove, .PLANE GEOMETRY PROPOSITION XLIII. THEOREM 260. The perpendicular bisectors of the sides of a tri- angle are concurrent in a point which is equidistant from the three vertices of the triangle. 1. Given A ABC with FG, HK, ED, the J_ bisectors of AB, BC, CA. To prove : (a) FG, HK, ED concurrent in some point as ; (6) the point equidistant from A, B, and C. REASONS 1. Two lines J_ respectively to two intersecting lines also intersect. 195. Draw OA, OB, and OC. v is in FG, the _L bisector of AB, OB = OA ; and v is in DE, the _L bi- ARGUMENT FG and ED will intersect at some point as 0. G. sector of CA, OC = OA. OB = OC. HK, the J_ bisector of BC, passes through 0. FG, HK, and ED are con- current in 0. 1. Also O is equidistant from A, B, and (7, Q.E.D. 2. Str. line post. I. 54, 15. 3. Every point in the _L bi- sector of a line is equi- distant from the ends of that line. 134. 4 Ax. 1. 54, 1. 5. Every point equidistant from the ends of a line, lies in the J_ bisector of that line. 139. 6. By def . of concurrent lines t 196. 7. By proof, OA = OB = OC. BOOK I 103 261. Cor. The point of intersection of the perpendic- ular bisectors of the three sides of a triangle is the locus of all points equidistant from the three vertices of the triangle. Ex. 337. Is it always possible to find a point equidistant from three given points ? from four given points ? Ex. 338. Construct the perpendicular bisectors of two sides of an acute triangle, and then construct a circle whose circumference shall pass through the vertices of the triangle. Ex. 339. Construct a circle whose circumference shall pass through the vertices of a right triangle. Ex. 340. Construct a circle whose circumference shall pass through the vertices of an obtuse triangle. PROPOSITION XLIV. THEOREM 262. The altitudes of a triangle are concurrent. K B L Given A ABC with its altitudes AD, BE, and CF. To prove AD, BE, and CF concurrent. OUTLINE OP PROOF Through the vertices A, B, and C, of triangle ABC, draw lines || BC, AC, and AB, respectively. Then prove, by means of UJ, that AD, BE, and CF are the _L bisectors, respectively, of the sides of the auxiliary A HKL. Then, by Prop. XLIII, AD, BE, and CF are concurrent. Q.E.D. 104 PLANE GEOMETRY PROPOSITION XLV. THEOREM 263. Any two medians of a triangle intersect each other in a trisection point of each. B Given A ABC with AD and CE any two of its medians. To prove that AD and CE intersect in a point O such that OD = 1 AD and OE = 1 CE. OUTLINE OF PROOF 1. AD and CE will intersect at some point as 0. 194. 2. Let R and 5 be the mid-points of AO and CO respectively. 3. Quadrilateral REDS is a O. 4. .-. ^ff = #0 = OD and CS = SO = OE. 5. That is, OD = J- .4Z) and OE*=\ CE. Q.E.D. 264. Cor. J7ie three medians of a triangle are con- current. 265. Def. The point of intersection of the medians of a triangle is called the median center of the triangle. It is also called the centroid of the triangle. This point is the center of mass or center of gravity of the triangle. Ex. 341. Draw a triangle whose altitudes will intersect on one of its sides, and repeat the proof for Prop. XLIV. Ex. 342. Draw a triangle whose altitudes will intersect outside of the triangle, and repeat the proof for Prop. XLIV. Ex. 343. Prove Prop. XLV by prolonging OD its own length and drawing lines to B and C from the end of the prolongation. Ex. 344. Construct a triangle, given two of its medians and the angle between them. BOOK I 105 PROPOSITION XLVI. THEOREM 266. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter side . B v 1\ A C Given rt. A ABC with Z A double Z B. To prove AB = 2 AC. ARGUMENT ONLY 1. Draw'C% making Zl = 2. Z ^ = 60. 3. .-.Zl=60. 5. .\AE = AC=EC. 6. In A EBC, Z 2 = 30. 7. Z z? = 30. 8. Z A. .\EB-EC. 9. 10. = AE= AC. Q.E.D. 267. Prop. XLVI is sometimes stated : In a thirty-sixty de- gree right triaiujle, the hypotenuse is double the shorter side. 268. Historical Note. Such a triangle (a 30-60 right triangle) is spoken of by Plato as "the most beautiful right-angled scalene triangle/' Ex. 345. Prove Prop. XLVI by drawing CE = CA. Ex. 346. Prove Prop. XLVI by drawing lines through the ends of the hypotenuse parallel to the other two sides, thus forming a rectangle. 106 PLANE GEOMETRY CONSTRUCTION OF TRIANGLES 269. A triangle is determined, in general, when three parts are given, provided that at least one of the given parts is a line. The three sides and the three angles of a triangle are called its parts ; but there are also many indirect parts ; as the three medians, the three altitudes, the three bisectors, and the parts into which both the sides and the angles are divided by these lines. A A I B C 270. The notation given in the annexed figures may be used for brevity : A, B, C, the angles of the triangle ; in a right triangle, angle C is the right angle, a, b } c, the sides of the triangle; in a right triangle, c is the hypotenuse. m a) m b) m c , the medians to o, 6, and c respectively. h a , 7i b , 7i c , the altitudes to a, b, and c respectively. t n > tb> to the bisectors of A, B, and C respectively. 1 M d u , the segments of a made by the altitude to a. s ay r a , the segments of a made by the bisector of angle A. 271. The student should review the chief cases of construc- tion of triangles already given: viz. a, B, c; A,b,C', a, 6, c; right triangles : a, b ; 6, c ; b, B ; b, A ; c>A;. review also 152. Ex. 347. State in words the first eight cases given in 271. BOOK I 107 PROPOSITION XLVII. PROBLEM 272. To construct a triangle having two of its sides equal respectively to two given lines, and the angle oppo- site one of these lines equal to a given angle. a B /A X FIG. 1. Given lines a and b, and /- B. To construct A ABC. I. Construction 1. Draw any line, as BX. 2. At any point in BX, as B, construct Z XBY= to the given B. 125. 3. On B Y lay off BC = a. 4. With C as center and with b as radius, describe an arc cutting BX at A. 5. A ABC is the required A. II. Tfie proof is left as an exercise for the student. III. Discussion (1) b may be greater than a ; (2) b may equal a ; (3) b may be less than a. (1) If b > a, there will be one solution, i.e. one A and only one can be constructed which shall contain the .given parts. This case is shown in Fig. 1. (2) If b = a, the A will be isosceles. The construction will be the same as for case (1). 108 PLANE GEOMETRY (3) If b < a. Make the construction as for case (1). If b > the iL from G to 5-3T, there will be two solutions, since A AB<7 and 1)56', Fig. 2, both contain the required parts. If b equals the _L from C to BX there will be oe solution. The A will be a rt. A. If b < the _L from C to BX, there will be no solution. In the cases thus far considered, the given Z was acute. The discussion of the cases in which the given Z is a rt. Z. and in which it is an obtuse Z is left to the student. 273. Question. Why is (1) the only case possible when the given angle is either right or obtuse ? 274. The following exercises are given to A illustrate analysis of problems and to show the use of auxiliary triangles in constructions. M H B Ex. 348. Construct a triangle, given a, h a , m a - FIG. 1. ANALYSIS. Imagine the problem solved as in Fig. 1, and mark the given parts with heavy lines. The triangle AHM is determined and may be made the basis of the construction. Ex. 349. Construct a triangle, given 6, ra a , m c ANALYSIS. From Fig. 2 it will be seen 'that tri- v -" angle AGO may be constructed. Its three sides are known, since AO = f m a and CO = f m c . A Ex. 350. Construct a triangle, given a, h a ,h c . ANALYSIS. In Fig. 3, right triangle CHB is determined. The locus of vertex A is a line parallel to CB, so that the distance between r it and CB is equal to h a , FIG, 3. BOOK I 109 Ex. 351. Construct a triangle, given & ,c, ra a . ANALYSIS. Triangle ABK, Fig. 4, is deter- mined by three sides, 6, c, and 2w a . Since ABKC is a parallelogram, ylTf bisects (75. Ex. 352. Construct a triangle, given a, w, and the angle between m a and a. ANALYSIS. Triangle ^3f is determined, as shown in Fig. 5. A M y\ v C M B T S N FIG. 5. FIG- C. Ex. 353. Construct a trapezoid, given its four sides. ANALYSIS. Triangle R8T, Fig. 6, is determined. Ex. 354. Construct a parallelogram, given the perimeter, one base angle, and the altitude. ^T~ ~~7 CONSTRUCTION. The two parallels, CH and s' f\ I / AE, Fig. 7, may be drawn so that the distance A B E between them equals the altitude ; at any point FlG - 7 - B construct angle EBC equal to the given R M base angle ; draw CA, bisecting angle FOB ; /\ \ measure AE equal to half the given perimeter ; / \ \ complete parallelogram CHEB. 75 N Ex. 355. Construct a trapezoid, given the Fl - 8 - two non-parallel sides and the difference between the bases. (See Fig. 8.) Construct a triangle, given : Ex. 356. A, h a , t a . Ex. 360. B,h a b. Ex. 357. h a) l a , d a - Ex. 361. s a , t n , h a . Ex. 358. h a , m a , l a . Ex. 362. a, r 6 , C. Ex. 359. a, tb, C. Ex. 363. a, &, B. Construct an isosceles trapezoid, given : Ex. 364. The two bases and the altitude. Ex. 365. One base, the altitude, and a diagonal. Ex. 366. A base, a diagonal, and the angle between them. 110 PLANE GEOMETRY DIRECTIONS FOR THE SOLUTION OF EXERCISES 275. I. Make the figures clear, neat, accurate, and general in character. II. Fix firmly in mind hypothesis and conclusion with ref- erence to the given figure. III. Recall fundamental propositions related to the propo- sition in question. IV. If you can find no theorem which helps you, contradict the conclusion in every possible way (reductio ad absurdum) and try to show the absurdity of the contradiction. V. Make frequent use of the method of analysis, which con- sists in assuming the proposition proved, seeing what results follow until a known truth is reached, and then retracing the steps taken. VI. If it is required to find a point which fulfills two conditions, it is often convenient to find the point by the Intersection of Loci. By finding the locus of a point which satisfies each condition separately, it is possible to find the points in which the two loci intersect; i.e. the points which satisfy both conditions at the same time. VII. See 152 and exercises following 274 for method of attacking problems of construction. The method just described under V is a shifting of an uncertain issue to a certain one. It is sometimes called the Method of Successive Substitutions. It may be illustrated thus : 1. A is true if B is true. 3. But C is true. 2. B is true if C is true. 4. .-. A is true. This is also called the Analytic Method of proof. The proofs of the theorems are put in what is called the Synthetic form. But these were first thought through analytically, then rearranged in the form in which we find them. BOOK I 111 MISCELLANEOUS EXERCISES Ex. 367. The perpendiculars drawn from the extremities of one side of a triangle to the median upon that side are equal. Ex. 368. Construct an angle of 75 ; of 97 . Ex. 369. Upon a given line find a point such that perpendiculars from it to the sides of an angle shall be equal. Ex. 370. Construct a triangle, given its perimeter and two of its angles. Ex. 371. Construct a parallelogram, given the base, one base angle, and the bisector of the base angle. Ex. 372. Given two lines that would meet if sufficiently prolonged. Construct the bisector of their angle, without prolonging the lines. Ex. 373. Construct a triangle, having given one angle, one adjacent side, and the difference of the other two sides. Case 1 : The side oppo- site the given angle less than the other unknown side. Case 2 : The side opposite the given angle greater than the other unknown side. Ex. 374. The difference between two adjacent angles of a parallelo- gram is 90 ; find all the angles. Ex. 375. A straight railway passes 2 miles from a certain town. A place is described as 4 miles from the town and 1 mile from the railway. Represent the town by a point and find by construction how many places answer the description. Ex. 376. Describe a circle through two given points which lie out- side a given line, the center of the circle to be in that line. Show when no solution is possible. Ex. 377. Construct a right triangle, given the hypotenuse and the difference of the other two sides. Ex. 378. If two sides of a triangle are unequal, the median through their intersection makes the greater angle with the lesser side. Ex. 379. Two trapezoids are equal if their sides taken in order are equal, each to each. Ex. 380. Construct a right triangle, having given its perimeter and an acute angle. Ex. 381. Draw a line such that its segment intercepted between two given indefinite lines shall be equal and parallel to a given finite line. Ex. 382. One angle of a parallelogram is given in position and the point of intersection of the diagonals is given ; construct the parallelo- gram. 112 PLANE GEOMETRY Ex. 383. Construct a triangle, given two sides and the median to the third side. Ex. 384. If from- any point within a triangle lines are drawn to the three vertices of the triangle, the sum of these lines is less than the sum of the sides of the triangle, and greater than half their sum. Ex, 385. Repeat the proof of Prop. XIX for two cases at once, using Figs. 1 and 2. Ex. 386. If the angle at the vertex of an isosceles triangle is four times each base angle, the perpendicular to the base at one end of the base forms FIG. 1. FIG. 2. with one side of the triangle, and the prolongation of the other side through the vertex, an equilateral triangle. Ex. 387. The bisector of the angle C of a triangle ABC meets AB in Z>, and DE is drawn parallel to AC meeting BC in E and the bisector of the exterior angle at C in F. Prove DE = EF. Ex. 388. Define a locus. Find the locus of the mid-points of all the lines drawn from a given point to a given line not passing through the point. Ex. 389. Construct an isosceles trapezoid, given the bases and one angle. Ex. 390. Construct a square, given the sum of a diagonal and one side. Ex. 391. The difference of the distances from any point in the base prolonged of an isosceles triangle to the equal sides of the triangle is constant. Ex. 392. Find a point X equidistant from two intersecting lines and at a given distance from a given point. Ex. 393. When two lines are met by a transversal, the difference of two corresponding angles is equal to the angle between the two lines. Construct a triangle, given : Ex. 394. .4, h a , l a - Ex 398. a, m (l , B. Ex. 395. A, t a , s a . Ex. 399. wi,., 7t c , B. Ex. 396. a, h u , l a . Ex. 400. &, c, B + C. Ex. 397. a, b + c, A. Ex. 401. A, B, b + c. BOOK II THE CIRCLE 276. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. 277. Def. The curve which forms the boundary of a circle is called the circumference. 278. Def. The fixed point within is called the center, and a line joining the center to any point on the circum- ference is called a radius, as QR. 279. From the above definitions and from the definition of equal figures, 18, it follows that : (a) All radii of the same circle are equal. (6) All radii of equal circles are equal. (c) All circles having equal radii are equal. 280. Def. Any portion of a circumference is called an H arc, as DF, FC, etc. 281. Def. A chord is any straight line having its ex- tremities on the circumfer- ence, as DF. 282. Def. A diameter is a chord which passes through the center, as BC. 283. Since any diameter is twice a radius, it follows that All diameters of a circle are equal. 113 114 PLANE GEOMETRY PROPOSITION I. THEOREM 284. Every diameter of a circle bisects tlw circumfer- ence and the circle. M Given circle AMBN with center O, and AB, any diameter. To prove: (a) that AB bisects circumference (?>) that AB bisects circle AMBN. ARGUMENT 1. Turn figure AMB on AB as an axis until it falls upon the plane of ANB. Arc AMB will coincide with arc ANB. .\ arc AMB = arc ANB ; i.e. AB bisects circumference AMBN. Also figure AMB will coincide with figure ANB. .-.figure AMB = figure ANB; i.e. AB bisects circle AMBN. Q.E.D. REASONS 1. 54, 14. 2. 279, . 3. 18. 4. 279, a. 5. 18. Ex. 402. A semicircle is described upon each of the diagonals of a rectangle as diameters. Prove the semicircles equal. Ex. 403. Two diameters perpendicular to each other divide a circum- ference into four equal arcs. Prove by superposition. Ex. 404. Construct a circle which shall pass through two given points. Ex. 405. Construct a circle having a given radius r, and passing through two given points A and B. BOOK II 115 285. Def . A secant of a cir- cle is a straight line which cuts the circumference in two points, but is not terminated by the circumference, as MN. 286. Def. A straight line is tangent to, or touches, a cir- cle if, however far prolonged, it meets the circumference in but one point. This point is called the point of tangency. HK is tangent to circle at point T, and T is the point of tangency. 287. Def. A sector of a circle is a plane closed figure whose boundary is composed of two radii and their intercepted arc, as sector SOR. 288. Def. A segment of a circle is a plane closed figure whose boundary is composed of an arc and the chord joining its extremi- ties, as segment DCE. 289. Def. A segment which is one half of a circle is called a semi- circle, as segment AMB. 290. Def. An arc which is half of a circumference is called a semicircumference, as arc AMB. 291. Def. An arc greater than a semicircumference is called a major arc, as arc DME; an arc less than a semicir- cumference is called a minor arc, as arc DCE. 292. Def. A central angle, or angle at the center, is ail angle whose vertex is at the center of a circle and whose sides are radii. Ex. 406. The line joining the centers of two circles is 0, the radii are 8 and 10, respectively. What are the relative positions of the two circles ? Ex. 407. A circle can have only one center. 116 PLANE GEOMETRY PROPOSITION II. THEOREM 293. In equal circles, or in the same circle, if two cen- tral angles are equal, tfay intercept equal arcs on the cir- cumference; conversely, if two arcs are equal, the central angles that intercept them are equal. I. Given equal circles ABM and CDN, and equal central A and Q, intercepting arcs AB and CD, respectively. To prove A~B = CD. ARGUMENT 1. Place circle ABM upon circle CDN so that center shall fall upon center Q, and OA shall be collinear with QC. 2. A will fall upon <7. 3. OB will become collinear with QD. 4. .-. B will fall upon D. 5. .*. AB will coincide with CD. 6. .'. AB = CD. Q.E.D. RKASONS 1. 54,14. 2. 279, b. 3. By hyp. 4. 279, b. 5. 279,6. 6. 18. II. Conversely : Given equal circles ABM and CDN, and equal arcs AB and CD, intercepted by A and Q, respectively. To prove Zo - ZQ. BOOK II 117 ARGUMENT 1. Place circle ABM upon circle CDN so that center shall fall upon center Q. 2. Rotate circle AB M upon as a pivot until AB falls upon its equal cbj A upon C, B upon D. 3. O4 will coincide with QC'and OB with QD. 4. .-. Z = Z Q. Q.E.D. REASONS 1. 54, 14. 2. 54,14. 3. 17. 4. 18. 294. Cor. In equal circles, or in the same circle, if two central angles are unequal, the greater angle inter- cepts the greater arc ; conversely, if two arcs are unequal, the central angle that intercepts the greater arc is the greater. (HINT. Lay off the smaller central angle upon the greater.) 295. fcef. A fourth part of a circumference is called a quadrant. From Prop. II it is evident that a right angle at the center intercepts a quadrant on the cir- cumference. Thus, two _L diam- eters AB and CD divide the circumference into four quadrants, AC, CB, BD, and DA. 293. Def. A degree of arc, or an arc degree, is the arc intercepted by a central angle of one degree. 297. A right angle contains ninety angle degrees ( 71) ; therefore, since equal central angles intercept equal arcs on the circumference, a quadrant contains ninety arc degrees. Again, four right angles contain 360 angle degrees, and four right angles at the center of a circle intercept a complete cir- cumference ; therefore, a circumference contains 360 arc degrees. Hence, a semicircumference contains 180 arc degrees. Ex. 408. Divide a given circumference into eight equal arcs ; sixteen, equal arcs. 118 PLANE GEOMETRY Ex. 409. Divide a given circumference into six equal arcs ; three equal arcs ; twelve equal arcs. Ex. 410. A diameter and a secant perpendicular to it divide a cir- cumference into two pairs of equal arcs. Ex. 411. Construct a circle which shall pass through two given points A and B and shall have its center in a given line c. Ex. 412. If a diameter and another chord are drawn from a point in a circumference, the arc intercepted by the angle between them will be bisected by a diameter drawn parallel to the chord. Ex. 413. If a diameter and another chord are drawn from a point in a circumference, the diameter which bisects their intercepted arc will be parallel to the chord. PROPOSITION III. THEOREM 298. In equal circles, or in the same circle, if two chords are equal, they subtend equal arcs; conversely, if two arcs are equal, the chords that subtend them are equal. I. Given equal circles and Q, with equal chords AB and CD. To prove AB = CD. 1. 2 3. 4. 5. G. ARGUMENT Draw radii OA, OB, QC, QD. In A OAB and QCD, AB = CD. OA = QC and OB = QD. .. A OAB = A QCD. .-. ZO = ZQ. .-. AB= CD. Q.K.D. REASONS 54, 15. By hyp. 279, b. 116. 110. 293, I. BOOK II 119 II. Conversely : Given equal circles O and Q, and equal arcs AB and CD. To prove chord AB = chord CD. ARGUMENT 1. Draw radii OA, OB, QC, QD. 2. AB = CD. 4. OA = QC and OB = QD. 5. .'. A OAB = A QCD. 6. .-. chord AB = chord CD. Q.E.D. REASONS 54, 15. 2. By hyp. 3. 293,11. 4. 279, b. 5. 107. 6. 110. 1. Ex. 414. If a circumference is divided into any number of equal arcs, the chords joining the points of division will be equal. Ex. 415. A parallelogram inscribed in a circle is a rectangle. Ex. 416. If two of the opposite sides of an inscribed quadrilateral are equal, its diagonals are equal. Ex. 417. State and prove the converse of Ex. 416. 299. Def. A polygon is inscribed in a circle if all its vertices are on the circumference. Thus, polygon ABCDE is an inscribed polygon. 300. Def. If a polygon is in- scribed in a circle, the circle is said to be circumscribed about the poly- gon. Ex. 418. Inscribe an equilateral hexagon in a circle ; an equilateral triangle/ Ex. 419. The diagonals of an inscribed equilateral pentagon are equal. Ex. 420. If the extremities of any two intersecting diameters are joined, an inscribed rectangle will be formed. Under what conditions will tho rectangle be a square ? Ex. 421. State the theorems which may be used in proving arcs equal. State the theorems which may be. used in proving chords equal. 120 PLANE GEOMETRY PROPOSITION IV. THEOREM 301. In equal circles, or in tlie same circle, if two chords are unequal, the greater chord subtends tlie greater minor arc ; conversely, if two minor arcs are unequal, the chord that subtends the greater arc is tlie greater. I. Given circle 0, with chord AB > chord CD. To prove AB > CD. ARGUMENT REASONS 1. Draw radii OA, OB, OC, OD. 1. 54, 15. 2. In A OAB and OCD, OA = OC, OB = OD. 2. 279, a. 3. Chord AB > chord CD. 3. By hyp. 4. .-. Zl > Z2. 4. 173. 5. .-. AB > CD. Q.E.D. 5. 294. II. Conversely: Given circle O, with AB > CD. To prove chord AB > chord CD. ARGUMENT REASONS 1. Draw radii OA, OB, OC, OD. 1. 54, 15. 2. In A OAB and OCD, OA = OC, OB = OD. 2. 279, a. 3. AB > CD. 3. By hyp. 4. .-. Zl > Z2. 4. 294. 5. .-. chord AB > chord OD. Q.E.D. 5. 172. Ex. 422.. Prove the converse of Prop. IV by the indirect method. BOOK II 121 PROPOSITION V. THEOREM 302. The diameter perpendicular to a chord bisects w chord and also its subtended arcs. REASONS 1. 54, 15. 2. 279, a. a 94. 4. 212. 5. 212. 6. 75. 7. 293, I. Given chord AB and diameter CD J_ AR at E. To prove AE = EB, AC = CB, and AD = DB. ARGUMENT 1. Draw radii OA and OB. 2. In A OAB, OA = OB. 3. .-.A OAB is an isosceles A. 4. .-. OE bisects AB, and AE = EB. 5. Also OE bisects /. BOA, and Zl = Z2. 6. .-. ^AOD = ZDOB. 1. .-. AC= CB and AD = DB. Q.E.D. 303. Cor. I. The perpendicular bisector of a chord passes through the center of the circle. 304. Cor. II. The locus of the centers of all circles ivhich pass through two given points is the perpendicu- lar bisector of the line which joins the points. 305. Cor. in. Tlw locus of the mid-points of all chords of a circle parallel to a given line is the diameter per- pendicular to the line. Ex. 423. If the diagonals of an inscribed quadrilateral are unequal, its opposite sides are unequal. 122 PLANE GEOMETRY Ex. 424. Through a given point within a circle construct a chord which shall be bisected at the point. Ex. 425. Given a line fulfilling any two of the five following condi- tions, prove that it fulfills the remaining three: 1. A diameter. 2. A perpendicular to a chord. 3. A bisector of a chord. 4. A bisector of the major arc of a chord. 5. A bisector of the minor arc of a chord. Ex. 426. Any two chords of a circle are given in position and magni- tude ; find the center of the circle. Ex. 427. The line passing through the middle points of two parallel chords passes through the center of the circle. Ex. 428. Given an arc of a circle, find the center of the circle. PROPOSITION VI. PROBLEM 306. To bisect a given arc. A Given AB, an arc of any circle. To bisect AB. The construction, proof, and discussion are left as an exercise for the student. Ex. 429. Construct an arc of 45 ; of 30. Construct an arc of 30, using a radius twice as long as the one previously used. Are these two 30 arcs equal ? Ex. 430. Distinguish between finding the " mid-point of an arc '"and the " center of an arc." BOOK II 123 PROPOSITION VII. THEOREM 307. In equal circles, or in the same circle, if two chords are equal, they are equally distant from the center; con- versely, if two chords are equally distant from the center, they are equal. B I. Given circle with chord AB = chord CD, and let OE and OF be the distances of AB and CD from center O, respectively. To prove OE = OF. ARGUMENT 1. Draw radii OB and 0(7. 2. E and F are the mid-points of AB and CD, respectively. .'. in rt. A OEB and OCF, EB = CF. OB = OC. . .-. A OEB = A OCF. .'. OE OF. Q.E.D. REASONS 54, 15. 302. 3. 54, 8 a. 4. 279, a. 5. 211. 6. 110. II. Conversely : Given circle O with OE, the distance of chord AB from center 0, equal to OF, the distance of chord CD from center O. To prove chord AB = chord CD. HINT. Prove A OEB = A OCF. Ex. 431. If perpendiculars from the center of a circle to the sides of an inscribed polygon are equal, the polygon is equilateral. Ex. 432. If through any point in a diameter two chords are drawn making equal angles with the diameter, the two chords are equal. 124 PLANE GEOMETRY PROPOSITION VIII. THEOREM 308. In equal circles, or in the* same circle, if two chords are unequal, the greater chord is at the less dis- tance from the center. Given circle O with chord AB > chord CD, and let GF and OH be the distances of AB and CD from center 0, respectively. To prove OF < OH. ARGUMENT 1. From A draw a chord AE, equal to DC. 2. From draw OG AE. 3. Draw FG. 4. AB > CD. 5. .-. AB > AE. 6. F and G are the mid-points of AB and .4J?, respectively. .'. AF> AG. .-.41 >Z2. Z JFO = Z .-. Z3 ," BOOK II 125 PROPOSITION IX. THEOREM (Converse of Prop. VIII) 310. In equal circles, or in the same circle, if two chords are unequally distant from the center, the chord at the less distance is the greater. Given circle with OF, the distance of chord AB. from center 0, less than Off, the distance of chord CD from center 0. To prove chord AB > chord CD. The proof is left as an exercise for the student. HINT. Begin with A OGF. 311. Cor. I. A diameter is greater than any other chord. 312. Cor. II. The locus of the mid-points of all chords of a circle equal to a given chord is the circumference hav- ing the same center as the given circle, and having for ra- dius the perpendicular from the center to tfo given chord. Ex. 433. Prove Prop. IX by the indirect method. Ex. 434. Through a given point within a circle construct the mini- mum chord. Ex. 435. If two chords are drawn from one extremity of a diameter, making unequal angles with it, the chords are unequal. Ex. 436. The perpendicular from the center of a circle to a side of an inscribed equilateral triangle is less than the perpendicular from the center of the circle to a side of an inscribed square. (See 308.) 126 PLANE GEOMETRY PROPOSITION X. THEOREM 313. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. A M T B Given line AB, tangent to circle at T, and OT, a radius drawn to the point of tangency. To prove AB J_ OT. ARGUMENT 1. Let M be any point on AB other than T\ then M is outside the circumference. 2. Draw OM, intersecting the circumference at S. 3. OS < OM. 4. OS = OT. 5. .'. OT < OM. 6. .-. or is the shortest line that can be drawn from to AB. 7. .'. OT.AB ; i.e. AB. OT. Q.E.D. 314. Cor. I. (Converse of Prop. X). A straight line perpendicular to a radius at its outer extremity is tangent to the circle. HINT. Prove by the indirect method. In the figure for Prop. X, suppose that AB is not tangent to circle O at point T] then draw CD through T, tangent to circle O. Apply 63. 315. Cor. H. A perpendicular to a tangent at the point of tangency passes through the center of the circle. 1. REASONS 286. 2. 54, 15. 3. 4. 5. 6. 54, 12. 279, a. 309. Arg. 5. 7. 165. BOOK II 127 316. Cor. HI. A line drawn from the center of a circle perpendicular to a tangent passes through the point of tangency. 317. Def. A polygon is circum- scribed about a circle if each side of the polygon is tangent to the circle. In the same figure the circle is said to be inscribed in the polygon. Ex. 437. The perpendiculars to the sides of a circumscribed polygon at their points of tangency pass through a com- mon point. Ex. 438. The line drawn from any vertex of a circumscribed polygon to the center of the circle bisects the angle at that vertex and also the angle between radii drawn to the adjacent points of tangency. Ex. 439. If two tangents are drawn from a point to a circle, the bisector of the angle between them passes through the center of the circle. Ex. 440. The bisectors of the angles of a circumscribed quadrilateral pass through a common point. Ex. 441. Tangents to a circle at the extremities of a diameter are parallel. PROPOSITION XI. PROBLEM 318. To construct a tangent to a circle at any given point in the circumference. The construction, proof, and discussion are left as an exer- cise for the student. (See 314.) Ex. 442. Construct a quadrilateral which shall be circumscribed about a circle. What kinds of quadrilaterals are circumscriptible ? Ex. 443. Construct a parallelogram which shall be inscribed in a circle. What kinds of parallelograms are inscriptible ? Ex. 444. Construct a line which shall be tangent to a given circle and parallel to a given line. Ex. 445. Construct a line which shall be tangent to a given circle and perpendicular to a given line. 128 PLANE GEOMETRY 319. Def. The length of a tangent is the length of the seg- ment included between the point of tangency and the point from which the tangent is drawn; as TP in the following figure. PROPOSITION XII. THEOREM 320. If two tangents are drawn from any given point to a circle, these tangents are equal. Given PT and PS, two tangents from point P to circle 0. To prove PT=PS. The proof is left as an exercise for the student. Ex. 446. The sum of two opposite sides of a circumscribed quadrilat- eral is equal to the sum of the other two sides. Ex. 447. The median of a circumscribed trapezoid is one fourth the perimeter of the trapezoid. Ex. 448. A parallelogram circumscribed about a circle is either a rhombus or a square. Ex. 449. The hypotenuse of a right triangle circumscribed about a circle is equal to the sum of the other two sides minus a diameter of the circle. Ex. 450. If a circle is inscribed in any triangle, and if three triangles are cut from the given, triangle by drawing tangents to the circle, then the sum of the perimeters of the three triangles will equal the perimeter of the given triangle. BOOK II 129 PROPOSITION XIII. PROBLEM 321. To inscribe a circle in a given triangle. B H Given A ABC. To inscribe a circle in A AB C. I. Construction 1. Construct AE and (7/>, bisecting A CAB and BCA, respec- tively. 127. 2. AE and CD will intersect at some point as 0. 194. 3. From draw OF AC. 149. 4. With as center and OF as radius construct circle FGH. 5. Circle FGH is inscribed in A ABC. II. The proof and discussion are left for the student. 322. Def. A circle which is tangent to one side of a triangle and to the other two sides prolonged is said to be escribed to the triangle. Ex. 451. Problem. To escribe a circle to a given triangle. Ex. 452. (a) Prove that if the lines that bi- sect three angles of a quadrilateral meet at a com- mon point P, then the line that bisects the remaining angle of the quadri- lateral passes through P. (&) Tell why a circle can be inscribed in this particular quadrilateral. Ex. 453. In triangle ABC, draw XY parallel to BC so that XY + BC=BX+ CT. Ex. 454. Inscribe a circle in a given rhombus. 130 PLANE GEOMETRY PROPOSITION XIV. PROBLEM 323. To circumscribe a circle about a given triangle. \ Given AAHC. To circumscribe a circle about A ABC. The construction, proof, and discussion are left as an exercise for the student. 324. Cor. Three points not in tlw same straight line determine a circle. Ex. 455. Discuss the position of the center of a circle circumscribed about an acute triangle ; a right triangle ; an obtuse triangle. Ex. 456. Circumscribe a circle about an isosceles trapezoid. Ex. 457. Given the base of an isosceles triangle and the radius of the circumscribed circle, to construct the triangle. Ex. 458. The inscribed and circumscribed circles of an equilateral triangle are concentric. Ex. 459. If upon the sides of any triangle equilateral triangles are drawn, and circles circumscribed about the three triangles, these circles will intersect at a common point. Ex. 460. The two segments of a secant which are between two con- centric circumferences are equal. Ex. 461. The perpendicular bisectors of the sides of an inscribed quadrilateral pass through a common point. Ex. 462. The bisector of an arc of a circle is determined by the center of the circle and another point equidistant from the extremities of the chord of the arc. Ex. 463. If two chords of a circle are equal, the lines which connect their mid-points with the center of the circle are equal. BOOK II 131 TWO CIRCLES 325. Def. The line determined by the centers of two circles is called their line of centers or center-line. 326. Def. Concentric circles are circles which have the same center. PROPOSITION XV. THEOREM 327. If two circumferences meet at a point which is not on their line of centers, they also meet in one other point. M N Given circumferences M and N meeting at P, a point not on their line of centers OQ. To prove that the circumferences meet at one other point, as R. ARGUMENT REASONS 1. Draw OP and QP. 1. 54, 15. 2. Rotate A OPQ about OQ as an axis until 2. 54, 14. it falls in the position QRO. 3. OR = OP a radins of circle M. 4. .-. R is on circumference M. 5. Also QR = QP = a radius of circle N. 6. .*. R is on circumference N. 7. .*. R is on both circumference M and circumference JV; i.e. circumferences M and N meet at R. Q.E.D. 3. By cons. 4. 279, a. 5. By cons. 6. 279, a. 7. Args. 4 and 6. 328. Cor. I. If two circumferences intersect, their line of centers bisects their common chord at right angles. 132 PLANE GEOMETRY 329. Cor. II. If two circumferences meet at one point only, tJicut point is on their line of centers. HINT. If they meet at a point which is not on their line of centers, they also meet in another point ( 327). This contradicts the hypothesis. 330. Def . Two circles are said to touch or be tangent to each other if they have one and only one point in common. They are tangent internally or externally according as one circle lies within or outside of the other. Tangent externally. Tangent internally. . Concentric. 331. From 330, Cor. II may be stated as follows : If two circles are tangent to each other, their common point lies on their line of centers. 332. Cor. III. If two circles are tangent to each other, they have a common tangent line at their point of contact. HINT. Apply 314. 333. Def. A line touching two circles is called an external common tangent if both circles lie on the same side of it ; the line is called an internal common tangent if the two circles lie on opposite sides of it. FIG. 1. FIG. 2. Thus a belt connecting two wheels as in Fig. 1 is an illus- tration of external common tangents, while a belt arranged as in Fig. 2 illustrates internal common tangents. BOOK II 133 334. Questions. In case two circles are tangent internally how many common tangents can be drawn ? in case their circumferences intersect ? in case they are tangent externally ? in case they are wholly outside of each other ? in case one is wholly within the other ? Ex. 464. If two circles intersect, their line of centers bisects the angles between the radii drawn to the points of intersection. Ex. 465. If the radii of two intersecting circles are 5 inches and 8 inches, what may be the length of the line joining their centers? Ex. 466. If two circles are tangent externally, tangents drawn to them from any point in their common internal tangent are equal. Ex. 467. Two circles are tangent to each other. Construct their com- mon tangent at their point of contact. Ex. 468. Construct a circle passing through a given point and tan- gent to a given circle at another given point. Ex. 469. Find the locus of the centers of all circles tangent to a given circle at a given point. MEASUREMENT 335. Def. To measure a quantity is to find how many times it contains another quantity of the same kind. The result of the measurement is a number and is called the numerical measure, or measure-number, of the quantity which is measured. The measure employed is called the unit of measure. Thus, the length or breadth of a room is measured by find- ing how many feet there are in it ; i.e. how many times it con- tains a foot as a measure. 336. It can be shown that to every geometric magnitude there corresponds a definite number called its measure-number. The proof that to every straight line segment there belongs a measure-number is found in the Appendix, 595. The method of proof there used shows that operations with measure-numbers follow the ordinary laws of algebra. 337. Def. Two quantities are commensurable if there exists a measure that is contained an integral number of times in each. Such a measure is called a common measure of the two quantities, 134 PLANE GEOMETRY Thus, a yard and a foot are commensurable, each containing an inch a whole number of times; so. too, 12| inches and 18} inches are commensurable, each containing a fourth of an inch a whole number of times. 338. Questions. If two quantities have a common measure, hoio many common measures have they ? Name some common measures of 12^ inches and 18| inches. What is their greatest common measure ? What is their least common measure ? \ 339. Def. Two quantities are incommensurable if there exists no .measure that is contained an integral number of times in each. It will be shown later that a diagonal and a side of the same square cannot be measured by the same unit, without a re- mainder; and that the diagonal is equal to V2 times the numerical measure of the side. Now V2 can be expressed only approximately as a simple fraction or as a decimal. It lies between 1.4 and 1.5, for (1.4) 2 = 1.96, and (1.5) 2 = 2.25. Again, it lies between 1.41 and 1.42,* between 1.414 and 1.415, between 1.4142 and 1.4143, and so on. By repeated trials values may be found approximating more and more closely to V2, but no decimal number can be obtained that, taken twice as a factor, will give exactly 2. 340. When we speak of the ratio of one quantity to another, we have in mind their relative sizes. By this is meant not the difference between the two, but how many times one con- tains the other or some aliquot part of it. In algebra the ratio of two numbers has been defined as the indicated quotient of the first divided by the second. Since to each geometric mag- nitude there corresponds a number called its measure-number ( 336), therefore : 341. Def. The ratio of two geometric magnitudes may be defined as the quotient of their measure-numbers, when the same measure is applied to each. * The student should multiply to get the successive approximations. BOOK II 135 Thus, if the length of a room is 36 feet and the width 27 feet, the ratio of the length to the width is said to be the ratio of 36 to 27 ; i.e. f f , which is equal to f . The ratio of the width to the length is f J, which is equal to J. The term ratio is never applied to two magnitudes that are unlike. 342. Def. If the two magnitudes compared are commen- surable, the ratio is called a commensurable ratio and can always be expressed as a simple fraction. 343. Def. If the two magnitudes compared are incommen- surable, the ratio is called an incommensurable ratio and can be expressed only approximately as a simple fraction. Closer and closer approximations to an incommensurable ratio may be obtained by repeatedly using smaller and smaller units as measures of the two magnitudes to be compared and by finding the quotient of the numbers thus obtained. Two magnitudes, e.g. two line segments, taken at random are usually incommensurable, commensurability being compara- tively rare. 344. Historical Note. The discovery of incommensurable magni- tudes is ascribed to Pythagoras, whose followers for a long time kept the discovery a secret. It is believed that Pythagoras was the first to prove that the side and diagonal of a square are incommensurable. A more complete account of the work of Pythagoras will be found in 510. Ex. 470. What is the greatest common measure of 48 inches and 18 inches ? Will it divide 48 inches 18 inches ? 48 inches 2 x 18 inches ? Ex. 471. Draw any two line segments which have a common meas- ure. Find the sum of these lines and, by laying off the common measure, show that it is a measure of the sum of the lines. Ex. 472. Given two lines, 5 inches and 4 inches long, respectively. Show by a diagram that any common measure of 5 inches and 4 inches is also a measure of 15 inches plus or minus 8 inches. Ex. 473. Find the greatest common divisor of 728 and 844 by division and point out the similarity of the process to that used in Prop. XVI. 136 PLANE GEOMETRY PROPOSITION XVI. PROBLEM 345. To determine whether two given lines are commen- surable or not; and if they are commensurable, to find their common measure and their ratio. F B E G D Given lines AB and CD. To determine: (a) whether AB and CD are commensurable and if so, (6) what is their common measure; and (c) what is the ratio of AB to CD. I. Construction 1. Measure off AB on CD as many times as possible. Sup- pose it is contained once, with a remainder ED. 2. Measure off ED on AB as many times as possible. Sup- pose it is contained twice, with a remainder FB. 3. Measure off FB on ED as many times as possible. Sup- pose it is contained three times, Avith a remainder GD. 4. Measure off GD on FB as many times as possible, and so on. 5. It is evident that this process will terminate only when a remainder is obtained which is a measure of the remainder immediately preceding. 6. If this process terminates, then the two given lines are commensurable, and the last remainder is their greatest com- mon measure. 7. For example, if GD is a measure of FB, then AB and CD are commensurable, GD is their greatest common measure, and the ratio of AB to CD can be found. II. Proof ARGUMENT 1. Suppose FB = 2 GD. 2. ED = EG + GD. REASONS 1. See I, 7. 2. 54,11. BOOK II 137 3. /. ED = 3 FB + GD = 1 GD. 4. AB 2 ED "+ FB. 5. .'. AB = 14 GD + 2 GD = 16 S.D. 6. CD == .45 -f 7). 7. .. C/> =16 G/> 4- 7 (?Z) = 23 GZ>. 8. .-. ^45 and CD are commensurable. 9. Also, GD is a common measure of AB 9. Args. 5 and and CD. LO. The ratio of AB to CD = the ratio of 16 GD to 23 GD = if. Q.E.D. III. A full discussion of this problem will be found in the Appendix, 598. 3. 309. 4. 54, 11. 5. 309. 6. 54, 11. 7. 309. 8. 337. 9. Args. 5 7. 10. 341. CONSTANTS AND VAKIABLES. LIMITS 346. Consider an isosceles triangle ABC, whose base is AC and whose altitude is LB. Keeping the base AC the same (con- stant), suppose the altitude to change (vary). If LB increases, what will be the B effect upon the lengths of AB and CB ? what the effect upon the base angles ? upon the vertex angle ? Will the base angles always be equal to each other ? What limiting value have they ? Is the base angle related to half the vertex angle or are the two independent ? What relation is there ? Is this relation A L C constant or does it change ? Imagine the altitude of the triangle to diminish. Repeat the questions given above, considering the altitude as decreasing. What is now the limiting value for the altitude ? what for the length of one of the equal sides ? for the base angles ? for the angle at the vertex ? Ex. 474. Consider an isosceles triangle with a constant altitude and a variable base. Repeat the questions given above. 138 PLANE GEOMETRY Ex. 475. Consider an isosceles triangle with constant base angles, but variable base. Tell what other constants and what other variables there would be in this case. Ex. 476. If through any point in the base of an isosceles triangle lines are drawn parallel to the equal sides of the triangle, a parallelo- gram will be formed whose perimeter will be constant ; i.e. the perimeter will be independent of the position of the point. 347. Def. A magnitude is constant if it does not change throughout a discussion. 348. Def. A magnitude is variable if it takes a series of different successive values during a discussion. 349. Def. If a variable approaches a constant in such a way that the difference between the variable and the constant may be made to become and remain smaller than any fixed number previously assigned, however small, the constant is called the limit of the variable. 350. The variable is said to approach its limit as it becomes more and more nearly equal to it. Thus, suppose a point to move from A toward B, by successive steps, under the A ?* ? restriction that at each step it must go over one half the segment between it and B. At the first step it reaches C, whereupon there remains the segment CB to be traveled over ; at the next step it reaches D, and there remains an equal segment to be covered. Whatever the number of steps taken, there must always remain a segment equal to the segment last covered. But the segment between A and the moving point may be made to differ from AB by as little as we please, i.e. by less than any previously assigned value. For assign some value, say, half an inch. Then the point, continuing to move under its governing law, may approach B until there remains a segment less than half an inch. Whatever be the value assigned, the variable segment from A to the moving point may be made to differ from the constant segment AB by less than the assigned value. BOOK II 139 Again, the numbers in the series 4, 2, 1, ^, J, i, etc., in which each term is one half of the preceding term, approach as a limit as the number of terms in the series is increased. For if we assign any value, as Twrnr* ^ * s evident that a term of the series may be found which is less than 10 j, 00 ; it is also evident that no term of the series can become 0. 351. In elementary geometry the variables that approach limits are usually such that they cannot attain their limits. There are, however, variables that do attain their limits. The limiting values of algebraic expressions are frequently of this kind ; thus, the expression - approaches 1 as x approaches ar -f- 1 0, and has the limit 1 when x becomes zero. 352. Def. Two variables are said to be related when one depends upon the other so that, if the value of one variable is known, the value of the other can be obtained. For example, the diagonal and the area of a square are re- lated variables, for there is a value for the area for any value which may be given to the diagonal, and vice versa. 353. Questions. On the floor is a bushel of sand. If we keep adding to this pile forever, how large will it become ? Does it depend upon the law governing our additions ? If we add one quart each hour, how large will it become ? If we add one quart the first hour, a half quart the second hour, a fourth quart the third hour, etc., each hour adding one half as much as the preceding hour, how large will the pile become ? 354. Historical Note. Achilles and the Tortoise. One of the early Greek schools of mathematics, founded during the fifth century B.C., at Elea, Italy, and known as the Eleatic School, was famous for its investigations of problems involving infinite series. Zeno, one of the most prominent members, proposed this question: He "argued that if Achilles ran ten times as fast as a tortoise, yet if the tortoise had (say) 1000 yards start, it could never be overtaken : for, when Achilles had gone the 1000 yards, the tortoise would still be 100 yards in front of him ; by the time he had covered these 100 yards, it would still be 10 yards in front of him ; and so on forever : thus Achilles would get nearer and nearer to the tortoise but never overtake it." Was Zeno right ? If not, can you find the fallacy in his argument ? 140 PLANE GEOMETRY PROPOSITION XVII. THEOREM 355. If two variables are always equal, and if each ap- proaches a limit, then their limits are equal. Given two variables, V and F', which are always equal and which approach as limits L and L', respectively. To prove L = L 1 . ARGUMENT 1. Either L = L', or L = L 1 . 2. Suppose that one limit is greater than the other, say L > L' ; then F, in ap- proaching L, may assume a value be- tween L' and L\ i.e. V may assume a value > L'. 3. But F' cannot assume a value > L'. 4. .-. F may become > F'. 5. But this is impossible, since F and F' are always equal. L = L'. Q.E.D. 6. REASONS 1. 161, a. 2. 349. 349. Args. 2 and 3. 5. By hyp. 6. 161, 6. 356. Question. In the above proof are V and V increasing or de- creasing variables ? The student may adapt the argument above to the case in which V and V are decreasing variables. Ex. 477. Apply Prop. XVII to the accompanying figure, where variable V is represented by the line AB, variable V by the line CD, limit L by line AE, and limit L' by line CF. B H 357. Note. It will be seen that, in the application of Prop. XVII, there are three distinct things to be considered : (1) Two variables that' are always equal; (2) The limits of these two variables ; (3) The equality of these limits themselves,. BOOK II 141 PROPOSITION XVIII. THEOREM 358. An angle at the center of a circle is measured its intercepted arc. Given central /-AOB and AB intercepted by it; let /.COE be any unit Z (e.g. a degree), and let CE, intercepted by the unit Z, be the unit arc. To prove the measure-number of /.AOB, referred to /.COE, equal to the measure-number of AB, referred to CE. I. If /.AOB and /.COE are commensurable. (a) Suppose that /.COE is contained in /.AOB an integral number of times. ARGUMENT 1. Apply Z COE to /.AOB as a measure. Suppose that /iCOE is contained in Z AOB r times. 2. Then r is the measure-number of Z ^10,8 referred to /.COE as a unit. 3. Now the r equal central A which com- pose /.AOB intercept r equal arcs on the circumference, each equal to CE. 4. .-. r is the measure-number of AB re- ferred to CE as a unit. 5. .-. the measure-number of /.AOB, re- ferred to /.COE as a unit, equals the measure-number of AB, referred to CE as a unit. Q.E.D. REASONS 1. 335. 2. 335. 3. 293, I. 4. 335. 5. 54,1. 142 PLANE GEOMETRY I. It Z.AOB and Z COE are commensurable. (6) Suppose that Z COE is not contained in /.AOB an integral number of times. The proof is left as an exercise for the student. HINT. Some aliquot part of ZCOE must be a measure of /.AOB. (Why ?) Try \ FIG. 1. FIG. 2. FIG. 3. 366. Cor. I. All angles inscribed in the same segment are equal. (See Fig. 1.) 367. Cor. II. Any angle inscribed in a semicircle is a right angle. (See Fig. 2.) 368. Cor. III. The locus of the vertex of a right tri- angle having a given hypotenuse as base is the circuin- ference having the same hypotenuse as diameter. 369. Cor. IV. Any angle inscribed in a segment less than a semicircle -is an obtuse angle. (See Fig. 3.) 370. Cor. V. Any angle inscribed in a segment greater than a semicircle is an acute angle. (See Fig. 3.) Ex. 480. If an inscribed angle contains 24 angle degrees, how many arc degrees are there in the intercepted arc ? how many in the rest of the circumference ? Ex. 481. If an inscribed angle intercepts an arc of 70, how many degrees are there in the angle ? Ex. 482. How many degrees are there in an angle inscribed in a seg- ment whose arc is 140 ? Ex. 483. Construct any segment of a circle so that an angle inscribed in it shall be an angle ^of : (a) 60 ; (ft) 45 ; (c) 30. Ex. 484. Repeat Ex. 483, using a given line as chord of the segment. How many solutions are there to each case of Ex. 483 ? how many to each case of Ex. 484 ? Ex. 485. The opposite angles of an inscribed quadrilateral are sup- plementary. BOOK II 147 Ex. 486. If the diameter of a circle is one of the equal sides of an isosceles triangle, the circumference will bisect the base of the triangle. Ex. 487. By means of a circle construct a right triangle, given the hypotenuse and an arm. Ex. 488. By means of a circle construct a right triangle, given the hypotenuse and an adjacent angle. Ex. 489. Construct a right triangle, having given the hypotenuse and the altitude upon the hypotenuse. 371. Historical Note. Thales (640-546 B.C.), the founder of the earliest Greek school of niathe- matics, is said to have discov- ered that all triangles having a diameter of a circle as base, with their vertices on the cir- cumference, have their vertex angles right angles. Thales was one of the Seven Wise Men. He had much business shrewdness and sagacity, and was renowned for his practi- cal and political ability. He went to Egypt in his youth, and while there studied geom- etry and astronomy. The story is told that one day while viewing the stars, he fell into a ditch ; whereupon an old woman said, " How canst thou know what is doing in the heavens, when thou seest not what is at thy feet ? " According to Plutarch, Thales computed the height of the Pyramids of Egypt from measurements of their shadows. Plutarch gives a dialogue in which Thales is addressed thus, "Placing your stick at the end of the shadow of the pyramid, you made by the same rays two triangles, and so proved that the height of the pyramid was to the length of the stick as the shadow of the pyramid to the shadow of the stick." This compu- tation was regarded by the Egyptians as quite remarkable, since they were not familiar with applications of abstract science. The geometry of the Greeks was in general ideal and speculative, the Greek mind being more attracted by beauty and by abstract relations than by the practical affairs of everyday life. THALES 148 PLANE GEOMETRY PROPOSITION XX. PROBLEM 372. To construct a perpendicular to a given straight line at a given point in the line. (Second method. For another method, see 148.) \ M / ^ ++' 1 \ ^ 1 1 \ ^"" 1 \ Given line AB and P, a point in AB. To construct a _L to AB at P. I. Construction 1. With O, any convenient point outside of AB, as center, and with OP as radius, construct a circumference cutting AB at P and Q. 2. Draw diameter QM. 3. Draw PM. 4. PM is J_ AB at P. II. The proof and discussion are left as an exercise for the student. The method of 372 is useful when the point P is at or near the end of the line AB. Ex. 490. Construct a perpendicular to line AB at its extremity B, without prolonging AB. Ex. 491. Through one of the points of intersection of two circumfer- ences a diameter of each circle is drawn. Prove that the line joining the ends of the diameters passes through the other point of intersection. BOOK II 149 PROPOSITION XXI. PROBLEM 373. To construct a tangent to a circle from a point out- side. Given circle and point P outside the circle. To construct a tangent from P to circle 0. I. Construction 1. Draw PO. 2. With Q, the mid-point of PO, as center, and with QO as radius, construct a circumference intersecting the circumference of circle in points T and F. 3. Draw PTand PF. 4. PT and PFare tangents from P to circle 0. II. The proof and discussion are left as an exercise for the student. HINT. Draw OT and OF and apply 367. Ex. 492. Circumscribe an isosceles triangle about a given circle, the base of the isosceles triangle being equal to a given line. What restric- tion is there on the length of the base ? Ex. 493. Circumscribe a right triangle about a given circle, one arm of the triangle being equal to a given line. What is the least length possible for the given line, as compared with the diameter of the circle ? Ex. 494. If a circumference M passes through the center of a circle B, the tangents to B at the points of intersection of the circles intersect on circumference M. Ex. 495. Circumscribe an isosceles triangle about a circle, the altitude upon the base of the triangle being given. 150 PLANE GEOMETRY Ex. 496. Construct a common external tangent to two given circles. r' X " v M Given circles M TE and NV8. To construct a common external tangent to circles MTE and N VS. I. Construction 1. Draw the line of centers OQ. 2. Suppose radius OM > radius QN. Then, with as center and with OL = OM $xVas radius, construct circle LPK. 3. Construct tangent QP from point Q to circle LPK. 373. 4. Draw OP and prolong it to meet circumference M TE at T. 5. Draw QV\\ OT. 188. 6. Draw TV. 7. TV is tangent to circles M TE and NVS. II. The proof and discussion are left as an exercise for the student. Ex. 497. Construct a second common external tangent to circles M TE and NVS by same method. How is the method modified if the two circles are equal ? Ex. 498. Construct a common internal tangent to two circles. HINT. Follow steps of Ex. 490 except step 2. Make OL = OM + QN. Ex. 499. By moving Q toward O in the preceding figure, show when there are four common tangents ; when only three ; when only two ; when only one ; when none. BOOK II 151 PROPOSITION XXII. PROBLEM 374. With a given line as chord, to construct a segment of a circle capable of containing a given angle. B Given line AB and Z M. To construct, with AB as chord, a segment of a circle capable of containing Z M. I. Construction 1. Construct an Z, as /.CDE, equal to the given Z M. 125. 2. With H, any convenient point on DE, as center and with AB as radius, describe an arc cutting DC at some point, as K. 3. Draw HK. 4. Circumscribe a circle about A KDH. 323. 5. Segment KDH is the segment capable of containing Z Jf. II. The proof -and discussion are left to the student. 375. Questions. Without moving AB, can you construct a A with AB as base and a vertex Z = Z M? What is the sum of the base A ? 376. Cor. The locus of the vertices of all triangles hav- ing a given base and a given angle at their vertices is the arc ivhich forms, with the given base, a segment capable of containing the given angle. Ex. 500. On a given line construct a segment that shall contain an angle of 105 ; of 135. Ex. 501. Find the locus of the vertices of all triangles having a com- mon base 2 inches long and having their vertex angles equal to 60. Ex. 502. Construct a triangle, having given 6, 7i 6 , and B. 152 PLANE GEOMETRY PROPOSITION XXIII. THEOREM 377. An angle formed by two chords which intersect within a circle is measured by one half the sum of the arc intercepted between its sides and the arc intercepted be- tween the sides of its vertical angle. r B Given two chords AB and CD, intersecting at E. To prove that Z 1 oc ^ (#/) -j- Jic). ARGUMENT Draw AD. 1. 2. 3. Z 2 oc i. BD. 4. Z 3 oc | AC. 5. .-. Z 2 + ^ 3 6. .-. Z 1 oc i. (D 4. + Q.E.D. REASONS 54, 15. 215. 365. 365. 362,6. 309. Ex. 503. One angle formed by two intersecting chords intercepts an arc of 40. Its vertical angle intercepts an arc of 60. How large is the angle ? Ex. 504. If an angle of two intersecting chords is 40 and its inter- cepted arc is 30, how large is the opposite arc ? Ex. 505. If two chords intersect at right angles within a circumfer- ence, the sum of two opposite intercepted arcs is equal to a semicircum- ference. Ex. 506. If M is the center of a circle inscribed in triangle ABC and if AM is prolonged to meet the circumference of the circumscribed circle at D, prove that BD = DM = DC. BOOK II 153 PROPOSITION XXIV. THEOREM 378. An angle formed by a tangent and a chord is measured by one half its intercepted arc. B E Given Z ABC formed by tangent AB and chord BC. To prove that Z AB C 9? 1 BC. ARGUMENT 1. Draw diameter BD. 2. Z.ABD is a rt. Z. 3. .-. Z.ABD ; i a semicircumference ; i.e. Z ^5D QC i arc #<7. 4. Z CBD QC i. cz). 5. .'. /-ABD Z C/?Z> Q? -J- (arc .BCD CD). .1 &. Q.E.D. REASONS 1. 64, 15. 2. 313. 3. 297, 4. 365. 5. 362, 6. 6. 309. Ex. 507. In the figure of 378, if arc BC = 100, find the number of degrees in angle ABC ; in angle CBD ; in angle CBE. Ex. 508. If tangents are drawn at the extremities of a chord which subtends an arc of 120, what kind of triangle is formed ? Ex. 509. If a tangent is drawn to a circle at the extremity of a chord, the mid-point of the subtended arc is equidistant from the chord and the tangent. Ex. 510. Solve Prop. XXII by means of Prop. XXIV. HINT. Observe that in the figure for Prop. XXIV any angle inscribed in segment BDC would be equal to angle ABC. 154 PLANE GEOMETRY PROPOSITION XXV. THEOREM 379. OQ. OP= OM. .'.PQ>MQ. /. P is not on circumference S. II. 3/not between O and Q, Fig. 2. OQ+ QP>OP. .'. OQ+QP>OM. .:QP> QM. .'. P is not on circumference S. BOOK II 159 Ex. 555. If two circumferences intersect, neither point of intersection is on their line of centers. Ex. 556. In any right triangle ABC, right-angled at C, the radius of the inscribed circle equals \(a + b c) and the radius of the escribed circle tangent to c equals |( + b + c). Ex. 557. In the accom- . ? B P panying figure a, &, c, are the sides of triangle ABC. Prove : 1. a = TP; 2. AP= 3. TB = ( Ex. 558. Trisect a quadrant ; a semicircumference ; a circumference. Ex. 559. Describe circles about the vertices of a given triangle as centers, so that each shall touch the other two. Ex. 560. Construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. Construct a triangle, having given : Ex. 561. h a , h c , C. Ex. 562. A, B, and JR, the radius of the circumscribed circle. Ex. 563. a, B, It. Ex. 564. C and the segments of c made by t c . Ex. 565. C and the segments of c made by h c . Ex. 566. r, the radius of the inscribed circle, and the segments of c made by t c . Construct a right triangle ABC, right-angled at (7, having given : Ex. 567. c, h e . Ex. 568. c and one segment of c made by h c . Ex. 569. The segments of c made by h c . Ex. 570. The segments of c made by t c . Ex. 571. c and a line I, in which the vertex of C must lie. Ex. 572. c and the perpendicular from vertex C to a line I. Ex. 573. c and the distance from C to a point P. Construct a square, having given : Ex. 574. The perimeter. Ex. 575. A diagonal. Ex. 576. The sum of a diagonal and a side. 160 PLANE GEOMETRY Construct a rectangle, having given : Ex. 577. Two non-parallel sides. Ex. 578. A side and a diagonal. Ex. 579. The perimeter and a diagonal. Ex. 580. A diagonal and an angle formed by the diagonals. Ex. 581. A side and an angle formed by the diagonals. Ex. 582. The perimeter and an angle formed by the diagonals. Construct a rhombus, having given : Ex. 583. A side and a diagonal. Ex. 584. The perimeter and a diagonal. Ex. 585. One angle and a diagonal. Ex. 586. The two diagonals. Ex. 587. A side and the sum of the diagonals. Ex. 588. A side and the difference of the diagonals. Construct a parallelogram, having given : Ex. 589. Two non-parallel sides and an angle. Ex. 590. Two non-parallel sides and a diagonal. Ex. 591. One side and the two diagonals. Ex. 592. The diagonals and an angle formed by them. Construct an isosceles trapezoid, having given : Ex. 593. The bases and a diagonal. Ex. 594. The longer base, the altitude, and one of the equal sides. Ex. 595. The shorter base, the altitude, and one of the equal sides. Ex. 596. Two sides and their included angle. Construct a trapezoid, having given : Ex. 597. The bases and the angles adjacent to one base. Ex. 598. The bases, the altitude, and an angle. Ex. 599. One base, the two diagonals, and their included angle. Ex. 600. The bases, a diagonal, and the angle between the diagonals. Construct a circle which shall : Ex. 601. Touch a given circle at P and pass through a given point Q. Ex. 602. Touch a given line I at P. Ex. 603. Touch three given lines two of which are parallel. Ex. 604. Touch a given line I at P and also touch another line m. Ex. 605. Have its center in line Z, cut I at P, and touch a circle K. BOOK III PROPORTION AND SIMILAR FIGURES 382. Def. A proportion is the expression of the equality of two ratios. EXAMPLE. If the ratio - is equal to the ratio -, then the equation b d - = - is a proportion. This proportion may also be written a : b = c : d or 6 d a : b : : c : d, and is read a is to b as c is to d. 383. Def. The four numbers a, b, c, d are called the terms of the proportion. 384. Def. The first term of a ratio is called its antecedent and the second term its consequent ; therefore : The first and third terms of a proportion are called ante- cedents, and the second and fourth terms, consequents. 385. Def. The second and third terms of a proportion are called its means, and the first and fourth terms, its extremes. 386. Def. If the two means of a proportion are equal, this common mean is called the mean proportional between the two extremes, and the last term of the proportion is called the third proportional to the first and second terms taken in order ; thus, in the proportion a : b = b : c, b is the mean proportional between a and c, and c is the third proportional to a and b. 387. Def. The fourth proportional to three given numbers is the fourth term of a proportion the first three terms of which are the three given numbers taken in order ; thus, if a : b = c : d, d is called the fourth proportional to a, b, and c. 161 162 PLANE GEOMETRY PROPOSITION I. THEOREM 388. If four numbers are in proportion, the product of the extremes is equal to the product of the means. Given a : b = c : d. To prove ad = be. 1. ARGUMENT a __ G b~d' 2. bd = bd. 3. .-.ad = bc. Q.E.D. REASONS 1. By hyp.* 2. By iden. 3. 54, la. 389. Note. The student should observe that the process used here is merely the algebraic "clearing of fractions," and that as fractional equations in algebra are usually simplified by this process, so, also, pro- portions may be simplified by placing the product of the means equal to the product of the extremes. 390. Cor. I. The mean proportional between two num- bers is equal to the square root of their product. 391. Cor. II. If two proportions have any three terms of one equal respectively to the three corresponding terms of the other, then the remaining term of the first is equal to the remaining term of tfa second. Ex. 606. Given the equation m : r = d : c : solve (1 ) for d, (2) for r, (3) for m, (4) for c. Ex. 607. Find the fourth proportional to 4, 6, and 10 ; to 4, 10, and 0; to 10, 6, and 4. Ex. 608. Find the mean proportional between 9 and 144 ; between 144 and 9. Ex. 609. Find the third proportional to f and f ; to f and $. 392. Questions. What rearrangement of numbers can be made in Ex. (507 without affecting the required term ? in Ex. 008 ? in Ex. 009 ? Ex. 610. Find the third proportional to a 2 & 2 and a b. Ex. 611. Find the fourth proportional to a 2 6 2 , a ft, and a + b. * See 382 for the three ways of writing a proportion. BOOK III 163 Ex. 612. If in any proportion the antecedents are equal, then the consequents are equal and conversely. Ex. 613. If a : b = c : d, prove that ma : kb = me : kd. Ex. 614. If I : k = b : m, prove that Ir : kr = be : me. Ex. 615. If x : y = b : c, prove that dx : y = bd : c. Ex. 616. If x : y = b : c, is dx : y = 6 : cd a true proportion ? PROPOSITION II. THEOREM (Converse of Prop. I) 393 If the product of two numbers is equal to the prod- uct of two other numbers, either pair may be made the means and the other pair the extremes of a proportion. Given ad = be. To prove a:b = c:d. ARGUMENT 1. ad = bc. 2. bd = bd. 3. /.- = -; i.e. a : b = c : d. Q.E.D. 6 d REASONS 1. By hyp. 2. By iden. 3. 54, 8 a. The proof that a and d may be made the means and b and c the extremes is left as an exercise for the student. 394. Note. The pupil should observe that the divisor in Arg. 2 above must be chosen so as to give the desired quotient in the first mem- ber* of the equation : thus, if hi = kf, and we wish to prove that- = -, we must divide by fl ; then M = &, i.e. \ =* fl fl J I Ex. 617. Given pt = cr. Prove p : r = c : t ; also, c:p = t:r. Ex. 618. From the equation rs = Im, derive the following eight pro- portions : r : I = m : s, s:l = m:r, l:r = s: , m : r = s : I r : m = I : s, s : m I : r, I : s = r : w, m : s = r : I. Ex. 619. Form a proportion from 7 x 4 = 3 x a ; from ft = gb. How can the proportions obtained be verified ? 164 PLANE GEOMETRY Ex. 620. Form a proportion from (a + c)(a b) = de. Ex. 621. Form a proportion from m' 2 2 mil + n 2 = db. Ex. 622. Form a proportion from c' 2 + 2 cd + d' 2 = a + 6. Ex. 623. Form a proportion from (a + &)( 6) =4se, making x (1) an extreme ; (2) a mean. Ex. 624. If7 x + By :l2 = 2x+y : 3, find the ratio x : y. PROPOSITION III. THEOREM 395. If four numbers are in proportion, they are in proportion by inversion; that is, the second term is to the first as tlie fourth is to the third. Given a : b = c : d. To prove b:a = d:c. ARGUMENT a : b = c : d. .-. ad = bc. Q.E.D. REASONS 1. By hyp. 2. 388. 3. 393. PROPOSITION IV. THEOREM 396. Tf four numbers are in proportion, they are in proportion by alternation; that is, the first term is to third as the second is to the fourth. Given a:b = c:d. To prove a : c = b : d. 1. 2. 3. ARGUMENT a : b = c : d. .. a: c = b :d. Q.E.D. REASONS 1. By hyp. 2. 388. 3. 393. Ex. 625. If x : J / = y : |, what is the value of the ratio x : y ? Ex. 626. Transform a : x = 4 : 3 so that x shall occupy in turn every place in the proportion. BOOK III 165 397. Many transformations may be easily brought about by the following method : (1) Reduce the conclusion to an equation in its simplest form ( 388), then from this derive the hypothesis. (2) Begin with the hypothesis and reverse the steps of (1). This method is illustrated in the analysis of Prop. V. PROPOSITION V. THEOREM 398. If four numbers are in proportion, they are in proportion by composition; that is, the sum of the first two terms is to the first (or second) term as the sum of the last two terms is to the third (or fourth) term. Given a:b = c:d. To prove : (a) a + &:o = c4-d:c; (b) a + b : b = c + d : d. I. Analysis (1) The conclusions required above, when reduced to equa- tions in their simplest forms, are as follows: (a) (b) 1. ac -\-ad = ac + bc. 1. bc + bd = ad + bd. 2. Whence ad = be. 2. Whence be = ad. 3. .: a:b = c:d. 3. .: a:b = c:d. (2) Now begin with the hypothesis and reverse the steps. II. Proof (d) ARGUMENT 1. a:b = c:d. 2. .-. ad = be. 3. ac = ac. 4. ..ac+ad=ac-)-bc] i.e. a(c + d) = c(a + 6). 5. .-.a +b: a = c + d : e. Q.E.D. REASONS 1. By hyp. 2. 388. 3. By iden. 4. 54, 2. 5. 393. (b) The proof of (b) is left as an exercise for the student. 166 PLANE GEOMETRY Ex. 627. If - = -, fin 2/5 x y PROPOSITION VI. THEOREM 399. If four numbers are in proportion, they are in proportion by division; that is, the difference of the first two terms is to the first (or second) term as the differ- ence of the last two terms is to the third (or fourth) term. Given a : b = c : d. To prove : (a) a b:a = c d : c ; (6) a b : b == c d : d. I. The analysis is left as an exercise for the student. II. Proof (a) ARGUMENT 1. a: b = c: d. 2. .-. ad = be. 3. ac = ac. 4. .-. ac ad = ac be, i.e. a(c d) = c(a 5). 5. .'. a b:a = c d: c. Q.E.D. REASONS 1. By hyp. 2. 388. 3. By iden. 4. 54,3. 5. 393. (b) The proof of (b) is left as an exercise for the student. Ex. 628. H?=|, find^rJ!; y 3 x y PROPOSITION VII. THEOREM 400. If four numbers are in proportion, tlwy are in proportion by composition and division; that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference. Given a : b = c : d. To prove d-}- b: a 6 = c + d:c d. BOOK III 167 ARGUMENT 1. a:b = c:d. o a -\-b c + d a c -? Anrl a-b c-d a c 4. a + b c + d. REASONS 1. By hyp. 2. 398. 3. 399. 4. 54, 8 a. PROPOSITION VIII. THEOREM 401. In a series of equal ratios the sum of any number of antecedents is to the sum of tfo corresponding conse- quents as any antecedent is to its consequent. Given a : b = c : d = e :/ = g : h. To prove a + c -f- e + g :b -\- d -f- / -f h = a:b. I. Analysis Simplifying the conclusion above, we have : ab -f- be -{- be -f- bg = ab + ad -f- of + ah. The terms required for the first member of this equation, and their equivalents for the second member, may be obtained from the hypothesis. II. Proof ARGUMENT REASONS ab = ab. 1. By iden. bc = ad. 2. 388. be = af. 3. 388. bg = ah. 4. 388.. 5. .-. ab + be -f be + bg = ab + ad +af+ ah ; 5. 54, 2. i.e. b (a +c -h e + r/) = a (b + d -f-/+ 7t). 6. .-. a+ c + e + 0:& + d+/+7i = a:6. 6. 393. Q.E.D. 168 PLANE GEOMETRY Ex. 629. With the hypothesis of Prop. VIII, prove a + c+e:b + d+f=g:h. Ex. 630. If - = - = - = , prove m ~ s + k ~P _ T t g q r t + g q Ex. 631. If = -, find -. HINT. Use Prop. VI. Ex. 632. If a : b c : d, show that b a: b + a = d c : d + c. Ex. 633. Given the proportion a : b = 11 : 6. Write the proportions that result from taking the terms (1) by inversion ; (2) by alternation ; (3) by composition ; (4) by division ; (5) by composition and division. PROPOSITION IX. THEOREM 402. The products of the corresponding terms of any number of proportions form a proportion. Given a : b = c : c7, e:f=g:h, i:j = k:l. To prove aei : bfj = cgk : dhl. 1. ARGUMENT a c / 2. i _ cgk t ' .e. aei : = Q.E.D. REASONS 1. By hyp. 2. 54, 7 a. 403. Cor. // four numbers are in proportion, equi- multiples of the first two and equimultiples of the last two are also in proportion. HINT. Given a : b = x : y. To prove am :bm = nx: ny. Ex. 634. ltr:s = m:t, is fr : qs = qm :ft ? Prove your answer. Ex. 635. If a : b 3 : 4 and x : y = 8 : 0, find the value of ax : by. BOOK III 169 Ex. 636. If four numbers are in proportion, equimultiples of the antecedents and equimultiples of the consequents are also in proportion. Ex. 637. If r : s = t : m, prove 3a + 2w:4s = 3r + 2:4r. Ex. 638. Iiw:x = y:z, prove ax + bz : ex -\- dz = aw + by : cw + dy. Ex. 639. If = - and - = - , prove that : - = - : - , and state the r v s w n s I w theorem thus derived. PROPOSITION X. THEOREM 404. If four numbers are in proportion, like powers of these numbers are in proportion, and so also are like roots. Given a : b c : d. To prove : (a) a p :b p = c p : d p . (b) Va : V& = Vc : - REASONS 1. By hyp. = i- e - v t Q.E.D. 3. 54, 13. Ex.640. s q Vs 3s Ex. 641. If =P = r - , prove that n q s n 405. Def. A continued proportion is a series of equal ratios in which the consequent of any ratio is the same number as the antecedent of the following ratio ; thus, a : b = c : d e : f= g:h\s merely a series of equal ratios, while a : 6 = 6 : c = c : d = d : is not only a series of equal ratios but a continued proportion as well. 170 PLANE GEOMETRY Ex. 642. In the continued proportion a :b = b : c = c:d = d:e t prove that : = !. = ; and =2!. c b 2 ' d 6 3 e 6* Ex. 643. If x 3 : ?/ 3 = 8 : 27, find - . y Ex.644. If ^/m:l = Vn : 16, find -. 406. Def. The segments of a line are the parts into which it is divided. The line AE is divided internally at C if this point is between the extremities of the line. The segments into which it is divided are A C and CB. D A C B AB is divided externally at D if this point is on the prolonga- tion of the line. The segments are AD and DB. It should be noted that in either case the point of division is one end of each segment. 407. Def. Two straight lines are divided proportionally if the ratio of one line to either of its segments is equal to the ratio of the other line to its corresponding segment. 408. In Prop. XT, II, the following theorems (Appendix, 586 and 591) will be assumed : (a) The quotient of a variable by a constant is a variable. (b) The limit of the quotient of a variable by a constant is the limit of the variable divided by the constant. Thus, if x is a variable and k a constant : (1) - is a variable. K 1* ?/ (2) If the limit of x is ?/, then the limit of - is ^. /C K Ex. 645. Tn the figure of 409, name the segments into which AB is divided by D ; the segments into which AD is divided bv B. BOOK III 171 PROPOSITION XI. THEOREM 409. s 3. Through the several points of division on AB, as F, G, etc., draw lines || BC. 4. These lines are || DE and to each other. 5. .. AC is divided into r equal parts and AE into s equal parts. 6. .:^= r -. AE s 7. ... ^? = ^. Q.E.D. AD AE REASONS 1. 335. 2. 341. 3. 179. 4. 180. 5. 244. 6. 341. 7. 54,1, 172 PLANE GEOMETRY II. If AB and AD are incommensurable (Fig. 2). 1. 6. E FIG. 2. KC ARGUMENT Let m be a measure of AD. Apply m as a measure to AB as many times as possible. There will then be a remainder, HB, less than m. AH and AD are commensurable. Draw HK\\ BC. AH = AK ' AD~ AE' Now take a smaller measure of AD. No matter how small a measure of AD is taken, when it is applied as a measure to AB, the remainder, HB, will be smaller than the measure taken. .-. the difference between AH and AB may be made to become and remain less than any previously assigned line, however small. .-. AH approaches AB as a limit. ., approaches as a limit. AD AD 9. Likewise the difference between AK and AC may be made to become and remain less than any previously as- signed line, however small, REASONS 1. 339. 2. 337. 3. 179. 4. 409, I. 5. 335. 6. Arg. 5. 7. 349. 8. 408, b. 9. Arg. 6. BOOK III 173 ARGUMENT 10. .. AK approaches AC as a limit. AK A C 1 11. /. - - approaches - - as a limit. AE AE Apr A~K 12. But - - is always equal to . AD AE 1Q AB AC 10. .*. -- = - . Q.E.D. AD AE REASONS 10. 349. 11. 408,6. 12. Arg. 4. 13. 355. 410. Cor. A straight line parallel to one side of a tri- angle divides the other two sides into segments which are proportional. Thus, in the figures for Prop. XI, AD : DB = AE : EC. HINT. Prove by using division. Ex. 646. Using Fig. 2 of Prop. XI, prove (1) AB:AC=AD:AE; (2) AB : AC = DB : EC. Ex. 647. In the diagram at the right, w, n, and r are parallel to each other. Prove that a : b \ m = c : d ; also that a : c = b : d. \ Ex. 648. If two sides of a triangle are 12 P\ inches and 18 inches, and if a line is drawn paral- \ lei to the third side and cuts off 3 inches from the vertex on the 12-inch side, into what segments will it cut the 18-inch side ? Ex. 649. If two lines are cut by any number of parallels, the two lines are divided into segments which are proportional : (a) if the two lines are parallel ; (6) if the two lines are oblique. Ex. 650. Jf through the point of intersection of the medians of a triangle a line is drawn parallel to any side of the triangle, this line divides the other two sides in the ratio of 2 to 1. Ex. 651. A line can be divided at but one point into segments which have a given ratio (measured from one end). Ex. 652. A line parallel to the bases of a trapezoid divides the other two sides and also the two diagonals proportionally. Ex. 653. Apply the proof of Prop. XI to the case in which the par- allel to the base cuts the sides prolonged : (a) through the ends of the base ; (&) through the vertex. 174 PLANE GEOMETRY PROPOSITION XII. PROBLEM 411. To construct the fourth proportional to three given lines. R \M b \L S Given lines a, b, and c. To construct the fourth proportional to a, b, and c. I. Construction 1. From any point, as R, draw two indefinite lines RS and RT. 2. On RS lay off RM = a and ML = b. 3. On RT lay off RF = c. 4. Draw MF. 5. Through L construct LG II MF. 188. 6. FG is the fourth proportional to a, b, and c. II. The proof and discussion are left as an exercise for the student. 412. Question. Could the segments a, ft, and c be laid off in any other order ? 413. Cor. I. To construct the third proportional to two given lines. 414. Cor. II. To divide a given line into segments pro- portional to two or more given lines. Ex. 654. Divide a given line into segments in the ratio of 3 to 5. Ex. 655. Divide a given line into segments proportional to 2, 3, and 4. BOOK III 175 Ex. 656. Construct two lines, given their sum and their ratio. Ex. 657. Construct two lines, given their difference and their ratio. Ex. 658. If a, &, and c are three given lines, construct x so that : (a) x:a = b:c; (6) x = - PROPOSITION XIII. THEOREM (Converse of Prop. XI) 415. If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. F E C Given A ABC. and DE so drawn that = . AD AE To prove DE II EG. ARGUMENT 1. DE and EC are either II or not II. 2. Suppose that DE is not II BC, but that some other line through D, as DF, is II BC. 3rp-i AE AC> . Inen = . AD AF 4. But ^ = ^. AD AE 5. .'. AF = AE. 6. This is impossible. 7. .*. DE II BC. Q.E.D. REASONS 1. 161, a. 2. 179. 3. 409. 4. By hyp. 5. 391. 6. -54,12. 7. 161, b. 416. Cor. If a straight line divides two sides of a tri- angle into segments which are proportional;, it is parallel to the third side. Thus, if AD : DB = AE : EC, DE is II BC. 176 PLANE GEOMETRY Ex. 659. In the diagram at the right, if AB = 15, AC - 12, AD = 10, and AE = 8, prove DE parallel to BC. Ex. 660. If AB = 60, DB = 15, AE = 28, and EC = 12, is DE parallel to BC ? Prove. ^1 J* Ex. 661. If DE || BC, AB = 25, DB = 5, and AC = 20, find AE. Ex. 662. If DE || BC, AB= 80, ^Z> = 25, and EC = 4, find SIMILAR POLYGONS 417. Def. If the angles of one polygon, taken in order, are equal respectively to those of another, taken in order, the poly- gons are said to be mutually equiangular. The pairs of equal angles in the two polygons, taken in order, are called homolo- gous angles of the two polygons. 418. Def. If the sides of one polygon, taken in order as antecedents, form a series of equal ratios with the sides of another polygon, taken in order as consequents, the polygons are said to have their sides proportional. Thus, in the accom- panying figure, if a : I = b : m = c : n d : o = e : p, the two polygons have their sides proportional. The lines forming any ratio are called homologous lines of the two polygons, and the ratio of two such lines is called the ratio of similitude of the polygons. 419. Def. Two polygons are similar if they are mutually equiangular and if their sides are proportional. BOOK III 177 PROPOSITION XIV. THEOREM 420. Two triangles which are mutually equiangular are similar. H A K C Given A ABO and DEF with = Zz>, Z = Z #, and To prove A ABC ~ A DEF. ARGUMENT 1. Place A DEF on A AB c so that Z D shall coincide with Z^4, DE falling on AB and DF on .4(7. Represent A DEF in its new position by A AHK. 2. Z = Z = Z 3. .\HK\\BC. 5. 6. By placing A D^ on A ABC so that Z.E shall coincide with Z.B, it may be shown that AB BC 7. AB__BC!__ AC ' DE EF DF 8. .'. A ABC~ A DEF. Q.E.D. REASONS 1. 54,14. 2. By hyp. 3. 184. 4. 409. 5. 309. 6. By steps sim- ilar to 1-5. 7. 54, 1, 8. 419. 421. Cor. i. if two triangles have two angles of one equal respectively to two angles of the other, the triangles are similar. 178 PLANE GEOMETRY 422. Cor. II. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. 423. Cor. in. If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms * a triangle which is similar to the given triangle. Ex. 663. Upon a given line as base construct a triangle similar to a given triangle. Ex. 664. Draw a triangle ABC. Estimate the lengths of its sides. Draw a second triangle DEF similar to ABC and having DE equal to two thirds of AB. Compute DF and EF. Ex. 665. Any two altitudes of a triangle are to each other inversely as the sides to which they are drawn. Ex. 666. At a cer- tain hour of the day a tree, BC, casts a shadow, CA. At the same time a vertical pole, ED, casts a shadow, DF. What measurements are necessary to determine the height of the tree ? Ex. 667. If CA is found to be 64 feet ; DF, 16 feet ; ED, 10 feet ; what is the height of the tree ? Ex. 668. To find the distance across a river from A to B, a point C was located so that BC was perpendicular to AB at B. CD was then measured off 100 feet in length and perpendicular to BC at C. The line of sight from D to A intersected BC at E. By measurement CE was found to be 90 feet and EB 210 feet. What was the distance across the river ? B BOOK III 179 Ex. 669. Two isosceles triangles are similar if the vertex angle of one equals the vertex angle of the other, or if a base angle of one equals a base angle of the other. 424. It follows from the definition of similar polygons, 419, and from Prop. XIV that: (1) Homologous angles of similar triangles are equal. (2) Homologous sides of similar triangles are proportional. (3) Homologous sides of similar triangles are the sides opposite equal angles. 425. Note. In case, therefore, it is desired to prove four lines pro- portional, try to find a pair of triangles each having two of the given lines as sides. If, then, these triangles can be proved similar, their homologous sides will be proportional. By marking with colored crayon the lines required in the proportion, the triangles can readily be found. If it is desired to prove the product of two lines equal to the product of two other lines, prove the four lines proportional by the method just sug- gested, then put the product of the extremes equal to the product of the means.* 426. Def. The length of a secant from an external point to a circle is the length of the segment included between the point and the second point of intersection of the secant and the circumference. Ex. 670. If two chords intersect within a circle, es- tablish a proportionality among the segments of the chords. Place the product of the extremes equal to the product of the means, and state your result as a theorem. Ex. 671. If two secants are drawn from any given point to a circle, what are the segments of the secants? Does the theorem of Ex. 670 still hold with regard to them? Ex. 672. Kotate one of the secants of Ex. 671 about the point of intersection of the two until the rotating secant becomes a tangent. What are the segments of the secant which has become a tangent ? Does the theorem of Ex. 670 still hold ? Prove. * By the product of two lines is meant the product of their measure-numbers. This will be discussed again in Book IV. 180 PLANE GEOMETRY PROPOSITION XV. THEOREM 427. Two triangles which have their sides proportional are similar. H K Given A DEF and ABC such that = AB To prove A DEF ~ A ABC. ARGUMENT 1. On DE lay off DH AB, and on DF lay off DK=AC. 2. Draw HK. DE _DF AB AC DE ^ DF DH~ DK 5. .'. HK II EF. 6. .'. A DEF ~ A DHK. It remains to prove A DHK = A ABC. 7. ^- , . V-^- , ^ 3. 4 . VE _ DF 8. BJ3* ' pjf' BC^' 9. .:HK = BC. 10. Kow J9^= AB and DK= AC. 11. .-. A J)HK=A ABC. 12. But A 7)F~ A Dfl-iT. 13. .-. A DEF ~ A ^#6". Q.E.D. ^(7 J:C7 REASONS 1. 54,14. 2. 54, 15. 3. By hyp. 4. 309. 5. 415. 6. 423. 7. 424, 2. 8. By hyp. 9. 391. 10. Arg. 1. 11. 116. 12. Arg. 6. 13. 309. C Ex. 673. Tf the sides of two triangles are 0, 12, 15, and 6, 8, 10, respectively, are the triangles similar ? Explain. BOOK III 181 Ex. 674. Construct a triangle that shall have a given perimeter and shall be similar to a given triangle. Ex. 675. Construct a trapezoid, given the two bases and the two diagonals. HINT. How do the diagonals of a trapezoid divide each other ? PROPOSITION XVI. THEOREM 428. If two triangles have an angle of one equal to an angle of the other, and the including sides propor- tional, the triangles are similar. A K C D Given A ABC and DEF with /. A = * To prove A AB C ~ A DJF. _ -, ^B ^C D and = DE DF 3. 4. 5. 6. 7. ARGUMENT Place A DEF on A 45(7 so that Z /> shall coincide with Z ^4, D-E falling on AB, and D-F on .4(7. Represent A DEF in its new position by A AHK. AB _AC DE DF AB _AC ' AH~~ AK .'. HK || BC. .'. A ABC ~ A AHK. But A AHK is A DEF transferred to a different position. .-. A ABO ~ A DEF. Q.E.D. REASONS 1. 54, 14. 2. By hyp. 3. 309. 415. 423. Arg. 1. 7. 309. Ex. 676. Two triangles are similar if two sides and the median drawn to one of these sides in one triangle are proportional to two sides and the, corresponding median in the other triangle. 182 PLANE GEOMETRY B B D C FIG. 1. E FIG. 2. AD C FIG. 3. Ex. 677. In triangles AB C and DB (7, Fig. 1,AB = AC and BD = BC. Prove triangle ABC similar to triangle DU<7. Ex.678. InVlg.2,AB:AC = AE:AD. Prove triangle AB (7 simi- lar to triangle ^IDJE 1 . Ex. 679. If in triangle ABC, Fig. 3, (L4 = BC, and if Z) is a point such that CA : AB = AB : AD, prove AB = BD. Ex. 680. Construct a triangle similar to a given triangle and having the sum of two sides equal to a given line. PROPOSITION XVII. THEOREM 429. Two triangles that have their sides parallel each to each, or perpendicular each to each, are similar. FIG. 1. Given A ABC and A'B'C', with AB, BC, and CA II (Fig. 1) or _L (Fig. 2) respectively to A'B', B'C', and C'A', To prove A ABC ~ A A'B'C'. ARGUMENT 1. AB, BC, and CA are II or _L respectively to ^'7?', B'C', and C y '.4 f . 2. .*. A A, B, and C are equal respectively or are sup. respectively to A A', B', and c'. REASONS 1. By hyp. 2. 198,201. BOOK III 183 ARGUMENT 3. Three suppositions may be made, there- fore, as follows : = 2 rt. A, Z C + Z c' = 2 rt. A. Z c + Z (7' = 2 rt. A (3) Z J. = Z ^', Z = Z '; hence, also, Z c y == Z <7'. 4. According to (1) and (2) the sum of the A of the two A is more than four rt. A. 5. But this is impossible. 6. .. (3) is the only supposition admissible; i.e. the two A are mutually equi- angular. 7. .-. A ABC~ AA'B'C'. Q.E.D. REASONS 3. 161, a. 4. 54, 2. 5. 204. 6. 161, b. 7. 420. 430. Question. Can one pair of angles in Prop. XVII be supple- mentary and the other two pairs equal ? SUMMARY OF CONDITIONS FOB SIMILARITY OF TRIANGLES 431. I. Two triangles are similar if they are mutually equi- angular. (a) Two triangles are similar if two angles of one are equal respectively to two angles of the other. (6) Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. (c) If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle. II. Two triangles are similar if their sides are proportional. III. Two triangles are similar if they have an angle of one equal to an angle of the other, and -the including sides pro- portional. IV. Two triangles are similar if their sides are parallel each to each, or perpendicular each to each. 184 PLANE GEOMETRY Ex. 681. Inscribe a triangle in a circle and circumscribe about the circle a triangle similar to the inscribed triangle. Ex. 682. Circumscribe a triangle about a circle and inscribe in the circle a similar triangle. Ex. 683. The lines joining the mid-points of the sides of a triangle form a second triangle similar to the given triangle. Ex. 684. ABC is a triangle inscribed in a circle. A line is drawn from A to P, any point of BC, and a chord is drawn from B to a point Q in arc BC so that angle ABQ equals angle APC. Prove AB x AC = AQ x AP. PROPOSITION XVIII. THEOREM 432. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the other two sides. A PC Given A ABC with BP the bisector of Z ABC. To prove AP : PC = AB : B C. ARGUMENT 1. Through C draw CE II PB, meeting AB prolonged at E. 2. In A AEG, AP : PC AB : BE. 3. Now Z 3 = Z 1. 4. AndZ4 = Z2. 5. ButZl=Z2. 6. .-. Z3 = Z4. 7. .'. BC= BE. 8. .'. AP : PC = AB : BC. Q.E.D. REASONS 1. 179. 2. 410. 3. 190. 4. 189. 5. By hyp. 6. 54, 1. 7. 162. 8. 309. BOOK III 185 "El*-. 685. The sides of a triangle are 8, 12, and 15. Find the seg- ments of side 8 made by the bisector of the opposite angle. Ex. 6860 In the triangle of Ex. 685, find the segments of sides 12 and 15 made by the bisectors of the angles opposite. PROPOSITION XIX. THEOREM 433. The bisector of an exterior angle of a triangle di- vides the opposite side externally into segments which are proportional to the other two sides. Given A ABC, with BP the bisector of exterior Z CBF. To prove AP : PC AB : BC. ARGUMENT 1. Through C draw CE II PB, meeting AB a**. 2. Then in A ABP, AP : PC = AB : BE. Now Z3 = Z1. And Z4 = Z2. But Z1=Z2. .-. Z3 = Z4. .'. BC = BE. /. AP : PC = AB : BC. Q.E.D. REASONS 1. 179. 2. 409. 3. 190. 4. 189. 5. By hyp. 6. 54, 1. 7. 162. 8. 309. Ex. 687. Compare the lettering of the figures for Props. XVIII and XIX, and also the steps in the argument. Could one argument serxe for the two cases ? Ex. 688. The sides of a triangle are 9, 12, and 16. Find the segments of side 9 made by the bisector of the exterior angle at the opposite vertex. 186 PLANE GEOMETRY 434. Def. A line is divided harmonically if it is divided internally and externally into segments whose ratios are numeri- cally equal ; thus, if line AC is divided internally at P and ex- ternally at Q so that the ratio of ( f t AP to PC is numerically equal to ^ PC Q the ratio of AQ to QC, AC is said to be divided harmonically. Ex. 689. The bisectors of the interior and exterior angles at any vertex of a triangle divide the opposite side harmonically. Ex. 690. Divide a given straight line harmonically in the ratio of 3 to 6 ; in the ratio of a to &, where a and b are given straight lines. PROPOSITION XX. THEOREM 435. In two similar triangles any two homologous alti- tudes have the same ratio as any two homologous sides. A Given two similar A ABC and DEF, with two corresponding altitudes AH and DK. _ AH AB BC CA To prove = = = -r- , DK DE EF FD ARGUMENT 1. In rt. A ABH and DEK, Z. B = Z.E. 2. .-. AABH~ A DEK. o AH, opposite Z B_AB, opposite /.BHA ' ' DK, opposite Z.E DE, opposite s/EKD ^ , AB BC CA 4. But = = . EF FD DE AH _ AB DK ~ DE BC EF CA Q.E.D. REASONS 1. 424,1. 2. 422: 3. 424, 2. 4. 424,2. 5. 54, 1. BOOK HI 187 PROPOSITION XXI. THEOREM 436. If three or more straight lines drawn through a common point intersect two parallels, the corresponding segments of the parallels are proportional. / I \ \ / I \ \ Given lines PA, PB, PC, PD drawn through a common point P and intersecting the II lines AD and A'D' at points Aj B, (7, D and A'j B', C 1 , D', respectively. SOL C'D'' A 7? To prove - A'B' BC B'C' ARGUMENT 1. AD II A'D'. 2. .'.A APE ~ A A'PB'. AB _PB A'B'~ PB 1 ' 3. 4. Likewise A BPC 5. And -*= A J?'PC'. B'C' AB PB' BC A'B' B'C'' 1. Likewise it can be proved that BC _ CD B'C'~~ C'D'' 8. AB 'A T B' BC B'C' CD C'D' Q.E.D. REASONS 1. By hyp. 2. 423. 3. 424, 2. 4. Args. 1-2. 5. 424, 2. 6. 54, 1. 7. By steps sim- ilar to 1-6. 8. 54, 1. Ex. 691. If three or more non-parallel straight lines intercept pro- portional segments on two parallels, they pass through a common point. 188 PLANE GEOMETRY Ex. 692. A man is riding in an automobile at the uniform rate of 30 miles an hour on one side of a road, while on a footpath on the other side a man is walking in the opposite direction. If the distance between the footpath and the auto track is 44 feet, and a tree 4 feet from the footpath continually hides the chauffeur from the pedestrian, does the pedestrian walk at a uniform rate ? If so, at what rate does he walk ? Ex. 693. Two sides of a triangle are 8 and 11, and the altitude upon the third side is 6. A similar triangle has the side homologous to 8 equal to 12. Compute as many parts of the second triangle as you can. Ex. 694. In two similar triangles, any two homologous bisectors are in the same ratio as any two homologous sides. Ex. 695. In two similar triangles, any two homologous medians are in the same ratio as any two homologous sides. 437. Drawing to Scale. Measure the top of your desk. Make a drawing on paper in which each line is y 1 ^ as long as the corresponding line of your desk. Check your work by measuring the diagonal of your drawing, and the corresponding line of your desk. This is called drawing to scale. Map draw- ing is a common illustration of this principle. The scale of the drawing may be represented : (1) by saying, " Scale, yL " or " Scale, 1 inch to 12 inches " ; (2) by actually drawing the scale as indicated. Ex. 696. Using the scale above, draw lines on paper to represent 24 inches ; 3 feet 3 inches. Ex. 697. On the black- board draw, to the scale above, a circle whose diame- ter is 28 feet. Ex. 698. The figure represents a farm drawn to the scale indicated. Find the cost of putting a fence around the farm, if the fenc- ing costs $2.50 per rod. B BOOK III 189 PROPOSITION XXII. THEOREM 438. If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. H Given polygons ABODE and FGHIJ with A ABC ~ A FGHj AACD ~ A FHI, A ADE ~ A FIJ. To prove polygon ABODE ~ polygon FGHIJ. ARGUMENT REASONS 1. In A ABC and FGH, Z B = Z 0. 1. 424, 1. 2. AlsoZl = Z2. 2. 424,1. 3. In A ACD and OT7, Z 3 = Z4. 4. .-. Z1 + Z3 = Z2 + Z4. 5. .'. Z 5CZ* = Z (?#/. 6. Likewise Z (me = Z HIJ, Z ^ = Z J, 7. .*. polygons ABODE and FGHIJ are mutually equiangular. 8. In 2! 9. HF HI IF 10. And in A ADE and F/e7, = = . IF IJ JF EA 11 AB BC CD = DE FG ~~ GH ~~ HI~ IJ ~ JF' 12. .-. polygon ABODE ~ polygon FGHIJ. Q.E.D. 3. 424, 1. 4. 54, 2. 5. 309. 6. By steps sim- ilar to 1-5. 7. By proof. 8. 424, 2. 9. 424,2. 10. 424, 2. 11. 54,1. 12. 419. 190 PLANE GEOMETRY 439. Cor. Any two similar polygons may be divided into the same number of triangles similar each to each and similarly placed. PROPOSITION XXIII. PROBLEM 440. Upon a line homologous to a side of a given poly- gon, to construct a polygon similar to the given polygon. C "--* ^ D /O P.-L Given polygon AD and line JI/Q homol. to side AE. To construct, on MQ, a polygon ~ polygon AD. I. Construction 1. Draw all possible diagonals from A, as AC and AD. 2. At M, beginning with MQ as a side, construct A 7, 8, and 9 equal respectively to A 1, 2, and 3. 125. 3. At Q, with MQ as a side, construct Z 10 equal to Z 4, and prolong side QP until it meets ML at P. 125. 4. At P, with PM as a side, construct Zll equal to Z5, and prolong side PO until it meets MR at 0. 125. 5. At O, with OM as a side, construct Z12 equal to Z6, and prolong side ON until it meets MF at N. 125. 6. MP is the polygon required. II. Proof ARGUMENT 1. A ADE ~ A MPQ, A ACD ~ A MOP, and A ABC ~ A MNO. 2. .-. polygon MP ~ polygon AD. Q.E.D. REASONS 1. 421. 2. 438. BOOK III 191 III. The discussion is left as an exercise for the student. PROPOSITION XXIV. THEOREM 441. The perimeters of two similar polygons are to each other as any two homologous sides. d e Given ~ polygons P and Q, with sides a, b, c, d, e, and / hoinol. respectively to sides k, I, in, n, o, and p. perimeter of P _a To prove perimeter of Q k ARGUMENT a k I m n o p o . a + &+ c -f-d+e+/_a fc-f- I -f- m H- n -f- o -\-p k o mi 4. perimeter of P a 3. I hat is, i = -. perimeter oi Q k Q.E.D. REASONS 1. 419. 2. 401. 3. 309. Ex. 699. The perimeters of two similar polygons are 152 and 138 ; a side of the first is 8. Find the homologous side of the second. Ex. 700. The perimeters of two similar polygons are to each other as any two homologous diagonals. Ex. 701. The perimeters of two similar triangles are to each other as any two homologous medians. Ex. 702. If perpendiculars are drawn to the hypotenuse of a right triangle at its extremities, and if the other two sides of the triangle are prolonged to meet these perpendiculars, the figure thus formed contains five triangles each of which is similar to any one of the others. 192 PLANE GEOMETRY PROPOSITION XXV. THEOREM 442. In a right triangle, if the altitude upon the hypotenuse is drawn: I. The triangles thus formed are similar to the given triangle and to each other. II. The altitude is a mean proportional between the segments of the hypotenuse. III. Either side is a mean proportional between the whole hypotenuse and the segment of the hypotenuse adjacent to that side. f\ D Given rt. A ABC and the altitude CD upon the hypotenuse. To prove : I. A ^CR^^ABC, A ADC ~ A ABC, and A BCD ~ A ADC. II. BD : CD = CD : DA. III. AB : B C = BC : DB and AB : AC = AC : AD. I. ARGUMENT 1. In rt. A B CD and ABC, Z.B = Z.B. 2. .\A BCD ~ A ABC. N 3. In rt. A ^Z>(7 and ^B<7, Z.A/.A. 4. .-. A ^Z>, as center and with a radius equal to 07), describe a semicircumference. 3. At C construct CEA.AD, meeting the semicircumference at E. 148. 4. CE is the required line I. II. The proof and discussion are left as an exercise for the student. Ex. 706. By means of 444, II, construct a mean proportional be- tween two given lines by a method different from that given in Prop. XXVI. Ex. 707. Use the method of Prop. XXVI to construct a line equal to \/3 6, a and b being given lines. ANALYSIS. Let x = the required line. Then x = V3 ab. .-. x 2 = 3 ab. .'. 3 a : x = x : b. Ex. 708. Construct a line equal to a\/3, where a is a given line. Ex. 709. Using a line one inch long as a unit, construct a line equal to V3 ; \/5 ; V(>. Choosing your own unit, construct a line equal to 3v% 2V3, 5V6. BOOK in 195 PROPOSITION XXVII. THEOREM 446. In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. C Given rt. A ABC, with its rt. Z at (7. To prove a 2 -f- b 2 = c 2 . ARGUMENT 1. From C draw CD J_ AB forming seg ments I and d. a 2 = c L + b 2 = c c = c 2 . Q.E.D. REASONS 1. 155. 2. 443, II. 3. 443,11. 4. 54,2. 5. 309. 447. Cor. I. The square of either side of a right tri- angle is equal to the square of the hypotenuse minus the square of the otJier side. 448. Cor. II. The diagonal of a square is equal to its side 1 , multiplied by tlie square root of two. OUTLINE OF PROOF 2. Q.E.D. 449. Historical Note. The property of the right triangle stated in Prop. XXVII was known at a very early date, the ancient Egyptians, 2000 B.C., having made a right triangle by stretching around three pegs a cord measured off into 3, 4, and 5 units. See Note, Book IV, 510. 196 PLANE GEOMETRY Ex. 710. By means of 448, construct a line equal to V2 inches. Ex. 711. If a side of a square is 6 inches, find its diagonal. Ex. 712. The hypotenuse of a right triangle is 15 and one arm is 9. Find the other arm and the segments of the hypotenuse made by the perpendicular from the vertex of the right angle. Ex. 713. Find the altitude of an equilateral triangle whose side is 6 inches. Ex. 714. Find a side of an equilateral triangle whose altitude is 8 inches. Ex. 715. Divide a line into segments which shall be in the ratio of 1 to V2. Ex. 716. The radius of a circle is 10 inches. Find the length of a chord 6 inches from the center ; 4 inches from the center. Ex. 717. The radius of a circle is 20 inches. How far from the cen- ter is a chord whose length is 32 inches ? whose length is 28 inches ? Ex. 718. In a circle a chord 24 inches long is 5 inches from the center. How far from the center is a chord whose length is 12 inches ? 450. Def. The projection of a point upon a line is the foot of the perpendicular from the point to the line. 451. Def. The projection of a line segment upon a line is the segment of the second line included between the projections of the extremities of the first line upon the second. Thus, C is the projection of A upon MN, D is the projection of B upon MN, and CD is the projection of AB upon MN. K B c / \D MC D N M C D N M Ex. 719. In the figures above, under what condition will the projec- tion of AB on MN be a maximum ? a minimum ? Will the projection CD ever be equal to AB ? greater than AB ? Will the projection ever be a point ? Ex. 720. In a right isosceles triangle the hypotenuse of which is 10 inches, find the length of the projection of either arm upon the hypotenuse. Ex. 721. Find the projection of one side of an equilateral triangle upon another if each side is 6 inches. BOOK ITT 197 Ex. 722. Draw the projections of the shortest side of a triangle upon each of th3 other sides : (1) in an acute triangle ; (2) in a right triangle ; (3) in an obtuse triangle. Draw the projections of the longest side in each case. Ex. 723. Two sides of a triangle are 8 and 12 inches and their included angle is 60. Find the projection of the shorter upon the longer. Ex. 724. In Ex. 723, find the projection of the shorter side upon the longer if the included angle is 30 ; 45. Ex. 725. Parallel lines that have equal projections on the same line are equal. PROPOSITION XXVIII. PROBLEM 452. In any triangle to find the value of the square of the side opposite an acute angle in terms of the other two sides and of the projection of either of these sides upon the other. X D FIG. 1. P FIG. 2. X Given ABAX, with Z.X acute; and p, the projection of b upon a. To find the value of y? in terms of a, 6, and p. ARGUMENT 1. In rt. A BAD, x? AD 2 + D~B 2 . 2. But AD 2 = b 2 - p 2 . 3. And DB = a p (Fig. 1) or p a (Fig. 2). = a 2 - 2 ap + p 2 . REASONS 1. 446. 2. 447. 3. 54,11. 4. 54, 13. 5. 309. i.e. x 2 = a 2 -f b 2 2 ap. Q.E.F. 453. Question. Why is it not necessary to include here the figure and discussion for a right triangle ? 198 PLANE GEOMETRY 454. Prop. XXVIII may be stated in the form of a theorem as follows : In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides dimin- ished by twice the product of one of these sides and the projec- tion of the other side upon it. Ex. 726. If the sides of a triangle are 7, 8, and 10, is the angle opposite 10 obtuse, right, or acute ? Why ? Ex. 727. Apply the statement of Prop. XXVIII to the square of an arm of a right triangle. Ex. 728. Find x (in the figure for Prop. XXVIII) in terms of a and b and the projection of a upon b. Ex. 729. If the sides of a triangle are 13, 14, and 15, find the pro- jection of the first side upon the second. Ex. 730. If two sides of a triangle are 4 and 12 and the projection of the first side upon the second is 2, find the third side of the triangle. PROPOSITION XXIX. THEOREM 455. In any obtuse triangle, the square of the side oppo- site the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of these sides and the projection of the other side upon it. Given A BAX with /.X obtuse, and p, the projection of b upon a. To prove x* = a 2 + b 2 + 2 ap. BOOK III 199 ARGUMENT 1. In rt. A BAD, of = AD 2 -f- I)ff. 2. But AD 2 = b 2 -p 2 . 3. And DB = a+p. 5. .-. 2 = 6 2 -p 2 +a 2 + i.e. a^ = a 2 + 6 2 + 2 op. Q.E.D. REASONS 1. 446. 2. 447. 3. 54, 11. 4. 54,13. 5. 309. 456. From Props. XXVIII and XXIX, we may derive the following formulas for computing the projection of one side of a triangle upon another ; thus if a, 6, and c represent the sides of a triangle : From Prop. XXVIII, p = a ' + W ~ c \ (1) From Prop. XXIX, - p = a * + b * ~ c \ (2) a It is seen that the second members of these two equations are identical and that the first members differ only in sign. Hence, formula (1) may always be used for computing the length of a projection. It need only be remembered that if p is positive in any calculation, it indicates that the angle oppo- site c is acute; while if p is negative, the angle opposite c is obtuse. It can likewise be shown, (see Prop. XXVII) that if j> = 0, the angle opposite c is a right angle. Ex. 731. Write the formula for the projection of a upon 6. Ex. 732. In triangle ABC, a = 15, b = 20, c = 25 ; find the projec- tion of 6 upon c. Is angle A acute, right, or obtuse ? Ex. 733. In the triangle of Ex. 732, find the projection of a upon b. Is angle C acute, right, or obtuse ? Ex. 734. The sides of a triangle are 8, 14, and 20. Is the angle opposite the side 20 acute, right, or obtuse ? Ex. 735. If two sides of a triangle are 10 and 12, and their included angle is 120, what is the value of the third side ? Ex. 736.* If two sides of a triangle are 12 and 16, and their included angle is 45, find the third side. 200 PLANE GEOMETRY Ex. 737. If in triangle ABC, angle C = 120, prove that AB 2 = BC 2 + AC* + AC-BC. Ex. 738. If a line is drawn from the vertex C of an isosceles tri- angle ABC, meeting base AB prolonged at D, prove that ~CD 2 _ CB 2 = AD BD. PROPOSITION XXX. THEOREM 457. In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side increased by twice the square of the median upon that side. D B Given A ABC with m a , the median to side a. To prove b 2 + c 2 = 2f ^Y+ 2 m a 2 . AKGUMENT 1. Suppose b > c ; then Z ADC is obtuse and Z BDA is acute. 2. Let p be the projection of m a upon a. 3. Then from AJZ)C', 4. And from C 2 = REASONS 1. 173. 2. 451. 3. 455. 4. 454. Q.E.D. 5. 54, 2. BOOK III 201 458. Cor. The difference of the squares of two sides of any triangle is equal to twice the product of the third side and the projection of the median upon the third side. HINT. Subtract the equation in Arg. 4 from that in Arg. 3 ( 457) member from member. Ex. 739. Write the formula involving the median to 6 ; to c. Ex. 740. Apply Prop. XXX to a triangle right-angled at A ; at B ; at. C. 459. It will be seen that the formula of Prop. XXX con- tains the three sides of a triangle and a median to one of these sides. Hence, if the three sides of a triangle are given, the median to any one of them can be found ; also, if two sides and any median are given, the third side can be found. Ex. 741. If the sides of a triangle ABC are 5, 7, and 8, find the lengths of the three medians. Ex. 742. If the sides of a triangle are 12, 16, and 20, find the median to side 20. How does it compare in length with the side to which it is drawn ? Why ? Ex. 743. In triangle ABC, a - 16, b - 22, and m c = 17. Find c. Ex. 744. In a right triangle, right-angled at C, m c = 8 ; what is c ? Find one pair of values for a and b that will satisfy the conditions of the problem. Ex. 745. The sum of the squares of the four sides of any parallel- ogram is equal to the sum of the squares of its diagonals. Ex. 746. The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of its diagonals increased by four times the square of the line joining their mid-points. Ex. 747. Construct a triangle ABC, given &, c, and m a . Ex. 748. Construct a triangle ABC, given a, 6, and the projection of b upon a. Ex. 749. Compute the side of a rhombus whose diagonals are 12 and 10. Ex. 750. If the square of the longest side of a triangle is greater than the sum of the squares of the other two sides, is the triangle obtuse, right, or acute ? Why ? 202 PLANE GEOMETRY PROPOSITION XXXI. THEOREM 460. // through a point within a circle two chords are drawn, the product of the two segments of one of these chords is equal to the product of the two segments of the other. Given P, a point within circle 0, arid AB and CD, any two chords drawn through P. To prove PA PB PC - PD. The proof is left as an exercise for the student. HINT. Prove A APG ~ A PDB. Ex. 751. In the figure for Prop. XXXI, if PA = 5, PB = 12, and PD = 6, find PC. Ex. 752. In the same figure, if PC = 10, PD = 8, and AB = 21, find PA and PS. Ex. 753. In the same figure, if PC = 6, DC = 22, and AB = 20, find AP and PS. Ex. 754. In the same figure, if PA = m, PC = n, and PD = r, find PS. Ex. 755. If two chords intersecting within a circle are of lengths 8 and 10, and the second bisects the first, what are the segments of the second ? Ex. 756. By means of Prop. XXXI construct a mean proportional between two given lines. Ex. 757. If two chords intersect within a circle and the segments of one chord are a and b inches, while the second chord measures d inches, construct the segments of the second chord. HINT. Find the locus of the mid-points of chords equal to d. BOOK III 203 Ex. 758. If two lines AS and CD intersect at E so that AE - EB = CE ED, then a circumference can be passed through the four points A, B, C, D. PROPOSITION XXXII. THEOREM 461. // a tangent and a secant are drawn from any given point to a circle, the tangent is a mean propor- tional between the whole secant and its external segment. Given tangent PT and secant PB drawn from the point P to the circle 0. m To prove : PT PT PC 1. 2. 3. 4. 5. ARGUMENT Draw CT and BT. In A PET and CTP, Z P = Z P. Z2 = Z2'. .-. APBT~ A CTP. PB, opposite Z 3 in A PBT ' PT, opposite Z 3' in A CTP _ PT, opposite Z 2 in A PBT PC, opposite Z 2' in A CTP Q.E.D. REASONS 1. 54, 15. 2 By iden. 3. 362, a. 4. 421. ~ 5. 424, 2. 462. Cor. I. If a tangent and a secant are drawn from any given point to a circle-, the square of the tangent is equal to the product of tJie whole secant and its external segment. 204 PLANE GEOMETRY 463. Cor. II. If two or more secants are drawn from any given point to a circle, the product of any secant and its external segment is constant. Given secants PD, PE, PF, drawn from point P to circle 0, and let their external segments be denoted by PA, PB, P<7, respectively. To prove PD PA = PE PB = PF PC. ARGUMENT 1. From P draw a tangent to circle 0, as PT. 2. PT 2 = PD PA ; PT 2 = PE PB ; PT 2 = PF PC. 3. .'. PD PA = PE- PB PF PC. Q.E.D. D 1. 286. 2. 462. 3. 54,1. Ex. 759. If a tangent and a secant drawn from the same point to a circle measure 6 and 18 inches, respectively, how long is the external segment of the secant ? Ex. 760. Two secants are drawn to a circle from an outside point. If their external segments are 12 and 9, and the internal segment of the first secant is 8, what is the length of the second secant ? Ex. 761. The tangents to two intersecting circles from any point in their common chord (prolonged) are equal. Ex. 762. If two circumferences intersect, their common chord (pro- longed) bisects their common tangents. 464. Def. A line is said to be divided in extreme and mean ratio if it is divided into two parts such that one part is a mean proportional between the whole line and the other part. Thus, AB is divided in extreme PR and mean ratio at P if AB \ AP = AP ; PB. This division is known as the golden section. BOOK III 205 PROPOSITION XXXIII. PIOBLEM 465. To divide a line internally in extreme and mean ratio. XX / -.*' \ \\ i iV ' A P B Given line AB. To find, in AB, a point P such that AB : AP = AP : PB. I. Construction 1. From B draw SO _L .4B and = \AB. 148. 2. With as center and with BO as radius describe a cir- cumference. 3. From A draw a secant through 0, cutting the circumfer- ence at C and Z). 4. With A as center and with AC as radius draw CP, cutting ^45 at P. 5. AB : AP = AP : P. II. Proof ARGUMENT 1. AB is tangent to circle O. 2. .-. AD : AB = AB : AC. 3. .'. AD AB : AB = AB AC : AC. 4. .'. AD CD : AB = AB ^P : AP. 5. .'. AP: ^/? = PJ3~: AP. 6. .'. ^: ^P = ^P: PB. REASONS 1. 314. 2. 461. 3. 399_ 4. 809. 5. 309. 6. 395. Q.E.D. III. The discussion is left as an exercise for the student. 206 PLANE GEOMETRY Ex. 763. Divide a line AB externally in extreme and mean ratio. HINT. In the figure for Prop. XXXIII prolong BA to J", making P'A = AD. Then prove AB -. P'A = P'A : P'B. Ex. 764. If the line I is divided internally in extreme and mean ratio, and if s is the greater segment, find the value of s in terms of I. HINT. I : s = s : I s. Ex. 765. A line 10 inches long is divided internally in extreme and mean ratio. Find the lengths of the two segments. Ex. 766. A line 8 inches long is divided externally in extreme and mean ratio. Find the length of the longer segment. MISCELLANEOUS EXERCISES Ex. 767. Explain how the accompanying figure can be used to find the distance from A to B on opposite sides of a hill. CE = \BC, CD = \ AC. ED is found by measurement to be 125 feet. What is the distance AB ? Ex. 768. A little boy wished to obtain the height of a tree in his yard. He set up a ver- tical pole 6 feet high and watched until the shadow of the pole measured exactly 6 feet. He then measured quickly the length of the tree's shadow and called this the height of the tree. Was his answer correct ? Draw figures and explain. Use this method for measuring the height of your school building and flag pole. Ex. 769. If light from a tree, as AB, is allowed to pass through a small aperture 0, in a window shutter W, and strike a white screen or wall, an inverted image of the tree, as CD, is formed on the screen. If the dis- tance OE= 30 feet, OF = 8 feet, and the length of the tree AB= 35 feet, find the length of the image CD. Under what condition will the length 'of the image equal the length of the tree ? This exercise illustrates the principle of the photographer's camera. Ex. 770. By means of Prop. XXXII construct a mean proportional between two given lines. Ex. 771. In a certain circle a chord 5\/5 inches from the center is 20 inches in length. Find the length of a chord 9 inches from the center. BOOK III 207 Ex. 772. Compute the length of : (1) the common external tangent, (2) the common internal tangent, to two circles whose radii are 8 and 6, respectively, and the distance between whose centers is 20. Ex. 773. If the hypotenuse of an isosceles right triangle is 16 inches, what is the length of each arm ? Ex. 774. If from a point a tangent and a secant are drawn and the segments of the secant are 4 and 12, how tong is the tangent ? Ex. 775. Given the equation m + n = ; solve for x. X + C C Ex. 776. Find a mean proportional between a 2 + 2 ab + 6 2 and a 2 - 2 ab + 6 2 . Ex. 777. The mean proportional between two unequal lines is less than half their sum. Ex. 778. The diagonals of a trapezoid divide each other into segments which are proportional. Ex. 779. ABC is an isosceles triangle inscribed in a circle. Chord BD is drawn from the vertex B, cutting the base in any point, as E. Prove BD : ABAB : BE. "" ""/) Ex. 780. In a triangle ABO the side AB is 305 feet. If a line parallel to B C divides AC in the ratio of 2 to 3, what are the lengths of the segments into which it divides AB ? Ex. 781. Construct, in one figure, four lines whose lengths shall be that of a given unit multiplied by V2, \/3, 2, Vs"", respectively. Ex. 782. Two sides of a triangle are 12 and 18 inches, and the perpen- dicular upon the first from the opposite vertex is 9 inches. What is the length of the altitude upon the second side ? Ex. 783. If a : b = c : d, show that . Also translate this fact into a verbal statement. Ex. 784. If a constant is added to or subtracted from each term of a proportion, will the resulting numbers be in proportion ? Give proof. Ex. 785. If r : s = t : g, is 3 r + - : s = 1 1 : 2 q ? Prove your answer. Ex. 786. One segment of a chord drawn through a point 7 units from the center of a circle is 4 units long. If the diameter of the circle is 15 units, what is the length of the other segment ? Ex. 787. The non-parallel sides 'of a trapezoid and the line joining the mid-points of the parallel sides, if prolonged, are concurrent. 208 PLANE GEOMETRY Ex. 788. Construct a circle which shall pass through two given points and be tangent to a given straight line. Ex. 789. The sides of a triangle are 10, 12, 15. Compute the lengths of the two segments into which the least side is divided by the bisector of the opposite angle. . Ex. 790. AB is a chord of a circle, and CE is any chord drawn through the middle point C of arc AB, cut- ting chord AB at D. Prove GE : CA = CA : CD. Ex. 791. Construct a right triangle, given its perime- ter and an acute angle. Ex. 792. The base of an isosceles triangle is a, and the perpendicular let fall from an extremity of the base to the opposite side is 6. Find the lengths of the equal sides. Ex. 793. AD and BE are two altitudes of triangle CAB. Prove that AD: BE= CA-.BC. Ex. 794. If two circles touch each other, their common external tangent is a mean proportional between their diameters. Ex. 795. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Ex. 796. If three circles intersect each other, their common chords pass through a common point. Ex. 797. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments of the third side made by the bisector. Given A ABC with t, the bisector of Z J?, dividing side 6 into the two segments s and r. To prove & = ac rs. OUTLINE OF PROOF 3. Prove that a : t = t-\- m : c. Then ac = i 2 + tm = t 2 + rs. .'. t 2 = ac rs. Ex. 798. In any triangle the product of two sides is equal to the product of the altitude upon the third side and the diameter of the circumscribed circle. HINT. Prove A ABD ^ l^EBC. Then prove ac = hd. BOOK IV AREAS OF POLYGONS 466. A surface may be measured by finding how many times it contains a unit of surface. The unit of surface most fre- quently chosen is a square whose side is of unit length. If the unit length is an inch, the unit of surface is a square whose side is an inch. Such a unit is called a square inch. If the unit length is a foot, the unit of surface is a square whose side is a foot, and the unit is called a square foot. 467. Def. The result of the measurement is a number, which is called the measure-number, or numerical measure, or area of the surface. B C A D FIG. 1. Rectangle AC = 15 U. 468. Thus, if the square U is contained in the rectangle ABCD (Fig. 1) 15 times, then the measure-number or area of rectangle ABCD, in terms of U, is 15. If the given square is not contained in the given rectangle an integral number of 209 210 PLANE GEOMETRY G E H FIG. 2. Rectangle EG = 8U+. times without a remainder (see Fig. 2), then by taking a square which is an aliquot part of U, as one fourth of u, and applying M E FIG. 3. Rectangle EG = H U+= 11J U+. it as a measure to the rectangle (see Fig. 3) a number will be obtained which, divided by four,* will give another (and G FIG. 4. Rectangle A G = *?$ > 7+ = 111 U+ . usually closer) approximate area of the given rectangle. By proceeding in this way (see Fig. 4), closer and closer approxi- mations of the true area may be obtained. * It takes four of the small squares to make the unit itself. BOOK IV 211 469. If the sides of the given rectangle and of the unit square are commensurable, a square may be found which is an aliquot part of U, and which is contained in the rectangle also an integral number of times. 470. If the sides of the given rectangle and of the unit square are incommensurable, then closer and closer approxima- tions to the area may be obtained, but no square which is an aliquot part of U will be also an aliquot part of the rectangle (by definition of incommensurable magnitudes). There is, how- ever, a definite limit which is approached more and more closely by the approximations obtained by using smaller and smaller subdivisions of the unit square, as these subdivisions approach zero as a limit.* 471. Def. The measure-number, or area, of a rectangle which is incommensurable with the chosen unit square is the limit which successive approximate measure-numbers of the rec- tangle approach as the subdivisions of the unit square approach zero as a limit. For brevity the expression, the area of a figure, is used to mean the measure-number of the surface of the figure with respect to a chosen unit. 472. Def. The ratio of any two surfaces is the ratio of their measure-numbers (based on the same unit). 473. Def. Equivalent figures are figures which have the same area. The student should note that : Equal figures have the same shape and size; such figures can be made to coincide. Similar figures have the same shape. Equivalent figures have the same size. Ex. 799. Draw two equivalent figures that are not equal. * For none of these approximations can exceed a certain fixed number, for example (h + \)(b + \), where the measure applied is contained in the altitude h times with a re- mainder less than the measure, and in the base b times with a remainder less than the measure. 212 PLANE GEOMETRY Ex. 800. Draw two equal figures on the blackboard or cut them out of paper, and show that equal figures may be added to them in such a way that the resulting figures are not equal. Are they the same size ? Ex. 801. Draw figures to show that axioms 2, 7, and 8, when applied to equal figures, do not give results which satisfy the test for equal figures. s ni FIG. 1. FIG. 2. Ex. 802. Fig. 1 above represents a card containing 64 small squares, cut into four pieces, I, II, III, and IV. Fig. 2 represents these four pieces placed together in different positions forming, as it would seem, a rectangle containing 65 of these small squares. By your knowledge of similar triangles, try to explain the fallacy in the construction. 474. Historical Note. Geometry is supposed to have had its origin in land surveying, and the earliest traditions state that it had its beginning in Egypt and Babylon. The records of Babylon were made on clay tablets, and give methods for finding the approximate areas of several rectilinear figures, and also of the circle. The Egyptian records were made on papyrus. Herodotus states that the, fact that the inundations of the Nile caused changes in the amount of taxable land, rendered it neces- sary to devise accurate land measurements. This work was done by the Egyptian priests, and the earliest manu- script extant is that of Ahmes, who lived about 1700 B.C. This manu- script, known as the Rhind papyrus, is preserved in the British Museum. It is called "Directions for knowing all dark things," and is thought to be a copy of an older manuscript, dating about 3400 B.C. In addition to problems in arithmetic it contains a discussion of areas. Problems on pyramids follow, which show some knowledge of the properties of similar figures and of trigonometry, and which give dimensions, agreeing closely with those of the great pyramids of Egypt. The geometry of the Egyptians was concrete and practical, unlike that of the Greeks, which was logical and deductive, even from its beginning. BOOK IV 213 PROPOSITION I. THEOREM 475. The area of a rectangle is equal to the product of its base and its altitude. (See 476.) B C A D Given rectangle ABCD, with base AD and altitude AB, and let U be the chosen unit of surface, whose side is u. To prove the area of ABCD = AD AB. I. If AD and AB are each commensurable with u. (a) Suppose that u is contained in AD and AB each an in- tegral number of times. ARGUMENT 1. Lay off u upon AD and AB, respectively. - Suppose that u is contained in AD r times, and in AB s times. 2. At the points of division on AD and on AB erect J to AD and AB, respectively. 3. Then rectangle A BCD is divided into unit squares. 4. There are r of these unit squares in a row on AD, and s rows of these squares in rectangle A BCD. 5. .-. the area of ABCD = r . 6. But r and s are the measure-numbers of AD and AB, respectively, referred to the linear unit u. 7. .*. the area of ABCD = AD-AB, Q.E.D. REASONS 1. 335. 2. 63. 3. 466. 4. Arg. 1. 5. 467. 6. Arg. 1. 7. 309. 214 PLANE GEOMETRY (6) If u is not a measure of AD and AB, respectively, but if some aliquot part of u is such a measure. The proof is left to the student. B C M A QD II. If AD and AB are each incommensurable with u. ARGUMENT 1. Let m be a measure of u. Apply m as a measure to AD and AB as many times as possible. There will be a remainder, as QD, on AD, and a re- mainder, as PB, on AB, each less than m. 2. Through Q draw QM _L AD, and through P draw PM _L AB. 3. Now .4 Q and 4P are each commensu- rable with the measure m, and hence commensurable with u. 4. .-. the area of rectangle APM Q= A Q AP. 5. Now take a smaller measure of u. No matter how small a measure of u is taken, when it is applied as a measure to AD and .4Z?, the remainders, QD and PJ5, will be smaller than the measure taken. 6. .-. the difference between A Q and AD may be made to become and remain less than any previously assigned segment, however small. REASONS 1. 339. 2. 63. 3. 337. 4. 475, I. 5. 335. 6. Arg. 5. BOOK IV 215 ARGUMENT 7. Likewise the difference between AP and AB may be made to become and remain less than any previously as- signed segment, however small. 8. .. A Q approaches AD as a limit, and AP approaches AB as a limit. 9. .. AQ'AP approaches AD-AB as a limit. 10. Again, the difference between APMQ and A BCD may be made to become and remain less than any previously assigned area, however small. 11. .*. APMQ approaches ABCD as a limit. 12. But the area of APMQ is always equal to AQ - AP. 13. .-. the area of ABCD AD-AB. Q.E.D. 9. 10. 11. 12. REASONS Arg. 5. 349. 477. Arg. 5. 349. Arg. 4. 13. 355. III. If AD is commensurable with u but AB incommensu- rable with u. The proof is left as an exercise for the student. 476. Note. By the product of two lines is meant the product of the measure-numbers of the lines. The proof that to every straight line segment there belongs a measure-number is given in 595. 477. If each of any finite number of variables approaches a finite limit, not zero, then the limit of their product is equal to the product of their limits. (See 593.) 478. Cor. I. The area of a square is equal to the square of its side. 479. Cor. II. Any two rectangles are to each other as the products of their bases and their altitudes. OUTLINE OF PROOF. Denote the two rectangles by R and R', their bases by b and b', and their altitudes by h and h', re- spectively. Then R = b-h and R' = b'- h'. .-. f = -^A. 216 PLANE GEOMETRY 480. Cor. III. (a) Two rectangles having equal bases are to each other as their altitudes, and (b) two rectangles having equal altitudes are to each other as their bases. OUTLINE OF PROOF s R b h h v R b h b Ex. 803. Draw a rectangle whose base is 7 units and whose altitude is 4 units and show how many unit squares it contains. Ex. 804. Find the area of a rectangle whose base is 12 inches and whose altitude is 5 inches. Ex. 805. Find the area of a rectangle whose diagonal is 10 inches, and one of whose sides is 6 inches. Ex. 806. If the area of a rectangle is 60 square feet, and the base, 5 inches, what is the altitude ? Ex. 807. If the base and altitude of a rectangle are 2J inches and l\ inches, respectively, find the area of the rectangle. Ex. 808. Find the area of a square whose diagonal is 8 \/2 inches. Ex. 809. Find the successive approximations to the area of a rec- tangle if its sides are V 10 and V 5, respectively, using 3 times 2 for the first approximation, taking the square roots to tenths for the next, to hundredths for the next, etc. Ex. 810. Compare two rectangles if a diagonal and a side of one are d and s, respectively, while a diagonal and side of the other are d' and s 1 . Ex. 811. Construct a rectangle whose area shall be three times that of a given rectangle. Ex. 812. Construct a rectangle which shall be to a given rectangle in the ratio of two given lines, m and n. Ex. 813. Compare two rectangles whose altitudes are equal, but whose bases are 15 inches and 3 inches, respectively. Ex. 814. From a given rectangle cut off a rectangle whose area is two thirds that of the given one. Ex. 815. If the base and altitude of a certain rectangle are 12 inches and 8 inches, respectively, and the base and altitude of a scond rectangle are 6 inches and 4 inches, respectively, compare their areas. BOOK IV 217 PROPOSITION II. THEOUEM 481. The area of a parallelogram equals the product of its base and its altitude. R b C A E Given HH ARCD, with base 6 and altitude h. To prove area of A ROD, = b - h. ARGUMENT 1. Draw the rectangle ERCF, having b as base and h as altitude. 2. In rt. A DCF and ARE, DC = AR. Also CF = RE. .-. A DCF = A ARE. Now figure ARCF = ngure^flC^. .-. area of ARCD = area of ERCF. But area of ERCF = b - h. .-. area of ARCD = b - h. Q.E.D. REASONS 1. 223. 232. 232. 211. By iden. 54, 3. 475. 54, 1. 482. Cor. I. Parallelograms having equal bases and equal altitudes are equivalent. 483. Cor. n. Any two parallelograms are to each other as the products of their bases and their altitudes. HINT. Give a proof similar to that of 479. 484. Cor. III. (a) Two parallelograms having equal bases are to each other as their altitudes, and (b) two par- allelograms having equal altitudes are to each other as their bases. (HINT. Give a proof similar to that of 480.) 218 PLANE GEOMETRY PROPOSITION III. THEOREM 485. Tlw area of a triangle equals one half the product of its base and its altitude. 1. REASONS CB , and 1. 179. CA. Let 2. 220. f J. 3. 236. 4. 481. Q.E.I). o. 54, 8 a. A b C Given A ABC, with base b and altitude h. To prove area of A ABC = ^b- h. ARGUMENT Through A draw a line through B draw a line these lines intersect at X. 2. Then JA r (7 is a O. 3. /. A 4. But area of 5. .-. area of A 486. Cor. I. Triangles having equal bases and equal altitudes are equivalent. 487. Cor. II. Any two triangles arc to each other as the products of their bases and their altitudes. OUTLINE OF PROOF Denote the two A by T and T', their bases by b and b', and their altitudes by h and h', respectively. Then T = i b h and 'b^h 1 ' 488. Cor. III. (a) Two triangles having equal bases are to each other as their altitudes, and (b) two triangles having equal altitudes are to each as their bases. BOOK IV 219 OUTLINE OF PROOF ?~b.h'Ji T'b'.hb' 489. Cor. IV. A triangle is equivalent to one half of a parallelogram having the same base and altitude. Ex. 816. Draw four equivalent parallelograms on the same base. Ex. 817. Find the area of a parallelogram having two sides 8 inches and 12 inches, respectively, and the included angle 60. Find the area if the included angle is 45. Ex. 818. Find the ratio of two rhombuses whose perimeters are 24 inches and 16 inches, respectively, and whose smaller base angles are 30. Ex. 819. Find the area of an equilateral triangle having a side equal to 6 inches. Ex. 820. Find the area of an equilateral triangle whose altitude is 8 inches. Ex. 821. Construct three or more equivalent triangles on the same base. Ex. 822. Find the locus of the vertices of all triangles equivalent to a given triangle and standing on the same base. Ex. 823. Construct a triangle equivalent to a given triangle and hav- ing one of its sides equal to a given line. Ex. 824. Construct a triangle equivalent to a given triangle and hav- ing one of its angles equal to a given angle. Ex. 825. Construct a triangle equivalent to a given triangle and hav- ing two of its sides equal, respectively, to two given lines. Ex. 826. Divide a triangle into three equivalent triangles by drawing lines through one of its vertices. Ex. 827. Construct a triangle equivalent to f of a given triangle ; | of a given triangle. Ex. 828. Construct a triangle equivalent to a given square. Ex. 829. The area of a rhombus is equal to one half the product of its diagonals. Ex. 830. If from any point in a diagonal of a parallelogram lines are drawn to the opposite vertices, two pairs of equivalent triangles are formed. Ex. 831. Two lines joining the mid-point of the diagonal of a quadri- lateral to the opposite vertices divide the figure into two equivalent parts. Ex. 832. Find the area of a triangle if two of its sides are 6 inches and 9 inches, respectively, and the. included angle is 60. 220 PLANE GEOMETRY PROPOSITION IV. PROBLEM 490. To derive a formula for the area of a triangle in terms of its sides. C B Given A ABC, with sides ft, 6, and c. To derive a formula for the area of A ABC in terras of a, 6, and c. ARGUMENT KEASONS 1. Let h a denote the altitude upon , p the pro- 1. 485. jection of b. upon ft, and T the area of A ABC. Then To L 2. h*=V-p> = 3. (Fig. 2). Va . '2a _(ft-|- 6 + c)(ft-f- b c)(c-\-a 6)(c ft 4ft 2 ^ /(g-f-64.c)(ft-f6-c)(c+ft-'^)(c-ft 4ft 2 6. Now let ft + & -f- c = 2 s. Then a + b c = 2s 2c = 2(s c"); 2. 447. 3. 456. 4. 309. 5. 54, 13. 6. 54, 3, BOOK IV 221 ARGUMENT -6)2(.-c) 7.ThenA. = > /2i2)= i(2*)r = w. a) (a ft) < s c) Q.E.F. Ex. 838. Derive a formula for the radius of a circle circumscribed about a triangle, in terms of the sides of the triangle. OUTLINE OF SOLUTION (See figure for Ex. 798.) 1. dh = ac; i.e. dor2It=. h abc Q.E.F. 2 h 4V s(s a) (s 6) (s - c) Ex. 839. If the sides of a triangle are 9, 10, and 11, find the radius of the inscribed circle ; the radius of the circumscribed circle. Ex.840. The sides of a triangle are 3 a, 4 a, and 5 a. Find the radius of the inscribed circle ; the radius of the circumscribed circle. What kind of a triangle is it ? Verify your answer by comparing the radius of the circumscribed circle with the longest side. Ex. 841. Derive formulas for the bisectors of the angles of a triangle in terms of the sides of the triangle. OUTLINE OF SOLUTION (See figure for Ex. 797.) 1. tb 2 = ac rs. 2. Buta:c=r:s. 3. .. a -\- c : a =b : r. 4:. . ac (a ', + b + c)(a 6 -f c" K.ewise o. . . t ( a + c (a-fc) 2 a ) A (ZC6(^5 OJ M ^ . 8 T, ( + c) 2 ikp.wise t x// ( + c) 2 a + c :f>af r^ . Q.E.F. Find r, E, T, h a , w a , and , having given : Ex. 842. a = 11, b = 9, c = 1(5. What kind of an angle is C ? Ex. 843. a = 13, b = 15, c = 20. What kind of an angle is C ? Ex. 844. a = 24, b = 10, c = 26. What kind of an angle is C ? BOOK IV 223 TRANSFORMATION OF FIGURES 493. Def. To transform a figure means to find another figure which is equivalent to it. PROPOSITION VI. PKOBLEM 494. To construct a triangle equivalent to a given polygon. Given polygon ABCDEF. To construct a A =0= polygon ABCDEF. (a) Construct a polygon o ABCDEF, but having one side less. I. Construction 1. Join any two alternate vertices, as C and A. 2. Construct BG II CA, meeting FA prolonged at G. 188. 3. Draw CG. 4. Polygon GCDEF =0= polygon ABCDEF and has one side less. II. Proof ARGUMENT 1. A AGC and ABC have the same base CM, and the same altitude, the J_ be- tween the Us CA and BG. 2. .'. A AGC o A ABC. 3. -But polygon A CDEF = polygon A CDEF. 4. .-. polygon GCDEF =0= polygon ABCDEF. Q.E.D. REASONS 1. 235. 2. 486. 3. By iden. 4. 54, 2. 224 PLANE GEOMETRY G F H (b) In like manner, reduce the number of sides of the new polygon GGDEF until A DHK is obtained. The construction, proof, and discussion are left as an exercise for the student. Ex. 845. Transform a scalene triangle into an isosceles triangle. Ex. 846. Transform a trapezoid into a right triangle. Ex. 847. Transform a parallelogram into a trapezoid. Ex. 848. Transform a pentagon into an isosceles triangle. Ex. 849. Construct a triangle equivalent to f of a given trapezium. Ex. 850. Transform | of a given pentagon into a triangle. Ex. 851. Construct a rhomboid and a rhombus which are equivalent, and which have a common diagonal. PROPOSITION VII. THEOREM 495. The area of a trapezoid equals the product of its altitude and one half tfa sum of its bases. V D Given trapezoid AFED, with altitude h and bases b and b'. To prove area of AFED = | (b + b')h. BOOK IV 225 ARGUMENT 1. Draw the diagonal AE. 2. The altitude of A AED, considering b as base, is equal to the altitude of A AFE, considering b' as base, each being equal to the altitude of the trapezoid, h. 3. .'. area of A AED = ^ b - h. 4. Area of A AFE = *- b' - h. 5. .-. area of trapezoid AFED = %(b + b')h. Q.E.D. REASONS 1. 54, 15. 2. 235. 485. 485. 54, 2. 496. Cor. The area of a trapezoid equals the product of its altitude and its median. 497. Question. The ancient Egyptians, in attempting to find the area of a field in the shape of a trapezoid, multiplied one half the sum of the parallel sides by one of the other sides. For what figure would this method be correct ? Ex. 852. Find the area of a trapezoid whose bases are 7 inches and 9 inches, respectively, and whose altitude is 5 inches. Ex. 853. Find the area of a trapezoid whose median is 10 inches and whose altitude is 6 inches. Ex. 854. Through a given point in one side of a given parallelogram draw a line which shall divide the parallelogram into two equivalent parts. Will these parts be equal ? Ex. 855. Through a given point within a parallelogram draw a line which shall divide the parallelogram into two equivalent parts. Will these parts be equal ? Ex. 856. If the mid-point of one of the non-parallel sides of a trape- zoid is joined to the extremities of the other of the non-parallel sides, the area of the triangle formed is equal to one half the area of the trapezoid. Ex. 857. Find the area of a trapezoid whose bases are b and b' and whose other sides are each equal to s. Ex. 858. If the 'sides of any quadrilateral are bisected and the points of bisection joined, the included figure will be a parallelogram equal in area to half the original figure. 226 PLANE GEOMETRY PROPOSITION VIII. THEOREM 498. Two triangles which have an angle of one equal to an angle of the other are to each other as tJie products of the sides including the equal angles. B Given A AB C and DEF, with Z A = A ABC AC AB To prove A DEF DF DE ARGUMENT 1. Let h be the altitude of A ABC upon side AC, and h' the altitude of A DEF upon side DF. Then, A ABC AC h AC REASONS 487. A DEF DF h 1 DF h 1 ' 2. 3. In rt. A AB G and DEK, /. A = /. D. .-. AABG ~ ADEK. 2. 3. By hyp. 422. 4. h' DE 4. 424, 2. 5. AABC AC AB AC - AB n TF n 5. 309. A DEF DF DE DF -DE Ex. 859. Draw two triangles upon the blackboard, so that an angle of one shall equal an angle of the other. Give a rough estimate in inches of the sides including the equal angles in the tw6 triangles, and compute the numerical ratio of the triangles. Ex. 860. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides including the supplementary angles. BOOK IV 227 PROPOSITION IX. PROBLEM 499. To construct a square equivalent to a given tri- B angle. \ D S / / / / \ V / Given A ABC, with base b and altitude h. To construct a square =c= A AB C. I. Analysis 1. Let # = the side of the required square; then # 2 of required square. 2. i b h area of the given A ABC. 3. .'.a? = b-h. 4. /. J -b : x = x : h. 5. .-. the side of the required square will be a mean pro- portional between |- b and h. II. Construction 1. Construct a mean proportional between 7 ] b and h. Call it x. 445. 2. On a?, as base, construct a square, S. 3. S is the required square. III. Proof ARGUMENT 1. l 6 : x = x : h. o 2 177 w. . . X -Q Oil. 3. But x 2 = area of S. 4. And ^ b h = area of A ABC. 5. .'. So A ABC. Q.E.D. REASONS 1. By cons. 2. 388. 3. 478. 4. 485. 5. 54, 1. 228 PLANE GEOMETRY IV. The discussion is left as an exercise for the student. 500. Question. Could x be constructed as a mean proportional be- tween b and h ? 501. Problem. To construct a square equivalent to a given parallelogram. Ex. 861. Construct a square equivalent to a given rectangle. Ex. 862. Construct a square equivalent to a given trapezoid. Ex. 863. Upon a given base construct a triangle equivalent to a given parallelogram. Ex. 864. Construct a rectangle having a given base and equivalent to a given square. 502. Props. VI, VIII, and IX form the basis of a large class of important constructions. (a) Prop. VI enables us to construct a triangle equivalent to any polygon. It is then an easy matter to construct a trapezoid, an isosceles trapezoid, a parallelogram, a rectangle, or a rhombus equivalent to the triangle and hence equivalent to the given polygon. (6) Prop. VIII gives us a method for constructing an equi- lateral triangle equivalent to any given triangle. (See Ex. 865.) Hence Prop. VIII, with Prop. VI, enables us to construct an equilateral triangle equivalent to any given polygon. (c) Likewise Prop. IX, with Prop. VI, enables us to con- struct a square equivalent to any given polygon or to any fractional part or to any multiple of any given polygon. Ex. 865. (a) Transform triangle ABC into triangle DBG, retaining base BC and making angle DBC= 60. (6) Transform triangle DBC into triangle EBF, retaining angle DBC 60 and making sides EB and BF equal. (Each will be a mean proportional between DB and BC.} (c) What kind of a triangle is EBF? Ex. 866. Transform a parallelogram into an equilateral triangle. Ex. 867. Construct an equilateral triangle equivalent to f of a given trapezium. BOOK IV 229 Ex. 868. Construct each of the following figures equivalent to f of a given irregular pentagon : (1) a triangle; (2) an isosceles triangle ; (3) a right triangle ; (4) an equilateral triangle ; (5) a trapezium ; (6) a trape- zoid; (7) an isosceles trapezoid ; (8) a parallelogram; (9) a rhombus; (10) a rectangle ; (11) a square. Ex. 869. Transform a trapezoid into a right triangle having the hypotenuse equal to a given line. What restrictions are there upon the given line ? Ex. 870. Construct a triangle equivalent to a given trapezoid, and having a given line as base and a given angle adjacent to the base. PROPOSITION X. THEOREM 503. Two similar triangles are to each other as the squares of any two homologous sides. B b 1 Given two similar A ABC and A'B'C', with, b and b' two homol. sides. A ABC b 2 To prove A A'B'C' b' 2 ARGUMENT 1. A ABC ~ A A'B'C 1 . . A ABC b - c .>. X lltUl A A'B'C' b 1 c' b 1 c' 4. But c - = i r c' 6' K A ABC b b b 2 ' AA'B'C'~b' b 1 ~ b' 2 ' REASONS 1. By hyp. 2. 424, 1. 3. 498. 4. 424, 2. 5. 309. 230 PLANE GEOMETRY 504. Cor. Two similar triangles are to each other as the squares of any two homologous altitudes. (See 435.) Ex. 871. Two similar triangles are to each other as the squares of two homologous medians. Ex. 872. Construct a triangle similar to a given triangle and having an area four times as great. Ex. 873. Construct a triangle similar to a given triangle and having an area twice as great. Ex. 874. Divide a given triangle into two equivalent parts by a line parallel to the base. Ex. 875. Prove Prop. X by using 487. Ex. 876. Draw a line parallel to the base of a triangle and cutting off a triangle that shall be equivalent to one third of the given triangle. Ex. 877. In two similar triangles a pair of homologous sides are 10 feet and 6 feet, respectively. Find the homologous side of a similar tri- angle equivalent to their difference. - Ex. 878. Construct an equilateral triangle whose area shall be three fourths that of a given square. PROPOSITION XI. THEOREM \ 505. Two similar polygons are to each other as the squares of any two homologous sides. Given two similar polygons P and P f in which a and a', b and b', etc., are pairs of homol. sides. P a 2 To prove -=-. BOOK IV 231 ARGUMENT 1. Draw all possible diagonals from any two homol. vertices, as y and V'. 2. Then the polygons Will be divided into the same number of A ^ each to each and similarly placed, as A 1 and A I', A II and A II', etc. 3. Then AT a' 2 *. Air e' 2 ' A III _d 2 5. Aiir d' 2 6. But a _ e d_ I 1 ' = ( *L- A I A II A III a' e* 9. .'. 10. A I' AH' A III' A I + A II -f A III = A I _ a 2 A r + A ir + A in' ~~ A i' ~~ ^ 2 ' REASONS 1. 54, 15. 2. 439. 3. 503. 4. 503. 5. 503. 6. 419. 7. 54,13. 8. 54,1. 9. 401. 10. 309. 506. Cor. Two similar polygons are to each other as the squares of any two homologous diagonals. Ex. 879. If one square is double another, what is the ratio of their sides ? Ex. 880. Divide a given hexagon into two equivalent parts so that one part shall be a hexagon similar to the given hexagon. Ex. 881. The areas of two similar rhombuses are to each other as the squares of th^ir homologous diagonals. Ex. 882. One side of a polygon is 8 and its area is 120. The homol- ogous side of a similar polygon is 12 ; find its aiva. 232 PLANE GEOMETRY PROPOSITION XII, THEOREM 507. Tlw square described on the hypotenuse of a right triangle is equivalent to the sum of the squares described on the otlwr two sides. E Given rt. A ABC, right-angled at C, and the squares described on its three sides. To prove square AD =0= square BF + square CH. ARGUMENT 1. From C draw CM J_ AB, cutting AB at L and KD at M. 2. Draw CK and BH. 3. A ACG, BCA, and FOB are all rt. A. 4. .'. ACF and GCB are str. lines. 5. Ill A CAK and HAS, CA = HA, AK = AB. 6. Z CAB = Z CAB. 8. .-. ZCAK= 9. /. ACAK = AHAB. 10. A GMiTand rectangle AM have the same base AK and the same altitude, the JL between the Us AK and C y 3f. REASONS 1. 155. 2. 54, 15. 3. By hyp. 4. 76. 5. 233. 6. By iden. 7. 64. 8. 54, 2. 9. 107. 10. 235. BOOK IV 233 ARGUMENT 11. /.A CAK o= 1 rectangle AM. 12. Likewise A HAB and square CH have the same base HA and the same alti- tude, the _L between Us HA and GB. 13. .'.A HAB =0= 1 square Ctf, 14. But A CAK A #J. 15. .'. rectangle ^13f=c= ^ square (7#. 16. .'. rectangle AM =0 square CH. 17. Likewise, by drawing CD and AE, it may be proved that rectangle LD =c= square .W. 18. .*. rectangle JJf-f- rectangle LDo square CH -h square BF. 19. .'. square AD=c= square CH+ square BF. Q.E.D. REASONS 11. 489. 12. 235. 13. 489. 14. Arg. 9. 15. 54, 1. 16. 54, 7 a. 17. By steps simi- lar to 5-16. 18. 54,2. 19. 309. 508. Cor. I. The square described on either side of a right triangle is equivalent to the square described on the hypotenuse minus the square described on tlie other side. 509. Cor. n. If similar polygons are described on the three sides of a right triangle as homologous sides, the polygon described, on the hypotenuse is equivalent to the sum of the polygons described on the other two sides. Given rt. A ABC, right-angled at C, and let P, Q, and R be ~ polygons described on a, b, and c, respectively, as hornol. sides. To prove R o P -f Q. ARGUMENT ONLY 2. ^ = ^. 3. R c 2 P+ Q c 1 P-hQ. Q.E.D. Ex. 883. The square on the hypotenuse of an isosceles right triangle is equivalent to four times the triangle. 234 PLANE GEOMETRY 510. Historical Note. Prop. XII is usually known as the Pythago- rean Proposition, because it was discovered by Pythagoras. The proof given here is that of Euclid (about 300 B.C.). Pythagoras (569-500 B.C.), one of the most famous mathe- maticians of antiquity, was born at Samos. He spent his early years of manhood study- ing under Thales and traveled in Asia Minor and Egypt and probably also in Babylon and India: He returned to Samos where he established a school that was not a great success. Later he went to Crotona in Southern Italy and there gained many adherents. He formed, with his closest fol- lowers, a secret society, the members of which possessed PYTHAGORAS all things in common. They used as their badge the five-pointed star or pentagram which they knew how to construct and which they considered symbolical of health. They ate simple food and practiced severe discipline, having obedience, temper- ance, and purity as their ideals. The brotherhood regarded their leader with reverent esteem and attributed to him their most important dis- coveries, many of which were kept secret. Pythagoras knew something of incommensurable numbers and proved that the diagonal and the side of a square are incommensurable. The first man who propounded a theory of incommensurables is said to have suffered shipwreck on account of the sacrilege, since such numbers were thought to be symbolical of the Deity. Pythagoras, having incurred the hatred of his political opponents, was murdered by them, but his school was reestablished after his death and it flourished for over a hundred years. Ex. 884. Use the adjoining figure to prove the Pythagorean theorem. Ex. 885. Construct a triangle equivalent to the sum of two given triangles. B BOOK IV 235 Ex. 886. The figure represents a farm drawn to the scale indicated. Make accurate measure- ments and calculate ap- proximately the number of acres in the farm. Ex. 887. A farm XYZW, in the form of a trapezium, has the follow- ing dimensions : XY = 60 rods, YZ = 70 rods, ZW = 90 rods, WX = 100 rods, and XZ = 66 rods. Draw a plot of the farm to the i.* F~~ 1 1 1 calculate the area of the Q 20 40 farm in acres. 80 160 511. Def. By the rectangle of two lines is meant the rectangle having these two lines as adjacent sides. a 2 ab ab 6 2 Ex. 888. The square described on the sum two lines is equivalent to the sum of the squa described on the lines plus twice their rectangl Ex. 889. The square described on the difference of two lines is equivalent to the sum of the squares described on the lines diminished by twice their rectangle. HINT. Let AB and CB be the given lines. of res e. a b Ex. 890. The rectangle whose sides are the sum and difference respectively of two lines is equivalent to the difference of the squares described on the lines. HINT. Let AB and BC be the given lines. B C Ex. 891. Write the three algebraic formulas corresponding to the last three exercises. 236 PLANE GEOMETRY PROPOSITION XIII. PROBLEM 512. To construct a square equivalent to the sum of two given squares. Given squares P and Q. To construct a square =0= the sum of P and Q. I. Construction 1. Construct the rt. A ABC, having 'for its sides p and q, the sides of the given squares. 2. On r, the hypotenuse of the A, construct the square R. 3. R is the required square. II. The proof and discussion are left to the student. Ex. 892. Construct a square equivalent to the sum of three or more given squares. Ex. 893 . Construct a square equivalent to the difference of two squares. Ex. 894. Construct a square equivalent to the sum of a given square and a given triangle. Ex. 895. Construct a polygon similar to two given similar polygons and equivalent to their sum. (See 509.) Ex. 896. Construct a polygon similar to two given similar polygons and equivalent to their difference. Ex. 897. Construct an equilateral triangle equivalent to the sum of two given equilateral triangles. Ex. 898. Construct an equilateral triangle equivalent to the differ- ence of two given equilateral triangles, BOOK IV 237 PROPOSITION XIV. PROBLEM 513. To construct a polygon similar to one of two given polygons and equivalent to the other. Given polygons P and Q, with s a side of P. To construct a polygon ~ P and =0= Q. I. Analysis 1. Imagine the problem solved and let R be the required polygon with side x homol. to s, a side of P. 2. Then P : R = s 2 : 3? ; i.e. P : Q = s 2 : x~, since Q =0= R. (1) 3. Now to avoid comparing polygons which are not similar, we may reduce P and Q to =o squares. Let the sides of these squares be m and n, respectively ; then m 2 =0= P and n 2 =0= Q. 4. .-. m 2 : w 2 = s 2 : or, from (1). 5. .. m : n = s : x. 6. That is, x is the fourth proportional to ra, n, and s. II. The construction, proof, and discussion are left as an exercise for the student. 514. Historical Note. This problem was first solved by Pythagoras about 550 B.C. Ex. 899. Construct a triangle similar to a given triangle and equiva- lent to a given parallelogram. Ex. 900. Construct a square equivalent to a given pentagon. Ex. 901. Construct a triangle, given its angles and its area (equal to that of a given parallelogram). HINT. See Prop. XIV. Ex. 902. Divide a triangle into two equivalent parts by a line drawn perpendicular to the base. HINT. Draw a median to the base, then apply Prop. XIV. 238 PLANE GEOMETRY Ex. 903. Fig. 1 represents maps of Utah and Colorado drawn to the scale indicated. By carefully measuring the maps: (1) Calculate the perimeter of each state. (2) Calculate the area of each state. (3) Check your results for (2) by comparing with the areas given for these states in your geography. U T COLORADO SCALE OF MILES 200 300 400 500 PENNSYLVANIA . Pittsburg Harrisburg* Philadelphia , FIG. 2. Ex. 904. Fig. 2 repre- / -+-=+. i : 8 JJE^v / D '1C" h%i^*''' b C B~ &- Ex. 938. Construct a rectangle equivalent to a given rhombus, the difference of the base and altitude of the rectangle being equal to a given line. Ex. 939. Construct two lines, having given their difference and their product. Ex.940. Construct a triaijgle, given its three altitudes. HINT. Con- struct first a triangle whose sides are proportional to the three given alti- tudes. Ex. 941. Through the vertices of an equilateral triangle draw three lines which shall form an equilateral triangle whose side is equal to a given line. 242 PLANE GEOMETRY Ex. 942. The feet of the perpendiculars dropped upon the sides of a triangle from any point in the circumference of the circumscribed circle are collinear. rs OUTLINE OF PROOF. The circle having ^ -""Tt AP as diameter will pass through M and Q. *-^^ . / .-. Zl -Z.V and Z2 = Z2'. Similarly Z PKM = Z PB M. and .. QM and Jf/f form one str. line. Ex. 943. Given the base, the angle at the vertex, and the sum of the other two sides of a triangle; construct the triangle. ANALYSIS. Imagine the problem solved and draw A ABC. Prolong AB, making BE = BC, since the line c + a is given. Since ZE^ZABC, A AEC can be constructed. Ex. 944. The hypotenuse of a right triangle is given in magnitude and position ; find the locus of the center of the inscribed circle. Ex. 945. Prove Prop. VIII, Book IV, by using the following figure, in which A'D'E' is placed in the position ADE. E' Ex. 946. Prove Prop. VIII, Book IV, using two triangles such that one will not fall wholly within the other. Ex. 947. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides including the supplementary angles. (Prove by method similar to that of Ex. 945.) BOOK IV 243 Ex. 948. Given base, difference of sides, and difference of base angles ; construct the triangle. ANALYSIS. In the accompanying figure suppose c and EA, (/> a), to be given. A consideration of the figure will show that Z2 = Zl-f-ZA Add Z 1 to both members of the equation ; then Zl +Z2 = 2Z1 + ZA ButZl + Z2 = Z. B c ~~ A .'. Z7? = 2Z1 + ZA .-. Zl = \(Z.B - Z.A). The A BE A may now be constructed. The rest of the construction is left for the student. Ex. 949. Given base, vertex angle, and difference of sides, construct the triangle. ANALYSIS. AE = AB- EC. Z.AEG=W B + \ Z B. . . A AEG can be constructed. Ex. 950. If upon the sides of any triangle equilateral triangles are constructed, the lines joining the centers of the equilateral triangles form an equilateral triangle. HINT. Circumscribe circles about the three A C equilateral A. Join 0, the common point of intersection of the three circles, to A, B, and (7, the vertices of the given A. Prove each Z at the supplement of the Z opposite in an equilateral A, and also the supplement of the Z opposite in the A to be proved equi- lateral. Ex. 951. To inscribe a square in a semicircle. ANALYSIS. Imagine the problem solved and ABCD the required square. Prove OA = . Draw EF _L EO 2 FE : EO = BA : AO. meeting OB prolonged at F. .'. EF=2 OE. E A D Ex. 952. To inscribe a square in a given triangle. OUT LINK OF SOLUTION. Imagine the prob- lem solved and ABCD the required square-. Draw >V/'|I ET and construct square KSFJL Draw 7?F, thus determining point (7. The cons, will be evident from the figure. To prove ABCD a square, prove BC = CD. BC: SF = CD : FH. But SF = FH. . : BC = CD. R A KD T H 244 PLANE GEOMETRY E K Ex. 953. In a given square construct a square, having a given side, so that its vertices shall lie in the sides of the given square. HINT. Construct a rt. A, given the hypotenuse and sum of arms. Ex. 954. Construct a triangle, given m a , w c , h b . ANALYSIS. Imagine the problem solved and R that ABC is the required A. If BF were moved II to itself till it contained K, the rt. A AKT would be formed and KT would equal \ BF. Then A AKT can be made the basis of the ^ ~~p j> required construction. Ex. 955. Construct a quadrilateral, given two of its opposite sides, its two diagonals, and the angle between them. OUTLINE OF CONSTRUCTION. Imagine the problem solved and that ABCD is the required quadrilateral, s, s', d, d', and /.DOC being given. By II motion of d and d' the parallelogram BKFD may be obtained. The required construction may be begun by draw- ing O BKFD, since two sides and the in- cluded Z are known. With K as center and s as radius, describe an arc ; with D as center and arc intersecting the first, as at C, to locate A. Ex. 956. Between two circles draw a line which shall be parallel to the line of centers and equal to a given line I. -^.. K ' as radius, describe an Construct EH ABKG, or EH DACF, Ex. 957. Find x = , where , ft, c, and d represent given lines. d 1 HINT. Here x = - Let =y and construct d d d Then construct y. Ex. 958. Transform any given triangle into an equilateral triangle by a method different from that used in Ex. 865. ANALYSIS. Call the base of the given triangle ft and its altitude a. Let x = the side of the required equilateral triangle. Then V3 = -6-^. .'. \ b : x = 3 4 2 BOOK Y REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 515. Def. A regular polygon is one which is both equilateral and equiangular. Ex. 959. Draw an equilateral triangle. Is it a regular polygon ? Ex. 960. Draw a quadrilateral that is equilateral but not equiangular ; equiangular but not equilateral ; neither equilateral nor equiangular ; both equilateral and equiangular. Which of these quadrilaterals is a regular polygon ? Ex. 961. Find the number of degrees in an angle of a regular do- decagon. 516. Historical Note. The following theorem presupposes the pos- sibility of dividing the circumference into a number of equal arcs. The actual division cannot be obtained by the methods of elementary geom- etry, except in certain special cases which will be discussed later. As early as Euclid's time it was known that the angular magnitude about a point (and hence a circumference) could be divided into 2 n , 2 n 3, 2" 5, 2 n 15 equal angles. In 1706 it was discovered by Gauss, then nineteen years of age, that a regular polygon of 17 sides can be con- structed by means of ruler and compasses, and that in general it is possi- ble to construct all polygons having (2" + 1) sides, n being an integer and (2"+ 1) a prime number. The first four numbers satisfying this condition are 3, 5, 17, 257. Gauss proved also that polygons having a number of sides equal to the product of two or more different numbers of this series can be constructed. Gauss proved, moreover, that only a limited class of regular polygons are constructible by elementary geometry. For a note on the life of Gauss, see 520. 245 246 PLANE GEOMETRY PROPOSITION I. THEOREM 517. If the circumference of a circle is divided into any number of equal arcs: (a) the chords joining the points of division form a regular polygon inscribed in the circle ; (b) tangents drawn at the points of division form a regular polygon circumscribed about the circle. Given circumference AGE divided into equal arcs AB, BC, CD, etc., and let chords AB, BC, CD, etc., join the several points of division, and let the tangents GH, HK, KL, etc., touch the circum- ference at the several points of division. To prove A BCD - and GHKL regular polygons. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. ARGUMENT AB = BC= CD = - - . .-. arc CEA = arc DFB = arc EAC = Also AB = BC= CD = . .-. ABCD is a regular polygon. Again, ZBAG = Z.GBA = ZCBH Z.HCB = - . . And AB = BC= CD=- . .-. AAGB = And AG = GB = BH= REASONS 1. By hyp. 2. 54, 7 a. 3. 362, a. 4. 298. 5. 515. 6. 362, a. 7. Arg. 4. 8. 105. 9. 110. 10. 110. BOOK V 247 ARGUMENT REASONS 11. .'. GH=HK=KL=-.. 11. 54, 7 a. 12. .-. GHKL- is a regular polygon. Q.E.D. 12. 515. 518. Questions. If, in the figure of 517, the circumference is di- vided into six equal parts, how many arcs, each equal to arc AB, will arc CEA contain ? arc DFB ? How many will each contain if the circum- ference is divided into n equal parts ? In step 10, why does ACr = GB ? 519. Cor. // the vertices of a regular inscribed poly- gon are joined to the jnid-points of the arcs subtended by the sides of the polygon, tlie joining lines will form a regular inscribed polygon of double tlxe number of sides. Ex. 962. An equilateral polygon inscribed in a circle is regular. Ex. 963. An equiangular polygon circumscribed about a circle is regular. 520. Historical Note. Karl Friedrich Gauss (1777-1855) was born at Brunswick, Germany. Although he was the son of a bricklayer, he was enabled to receive a liberal ed- ucation, owing to the recogni- tion of his unusual talents by a nobleman. He was sent to the Caroline College but, at the age of fifteen, it was admitted both by professors and pupils that Gauss already knew all that they could teach him. He became a student in the Uni- versity of Gottingen and while there did some important work on the theory of numbers. On his return to Brunswick, he lived humbly as a private tutor, until 1807, when he was appointed professor of astronomy and director of the observatory at Gottingen. While there he did important work in physics as well as in astronomy. He also invented the telegraph independently of S. F. B. Morse. His lectures were unusually clear, and he is said to have given in them the analytic steps by which he developed his proofs ; while in his writings there is no hint of the processes by which he discovered his results. GAUSS 248 PLANE GEOMETRY PROPOSITION II. PROBLEM 521. To inscribe a square in a given circle. Given circle 0. To inscribe a square in circle 0. I. The construction is left as an exercise for the student. II. Proof ARGUMENT REASONS 1. AB _L CD. 1. By cons. 2. .-. Z AOC= 90. 2. 71. 3. .-. AC = 90, i.e. one fourth of the cir- 3. 358. cu inference. 4. .. the circumference is divided into four 4. Arg. 3. equal parts. 5. .-. polygon ACBD, formed by joining the 5. 517, a. points of division, is a square. Q.E.D. III. The discussion is left as an exercise for the student. 522. Cor. The side of a square inscribed in a circle is equal to the radius multiplied by ~\T2 ,' the side of a square circumscribed about a circle is equal to twice the radius. Ex. 964. Inscribe a regular octagon in a circle. Ex. 965. Inscribe in a circle a regular polygon of sixteen sides. Ex. 966. Circumscribe a square about a circle. BOOK V 249 Ex. 967. Circumscribe a regular octagon about a circle. Ex. 968. On a given line as one side, construct a square. Ex. 969. On a given line as one side, construct a regular octagon. Ex. 970. If a is the side of a regular octagon inscribed in a circle whose radius is S, then a = E V 2 \/2. PROPOSITION III. PROBLEM 523. To inscribe a regular hexagon in a given circle. Given circle 0. To inscribe in circle O a regular hexagon. I. The construction is left as an exercise for the student. HINT. AB = radius OA: II. Proof ARGUMENT 1. Draw OB. 2. Then A AB is equilateral. .-. Zo=60. .-. AB = 60, i.e. one sixth of the cir- cumference. .*. the circumference may be/ divided into .six equal parts. .-. polygon ABCDEF, formed by joining the points of division, is a regular inscribed hexagon. Q.E.D. III. The discussion is left as an exercise for the student. 6. REASONS 1. 54, 15. 2. By cons. 3. 213. 4. 358. 5. Arg. 4. 6. 517, a. 250 PLANE GEOMETRY 524. Cor. I. A regular inscribed triangle is formed by joining the alternate vertices of a regular inscribed liexagon. 525. Cor. II. A side of a regu- lar inscribed triangle is equal to the radius of the circle multi- plied by V 3. HINT. A ABD is a right triangle whose hypotenuse is 2 R and one side B. Ex. 971. Inscribe a regular dodecagon in a circle. Ex. 972. Divide a given circle into two segments such that any angle inscribed in one segment is five times an angle inscribed in the other. Ex. 973. Circumscribe an equilateral triangle about a circle. Ex. 974. Circumscribe a regular hexagon about a circle. Ex. 975. On a given line as one side, construct a regular hexagon. Ex. 976. On a given line as one side, construct a regular dodecagon. Ex. 977. If a is the side of a regular dodecagon inscribed in a circle whose radius is H, then a = R\2 V3. PROPOSITION IV. PROBLEM 526. To inscribe a regular decagon in a circle. ^E X FIG. 1. Given circle 0. To inscribe in circle a regular decagon. FIG. 2. BOOK V 251 I. Construction 1. Divide a radius R, of circle 0, in extreme and mean ratio. 465. (See Fig. 2.) 2. In circle O draw a chord AB, equal to s, the greater seg- ment of R. 3. AB is a side of the required decagon. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. II. Proof ARGUMENT Draw radius OA. On OA lay off OM = s. Draw OB and BM. Then OA : OM = OM : MA. .'. OA : AB = AB : MA. Also in A OAB and MAB, Z A = Z ^4. .'.A OAB ~ A MAB. v A OAB is isosceles, A MAB is isosceles, and -W = AB. But ^B = s = OM. .-. BM= OM. .-. A 50J/ is isosceles and Z.MBO = /-O. But Z ,47? J/ = Z o. GT /-AKO =2ZO. In A ABO, Z.ABO -f /.MAB 180. /. 2 Z + 2 Z + Z 0, or 5 Z 0, = 180. .-. Zo = 36. .-. AB = 36, i.e. one tenth of the cir- cumference. /. the circumference may be divided into ten equal parts. /. polygon ABCD , formed by join- ing the points of division, is a regular inscribed decagon. Q.E.D. 11. 12. 14. 15. REASONS 54, 15. 54,14. 54, 15. By cons. 309. By iden. 428. 94. 9. By cons. 10. 54,1. 111. 424,1. 13. 54, 2. 111. 204. 16. 309. 17. 54, 8 a. 18. 358. 19. Arg. 18. 20. 517, a. 252 PLANE GEOMETRY III. The discussion is left as an exercise for the student. 527. Cor. I. A regular pentagon is formed by joining the alternate vertices of a regular inscribed decagon. Ex. 978. Construct a regular inscribed polygon of 20 sides. Ex. 979. The diagonals of a regular inscribed pentagon are equal. Ex. 980. Construct an angle of 36 ; of 72. Ex. 981. Divide a right angle into five equal parts. Ex. 982. The eight diagonals of a regular decagon drawn from any vertex divide the angle at that vertex into eight equal angles. Ex. 983. Circumscribe a regular pentagon about a circle. Ex. 984. Circumscribe a regular decagon about a circle. Ex. 985. On a given line as one side, construct a regular pentagon. Ex. 986. On a given line as one side, construct a regular decagon. Ex. 987. The side of a regular inscribed decagon is equal to | B (V5 1), where E is the radius of the circle. HINT. By cons., H : s s : K s. Solve this proportion for s. PROPOSITION V. PROBLEM 528. To inscribe a regular pentedecagon in a circle. Given circle O. To inscribe in circle O a regular pentedecagon. I. Construction 1. Prom A, any point in the circumference, lay off chord AK equal to a side of a regular inscribed hexagon. 523. BOOK V 253 2. Also lay off chord AF equal to a side of a regular inscribed decagon. 526. 3. Draw chord FE. 4. FE is a side of the required pentedecagori. II. Proof ARGUMENT 1. Arc AE = ^ of the circumference. 2. Arc AF = T L of the circumference. 3. .'. arc FE = i T V> '. y 1 ^ of the cir- cumference. 4. .-. the circumference may be divided into fifteen equal parts. 5. .. the polygon formed by joining the points of division will be a regular inscribed pentedecagon. Q.E.D. REASONS 1. By cons. 2. By cons. 3. 54, 3. 4. Arg.3. 5. 517, a. III. The discussion is left as an exercise for the student. 529. Note. It has now been shown that a circumference can be divided into the number of equal parts indicated below : 2, 4, 8, 16,.-. 2n 3, 6, 12, 24, 3 x 2 n I [w being any positive 5, 10, 20, 40, ... 5 x 2' 1 | integer]. 15, 30, 60, 120, ... 15 x 2J Ex. 988. Construct an angle of 24. Ex. 989. Circumscribe a regular pentedecagon about a given circle. Ex. 990. On a given line as one side, construct a regular pente- decagon. . Ex. 991. Assuming that it is possible to inscribe in a circle a regular polygon of 17 sides, show how it is possible to inscribe a regular polygon of 51 sides. Ex. 992. If a regular polygon is inscribed in a circle, the tangents drawn at the mid-points of the arcs subtended by the sides of the in- scribed polygon form a circumscribed regular polygon whose sides are parallel to the sides of the inscribed polygon, and whose vertices lie on the prolongations of the radii drawn to the vertices of the inscribed polygon. 254 PLANE GEOMETRY PROPOSITION VI. THEOREM 530. A circle may be circumscribed about any regular polygon ; and a circle may also be inscribed in it. At- Given regular polygon ABCD To prove : (a) that a circle may be circumscribed about it ; (6) that a circle may be inscribed in it. (a) ARGUMENT 1. Pass a circumference through points A, B, and C. 2. Connect 0, the center of the circle, with all the vertices of the polygon. 3. Then OB '= OC. 4. .-. Z1 = Z2. 5. I Jut Z ABC = Z BCD. 6. .-. Z3 = Z4.~ 7. Also AB = CD. 8. .'. AABO= A OCD. 9. .-.OA = ^)D and circumference ABC passes through D. 10. In like manner it may be proved that circumference ABC passes through each of the vertices of the regular polygon ; the circle will then be circumscribed about the polygon. Q.E.D. KEASONS 1. 324. 2.. 54, 15. 3. 279, a. 4. 111. 5. 515. 6. 54, 3. 7. 515. 8. 107. 9. no. 10. By steps similar to 1-9. BOOK V 255 b\ ARGUMENT 1. Again AB, BC, CD, etc., the sides of the given polygon, are chords of the circumscribed circle. 2. Hence J from the center of the circle to theg_ chords are equal. 3. .-. with as center, and with a radius equal to one of these Js, as OH, a circle may be described to which all the sides of the polygon will be tangent^ 4. .-. this circle will be inscribed in the polygon. Q.E.D. REASONS 1. 281. 2. 307. 3. 314. 4. 317. 531. Def. The center of a regular polygon is the common center of the circumscribed and inscribed circles ; as 0, Prop. VI. 532. Def. The radius of a regular polygon is the radius of the circumscribed circle, as OA. 533. Def. The apothem of a regular polygon is the radius of the inscribed circle, as OH. 534. Def. In a regular polygon the angle at the center is the angle between radii of the polygon drawn to the extremities of any side, as Z AOF. 535. Cor. I. ffhe angle at the center is equal to four right angles Divided by the number of sides of the polygon. 536. Cor. II. An angle of a regular polygon is the supplement of the angle at the center. Ex. 993. Find the number of degrees in the angle at the center of a regular octagon. Find the number of degrees in an angle of the octagon. Ex. 994. If the circle circumscribed about a triangle and the circle inscribed in it are concentric, the triangle is equilateral. Ex. 995. How many sides has a regular polygon whose angle at the center is 30 ? 256 PLANE GEOMETRY PROPOSITION VII. THEOREM 537. Regular polygons of the same number of sides are similar. C Given two regular polygons, ABCDE and A'B'C'D'E', of the same number of sides. To prove polygon ABCDE ~ polygon A'B'C'D'E'. ARGUMENT 1. Let n represent the number of sides of each polygon; then each angle of each polygon equals (rt-2)2rt. A n 2. .'. the polygons are mutually equi- angular. 3. AB = BC = CD = - . 4. A'B' = B'C' = C'D' = . AB __ BC __ CD_ _ A \ 7 >f pf f ,i ft 1 /t 1 A 11 x> O C 1) 5. 6. .-. polygon ABCDE ~ polygon A'B'C'D'E'. Q.E.D. REASONS 1. 217. 2. Arg. 1. 3. 515. 4. 515. 5. 54, 8 a. 6. 419. Ex. 996. Two homologous sides of two regular pentagons are 3 inches and 5 inches, respectively ; what is the ratio of their perimeters ? of their areas ? Ex. 997. The perimeters of two regular hexagons are 30 inches and 72 inches, respectively ; what is the ratio of their areas ? BOOK V 257 PROPOSITION VIII. THEOREM 538. The perimeters of two regular polygons of the same number of sides are to each other as their radii or as their apothems. * E F' E' Given regular polygons ABODE and A'B'C'D'E' having the same number of sides. Let OA and O'A' be radii, OF and O'F' apothems, P and p' perimeters, of the two polygons, respectively. P OA OF To prove - = -. p' O'A' O'F' The proof is left as an exercise for the student. HINT. From 537 and 441, = _ (535 and 428). Then AE A'E< OA O'A 1 'E . Of O'F' Prove &AOE~&A'0'E' ( 435). 539. Cor. The areas of two regular polygons of tlie same number of sides are to each other as the squares of their radii or as the squares of their apothems. Ex. 998. Two regular hexagons are inscribed in circles whose radii are 7 inches and 8 inches, respectively. Compare their perimeters. Com- pare their areas. Ex. 999. The lines joining the mid-points of the radii of a regular pentagon form a regular pentagon whose area is one fourth that of the first pentagon. 258 PLANE GEOMETRY MEASUREMENT OF THE CIRCUMFERENCE AND OF THE CIRCLE 540. The measure of a straight line, i.e. its length, is obtained by laying off upon it a straight line taken as a stand- ard or unit ( 335). Since a straight line cannot be made to coincide with a curve, it is obvious that some other system of measurement must be adopted for the circumference. The following theorems will develop the principles upon which such measure- ment is based. PROPOSITION IX. THEOREM 541. I. The perimeter and area of a regular polygon inscribed in a circle are less, respectively, than the perime- ter and area of the regular inscribed polygon of twice as many sides. II. The perimeter and area of a regular polygon cir- cumscribed about a circle are greater, respectively, than the perimeter and area of the regular circumscribed poly- gon of twice as many sides. The proof is left as an exercise for the student. HINT. The sum of two sides of a triangle is greater than the third side. Ex. 1000. A square and a regular octagon are inscribed in a circle whose radius is 10 inches ; find : () The difference between their perimeters. (6) The difference between their areas. BOOK V 259 ARCHIMEDES 542. Historical Note. Archimedes (285?-212 B.C.) found the circumference and area of the circle by a method similar to that given in this text. He was born in Syracuse, Sicily, but studied in Egypt at the University of Alexandria. Although, like Plato, he re- garded practical applications of mathematics as of minor im- portance, yet, on his return to Sicily he is said to have won the admiration of King Hiero by applying his extraordinary mechanical genius to the con- struction of war-engines with which great havoc was wrought on the Roman army. By means of large lenses arid mir- rors he is said to have focused the sun's rays and set the Roman ships on fire. Although this story may be untrue, never- theless such a feat would be by no means impossible. Archimedes invented the Archimedes screw, which was used in Egypt to drain the fields after the inundations of the Nile. A ship which was so large that Hiero could not get it launched was moved by a system of cogwheels devised by Archimedes, who remarked in this connection that had he but a fixed fulcrum, he could move the world itself. The work most prized by Archimedes himself, however, and that which gives him rank among the greatest mathematicians of all time, is his in- vestigation of the mechanics of solids and fluids, his measurement of the circumference and area of the circle, and his work in solid geometry Archimedes was killed when Syracuse was captured by the Romans. The story is told 'that he was drawing diagrams in the sand, as was the custom in those days, when the Roman soldiers came upon him. He begged them not to destroy his circles, but they, not knowing who he was, and thinking that he presumed to command them, killed him with their spears. The Romans, directed by Marcellus, who admired his genius and had given orders that he should be spared, erected a monument to his memory, on which were engraved a sphere inscribed in a cylinder. The story of the re-discovery of this tomb in 75 B.C. is delightfully told by Cicero, who found it covered with rubbish, when visiting Syracuse. Archimedes is regarded as the greatest mathematician the world has known, with the sole exception of Newton. 260 PLANE GEOMETRY PROPOSITION X. THEOREM 543. By repeatedly doubling the number of sides of a regular polygon inscribed in a circle, and making the polygons always regular: I. The apothem can be made to differ from the ra- dius by less than any assigned value. II. The square of the apothem can be made to differ from the square of the radius by less than any assigned value. Given AB the side, OC the apothem, and OB the radius of a regular polygon inscribed in circle AMB. To prove that by repeatedly doubling the number of sides of the polygon : I. OB OC can be made less than any assigned value. II. OB* OC 2 can be made less than any assigned value. I. ARGUMENT 2. By repeatedly doubling the number of sides of the inscribed polygon and making the polygons always regular, AB } subtended by one side AB of the polygon, can be made less than any previously assigned arc, however small. .. chord AB can be made less than any previously assigned line segment, how- ever small. 1. REASONS 519. 301. BOOK V 261 4. ARGUMENT . CB, which is %AB, can be made less than any previously assigned value, however small. But OB OO < CB. .-. OB OC, being always less than CB, can be made less than any previously assigned value, however small. Q.E.D. II. Again, "off oc 2 CB 2 . But CB can be made less than any pre- viously assigned value, however small. .. CB can be made less than any previ- ously assigned value, however small. OB 2 OC 2 , being always equal to CB 2 , 3. REASONS 544. 4. 168. 5. 59, 10. 447. I, Arg. 3. 3. 545. 4. 309. can be made less than any previously assigned value, however small. Q.E.D. 544. If a variable can be made less than any assigned value, the quotient of the variable by any constant, except zero, can be made less than any assigned value. 545. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value. (For proofs of these theorems see Appendix, 586 and 589.) Ex. 1001. Construct the following designs: (1) on the blackboard, making each line 12 times as long as in the figure ; (2) on paper, making each line 4 times as long : 262 PLANE GEOMETRY PROPOSITION XI. THEOREM 546. By repeatedly doubling the number of sides of regular circumscribed and inscribed polygons of the same number of sides, and making the polygons always regular : I. Their perimeters approach a common limit. II. Their areas approach a common limit. Given P and p the perimeters, R and r the apothems, and K and k the areas respectively of regular circumscribed and inscribed polygons of the same number of sides. To prove that by repeatedly doubling the number of sides of the polygons, and making the polygons always regular : I. P and p approach a common limit. II. K and k approach a common limit. I. ARGUMENT 1. Since the two regular polygons have the 2. 3. P R same number of sides, = . p r Pp_Rr P R Rr Pp = K 4. But by repeatedly doubling the number of sides of the polygons, and making them always regular, R r can be KEASONS 1. 538. 2. 399. 3. 54, 7 a. 4. 543, I. BOOK V 263 5. ARGUMENT made less than any previously as- signed value, however small. R r R can be made less than any pre- viously assigned value, however small. T5 n* P can be made less than any R previously assigned value, however small, P being a decreasing variable. P p, being always equal to P - , can be made less than any previously assigned value, however small. P and p approach a common limit. Q.E.D. REASONS 5. 544. 6. 547. 7. 309. 8. 548. II. The proof of II is left as an exercise for the student. HINT. Since the two regular polygons have the same number of sides, K = *P. ( 539). The rest of the proof is similar to steps 2-8, 546, I. k r' 2 547. If a variable can be made less than any assigned value, the product of that variable and a decreasing value may be made less than any assigned value. 548. If two related variables are such that one is always greater than the other, and if the greater continually decreases while the less continually increases, so that the difference between the two may be made as small as ive please, then the two variables have a common limit tuhich lies between them. (For proofs of these theorems see Appendix, 587 and 594.) 549. Note. The above proof is limited to regular polygons, but it can be shown that the limit of the perimeter of any inscribed (or circum- scribed) polygon is the same by whatever method the number of its sides is successively increased, provided that each side approaches zero as a limit. 264 PLANE GEOMETRY 550. Def. The length of a circumference is the common limit which the successive perimeters of inscribed and circum- scribed regular polygons (of 3, 4, 5, etc., sides) approach as the number of sides is successively increased and each side ap- proaches zero as a limit. The term " circumference " is frequently used for " the length of a circumference." (See Prop. XII.) 551. The length of an arc of a circumference is such a part of the length of the circumference as the central angle which intercepts the arc is of 360. (See 360.) 552. The approximate length of a circumference is found in elementary geometry by computing the perimeters of a series of regular inscribed and circumscribed polygons which are ob- tained by repeatedly doubling the number of their sides. The perimeters of these inscribed and circumscribed polygons, since they approach a common limit, may be made to agree to as many decimal places as we please, according to the number of times we double the number of sides of the polygons. PROPOSITION XII. THEOREM 553. The ratio of the circumference of a circle to its diameter is the same for all circles. Given any two circles with circumferences (7 and (/, and with radii R and R', respectively. To prove = . . 2R 2R' BOOK V 265 ARGUMENT 1. Inscribe in the given circles regular polygons of the same number of sides, and call their perimeters .Pand P'. P R 2R 2. p' 4. As the number of sides of the two regu- lar polygons is repeatedly doubled, P approaches C as a limit, and p' approaches C 1 as a limit. p ' c 1 5. .-. approaches as a limit. 9;? 9 J? *-' jt *_ /t P' C Y ' 6. Also | approaches t - as a limit. p p' 7. But --is always equal to -. 2 R 2 R ' '2R~ 2R*' Q.E.D. REASONS 1. 517, a. 2. 538. 3. 396. 4. 550. 5. 408, b. 6. 408, b. 7. Arg/3. 8. 355. 554. Def. This constant ratio of the circumference of a circle to its diameter is usually represented by the Greek let- ter TT. It will be shown ( 568) that its value is approxi- mately 3-J-j or, more accurately, 3.1416. 555. Cor. I. Tlxe circumference of a circleis equal to 2 * R. 556. Cor. II. Any two circumferences are to each other as their radii. Ex. 1002. If the radius of a wheel is 4 feet, how far does it roll in two revolutions ? Ex. 1003. How many revolutions are made by a wheel whose radius is 3 feet in rolling 44 yards ? Ex. 1004. (a) Find the width of the ring between two concentric circumferences whose lengths are 2 feet and 3 feet, respectively. (6) Assuming the earth's equator to be 25,000 miles, find the width of the ring between it and a concentric circumference 1 foot longer. (c) Write your inference in the form of a general statement. 266 PLANE GEOMETRY PROPOSITION XIII. THEOREM 557. Tlie area of a regular polygon is equal to one half the product of its perimeter and its apothem. C Given regular polygon ABCD apothem. To prove area of ABCD - - = 1 P r. ARGUMENT 1. In polygon ABCD , inscribe a circle. 2. Then r, the apothem of regular polygon ABCD , is the radius of circle 0. 3. .-. area of ABCD = 1 P r. Q.E.D. , P its perimeter, and r its REASONS 530. G 33. 3. 492. Ex. 1005. Find the area of a regular hexagon whose side is inches. Ex. 1006. The area of an inscribed regular hex- agon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. Ex. 1007. The figure represents a flower bed drawn to the scale of 1 inch to 20 feet. Find the number of square feet in the flower bed. 558. Def. The area of a circle is the common limit which the successive areas of inscribed and circumscribed regular polygons approach as the number of sides is successively in- creased and each side approaches zero as a limit. BOOK V 267 PROPOSITION XIV. THEOREM 559. The area of a circle is equal to one half the prod- uct of its circumference and its radius. ircle 0, with radius R, circumference C, and area K. Given cir To prove K = OR. ARGUMENT REASONS 1. Circumscribe about circle a regular 1. 517, b. polygon. Call its perimeter P and its area S. 2. Thens = p*. 2. 557. 3. As the number of sides of the regular 3. 550. circumscribed polygon is repeatedly doubled, P approaches C as a limit. 4. .*. \PR approaches J CR as a limit.- 4. 561. 5. Also S approaches .K" as a limit. 5. . 6. But S is always equal to 1 PR. 6. Arg. 2. 7. ,'.K\CR. Q.E.D. 7. 355. \ 560. TJie product of a variable and a constant is a variable. 561. The limit of the product of a variable omd a constant, not zero, is the limit of the variable multiplied by the constant. (Proofs of these theorems will be found in the Appendix, 585 and 590.) 562. Cor. I. The area of a circle is equal to irR~. HINT. K= \ C R = | 2 irH . E = -rrT? 2 . 268 PLANE GEOMETRY 563. Cor. II. The areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. 564. Cor. m. The area of a sector whose angle is a is 551.) Ex. 1008. Find the area of a circle whose radius is 3 inches. Ex. 1009. Find the area of a sector the angle of which is (a) 45, (6) 120, (c) 17, and the radius, 5 inches. Find the area of the segment corresponding to (&). HINT. Segment = sector triangle. Ex. 1010. If the area of one circle is four times that of another, and the ra = ^ ; (b)x = aVb. Ex. 1104. Find the area included between a circumference of radius 7 and an inscribed square. Ex. 1105. What is the locus of the center of a circle of given radius whose circumference cuts at right angles a given circumference ? Ex. 1106. Two chords of a certain circle bisect each other.. One of them is 10 inches long ; how far is it from the center of the circle ? 280 PLANE GEOMETRY Ex. 1107. Show how to find on a given straight line of indefinite length a point which shall be equidistant from two given points A and B in the plane. If A and E lie on a straight line which cuts the given line at an angle of 45 at a point 7 inches distant from A and 17 inches from J5, show that OA will be 13 inches. Ex. 1108. A variable chord passes, when prolonged, through a fixed point outside of a given circle. What is the locus of the mid-point of the chord ? Ex. 1109. A certain parallelogram inscribed in a circle has two sides 20 feet in length and two sides 15 feet in length. What are the lengths of the diagonals ? Ex. 1110. Upon a given base is constructed a triangle one of the base angles of which is double the other. The bisector of the larger base angle meets the opposite side at the point P. Find the locus of P. Ex. 1111. What is the locus of the point of contact of tangents drawn from a iixed point to the different members of a system of con- centric circles ? Ex. 1112. Find the locus of all points, the perpendicular distances of which from two intersecting lines are to each other as 3 to 2. Ex. 1113. The sides of a triangle are a, 6, c. Find the lengths of the three medians. Ex. 1114. Given two triangles; construct a square equivalent to their sum. Ex. 1115. In a circle whose radius is 10 feet, two parallel chords are drawn, each equal to the radius. Find the area of the portion be- tween these chords. Ex. 1116. A has a circular garden and B one that is square. The distance around each is the same, namely, 120 rods. Which has the more land, A or B ? How much more has he ? Ex. 1117. Prove that the sum of the angles of a pentagram (a five- pointed star) is equal to two right angles. Ex. 1118. AB and A' B' are any two chords of the outer of two con- centric circles ; these chords intersect the circumference of the inner circle in points P, Q and P', Q' respectively : prove that^lP. PB=A'P' P'B'. Ex. 1119. A running track consists of two parallel straight por- tions joined together at the ends by semicircles. The extreme length of the plot inclosed by the track is 176 yards. If the inside line of the track is a quarter of a mile in length, find the cost of seeding this plot at \ cent a square yard. (TT - 2 r 4 .) BOOK V 281 Ex. 1120. If two similar triangles, ABC and DEF, have their homologous sides parallel, the lines AD, BE, and CF, which join their homologous vertices, meet in a point. Ex. 1121. In an acute triangle side AB = 10, AC=T, and the pro- jection of AC on AB is 3.4. Construct the triangle and compute the third side BC. Ex. 1122. Divide the circumference of a circle into three parts that shall be in the ratio of 1 to 2 to 3. Ex. 1123. The circles having two sides of a triangle as diameters intersect on the third side. Ex. 1124. Construct a circle equivalent to the sum of two given circles. Ex. 1125! Assuming that the areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles, prove that the bisector of an angle of a triangle divides the opposite side into segments propor- tional to the adjacent sides. Ex. 1126. In a circle of radius 5 a regular hexagon is inscribed. Determine (a) the area of one of the segments of the circle which are exterior to the hexagon; '(6) the area of a triangle whose vertices are three successive vertices of the hexagon ; (c) the area of the ring bounded by the circumference of the given circle and that of the circle inscribed in the hexagon. Ex. 1127. Find the locus of the extremities of tangents to a given circle, which have a given length. Ex. 1128. A ladder rests with one end against a vertical wall and the other end upon a horizontal floor. If the ladder falls by sliding along the floor, what is the locus of its middle point ? Ex. 1129. An angle moves so that its magnitude remains constant and its sides pass through two fixed points. Find the locus of the vertex. Ex. 1130. The lines joining the feet of the altitudes of a triangle form a triangle whose angles are bisected by the altitudes. Ex. 1131. Construct a triangle, given the feet of the three altitudes. Ex. 1132. If the radius of a sector is 2, what is the area of a sector whose central angle is 152 ? Ex. 1133. The rectangle of two lines is a mean proportional between the squares on the lines. Ex. 1134. Show how to inscribe in a given circle a regular polygon similar to a given regular polygon. 282 PLANE GEOMETRY FORMULAS OF PLANE GEOMETRY 570. In addition to the notation given in 270, the follow- ing will be used : a = side of polygon in general. b = base of a plane figure. &, 6' bases of a trapezoid. C = circumference of a circle. D'= diameter of a circle. E = sum of exterior angles of a polygon. h = altitude of a plane figure. / = suin of interior angles of a polygon. K = area of a figure in general. I = line in general. P= peri meter of polygon in general. p = projection of 6 upon a. _R = radius of circle, or radius of regular polygon. r = apothem of regular polygon, or radius of inscribed circle. s = the longer of two segments of a line ; or 8 = \(a + b + c). X = angle in general. x a = side of a triangle opposite an acute angle. x<> = side of a triangle opposite an obtuse angle. FIGURE Any triangle. Polygon. Central angle. Inscribed angle. Angle formed by two chords. Angle formed by tangent and chord. Angle formed by two secants. Angle formed by secant and tangent. Angle formed by two tan- gents. Similar polygons. Right triangle. Any triangle. Obtuse triangle. FORMULA REFERENCE A+ B+ 0=180. 204. 1= (n 2)180/i< V 216, 219. E = 4 rt. A. 218. X& intercepted arc. 358. X <* i intercepted arc. 3(55. JTc i sum of arcs. 377. X<\ intercepted arc. X cc i difference of arcs. JT5? \ difference of arcs. X^ I difference of arcs. P = a_ P' a' -2 ap. = a 2 + 6 2 + 2 ap. 378. 379. 379. . 379. 441. 446. 452. 455. FORMULAS 283 FIGURE FORMULA REFERENCE Any triangle. P+*. 2(^ + 2 w (l 2 . 457. Line divided in extreme l:s = s : I - s. 465 and Ex. 763. and mean ratio. '. Rectangle. K = b ; - h. 475. Square. K- a 2 . 478. Parallelogram. K = '6 h. 481. V Triangle. A K = 1 b - h. 485. L & h a = ^ ^Vs(s - a) (s - b) (s - c). 490. .a , J g' = Vs(s-)(s-6)(s-c). 490. * K = i(a + b + c)r. 491. Polygon. K = iP-r. 492. Circle inscribed in tjritin or le Vs(s a)(s b~)(s c) Fx g) (s c) Bisector of angle of tri- angle. t a - o \/7i/ i o/'o /v\ TTv Q41 6 + C Trapezoid. ;, K = |(fi + ft')'*- 495. Regular polygons of same P _ E r number of sides. ^, = V' 538 - K_ ^2 = ^2 539 ' Circles. _d_ = r" 1 2^' 558 ' c = 2 TT.B. 555. c__ p * J-it (\ + C K CP ^/' /o * * Regular polygon. K = |P-r. N irT? 2 . \^ ^ X^" 562 ' Circles. K _ ^ = ^. '563. K 1 R* D' 2 oector. if central Z ^^ ~ ^ j\. 360 D Segment. K = sector ^ triangle. Ex. 1009. APPENDIX TO PLANE GEOMETRY MAXIMA AND MINIMA I 571. Def. Of all geometric magnitudes that satisfy given conditions, the greatest is called the maximum., and the least is called the minimum.* 572. Def. Isoperimetric figures are figures which have the same perimeter. PROPOSITION I. THEOREM 573. Of all triangles having two given sides, that in which these sides include a right angle is the maximum. B C Given A ABC and AEC, with AB and AC equal to AE and AC respectively. Let Z CAB be a rt. Z. and /. CAE an oblique /.. To prove A ABC > A AEC. Draw the altitude EF. A ABC and AEC have the same base, AC. Altitude AB > altitude EF. .'. A ABC > A AEC. Q.E.D. 574. Cor. I. Conversely, if two sides are given, and if the triangle is a maximum, then the given sides include a right angle. HINT. Prove by reductio ad absurdum. * III totfl: mathematics a 9Wwhat bro APPENDIX 285 575. Cor. II. Of all parallelograms Tiaving given sides, the one that is rectangular is a maximum, and con- versely. Ex. 1135. Construct the maximum parallelogram having two lines of given lengths as diagonals. Ex. 1136. What is the minimum line from a given point to a given line? Ex. 1137. Of all triangles having the same base and altitude, that which is isosceles has the minimum perimeter. PROPOSITION II. THEOREM 576. Of all equivalent triangles having the same base, that which is isosceles has the least perimeter. G A A C Given equivalent A ABC and AKC with the same base AC, and let AB = BG and AE^EC. To prove AB + BC + CA < AE + EC + CA. Draw CFA.AC and let CF meet the prolongation of AB at G. Draw EG and BE and prolong BE to meet GC at F. BF II AC. Z CBF = ^FBG. BF bisects CG and is _L CG. .-, BG = BG and EC = EG. AB + BG < AE + EG. .'. AB + BC < AE-\-EC. .-. AB + BC + CA < AE + #G' 4- CA. Q.E.D. 286 PLANE GEOMETRY Ex. 1138. Of all equivalent triangles having the same base, that which has the least perimeter is isosceles. (Prove by reductio ad absur- dum.} Ex. 1139. Of all equivalent triangles, the one that has the minimum perimeter is equilateral. Ex. 1140. State and prove the converse of Ex. 1139. PROPOSITION III. THEOREM 577. Of all isoperimetric triangles on the same base, the isosceles triangle is the maximum. A . H C Given isosceles A ABC and any other A as AEC having the same base and the same perimeter as A ABC. To prove A ABC > A AEC. Draw BH _L AC, EF from E II AC, and draw AF and FC. A AFC is isosceles. .-. perimeter of A AFC < perimeter of A AEC. .\ perimeter of A AFC < perimeter of A ABC. .'. AF < AB. .'. FH < BH. .'. A AFC < A ABC. .-. A AEC < A ABC', i.e. A ABC > A AEC. Q.E.D. Ex. 1141. Of all triangles having a given perimeter and a given base, the one that has the maximum area is isosceles. Ex. 1142. What is the maximum chord of a circle ? What is the maximum and what the minimum line that can be drawn from a given exterior point to a given circumference ? APPENDIX 287 Ex. 1143. Of all triangles having a given perimeter, the one that has the maximum area is equilateral. PROPOSITION IV. THEOREM 578. Of all polygons having all their sides but one equal, respectively, to given lines taken in order, the maximum can be inscribed in a semicircle having tlw undetermined side as diameter. C E A H Given polygon ABCEFH, the maximum of all polygons sub- ject to the condition that AB, BC, CE, EF, FH, are equal respec- tively to given lines taken in order. To prove that the semicircumference described with AH as diameter passes through B, C, E, and F. Suppose that the semicircumference with AH as diameter does not pass through some vertex, as E. Draw AE and EH. Then Z AEH is not a rt. Z. Then if the figures ^.BC^and EFHaxe revolved about E until AEH becomes a rt. Z, A AEH will be increased in area. .. polygon ABCEFHc&n be increased in area without chang- ing any of the given sides. But this contradicts the hypothesis that polygon ABCEFH is a maximum. /. the supposition that vertex E is not on the semicircumfer- ence is false. .-. the semicircumference passes through E. In the same way it may be proved that every vertex of the polygon lies on the semicircumference. Q.E.D. Ex. 1144. Given the base and the vertex angle of a triangle, con- struct the triangle so that its area shall be a maximum. 288 PLANE GEOMETRY Ex. 1145. Find the point in a given straight line such that the tan- gents drawn from it to a given circle contain a maximum angle. PROPOSITION V. THEOREM 579. Of all polygons that have their sides equal, re- spectively, to given lines taken in order, the polygon that can be circumscribed by a circle is a maximum. A'/ Given polygon ABCD which is circumscribed by a O, and polygon A'B'C'D' which cannot be circumscribed by a O, with AB = A'B', BC = B'C', CD=C'D', and DA = D' A'. To prove ABCD > A'B'c'D 1 . From any vertex as A draw diameter AE\ draw EC and ED. On C'D', which equals CD, construct AD'C'E' equal to ADCE-, draw A'E'. The circle whose diameter is A'E' does not pass through all the points B 1 , C 1 , D'. (Hyp.) .-. either ABCE or EDA or both must be greater, and neither can be less, than the corresponding part of polygon A'B'C'E'D' ( 578). .-. ABCED > A'B'C'E'D'. But A DCE = A D'C'E'. .-. ABCD > A'B'C'D'. Q.E.D. Ex. 1146. In a given semicircle inscribe a trapezoid whose area is a maximum. Ex. 1147. Of all equilateral polygons having a given side and a given number of sides, the one that is regular is a maximum. APPENDIX 289 PROPOSITION VI. THEOREM 580. Of all isoperimetric polygons of the same number of sides, the maximum is equilateral. Given polygon ABODE the maximum of all isoperimetric polygons of the same number of sides. To prove AB = B C = CD = DE = EA. Suppose, if possible, EG > CD. On BD as base construct an isosceles A BFD isoperimetric with A BCD. ABFD>ABCD. .-. polygon ABFDE > polygon ABODE. But this contradicts the hypothesis that ABODE is the maxi- mum of all 1 isoperimetric polygons having the same number of sides. .-. BC = CD. In like manner any two adjacent sides may be proved equal. .-. AB = BO = CD = DE == EA. Q.E.D. 581. Cor. Of all isoperimetric polygons of the same number of sides, the maximum is regular. Ex. 1148. In a given segment inscribe a triangle whose perimeter is a maximum, 290 PLANE GEOMETRY PROPOSITION VII. THEOREM 582. Of two isoperiinetric regular polygons, that which has the greater number of sides has tlie greater area. F C H E Given the isoperimetric polygons P and Q, and let P have one more side than Q. To prove P > Q. In one side of Q take any point as H. EFHG may be considered as an irregular polygon having the same number of sides as P. .-. P > EFHG ; i.e. P > Q. Q.E.D. PROPOSITION VIII. THEOREM 583. Of two equivalent regular polygons, that which has the greater number of sides has the smaller pe- rimeter. H Given square S =c= regular hexagon H. To prove perimeter of S > perimeter of H. Construct square R isoperimetric with H. APPENDIX 291 Area of H > area of R ; i.e. area of s > area of R. .-. perimeter of S > perimeter of R. .-. perimeter of S > perimeter of H. Q.E.D. 584. Cor. Of all polygons having a given number of sides and a given area, that which has a minimum perimeter is regular. Ex. 1149. Among the triangles inscribed in a given circle, the one that has a maximum perimeter is equilateral. Ex. 1150. Of all polygons having a given number of sides and in- scribed in a given circle, the one that has a maximum perimeter is regular. VARIABLES AND LIMITS. THEOREMS PROPOSITION I. THEOREM 585. If a variable can be made less than any assigned value, the product of the variable and any constant can be made less than any assigned value. Given a variable V, which can be made less than any previ- ously assigned value, however small, and let K be any constant. To prove that v K may be made as small as we please, i.e. less than any assigned value. Assign any value, as a, no matter how small. Now a value for v may be found as small as we please. Take V < . Then v K < a ; i.e. V K may be made less K than any assigned value. Q.E.D. 586. Cor. I. If a variable can be made less than any assigned value, the quotient of the variable by any con- stant, except zero, can be made less than any assigned value. HINT. - F, which is the product of the variable and a constant. K K 587. Cor. II. If a variable can be made less than any assigned value, the product of that variable and a de- 292 PLANE GEOMETRY creasing value may be made less than any assigned value. HINT. Apply the preceding theorem, using as K a- value greater than any value of the decreasing multiplier. 588. Cor. m. The product of a variable and a variable may be a constant or a variable. 589. Cor. IV. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value. (Apply Cor. II.) Ex. 1151. Which of the corollaries under Prop. I is illustrated by the theorem: " The product of the segments of a chord drawn through a fixed point within a circle is constant ' ' ? PROPOSITION II. THEOREM 590. The limit of the product of a variable and a con- stant, not zero, is the limit of the variable multiplied by the constant. Given any variable F which, approaches the finite limit L, and let K be any constant not zero. To prove the limit of K V= K - L. Let# = F; then v = L R. .'. K- V=K> L K - R. But the limit of K - R = 0. .-. the limit of K v the limit of (K - L K R ) = K L. Q.E.D. 591. Cor. The limit of the quotient of a variable by a constant is the limit of the variable divided by the constant. HINT. = "F, which is the product of the variable and a constant. K K PROPOSITION III. THEOREM 592. If two variables approach finite limits, not zero, then the limit of their product is equal to tJie product of their limits. APPENDIX 293 Given variables F and F' which approach the finite limits L and L', respectively. To prove the limit of F F' = L L 1 . Let R = L F and X 1 = L' F'. Then V L E and F' = ' R'. .*. F F' = L L' (L 1 R -f i . 72' . R 1 ). But the limit of (L 1 R + - R 1 R - R')= 0. .-. the limit of F F' = the limit of \_L L' (L 1 - R -f L R' R- ')] = ' .-. the limit of F - F' = L - L'. Q.E.D. 593. Cor. If each of any finite number of variables approaches a finite limit, not zero, then the limit of their product is equal to the product of their limits. PROPOSITION IV. THEOREM 594. If two related variables are such that one is always greater than the other, and if the greater con-? tinually decreases while the less continually increases, so that the difference between the two may be made as small as we please, then the two variables have a com- mon limit which lies between them. A P Q R L R' Q' P' Given the two related variables AP and AP', AP' greater than AP, and let AP and AP 1 be such that as AP increases AP 1 shall decrease, so that the difference between AP and AP' shall ap- proach zero as a limit. To prove that AP and AP 1 have a common limit, as AL, which lies between AP and AP'. Denote successive values of AP by AQ, AR, etc., and denote the corresponding values of AP' by AQ\ AR 1 , etc. Since every value which AP assumes is less than any value which AP' assumes (Hyp.) .-. AP < AR', ^ i continually increasing, 294 PLANE GEOMETRY A P Q R L R' Q' P' Hence AP has some Jimit. (By def. of a limit, 349.) Since any value which AP f assumes is greater than every value which AP assumes (Hyp.) .'. AP' > AR. But AP' is continually decreasing. Hence AP' has some limit. (By def. of a limit, 349.) Suppose the limit of AP = the limit of AP 1 . Then let the limit of AP be AK, while that of AP' is AK 1 . Then AK and AK' have some finite values, as m and m', and their difference is a finite value, as d. But the difference between some value of AP and the cor- responding value of AP 1 cannot be less than the difference of the two limits AK and AK'. This contradicts the hypothesis that the difference between AP and AP 1 shall approach zero as a limit. .. the limit of AP =the limit of AP 1 and lies between AP and AP', as AL. Q.E.D. 595. THEOREM. With every straight line segment there is associated a number which may be called its measure- number. a For line segments commensurable with the unit this theo- rem was considered in 335 and 336; we shall now consider the case where the segment is incommensurable with the chosen unit. Given the straight line segment a and the unit segment u; to express a in terms of u. Apply u (as a measure) to a as many times as possible, sup- pose t times, then t u < a < (t + 1) u. Now apply some fractional part of u, say -, to a, and sup- P APPENDIX 295 pose it is contained ^ times, then p p Then apply smaller and smaller fractional parts of u to a, say , , , , and suppose them to be contained t>, 3 , t 4 , times respectively, then p 2 [) P P Now the infinite series of increasing numbers t, J , , -, p p* none of which exceeds the finite number t + 1, defines a num- ber n (the limit of this series) which we shall call the measure- number of a with respect to u. Moreover, this number n is unique, i.e. independent of p (the number of parts into which the unit was divided), for if m is any number such that m < n, then m u < a, and if m > n, then m u > a ; we are therefore justified in associating the number n with a, and in saying that n u = a. 596. Note. Manifestly, the above procedure may be applied to any geometric magnitude whatever, i.e. every geometric magnitude has a unique measure-number. 597. Cor. If a magnitude is variable and approaches a limit, then, as the magnitude varies, the successive measure-numbers of the variable approach as their limit the measure-number of the limit of the magnitude. 598. Discussion of the problem : To determine whether two given lines are commen- surable or not ; and if they are commensurable, to find their common measure and their ratio ( 345). Moreover, GD is the greatest common measure of AB and CD. For every measure of AB is a measure of its multiple CE. Hence, every common measure of AB and CD is a common measure of CE and CD and therefore a measure of their differ- I 296 PLANE GEOMETRY ence ED, and therefore of AF, which is a multiple of ED. Hence, every common measure of AB and CD is a common measure of AB and AF and therefore a measure of their differ- ence FB. Again, every common measure of ED and FB is a common measure of ED and EG (a multiple of FB) and hence of their difference GD. Hence, no common measure of AB and CD can exceed GD. Therefore, GD is the greatest common measure of AB and CD. Now, if AB and CD are commensurable, the process must ter- minate; for any common measure of AB and CD is a measure of each remainder, and every segment applied as a measure is less than the preceding remainder. Now, if the process did not terminate, a remainder could be reached which would be less than any assigned value, however small, and therefore less than the greatest common measure, which is absurd. If AB and CD are incommensurable, the process will not ter- minate; for, if it did, the last remainder obtained would be a common measure of AB and CD, as shown above. 599. THEOREM. An angle can be bisected by only one line. Given Z ARC, bisected by BT. To prove that no other bisector of /. ABC exists. Suppose that another bisector of /.ABC exists, e.g. BF. Then Z ABF Z. ABT. This is impossible. .-. no other bisector of /.ABC exists. Q.E.D. 600. Note on Axioms. The thirteen axioms ( 54) refer to num- bers and may be used when referring to the measure-numbers of geometric magnitudes. Axioms 2-9 are not applicable always to equal figures. (See Exs. 800 and 801.) Axioms 7 and 8 hold for positive numbers only, but do not hold for negative numbers, for zero, nor for infinity ; axioms 11 and 12 hold only when the number of parts is finite. INDEX (The numbers refer to articles.) ART. Acute triangle 98 Adjacent angles 42 Alternation 396 Altitude .... 100, 228, 229 Analysis of a proportion . . 397 of constructions .... 151, 152, 170 (Ex. 187), 274, 499 of exercises 274 Angle 38 acute 47 at center of circle . . . 292 at center of regular polygon 534 bisector of ... 53, 127, 599 central 292 degree of 71 inscribed in circle . . . 363 inscribed in segment . . 364 magnitude of 41 obtuse 48 of one degree 71 reflex 49 right 45, 71 right and left side of . . 199 sides of 38 straight 69 vertex of 38 Angles 38 added 43 adjacent 42 alternate exterior . 182 ART. Angles, alternate interior . . 182 complementary .... 68 corresponding 182 designation of .... 39, 50 difference of 44 equal 44 exterior 87, 182 exterior interior .... 182 homologous . . . 417, 424 interior 86, 182 oblique 51 of a polygon 8(5 subtracted 44 sum of 43 supplementary .... 69 supplementary-adjacent . 72 vertical 70 Antecedents 384 Apothem 533 Application 79 Arc 121, 280 degree of 296 length of 551 major 291 minor 291 Arcs, similar ...... 565 Area of a figure . . . . . 471 of circle 558 of rectangle . . ... 468-471 of surface 407 Argument only . . . . 171 297 298 INDEX ART. Arms of isosceles triangle . . 94 of right triangle .... 96 Assumptions 12, 54 Axiom 55 Axioms 54 note on 600 Rase of isosceles triangle . . 94 of polygon 99 Bisector of angle . .53, 127, 599 of line 52, 145 Bisectors of angles of triangle 126 Center-line . . . " . . . .325 Center of circle . . . 119,278 of regular polygon . . .531 Central angle 292 Centroid . .205 Chord 281 Circle . . . . . . 119, 276 arc of 121, 280 area of 558 center of .... 119, 278 chord of 281 circumference of . 119, 277, 550 circumscribed 300 determined 324 diameter of .'.... 282 escribed 322 inscribed 317 radius of .... 119, 278 secant of ....... 285 sector of .... 287, 564 segment of 288 tangent to 286 Circles, concentric .... 326 tangent, externally . . . 330 tangent, internally . . . 330 Circumference . . .119, 277, 550 length of . . . . 550, 552 measurement of .... 540 Circumscribed circle .... 300 Circumscribed polygon . Clockwise motion . . . Closed figure .... Closed line Coincidence .... Collinear segments . . Commensurable quantities 337, Common measure Compasses, use of Complementary angles . Complete demonstration Composition .... Composition and division Concentric circles . . Conclusion . . . . . Concurrent lines . . . Consequents .... Constant Construction .... of triangles .... Continued proportion . Converse theorem . . Corollary Counter-clockwise motion ART. 317 86 83 82 17 28 342, 469 . . 337 . . 31 . . 68 . . 171 . . 398 . . 400 . . 326 . . 57 . . 190 . . 334 346, 347 . . 123 . . 269 . . 405 . . 135 . . 58 50 Definition, test of .... 138 Degree, of angle 71 of arc 296 Demonstration 56, 79 complete 171 Determined circle .... 324 Determined line and point . 25, 26 Determined triangle . . 133, 209 Diagonal 88 Diameter . 282 Dimensions 10 Discussion 123 Distance between two points . 128 from point to line . . .166 Division . 399 Drawing to scale 437 INDEX 299 ART. Equal figures 18, 473 homologous parts of . 109, 110 Equiangular polygon . *. . 90 Kquidistance 128 Equilateral polygon .... 89 Equilateral triangle .... 95 Equivalent figures .... 473 Escribed circle 322 Euclid's Elements .... 114 Exterior angle of polygon . . 87 Extreme and mean ratio 464, 465 (Ex. 763), 526 Extremes 385 Figure, closed 83 geometric 13 plane 36 rectilinear 37 Figures, equal 18, 473 equivalent 473 isoperimetric 572 similar 419, 473 transformation of ... 493 Formulas of plane geometry . 570 Fourth proportional .... 387 Geometric figure 13 Geometric figures, equal . . 18 Geometric solid . . . 2, 9, 11 Geometry 14 plane 36 subject matter of .... 1 Golden section 464 Harmonical division .... 434 Hexagon 92 Historical notes Achilles and the tortoise . 354 Ahmes 474, 569 Archimedes .... 542, 569 Area of a circle .... 569 Cicero . . 542 ART. Descartes 237 Division of circumference . 516 Egyptians .,...-. 371, 449, 474, 497, 569 Euclid . . . .114, 510, 516 Gauss 516, 520 Herodotus . . . . . . 474 Hero of Alexandria . . . 569 King Hiero 542 Lambert 569 Lindemann 569 Metius of Holland . . . 569 Morse, S. F. B 520 Newton 542 Origin of geometry ... 474 Plato . . . . . 268, 542 Plutarch 371 Pons asinorum 114 Ptolemy 569 Pythagoras . . 344, 510, 514 Richter . 569 Seven Wise Men . . . .371 Shanks . 569 Squaring the circle . . .569 Stobseus 114 Thales 371, 510 Zeno . . . v . . . . .354 Homologous parts 109, 110, 417, 418, 424 Hypotenuse 96 Hypothesis ....... 57 Incommensurable quantities . 339, 343, 344, 470 Indirect proof .... 159, 161 Inscribed angle 363 Inscribed circle 317 Inscribed polygon 299 Instruments, use of .... 31 Inversion 395 Isoperimetric figures 572 Isosceles trapezoid .... 227 300 INDEX ART Isosceles triangle 94 arms of 94 base of 94 sides of 94 vertex angle of .... 94 Length of arc "551 of circumference . . 550, 552 Length of perpendicular . . 166 of secant 426 of tangent 319 Limits . 346, 349^351, 590-594 Line 4, 7, 11, 27 bisected 52, 145 broken 30 closed 82 curved ....... 29 determined 25 divided externally . . . 406 divided harmonically . . 434 divided in extreme and mean ratio . ... . 464, 465 (Ex. 763), 526 divided internally . . . 406 of centers 325 right 23 segments of . /. . . 27,406 straight 21, 23 Line segment ...... 27 Lines 20 concurrent 196 difference of 31 divided proportionally . . 407 homologous 109, 110, 418, 424 oblique 46 parallel 177 perpendicular 45 product of . 425, 475, 476, 511 rectangle of . 425, 475, 476, 611 Locus ..... 129, 130, 131 as an assemblage .... 130 as a path 130 ART. Locus, finding of a .... 144 of a point 130 of alf points 130 problem, solution of . . . 143 Magnitude of an angle . . . 41 Major arc 291 Maximum 571 Mean proportional 386 Means 385 Measure, commoi^ 337 numerical , V. . . 335, 467 unit of 335, 466 Measure-number, 335, 467, 471, 595 Measurement 335 of angle ....... 358 of arc 358 of circle . , . . . 540, 558 of circumference . . 540, 552 of distances by means of triangles 115 of line. 335, 595 of rectangle. . . . 468-471 of surface 466 Median, of trapezoid .... 25t of triangle 102 Median center of triangle . . 265 Methods of attack 103, 104, 106, 110 (Ex. 62), 115, 151, 152, 158, 159, 176 (Ex. 187), 269-274, 275, 397, 398, 425, 499, 502, 513. Minimum .^.^11 Minor arc . . . .291 Numerical measure . 335, 467 Oblique angles 51 Oblique lines 46 Obtuse triangle 97 One to one correspondence 595, 596 Opposite theorem. , % . 136 INDEX 301 ART. Optical illusions 81 Outline of proof 171 Parallel lines ' 177 Parallelogram 220 altitude of 228 base of 228 Parallelograms classified 223, 243 Pentagon 92 Perimeter of polygon ... 85 Perpendicular .... 45, 166 Pi . 554 evaluation of 568 Plane 32, 33, 34 Plane figure . 36 Plane geometry 36 Point 5, 6, 11 determined 26 of tangency 286 Polygon 84 angles of 86 base of . 00 circumscribed 317 diagonal of . . . " . . . 88 equiangular ...... 90 ' equilateral . . . . . . 89 exterior angles of .... 87 inscribed 299 perimeter of 85 regular 91, 515 sides of 84 vertices of 84 Polygons, mutually equiangular 417 sides proportional . . .418 similar 419 Pons asinorum 114 Portraits Archimedes 542 Descartes 237 Euclid 114 Gauss 520 Pythagoras 510 , ART. Thales 371 Postulate 15 circle 122 parallel line . . . 178, 179 revolution 40, 54 straight line . . . . 24, 54 transference . . . . 16, 64 Problem defined 59 Problems of computation . . 59 Problems of construction . . 59 analysis of 151, 152, 176 (Ex. 187), 274, 499 discussion of 123 proof of . 123 solution of 123 Product of lines 425, 475, 476, 511 Projection of line 451 of point 450 Proof 79, 123 analytic method .... 275 by exclusion . . . 159, 161 by successive substitutions 275 indirect 159, 161 necessity for 81 outline of 171 reductio ad absurdum 159, 161 synthetic method . . . 275 Proportion 382 analysis of 397 antecedents of 384 by alternation 396 by composition . . . 398 by composition and division 400 by division 399 by inversion . . . . . 395 consequents of .... 384 continued 405 extremes of 385 means of 385 terms of 383 ways of writin .... 382 302 INDEX ART;. Proportional, fourth .... 387 mean 380 third . . . 386 Proportions simplified . 380, 397 Proposition 56 Quadrant 295 Quadrilateral ...... 92 Quadrilaterals classified 220, 243 Radius of circle . . of regular polygon 119, 278 . . 532 Ratio .... 340-343, 382, 384 antecedent of 384 consequent of 384 extreme and mean . . . 464, 465 (Ex. 763), 526 of any two surfaces . . . 472 of similitude 418 of two commensurables . 342 pf two incommensurables . 343 of two magnitudes . . . 341 Rectangle 223 area of 468-471 of two lines 425, 475, 476. 511 Rectilinear figure .... 37 Reductio ad absurdum . 159, 161 Regular polygon . . . 91, 515 angle at center of ... 534 apothem of 533 center of 531 radius of 532 Related variables . . . 352, 594 Rhomboid . 224 Rhombus ....... 226 Right triangle 95 arms of 96 hypotenuse of .... 96 sides of ....... 96 Ruler, use of 31 Scale, drawing to . 437 ART. Scalene triangle 93 Secant 285 length of 426 Sector ...... 287, 564 Segment, of circle .... 288 of line 27 Segments, added 31 collinear 28 difference of 31 of a line 406 similar ........ 565 subtracted 31 sum of 31 Semicircle 289 Semicircumference .... 290 Series of equal ratios . 401, 405 Sides, of angle. 38 of polygon 81 Similar arcs 5(55 Similar polygons . . . . .419 Similar sectors 565 Similar segments .... 565 Similarity of triangles . . . 431 Similitude, ratio of .... 418 Solid, geometric . . . 2, 9, 11 Solution of exercises . . . 275 of locus problems . . . 143 of problems 123 of theorems 79 Square 225 Squaring the circle .... 569 Straight angle 69 Straightedge 21 Straight line 21, 23 determined 25 Summary, of divisions of cir- cumference ' . . . . 529 of equal triangle theorems 118 of formulas of plane geom- etry 570 of parallel line theorems . 197 of quadrilaterals .... 243 INDEX . 303 Summary, of similar triangle theorems 431 of trapezoids . Exs. 320, 321 of unequal angle theorems 174 of unequal line theorem.8 . 175 Superposition . . . lit, 103, 106 Supplementary angles ... 69 Supplementary -adjacent angles 72 Surface . . . . . . 3, 8, 11 curved 35 plane .... 32, 33, 34 Surfaces 32 Tangent 286 external common .... 333 internal common ...-. 333 length of 319 Theorem 57 converse of 135 opposite of 136 solution of . . . . . . 79 Third proportional .... 386 Transformation 493 Transversal . 181 Trapezium 222 Trapezoid 221 altitude of 229 bases of 229 isosceles 227 median of 251 Triangle 92 acute . 98 ART Triangle, altitudes of ... 100 bisectors of angles of . .126 centroid of 265 construction of .... 269 determined . . . . 133, 269 equilateral 95 isosceles 94 median center of .... 265 medians of 102 notation of 270 obtuse ....... 97 parts of 269 right .96 scalene ....... 93 vertex of 99 vertex angle of ' . . . 94, 99 Triangles, classified . . . 93-98 similarity of 431 Unit of measure . . . 335. 466 Variable . . 346, 348, 585-594 approaches its limit . . . 350 independent 346 limit of 349 reaches its limit .... 351 related ..... 352, 594 Vertex of angle . . . . . 38 of polygon 84 of triangle 99 Vertex angle of triangle . 94, 99 Vertical angles 70 ELEMENTS OF TRIGONOMETRY By ANDREW W. PHILLIPS, Ph.D., and WENDELL M. STRONG, Ph.D., Professors in Yale University Plane and Spherical Trigonometry. With Tables $1-40 The same. Without Tables . .90 Logarithmic and Trigonometric Tables i.oo IN this text-book full recognition is given to the rigorous ideas of modern mathematics in dealing with the funda- mental series of trigonometry. Both plane and spherical trigonometry are treated in a simple, direct manner, free from all needless details. 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