mm . I ^ J.. --^— -^.—^ «-.- ^« ^*i Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofplanegOOsandrich ELEMENTS OF PLANE GEOMETRY BY ALAN SANDERS HUGHES HIGH SCHOOL, CINCINNATI, OHIO >>•?€ NEW YORK . : . CINCINNATI • : . CHICAGO AMERICAN BOOK COMPANY COPYEIGHT, 1901, BY AMEEICAN BOOK COMPANY. Entered at Stationers' Hall, London. BANDBES' PLANE GEOM. E-P 2 PURPOSE AND DISTINCTIVE FEATURES This work has been prepared for the use of classes in high schools, academies, and preparatory schools. Its distinctive features are : — 1. The omission of parts of demonstrations. By this expedient the student is forced to rely more on his own reasoning powers, and is prevented from acquiring the detrimental habit of memorizing the text. As it is necessary for the beginner in Geometry to learn the form of a geometrical demonstration, the demon- strations of the first few propositions are given in full. In the succeeding propositions only the most obvious steps are omitted, the omission in each case being indi- cated by an interrogation mark (?). In no case is the student expected to originate the plan of proof. 2. The introduction^ after each proposition, of exercises bearing directly upon the principle of the proposition. As soon as a proposition has been mastered, the stu- dent is required to apply its principle in the solution of a series of easy exercises. Hints or suggestions are given to aid the pupil in the solution of the more difficult exercises. 3 4 PURPOSE AND DISTINCTIVE FEATURES 8. All constructions^ such as draiving parallels^ erecting perpendiculars, etc., are given before they are required to he used in demonstrations. 4. Exercises in Modern Greometry, Exercises involving the principles of Modern Geometry are .given under their proper propositions. As the omis- sion of these exercises cannot affect the sequence of propositions, they may be disregarded at the discretion of the teacher. 5. Propositions and converses. Whenever possible, the converse of a proposition is given with the proposition itself. 6. Number of exercises. Besides the exercises directly following each proposi- tion, miscellaneous exercises are given at the end of each book. It may be found that there are more exercises given than can be covered by a class in the time allotted to the subject of Plane Geometry ; in which case the teacher will have to select from the lists given. While the exercises have been drawn from many sources, the author has availed himself in particular of the recent entrance examination papers of the best Ameri- can colleges and scientific schools. The author wishes to express his obligations to his colleagues in the Cincinnati Higli Schools for their criti- cism and encouragement, and especially to Miss Celia Doerner of Hughes High School for valuable suggestions and for her painstaking reading of the proof. CONTENTS PAGE Preliminary Definitions .9 Axioms 12 Postulates 13 Symbols and Abbreviations , . 14 BOOK I Rectilinear Figures 15 Additional Exercises 71 BOOK IT Circles 79 Additional Exercises , . . 122 BOOK III Proportions — Similar Polygons ...... 129 Additional Exercises <, 173 BOOK IV Areas of Polygons 181 Additional Exercises » 213 BOOK Y Measurement of the Circle . . . . o . . 220 Additional Exercises » . 243 5 INDEX OF MATHEMATICAL TERMS [The references are to articles] Abbreviations, list of, 29 Acute angle, 17 Adjacent angles, 15 Alternate exterior angles, 113 Alternate interior angles, 113 Alternation, in proportion, 416 Altitude, of parallelogram, 561 of trapezoid, 561 of triangle, 194 Analysis, 246 Angle, 13 acute, 17 at center of a regular polygon, degree of, 347 inscribed, 354 oblique, 17 obtuse, 17 right, 16 sides of, 13 vertex of, 13 Angles, adjacent, 15 alternate exterior, 113 alternate interior, 113 complementary, (yQ corresponding, 113 exterior, 113 homologous, 544 interior, 113 opposite, 75 supplementary, 65 vertical, 75 Antecedents, in proportion, 404 Apothem, 716 Arc, 21 degree of, 347 Area, 561 Axiom, 27 Axioms, list of, 27 716 Base, of isosceles triangle, 50 of parallelogram, 561 of polygon, 561 of trapezoid, 561 Center, of circle, 19 of regular polygon, 716 of similitude, 556 Chord, 247 Circle, 19 angle inscribed in, 354 center of, 19 circumference of, 19 diameter of, 20 inscribed in polygon, 247 radius of, 20 sector of, 343 segment of, 354 tangent to, 306 Circumscribed circle, 247 polygon, 247 Commensurable quantities, 342 Complementary angles, 66 Composition and division, 430 Composition, in proportion, 421 Conclusion, 22 Concurrent lines, 601 Consequents, in proportion, 404 Constant, 340 Continued proportion, 442 Converse, 72 Corollary, 25 Corresponding angles, 113 Curved line, 7 Decagon, 152 Degree, of angle, 347 of arc, 347 6 INDEX OF MATHEMATICAL TERMS Determination of straight line, 246 Diagonal, 152 Diameter, 20 Direct tangent, 656 Distance; from point to line, 223 from point to point, 223 Division, external, 506 in proportion, 427 internal, 506 Dodecagon, 152 Equiangular polygon, 152 triangle, 18 Equilateral polygon, 152 triangle, 18 Equivalent polygons, 561 Exterior angles, 113 External division, 506 Extreme and mean ratio, 551 Extremes, in proportion, 404 Fourth proportional, 404 Geometrical figures, 9 Geometry, 11 plane, 12 Harmonical division, 509 pencil, 512 Hexagon, 152 Homologous angles, 544 sides, 544 Hypotenuse, 43 Hypothesis, 22 Incommensurable quantities, 342 Indirect proof, 39 Inscribed angle, 354 circle, 395 Interior angles, 113 Inversion in proportion, 419 Isosceles triangle, 18 Isosceles triangle, base of, 50 vertex of, 50 Left side, of angle, 130 Legs, of right triangle, 43 Limit, 340 Line, 4 curved, 7 straight, 6 Lines, parallel, 107 perpendicular, 16 Locus, 233 Material body, 1 Mean proportional, 404 Means, of a proportion, 404 Median of a triangle, 173 Minutes, of arc, 347 Mutually equiangular triangles, 137 Oblique angle, 17 Obtuse angle, 17 Octagon, 152 Parallel lines, 107 Parallelogram, 194 altitude of, 561 bases of, 561 Pentagon, 152 Pentedecagon, 152 Perimeter of polygon, 152 Perpendicular lines, 16 Plane surface, 8 angle, 13 figure, 10 geometry, 12 Point, 5 Polar, 521 Pole, 521 Polygon, 152 circumscribed, 395 diagonal of, 152 inscribed, 247 perimeter of, 152 regular, 152 Polygons, similar, 477 8 INDEX OF MATHEMATICAL TERMS Postulate, 28 Problem, 23 Projection, 657 Proportion, 404 antecedents of, 404 consequents of, 404 couplets, 404 extremes of, 404 means of, 404 Proportional, fourth, 404 mean, 404 third, 404 Proposition, 24 Quadrarft, 346 Quadrilateral, 152 Quantities, commensurable, 342 constant, 340 incommensurable, 342 variable, 340 Radius, 20 Ratio, 340 extreme and mean, 551 Rays, of pencil, 512 Rectangle, 194 Regular polygon, 707 angle at center of, 7 16 apothem of, 716 center of, 716 Rhomboid, 194 Rhombus, 194 Right angle, 16 Right-angled triangle, 18 Right side of angle, 130 Scalene triangle, 18 Scholium, 26 Secant, 262 Seconds, of arc, 347 Sector, 343 Segment, 354 angle inscribed in, 354 Sides of angle, 13 of triangle, 18 Similar arcs, 787 polygons, 477 sectors, 787 segments, 787 Similitude, center of, 556 Solid, geometrical, 2 Square, 194 Straight line, 6 determined, 246 divided externally, 506 divided harmonically, 509 divided internally, 506 Supplementary angles, 65 Surface, 3 Symbols, list of, 29 Tangent circles and lines, 306 Tangent, direct, 556 transverse, 556 Theorem, 22 Third proportional, 404 Transversal, 113 Transverse tangent, 556 Trapezium, 194 Trapezoid, 194 altitude of, 561 bases of, 561 Triangle, 18 altitude of, 194 equiangular, 18 equilateral, 18 isosceles, 18 median of, 173 right-angled, 18 scalene, 18 Variable, 340 Vertex of angle, 13 Vertical angles, 75 in isosceles triangle, 50 PLANE GEOMETRY DEFINITIONS 1. Every material body occupies a limited portion of space. If we conceive the body to be removed, the space that is left, which is identical in form and magnitude with the body, is a geometrical solid. 2. A geometrical solid, then, is a limited portion of space. It has three dimensions : length, breadth, and thickness. 3. The boundaries of a solid are surfaces. A surface has but two dimensions : length and breadth. 4. The boundaries of a surface are lines. A line has length only. 5. The ends of a line are points. A point has position, but no magnitude. 6. A straight line is one that does not change its direction at any point. 7. A curved line changes its direction at every point. 8. A plane surface is a surface, such that a straight line joining any two of its points will lie wholly in the surface. • 9. Any combination of points, lines, surfaces, or solids, is a geometrical figure. 10. A figure formed by points and lines in a plane is a plane figure. 10 PLANE GEOMETRY 11. Geometry is the science that treats of the properties, the construction, and the measurement of geometrical figures. 12. Plane Geometry treats of plane figures. 13. A plane angle is the amount of divergence of two lines that meet. The lines are the sides of the angle, and their point of meeting is the vertex. One way to indicate an angle is by the use of three letters. Thus, the angle in the accompanying fig- ure is read angle ABC or angle CBA, the letter at the vertex being in the middle. If there is only one angle at the ver- tex B, it may be read angle B. Another way is to place a small figure or letter within the angle near the vertex. The above angle may be read angle S. The size of an angle in no way depends upon the length of its sides, and is not altered by either increasing or diminishing their length. 14. Two angles are equal if they can be made to coincide. Thus, angles ABC and DEF are equal, whatever may be the length of each side, if angle ABC can be placed upon angle DEF so that the vertex B shall fall upon vertex E,BC fall upon EF, and BA fall upon ED. [It should be noticed that angle ABC can be made to coincide with angle DEF in another way, i.e. ABC may be turned over and then placed upon DEF, EG falling upon ED, and BA upon EF.-\ 15. Two angles that have the same ver- tex and a common side are adjo/cent angles. Angles 1 and 2 are adjacent angles. DEFINITIONS 11 16. If a straight line meets another straight line so as to make the adjacent angles that they form equal to each other, the angles formed are right angles. Angles ABC and ABD are right angles. In this case each line is perpendicular to the other. 17. An angle that is less than a right angle is acute, and one that is greater than a right angle is obtuse. An angle that is not a right angle is oblique. 18. A triangle is a portion of a plane bounded by three straight lines. The lines are called the sides of the tri- angle, and their angles the angles of the triangle. An equilateral triangle has three equal sides. An isosceles triangle has two equal sides. A scalene triangle has no two sides equal. An equiangular triangle has three equal angles. A right-angled triangle contains one right angle, 19. A circle is a portion of a plane bounded by a curved line, all points of which are equally distant from a point within, called the center. The bounding line is called the circumference. 20. The distance from the center to any point on the circumference is a radius. 21. Any portion of a circumference is an arc. 12 PLANE GEOMETRY 22. A theorem is a truth requiring demonstration. The statement of a theorem consists of two parts, the hypothesis and the conclusion. The hypothesis is that part which is assumed to be true; the conclusion is that which is to be proved. 23. A problem proposes to effect some geometrical construc- tion, such as to draw some particular line, or to construct some required figure. 24. Theorems and problems are called propositions. 25. A corollary is a truth that may be readily deduced from one or more propositions, 26. A scholium is a remark made upon one or more proposi- tions relating to their use, connection, limitation, or extension. 27. An axiom is a self-evident truth. Axioms 1. Things that are equal to the same thing 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 equal. 4. If equals are multiplied by equals, the products are equal. . 5. If equals are divided by equals, the quotients are equal. 6. If equals are added to unequals, the sums are unequal in the same order. 7. If equals are subtracted from unequals, the remainders are unequal in the same order. 8. If unequals are multiplied by positive equals, the prod- ucts are unequal in the same order. DEFINITIONS 13 . 9. If unequals are divided by positive equals, the quotients are unequal in the same order. 10. If unequals are added to unequals, the greater to the greater, and the less to the less, the sums are unequal in the same order. 11. The whole is greater than any of its parts. 12. The whole is equal to the sum of all its parts. 13. Only one straight line can be drawn joining two points. [It follows from this axiom that two straight lines can intersect in only one point.] 14. The shortest distance from one point to another is meas- ured on the straight line joining them. 15. Through a point only one line can be drawn parallel to another line. 16. Magnitudes that can be made to coincide with each other are equal. [This axiom affords the ultimate test of the equality of geometrical magnitudes. It implies that a figure can be taken from its position, with- out change of form or size, and placed upon another figure for the purpose of comparison.] Of the foregoing, the first twelve axioms are general in their nature, and the student has probably met with them before in his study of algebra. The last four are strictly geometrical axioms. 28. A postulate is a self-evident problem. Postulates 1. A straight line can be drawn joining two points. 2. A straight line can be prolonged to any length. 3. If two lines are unequal, the length of the smaller can be laid off on the larger. 4. A circumference can be described with any point as a center, and with a radius of any length. 14 PLANE GEOMETRY 29. SYMBOLS AND Z Angle. A Angles. R.A. Right angle. R.A.'s. Right angles. A Triangle. k. Triangles. O Circle. © Circles. ± Perpendicular. Js Perpendiculars. 11 Parallel. lis Parallels. ABBREVIATIONS .-. Therefore. = Equals or equal. > Is (or are) greater than. < Is {or are) less than. ~ Is (or are) measured by. Prop. Proposition. Cor. Corollary. Schol. Scholium. Q.E.D. Quod erat demonstrandum, which was to be proved. Q.E.r. Quod erat faciendum, which was to be done. BOOK I Proposition I. Theorem 30. If two triangles have two sides and the included angle of one equal respectively to two sides and the included angle of the other, the triangles are equal in all respects. Let the A ABC and DBF have AB = DE^ BC = EF^ and Z.B = /.E. To Prove the A ABC and DEF equal in all respects. Proof. Place the A ABC upon the A DEF so that Z B shall coincide with its equal Z E, BA falling upon ED, and BC upon EF. Since, by hypothesis, BA = ED, the vertex A will fall upon the vertex D. Since, by hypothesis, BC = EF, the vertex c will fall upon the vertex F. Since, by Axiom 13, only one straight line can be drawn joining two points, ^(7 will coincide with DF. ,\ the A coin- cide throughout and are equal in all respects. q.e.d. 15 16 PLANE GEOMETRY 31. Scholium. By showing that the A coincide, we have not only proved that they are equal in area, but also that ZA = Z.D, /.C=/.F, and AC = DF. It should be noticed that the sides AC and DF, which have been proved equal, lie opposite respectively to the equal angles B and E. Also, that the equal angles A and D lie opposite respec- tively to the equal sides BC and EF, and that the equal angles C and F lie opposite respectively to the equal sides AB and DE. Principle. In triangles that have been proved equal in all respects, equal sides lie opposite equal angles, and equal angles lie opposite equal sides. D E 32. Exercise. Prove Prop. I. , using this pair of triangles. 33. Exercise. In the triangle ABC, AB = AC, and AD bisects the angle BAG. Prove that AD also bisects BG. Suggestion. Show by § 30 that the A ABD and ADC are equal in all respects. Then, by the principle of § 31, BD = DG. 34. Exercise. ABC is a triangle having AB = BC. BE is laid off equal to BD. AD = CE. Prove ion. Show that AABD = AEBC. BOOK I 17 Proposition II. Theorem 35. If two triangles have two angles and the included side of one equal respectively to two angles and the included side of the other, the triangles are equal in all respects. Let the A ABC and DBF have Z.A=/.D, /.c = Z.F, and AC = DF. To Prove the /^ ABC and DEF equal in all respects. Proof. Place the A ABC upon the A DEF, so that Z A shall coincide with its equal Z.D, AB falling upon DE, and AC falling upon DF. Since, by hypothesis, AC = DF, the vertex c will fall upon vertex F. Since, by hypothesis, Z C = Zf, the side CB will fall upon FE, and the vertex B will be on FE or its prolongation. Since AB falls upon DE, the vertex B will be upon DE or its prolongation. The vertex B, being at the same time on DE and FE, must be at their point of intersection ; and since two straight lines have only one point of intersection (Axiom 13), the vertex B must fall at E. .'. the A ABC and DEF coincide throughout, and are equal in all respects. q.e.d. SANDERS^ GEOM. — 2 18 PLANE GEOMETRY AwD 36. Exercise. Prove Prop. II., using this pair of triangles. 37. Exercise. In the A ^50, BD bisects /.ABC and is perpendicular to AC. Prove that BD bisects AC and that AB = BC. 38. Exercise. ABC is a A having Z BAG ^ = BCA. AD bisects ZBAC and CU bisects ZBCA. Prove AD = CE. Suggestion. Prove ^ ^DC and AEC equal in all respects by § 35. Then by the Principle of § 31, AD = EC. C 39. The next proposition is an example of what is called the indirect proof. The reasoning is based on the following Principle : If the direct consequences of a certain supposition are false, the supposi- tion itself is false. To prove a theorem by this plan, the following steps are necessary : 1. The theorem is supposed to be untrue. 2. The consequences of this supposition are shovru to be false. 3. Then, by the above Principle, the supposition (that the theorem is untrue) is false. 4. The theorem is therefore true. BOOK I 19 Proposition III. Theorem 40. At a given point in a line only one perpen- dicular can he erected to that line. Let CD be ± to ab at the point D. To Prove CD is the only _L that can be erected to AB at D. Proof. Suppose a second ±, as DE, could be erected to AB at D. By hypothesis and § 16, Z. CD A = Z. CDB. By supposition and § 16, /. EDA = Z EDB. But Z EDA > Z CD A, and Z EDB < Z CDB. .-, Z EDA cannot equal Z EDB, and DE cannot be ± to AB. The supposition that a second ± could be erected to AB at D is therefore false, and only one _L can be erected to AB at that point. Q.E.D. Note. The points and lines of the above figure, and of all figures given in the first five books of this geometry, are understood to be in the same plane. The term "line" is used in this work for "straight line." 20 PLANE GEOMETRY 41. CoBOLLARY. All right angles are equal. A D G Let /.ABC and Z def be 2 R.A/s. To Prove /.ABG = A DEF. Proof. Suppose them to be unequal and that /.ABC, when superimposed upon / I)EF, takes the position GEF. Then at E there would be two perpendiculars to EF, which contradicts § 40. Therefore the supposition that the right angles ABC and DEF are unequal is false, and they are equal. q.e.d. 42. Scholium. The right angle is the unit of measure for angles. An angle is generally expressed in terms of the right angle. Thus, Z ^ = | E.A., or / B = 1\ R.A., etc. 43. Definitions. In a right-angled triangle the side opposite the right angle is called the hypotenuse. The other two sides are the legs of the triangle. 44. Exercise. If two B.A. ^ have the legs of one equal respectively to the legs of the other, the A are equal in all respects. 45. Exercise. A is 40 miles west of B. C is 30 miles north of A^ and D is 30 miles south of A. From C to 5 is 50 miles. How far is it from DtoB? 46. Exercise. A is m yards north of B. G is n yards west of A, and Disn yards east of B. Prove that the distance from ^ to O is the same as the distance from A to D. BOOK I 21 Proposition IV. Theorem 47. If a perpendicular is drawn to a line at its middle point, I. Any point on the perpendicular is equally dis- tant from the extremities of the line. II. Any point without the perpendicular is un- equally distant from the extremities of the line. I. Let CD be ± to AB at its middle point D, and P be any point on CD. To Prove P equally distant from A and B. Draw PA and PB. [It is required to prove PA = PB, for PA and PB measure the distance from P to A and B respectively.] Proof. The A PAD and PBD have AD = DB (Hypothesis), Z 1 = Z 2 (Right Angles), PD = PD (Common). The A are equal in all respects by § 30. ,-. PA = PB, and P is equally distant from A and B. q.e.d. 22 ' PLANE GEOMETRY C II. Let , will CD bisect AB ? 58. Exercise. When will the lines AB and CD bisect each other ? 59. Exercise. In a given line find a point that is equally distant from two given points. When is this problem impossible ? 26 PLANE GEOMETRY Proposition YII. Theorem 60. The sum of the adjacent angles formed hy one line meeting another, is two right angles. B Let AB meet CD at 5. To Prove /.ABC -\-/. abb = 2 R.A.'s. Proof. Erect BE perpendicular to CD at B. (§ 53.) By construction /.EBC and Z.EBD are R. A.'s. /.ABC=1 R.A. -\-/EBA. Z ABD = 1 R.A. — Z EBA. Adding (1) and (2), Aabc-\-Z abd = 2 R.A's. (1) (2) Q.E.D. 61. Corollary I. If one of two adjacent angles formed by one line meeting another is a right angle, the other is also a right angle. 62. Corollary II. If two straight lines intersect each other, and one of the angles formed is a right angle, the other three angles are also right angles. 63. Corollary III. The sum of all the angles formed at a point in a line, and on the same side of the line, is two right angles. p Suggestion. Show that the sum of all D». the angles at C equals /.FCA + Z.FCB A ^¥f^ B or ZGCA + ZGCB, etc. BOOK I 27 64. Corollary IV. TJie sum of all the angles formed about a j)oint is four right angles. Suggestion. Prolong one of the lines, as OE, to a. Then apply § 63 to the angles on each side of GE. 65. Definition. If two angles are together equal to two right angles, they are called supplemen- tary angles. Each angle is the sup- plement of the other. Adjacent angles formed by one line meeting another are supplementary adjacent angles. 66. Definition. If two angles are together equal to one right angle, they are called complementary angles. Each angle is the complement of the other. 67. Exercise. Find the supplement and also the complement of each of the following angles : | R.A., ^ R. A,, | R.A. Find the value of each of two supplementary angles, if one is five times the other. 68. Exercise. complement. Given an angle, construct its supplement and also its 69. Exercise. Prove that the bisectors of two supplementary adjacent angles are perpendicular to each other. 70. Exercise. Through the vertex of a right angle a line is drawn outside of the angle. What is the sum of the two acute angles formed ? [Zl+Z2 = ?] 71. Exercise. Find the supplement of the complement of | R.A., also the complement of the supplement of If R. A. 28 PLANE GEOMETRY 72. Definition. One proposition is the converse of another, when the hypothesis and conclusion of one are respectively the conclusion and hypothesis of the other. The converse of a proposition is not necessarily true. We shall prove later (see § 85) that "if the sides of one triangle are equal respectively to the sides of another, the angles of the first triangle are equal respectively to those of the second." Show, by drawing triangles, that the converse of this propo- sition, i.e. " if the angles of one triangle are equal respectively to the angles of another, the sides of the first triangle are equal respectively to those of the second," is not necessarily true. Proposition VIII. Theorem (Converse of Prop. VII.) 73. If the sum of two adjacent angles is two right angles, their exterior sides form a straight line. ,0 Let Z CD A + Z CDB = 2 K.A.'s. To Prove AD and DB form a straight line. Proof. Suppose DB is not the prolongation of AD, and that some other line, as DE, is. By § 60 Z CDA + Z CDE would equal 2 R.A.'s. By hypothesis Z CDA + Z CDB = 2 R.A.'s. By Axiom 1., Z CDA -\- Z CDE would equal Z CDA + Z CDB. Whence Z CDE would equal Z CDB. This contradicts Axiom 11. Therefore the supposition that DB is not the prolongation of AD is false, and AD and DB form a straight line. q.e.d. BOOK 1 29 74. Exercise. ABO and DEF are R.A. A equal in all respects, right angled at B and D. (^ Place A ABC in the position of A GED. Prove that GD and DF form a straight line. 75. Definition. If two lines inter- sect each other, the opposite angles formed are called vertical angles. Z. 1 and Z 3 are vertical angles, as are also Z2 and Z4. 76. Exercise. The bisectors of two opposite angles form a straight line. Let FE, HE, GE, and JE be the bi- sectors of A AEG, CEB, BED, and BE A respectively. To Prove that FE and EG form a straight line, and HE and EJ form a straight line. Suggestion. Use §§ 69 and 73. Proposition IX. Theorem 77. If two straight lines intersect, the opposite or ver- tical angles are equal. Let AB and CD intersect. To Prove Z 1 = Z 3 and Z 2 = Z 4. Proof. Zl+Z2=2E.A.'s. (authority ?) Z2-f-Z3 = 2R.A.'s. (?) Z1+Z2 = Z2+Z3. (?) Zlz=Z3. (?) In the same manner prove Z 2 = Z 4. q.e.d. 30 PLANE GEOMETRY 78. Exercise. One angle formed by two intersecting lines is | R.A. Find the other three. 79. Exercise. The bisector of an angle bisects its vertical angle. 80. Exercise. Two lines intersect, making the sum of one pair of vertical angles equal to five times the sum of the other pair of vertical angles. Find the values of the four angles. Proposition X. Theorem 81. In an isosceles triangle, the angles opposite the equal sides are equal. Let ABC be an isosceles A, having AB =BC. To Prove /.A = Z.C. Proof. Draw BD bisecting AC. (§ 55.) B and D are each equally distant from A and C; .-. Zl and Z2 are E.A.'s. (?) " d Show that A ABD and BBC are equal in all respects. Whence /-A = /.c. q.e.d. 82. Corollary. An equilateral triangle is equiangular. 83. Exercise. ABC is an isosceles triangle. D is the middle point of the base AC E and F are the middle points of the equal sides ^5 and ^C. Prove DE = DF. 84. Exercise. ABC is an isosceles triangle having AB=BC. BD and BE are drawn making Zl = Z2. Prove Z3 = Z4. BOOK I 81 Proposition XI. Theorem 85. If two triangles havo three sides of the one equal respectively to three sides of the other, the triangles are equal in all respects. Let ABC and DEF be two A, having AB = DE^ BC = EF, and AC = DF. To Prove A ABC and DEF equal in all respects. Proof. Place A ABC so that AC shall coincide with DF, A falling on D and C on F, and the vertex B falling at G, on the opposite side of the base from the vertex E. Draw EG. Prove Zl = Z2andZ3 = Z4. Adding, Z1 + Z3 = Z2 + Z4, orZ DEF = Zdgf. Prove A DEF and DGF equal in all respects. .*. A DEF and ABC are equal in all respects. q.e.d. 86. Exercise. Construct a triangle having given its three sides. 87. Exercise. Construct a triangle equal to a given triangle. 88. Exercise, Construct a triangle whose sides are in the ratio of 3, 4, and 5. 32 PLANE GEOMETRY Propositiox XII. Problem 89. To draw a perpendicular to a line from a point without. C^-- .-^-'D Let AB be the given line and P the point without. Required to draw a perpendicular from P to the line AB. Let s be any point on the opposite side of AB from P. With P as a center, and Ps as a radius, describe an arc intersecting AB ?X C and D. With c as a center, and with a radius greater than one half of CD, describe an arc ; with i) as a center, and with the same radius, describe an arc intersecting the first arc at E. Draw PE. Show that PE is perpendicular to CD. q.e.f. 90. Exercise. from the point G. Draw a perpendicular to AB 91. Exercise. If the line AB (see § 89) were situated at the bottom of this page, and there were no room below it for the points, how could the perpen- dicular be drawn ? •C BOOK I 33 Proposition XIII. Theorem 92. From a point without a line only one perpendicu- lar can be draivn to the line. \ \ 21 E fV Let CD be a ± from C to AB. To Prove that CD is the only J_ that can be drawn from C to AB. Proof. Suppose a second J_, as CE, could be drawn. Prolong CD until DF = CD, and draw EF. Prove A CDE and FDE equal in all respects. Whence Z1 = Z2. But Z 1 = 1 R.A. by supposition. Show that Z 1 + Z 2 = 2 R. A.'s. If the sum of angles 1 and 2 is two E-.A.'s, CE and EF form, a straight line. (§ 73.) The points C and F are therefore connected by two straight lines (CDF and CEF), which contradicts (?). Therefore the supposition that a second _L could be drawn from C to the line AB is false, and only one _L can be drawn. Q.E.D. 93. Exercise. Show that a triangle cannot have two right angles. SANDERS' GEOM. 3 34 PLANE GEOMETRY Proposition XIV. Problem 94. To bisect a given angle. Let ABC be any angle. Required to bisect it. With ^ as a center, and with any convenient radius, describe an arc intersecting the sides of the angle at D and E. With Z) as a center, and with a radius greater than one half of DEj describe an arc ; with ^ as a center, and with the same radius, describe an arc intersecting this arc at F. Join B and F. Then will BF bisect Z.ABG. Draw FE and FD. Prove A BEF and BDF equal in all respects. Whence Z 1 = Z 2, and /. ABC i^ bisected. q.e.f. 95. Exercise. At a given point on a line construct an angle equal to I R.A. 96. Exercise. Divide a given angle into quarters. 97. Exercise. At a given point on a line construct an angle equal to \l R.A.'s. 98. Exercise. Prove § 81 by drawing BD (see figure of § 81) bisect- ing angle ABC. 99. Exercise. Construct a triangle ABC, making the side AB two inches long, Z ^ = 1 R. A. and Z 5 = ^ R. A. BOOK I 35 Proposition XV. Problem 100. At a point on a line to construct an angle equal to a given angle. C D Let /.ABC be the given angle, and F the point on the line BE. Required to construct an angle at F on the line BE that shall equal A ABC. With 5 as a center, and with any radius, describe the arc MG. With i^ as a center, and with the same radius, describe the indefinite arc LK^ intersecting BE at K. With iT as a center, and with the distance MG as a radius, describe an arc intersecting the arc LK afc H. Draw HF. Then will Z HFK =/.ABC. Draw MG and HE. Prove A MB G and HFK equal in all respects. Whence /.B=/.F. q.e.p. 101. Exercise. Construct a triangle having given two sides and the included angle. 102. Exercise. Construct a triangle having given two angles and the included side. 103. Exercise. Construct an angle equal to the sum of two given angles. 104. Exercise. Construct an angle that is double a given angle. 36 PLANE GEOMETRY 105. ExERcisK. Construct an angle equal to the difference between two given angles. 106. Exercise. Draw any triangle. Construct an angle equal to the sum of the angles of this triangle. From your drawing what do you infer the sum of the angles to be ? See § 138. 107. Definition. Parallel lines are lines lying in the same plane, which do not meet, how far soever they may be prolonged. Proposition XVI. Theorem 108. If two lines are parallel to a third line, they are parallel to each other. A C E Let AB and CD be II to EF. To Prove AB and CD II to each other. Proof. Since AB and CD are in the same plane, if they are not parallel they must meet. If they do meet we should have two lines drawn through the same point parallel to EF. This contradicts (?). Therefore they cannot meet, and, by definition (§ 107), are parallel. q.e.d. 109. Exercise. If a line be drawn on this page parallel to the upper edge, show that it is also parallel to the lower edge. 110. Exercise. Give an example of two lines that never meet, how far soever they be prolonged, and yet are not parallel. \_Note. — To do this the student must leave the province of plane geometry and think of lines in different planes.] BOOK I 37 Proposition XVII. Theorem 111. If two lines are perpendicular to the same line, they are parallel. E F Let AB and CD be ± to ef. To Prove AB and CD II to each other. Proof. If AB and CD are not parallel, they will meet at some point. (?) Then we should have two perpendiculars drawn from that point to EF. This contradicts (?). .-. AB and CD are parallel. q.e.d. 112. Problem. Through a given point to draw a line parallel to a given line. r Let P be the given point \ ,/ ; \ / and AB the given line. D — }{^ • )( E Required to draw through ' '^ I F a parallel to AB. \ :c A ^-r^ ^^-r^ B Draw PC ± to AB. ^~~r-"" Through P draw D^ J_ to C. Prove DE and AB parallel, q.e.f. PC. '-Jy 113. Definitions. A straight line that cuts two or more lines is called a transversal. 38 PLANE GEOMETRY If two lines are cut by a transversal, eight angles are formed, which are named as follows : The four angles [Zl, Z2, Z7, and Z8], lying without the two lines, are called exterior angles. The four angles [Z3, Z4, Z.5, and Z 6], lying within the two lines, are called interior angles. The two pairs of exterior angles [Z 1 and Z 7, Z 2 and Z 8], lying on the same side of the transversal, are called exterior angles on the same side. The two pairs of interior angles [Z3 and Z.6, Z4 and Z^, lying on the same side of the transversal, are called interior angles on the same side. The four pairs of angles [Z 1 and Z 5, Z 2 and Z 6, Z 3 and Z 7, Z4 and Z8], lying on the same side of the transversal, one an exterior and the other an interior angle, are called corresponding angles. The two pairs of exterior angles [Z 1 and Z 8, Z 2 and Z 7], lying on opposite sides of the transversal, are called alternate exterior angles. The two pairs of interior angles [Z 3 and Z 6, Z 4 and Z 5], lying on opposite sides of the transversal, are called alternate interior angles. The four pairs of angles [Z 1 and Z 6, Z 2 and Z 5, Z 3 and Z 8, Z 4 and Z 7], lying on opposite sides of the transversal, one an exterior and the other an interior angle, are called alternate exterior and iyiterior angles. 114. Exercise. Show that if any one of the following sixteen equations is true, the other fifteen equations are also true. 7 A BOOK I 39 1. Z3 = Z6. 9. Z3 + Z5 = 2R.A.'s. 2. Z4 = Z5. 10. Z4 + Z6 = 2R.A.'s. 3. Z1=Z8. 11. Zl + Z7.= 2R.A.'s. 4. Z2=Z7. 12. Z2 + Z8 = 2 R.A.'s. 5. Z1=Z6. 13. Zl + Z6 = 2 R.A.'s. 6. Z2 = Z6. 14. Z2 + Z5 = 2 R.A.'s. 7. Z3 = Z7. 15 Z3 + Z8 = 2 R.A.'s. 8. Z4 = Z8. 16. Z4 + Z7= 2 R.A.'s. Proposition XVIII. Theorem 115. If two lines are cut hy a transversal, mahing the alternate interior angles equal, the lines are parallel. Let AB and CD be cut by the transversal EF, making Z1=Z2. To Prove AB and CD parallel. Proof. From M, the middle point of SO, draw MH A- to CD, and prolong MH until it meets AB in some point G. Prove the A GMO and MSH equal in all respects. Whence Z.h = Ag. Z-H\s by construction a R.A. .-. Z (? is a R.A. AB and CD are parallel. (?) q.e.d. 40 PLANE GEOMETRY 116. Corollary. If two lines are cut by a transversal^ making any one of the following six cases true, the lines are parallel. 1. The alternate interior angles equal. 2. The alternate exterior angles equal. 3. The corresponding angles equal. 4. The sum of the interior angles on the same side equal to two E.A.'s. 5. The sum of the exterior angles on the same side equal to two E-.A.'s. 6. The sum of the alternate interior and exterior angles equal to two E-.A.'s. 117. Exercise. FE intersects AB and CD, making Z m = f R.A. What value must /.n have in order that AB and CD shall be i)arallel ? 118. Exercise. Through a given point to draw a parallel to a given line. (This exercise is to be based on § 115. Another solution was given in § 112.) [Through the given point P draw any line PM to the given line AB. Through in g P draw CD, making Z2 = Z1. Prove ^ CD parallel to AB. Work this exercise by making the alternate exterior angles equal; also by making the corresponding angles equal.] 119. Exercise. The sum of two angles of a triangle cannot equal two right angles. 120. Exercise. The bisectors of the equal angles 1 and 2 in the figure of § 118, are parallel. BOOK 1 41 Proposition XIX. Theorem 121. If two parallels are cut by a transversal, the alternate interior angles are eqiial . Let the parallel lines AB and CB be cut by the trans- versal EF. To Prove Aaos=Z. OSD. Proof. Suppose Zaos is not equal to Z OSD. Draw GH through 0, making Z GOS = Z OSD. GH and CD are parallel. (?) AB and CD are parallel. (?) Through there are two parallels to CD, which contra- dicts (?). .*. The supposition that ZAOS and Z OSD are unequal, etc. Q.E.D. 122. Corollary I. If two parallels are cut by a transversal, the six cases of § 116 are true. 123. Corollary II. If a line is perpendicular to one of two parallels, it is perpendicular to the other also. 124. Exercise, The bisectors of two alternate exterior angles, formed by a transversal cutting two parallel lines, are parallel. 125. Exercise. If a line joining two parallels is bisected, any other line through the point of bisection, and joining the parallels, is also bisected. 126. Exercise. If AB and CD are parallel (§ 117), and Zw = If R.A., find the values of the other seven angles. 42 PLANE GEOMETRY Proposition XX. Theorem 127. If two lines are cut hy a transversal, making the sum of the interior angles on the same side less than two right angles, the lines will meet if sufficiently produced. Let AB and CD be cut by EF, making Z 1 4- Z 2 < 2 R. A.'s. To Prove that AB and CD will meet. Proof. If AB and CD do not meet, they are parallel. (?) If they are parallel, Z 1 + Z 2 = 2 R.A.'s. (?) This contradicts (?). .*. they cannot be parallel and must meet. Q.E.n 128. Corollary. If two lines are cut by a transversal,- mak- ing any one of the six cases of § 116 untrue, the lines will meet if sufficiently produced. 129. Exercise. The bisectors of any two exterior angles of a tri- angle will meet. Prove that DA and FC meet. Suggestion. ZEAB<2 E.A.'s. (?) ZKIR.A. Z3<1 R.A. Similarly, Z4<1 R.A. Whence, /3 +Z 4<2 R.A.'s. BOOK I 43 130. Definition. Each angle, viewed from its vertex, has a right side and a left side. AB is the right side of /.ABC, and £C is its left side. Proposition XXI. Theorem 131. If two angles have their sides parallel, right side to right side, and left side to left side, the angles are equal. a Let Z 1 and Z. 2 have their sides parallel, right side to right side, and left side to left side. To Prove Z1 = Z2. Proof. Prolong AB and EF until they intersect. Z 1 = Z 3. (?) Z3 = Z2. (?) Z 1 = Z 2. (?) Q.E.D. 44 PLANE GEOMETRY 132. Corollary. If two angles have their sides parallel, right side to left side, and left side to right side, the angles are supplemeyitary. Let Zl and Z2 have their sides parallel, right side to left side and left side to right side. To Prove Zl+Z 2=2 R.A.'s. d— Proof. Z 2 = Z 3, Z 1+Z 3=2 E.A.'s. (?) .-. ZH-Z2 = 2R.A.'s. Q.E.D. 133. Exercise. ZlandZ2 have their sides parallel, right side to right side, etc. Z 2 and Z. 3 have their sides parallel, right side to right, etc. Prove that Z 1 = Z 3. Proposition XXII. Theorem 134. If the sides of one angle are perpendicular to those of another, right side to right side and left side to left side, the angles are equal. J E H Let Z 1 and Z 2 have DE ± to BC and FE A. to AB. To Prove Z 1 = Z 2. Proof. Draw BH II to ED and BJ II to FE. Z 3 = Z 2. (?) BE \^ 1. to BC (?) and JB is .L to AB. (?) Z3 + Z4 = 1R.A. and Zl+Z4 = l R.A. Z 3 = Z 1. (?) .-. Z 2 = Z 1. (?) Q.E.D. BOOK I 45 135. Corollary. If the of one angle are perpendicular to those of another, right side to left side and left side to right side, the angles are supplementary. To Prove Z 1 + Z 2 = 2 E.A.'s. Proof. Prolong AB to G. Show that Z3 = Z2. Zl + Z3==2E.A.'s. .-. Z l-f Z 2 = 2 E.A.'s. 136. Exercise. In A^JSC, AD is 1. to BC and CE ± to AB. Compare Zl and Z2. 137. Definition. Two triangles are mutually equiangular when ths angles of one are equal re- spectively to the angles of the other. Proposition XXIII. Theorem 138. The Slim of the interior angles of a triangle is two right angles. Let ABC be any A. To Prove Zl+Z2-hZ3 = 2 R. A.'s. Proof o Draw DE through the vertex B, parallel to AC. Z 4 = Z 2 and Z 5 = Z 3. (?) Z44-Z1 + Z5 = 2 R.A.'s. (?) Z2 + Zl+Z3 = 2R.A.'s. (?) Q.E.D. 46 PLANE GEOMETRY 139. Corollary I. If two aiigles of a triangle are knowrij the third can be found by subtracting their sum from two right angles. 140. Corollary II. If two angles of one triangle are equal respectively to two angles of another, the third angles are equal, and the triangles are mutually equiangular. 141. Corollary III. A triangle can contain only one right angle; and it can contain only one obtuse angle. 142. Corollary IV. In a right-angled triangle, the sum of the acute angles is one right angle. 143. Corollary V. Since an equilateral triangle is also equiangular, each angle is two thirds of a right angle. 144. Corollary VI. An exterior angle of a triangle (formed by prolongiyig a side) is equal to the sum of the two opposite interior angles of the triangle. 145. Exercise. One of the acute angles of a R.A. A is f R.A. What is the other ? 146. Exercise. Find the angles of a A, if the second is twice the first, and the third is three times the second. 147. Exercise. Find the angles of an isosceles A, if a base angle is one half the vertical angle. 148. Exercise. Given two angles of a triangle, construct the third. 149. Exercise. Prove that the bisectors of the acute angles of an isosceles right-angled triangle make with each other an angle equal to H R.A.'s. 150. Exercise. Prove that the bisector of an exterior vertical angle of an isosceles triangle is parallel to the base. BOOK I 47 151. Exercise. Prove § 138, using these figures. 152. Definitions. A portion of a plane bounded by straight lines is called a polygon. The bounding line of a polygon is its perimeter. A diagonal of a polygon is a straight line joining any two of its vertices that are not consecutive. A three-sided polygon is a triangle; a four-sided polygon is a quadrilateral; a five-sided polygon is a pentagon; a six- sided polygon is a hexagon; an eight-sided polygon is an octago7i; a ten-sided polygon is a decagon; and a fifteen-sided polygon is a. pentedecagon. A polygon whose angles are equal is an equiangular polygon. A polygon whose sides are equal is an equilateral polygon. A polygon that is both equilateral and equiangular is a regular polygon. 153. Exercise. Show that an equilateral triangle is regular. 154. Exercise. Show, by drawings, that an equilateral quadrilateral is not necessarily regular. 155. Exercise. How many diagonals can be drawn in a triangle ? In a quadrilateral ? In a hexagon ? 156. Exercise. How many diagonals can be drawn from one vertex in a polygon of n sides ? How many from all the vertices ? 48 PLANE GEOMETRY Proposition XXIV. Theorem 157. Tixe sum of the interior angles of a polygon is twice as many right angles as tJie polygon has sides, less four right angles Let ABC ... J^' be a polygon of n sides. To Prove that the sum of its interior angles is (2 n— 4) E.A.'s. Proof. From any point within the polygon, as 0, draw lines to all the vertices. The polygon is now divided into n A. (?) The sum of the angles of each A is 2 E.A.'s. (?) The sum of the angles of the w A is 2 w R. A.'s. (?) The sum of the angles of the polygon is equal to the sum of the angles of the A, diminished by the sum of the angles about O ; that is, by 4 R.A.'s. .-. the sum of the angles of the polygon is (2 n — 4) E.A.'s. Q.E.D. 158. Corollary. The value of each angle of an equiangular polygon of n sides is M.AJs. 159. Exercise. What is the sum of the interior angles of a quadri- lateral ? Of a pentagon ? Of a hexagon ? Of a polygon of 100 sides ? 160. Exercise. How many sides has the polygon in which the sum of the interior angles is 20 R.A.'s ? 26 R.A.'s ? 98 R.A.'s ? (2 s - 4) R.A.'s? BOOK I 49 161. Exercise. How many sides has the equiangular polygon in which one angle is f R.A. ? 1 R.A. ? If R.A. ? 114 R.A. ? 162. Exercise. How many sides has the equiangular polygon in which the sum of four angles is 6 R.A.'s ? C D 163. Exercise. Prove § 157, using this figure. Show that the polygon is divided into w — 2 triangles, the sum of the angles of which is equal to the sum of the angles of the polygon. Proposition XXV. Theorem 164. The sum of the exterior angles of a polygon, formed hy prolonging one side at eaeh vertex, is four R.A.'s. Let ^5 ... ^ be a polygon of n sides. To Prove that the sum of its exterior angles 1, 2, 3, etc., is 4 E.A.'s. Proof. The sum of each exterior angle and its adjacent interior angle is 2 R.A.'s. (?) 2n R.A.'s is the sum of all exterior and interior angles. (?) (2 n — 4) R.A.'s is the sum of the interior angles. (?) 4 R.A.'s is the sum of the exterior angles. (?) q.e.d. SANDERS' GEOM. — 4 60 PLANE GEOMETRY 165. Scholium. It is indifferent which side is prolonged at any vertex, as the exterior angles formed at any vertex by prolonging both sides are equal. 166. Exercise. How many sides has the polygon in which the sum of the interior angles is five times the sum of the exterior angles ? 167. Exercise. Complete the following table. The polygons are equiangular. No. of Sides. Value of each Interior Angle. Value of each Exterior Angle. 3 4 6 12 f R.A. IR.A. If R.A. f R.A. IR.A. |R.A. Proposition XXVI. Theorem 168. The sum of two sides of a triangle is greater than the third side, hut the difference of two sides of a triangle is less than the third side. Let ABC be any A. To Prove AB + bOAC, Proof. Apply axiom 14. Let BEF be any A. To Prove BE — EF< BF. Proof. BEUAB-\-BC-\-CA\ X-^"^^ k^^ ^C 173. Definition. A medial line of a triangle (or simply a median) is a line drawn from any vertex of the triangle to the middle point of the opposite side. 174. Exercise. A median to one side of a triangle is less than one half the sum of the other two sides. To prove BD BE. (?) Divide both members by 2, recol- E lecting that BD = DE and EG = AB. 175. Exercise. The sum of the three medians of a triangle is less than its perimeter. Suggestion. Use the preceding exercise. A^: i:^^B 176. Exercise. The lines AB and CD have their extremities joined by CB and AD. C Prove CB + AD>AB-\-CD. 62 PLANE GEOMETRY Proposition XXVII. Theorem 177. // froiiv a point within a triangle two lines are drawn to the ejctremities of a side, their sum is less than that of the two remaining sides of the triangle. Let ABC be any A, O any point within, and OA and OC lines drawn to the extremities of A C. To Prove OA + OCAO-\-OD. (?) OD-{-DC>OC. (?) Add these inequalities and show that AB -\-BC>AO -\- OC. Q.E.D. 178. Exercise. FroYe ZA0C> Z ABC. Suggestion. Show that ZAOC>ZODC and Z ODC> ZABC. Give another proof for this exercise without prolonging AO. 179. Exercise. The sum of the Hues drawn from a point within a triangle to the three vertices is less than the perimeter of the triangle. 180. Exercise. Prove that the perimeter of the star is greater than that of the polygon ABCDEF. BOOK 1 53 Proposition XXVIII. Theorem 181. If two triangles have two sides of the one equal respectively to two sides of the other, and the included angles unequal, the third sides are unequal, and the greater third side belongs to the triangle having the greater included angle. Let and To Prove C D the A ABC and DBF have ab = de, bc = ef Zb>Z.e. AC>DF. Proof. Of the two sides, AB and BC, let AB be the one which is not the larger. Draw B G, making Zabg=ZF; prolong B G, making BG= EF. Draw AG. Prove Aabg = A DEF, whence AG = DF. Draw BH bisecting Z GBC. Draw GH. Prove Agbh = Ahbc. Whence HG=HC. AH+HG>AG. (?) AOAG. (?) AC>J)F. (?) Q.E.D 54 PLANE GEOMETRY 182. Converse. If two triangles have two sides of tlie one equal respectively to two sides of the other, and the third sides unequal, the included angles are unequal, and the greater included angle belongs to the triangle having the greater third side. Let A ABC and DEF have AB = DE, BG = EF, and AC> DF. To Prove Zb>E. Proof. Zb= Ze, ov ZbZ.E. Show that Z B cannot equal Z E. Show that Z B cannot be less than Z E. .-. Zb>Ze. Q.E.D. 183. Exercise. B is fifty miles west of A. C is forty miles north of B, and D is forty miles southeast of B. Show that O is a greater distance from A than D is. 184. Exercise. In the isosceles triangle ABC, BD is drawn to a point D on the base AG s,o that AD>DC. Prove Z ADB >ZBDC. Suggestion. Compare A ABD and DBC, using § 182. Then compare A ADB and DBC, using §144. BOOK I 55 Proposition XXIX. Theorem 185. // two angles of a triangle are equal, the sides opposite them are equal. Let ^5C be a A having Z .4 = Z C. To Prove AB=BC. Proof. Draw BD bisecting Zb. Prove ^ABD and BDG mutually equiangular. Prove Aabd and BDC equal in all respects. Whence AB=BC. q.e.d. 186. Corollary. An equiangular tri- angle is equilateral. 187. Exercise. ABC is an isosceles triangle hSiYmgAB = BC. AD and DC bisect Z A and Z C respectively. Prove AD = DC. 188. Exercise. If the bisector of an angle of a triangle bisects the opposite side, it is also per- pendicular to that side, and the triangle is isosceles. Let BB bisect Z B and also bisect AC. To Prove BD ± to AC, and A ^5(7 isosceles. Suggestion. Prolong BD until DE = BD. Prove A ABD = ADEC. Whence Zl = Z6. Prove IlBCE isosceles. 56 PLANE GEOMETRY Proposition XXX. Theorem 189. If two sides of a triangle are unequal, the angles opposite to them are unequal, the greater angle being opposite the greater side; and conversely, if two angles of a triangle are unequal, the sides opposite them are unequal, the greater side lying opposite the greater angle. Let ^^C be a A having AB>BG. To Prove ZOZ.A. Proof. On AB lay off BD=BG and draw DC. Z1 = Z2. (?) Z1>Z^. (?) Z2>Z^. (?) Ac>Aa. (?) Let je:2?'(? be a A having Zg'>Zjs;. To Prove EF::^FG. Proof. Draw CD making Z 1 = Z J?. HG+HF>FG. (?) HG = EH. (?) EF > FG. (?) Q.E.D. Q.E.D. 190. Exercise. Prove the converse to this proposition indirectly. Show that EF can neither be equal to FG nor less than FG, and must consequently be greater than FG. 191. Exercise. ABC is a triangle having ^O^C AD bisects A A and BD bisects Z. B. Prove AD>BD. BOOK I 57 Proposition XXXI. Theorem 192. If two right-angled triangles have the hypote- nuse and a side of one equal respectively to the hypote- nuse and a side of the other, the triangles are equal in all respects. S E Let ABC and DEF be two R.A.A having hypotenuse ab=z hypotenuse DE, and AC = DF. To Prove the A ABC and DEF equal in all respects. Proof. Place A ABC so that AC coincides with its equal DF, A falling at D, and C at F, and the vertex B falling at some point G on the opposite side of the base DF from E. Show that EF and FG form a straight line. Show (in the Agde) that Z.G =/.E. Z3 =Z4. (?) Adfg and DFE are equal in all respects. (?) A ABC and DFE are equal in all respects. q.e.d. 193. Exercise. If a line is drawn from the vertex of an isosceles triangle ± to the base, it bisects the base and the vertical angle. 194. Definitions. A quadrilateral having its opposite sides parallel is called a parallelogram. A quadrilateral with one pair of parallel sides is a trapezoid. 58 PLANE GEOMETRY A quadrilateral with no two of its sides parallel is a trapezium. A parallelogram whose angles are right angles is a rectangle. A parallelogram whose angles are oblique angles is a rhomboid. A square is an equilateral rectangle; and a rhombus is an equilateral rhomboid. Proposition XXXII. Theorem 195. The opposite sides of a parallelogram are equal; and conversely, if the opposite sides of a quadrilateral are equal, the figure is a parallelogram. Fi Let ABGB be a parallelogram. To prove ab = CB and BC = AB, Proof. Draw the diagonal BB. Z 1 = Z 2. (?) Z3 = Z4. (?) Show that A abb =Abcb. Whence ab = CB and BC = AB. q.e.d. Let EFGH be a quadrilateral having EF^ GH and FG = EH. To prove EFGH a parallelogram. Proof. Draw the diagonal FH. Prove A EFH = A FGH. Whence Z 1 = Z2 and Z 3 = Z4. Since Zl = Z2, FG and EH are parallel. (?) Similarly EF is parallel to GH. EFGH is a parallelogram. q.e.d. BOOK I 59 196. Corollary I. xi diagonal of a parallelogram divides it into two triangles equal in all respects. 197. Corollary II. Two parallelograms are equal if they have two adjacent sides and the included angle of one equal respec- tively to two adjacent sides and the included angle of the other. 198. Corollary III. Parallels included between two paral- lels and limited by them, are equal. Proposition XXXIII. Theorem 199. The opposite angles of a parallelogram are equal ; and conversely, if the opposite angles of a quadrilateral are equal, the figure is a parallelogram. B< rC Let ABCD be a parallelogram. To Prove Z ^ = Z C and /.B = Z.D. Proof. Show by § 131 that Aa=/.C and Zb=Z.D. Q.E.D. Conversely. In the quadrilateral ABCD let /. A = /. C and Z.B = Z.D. To Prove ABCD a parallelogram. Proof. Z^-hZj5+Zc + Zi) = 4 K. A.'s. (?) Z JL = Z c and Z £ = Z D. 2Z^ + 2Z5=4R.A.'s. (?) Z ^ + Z i5 = 2 R. A.'s. (?) BG and AD are parallel. (?) Similarly prove AB and CD parallel. ABCD is a parallelogram. (?) q.e.d. 60 PLANE GEOMETRY 200. Corollary. Tlie adjacent angles of a parallelogram are supplementai'y ; and conversely, if the adjacent angles of a quadn- lateral are supplementary, the figure is a parallelogram. 201. Exercise. If one of the angles of a parallelogram is a right angle, the other three are also right angles. 202. Exercise. If one angle of a parallelogram is f R.A., how large are the others ? 203. Exercise. If two sides of a quadrilateral are parallel, and a pair of opposite angles are equal, the figure is a parallelogram. 204. Exercise. If an angle in one parallelogram is equal to an angle in another, the remaining angles are equal each to each. Proposition XXXIV. Theorem 205. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. \ Let ABCD be a quadrilateral having BC and AD equal and parallel. To Prove ABCD o, parallelogram. Proof. Draw the diagonal BD. AABD =ABCD. (?) Whence AB = CD. Prove ABCD Si parallelogram. [§ 195. Converse.] q.e.d. 206. Exercise. The line joining the middle points of two opposite sides of a parallelogram is parallel to each of the other two sides and equal to either of them. BOOK I 61 Proposition XXXV. Theorem 207. The diagonals of a parallelogram bisect each other; and conversely ^ if the diagonals of a quadri- lateral bisect each other, the figure is a parallelogram. Let ABCB be a parallelogram, DB and AC its diagonals. To Prove BO = OD and AO = OC. Proof. I'rove Aboc=AAOD, whence BO = OD and A Conversely. In the quadrilateral ABCD, Let AO = OC and BO = OB. To Prove ABCD a parallelogram. Proof. Prove Aboc =AA OD, whence Z 1 = Z 2 and BC = AD, Prove ABCD a parallelogram. (§ 205.) 208. Corollary I. The diagonals of a square b 1. Are equal. 2. Bisect each other. 3. Are perpendicular to each other. 4. Bisect the angles of the square. 209. Corollary II. The diagonals of a rhombus 1. Are unequal. B 2. Bisect each other. 3. Are perpendicular to each other. 4. Bisect the angles of the rhombus. To prove the diagonals unequal. = 0C. Q.E.D. 62 PLANE GEOMETRY first show that Z A and Z D of the rhombus are unequaL (They are supplementary and oblique.) Then apply § 181 to A ABD and ^CD. 210. Corollary III. The diagonals of a rectangle that is not a square 1. Are equal. ^ 2. Bisect each other. 3. Are not perpendicular to each other. 4. Do not bisect the angles of the rec- tangle. To prove that the diagonals are not perpendicular to each other, apply § 182 to Aboc and COD. (BC and CB are unequal because the rectangle is not a square.) To prove that the diagonals do not bisect the angles of the rectangle, show that A 4 and 5 of Aacb are unequal, but Z 3 = Z4. (?) .'. Z 3 and Z 5 are unequal. 211. Corollary IY. The diagonals of a rhomboid that is not a rhombus 1. Are unequal. 2. Bisect each other. 3. Are not perpendicular to each other. 4. Do not bisect the angles of the ^i rhomboid. 212. Exercise. Any line drawn through the point of intersection of the diagonals of a parallelogram and limited by the sides is bisected at the point. 213. Exercise. If the diagonals of a parallelogram are equal, the figure is a rectangle. 214. Exercise. Given a diagonal, construct a square. 215. Exercise. Given the diagonals of a rhombus, construct the rhombus. BOOK I 63 Proposition XXXYI. Theorem 216. If from a point without a line a perpendicular is drawn to the line, and oblique lines are drawn to differ- ent points of it, I. The perpendicular is shorter than any oblique line. II. Two oblique lines that ineet the given line at points equally distant from the foot of the perpendicular are equal. III. Of two oblique lines that meet the given line at points unequally distant from the foot of the perpen- dicular j the one at the greater distance is the longer. D C E F I. Let ABhQ the given line and P the point without, PC the ±, and PD any oblique line. To Prove PCpd. Suggestion. Show that Z 1 is obtuse. Then apply § 189, converse, to APEF, recollecting that PE - PD. 64 PLANE GEOMETRY 217. Corollary I. The perpendicular is the shortest distance from a point to a line, and conversely. 218. Corollary II. From a point without a line only two equal lines can he drawn to the line. Note. The number of pairs of equal lines that can be drawn from a point to a line is of course infinite. 219. Corollary III. If from a point without a line a per- pendicular and two equal oblique lines he drawn, the ohlique lines meet the given line at points equally distant from the foot of the perpendicular. Suggestion. Use § 192. 220. Definition. An altitude of a triangle is a perpendic- ular drawn from the vertex of any angle to the opposite side. 221. Exercise. The sum of the altitudes of a triangle is less than the perimeter. Proposition XXXYII. Theorem 222. Two parallels are everywhere equally distant. A- E F \ B "^ G H Let AB and CD be two ll's. To Prove that they are everywhere equally distant. Proof. From any two points on AB, as E and F, draw EG and FH ± to CD. They are also ± to ^5 (?), and they measure the distance between the parallels at E and F. EG and FH are parallel. (?) EG and FH are equal. (?) Therefore the parallels are equally distant at E and F. Since E and F are any points on AB, the parallels are every- where equally distant. q.e.d. BOOK I 65 223. Scholium. The term distance in geometry means shortest distance. The distance from one point to another is measured on the straight line joining them. (Axiom 14.) The distance from a point to a line is the perpendicular drawn from that point to the line. (§ 216.) The distance between two parallels is measured on a line perpendicular to both. (§ 222.) The distance between two lines in the same plane that are not parallel is zero ; for distaiice means shortest distance, and the lines will meet if sufficiently produced. 224. Corollary. If two points are on the same side of a given line and equally distant from it, the line joining the points is parallel to the given line. 225. Exercise. If the two angles at the B C extremities of one base of a trapezoid are / \ equal, the two non-parallel sides are equal. / | • \ Suggestion. Draw BE and CF ± to AD. j \ I \ BE=CF (?). Prove k. ABE and CDF L — i i — 1 equal. Whence AB = CD. ^ ^ ^ 226. Exercise. 11 the two non-parallel sides of a trapezoid are equal, the angles at the extremities of either base are equal. Suggestion. In the figure of the preceding exercise, prove A ABE and CFD equal. Whence ZA = ZD. 227. Exercise. If a quadrilateral has one pair of opposite sides equal and not parallel, and the angles made by these sides with the base equal, the quadrilateral is a trapezoid. Suggestion. In the figure of § 225, let AB = CD and ZA= ZD. Prove /S^ ABE and CFD equal, and then use § 224. 228. Exercise. If two points are on opposite sides of a line, and are equally distant from the line, the line joining them is bisected by the given line. 229. Exercise. If a rectangle and a rhomboid have equal bases and equal altitudes, the perimeter of the rectangle is less than that of the rhomboid. SANDERS' GEOM. 6 66 PLANE GEOMETRY Proposition XXXVIII. Theorem 230. Anj/ point on the bisector of an angle is equally distant from the sides of the angle; and any point not on the bisector is unequally distant from the sides. Let ABC be any angle, BD its bisector, and P any point on BD. To Prove P equally distant from AB and BC. Proof. Draw PE and PF perpendicular to AB and BC respec- tively. Prove A EPB = A PBF. Whence PE = PF. q.e.d. Let ABC be any angle, BD its bisector, and P any point with- out BD. To Prove P unequally distant from AB and BC. Proof. Draw PE and P// ± to AB and BC respectively. From F (where PE intersects BD) draw FG ± to BC. Draw./"i. FP -\- FG> PG. (?) PG > PH. (?) FP + ^'G > PH. (?) FE = FG. (?) FP -\- FE > PH. (?) PE > PH. (?) Q.E.J BOOK I 67 231. Corollary. Any point that is equally distant from the sides of an angle is on the bisector. 232. Exercise. Prove the second part of § 230 indirectly. Suppose PE= PG. Draw PB. q Prove A PEB = A PBG. Whence Z PBE = Z PBG. .: PB must bisect ZABC. Definition^. The locus of a point satisfying a certain condition is the line, lines, or part of a line to which it is thereby restricted ; provided, however, that the con- dition is satisfied by every point of such line or lines, and by no other point. The bisector of an angle is the locus of points that are equally distant from its sides; for by § 230, all the points on the bisector are equally distant from the sides, and all points with- out the bisector are unequally distant from the sides. 234. Exercise. What is the locus of points that are equally distant from a given point ? From two given points ? 235. Exercise. What is the locus of points that are equally distant from a given line ? 236. Exercise. What is the locus of points that are equally distant from a given circumference ? 237. Exercise. The bisectors of the interior angles of a triangle meet in a common point. To Prove that the bisectors AD, BF, and EC meet thus the chord AB subtends the arc ANB, and also the arc AMB. Unless the contrary is specially stated, we shall assume the chord to belong to the smaller arc. An inscribed polygon is a polygon whose vertices are in the circumference and whose sides are chords. [The polygon ABCD is inscnbed m the circle; the circle is also said to be circum- scribed about the polygon.] 79 80 PLANE GEOMETRY Proposition I. Problem 248. To find the center of a given circle. D Let xyz be the given circle. Required to find its center. Join any two points on the circumference, as A and B, by the line AB. Bisect AB by the perpendicular DC, Bisect DC. Then is the center of the circle. By definition, the center of the circle is equally distant from A and B. By § 48 the center is on DC. By definition the center of the circle is equally distant from D and C. Since the center is on DC, and is also equally distant from D and C, it must be at the middle point of DC, that is, at 0. Therefore, is the center of the circle xyz. q.e.f. 249. Corollary. A line that is perpendicular to a chord and bisects it, passes through the center of the circle. Note. It follows from § 249 that the only chords in a circle that can bisect each other are diameters. BOOK II 81 250. Exercise. Describe a circumference passing through two given points. How many different circumferences can be described passing through two given points ? 251. Exercise. Describe a circumference, with a given radius, and passing through two given points. How many circumferences can be described in this case ? What limit is there to the length of the given radius ? Proposition II. Theorem 252. A diameter divides a circle and also its circum- ference into two equal parts. Let AB be a diameter of the circle whose center is 0. To Prove that AB divides the circle and also its circumfer- ence into two equal parts. Proof. Place ACB upon ADB so that AB is common. Then will the curves ACB and ADB coincide, for if they do not there would be points in the two arcs unequally distant from the center, which contradicts the definition of circle. Therefore AB divides the circle and also its circumference into two equal parts. q.e.d. 253. Exercise. Through a given point draw a line bisecting a given circle. When can an infinite number of such lines be drawn ? SANDEKS' GEOM. — 6 82 PLANE GEOMETRY Propositiox III. Theorem 254. A diameter of a circle is greater than any other chord. Let AB be a diameter of the O whose center is O, and CD be any other chord. To Prove AB > CD. Proof. Draw the radii OC and OD. Apply § 168 to A OCD, recollecting that AB = OC + OD. Q.E.D. 255. Exercise. Prove this Proposition (§ 254), using a figure in which the given chord CD intersects the diameter AB. 256. Exercise. Through a point witliin a circle draw the longest possible chord. 257. Exercise. The side AC of an inscribed triangle ABC is a diameter of the circle. Compare the angle B with angles A and C. 258. Exercise. AB is perpendicular to the chord CD. and bisects it. Prove AB > CD. 259. Exercise. The diameter AB and £ the chord CD are prolonged until they meet at ^. Prove EA < EC and EB > ED. BOOK II 83 Proposition IV. Theorem 260. A straight line cannot intersect a circumference in more than two points. Let CDR be a circumference and AB d^ line intersecting it at C and D. To Prove that AB cannot intersect the circumference at any- other point. Proof. Suppose that AB did intersect the circumference in a third point E. Draw the radii to the three points. Now we have three equal lines (why equal?) drawn from the point to the line AB, which contradicts (?). Therefore the supposition that AB could intersect the circum- ference in more than two points is false. q.e.d. 261. Exercise, Show by §§ 249 and 92 that AB cannot intersect the circumference in three points ( O, Z>, and E ) . A, 262. Definition. A secant is a straight line that cuts a circumference. 84 PLANE GEOMETRY , Proposition Y. Theorem 263. Circles having equal radii are equal; and con- versely, equal circles have equal radii. Let the (D whose centers are and G have equal radii. To Prove the © equal. Proof. Place the O whose center is upon the O whose center is C, so that their centers coincide. Then will their circumferences also coincide, for if they do not, they would have unequal radii, which contradicts the hypothesis. Since the circumferences coincide throughout, the circles are equal. q.e.d. Conversely. Let the circles be equal. To Prove that their radii are equal. Proof. Since the circles are equal, they can be made to coincide. Therefore their radii are equal. q.e.d. 264. Exercise. Circles having equal diameters are equal ; and con- versely, equal circles have equal diameters. 265. Exercise. Two circles are described on the diagonals of a rec- tangle as diameters. How do the circles compare in size ? 266. Exercise. If the circle described on the hypotenuse of a right- angled triangle as a diameter is equal to the circle described with one of the legs as a radius, prove that one of the acute angles of the triangle is double the other. BOOK II 85 Proposition VI. Theorem 267. In the same circle or in equal circles, radii form- ing equal angles at the center intercept equal arcs of the circumference; and conversely, radii intercepting equal arcs of the circumference form equal angles at the center. Let ABC and BEF be two equal angles at the centers of equal circles. To Prove arc CA = arc DF. Proof. Place the circle whose center is B upon the circle whose center is E, so that Z B shall coincide with its equal Ze. Since the radii are equal, A will fall upon D and C upon F. The arc AC will coincide with the arc BF. (Why ?) Therefore the arc AC = arc DF. q.e.d. Conversely. Let arc CA = arc DF. To Prove Zabc = Zdef. Proof. Place the circle whose center is B upon the circle whose center is E, so that the circles coincide, and the arc AC coincides with its equal arc DF. BC will then coincide with EF (?) and AB with DE. (?) Consequently the angles ABC and DEF coincide and are equal. q.e.d. 268. Exercise. Two intersecting diameters divide a circumference into four arcs which are equal, two and two. 86 PLANE GEOMETRY Proposition VII. Theorem 269. In the same circle, or in equal circles, if tiuo arcs are equal, the chords that subtend them are also equal; and conversely, if two chords are equal, the arcs that are subtended by them are equal. Let ABC and DEF be two equal arcs in the equal © whose centers are x and y. To Prove chord AC == chord DF. Proof. Draw the radii xA, xC, yD, and yF, Show that Z1=Z2. Prove A AxC and DyF equal. Whence AC = DF. Conversely. Let chord AC = chord I)F. To Prove arc ABC = arc DFF. Proof. Draw the radii xA, xC, yD, and yF. Prove A AxC and DyF equal. Whence Z1 = Z2. .-. arc ABC = arc DFF. (?) q.e.d. q.e.d. 270. Exercise. If the circumference of a circle is divided into four equal parts and their extremities are joined by chords, the resulting quadrilateral is an equilateral parallelogram. BOOK II 87 Proposition VIII. Theorem 271. In the same circle, or in equal circles, if two arcs are unequal and each is less than a semi-circumfer- ence, the greater arc is subtended by the greater chord ; and conversely, the greater chord subtends the greater arc. Let M and N be the centers of equal circles in which arc ABOdiVC DEF. To Prove chord J C > chord DF. Proof. Draw the diameters AG and DH. Place the semicircle ACG so that it shall coincide with the semicircle DFH, A falling on D and G on H. Because the arc ABC is greater than the arc DEF, the point C will fall beyond F at some point R, the chord AC taking the position DR. Draw the radii NF and NR. Apply § 181 to A DNF and DNR, proving DR>DF. .: AC> DF. Q.E.D. Conversely. Let chord AC> chord DF. To Prove arc ^5C> arc DEF. Proof. Show that the arc ABC can neither be equal to the arc DEF nor less than it, .-. the arc ABC must be greater than the arc DEF. q.e.d. 88 PLANE GEOMETRY 272. Exercise. ^jBC is a scalene triangle. How do the arcs AB^ BC, and AC compare ? 273. Exercise. Give a direct proof for the converse of Prop. VIII. [Draw the radii and show that ZAMC is less than ZBND. Then place one circle upon the other, etc.] Propositio:^^ IX. Theorem 274. A diameter that is perpendicular to a chord bisects the chord and also the arc subtended by it. / 4\ \ \./ 1 2 y ^^ B Let AB be a diameter ± to CD. To Prove CE = ED and arc CB = arc BB. Proof. Draw the radii OG and OB. Prove A GOE and OEB equal. Whence CE — ED and Z.3 = Z 4. Show that arc GB = arc BD. Q.E.D. BOOK II 89 275. Corollary I. TJie diameter AB also bisects the arc CAD. 276. Corollary II. Prove the six propositions that can be formulated from the following, data, using any two for the hypothesis and the remaining two for the conclusion. A line that 1. Passes through the center of the O. 2. Bisects the chord. 3. Is perpendicular to the chord. 4. Bisects the arc. [Prop. IX. itself is one of the six proposi- tions, and is formed by using 1 and 3 as hypothesis, and 2 and 4 as conclusion ; and the statement of § 249 uses 2 and 3 for its hypothesis and 1 for its conclusion.] 277. Corollary III. Bisect a given arc. 278. Exercise. What is the locus of the centers of parallel ch'^rds in a circle ? 279. Exercise. Perpendiculars erected at the middle points of the sides of a quadrilateral in- scribed in a circle pass through a common point. Is this true for inscribed polygons of more than four sides ? 280. Exercise. Through a given point in a circle draw a chord that shall be bisected at the point. 281. Exercise. If the line joining the middle points of two chords in a circle passes through the center of the circle, prove that the chords are parallel. 282. Exercise. The chord AB divides the circumference into two arcs ACB and ADB. (See figure of § 276.) If CD is drawn connecting the middle points of these arcs, prove that it is perpendicular to AB and bisects it. 90 PLANE GEOMETRY Proposition X. Theorem 283. In the same circle or in equal circles equal chords are equally distant from the center; and conversely, chords that are equally distant from the center are equal. Let AB and CD be equal chords in the equal circles whose centers are M and N. To Prove AB and CD equally distant from the centers. Proof. Draw MR and NS 1. to AB and CD respectively. MR and NS measure the distance of the chords from the centers. (§ 223.) Draw the radii MB and ND. Prove the A MRB and NSD equal. Whence MR = NS. q.e.d. Conversely. Let AB and CD be equally distant from the centers (MR — NS). To Prove AB = CD. Proof. Prove A MEB and NSD equal. Whence RB = SD. Therefore AB = CD. (?) q.e.d. 284. Exercise. What is the locus of the centers of equal chords in a circle ? BOOK II 91 285. Exercise. AB and CD are two intersect- iug chords, and they make equal angles with the line joining their point of intersection with the center of the circle. How do AB and CD compare in length ? D 286. Exercise. If two equal chords intersect in a circle, the seg- ments of one chord are equal respectively to those of the other. 287. Exercise. If from a point without a circle two secants are drawn terminating in the concave arc, and if the line joining the center of the circle with the given point bisects the angle formed by the secants, the secants are equal. 288. Exercise. If two chords intersect in a circle and a segment of one of them is equal to a segment of the other, the chords are equal. 289. Exercise. The line joining the center of a circle with the point of intersection of two equal chords, bisects the angle formed by the chords. 290. Exercise. Through a given point of a chord to draw another chord equal to the given chord. \^Suggestion. — A^^\y § 285.] 291. Exercise. Through a given point in a circle only two equal chords can be drawn. For what point in the circle is this statement untrue ? 292. Exercise. If two equal chords be prolonged until they meet at a point without the circle, the secants formed are equal. 293. Exercise. Given three points A^ B, and C on a circumference, to determine a fourth point X on that circumference, such, that if the chords AB and CX be prolonged until they meet at a point without the circle, the secants formed are equal. 294. Exercise. An inscribed quadrilateral ABCD has its sides AB and CD parallel, and angles D and C equal. Prove that the sides AD and BC are equally distant from the center of the circle. 92 PLANE GEOMETRY Proposition XI. Theorem 295. In the same circle or in equal circles, the smaller of two unequal chords is at the greater distance from the center ; and conversely, of two unequal chords, the one at the greater distance from the center is the smaller. Let M and N be the centers of equal (D, and let AB < CD. To Prove that AB is at a greater distance from M than GB is from N. Proof. Place O xAB so that it coincides with O yCB^ B fall- ing on C and the chord AB taking the position CG. Draw NS and NF _L to GC and CD respectively. Draw 8F. Prove Z1>Z2. Whence Z 3 < Z 4. (?) Whence NS > I^F. (?) q.e.d. Conversely. Let JV5f > J^F. To Prove GC Z2. (?) CF>SC. (?) CD> GC. (?) Q.E.D 296. Exercise. Prove the converse to Prop. XI. indirectly. fShow that AB can neither be equal to nor greater than CD.'] BOOK II 93 297. Exercise. Through a point within a circle draw the smallest possible chord. Proposition XII. Theorem 298. Through three points not in the same straight line, one circumference, and only one, can be passed. (?) (?) Let A, B, and C be three points not in the same straight line. To Prove that a circumference, and only one, can be passed through A, B, and C. Proof. Draw AB and BC. Bisect AB and BChj the Js DE and FG. Draw DF. Show that Z 1 + Z 2 < 2 E.A.'s. Whence DE and FG meet. (?) is equally distant from A and B. is equally distant from B and C. Therefore is equally distant from A, B, and C. Therefore a circumference described with as a center, and with OA, OB, or OC as a radius, will pass thi*ough A, B, and C. The line DE contains all the points that are equally distant from A and B. (?) The line GF contains all the points that are equally distant from B and C. (?) Therefore their point of intersection is the only point that is equally distant from A, B, and C. Therefore only one circumference can be passed through A, By and C. Q.E.D. 94 PLANE GEOMETRY 299. Corollary. Two circumferences can intersect in only two points. 300. Exercise. Why cannot a circumference be passed through three points that are in a straight line ? 301. Exercise. Circumscribe a circle about a given triangle. 302. Exercise. Show, by using §§ 298 and 249, that the perpendicu- lars erected to the sides of a triangle at their middle points pass through a common point. 303. Exercise. Find the center of a given circle by using § 298. 304. Exercise. From a given point without a circle only two equal secants, terminating in the circumference, can be drawn. Suggestion. — Suppose that three equal secants could be drawn. Using the given point as a center and the length of the secant as a radius, describe a circle. Apply § 299. 305. Exercise. Circumscribe a circle about a right-angled triangle. Show that the center of the circle lies on the hypotenuse. 306. Definitions. A straight line is tangent to a circle when it touches the cir- cumference at one point only. The point at which the straight line meets the circum- ference is called the point of tangency. All other points of the straight line lie without the circumference. ^The circle is also said to be tangent to the line. Two circles are tangent to each other when their circumfer- ences touch at one point only. If one circle lies outside of the other, they are tangent externally; if one circle is within the other, they are tangent internally. BOOK II 95 Proposition XIII. Theorem 307. If a line is perpendicular to a radius at its outer extremity it is tangent to the circle at that point; and conversely, a tangent to a circle is perpendicular to the radius drawn to the point of tangency. Let ab\>q 1. to the radius CD at Z). To Prove AB tangent to the circle. Proof. Connect C with any other point oi AB as E. CE > CD. (?) Since CE is longer than a radius, E lies without the circumference. E is any point on AB (except D). Therefore every point on AB (except 7)) lies without the cir- cumference, and AB touches the circumference at D only. Q.E.D. Conversely. Let AB be tangent to the O at D. To Prove AB JL to CD. Proof. Connect C with any other point of AB as E. Since AB is tangent to the circle at D, E lies without the circumference. CE>CD. (?) CE is the distance from C to any point of AB (except D). CD is therefore the shortest distance from C to AB. :. CD is perpendicular to AB. q.e.d. 96 PLANE GEOMETRY CoROLLAKY I. At a giveu point on a circumference draw a tangent to the circle. Corollary II. At a point on a circumference only one tan- gent can be drawn to the circle. 308. Exercise. A perpendicular erected to a tangent at the point of tangency will pass through the center of the circle. 309. Exercise. If two tangents are drawn to a circle at the ex- tremities of a diameter, they are parallel. 310. Exercise. The line joining the points of tangency of two par- allel tangents passes through the center of the circle. 311. Exercise. If two unequal circles have the same center, a line that is tangent to the inner circle, and is a chord of the outer, is bisected at the point of tangency. 312. Exercise. Draw a line tangent to a circle and parallel to a given line. 313. Exercise. Draw a line tangent to a circle and perpendicular to a given line. 314. Exercise. If an equilateral polygon is inscribed in a circle, prove that a second circle can be inscribed in the polygon. 315. Exercise. Circumscribe about a given circle a triangle whose sides are parallel to the sides of a given triangle. 316. Exercise. To construct a triangle having given two sides and an angle opposite one of them. Let m and n be the two given sides, and Z s the angle oppo- site side n. ^ Required to construct the A. Lay off an indefinite line AD. At A construct Za=Zs. Make AB = m. With B as a center, and n as a radius, describe an arc intersecting AD at C. Draw BC. Show that A ^J5C is the required A. A- ^^^ BOOK II 97 Scholium. When the given angle is acute, and the side opposite the given angle is less than the perpen- dicular from B to AD J there is no construction. When the given angle is acute, and the side opposite the given angle is equal to the perpendicular from B to AD, there is one construction, and the A is right-angled. When the given angle is acute, and the side opposite the given angle is greater than the perpendicular from B to AD and is less than AB, there are two constructions. Both A ABC and A AB& fulfill the required conditions. When the given angle is acute, and the side opposite the given angle is equal to AB, there is one construction. When the given angle is acute, and the side opposite the given angle is greater than AB, there is one construction. A ABC fulfills the required conditions, but A ABC' does , not. If the given angle is ob- tuse, the opposite side must "~~~ — '"" be greater than AB (?), and there never can Lg more than one construction. yo 317. Exercise. Construct a triangle ABC in which AB = 5 inches, ZA = ^ BA, and side BC = 1, 2, 3, 4, and 5 inches in turn. State the number of solutions in each case. How long must ^C be in order to form a right-angled triangle ? SANDERS' GEOM. 7 98 PLANE GEOMETRY Proposition XIV. Theorem 318. Parallel lines intercept equal arcs of a circum- ference; and conversely, lines intercepting equal arcs of a circumference are parallel. I. Let AB and CD be parallel chords. To Prove arc ^C = arc BD. Proof. Draw the diameter EF A. to AB. EF is A- to CD. (?) EA = EB and EC = ED. (?) Whence AC = BD. Conversely. Let AC = BD. To Prove AB and CD parallel. Draw the diameter EF ± to AB. AE = EB. (?) AC = BD. (?) EC = ED. (?) EF is ± to CD. (?) AB and CD are parallel. (?) q.e.d. q.e.d 99 F F II. Let the tangent AB and tlie chord CD be parallel. To Prove CE = ED. Proof. Draw the diameter FE to the point of tangency E. FE is ± to AB. (?) FE is ± to Ci>. (?) CE = ED. (?) Q.E.D. Conversely. Let CE = ED. To Prove AB and CD parallel. Proof. Draw the diameter FE to the point of tangency E. Prove AB and CD each ± to EF. q.e.d. III. Let the tangents AB and CD be parallel. To Prove EMF = ENF. Proof. Draw the chord AT II to AB. XT is 11 to CD. (?) EX = EY and XF = YJ?^. (?) EMF = ^JV^F. Q.E.D. Conversely. Let EMF = ENF. To Prove the tangents AB and CD parallel. [The proof is left to the student.] 100 PLANE GEOMETRY 319. Exercise. ABCD is a trapezoid inscribed in the circle whose center is 0. Prove tliat the non-parallel sides AB and CD are equal. 320. Exercise. Prove the converse of the pre- ^ ceding exercise, i.e. if tw^o opposite sides of an in- scribed quadrilateral are equal, the quadrilateral is a trapezoid. 321. Exercise. The diagonals of an inscribed trapezoid are equal. 322. Exercise. The side AB of the inscribed angle ABC is a diameter. Prove that the diameter DE drawn parallel to BC bisects the arc AC. Proposition XY. Theorem 323. If two circumferences intersect each other, the line Joining their centers bisects at right angles their common chord. Let AB be the line joining the centers of two circumferences intersecting at C and D. ^^— --^c To Prove AB bisects CB at right angles. Proof. Use § 49. 324- Exercise. Prove § 32.3, using this figure. 325. Exercise. The centers of all circles that pass through C and D (figure of § 323) are on AB or its prolongation. BOOK II 101 Proposition XVI. Theorem 326. // two circles are tangent, either eocternally or internally, their centers and the point of tangency are in the same straight line. Let A and B be the centers of two (D tangent externally at C. To Prove that A, C, and B are in the same straight line. Proof. Draw the radii AC and BC to the point of tangency. It is required to prove that ACB is a straight line. If it can be shown that A CB is shorter than any other line joining A and B, then, by Axiom 14, ACB is a straight line. I. To show that ACB is shorter than any other line joining A and B and passing through C. Let AinnB be any other line joining A and B and passing through C. AC-\-CB < AmC + CnB. (?) or ACB < AmnB. II. To show that ACB is shorter than any line joining A and B and not passing through C. Join A and B by any line ADB not passing through C. Since the circles touch at C only, any line joining the centers and not passing through C must pass outside of the circles, and must be greater than the sum of the radii. .-. ACB < ADB. ACB is the shortest distance between A and B. .'. ACB \?> ^ straight line. q.e.d. 102 PLANE GEOMETRY Let A and B be the centers of two circles tangent inter- nally at C. To Prove that A, B, and C are in a straight line. Proof. At C draw DE tangent to the outer circle. (?) All the points of DE except C lie entirely without the outer circle, and consequently entirely without the inner circle. DE touches the inner circle at C only, and is tangent to it also. Draw the radii A C and ^ C to the point of tangency . AC and BC are each ± to DE. (?) A, B, and C are in a straight line. (?) q.e.d. 327. Corollary. If two circles are tangent, either exter- nally or internally, and if at their point of tangency a line is drawn tangeyit to one of the circles, it is tangent to the other also. 328. Exercise. Two circles are tangent, and the distance between their centers is 10 in. The radius of one circle is 4 in. What is the radius of the other ? (Two solutions.) 329. Exercise. Draw a common tangent to two circles tangent to each other. (§327.) How many common tangents can be drawn to two circles that are tangent internally ? Tangent externally ? [In the latter case the student is expected at present to draw only one of the three common tangents.] BOOK II 103 Propositiox XVII. Theorem 330. a. If two circles are entirely without each other and are not tangent, the distance between their centers is greater than the sum of their radii. h. If two circles are tangent externally, the distance between their centers is equal to the sum of their radii. c. If two circles intersect, the distance between their centers is less than the sum and greater than the dif- ference of their radii. d. If two circles are tangent internally, the distance between their centers is equal to the difference of their radii. e. If one circle lies wholly within another, and is not tangent to it, the distance between their centers is less than the difference of their radii. AB > sum of radii. (?) AB passes through C. (?) AB = sum of radii. (?) Draw the radii AC and BC. AB < sum of radii. (?) AB > difference of radii. (?) 104 PLANE GEOMETRY AB prolonged passes through C. (?) AB = difference of radii. (?) AD is the radius of the large O. -BC7 is the radius of the small O. What is tlie difference of the radii ? AB CD. Lay off AE = CD. Now, by the definition of limit, AM can be made to differ from AB by less than any assignable quantity, and therefore by less than EB. So ^i^/ may be greater than AE. By the definition of limit, CN < CD. But since AE = CD, CN AE and CN < AE-, but by hypothesis AM and CN are always equal. The result being absurd, the supposition that AB and CD are unequal is false. Therefore AB and CD are equal. q.e.d. 342. Definition. Two magnitudes are commensurable when they have a common unit of measure ; i.e. when they each con- tain a third magnitude a whole number of times. Two magnitudes are incommensurable when they have no common unit of measure ; i.e. when there ex- ists no third magnitude, however small, that is contained in each a whole number of times. A 343. Definition. A sector is that part of a circle included between two radii and their intercepted arc. BOOK II 107 Proposition XIX. Theorem 344. In the saiive circle or in equal circles, two angles at the center have tiis same ratio as their intercepted arcs. Case I When the angles are commensurable. Let ABC and DBF be commensurable angles at the centers of equal (D. To Prove Zabc DF Abef Proof. Since A ABC and BEF are commensurable, they have a common unit of measure. Let Z X he this unit, and suppose it is contained in /.ABC m times, and in Z BEF n times. Zabc m^ n Whence (1) Zbef The small angles into which A ABC and BEF are divided are equal, since each equals Z x. By § 267, the arcs into which AC and BF are divided by the radii are equal. Since AC is composed of m of these equal arcs, and BF of n of these equal arcs, j^c m BF n Apply Axiom 1 to (1) and (2). Zabc _Ac Z BEF ~ BF (2) Q.E.D 108 PLANE GEOMETRY F Case II When the angles are incommensurable. Let ABC and DEF be two incommensurable angles at the centers of equal ©. m T^ /.ABC AC To Prove — = Z.DEF DF Proof. Let Z. DEF be divided into a number of equal angles, and let one of these be applied to Z ABC as a unit of measure. Since A ABC and DEF are incommensurable, ABC will not contain this unit of measure exactly, but a certain number of these angles will extend as far as, say, ABG, leaving a remainder Z GBC, smaller than the unit of measure. Since A ABG and DEF are commensurable, (?) ^^^ = ^ by Case I. A DEF DF By increasing indefinitely the number of parts into which A DEF is divided, the parts will become smaller and smaller, and the remainder A GBC will also diminish indefinitely. Now is evidently a variable, as is also — , and these AdEF •" DF variables are always equal to each other. (Case I.) The limit of the variable —, is — -' A DEF A DEF The limit of the variable — is DF DF By §341, ^ABCl^AC, ^.^.^, '' A DEF DF BOOK II 109 345. Corollary. In the same circle, or in equal circles, sectors are to each other as their arcs. [The proof is analogous to that of the Proposition, substituting sector for angle.^ 346. Scholium. If two diameters are drawn perpendicular to each other, four right angles are formed at the center of the circle. By § 267, the circumference is divided into four equal arcs called quadrants. If one of these right angles were divided into any number of equal parts, it could be shown by § 267, that the quadrant subtending the right angle is also divided into the same number of equal parts. If, for example, the right angle at the center were divided into four equal parts, the arcs intercepted by the sides of these angles would each be one fourth of a quadrant ; and conversely, radii intercepting an arc that is one fourth of a quadrant, form an angle at the center which is one fourth of a right angle. If any angle as Z Z) OM be taken at random and compared with a right angle, B S S44- Z.DOM _ DM E. A. quadrant' i.e. the angle DOM is the same part of a right angle that its intercepted arc is of a quadrant. In this sense an angle at the center is said to be measured by its intercepted arc. 347. Scholium. A quadrant is usually conceived to be divided into ninety equal parts, each part called a degree of arc. The angle at the center that is measured by a degree of arc is called a degree of angle. The degree is divided into sixty equal parts called minides, and each minute is again subdivided into sixty equal parts called seconds. Degrees, minutes, and seconds are designated by the symbols °, ', " respectively. Thus, 49 degrees, 27 minutes, and 35 sec- onds, is written 49° 27' 35". 110 PLANE GEOMETRY 348. Exercise. Add 23° 46' 27" and 19° 21' 36". 349. Exercise. Subtract 15° 42' 39" from 93° 16' 25". 350. Exercise. How many degrees in an angle of an equilateral triangle ? 351. Exercise. Multiply 13° 27' 35" by 3, and add the product to one half of 12° 15' 10". 352. Exercise. How many degrees are there in each angle of an isosceles right-angled triangle ? 353. Exercise. Express in degrees, minutes, and seconds the value of one angle of a regular heptagon. 354. Definition. An iyiscrihed angle is an angle whose vertex is in the circumference and whose sides are chords. The symbol '^ is used for the phrase is measured by. Thus, Z ABC ~ arc ^C is read : The angle ABC is measured by the arc AG. A segment is that part of a circle which is include between an arc and its chord. [ACB and ADB are both segments.] An angle is inscribed in a segment when its vertex is in the arc of the segment and its sides terminate in the extremities of that arc. [Z ABC and Z.ADC are inscribed in the seg- ment AmC.'] BOOK II 111 Proposition XX. Theorem 355. An inscribed angle is measured by one half of the arc intercepted by its sides. Let Z.ABC be an inscribed angle having a diameter for one of its sides. To Prove Z.ABC <^\AC. Proof. Draw the radius 0(7. Prove A1 = 2Zb. Zl'^AC. (§346.) .-. Z J5, which is one half Z 1, is measured by one half the arc A C. Case II Let A ABC be an inscribed angle hav- ing the center between its sides. To Prove Zabc ^^AC, Draw the diameter BD. Z ABD ^ 1 AB. (Case I.) ZbbC'^^dc. (Case I.) Z ABC, which is the sum of A ABD and DBC, is measured by the sum of their measures (^ AD -^^DC), that is, hj^AC, q.e.d. Q.E.D. 112 PLANE GEOMETRY Case III Let Z.ABC be an inscribed angle hav- ing the center without its sides. To Prove Zabc ^ } A C. Proof. Draw the diameter BD. ZdBC ^^DC. (?) /.DBA '^^DA. (?) /.ABC, which is the difference be- tween A DBC and DBA, is measured by the difference of their measures {^DC — ^DA), that is, by i ^C. q.e.d. 356. Corollary I. An the same segment are equal. les inscribed 357. Corollary II. Angles inscribed in a semicircle are right angles. [/I'^^ABC. But 1 of the arc ABC is a quadrant. Therefore, by § 346, Z 1 is a right angle.] 358. Corollary III. A71 angle inscribed in a segment that is greater than a semicircle is acute. 359. Corollary IY. An angle inscribed in a segment that is less than a semicircle is obtuse. BOOK II 113 360. Corollary V. The opjyosite angles of an inscribed quadrilateral are supple- mentary. [Show that the sum of the measures of zi 1 and 2. is a semicircumference, or two quadrants.] 361. Exercise. The sides of an inscribed angle intercept an arc of 50°. What is the size of the angle ? 362. Exercise. How many degrees in an arc intercepted by the sides of an inscribed ajigle of 40° ? 363. Exercise. If the opposite angles of a quadrilateral are supple- mentary, a circle may be circumscribed about it. (Converse of Cor. V.) [Pass a circumference through three of the vertices. Then show that the fourth vertex can fall neither without nor within the circumference.] 364. Exercise. Show by § 355 that the sum of the angles of a triangle is two right angles. 365. Exercise. Any parallelogram inscribed in a circle is a rectangle. 366. Exercise. Two circles are tangent at A. AD and AE are drawn through the extremities of a diameter BG. Prove that DE is also a diameter. I ^/ 367. Exercise. Prove the preceding exercise when the two circles are tangent externally. ^ 368. Exercise. The angles of an inscribed trapezoid are equal two and two. 369. Exercise. Prove § 355, Case I, using the figure of § 322. 370. Exercise. Two chords AB and CD intersect in a circle at the point E. Their extremities are joined by the lines AC and DB. Prove the A ACE and BDE mutually equiangular. 371. Exercise. The sum of one set of alternate angles of an inscribed octagon is equal to the sum of the other set. Sanders' geom. — 8 114 PLANE GEOMETRY Proposition XXI. Theorem 372. An angle formed hy two intersecting chords is measured by one half the sum of the arc intercepted hy the sides of the angle and the arc intercepted hy the sides of its vertical angle. Let Zl be an angle formed by the intersecting chords AB and CD. To Prove Z 1 ~ i (^D + 5C). Proof. Draw the chord AC. Z1 = Z2 + Z3. (?) Z.2^^BC. (?) Z3~i^i). (?) Since Z 1 is the sum of ^ 2 and 3, it is measured by the sum of their measures, .'. Z\^\{AD -\- EC). Q.E.D. 373. Exercise. Derive the measure of Z 4 in the above figure. 374. Exercise. If in the above figure the arc C BG contains 124° and the arc AD contains 172°, how many degrees in Z 1 ? 375. Exercise. Prove /.l-^l^AC + BD), using this figure. {DE is drawn parallel to AB.~\ 376. Exercise. If angle 1 (figure § 375) contains 85° and arc BG contains 55°, how many degrees in the arc AD ? 377. Exercise. Four points A, B, G, and D are so taken in a cir- cumference that the arcs AB, BG, GD, and DA form a geometrical pro- gression (AB = 2 BG, BG = 2 GD, etc.). Find the values of each of the angles formed by the intersection of the chords AC and BD. BOOK II 115 Proposition XXII. Theorem 378. An angle formed hy a chord jneeting a tangent at the point of tangency is measured hy one half the arc intercepted hy its sides. Let Zl be an angle formed by the chord AB and the tan- gent CD. To Prove Z 1 ~ i amb. Proof. Draw the diameter EB to the point of tangency. Z^5(7 = 1R.A. (?) A A right angle is measured by a quadrant. (?) \ arc EMB is a quadrant. (?) Zebc^^emb. ZEBAr^\AE. (?) Zl, which is the difference between Zebc and Zeba, is measured by the difference of their measures. Zl^^EMB —^EA. Zl^^AMB. Q.E.D. Similarly, it may be shown that Z ABD, which is the sum of R.A. EBB and Z EBA, is measured by the sum of their meas- ures, which is I" arc AEB. 379. Exercise. A chord that divides a circumference into arcs con- taining 80"^ and 280", respectively, is met at one extremity by a tangent. What are tlie angles formed by the lines ? 380. Exercise. A chord is met at one extremity by a tangent, mak- ing with it an angle of 55°. Into what arcs does the chord divide the circumference ? 116 PLANE GEOMETRY 381. Exercise. If two circles are tangent either externally or in- ternally, and through the point of contact two lines are drawn meeting one circumference in B and D and the other in E and C, BD and EC are parallel. [Draw the common tangent mn. Show that Z.3 and Z2 each equals Zl.] 382. Exercise, If tangents be drawn to the two circles at the points B and C (see the figures of the preceding exercise), prove they are parallel. Propositiox XXIII. Theorem 383. An angle formed hy two secants meeting without the circle is measured by one half the difference of the arcs intercepted hy its sides. Let Z 1 be an angle formed by the two secants AB and CB. To Prove Z 1 ~ i (.ic - be). Proof. Draw the chord CE. Z1=Z2-Z3. (?) Z 1 is therefore measured by the difference of the measures of Z 2 and 3, i.e. by ^{AC — BE). q.e.d. BOOK II 117 384. Exercise. If the secants AB and CB in the figure of § 383 intercept arcs of 70° and 42°, what is the size of p. 385. Exercise. Prove § 383, this figure. [Di^is II to ^C] using Proposition XXIY. Theorem 386. An angle formed by a tangent and a secant meet- ing without the circle is measured by one half the dif- ference of the arcs intercepted by its sides. Let Zl be an angle formed by the tangent AB and the secant CB. To Prove Z 1 ~ i (J C — AD). Proof. Similar to that of § 383. Exercise. Prove § 386, using this figure. \_EA is II to ^C] 387. Exercise. A tangent and a secant meeting without a circle form an angle of 35°. One of the arcs intercepted by them is 15°. How many degrees in the other ? 388. A triangle ABC is inscribed in a circle. The angle B is equal to 50°, and the angle C is equal to 60°. What angle does a tangent at A make with BC produced to meet it ? 118 PLANE GEOMETRY Proposition XXV. Theorem 389. An angle formed hy two tangents is measured hy one half the difference of the arcs intercepted hy its sides. Let Z 1 be an angle formed by the tangents AB and CB. To Prove Z 1 ~ i {ANC — AMC). Proof. Similar to that of §§ 383 and 386. Exercise. Prove § 389, using this D^ figure. IAD is drawn parallel to ^0. ] N Exercise. Prove § 389, using this figure. [BD is any secant drawn from 2?.] 390. Exercise. The angle formed by two tangents is 74°. How many degrees in each of the two arcs intercepted by them ? BOOK II 119 Proposition XXVI. Problem 391. Through a given point to draw a tangent to a Hven circle. Case I When the given point is on the circumference. Let A be the given point on the circumference of the circle whose center is 0. Required to draw a tangent to the circle through A. See § 307. Case II When the given point is with- out the circumference. Let A be the given point with- out the circle whose center is 0. Required to draw a tangent to the circle through A. Draw OA. On OA as a diameter, describe a circumference, cutting the given circumference at B and C. Draw AB and AC. AB and AG are the required tangents. Draw the radii OB and OC. Z 1 is a right angle. (?) AB is tangent to the circle. (?) Similarly, ^C is tangent to the circle, q.e.f 120 PLANE GEOMETRY B E Case II. Second Method 392. With A as center and ^0 as a radius, describe the arc DE. With O as a center and the diameter of the given circle as a radius, describe an arc cutting DE 2A, B. Draw OB intersecting the given circle at C. Draw AC. Then ^C is the required tangent. [The proof is left for the student.] 393. Corollary. The two tangents drawn from a point to a circle are equal ; and the line joining the point with the center of the circle bisects the angle between the tangents, and also bisects the chord of contact (B C in the figure to first method) at right angles. 394. Scholium. When the given point is without the circle, two tangents can be drawn ; when it is on the circumference, one, and when it is within the circle, none. 395. Definition. A polygon is circumscribed about a circle when each of its sides is tangent to the circle. In this case the circle is said to be inscribed in the polygon. 396. Exercise. If a quadrilateral is circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. Suggestion. Use § 393. 397. Exercise. From the point A two tangents AB and AC are drawn to the circle whose center is 0. At any point D on the included arc BC^ a third tangent FE is drawn. Prove that the perimeter of the A AEF is constant, and equal to the sum of the tangents AB and AC, BOOK II 121 398. Exercise. To inscribe a circle in a given triangle. Bisect two of the angles. Show that their point of meeting is equally distant from the three sides. .-. the three perpendiculars 01, 02, and 03 are equal. With as a center and with 1 as a radius, describe the required circle. Proposition XXYII. Problem 399. On a given line to construct a segment that shall contain a given angle. Let AB hQ the given line and Z M the given angle. Required to construct on AB a segment that shall contain Z M. Draw CD through J5, making /.I = /. M.' Erect BE ± to CD and bisect AB hy the J_ FG. Prove that BE and FG meet at some point 0. Show that is equally distant from A and B. With as a center describe a circle passing through A and B. X>C is tangent to this circle. (?) Z 1 ~ ^ AB. (?) Inscribe any angle as /. ASB in the segment ARB. ZASB^^AB. (?) ZASB = Zl = Z3f. (?) The segment ARB is the required segment, since any angle inscribed in it is equal to Z M. q.e.f. 122 PLANE GEOMETRY 400. Exercise. On a given line construct a segment that shall con- tain an angle of 135°. 401. Exercise. What is the locus of the vertices of the vertical angles of the triangles having a common base and equal vertical angles ? 402. Exercise. Construct a triangle, having given the base, the vertical angle, and the altitude. 403. Exercise. Construct a triangle, having given the base, the vertical angle, and the medial line to the base. EXERCISES 1. Two secants, AB and AC^ are drawn to the circle, and AB passes through the center. A Prove AB>AC. 2. One angle of an inscribed triangle is 42°, and one of its sides subtends an arc of 110°. Find the angles of the triangle. 3. Two chords drawn perpendicular to a third chord at its extremities are equal. [Show tliat BC and AD are diameters, and that k^ABC and ABB are equal] 4. AB and CD are two chords intersecting at E, and CE=BE. Prove AB = CD. 5. ABC is a triangle inscribed in the circle, whose center is O. OD is drawn perpendicular to AC. Prove ADOC^-AB. BOOK II 123 6. What is the locus of the centers of circles tangent to a line at a given point ? 7. P is any point within the circle whose center is 0. Prove that PA is the shortest line and PB the longest line from P to the circumference. 8. If a circle is described on the radius of an- other circle as a diameter, any chord of the greater circle drawn from the point of contact is bisected by the circumference of the smaller circle. 9. If from a point on a circumference a number of chords are drawn, find the locus of their middle points. (Ex. 8.) 10. From two points on opposite sides of a given line, draw two lines meeting in the given line, and making a given angle with each other. (§399.) 11. Work Ex. 10, taking the two points on the same side of the given line. When is the problem impossible ? 12. One of the equal sides of an isosceles triangle is the diameter of a circle. Prove that the circumference bisects the base. [Show that BD is ± to AC] 13. What is the locus of the centers of circles having a given radius and tangent to a given line ? 14. Describe a circle having a given radius and tangent to two non- parallel lines. How many circles can be drawn ? 15. What is the locus of the centers of circles having a given radius and tangent to a given circle ? 16. Describe a circle having a given radius and tangent to two given circles. 124 PLANE GEOMETRY 17. Describe a circle having a given radius and tangent to a given line and also to a given circle. C 18. The base AB of the isosceles triangle ABG is a chord of a circle, the circumference of which intersects the two equal sides at D and E. Prove CD = CE. A [Z A and Z B are measured by equal arcs. ] 19. If an isosceles triangle is inscribed in a cir- cle, prove that the bisector of the vertical angle passes through the center of the circle. Al 20. The altitude of an equilateral triangle is one and a half times the radius of the circumscribed circle. [Use the preceding Exercise and § 245.] 21. If a triangle is circumscribed about a circle, the bisectors of its angles pass through the center of the circle. [§ 230.] 22. The altitude of an equilateral triangle is three times the radius of the inscribed circle. [Use Ex. 21 and §245.] 23. The angle between two tangents to a circle is 30°. Find the number of degrees in each of the intercepted arcs. BOOK II 125 24. From a given point draw a line cutting a circle and making the chord equal to a given line. [The chord B8 \q equal to the given line. The dotted circle is tangent to R8.-\ 25. Find the angle formed by two tangents to a circle, drawn from a point the distance of which from the center of the circle is equal to the diameter. 26. With a given radius describe a circle that shall pass through a given point and be tangent to a given line. 27. With a given radius describe a circle that shall pass through a given point and be tangent to a given circle. 28. From a point without a circle draw the shortest line to the circumference. 29. ABC is an inscribed equilateral triangle. DE joins the middle points of the arcs BC and CA. Prove that DE is trisected by the sides of the triangle. /V 30. Find a point within a triangle such that the angles formed by drawing lines from it to the three vertices of the triangle shall be equal to each other. (§ 399.) 126 PLANE GEOMETRY 31. A median BD is drawn from angle B in the triangle ABC. Show that angle 5 is a right angle when BD is equal to one half of the base ACy an acute angle when BD is greater than one half of AC, and an obtuse angle when BD is less than one half of AC. 32. In any right-angled triangle, the sum of the two legs is equal to the sum of the hypote- nuse and the diameter of the inscribed circle. [Tangents drawn from a point to a O are equal.] 33. Tangents CA and DB drawn at the ex- tremities of the diameter AB meet a third tangent CD at C and D. Draw CO and DO. Prove CD=CA + DB and Z COD = 1 R.A. 34. If from one point of intersection of two circles two diameters are drawn, the other extremities of the diameters and the other point of intersection of the circles are in a straight line. [Draw DE and EF. Show that /.DEC -Y ACEF=2n.A.'B.^ 35. Through the points of intersec- tion of two circles two parallel secants are drawn, terminating in the curves. Prove the secants equal. [Show that the quadrilateral ECDF has its opposite angles equal, each to each. ] 36. In a given circle draw a chord the length of which shall be twice its distance from the center. BOOK II 127 37. Three equal circles are tangent to each other. Through their points of contact three oommon tangents are drawn. Prove. 1. The three tangents meet in a common point. 2. The point of meeting is equally distant from the three points of contact. 38. The sum of the angles subtended at the center of a circle by two opposite sides of a circumscribed quadrilateral is equal to two right angles. [To prove /.AOB-^ Z. COD = 2 R.A.'s.] 39. Find the locus of points such that tangents drawn from them to a given circle shall equal a given . , line. 40. Inscribe a circle in a given quadrant. [OD bisects ZAOB. DE is ± to OB. DF bisects Z. ODE.'] 41. If the tangents to a circle at the four verti- ces of an inscribed rectangle (not a square) be prolonged, they form a rhombus. 42. From any point (not the center) within a circle only two equal straight lines can be drawn to the circumference. 43. Given a circle and a point within or without (not the center), using the given point as a center to describe a circle, the circumference of which shall bisect the circumference of the given circle. 44. In a given circle inscribe a triangle, the angles of which are respec- tively equal to the angles of a given triangle. [Draw a tangent to the O, and from the point of contact draw two chords, making the three A at the point of contact equal to the A of the A.] 45. Circumscribe about a given circle a triangle, the angles of which are respectively equal to the angles of a given triangle. [Inscribe a O in the given A.] 46. Of all triangles having a common base and an equal altitude, the isosceles triangle has the greatest vertical angle. 47. Given the base, the vertical angle, and the foot of the altitude, construct the triangle. 128 PLANE GEOMETRY 48. If the sum of one pair of opposite sides of a quadrilateral is equal to the sum of the other pair, a circle can be inscribed in the quadrilateral. [Describe a O tangent to three of the sides. Show, by § 396, that the fourth side can neither cut this circle nor lie without it.] 49. Any point on the circumference circum- scribing an equilateral triangle is joined with the three vertices. Prove that the greatest of the three lines is equal to the sum of the other two. [Lay off DE = DC. Prove ^ AEC and ^Z>C equal in all respects.] 50. Two equal circles intersect at A and B. On the common chord AB as a diameter a third circle is described. Through A any line CD is drawn ter- minating in the circumferences and in- tersecting the third circumference at E. Prove that CD is bisected at E. [Show that A BCD is isosceles, and that BE is ± to the base (7Z>.] 51. Two equal circles intersect at A and B. With ^ as a center, any circle is described cutting the two equal circumferences at G and D. Prove that J., C, and D are in a straight line. [Drawee. ZBAC'^^BC. But BC=BD. Draw ^2). ZBAD^^BD. .-. ZBAC = ZBAD.^ 62. If two circles intersect, the longest common secant that can be drawn through either point of inter- section is parallel to the line joining their centers. rShow that CD = 2AB, and that any otner secani, through E is less than 2AB.2 BOOK III 404. Definitions. A proportion is the equality of ratios. - = - is a proportion, and expresses the fact that the ratio of b d a to 6 is equal to the ratio of c to d. The proportion - = - may also be written a:b=c:d and a:b :: c:d. In the proportion = -, the first and fourth terms (a and d) Oj are called the extremes^ and the second and third terms (b and c) are called the means. The first and third terms (a and c) are the antecedents, and the second and fourth terms (6 and d) are the consequents. In the proportion - = -, d is called a fourth proportional to b d the three quantities a, b, and c. If the means of a proportion are equal, either mean is a mean proportional or a geometrical mean between the extremes. Thus in the proportion - = -, 6 is a mean proportional between b c a and c. In this same proportion, c is called a third pi'opor- tional to a and b. Proposition I. Theorem 405. In a proportion, the product of the extremes is equal to the product of the means. M- (^> To Prove ad = be. Proof. [Clear fractions in (1) by multiplying both members by 6d] Q.E.D. SANDERS' GEOM. — 9 129 130 PLANE GEOMETRY 406. Corollary. The mean proportional between two quan- tities is equal to the square root of their product. Let ^ = ^. (1) be ^ To Prove b=zVac. Proof. [Apply § 405 to (1), and extract the square root of both members.] q.e.d. 407. Exercise. Find x in — = —. 12 X 408. Exercise. What is the geometrical mean or mean proportional between 9 and 4 ? 409. Exercise. 12 is the geometrical mean between two numbers. One of them is 16. What is the other ? 410. Exercise. Find the mean proportional between a^ -{■ 2 ab + b^ and a^-2ab + b^. Proposition II. Theorem. (Converse of Prop. I.) 411. If the product of two factors is equal to the prod- uct of two other factors, the factors of either product may he made the means, and the factors of the other product the eoctremes of a proportion. Let ad = bc. (1) To Prove -=-. 6 d Proof. [Divide both members of (1) by bd."] q.e.d. 412. Exercise. From the equation ad = be, derive the following eight proportions. a_c ^ — b ^ = ^ ^ = ^ b cC c d! d b^ a b* ' b_d b_a ^ — b '^ — ^ a c d c c a b a BOOK III 131 413. Exercise. Form different proportions from 414. Exercise. Form a proportion from a2 + 2 a& + 62 ^ r^^^y^ Wiiat is a + 6 called in this proportion ? 415. Exercise. Form a proportion from a^ + 6^ = x^ — y^. 416. Definition. A proportion is arranged by alternation when antecedent is compared with antecedent and consequent with consequent. If the proportion - = - is arranged by alternation, it be- a b conies - = — c d Proposition III. Theorem. 417. If four quantities are in proportion, they are in proportion by alternation. Let - = -. (1) To Prove - = -. c d Proof. Apply § 405 to (1) ad = he. (2) Apply § 411 to (2) ^=^' 418. Exercise. Write a proportion that will not be altered when arranged by alternation. 419. Definition. A proportion is arranged by inversion when the antecedents are made consequents, and the conse- quents are made antecedents. If the proportion - = - is arranged by inversion, it be- b d ^ '^ comes — = — a c 132 PLANE GEOMETRY Proposition IV. Theorem 420. If four quantities are in proportion, they are in proportion by inversion. To Prove -=-. a c Proof. Apply § 405 to (1) ad = be. (2) Apply § 411 to (2) l = f 421. Definition. A proportion is arranged by composition when the sum of antecedent and consequent is compared with either antecedent or consequent. The proportion - = - arranged by composition becomes a + b _ c-\- d a + b _ c + d a c b d Proposition V. Theorem 422. If four quantities are in proportion, they are in proportion hy composition. 1=1- (^> To Prove a + b^c + d^ a c Proof. Apply § 405 to (1) ad = be. (2) Add ac to both members of (2) oc + ad = ac -f be. (3) Factor (3) a(c -\- d) = e(a -\- b). (4) Apply § 411 to (4) a-\-b _ c -\- d a "" C * Q.E.D. BOOK III • 133 423. Note. The student may discover for himself the steps of the solution of this and the succeeding propositions by studying the analysis of the theorem. In the analysis we assume the conclusion (the part to be proved) to be a true equation. Working upon this conclusion by algebraic transforma- tions, we produce the hypothesis. The solution of the theorem begins with the last step of the analysis dJi^L reverses the work, step by step, until the first step or conclusion is reached. In § 422 we have given ^ = -• (1) h d We are to prove ^-±-^ = ^^ti. (2) a c Analysis Clear fractions in (2) c{a -\- b) = a{c + d). (3) Expand (3) ac -{■ be = ac -\- ad. (4) Subtract ac from both members of (4). be = ad. (5) Apply § 411 to (5) l = t (6) b d Let the student show that the solution of Prop. V. as given on the pre- ceding page may be obtained by reversing the steps of this analysis. 424. Exercise. Let -=-• b d To Prove a±b^c±d^ b d 425. Exercise. Arrange ^ ~ ■ = ^-^^ — by composition. b d 426. Exercise. Arrange ^— = — — — by composition and then 4 X find the value of x. 427. Definition. A proportion is arranged by division when the difference between antecedent and consequent is com- pared with either antecedent or consequent. The proportion - = - arranged bv division becomes b d a — b c — d a — b c — d b — a d — c b — a d — c = or = or = or — = • 134 PLANE GEOMETRY Proposition VI. Theorem 428. If four quantities are in proportion, they are in proportion by division. Let - = -. m h d ^^ To Prove ^!^~^ = ^-^- (2) a c ^ Proof. [_Analysis. Clear fractions in (2) c(a — b)=a(c — d), (3) Expand (3) ac — bc = ac — ad. (4) Subtract ac from both members of (4) — bc = — ad. (5) Divide both members of (5) by — 1 be = ad. (6) Apply § 411 to (6) 1 = ^1. Q.E.D. Let the student derive the solution of Prop. VI. from the analysis. 42a. Exercise. If a + b-c ^a^:^ c-^-d + a 2d then * _±_^ a + c-d ^ a — c 2d 430. Definition. A proportion is arranged by composition and division, when the sum of antecedent and consequent is compared with the difference of antecedent and consequent. The proportion - = -, arranged by composition and division, b d becomes a^f) ^ c-\-d a — b c — d BOOK III 135 Proposition VII. Theorem 431. If four quantities are in proportion, they are in proportion hy composition and division. Let " ' To Prove h d a-\-h c + d a — h c ~ d Proof. [Analyze and solve.] 432. Exercise. If ^ = ^, b d ««^„^ c — a d — b prove a + c b + d Proposition VIII. Theorem 433. If four quantities are in proportion, like powers of those quantities are proportional. M <^) To Prove ^ = £l. Proof. [Raise both members of (1) to the nth power.] q.e.d. 434. Corollary. If four quantities are in proportion^ like roots of those quantities are propoi'tional. . Exercise. If a _ b~ 'd' show that ^ = a2 02. -62 -d^ I. Exercise. If a b c 'd' show that a3 + 63 _ . c3 + # qS — ^3 c^ — d* 136 PLANE GEOMETRY Proposition IX. Theorem 437. If four quantities are in proportion, equimul- tiples of the antecedents are proportional to equimul- tiples of the consequents. Let ^ = ^. (1) To Prove hy dy Proof. Multiply both members of (1) by -• q.,e.d. 438. Exercise. Let To Prove 439. Exercise. Let To Prove ab + cd^a^ + c\ ah — cd a^ — c^ a h' _ c ~d' ax by" ex 'dy embers c a h~ c ' d ac bd C2 a _ b~ ' d 440. Exercise. Let a_b b c" To Prove a + c _ 62 -j_ c2 a - c ~ 62 _ c2 441. Exercise. Let a c b~d To Prove ma^ + wc2 _ a2 W&2 + nd2 62 442. Definition. A continued proportion is a proportion made up of several ratios that are successively equal to each other. Example : « = ^= « =2, etc. b d f h' BOOK III 137 Propositiox X. Theorem 443. In a continued proportion the sum of the ante- cedents is to the sum of the consequents as any antecedent is to its consequent. (1) Let a _ _c _ f h To Prove a-[- c + e -\- g _e h+d+f+h f Proof a e (2) c e e e 7"7 (3) (4) ' From ( h f (5) af=be (6) cf=de ef=fe (8) From gf=he (9). From (2), (3), (4), and (5). Add (6), (7), (8), and (9), and factor. /(a + c + e + g)=eiib + d +/+ h). Apply § 411 to (10). a-^c + e + g ^e b + d-\-f+h f (10) Q.E.D. 444. Exercise, If then will x + y a + h _V_z y + z _ z + x ^ b -\- c c + a 445. Exercise. Let To Prove a _ c _ ^ _ flf b~d~ f~ h a — c-\- e — g _ c b-d-\-f-h~ d 138 PLANE GEOMETRY Proposition XI. Theorem 446. If the terms of one proportion are multiplied hy the corresponding terms of another proportion, the prod- ucts are proportional. Let ^=£ (1) and ? = ^ (2). m Tk-. ax cm To Prove — = — hy dn Proof. [The proof is left to the student.] q.e.d. 447. Exercise. If the terms of one proportion are divided by the corresponding terms of another proportion, the quotients are proportional. 448. Exercise. If - = -, h d show that ^' + «^^ + ^' = cl+^d+i^. a^ - ab -\- b'^ c- - cd + d^ Proposition XII. Theorem 449. // a number of parallels intercept equal distances on one of two transversals, they will intercept equal dis- tances on the otlver also. V r Let AB^ CD, EF, and G// be a number of parallels cut by the transversals xy and zr, making AC = CE = EG. To Prove BD = DF = FH. Proof. [Proof similar to that of § 240.] q.e.d. BOOK III 139 450. Corollary I. A line drawn from the middle point of one of the inclined sides of a trapezoid parallel to either base, bisects the other inclined side. 451. Corollary II. A line join- ing the middle points of the inclined sides of a trapezoid is parallel to the bases. Suggestion. Draw FG II AB. Prove A CFn = A GDn whence Fn = nG. Prove FG = AB Prove AmnG a parallelogram. 452. Exercise. A line joining the middle points of two opposite sides of a parallelogram, is parallel to the two remaining sides and passes through the point of intersection of the diagonals. 453. Exercise. A line joining the middle points of the inclined sides of a trapezoid is equal to one half the sum of the parallel sides. [In the figure of § 451 show mn = ^ {BF + AG) Sind CF=GD]. 454. Exercise. If from the extremities of a diameter perpendiculars are drawn to a line cutting the circle, the parts intercepted between the feet of the perpendiculars and the curve are equal. [To prove CE = FD-I 455. Exercise. If perpendiculars are drawn from the extremities of a diameter of a circle to a line lying without the circle, the feet of these perpendiculars are equally distant from the center of the circle. 456. Exercise. A line joining the middle points of the inclined sides of a trapezoid bisects the diagonals of the trapezoid, and also bisects any line whose extremities are in the parallel bases. 457. Exercise. The inclined sides of a trapezoid are Oft. and 15 ft. respectively. If on the shorter of these sides a point is taken .S ft. from one end, and through that point a parallel to either base is drawn, where does the parallel intersect the other inclined side ? 140 PLANE GEOMETRY Proposition XIII. Theorem. 458. A line drawn parallel to one side of a triangle divides the other two sides proportionally. iet Z)^ be parallel to BC. r^ ^ AD AE To Prove — = DB EC Proof. Case I. When the segments AD and DB are com- mensurable. Let the common unit of measure be contained in AD m times, and in DB n times. Whence i^ = !^. (1) DB n Divide AD into m equal parts, each equal to the unit of measure, and DB into n equal parts, and through the points of division draw parallels to BC. These parallels intercept equal distances on AC (?). Conse- quently AE \fi divided into m equal parts, and EC into n equal parts. AE m ^2) q.e.d. Whence EG n Compare (1) and (2). AD AE ^ DB EC BOOK III 141 Case II. When the segments AD and DB are incommen- surable. A Let DEhQ parallel to BC. To Prove ^_^ = 4£. DB EC Proof. Divide AD into a number of equal parts, and let one of these parts be applied to Z) 5 as a unit of measure. Since AD and DB are incommensurable, this unit of measure will not be exactly contained in DB, but there will remain over some distance MB smaller than the unit of measure. Draw MN parallel to BC. Since AD and DM are commensurable (why ?), AD AE 1 ^ T — = — by Case I. DM EN This proportion is true, no matter how many equal divisions are made in AD. If the number of divisions is increased, the size of each division is diminished, and MB is also diminished. AD As the number of divisions is increased, the ratio is approaching — as its limit, and the ratio - — is approaching AE ^^ ^^ — as its limit. Since the variables — and — are always equal, and are DM EN J 1 J each approaching a limit, their limits are equal (?). mi £ AD AE Therefore ^ — = q.e.d. DB EC 142 PLANE GEOMETRY 459. Corollary I. DE is parallel to BC. », T»_ AD AE T DB EC To Prove — = — and — = — ^• AB AC Suggestion. Apply § 422 to AB AC AD^AE DB EG 460. Corollary II. If two lines are cut by any number of parallels, they are divided proportionally. Case I. When the two lines are parallel. MR RW WY To Prove NS SX xz Case II. When the two lines are oblique to each other. - T» AM MR RW WY To Prove — = — = = AN NS Use § 458 and § 459. SX XZ 461. Corollary III. To construct a fourth proportional to three given lines. Let a, b, and c be the three given a lines. ^ c Required to construct a fourth pro- portional to them. Construct any convenient angle, XYZ. Lay off YD = a, DE = b, and YF=:c. DiSiW DF. Draw EG II to DF. BOOK III 143 FG is the required fourth proportional. c YD YF ,„. a — = — (?) or - = — . DE FG "^ h FG Q.E.F. Note, If h and c are equal, FG is a third proportional to a and &. D/- 462. Corollary IV. To divide a line into parts propor- tional to given lines. Let AB he the given line. Required to divide it into parts pro- portional to the lines 1, 2, 3, and 4. Draw AC, making any convenient angle with Jfi. Lay off ^D = 1, Z)£ = 2, EF = 3, and FG = 4. Connect G and B. Through F, Ey and D draw parallels to GB. AH _ HI _ IJ _ JB AD~ DE~~ EF~ FG E/- F/- c Then or AH 1 m ~2 3 ~ 4 ' Q.E.F. 463. Exercise. In the triangle ABC, AB is 10 in. and ^C is 8 in. From a point D on the line AB, DE is drawn parallel to BC, making AD = 3 in. Find the lengths of AE and EC. 464. Exercise, Through the point of intersection of the medians of a triangle, a line is drawn parallel to any side of the triangle. How does it divide each of the other two g sides of the triangle ? Suggestion. Use § 245. 465. Exercise. Through a point within an an- gle draW a line limited by the sides of the angle and bisected by the point. Through the given point, P, draw PD \\ to BC, and lay off DE = DB. 144 PLANE GEOMETBY 466. Exercise. ABC is any angle and P a point within. To draw througli P a line limited by the sides of the angle, and cutting off a tri- angle whose area is a minimum. Draw HD so that HP = PD. A HBD is the minimum A. Draw any other line through P, as EF. Draw DG' II to BA. A PEH = A PGD. .: A EBF exceeds area of A HBD by A DGF. 467. Exercise. Construct a fourth pro- portional to three lines in the ratio of 2, 3, and 4. 468. Exercise. Construct a third pro- portional to two lines whose lengths are 1 in. and 3 in. respectively. 469. Exercise. Through a point P with- out an angle ABC, draw PE so that PD—DE. 470. Exercise. In the triangle ABC^ D is the middle point of. BC and Cf is any other point on BC. Prove that the parallelogram DEAF is greater than the parallelogram GHAJ. Suggestion. Draw LK so that ^ LG = GK. AABOAALK, (?) DEAF=l/\ABC, (?) and GHAJ = | A ALK. (?) 471. Exercise. Divide a line into any number of equal parts, using the principle of this proposition. Compare the method with that used in §240. 472. Exercise. proposition. Prove § 239, using the principle established in this 473. Exercise. If an equilateral triangle is inscribed in a circle, and through the center of the circle lines are drawn parallel to the sides of the triangle, these lines trisect the sides of the triangle. BOOK III - 145 Propositiox XIV. Theorem (Converse of Prop. XIII.) 474. // a line divides two sides of a triangle propor- tionally, it is parallel to the third side. Let AD __AE DB~~ EC To Prove DE parallel to BC. Proof. Suppose DE \s not parallel to BC and that any other line through z>, as DM, is parallel to BC. AD _AM DB " MC AD _AE DB~ EC (?) (?) (?) AM _AE MC~ EC Show that this last proportion is absurd. Therefore the supposition that DE is not parallel to BC is false. Q.E.D. 475. Corollary. If — = — , DE and BC are parallel. A B A Ky 476. Exercise. DE is drawn, cutting the sides AB and AC of ?t triangle ABC at D and E. The segment BD is \ of AB, and AE is f of AC. Show that DE and BC are parallel. 477. Definition. Two polygons are similar when they are mutually equiangular, and have their sides about the equal angles taken in the same order proportional. SANDERS' GEOM. 10 146 PLANE GEOMETRY Proposition XV. Theorem 478. Triangles that are fnutually equiangular are uinilar. Let ABC and DEF be two A having /. A = Z.D, /. B = /.E^ and Z.C = /. F. To Prove A ABC and DEF similar. Proof. Lay off EM = BA, EN = BC. Draw MN. Prove A ABC and DEF equal in all respects. Whence /.M =^ Z D. MN and DF are II. (?) EM_EN_ ,r,^ AB ^BC ED~ EF "^ DE~ EF T . -1 AB AC . BC AC In a Similar manner prove — = — , and — = DE DF EF DF The triangles are by hypothesis mutually equiangular, and we have proved their sides proportional, therefore by definition they are similar. q.e.d. 479. Corollary. Two triangles are similar if they have two angles of one equal respectively to two angles of the other. 480. Exercise. All equilateral triangles are similar, 481- Exercise. Are all isosceles triangles similar ? Are right-angled isosceles triangles similar ? BOOK III 147 482. Exercise. If the sides of a tri- angle ABC be cut by any transversal, in the points Z), £', and F, to prove DB EC FA [From A^ B, and O, draw perpendic- A ulars to the transversal. Shovsr that A AxD and DyB are similar, whence AR^Ax^ DB By Similarly, and BE^By EC~ Cz' CF^Cz FA Ax (1) (2) (3) Multiply (1), (2), and (3) together, member by member.] Note. Prove this exercise when the points Z>, E^ and F are all external, i.e. are all on the prolonged sides of the triangle. (If the figure be lettered as above, the proportions in the proof of this case will be precisely like the foregoing.) 483. Exercise. If Z>, E and F are three points on the sides of a triangle, either all external, or two internal and one external, such that AD^BEx^=\ DB EC FA ' r. C G the three points are in the same line. [Draw DE and EF. Let any other line than EF as EG be the pro- longation of DE. By the preceding exercise DB EC GA By hypothesis From (1) and (2) we derive AD^BEy^QI=l DB EC FA CG GA FC FA 0) (2) (3) 148 PLANE GEOMETRY Arrange (3) by division, CG FC ^^ CG , or - : FC AC FC AC GA - CG FA Whence CG = FC which is absurd. . •. the supposition that any other line than EF is the prolongation of DE is absurd.] 484. Exercise. If from any point on the cir- cumference of a circle circumscribed about a triangle perpendiculars be drawn to the three sides of the triangle, the feet of these perpendiculars are in the same straight line. [To prove x, y, and z are in a straight line. Connect F with the three vertices. By means of similar triangles, show : Az _Pz Cx Px Cy^Py^ Bz Pz Bx_Px Ay Py Multiply (1), (2), and (3) together, member by member, Az^^Cy^Bx Cx Bz Ay Az^Bx^Cy^, zB xC yA By the preceding exercise, x, y, and z are in the same straight line. ] 485. Exercise. If a triangle ABC be inscribed in a circle, tangents to this circle at A, B, and C meet BC, CA, and AB respectively in three points that are in the same straight line. [Let the tangents meet BC, CA, and AB in the points x, y, and z respectively. Prove AAzC and BzC similar. Whence Az _ AC zG (1) and BC Bz AC Cz (2) Combining (1) and (2), ^ = 4^ zB BC^ Similarly, BOOK III 149 Proposition XVI. Theorem 486. Triangles that have their corresponding sides proportional are similar. Let ABC and DBF be two A having AB BC AC DE EF -DF To Prove A ABC and DEF similar. Proof. Lay off EM = BA and EN = BC. Draw MN. Show that EM^EN^ ED EF MN is parallel to DF. (?) Prove A EMN and EDF similar. Whence en^mn_ EF DF ^ ^ By hypothesis ^ = ^. (2) Compare (1) and (2), remembering that BC=EN, and show that AC= MN. Prove A ABC and MEN equal in all respects. A DEF and MEN have been proved similar, and since A ABC and MEN are equal in all respects, A DEF and ABC are similar. q.e.d 150 PLANE GEOMETRY 487. Exercise. The sides of a triangle are 6 in., 8 in., and 12 in. respectively. The sides of a second triangle are 6 in., 3 in., and 4 in. respectively. Are they similar ? 488. Scholium. Polygons must fulfill two conditions in order to be similar, i.e. they must be mutually equiangular, and must have their corresponding sides proportional. Proposi- tions XV. and XVI. show that in the case of triangles, either of these conditions involves the other. Hence to prove triangles similar, it will be sufficient to show either that they are mutu- ally equiangular, or that their corresponding sides are proportional. 489. Exercise. A piece of cardboard 8 in. square is cut into 4 pieces, A., B, C, and 2>, as shown in the first figure. These pieces, as placed in the second figure, apparently., form a rectangle whose area is 65 sq. in. Explain the fallacy by means of similar triangles. 490. Exercise. The sides of a triangle are 12, 16, and 24 ft. re- spectively. A similar triangle has one side 8 ft. in length. What is the length of the other tw^o sides ? (Three solutions.) 491. Exercise. On a given line as a side construct a triangle similar to a given triangle. [Construct in tw^o ways. Use § 478 and also § 486.] 492. Exercise. Construct a triangle that shall have a given perime- ter, and shall be similar to a given triangle. 493. Exercise. If the sides of one triangle are inversely proportional to the sides of a second triangle, the triangles are not necessarily similar. [Let the sides of the first triangle be in the ratio of 2, 3, and 4. Then the sides of the second triangle are in the ratio of |, ^, and ^, or y^^, j\, and y\ ; and these fractions are in the ratio of the integers 6, 4, and 3. Therefore the triangles are not similar.] 494. Exercise. Any. two altitudes of a triangle are inversely pro- portional to the sides to which they are respectively perpendicular. BOOK III 151 Propositiox XVII. Theorem. 495. Triangles that have an angle in each equal, and the including sides proportional, are similar. AD and^ = ^. BE DF Let A ABC and BEF have Z A To Prove A ABC and BEF similar. Proof. JjSijoE BM = AB Sind BN = AC. Draw J/i\r. Prove A ABC and BMN equal in all respects. BM BE MN and EF are parallel. (?) Z1 = Z2 and Z3 = Z4. (?) A BMN and BEF are similar. (?) A ABC 2iX\dL BEF are similar. (?) ^. (?) BF ^ Q.E.D. 496. Exercise, If a line is drawn parallel to the base of a triangle, and lines are drawn from the vertex to different points of the base, these lines divide the base and the parallel proportionally. ^ DBI and ABF are similar. (?) DI^BI ' ' AF BF A IB J and FBG are similar. IJ ^ BI " FG Bf' BI IT AF FG etc. 152 PLANE GEOMETRY m Proposition XVIII. Theorem. 497. Triangles that have their sides parallel, each to each, or perpendicular, each to each, are similar. Let A ABC and DEF have AB II to DE, BCW to EF, and ^C II to DF. To Prove A ABC and Z)Zi^ similar. Proof. The angles of the A ABC are either equal to the angles of A DEF, or are their supplements. (§ 131 and § 132.) There are four possible cases : 1. The three angles of A ABC may be supplements of the angles of A DEF. 2. Two angles of A ABC may be supplements of two angles of A DEF, and the third angle of A ABC equal the third angle of A DEF, 3. One angle of A ABC may be the supplement of an angle of A DEF, and the two remaining angles oiAABC be equal to the two remaining angles of A DEF. 4. The three angles oiAABC may equal the three angles of A DEF. Show that in the first case the sum of the angles of A ABC would be four right angles. Show that in the second case the sum of the angles oiAABC would be greater than two right angles. Show, by means of § 140, that the third case is impossible BOOK III 153 unless the angles that are supplementary are right angles, in which case they would also be equal, and the triangles would have three angles of the one equal to three angles of the other. Therefore if two triangles have their sides parallel, each to each, the triangles are mutually equiangular, and consequently similar. Let A ABC and DEF have ABA.de, BC±EF, and AC _L JJF. To Prove A ABC and DEF similar. Proof. The angles ofA^jSC ^ ^ are either equal to the angles of A DEF, or are their supple- ments. [Show, as was done in the first part of this proposition, that the angles of A^^C are equal to those of A DEF, and conse- quently A ABC and DEF are similar.] q.e.d. Note. The equal angles are those that are included between sides that are respectively parallel or perpendicular to each other. 498. Exercise. The bases of a trapezoid are 8 in. and 12 in., and the altitude is G in. Find the altitudes of the two triangles formed by producing the non-parallel sides until they meet. 499. Exercise. The angles ABC, DAE, and DBE are right angles. Prove that two triangles in the diagram arc similar. 500. Exercise. The lines joining the middle b points of the sides of a given triangle form a sec- ond triangle that is similar to the given triangle. 501. Exercise. The bisectors of the exterior angles of an equilateral triangle form by their intersection a triangle that is also equilateral. 154 PLANE GEOMETRY Proposition XIX. Theorem. 502. The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides of the angle. Let BD be tlie bisector of Z B of the A ABC, _ „ AD AB To Prove — = DC BC Proof. Prolong AB until BE = BC. Draw CE. Z3 + Z4 = Z1 + Z2. (?) Z 3 = Z 4 and Z 1 = Z 2. (?) Z 4 = Z 2. (?) BD and EC are parallel. (?) BE DC (?) AB AD BC DC (?) Q.E.D. Conversely. A line drawn through the vertex of an angle of a triangle, dividing the opposite side into segments pro- BOOK III 155 portional to the adjacent sides of the angle, bisects the angle. Let AliC be a A in which BD . T T . AD AB IS drawn making — = DC BC To Prove that BD bisects Z.B. Proof. Prolong AB until BE = BC. Draw ^'C. BD is parallel to EC. (?) Z3 = Z1, and Z4 = Z2. (?) Since Z1 = Z2. (?) Z3 = Z4. (?) Q.E.D. 503. Exercise. The triangle ABC has AB = 8 in., BC = 6 in., and AC = 12 in. BD bisects ZB. What are the lengths of the segments into which it divides AC? 504. Exercise. BD is the bisector ot Z B in the triangle ABC. The segments ot AC are ^Z> = 5 in. and DC = 2 in. The sum of the sides AB and BC is 14 in. Find the lengths oi AB and BC. 505. Exercise. Construct a triangle having given two sides and one of the two segments into which the third side is divided by the bisector of the opposite angle. (Two constructions.) 506. Definition. A point C, taken on the line AB between the points A and B, is said to divide the line AB internally into two segments, CA and CB. A point C', taken on AB pro- ^ c B — C' duced, is said to divide AB externally into two segments, C'A and C'B. In each case, the seg- ments are the distances from C (or C') to the extremities of AB. 156 PLANE GEOMETRY Proposition XX. Theorem 507. The bisector of an exterior angle of a triangle divides the opposite side externally into two segments that are proportional to the adjacent sides of the angle. A C Let BD bisect the exterior /. CBF oi the A ABC, ^ ^ AD ab' To Prove — = DC BC Proof. Lay oR BE = BC. Draw EC. Z3 + Z4 = Z1+Z2. (?) Z3 = Z4, and Z1=Z2. (?) Z 4 = Z 2. (?) EC and BD are parallel. (?) AB AD BE DC AB BC AD DC (?) Q.E.D. 508. Exercise. The lengths of the sides of a triangle are 4, 5, and 6 yards, respectively. Find the lengths of the segments into which the bisector of the angle exterior to the largest angle of the triangle divides the opposite side externally. BOOK III 157 Conversely. A line drawn through the vertex of an angle of a triangle dividing the opposite side externally into seg- ments proportional to the adjacent sides of the angle, bisects the exterior angle. Let BD be drawn so that — = — • DC BC To Prove that BD bisects Z CBF. O Proof. Lay of^ be = B c. Draw CE. DC BC "^ DC be' ^'^ EC is parallel to BD. (?) Z4 = Z 1, and Z3 = Z 2. (?) Z 1 = Z 2. (?) Z3 = Z4. (?) Q.E.D. 509. Definition. A line Jr ^i: j, ' is divided harmonically when it is divided internally and externally in the same ratio. If, in this figure, AC _ AD CB ~ DB then AB h divided harmonically. 510. Exercise. The bisector of an angle of a triangle and the bisec- tor of its adjacent exterior angle divide the opposite side harmonically. (§§ 502, 507.) 158 PLANE GEOMETRY 511. Exercise. To divide a line internally and externally so that its segments shall have a given ratio, i.e. to divide a line harmonically. Let AB be the given line, and m and n lines in the given ratio. Required to divide AB internally and externally into segments having E n ^^,^^ . Draw AE making any X \ ^"-^ i ^ i angle with AB., and equal / \ ^^^^ to m. y \ ^"-9 Draw BG parallel to / \ / ^^, and equal to 71. A G\ /-'B Prolong CB until BD V^- = n. Draw ED. Draw EC and prolong it until it meets AB prolonged at some point F. By means of similar triangles, show ^ =. ^, and ^ = H ; whence ^ = iZ. q.^.p. GB n BF n OB BF 512. Definition. If the line ^5 is divided harmonically at G and D, and the four points A, B, c, and D are connected with any other point 0, the resulting O figure is called a harmoyiic pencil. The point is called the vertex of the pen- cil, and the four lines OA, OG, OB, and OB are called rays. a ^ / j ^ P 513. Exercise. 0-ACBD is a harmonic pencil, through G parallel to OD, and limited by OB produced. Prove that EF is bisected at C. T\Tf (?) (?) (?) Multiply (1), (2), and (3) together member by member. q.e.d. OD CF DB BG EG AG OD AD AG GB AD DB BOOK III 159 514. Exercise. 0-^(7-BZ> is a harmonic pencil versal cutting the rays at £", G^ II, and F. Prove that the transversal EH is divided harmonically, that is, EG ^EF GH ~ FH Through C draw IJ II to OD. Through G draw MN II to IJ. IC=CJ. (?) .-. MG = GN. (?) EG EF .^. GM OF ^'^ EG _EF GN OF . EG _EF " GH FH (?) (?) and EF any trans- O GH GN HF OF vrxJ^ __ JLJ ^pv 515. Exercise. ABC is an inscribed triangle, DE is a diam- eter perpendicular to AC. The vertex B is connected with the extremities of the diameter. Prove that BE and DB (pro- longed) divide the base AC har- monically. Suggestion. Show that BE and BG are the bisectors of Z B and the exterior angle at B respectively. 516. Exercise. Any triangle having AC for its base (see figure of AB § 515), and its other two sides in the ratio , will have its vertex in BC the circumference described on FG as a diameter, 517. Exercise. The bisectors of the exterior angles of a triangle meet the opposite sides produced in three points that are in the same straight line. [Let the bisectors of the exterior angles at A., B, and C, of the tri- angle ABC, meet the opposite sides BC, AC, and AB in the points X, Y, and Z, respectively. AY^AB ,c.. CX ^AC YC BC' ^'^ XB AB (?) BZ ZA BC AC (?) Whence AY ^, CX ^^BZ YC XB ZA 1. 160 PLANE GEOMETRY Proposition XXI. Theorem 518. In a right-angled triangle, if a perpendicular is drawn from the vertex of the right angle to the hypote- nuse, I. The triangles on each side of the perpendicular are similar to the original triangle, and to each other. II. The perpendicular is a inean proportional between the segments of the hypotenuse. III. Either side about the perpendicular is a mean proportional between the hypotenuse and the adjacent segment of the hypotenuse. Let ABC be a R.A. A, AC its hypotenuse, and BD 1. to AC. I. To Prove Aabd and BBC similar to A ABC and to each other. Proof. Show that Aabd and ABC are mutually equiangu- lar, and consequently similar. In the same manner show that Abdc and ABC are similar. Aabd and BDC are also mutually equiangular and similar. TT ^ T. ^D BD Q-^-^- II. To Prove — = BD DC Use the similar AABD and BDC. III. To Prove ^ = ^, and ^ = ?^. AB AD BC DC Use the similar A ABC and ABD, and also A ABC and BDC. Q.E.D. BOOK III 161 519. Corollary To construct a mean proportio7ial between two given lines. Let m and n be two given ^-^ lines. ■^-^.E Required to construct a mean j^ proportional between them. B C On the indelinite line AD lay off A B = m and BC = n. On ^c as a diameter describe a semicircle. Erect BE ± to AC. Draw AE and EC. Show that A EC is B. R.A. A, and that AB BE BE or — = BE BC BE .'. BE is the required mean proportional Q.E.F. 520. Exercise. Construct a third proportional to two given lines by means of Prop. XXI. 521. Definition. If the radius OG and externally at A and B, so that OA X OB = Og\ is divided internally C and through A send B perpendicu- lars are drawn to OG, each perpen- dicular is called the polar of the other point, which is called in rela- tion to the perpendicular its pole. \^EF is the polar of A, and A is the pole of EF. CD is the polar of B, and B is the pole of CD. Notice that OB is a third proportional to OA and the radius, and 0^ is a third proportional to OB and the radius.] 522. Exercise. Given a point, within or without a circle, draw its polar. 623. Exercise. Given a line, find its pole with respect to a given circle. SANDERS' GEOM. 11 162 PLANE GEOMETRY 524. Exercise. If from a point without a circle two tangents are drawn to the circle, their chord of contact is the polar of the point^ [To prove BC the polar of A. OA is ± to BC. (?) A OB A is a R.A. A. (?) By Case III. of this Proposition, ^ = ^, or Oi>xO^ OB OA 0B^.-\ 525. Exercise. Any line through the pole is divided harmonically by the pole, its polar, and the circumference. [Let A be the pole of C-F, and EG be any line through A. To Prove ^ = ^. AD CD AO^OD .ps AO^OE OD OB ^"^ OE OB (?) .*. AAOD and ODB are similar, as are also A OAE and OBE. AD^DB ,p. OE^OB ,p. AD^DB ,p. OD OB ^'^ AE EB ^'^ AE EB ^' ^ :. BA bisects Z DBE. Since CB is ± to AB, CB bisects the exterior angle at B. Now apply § 510.] 526. Exercise, If two circles are tan- gent externally, the portion of their com- mon tangent included between the points of contact is a mean proportional between the diameters of the circles. [Show that AEB is a R.A. A, and that EF (the half of CD) is a mean proportional between the radii.] 527. Exercise. If two tangents are drawn to a circle at the extremities of a diameter, the portion of any third tan- gent intercepted between them is divided at the point of contact into segments whose product is equal to the square of the radius. fShow that OAB is a R.A. A.] BOOK III 163 Proposition XXII. Theorem 528. // two chords intersect within a circle, the prod- uct of the segments of one is equal to the product of the segments of the other. Let the chords AB and CD intersect at E. To Prove AE - EB = CE • ED. Proof. Draw AC and DB. Prove Aaec and EDB mutually equiangular and therefore similar. Whence AE^ED^ . CE EB .-. AE ' EB = CE ' ED. (?) Q.E.D. Conversely. If two lines AB and CD intersect at E, so that AE • EB =CE • ED, then can a circumference be passed through the four points A, B, C, and D. [Pass a circumference through the three points A, B, and C. Then show that the point D cannot lie without this circum- ference, nor within it.] 529. Exercise. C and D are respectively the middle points of a chord AB and its subtended arc. If AC is 8, and CD is 4, what is the radius of the circle ? 530. Exercise. Two chords AB and CD intersect at the point E. AE is 8, EB is 6, and CD is 19. Find the segments of CD. 531. Exercise. If a chord is drawn through a fixed point within a circle, prove that the product of its segments is constant in whatever direction the chord is drawn. 164 PLANE GEOMETRY Proposition XXTII. Theorem 532. // from a point without a circle two secants he drawn terminating in the concave arc, the product of one secant and its external segment is equal to the prod- uct of the other secant and its external segment. Let AB and BC be two secants drawn from B to the circle whose center is 0. To Prove AB - DB = CB - EB. Proof. Draw AE and DC. Prove A AEB and CDB mutually equiangular and similar. BC DB ' .-. AB 'DB = BC 'EB. Q.E.D. Converse. If on two intersecting lines AB and CB, four points, A, D, C, and E, be taken, so that AB x DB = BC x EB, then can a circumference be passed through the four points. [Pass a circumference through three of the points. A, D, and E. Show by means of Prop. XXIII. and the hypoth- esis of the converse, that C can lie neither without nor within the circumference.] BOOK III 165 533. Exercise. One of two secants meeting without a circle is 18 in., and its external segment is 4 in. long. The other secant is divided into two equal parts by the circumference. Find the length of the second secant. 534. Exercise. Two secants intersect without the circle. The exter- nal segment of the first is 5 ft., and the internal segment 19 ft. long. The internal segment of the second is 7 ft. long. Find the length of each secant. 535. Exercise. If A and B are two points such that the polar of A passes through B, then the polar of B passes through A. Let CS^ the polar of ^1, pass through B. To Prove that the polar of B passes through A. Proof. [Draw AD ± to OB. The quadrilateral ADBC has its opposite angles supplementary, .'. a circle can be circumscribed about it. ODy. 0B= OAxOG= 6(?. :. ^Z> is the polar of B.'\ 536. Exercise. The locus of the intersection of tangents to a circle, at the extremities of any chord that passes through a given point, is the polar of the point. Let CD be any chord passing through A, .^^ ^VP E and B be the point of intersection of the tangents at G and D. To Prove that 5 is a point of the polar of A. [C2> is the polar of B. (§ 524.) The polar of B therefore passes through A. By § 535, the polar of A passes through B. ] 537. Exercise. If from any point on a given line two tangents are drawn to a circle, their chord of contact passes through the pole of the line. [Apply § 535.] 538. Exercise. If from different points on a given straight line pairs of tangents are drawn to a circle, their chords of contact all pass through a common point. 166 PLANE GEOMETRY Proposition XXIV. Theorem 539. If from a point without a circle a secant and a tangent are drawn, the secant terminating in the con- cave arc, the square of the tangent is equal to the prod- uct of the secant and its external segment. Let ^ B be a tangent and 5 C a secant drawn from B to the circle whose center is O. To Prove AB^ = BC x db. Proof. Draw AC and AD. Prove A CAB and DAB similar. Whence BC _AB AB~ DB -. AB^ = BC X DB. Q.E.D. 540. Exercise. Tangents drawn to two intersecting circles from a point on their common chord produced, are equal. 541. Exercise. Given two circles, to find a point such that the tangents drawn from it to the two circles are equal. [Describe any circle intersecting the two given circles. Draw the two common chords. Prove that tangents drawn to the two circles from C, the point of intersection of the common chords (prolonged), are equal.] BOOK III 167 Proposition XXV. Theorem 542. Two polygons are similar if they are composed of the same number of triangles, similar each to each, and similarly placed* Let the ^ ABC, ADC, DEC, and EFC be similar respectively to the A GHI, GJI, JLI, and LMI, and be similarly placed. To Prove polygons ABCFED and GHIMLJ similar. Proof. Show that the angles of ABCFED are equal respec- tively to the corresponding angles of GHIMLJ. AB _AC GH~ Gl Whence AD GJ AB Gil Similarly prove — = AD GJ AB GH AD GJ DE JL AC GI AD GJ DE EF LM (?) (?)' (?) etc. _FC _ ~ MI~ CB^ III The polygons are mutually equiangular and have their corresponding sides proportional. They are therefore similar by definition. q.e.d. 168 PLANE GEOMETRY 543. Corollary. On a given line to construct a polygon similar to a given polygon. 544. Definition. In similar polygons the corresponding sides are called ho7nologous sides, and the equal angles are called homologous angles. Proposition XXVI. Theorem 545. Two similar polygons can be divided into the same number of simMar triangles, similarly placed. Let ABCDEF and GHIJLM be two similar polygons. To Prove that they can be divided into the same number of similar triangles, similarly placed. Proof. From the vertex F draw all the possible diagonals. From M, homologous with F, draw all the possible diagonals. Prove A FAB and AfGII similar (§ 495). Whence Z3=Z4. Z5=Z6. (?) AB__B_Fl ,c^. A^ = ^. (9\ GH~ HM GH Hi' BFl^BC ,^. HM HI A FBC and MHI are similar. (?) Show that A FCD and MIJ are similar, and also A FDE and MJL. Q.E.D. BOOK III 169 Propositiox XXVII. Theorem 546. The perimeters of similar polygons are to each other as any two homologous sides. H C Let ABODE and FGHIJ be two similar polygons. To Prove AB + BC + CD -{- etc. _ CD^ HI FG 4- GH -\- HI -\- etc. Proof. By definition AB _BC _CD _DE _ AE GF ~ GH ~ HI ~ IJ ~ FJ [Apply § 443.] 547. Corollary. Tlie perimeters of shniJar polygons are to each other as any two homologous diagonals. 548. Exercise. The perimeters of similar triangles are to each other as any homologous altitudes. 549. Exercise. The perimeters of similar triangles are to each other as any homologous medians. 550. Exercise. The perimeters of two similar polygons are 78 and 65 ; a side of the first is 9, find the homologous side of the second. 551. Definition. A line is divided into extreme and mean ratio when it is divided into two parts so that one segment is a mean proportional between the whole line and the other segment. 170 PLANE GEOMETRY Proposition XXVIII. Problem 552. To divide a line into extreme and mean ratio. c.--^ A F B Let AB hQ the given line. Required to divide AB into extreme and mean ratio. Draw BG ± to AB and equal to one half of AB. Draw AC. With C as a center and CB as a radius describe a circle cutting ^C at Z), and AC prolonged at E. Lay oE AF = AD. £^_^^ (^8 539^ AE — AB AB — AD • AB AD AD _ AB — AF AB~ AD (§ 539.) - (?) AB AD AF FB ,ox AB AF AB AF (?) AF FB (?) (?) Q.E.F. 553. Exercise. To determine the values of the segments of a line that has been divided into extreme and mean ratio. In the figure of § 552, let the length of AB be a ; AF = x, then FB = a-x. Substituting these values in the last proportion, we get - = — - — » whence a^ — ax = x^. X a — X Solving the equation, a; = ^av/5-^a=^(V5-l), a _ X = |a - 1 aVS = - (3 - \/5). 554. Exercise. Divide a line 5 in. long into extreme and mean ratio, and calculate the value of the segments, BOOK III 171 Proposition XXIX. Problem 555. To draw a cominon tangent to two §iven circles. Let A and B be the centers of the two given circles. Required to draw a common tangent to the two circles. Let R stand for the radius of circle A, and r for the radius of circle B. Draw AB. Divide AB (internally and externally) at G so AC that ^^ = BC~ r Draw CD tangent to circle B. Draw the radius BD. Draw AE A. to BC prolonged. [It is required to show that AE = R.] Aaec and CBD are similar (?), whence AC BC AE BD J? A f — = — , and AE = R, and ED is a common tangent, r r Q.E.F. 556. Defin-ition. The two tangents that pass through the internal point of division of AB are called the transverse tan- gents. The two tangents that pass through the external point of division are called the direct tangents. 172 PLANE GEOMETRY The points of division are called the centers of similitude of the two circles. 557. Exercise. The line joining the centers of two circles is divided harmonically by the centers of similitude. 558. Exercise, The line joining the extremities of parallel radii of two circles passes through their external center of similitude if the radii are turned in the same direction ; but through their internal center if they are turned in opposite directions. 559. Exercise. All lines passing through a center of similitude of two circles and intersecting the circles are divided by the circumferences in the same ratio. Draw the radius AD. Draw a line BE parallel to AD, and by means of similar triangles prove that BE is a radius. Then CE ^ r CD E Similarly, CE' ^r CD' B 560. Exercise. A, Bj and C are the centers of three cir- cles ; a, &, and c their respective radii ; D, E, and F their external centers of similitude ; and D', E', and F' their internal centers of similitude. Prove that D, E, and F are in a straight line. rAF^a BD b .„^ CE c _., AF =z -, and whence BD^CE^^l DC EA J IFB 6' DC c' EA a Similarly, show that D, E', and F' are in a straight line, also E, D', and F' and also F, D', and E'. BOOK III 173 EXERCISES 1. If • ^=:^ a -n h — a d — c h — a d — c a c — a Prove = i — : — = — c b d b d — b S a -r b 'S c + d 2, If b~ d~ / prove xa — ye + zc _ e^^ xb-yf+ zd~ f 3. If prove 4. If prove 5. If prove a^ + ab _ a &2 + be ~ c a _b b~c' a^ (a + by ^ c {b + c)^' 6. n ^ = A := ^, and a + 6 = c, x^ 2/^ ^'^ prove x2 + ?/2 — 2;2. 7. The shadow cast by a church steeple on level ground is 27 yd., v^hile that cast by a 5-ft. vertical rod is 3 ft. long. How high is the steeple ? n P 8. The line joining the middle points of h^^ ^""A the non-parallel sides of a trapezoid circum- / \ scribed about a circle is equal to one fourth the /f 1\ perimeter of the trapezoid. / V y \ [See § 396.] ^ / ^ ^ ' \ ^ 174 PLANE GEOMETRY 9. Two circles intersect at B and C. BA and BD are drawn tangent to the circles. Prove that 5(7 is a mean proportional between AG and CD. [Prove ^ ABC and BCD similar,] 10. Find the lengths of the longest and A the shortest chords that can he drawn through a point 10 in. center of a circle having a radius 26 in. 11. Tangents are drawn to a circle at the extremities of the diameters AB. Secants are drawn from A and _B, meet- ing the tangents at D and E and inter- secting at C on the circumference. Prove the diameter a mean propor- tional between the tangents AD and BE. [A ABD and ABC are similar. (?)] 12. If two circles are tangent internally, chords of the greater drawn from the point of tangency are divided proportionally by the circum- ference of the less. 13. If two circles are tangent externally, secants drawn through their point of contact and terminating in the circumferences are divided pro- portionally at the point of contact. 14. Given the two segments of the base of a triangle made by the bisector of the vertical triangle, and the sum of the other two sides, to construct the triangle. [§ 502.] 15. Determine a point P in the circumference, from which chords drawn to two given points A and B shall have the ratio — • [Divide AB so that A^ = ^. *- CB n middle point of the arc ADB. ] Join C with the 16. In the triangle ABC, BD is a medial line, and DE and DF bisect angles ADB and BD C respectively. Prove that EF is parallel to^C BOOK III 175 17. D is the point of intersection of the medians ; E is the point of intersection of the perpendiculars at the middle points of the sides ; DE is prolonged to meet the altitude BI at F. Prove ED = l DF. [A EDG and DBF are similar, and BD=2DG.'^ 18. The point of intersection of the medians, the point of intersection of the perpendiculars at the middle points of the sides, and the point of intersection of the altitudes of a triangle are in the same straight line. [See Ex. 17.] 19. The triangles ABC and ADC have the same base and lie betvreen the same parallels. EF is drawn parallel to AC. Prove EG = HF. 20. Two tangents are drawn at the extremi- ties of the diameter AB. At any other point C on the circumference a third tangent DE is drawn. Prove that OD is a mean proportional between AD and DE, and that OE is a mean proportional between BE and DE. [Prove Z DOE a R.A., and use § 518. J 21. The prolongation of the common chord of two intersecting circles bisects their common tangent. [§ 539. J 22. To draw a line AC intersect- ing two given circles so that the chords AD and 5C shall be of given lengths. [See Ex. 24, p. 126.] 176 PLANE GEOMETRY 23. xy is any line drawn through the vertex A of the parallelogram ABCD and lying without the parallelogram. Prove that the perpendicular to xy from the opposite angle C is equal to the sum of the perpen- diculars from B and D to xy. [§453.] 24. The sum of the perpen- diculars from the vertices of one pair of opposite angles to a line lying without a parallelogram is equal to the sum of the per- pendiculars from the vertices of the other pair of opposite angles. 25. Two circles are tangent externally at C. DE and Ci^are common tangents. Prove that ZDCE=l R.A., and also that Z AFB = 1 11. A. 26. Prove that ^ DEC and CBE (see figure of Ex. 25) are similar, as are also ABAC and FCE. 27. Describe a circle passing through two given points and tangent to a given line. C E D C E D v--\ /^ A\- '% (1) (2) [The line joining the two given points A and B may be parallel to the given line CD (see Fig. 1), or its prolongation may meet the given line (see Fig. 2). In the second case DE'^ = DA x DB. (?) DE may be laid off on either side of D, . •. two © can be described fulfilling the con- ditions of the problem.] BOOK III 177 28. Describe a circle tangent to two given lines and passing through a given point. [P is the given point. Find another point D through which the circumference must pass. Then solve as in Ex. 27.] 29. Describe a circle tangent to two given lines and tangent to a given circle. \^DE and BG are the lines, and A the center of the given circle. Use Ex. 28.] 30. Through a given point P draw a line cutting a triangle so that the sum of the perpendiculars to the line from the two vertices on one side of the line shall equal the perpen- dicular from the vertex on the other side of the line. [0 is the point of intersection of the medians.] 31. In the triangle ABC, BE is drawn parallel to AC. FG connects the middle points of ^O and BE. Prove that FG prolonged passes through B. 32. The line joining the middle point of the lower base of a trapezoid with the point of intersection of the diagonals bisects the upper base. SANDERS' GEOM. 12 178 PLANE GEOMETRY 33. In the triangle ABC, let two lines drawn from the extremities of the base AC and intersecting at any point D on the median through B, meet the opposite sides in JS and F. Show that EF is parallel to AC. 34. ABC is an acute-angled triangle. DEF {called the pedal triangle) is formed by joining the feet of the altitudes of triangle ABC. Prove that the altitudes of triangle ABC bisect the angles of the pedal triangle DEF. [A O can be de- scribed passing through F, 0, D, and B. (?) Z1=Z2. (?)] 35. Prove the triangles AFE, BFD, and DCE similar to triangle ABC and to each other. [See figure of Ex. 34.] [To prove i^FBD and ABC similar. Show that ZA = Z2.] 36. Prove that the sides of the triangle ABC [see Ex. 34] bisect the exterior angles of the pedal triangle DEF. 37. The three circles that pass through two vertices of a triangle and the point of intersection of the altitudes are equal to each other. [Show that each is equal to the circle circumscribed about the triangle.] 38. Describe a circle passing through £^^^ -^ two given points and tangent to a given circle. [A and B are the given points and C the given circle. DEAB is any O passing through A and B and cutting the given O C. The common chord ED meets AB at G. GF is tangent to O C. AFB is the required O.] G 39. If one leg of a right-angled triangle is double the other, a perpen- dicular from the right angle to the hypotenuse divides it into segments having the ratio of 1 to 4. 40. The triangle ABC is inscribed in a circle, and the bisector of angle B intersects AC 2X D and the circumference at E. Prove AB ~BD BE BC' BOOK III 179 4i. The perpendicular drawn to a chord from any point in the circumference is a mean proportional between the perpendicu- lars from that point to the tangents drawn at the extremities of the chord. 42. The perpendicular drawn from the point of intersection of the medians of a triangle to a line without the triangle is equal to one third the sum of the perpen- diculars from the vertices of the triangle to that line. [§ 453.] 43. Construct a right-angled trian- gle, having given an acute angle and the perimeter. 44. Inscribe in a given triangle another triangle, the sides of which are parallel to the sides of a second given triangle. A 45. CD is a line perpendicular to the diameter AB. AE is drawn from A to any point on CD. Prove that AE x AF is A constant. [A circle can be passed through F, B, Q, and E. (?)] 46. Given the vertical angle, the medial line to the base, and the angle that the medial line makes with the base, to con- struct the triangle. 180 PLANE GEOMETRY 47. Given the base of a tri- angle and the ratio of the other two sides, to find the locus of its vertex. [Divide the given base AB harmonically at D and E, in the ratio of the two given sides. On DE as a diameter construct aO.] 48. In the parallelogram ABCD, BE is drawn cutting the diagonal AC in E, CD in G, and AB prolonged in F. Prove that BE^ = GE x EF. 49. If three circles intersect each other, their common chords intersect in the same point. [§ 528.] 50. In any inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. 61. To inscribe a square in a given semicircle. 52. To inscribe a square in a given triangle. 53. ABCD is a parallelogram, E a point on BC such that BE is one fourth of BC. AE cuts the diagonal BD in F. Show that BF is one fifth of BD. 54. Two chords of a circle drawn from a common point A on the circumference and cut by a line parallel to a tangent through A, are divided proportionally. [Suggestion. Join the extremities of the chords and prove the triangles similar.] BOOK IV 561. Definitions. We measure a magnitude by comparing it with a similar magnitude that is taken as the unit of meas- ure. If we wish to find the length of a line, we find how many times a linear unit of measure, say a foot, is contained in the line. This number, with the proper denomination, is called the length of the line. Similarly, we measure any portion of a surface by comparing it with some unit of surface measure. We find how many times this unit, say a square yard, is contained in the portion of surface. This number, with the denomination square yards, we call the area or superficial content of the surface measured. Polygons that have the same areas are equivalent polygons. Equivalent polygons are not necessarily equal in all respects. They need not even have the same number of sides. For example, a triangle, a square, and a hexagon may be equivalent. The base of a polygon is primarily the side upon which the figure stands; but usage has sanctioned a more extended appli- cation of the term. Any side of a polygon may be considered the base. In a parallelogram, if two opposite sides are hori- zontal lines, they are frequently called the upper and lower bases of the parallelogram. In a trapezoid, the two parallel sides are called its bases. * The altitude of a parallelogram is the perpendicular distance between two opposite sides. A parallelogram may therefore have two different altitudes. The altitude of a trapezoid is the perpendicular distance between its bases. 181 182 PLANE GEOMETRY Proposition I. Theorem 562. Parallelograms having equal bases and equal altitudes are equivalent. B Let ABCB and EFGH be two parallelograms having equal bases and equal altitudes. To Prove ^5 CD and EFGH equal in area. Proof. Place EFGH upon ABCD so that their lower bases shall coincide. Because they have equal altitudes their upper bases are in the same line. Prove A AIB and DJC equal. The parallelogram AIJD is composed of the quadrilateral ABJB and the Aaib. The parallelogram ABCB is composed of the quadrilateral ABJB and the Abjc, ABCB = AIJB. ABCB = EFGH. Q.E.D. 563. Exercise. Rectangles having equal bases and altitudes are equal in all respects. '564. Exercise. Construct a rectangle equivalent to a given paral- lelogram. 565. Exercise. Prove Prop. I. , using this figure : 566. Exercise. Construct a rectangle whose, area is double that of a given equilateral triangle. BOOK IV 183 567. Exercise. A line joining the middle points of two opposite sides of a parallelogram divides the figure into two equivalent parallelo- grams. Proposition II. Theorem 568. Triangles having equal hases and equal altitudes are equivalent. A C ^ Let the ^ABC and BEF have equal bases and equal altitudes. To Prove the i\ABC and BEF equal in area. Proof. On each triangle construct a parallelogram having for its base and altitude the base and altitude of the triangle. These parallelograms are equivalent. (?) .-. the triangles are equivalent. (?) q.e.d. 569. Corollary I. If a triangle and a parallelogram have equal bases and equal altitudes^ the triangle is equivalent to one half the parallelogram. 570. Corollary II. To construct a triangle equivalent to a given polygon. To construct a triangle equivalent to ABC . . . G. Draw BB. Through C draw CX parallel to BB, meeting AB prolonged at X. Draw BX. Show that Abxb and BOB have a com- mon base and equal altitudes. .-. Abxb = Abcb, and the polygon AXBEFG = polygon ABCBEFG. 184 PLANE GEOMETRY We have therefore constructed a polygon equivalent to the given polygon and having one side less than the given polygon has. A new polygon may be constructed equivalent to this polygon and having one side less; and this process can be repeated until a triangle is reached. 571. Exercise. Two triangles are equivalent if they have two sides of the one equal respectively to two sides of the other, and the included angles supplementary. [Place the A so that the two supplementary A are adjacent and a side of one A coincides with its equal in the other.] 572. Exercise. Bisect a triangle by a line drawn from a vertex. 573. Exercise. Bisect a triangle by a line drawn from a point in the perimeter. [BD is a medial line, BE is drawn II to FD. Show that PE bisects AABC.^ 574. Exercise. The diagonals of a paral- lelogram divide it into four equivalent triangles, a 575. Exercise. The three medial lines of a triangle divide it into six equivalent tri- angles. 576. Exercise. In the triangle ABC, X is any point on the median CD. Prove that the triangles AXC and BXC are equivalent. ^ 577. Exercise. On the base of a given triangle construct a second triangle equal in area to the first, and having its vertex in a given straight line. Under what conditions is this exercise impossible ? 578. Exercise. Construct a right-angled triangle equivalent to a given equilateral triangle. 579. Exercise. From a point in the perimeter of a parallelogram draw a line that shall divide the parallelogram into two equivalent parts. 580. Exercise. Construct an isosceles triangle equivalent to a given square. BOOK IV 185 Propositiox III. Theorem 581. Rectangles having equal bases are to each other as their altitudes. B |s Case I. When the altitudes are commensurable. Let ABCD and EFGH be parallelograms having equal bases and commensurable altitudes. To Prove ABCD EFGH AB EF Proof. Let S be the unit of measure for the altitudes, and let it be contained in AB m times and in EF n times, whence AB EF (1) Divide the altitudes by the unit of measure and through the points of division draw parallels to the bases. ABCD is divided into m parallelograms and EFGH into n parallelograms, and these parallelograms are all equal by § 562. ABCD m EFGH n (2) Apply Axiom 1 to (1) and (2). ABCD EFGH ~ AB EF q.e.d 186 PLANE GEOMETRY Case II. When the altitudes are incommensurable. E H Let the parallelograms ABCD and EFGH have equal bases and incommensurable altitudes. To Prove ABCD AB 'ef' EFGH Proof. Let EF be divided into a number of equal parts, and let one of these parts be applied to AB as a unit of measure. Since AB and EF are incommensurable, AB will not contain the unit of measure exactly, but a certain number of these parts will extend as far as /, leaving a remainder IB smaller than the unit of measure. Through I draw IJ parallel to the base AD. AIJD Al EFGH EF by Case I. By increasing indefinitely the number of equal parts into which EF is divided, the divisions will become smaller and smaller, and the remainder IB will also diminish indefinitely. Now AIJD A I is evidently a variable, as is also , EFGH EF and these variables are always equal. (Case I.) The limit of the variable ■ is EFGH EFGH AI AB The limit of the variable —— is EF EF By § 341 ABCD fJFGH AB EF Q.E.D BOOK IV 187 582. Corollary. Rectangles having equal altitudes are to each other as their bases. 583. Exercise. The altitudes of two rectangles having equal bases are 12 ft. and 16 ft. respectively. The area of the former rectangle is 96 sq. ft. What is the area of the other ? Proposition IV. Theorem 584. Ani/ two rectangles are to each other as the prod- ucts of their bases and altitudes. A D E H X Let A BCD and EFGH be any two rectangles. ABCD AD X AB To Prove EFGH EH X EF Proof. Construct a third rectangle XYZR, having a base equal to the base of ABCD and an altitude equal to the altitude of EFGH. ABCD AB ,os (?) (?) (?) Q.E.D. EFGH EF X EB XYZR XY EFGH XYZR EH XR ABCD AB X XR EFGH XY XEH ABCD AB X AD 188 PLANE GEOMETRY 585. Exercise. The base and the altitude of a certain rectangle are 5 ft. and 4 ft, respectively. The base and the altitude of a second rec- tangle are 10 ft. and 8 ft. respectively. How do their areas compare ? [The student must not assume that the area of the first rectangle is 20 sq. ft., as that has not yet been established.] Proposition Y. Theorem 586. The area of a rectangle is equal to the product of its base and altitude. Let A BCD be any rectangle. To Prove ABCD =axb. Proof. Let the square U, each side of which. is a linear unit, be the unit of measure for surfaces. ABCB a X b (?) Whence U 1x1 ABCD =ab X U. or ABCD = ab X the surface unit. or ABCD = ab surface units. This is usually abbreviated into ABGD = a xb. (1) Q.E.D. BOOK IV 189 587. Scholium. The meaning to be attached to formula (1) is, that the number of surface units in a rectangle is the same as the product of the number of linear units in the base by the number of linear units in the altitude. If the base is 4 ft. and the altitude 3 ft., the number of square feet (surface units) in the rectangle is 4 x 3 or 12. The area then is 12 square feet. 588. Corollary. TJie area of any parallelogram is equal to the product of its base and altitude. Let ABCD be any parallelogram and DE be its altitude. To Prove ABCD = ad x de. Proof. Draw AF ± to AD, meet- ing EC prolonged at F. Prove ADEF a rectangle. ADEF = ADCB. (?) ADEF = AD X ED. (?) ABCD = AD X ED. (?) Q.E.D. 589. Corollary. Any two parallelograms are to each other as the prod^icts of their bases and altitudes; if their bases are equal the parallelograms are to each other as their altitudes; if the altitudes are equal the parallelograms are to each other as their 590. Exercise. Construct a square equivalent to a given parallelo- gram. 591. Exercise. Construct a rectangle having equivalent to a given parallelogram. a given base and 592. Exercise. Of all equivalent parallelograms having a common base, the rectangle has the least perimeter. Of all equivalent rectan- gles, the square has the least perimeter. 190 PLANE GEOMETRY Proposition YI. Theorem 593. The area of a triangle is one half the product of its base and altitude. A CD. Let ABC be any A, and BD its altitude. To Prove ABC=\ACXBD. Proof. Construct the parallelogram A QBE. ACBE = AC X BD. (?) A ABC = \ AC X BD. (?) Q.E.D. 594. Corollary I. Triangles are to each other as the products of their bases and altitudes ; if their bases are equal the triangles are to each other as their altitudes; if their altitudes are equal the triangles are to each other as their bases. 595. Corollary II. TJie area of a triangle is one half the product of its perimeter and the radius of the inscribed circle. [Draw radii to the points of tan- b gency. /^S.E Connect the center O with the three vertices. Show that OD is the altitude of A A OB, and that OE and OF are altitudes ^ oi A BOC and AOC. Call the radius of the inscribed circle r. AAOB = ^AB'r. (?) A BOC = i BC -r. (?) A AOC = ^ AC -r. (?) A ABC = J (AB -\-BC -\- CA) r, (?) Q.E.D. ] BOOK IV 191 596. Corollary III. Calling 2 s the perimeter of triangle ABC, A ABC = s r, whence r = A ABC TJie radius of the inscribed circle of a triangle is equal to the area of the triangle divided by one half its perimeter. 597. Exercise. The area of a rhombus is equal to one half the product of its diagonals. 598. Exercise. Construct a square equivalent to a given triangle. 599. Exercise. Construct a square equivalent to a given polygon. 600. Exercise. Two triangles having a common base are to each other as the segments into which the line joining their vertices is divided by the common base, or base produced. [The A ABC and A CD have the com- mon base -40; to prove A ABC ^ BE ED AADG Draw the altitudes BF and DG. BE^BF ,p. AABC ^BF ED DG A ADC DG A ABC, A ADC' BE ED Note. When the two triangles are on the same side of the common base, BD, the line joining their vertices is divided externally at E. Prove TIF = — , using these A ADC DE 601. Definition. Lines that pass through a common point are called concurrent lines. 192 PLANE GEOMETRY 602. Exercise. If three concurrent lines AO, BO, and CO. drawn from the vertices of the triangle ABC, meet the opposite sides m the points D, E, and F, prove M^^E AF^ ^ DC EA FB [The point may be v^ithin or vyithout the triangle. BD AAOB (?) DC AAOC AF AAOC CE^ ABOC EA FB BD ABOC AAOB (?) (?) CE AF^^^ DC EA FB ■-■ 7?/) C7^ A W Conversely, if x x = 1, to prove that the lines AD, BE, DC EA FB ' ^ ' and CF are concurrent. [Draw AD and CF. Call their point of in- tersection O. Draw BO. Suppose BO pro- longed does not go to E, but some other point of AC, as E'. BD^CEl^AF^^ DC E'A FB (?) DO EA FB CE' ^ CE E'A EA Show that this last proportion is absurd. .-. AD, BE, and Ci^are concurrent.] 603. Exercise. Show by means of the converse of the last exercise that the following lines in a triangle are concurrent. 1. The medial lines. 2. The bisectors of the angles. 3. The altitudes. BOOK IV 193 Proposition VII. Theorem 604. The area of a trapezoid is one half the product of its altitude and the sum of its parallel sides. A " n "D Let ABCB be a trapezoid, and mn be its altitude. To Prove ABCD = \mn{BC + AD). Proof. Draw the diagonal AC. Show that mn is equal to the altitude of each triangle formed. ^ABC = \mn>BC. (?) AACD = \mn' AD. (?) ABCD = \mn{BC -\- AD). (?) q.e.d. 605. Corollary. The area of a trapezoid is equal to the product of the altitude and the line joining the middle points of the non-parallel sides. [EF = ^(BC -^ AD) (?) .-. ABCD =mn • EF.^ 606. Exercise. In the figure for § 604 let ^0 = 8 in., AD = 12 in., and mn = EF. Find the area of the trapezoid. 607. Exercise. Construct a square equivalent to a given trapezoid. 608. Exercise. Construct a rectangle equivalent to a given trape- zoid and having its altitude equal to that of the trapezoid. 609. Exercise. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezoid with the extremities of the opposite side is equivalent to one half the trapezoid. 610. Exercise. A straight line joining the middle points of the par- allel sides of a trapezoid divides it into tv^^o equivalent figures. SANDERS' GEOM. 13 194 PLANE GEOMETRY 611. Exercise. The area of a trapezoid is 12 sq. ft. The upper and lower bases are 7 ft. and 5 ft. respectively. Find its altitude. 612. Exercise. The area of a trapezoid is 24 sq. in. The altitude is 4 in. , and one of its parallel sides is 7 in. What is the other parallel side ? Proposition VIII. Theorem 613. Triangles that have an angle in one equal to an angle in the other, are to each other as the products of the including sides. ^ Let To Prove ABC and DEF have /.B — Ae. AABC AB ' BC A DEF BE ' EF Proof. Lay off BG = ED and BH = EF. Draw GH and AH. Prove Aabh Abhg Aabc a gbh = a def. Aabc BA BG (?) AB . BC Abhg bg • bh 614. Exercise. Prove §613, using this pair of triangles. 615. Exercise. The triangle ABC has ZB equal to Z^ of tri- angle DEF. The area of ^JBC is double that of DEF. AB is 8 ft., ^C is 6 ft., and DE is 12 ft. How long is EF? BOOK IV 195 Proposition IX. Theorem 616. Similar triangles are to each other as the squares of their hoinologous sides. Let To Prove Proof. A ABC and BEF be similar. A ABC AB^ A BEF be' Zb^ = Ae. (?) A ABC A BEF AB ' BC BE ' EF (?) AB BC (?) BE EF AABC ab' m A BEF JJe' Q.E.D. 617. Exercise. Similar triangles are to each other as the squares of their homologous altitudes. 618. Exercise. In the triangle ABC, ED is parallel to AC, and CD = \DB. How do the areas of triangles ABC and BDE compare ? 619. Exercise. The side of an equi- lateral triangle is the radius of a circle. The side of another equilateral triangle is the diameter of the same circle. How do the areas of these triangles compare ? 196 PLANE GEOMETRY 620. Exercise. Two similar triangles have homologous sides 12 ft. and 13 ft. respectively. Find the homologous side of a similar triangle equivalent to their difference. 621. Exercise. The homologous sides of two similar triangles are 3 ft. and 1 ft. respectively. How do their area's compare ? 622. Exercise. Similar triangles are to each other as the squares of any two homologous medians. 623. Exercise. The base of a triangle is 32 ft., and its altitude is 20 ft. What is the area of a triangle cut off by drawing a line parallel to the base at a distance of 15 ft. from the base ? 624. Exercise. A line is drawn parallel to the base of a triangle dividing the triangle into two equivalent portions. In what ratio does the line divide the other sides of the triangle ? 625. Exercise. Draw a line parallel to the base of a triangle, and cutting off a triangle that shall be equivalent to one third of the remain- ing portion. 626. Exercise. Equilateral triangles are constructed on the sides of a right-angled triangle as bases. If one of the acute angles of the right- angled triangle is 30", how do the largest and smallest equilateral triangles compare in area ? 627. Exercise. In the triangle ABC, the altitudes to the sides AB and AC are 3 in. and 4 in. respectively. Equilateral triangles are con- structed on the sides AB and ^O as bases. Compare their areas. 628. Exercise. The homologous altitudes of two similar triangles are 5 ft. and 12 ft. respectively. Find the homologous altitude of a tri- angle similar to each of them and equivalent to their sum. 629. Exercise. Draw a line parallel to the base of a triangle, and cutting off a triangle that is equivalent to f of the remaining trapezoid. 630. Exercise. Through 0, the point of intersection of the altitudes of the equilateral triangle ABC, lines are drawn parallel to the sides AB and BC respectively and meeting AG aXx and y. Compare the areas of triangles ABC and Oxy. BOOK TV 197 Proposition X. Theorem 631. Similar polygons are to each other as the squares of their homologous sides. C A £ Let ABODE and FGIIIJ be two similar polygons. ABODE CD^ To Prove FGHIJ HI Proof. From the vertex A draw all the possible diagonals. From F, homologous with A, draw the diagonals in FGHIJ. Similarly prove Aabc AABC AFGH AACD AFHI AABC AFGH AACD A FHI ' AACD AC Flf ¥h^ AACD A FHI AADE AFIJ AADE AFGH AFHI AFIJ AABC -\- AACD + AADE AaCD (?) (?) (?) (?) A FGH + A FHI + A FIJ AACD AFHI ABODE FGHIJ AFHI (?),ov ABODE AACD ^. (?) HI g (?) HI FGHIJ A FHI Q.E.D. 198 PLANE GEOMETRY 632. Corollary I. Similar polygons are to each other as the squares of their homologous diagonals. 633. Corollary II. In similar polygoyis homologous tri- angles are like parts of the polygons. [This was shown in the proof of the proposition.] 634. Exercise. The area of a certain polygon is 2i times the area of a similar polygon. A side of the first is 3 ft. Find the homologous side of the second. 635. Exercise. The homologous sides of two similar polygons are 8 in. and 15 in. respectively. Find the homologous side of a similar polygon equivalent to their sum. 636. Exercise. The areas of two similar pentagons are 18 sq. yds. and 25 sq. yds. respectively. A triangle of the former pentagon contains 4 sq. yds. What is the area of the homologous triangle in the second pentagon ? 637. Exercise. If the triangle ADE [see figure of § 631] contains 12 sq. in., and triangle i^/J" contains 9 sq. in., how do the areas of ABODE and FGHIJ compare ? 638. Exercise. The homologous diagonals of two similar polygons are 8 in. and 10 in. respectively. Find the homologous diagonal of a similar polygon equivalent to their difference. 639. Exercise. Connect C with w, the middle point of AD^ and H with n, the middle point of FI [see figure of § 631], and prove ABODE JCm FGHIJ Hn:'' 640. Exercise. If one square* is double another square, what is the ratio of their sides ? 641. Exercise. Construct a hexagon similar to a given hexagon and equivalent to one quarter of the given hexagon. « 642. Exercise. Construct a square equivalent to f of a given square. BOOK IV 199 Proposition XI. Theorem 643. The square described on the hypotenuse of a right- angled triangle is equivalent to the sum of the squares described on the other two sides. G Let ^jBC be a right-angled triangle. To Prove BO^ = ab^ 4- ^^ Proof. Describe squares on the three sides of the triangle. Draw AJ J_ to BC, and prolong it until it meets GF at L. Draw AF and BD. Show that the A BCD and ACF are equal by § 30. Show that the A ACF and the rectangle CJLF have a common base and equal altitudes. Whence, AACF = ^ CJLF. Similarly prove A BCD = ^A CDE. ACDE = CJLF (?) In a similar manner prove ABHI = BGLJ. ACDE H- ABHI = CJLF + BGLJ, or AC -i-AB" bg\ Q.E.D. 200 PLANE GEOMETRY 644. Note. The discovery of the proof of this proposition is attributed to Pythagoras (550 b.c), and the proposition is usually called the Pythago- rean Proposition. The foregoing proof is given by Euclid (Book I., Prop. 47). A shorter proof follows : In the R.A. A ABC, AJis drawn ± to the hypotenuse. or By § 518 BC AC AC CJ BC AB AB BJ (1) (2) Whence, AC^=BCx CJ. AB' = BCx BJ (3) (4) Adding (3) and (4) A^ -\- AB^ = BC {CJ -{- BJ) AC^+AB^ = B&. Q.E.D. 645. Corollary I. AC = BC — AB and AB" = BC — AC , that is, the square described on either side about the right angle is equivalent to the square described on the hypotenuse, diminished by the square described on the other side. 646. Corollary II. If the three sides of a right-angled triangle are made homologous sides of three similar polygons, the polygon on the hypotenuse is equivalent to the sum of the poly- gons on the other two sides. Let polygons M, N, and R be similar. M=:N-{-R. To Prove Proof. AB AC^ (?) Whence iV^- AB -\- AC M r' B^ (?) la" (?) N-[-R M AB -{-AC^ BC AB^-\-J^=Bc\ : N-\-R=M. Q.E.D. BOOK IV '201 647. Corollary III. TJie squai-e described on the hypotenuse is to the square described on either of the other sides, as the hypotenuse is to the segment of the hypotenuse adjacent to that side. ^ BG^ BC ^ BC^ BC Prove -^=9 = — and — -^ = AC^ JC Ib' bj 648. Corollary IV. The squares described on the two sides about the right angle are to each other as the adjacent segments of the hypotenuse. ^ AB^ BJ rrove — s = AC" JC In Exercises 649-654 reference is made to the figure of § 643. 649. Exercise. Show that BI is parallel to CE. 650. Exercise. The points H, A, and D are in a straight line, 651. Exercise. AG and HC are at right angles, as are also AF and BD. 652. Exercise. If HG, FD, and IE are drawn, the three triangles HBG, FCD, and EAI are equivalent. [Use § 571.] 653. Exercise. The intercepts AM and AN are equal, [/k BAN and CAM are similar to A BED and CIH respectively.] 654. Exercise. The three lines J.L, BD, and HC pass through a common point. [By means of similar triangles, show : fi = #i W ff=7i (^> and by Cor. IV, ^=^ (3). MB HB AN AB JC j^(f Multiply (1), (2), and (3) together, member by member. ^^ X — X — = 1. .. AL, BD, and HC are concurrent.] MB JC AN -■ 655. Exercise. The square described on the diagonal of a square is double the original square. 656. Exercise. The diagonal and side of a square are incommen- surable. [See preceding exercise.] 202 PLANE GEOMETRY 657. Definition. The projec- tion of CD on AB I?, tliat part of AB between the perpendiculars from the extremities of CD to AB. EF is the projection of CD on AB. MR is the projection of MN on AB. 658. Exercise. The projection of a line upon a line parallel to it, is equal to tlie line itself. The projection of a line upon another line to which it is oblique is less than the line itself. Proposition XII. Theorem 659. In any triangle the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides, diminished by twice the product of one of these sides and the projection of the other side upon it. Let ^5(7 be a A in which BC lies opposite an acute angle, and ^z> is the projection of ^^ on ^C. To Prove BC^ ^A^ -\-AC^ — 2 AC ^ AD. Proof. In figure (1) DC = AC — AD. In figure (2) DC =^ AD — AC. In either case D& = AC^-\- ad^ — 2 AC -AD. Dcf + BD^ = Ac''-{-AD^-\-BD^-2AC'AD (?) BC^ == IC^ + AB^ — 2 AC ' AD. (?) Q.E.D. BOOK IV 203 660. Exercise, Prove this proposition, using the projection oi AC on AB. 661. Exercise. In a triangle ABC, AB = 6 it., AC =6 ft, and BC = 7 it. Find the projection oi AC upon BC. Proposition XIII. Theorem 662. In an obtuse-angled triangle the square of the side opposite the obtuse angle is equivalent 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. B Let ABC be an obtuse-angled A, and CD be the projection of BC on ^C (prolonged). To Prove AB^ = BC^ + Ilf + 2 AC - CD. Proof. AD = AC -\- CD. AD^ =AC^ -i-CD^ -\-2AC- CD. AD^ -^BD' = Zc^-f Cd'^ -\- BD^ + 2AC' CD. Tb^ =AC^ -{-BC^ + 2 AC • CD. Q.E.D. 663. Corollary I. Tlie right-angled triangle is the only one in which the square of one side is equivalent to the sum of the squares of the other two sides. 664. Exercise, The sides of a triangle are 6, 3, and 5. Is its greatest angle acute, obtuse, or right ? 204 PLANE GEOMETRY Proposition XIV. Theorem 665. In any triangle the sum of the squares of two sides is equivalent to twice the square of one half the third side increased by twice the square of the medial line to the third side. B Let ABC be any A and BD be a medial line to AC, To Prove AB^ H- B(? = 2 AD^ + 2 BD^. Proof. Case I. When BD is oblique to AC. AB^ = AD^ 4- BD^ -\-2AD' DE. (?) BC^ —BD' -{-BC^ — 2BC ' BE. (?) AB^ 4- BC^ = 2 AD^ + 2 BD^. (?) Q.E.D. Case II. When BD is perpendicular to AC. AB^ = AD^ + Bd\ (?) BG^'^DC" + Bd\ (?) AB^ + Bc''^2Ad'' + 2Bd\ (?) Q.E.D. 666. Corollary I. Tlie sum of the squares of the sides of a parallelogram is equivalent to the sum of the squares of the diagonals. [Apply § Q^^ to ^ABC and ADC J and add. the equations.] BOOK IV 205 667. Corollary II. The sum of the squares of the sides of any quadrilateral is equivalent to the sum of the squares of the diagonals, increased by four times the square of the line joining the middle points of the diagonals. [To prove AB^ -\-BC^ +Cff + DA^ = BD^ +IC^ -\-^ Mn\ Draw AM and CM. Zb' + i^' = 2 Zm' + 2 Md\ (?) BO" + Cd'' = 2 cm^ + 2 md\ (?) 2(am^ 4- cm') = 2(2 an" + 2 mn\ (?) Add these three equations, member by member, and simplify. (Remember that 4 Md"^ = Bd'^.) (?)] Show that Cor. I. is a special case of Cor. II. 668. Exercise. In any triangle the difference of the squares of two sides is equivalent to the difference of the squares of their projections on the third side. Exercise. In the diameter of a circle two points A and B are taken equally distant from the center, and joined to any point P on the cir- cumference. Show that AP -\' PB is constant for all positions of P. 670. Exercise. Two sides and a diagonal of a parallelogram are 7, 9, and 8 respectively. Find the length of the other diagonal. 671. Exercise. ABCD is a rectangle, and P any point from which lines are drawn to the four vertices. Prove AP^ + CP^ = J3P^ + DP^- 206 PLANE GEOMETRY 672. Exercise. If the side AC ol the triangle ABC be divided at D so that mAD = uDC, and BD be drawn, prove mAB^ + nBC^= mAD'^ + nDC^ -\-(m + n)BD^. [niAB^ = m^AD^ + ^^ + 2 ^2> . DE). (?) n(^^ + B& -2 DC- DE). (?) mAJD^+nDC^+im-]- ti)BD^. (?) Show that § 666 is a special case of this exercise. ] 673. Exercise. The diagonals of a parallelogram are a ft. and b ft. respectively, and one side is c ft. Find the length of the other sides. 674. Exercise. Inthetriangle^jB(7(seefigureof §672), if JjB=9in., BC = 6 in., AC = 10 in., and AD = 4 in., find the length of BD. 675. Exercise. Find the lengths of the medians of a triangle. [In the triangle ABC represent the lengths of the sides by a, &, and c. Show that Median to AC = ^y/2 a^ + 2 c'^ - b^ Median to BC= iV2 b^ + 2 c^ - a^ Median to AB=^ V2 a^ + 2 b'^ - c2. ] 676. Exercise. In the triangle ABC, the lengths of the sides are represented by a, 6, and c (a being the length of BC opposite /.A, etc.). The sum of the sides is called 2 8. a + b + c = 2 8. Show that ^-^^-^ = S-a, . g + 6 + c _ 2 /S. a — 6 + c = S-b, «+6-c^^_^ BOOK IV 207 Proposition XV. Theorem 677. The area of the triangle ABC is WS{s — a)(s — b){s — c), in whiah a, b, and c are the lengths of the three sides and 2 s their sum. Let ABC be any A. To Prove AABC= Vs{s — a){S — b)(s — c). Proof. Draw tlie altitude bd. By § 659, a^ = b^ + c^-2b'AD. Whence AD = 52 -I- C^ _ gg 26 In the R.A. AABD by § 645, Bd' = (^ (^' + ^' - ay ^ 4.b'c'-(b' + c^ - ay 4.W 4 62 ^ [2 6c - 6^ - c^ + an [^ 6c + 6^ + c^ - a'^ [(g — 6 + c)(ft + 6 — c)][(6 + c — a)(6 + c + a)] 4 6^ 4 /a -|-6_-j-_c\ /6 4- c — a\fa — 6 + c'\/'a + 6 — c^ aSf a — 6 + c\ /a + 6 — c\ "A^2— A~-2— / .-. W = |(s)(5-a)(S-6)(S-c). BD = ~ Vs{s — a){s — b){s — c). 6 The area of A^5C = ^b - BD. (a) Area A^^C = V'S(S — a)(S — 6)(s — c). Q.E.D. 208 PLANE GEOMETRY 678. Corollary I. Tlie area of an equilateral triangle is one fourth the square of a side, multiplied by VS. [In the formula for the area of any triangle, substitute a for b and also for c. Area = J a^V3.] 679. Corollary II. The altitude drawn to the side b in tri- angle ABC is [See (a) of §677.] -Vs(s — a){S — b){S — c\ b Write the values of the altitudes drawn to a and c respectively. 680. Exercise. Show that the altitude of an equilateral triangle is ^aVS. (a = length of a side of the A.) 681. Exercise. The sides of a triangle are 5, 6, and 7. Find its area, and its three altitudes. 682. Exercise. The area of an equilateral triangle is 25 V3. Find its side, and also its altitude. 683. Exercise. The sides AB, BC, CD, and DA of a quadrilateral ABCD are 10 in., 17 in., 13 in., and 20 in. respectively, and the diagonal ^C is 21 in. What is the area of the quadrilateral ? 684. Exercise. Two sides of a parallelogram are 6 in. and 7 in. respectively, and one of its diagonals is 8 in. Find its area. 685. Exercise. Two diagonals of a parallelogram are 6 in. and 8 in. respectively, and one of its sides is 5 in. Find its area, and the lengths of its altitudes. 686. Exercise. The parallel sides of a trapezoid are 6 ft. and 8 ft. respectively ; one of its non-parallel sides is 4 ft., and one of its diagonals is 7 ft. Find its area. 687. Exercise. The area of a triangle is 126 sq. ft., and two of its sides are 20 ft. and 21 ft. respectively. Find the third side. [The work of this problem can be reduced by using the formula, area =^V4 h'^c^ - (h'^ + c'^ - a^)^, and substituting 20 and 21 for h and c respectively.] BOOK IV 209 Proposition XVI. Theorem 688. The area of a triangle is equal to the product of its three sides divided by four times the radius of the circumscribed circle. Let ABC be any A and let the lengths of its sides be repre- sented by a, b, and c, and the radius of the circumscribed O be called R. abc To Prove Aacb Proof. Draw the altitude BD, the diameter BE, and the chord EC. AABC=\h'BD. (?) (1) Prove Aabd and BEC mutually equiangular and similar, whence BD BC AB BE c in (1). AABC = ''^'- 4.R (3) a 2e' (2) Q.E.D. 689. Corollary. From the conclusion of the proposition we have Aabc = abc whence R = abc 4/?' SANDE&S' OEOAI. — 14 A A ABC The radius of the 210 PLANE GEOMETRY circle circumscribed about a triangle is equal to the product of the three sides divided by four times the area of the triangle. 690. Exercise. The sides of a triangle are 24 ft., 18 ft., and 30 ft. respectively. Find the radius of the circumscribed circle. Proposition XVII. Problem 691. To construct a square equivalent to the suin of two given squares, or equivalent to the difference of two given squares. B a b Let A and B be two given squares and a and b a side of each. Show that c = A-\-B. Show that D = A — B. 692. Corollary I. To construct a square equivalent to the sum of several given squares. a, b, c, and d are the sides of the given squares. Zl, Z 2, and Z3 are R.A.'s. Show that Mn' = a' -[- b^ -\- c^ -\- d\ 693. Corollary II. Construct a square having a given ratio to a given square. BOOK IV 211 Let A be the given square, and m and n lines having the given ratio. [Represent a side of the re- quired square by x. ma Then oiy^ = — a^ n X a. ma Construct a line equal to — (§ 461). n Call this line c. Then x^ = ca. Find X. (§ 519.)] 694. Exercise. Construct a square equivalent to the sum or differ- ence of a rectangle and a square. [Construct a square equivalent to the rectangle, and then proceed as in the proposition itself.] 695. Exercise. Construct a square equivalent to the sum of the squares that have for sides 2, 4, 8, 12, and 16 units respectively. 696. Exercise. If a = 2 in., construct lines having the following values: aV2j aVZ, aVS, aVo, aV?, and aVll. 697. Exercise. If a, b, and c are given lines, construct a2 + 3 &c + 4 62 and also x / 3 a^b - ^ a + 2b 2a + Sc 698. Exercise. Construct a square whose area shall be two thirds of the area of a given square. 699. Exercise. Construct a right-angled triangle, having given the hypotenuse and the sum of the legs. Let a be the given hypotenuse and b be the a sum of the legs. [Let X and y represent the legs. Then jc + y = 6, x^ + y2 = a^. (§643.) Solving these equations, we get x = l(b + V2 a2 - 62), y=l(b- V2 a2 _ 62). Construct these values of x and y. Then the three sides of the triangle are known. ] 212 PLANE GEOMETRY 700. Exercise. Construct a right-angled triangle, having given one leg, and the sum of the hypotenuse and the other leg. Proposition XVIII. Problem 701. To construct a polygon similar to either of two given similar polygons and equivalent to their sum. b Let A and B be the two given similar polygons. Required to construct a third polygon similar to either A or B, and equivalent to their sum. [Construct a R.A. A having a and b (homologous sides of A and B) for legs. On the hypotenuse of this A construct a polygon similar to A. Show, by § 646, that this is the required polygon.] 702. Corollary I. Construct a polygon similar to either of two given ^similar polygons and equivalent to their difference. 703. Corollary II. Construct a polygon similar to a given polygon and having a given ratio to it. Let a be a side of the given polygon A and — be the given ratio. [Construct oi^ = — a^. '' (§693.) On a side of the square x^ construct a polygon R similar to A. l^a^ R _^ _n _m .^.- r? BOOK IV 213 704. Corollary III. Construct a polygon similar to one given polygon and equivalent to another. Let A and B be the two given polygons. Required to construct a polygon similar to A and equivalent to B. [Construct a square C equivalent to A, and a square D equivalent to B. Let c and d be sides of these squares. Construct a line m = - — . (§ 461.) On m, homologous with a, construct a polygon i? similar to A. a^ ' Since A = c^, R = d' = s.] 705. Exercise. Construct a quadrilateral similar to a given quadri- lateral and whose area shall be 3 sq. in. (§ 704.) 706. Exercise. Construct an equilateral triangle the area of which shall be three fourths of that of a given square. EXERCISES 1. The diagonal of a rectangle is 13 ft., one of its sides is 12 ft.* What is its area ? 2. The square on the hypotenuse of an isosceles right-angled tri- angle is four times the area of the triangle. 3. The base of an isosceles triangle is 14 in., and one of the other sides is 18 in. Find the lengths of its altitudes. 4. Find a point within a triangle such that lines drawn from it to the three vertices divide the tri- angle into three equal parts. 214 PLANE GEOMETRY 6. If a circle is inscribed in a tri- angle, tlie lines joining the points of tangency with the opposite vertices are concurrent. [Showthat4?x:??x^^ FB DC EA 1. 6. Given a triangle, to construct an equivalent parallelogram the perim- eter of which shall equal that of the triangle. [FE=l{AC-^ BC).'] 7. The sum of the three perpendiculars from a point within an equi- lateral triangle to the three sides is equal to the altitude of the triangle. 8. The bases of two equivalent triangles are 10 ft. and 15 ft. respec- tively. Find the ratio of their altitudes. _ 9. ABCD is any parallelogram, and is any point within. Prove that the sum of the areas of triangles OAB and OCD equals one half the area of the parallelogram. 10. ABC is a right-angled triangle, and JSD bisects ^C. Prove that BD^ = BC^ - 3 DG^. 11. In the right-angled triangle ABC, AD is perpendicular to the hypotenuse BC, and the segments BD and DC are 9 ft. and 16 ft. respectively. Find the lengths of the sides, the area of the triangle, and the length of AD. 12. A square is greater than any other rectangle inscribed in the same circle. [Show that both square and rectangle have diameters for diagonals. ] BOOK IV 215 13. ABCD is any quadrilateral, and AE and CF are drawn to the middle points of BC and AD respectively. Prove AECF equivalent to BEA+ CFD. 14. From any point within the tri- angle ABC, OX, or, and OZ are drawn perpendicular to BC, CA, and AB re- spectively. ' Prove AZ^ + BX^ + CY^ = ZB^ + XC^ + T^. [Draw OA, OB, and 00. Then use §643.] 15. In the parallelogram ABCD any point on the diagonal ^O is joined with the vertices B and D. Prove triangles ABE and AED equiva- lent. 16. Draw a line through the point of intersection of the diagonals of a trapezoid dividing it into two equivalent trapezoids. 17. The square described on the sum of two lines is equivalent to the sum of the squares of the lines increased by twice their rectangle. 18. The square described on the difference of two lines is equivalent to the sum of the squares of the lines diminished by twice their rectangle. 19. The rectangle having for its sides the sum and the difference of two lines is equivalent to the differ- ence of their squares. 20. A triangle and a rectangle having equal bases are equivalent. How do their altitudes compare ? 21. Draw a straight line through a vertex of a triangle dividing it into two parts having the ratio of w to n. 216 PLANE GEOMETRY 22. Through a given point within or without a parallelogram draw a line dividing the parallelogram into two equivalent parts. 23. If a and h are the sides of a triangle, show that its area = \ah when the included angle is 30° or 150°; \ah-\/2 when the included angle is 45° or 135° ; i a& V3 when the included angle is 60° or 120°. [Using either a or 6 for base, find the altitude of the A.] 24. If equilateral triangles are described on the three sides of a right- angled triangle, prove that the triangle on the hypotenuse is equivalent to the sum of the triangles on the other sides. 25. On a given line as a base construct a rectangle equivalent to a given rhombus. 26. Bisect a triangle by a line drawn parallel to one of its sides. [§616.] 27. The square of a line from the vertex of an isosceles triangle to the base is equivalent to the square of one of the equal sides dimin- ished by the rectangle of the segments of the base [i.e. B& = AJE^ - AD x DC]. [Draw the altitude to AC. Use § 643.] 28. If, in Exercise 27, BD is drawn to a point D on the prolonged base, then BD^ = AB^ + AD x DC. 29. Three times the sum of the squares on the sides of a triangle is equivalent to four times the sum of the squares on its medians. [§665.] 30. If the base a of a triangle is increased d inches, how much must the altitude b be diminished in order that the area of the triangle shall be unaltered. 31. OC is a line drawn from the center of the circle to any point of the chord AB. Prove that OC^ = OA^ - AC x CB. 32. The lengths of the parallel sides of a trapezoid are a ft. and h ft. respectively. The two inclined sides are each c ft. Find the area of the trapezoid. B 33. From the middle point D of the base of the right-angled triangle ABC, DE is drawn perpendicular to the hypotenuse BC. Prove that BE"^ - EG^ = AB^. BOOK IV 217 34. In any circle the sum of the squares on the segments of two chords that are perpendicular to each other is equivalent to the square on the diameter. [§ 64;].] 35. Construct a triangle having given its angles and its area. 36. In the triangle ABC, AD, BE, and GF are lines drawn from the vertices and passing through a common point 0. Prove that ^ + ^+^=1. V OE Vbe BE J\AOC AD CF (?) A ABC 37. From any point within a triangle ABC, OD, OE, and OF are drawn to the three sides. From the vertices AD', BE', and CF' are drawn parallel to OD, OE, and OF respectively. Find similar expressions for -^— and Prove that OE BE' OD AD' OF CF' 1. rOE \_BE' A /\AOC E E' /\ABC J 38. Given the altitude, one of the angles, and the area, construct a parallelogram. 39. The two medial lines AE and CD of the triangle ABC intersect at F. Prove the triangle AFC equivalent to the quadrilateral BDFE. 40. The diagonals of a trapezoid divide it into four triangles, two of which are similar, and the other two equivalent. 41. Any two points, C and D, in the semi- circumference ACB are joined with the ex- tremities of the diameter AB. AE and BF are drawn perpendicular to the chord DC pro- longed. Prove that CE'^ + CF'^ = DE'^ + DF'^. A B [Use § 643.] 42. Describe four circles each of which is tangent to three lines that form a triangle. 218 PLANE GEOMETRY [One of the four is the inscribed circle of the A, and its radius is denoted by r. The other three are called escribed circles of the tri- angle, and their radii are denoted hy ^aj Ti^ and re. (?*« is the radius of the escribed circle lying between the sides of Z J. of the A.)] 43. The area of triangle ABC = fa (S — a). [A ABC = A ABE + A ACE - A BEC, and ra is the altitude of each of these A.] Show that rj (S—b) and re iS—c) are also expressions for the area of triangle ABC. 44. The area of triangle ABC Vr X 7'ft X re. [Ex. 43.] 45. Prove that ra-\- n + re —r = 4 B [B = radius of the circle cir- cumscribed about A ABC]. [Ex. 43 and § 689.] 46. Prove that r Va Th Tc 47. The area of a quadrilateral is equivalent to that of a triangle having two of its sides equal to the diagonals of the quadrilateral and its included angle equal to either of the angles be- tween the diagonals of the quadrilateral. IDF = BE and CQ = AE. Show that A GDF =AABC and A GED = A ACD.] 48. Parallelograms A DEB and BFGC are described on two sides of the triangle ABC. DE and GF are prolonged until they meet at H. HB is drawn, A third parallelogram AIJC is constructed on AC, having AI equal to and parallel to BH. Prove that AIJC is equivalent to the sum of ADEB and BFGG. IADEB=ALKI and BFGC = LCJK.'\ BOOK IV 219 49. The lines joining the points of tangency of the escribed circles with the opposite vertices of the triangle ABC, are concurrent. [See Ex. 5.] 50. Deduce the Pythagorean Theo- rem (Prop. XI, Bk. IV) from Exercise 48. 51. Through a point P within an angle draw a line such that it and the parts of the sides that are inter- cepted shall contain a given area. [Construct parallelogram BDEF = required area (Ex. 38), DE passing through P. If HG is the required line, A PIE = A IFH + A PDG. The A are similar, DP, PE, and FH are homologous sides, and DP and PE are known.] i. Is there any limit to the " given area " in Exercise 51 ? BOOK V 707. Definition. A regular polygon is a polygon that is both equilateral and equiangular. Proposition I. Theorem 708. // the circumference of a circle is divided into three or more equal parts, the chords joining the succes- sive points of division form a regular inscribed polygon; and tangents drawn at the points of division form a regular circumscribed polygon. Let the arcs AB, BC, etc., be equal. To Prove the polygon ABCD ••• a regular inscribed polygon. [The proof is left to the student.] Let the arcs AB, BC, etc., be equal. To Prove the polygon xyz •••a regular circumscribed polygon. Proof. [Draw the chords AB, BC, etc. Show that the A AyB, BzC, etc., are isosceles A and are equal in all respects.] 220 BOOK V 221 709. Corollary I. If at the middle points of the arcs sub- tended by the sides of a regular inscribed poly- gon, tangents to the circle are drawn, L The circumscribed polygon formed is regular. II. Its sides are parallel to the sides of the inscribed polygon. III. A line connecting the center of the circle with a vertex of the outer polygon passes through a vertex of the inner polygon. [?/0 bisects /-HON, consequently bisects arc MN, and there- fore passes through bJ] 710. Corollary II. If the arcs subtended by the sides of a regular inscribed polygon are bisected, and the points of division are joined with the extremities of the arcs, the polygon formed is a regular inscribed polygon of double the number of sides; and if at the extremities of the arcs and at their middle jioiids tangents are drawn, the polygoyi formed is a regular circumscribed polygon of double the number of sides. 711. Corollary III. TJie area of a regular inscribed poly- gon is less than that of a regular inscribed polygon of double the number of sides; but the area of a regular circumscribed polygon is greater than that of a regular circumscribed polygon of double the number of sides. 712. Exercise, An equiangular polygon circumscribed about a circle is regular. 713. Exercise. An inscribed equiangular polygon is regular if the number of its sides is odd. 714. Exercise. A circumscribed equilateral polygon is regular if the number of its sides is odd. 222 PLANE GEOMETRY Proposition II. Theorem 715. A circle can he circumscribed about any regular polygon; and one can also be inscribed in it. Let ABC "• GhQ2i regular polygon. I. To Prove that a circle can be circumscribed about it. Proof. Pass a circumference through three of the vertices, A, B, and C, and let O be its center. Draw the radii OA, OB, and OC. Draw OB. ShowthatZl = iZ5andZ3 = iZc. Prove AOCB and OCB equal in all respects. Whence OD = OB. Therefore the circumference that passes through A^ B, and C will also pass through D. Similarly, it can be shown that this circumference passes through the remaining vertices. q.e.d. II. To Prove that a circle can be inscribed in the polygon. Proof. Describe a circle about the regular polygon AB -" G. The sides AB, BC, etc., are all equal chords of this circle, and are equally distant from the center (?). With O as a center and this distance for a radius describe a circle. Show that AB, BC, etc., are tangent to this circle, which is, therefore, a circle inscribed in the regular polygon. q.e.d. BOOK V 223 716. Definitions. The common center of the circles that are inscribed in and circumscribed about a regular polygon, is called the ceriter of the x>olygon. The angles formed by radii drawn from this center to the vertices of the polygon are called angles at the center. Each angle at the center is equal to 4 right angles divided by the number of sides in the polygon. A line drawn from the center of the polygon perpendicular to a side, is an apothem. The apothem of a regular polygon is equal to the radius of the inscribed circle. 717. Exercise. How many degrees in the angle at the center of an equilateral triangle ? Of a square ? Of a regular hexagon ? Of a regular polygon of n sides ? 718. Exercise. How many sides has the polygon whose angle at the center is 30° ? 18" ? 719. Exercise. In what regular polygon is the apothem one half the radius of the circumscribed circle ? 720. Exercise. In what regular polygon is the apothem one half the side of the polygon ? 721. Exercise. Show that an angle at the center of any regular polygon is equal to an exterior angle of the polygon. Proposition III. Theorem 722. Regular polygons of the same number of sides are similar. [Show that the polygons are mutually equiangular and have their homologous sides proportional.] 224 PLANE GEOMETRY Proposition IV. Theorem 723. The perimeters of similar regular polygons are to each other as the radii of their inscribed or of their cir- cumscribed circles; and the polygons are to each other as the squares of the radii. N Let ABC '•• F and MNR ••• -S be two similar regular polygons. To Prove that their perimeters are proportional to the radii of the inscribed and of the circumscribed circles, and that their areas are proportional to the squares of these radii. Proof. Let x and y be the centers of the regular polygons. Draw xB and yN, and the apothems xE and yL. xB and yN are the radii of the circumscribed circles and xE and yL are the radii of the inscribed circles. Perimeter ABC ••- F_ BC^ _Bx _xE S~ NR~~ Ny~ yL Perimeter MNR Area ABC -" F (?) Area J/iV^i?--- S BC NR^ Bx Ny 2 2 ■ -2 ^' ^ xE ■ -') yL- Q.E.D. 724. Exercise. Two squares are inscribed in circles, the diameters of wliich are 2 in. and 6 in. respectively. Compare their areas. 725. Exercise. A regular polygon, the side of which is 6 in., is circumscribed about a circle having a radius \/3 in. Find the side of a similar polygon circumscribed about a circle the radius of which is 6 in. BOOK V 225 726. Exercise. The perimeters of similar regular polygons are to each other as the diameters of their inscribed or of their circumscribed circles ; and the polygons are to each other as the squares of the diameters. Proposition V. Problem 727. To inscribe a square in a given circle, C Let be the center of the given circle. Required to inscribe a square in the circle. Draw the diameters AB and CD at right angles. Connect their extremities. Prove ACBD an inscribed square. (§ 708.) Q.E.F. 728. Corollary I. Tangents to the circle at the extremities of the diameters AB and CD form a circumscribed square. 729. Corollary II. The side of the inscribed square is R ^2. The side of the circumscribed square is 2r. The area of the inscribed square is 2r^. The area of the circumscribed square is 4i2^. 730. Corollary III. By bisecting the arcs and drawing chords and tangents as described in § 710, regidar polygons of 8, 16, 32, 64, etc., sides can be inscribed in and circumscribed about the circle. SANDERS' GEOM. — 15 226 PLANE GEOMETRY 731. ExKRCiSE. The radius of a circle is 5 ft. Find the side and the area of the inscribed square. 732. Exercise. Find the side and the area of a square circumscribed about a circle, having a diameter 6 in. long. 733. Exercise. The area of a square is 16 sq. in. Find the radius of the inscribed circle and also the radius of the circumscribed circle. Proposition VI. Problem 734. To inscribe a regular hexagon in a circle, D Let be the center of the given circle. Required to inscribe a regular hexagon in the circle. Draw the radius OA. Lay off the chord AB = OA. Draw OB. AOAB \^ equilateral, and angle contains 60°. .-. the arc ^Z? is i of the circumference, and the chord AB is one side of a regular hexagon. Complete the hexagon ABCDEF. q.e.f. 735. Corollary I. The chords joining the three alternate vertices form an inscribed equilateral triangle. 736. Corollary II. Tangents drawn at the vertices of the inscribed hexagon and of the triangle form a regular circum- scribed heocagon and a regular circumscribed triangle. BOOK V 227 737. Corollary III. If the arcs are bisected and chords and tangents are draivn according to § 710, regular j^olygons of 12, 24, 48, etc., sides will be inscribed in and circumscribed about the circle. 738. Exercise. The side of the inscribed equilateral triangle is i?V3, and its area is |i?2\/3. 739. Exercise, The side of the circumscribed equilateral triangle is 2 i? V3, and its area is 3 i?2 V3. 740. Exercise. The side of a regular inscribed hexagon is JK, and its area is f ^'^Vs. 741. Exercise. The side of a regular circumscribed hexagon is |i?\/3, and its area is 2B'^y/Z. 742. Exercise. The area of a regular inscribed hexagon is double that of an equilateral triangle inscribed in the same circle. [Show this in two ways : 1st, by comparing the values of their areas as derived in §§ 738 and 740 ; 2d, by a geometrical demonstration using the figure of § 734.] * 743. Exercise. What is the area of a regular hexagon inscribed in a circle, the radius of which is 4 in. ? 744. Exercise. The area of a regular inscribed hexagon is 10 sq. in. What is the area of a regular hexagon circumscribed about the same circle ? 745. Exercise. The area of an equilateral triangle is 48 V3 sq. ft. Find the radii of the inscribed and of the circumscribed circles. 746. Exercise. The area of a regular hexagon is 54a2-v/3. Find the radii of the inscribed and of the circumscribed circles. 747. Exercise. Show that the circumscribed equilateral triangle is 4 times the inscribed equilateral triangle ; that the circumscribed square is 2 times the inscribed square ; and that the circumscribed regular hexa- gon is I of the inscribed regular hexagon. 748. Exercise. Divide a circumference into quadrants by the use of compasses only. [Suggestion. The side of an inscribed square is the altitude of an isosceles triangle whose base is 2 jR and one of whose sides is B V'3. ] 228 PLANE GEOMETRY Proposition YII. Problem 749. To inscribe a regular decagon in a circle. Let be the center of the given circle. Required to inscribe a regular decagon in the circle. Draw the radius OA. Divide it into extreme and mean ratio, OB being the greater segment. Lay off ^C = OB. Draw BC and OC. By definition (Art. 551), M = ^. ^ ^ OB BA AC BA ^^ AOAC and BAC are similar. (§ 495.) .-. A 5^ C is isosceles, and ^G' = J5C. (?) A ^OC is isosceles. (?) A1=Z3 + Z0 (?) or Zl = 2Zo. (?) Aa = 2Zo (?) and Zaco = 2Zo. (?) Za+Zaco-{-Zo = 1S0°. (?) 2Zo + 2Zo-{-Zo = 180°. (?) .-. Zo = 86°. .'. the arc AC, the measure of Z 0, contains 36° of arc, and is jij^ of the circumference. The circumference can therefore be divided into ten parts, each equal to the arc AC, and the chords joining the points of division form a regular inscribed decagon. q.e.f. BOOK V 229 750. Corollary I. The cliords joining the alternate vertices of a regular inscribed decagon form a regular inscribed pentagon. 751. Corollary II. Tangents drawn at the vertices of the regular inscribed pentagon and decagon form a regular circwm- scribed pentagon and a regular circumscribed decagon. 752. Corollary III. If the arcs are bisected and chords and tangents are drawn according to § 710, regular inscribed and circumscribed polygons of 20, 40, 80, etc., sides will be formed. 753. Exercise. The length of the side of a regular inscribed decagon is Kv^ -!)»•• 754. Exercise. Find the length of a side of a regular inscribed pen- tagon. [In the R. A. A ADC (see the figure of § 749), ^O is the side of the decagon, and AD is one half the difference between the radius and the side of the decagon.] VlO - 2 V5 Arts. r. 2 755. Exercise, Show that the sum of the squares described on the sides of a regular inscribed decagon and of a regular inscribed hexagon equals the square described on the side of a regular inscribed pentagon. [Represent the sides of the pentagon, hexagon, and decagon by p, h, and d, respectively. In the figure of § 749, DC^ = AC^ - AD^, whence p"^ = n(P - h"^ + 2hd. (1) By § 551 ~ = — ^ , whence hd = h^ - d^. (2) d h — d From (1) and (2) p'^ = d^ + h\ Give also to algebraic proof.] 756. Exercise. What is the length of the side of a regular decagon inscribed in a circle having a diameter 4 in. long ? 757. Exercise. If the side of a regular pentagon is 2V5 in., show that the radius of the circumscribed circle is v 10 + 2 Vs in. 230 PLANE GEOMETRY Proposition VIII. Problem 758. To inscribe a regular pentedecagon in a circle. A Let be the center of the given circle. Required to inscribe a regular polygon of fifteen sides in the circle. Lay off the chord AB = side of regular inscribed hexagon, and the chord AC = side of regular inscribed decagon. The arc ^5 contains 60°, (?) and the arc AC, 36°. (?) .\ the arc J5(7 contains 24° and is -^-^ of the circumference. The circumference can therefore be divided into fifteen parts, each equal to BC; and the chords joining the points of division form a regular inscribed pentedecagon. q.e.f. 759. Corollary I. Tangents drawn at the vertices of the inscribed, pentedecagon form a regidar circumscribed pentedecagon. 760. Corollary II. If the arcs are bisected, and chords and tangents are drawn as described in § 710, regular inscribed and circumscribed polygons of 30, 60, 120, etc., sides icill be formed. 761. Scholium. In Propositions V., VL, VII., and VIII. we have seen that the circumference can be divided into the following numbers of equal parts : 2, 4, 8, 16 ... 2^ 3, 6, 12, 24 ... 3x2" 5, 10, 20, 40 ... 5x2" 15, 30, 60, 120 ... 15 X 2'* n being any positive BOOK V 231 The mathematician Gauss has shown that it is possible to divide the circumference into 2" + 1 equal parts, n being a positive integer and 2" + 1 a prime number. It is therefore possible, by the use of ruler and compasses, to divide the circumference into 2, 3, 5, 17, 257, etc., equal parts. [An elementary explanation of the division of the circum- ference into seventeen equal parts is given in Felix Klein's " Vortrage liber ausgewahlte Fragerfder Elementar Geometric."] Proposition IX. Theorem 762. The arc of a circle is less than any line that envelops it and has the same extremities. Let AMB be the arc of circle and ASB any other line envelop- ing it and passing through A and B. To Prove AMB < ASB, Proof. Of all the lines (a MB, ASB, etc.) that can be drawn through A and B, and including the segment or area AMB, there must be one of minimum length. ASB cannot be the minimum line, for draw the tangent CD to the arc AMB. CD < CSD. (?) ACDB OC. (?) OA—OC 3.1415877 2) 2048 3.14159512) 3.1415914 2) 4096 3.1415933 2) 3.1415923 2) 8192 3.1415928 D 3.1415926 2) . The circumference of the circle therefore lies between 3.14159262) and 3.1415928 2). For ordinary accuracy the value of tt is taken as 3.1416. Note. — The vahie of tt has been carried out over seven hundred decimal places. [See article on "Squaring the Circle" in the Encyclo- psedia Britannica. ] The value of ir to thirty-five decimal places is 3.14159265358979323846264338327950288. By higher mathematics, the diameter and circumference of the circle have been shown to be incommensurable, so no exact expression for their ratio can be obtained. 795. Exercise. The radius of a circle is 10 in. Find its circum- ference and its area. 796. Exercise. The area of a circle is 7854 sq. ft.. Find its cir- ' cumf erence. BOOK V 243 797. Exercise. The circumference of a circle is 50 in. What is its area ? 798. Exercise. The radius of a circle is 60 ft. What is the area of a sector whose arc contains 40° ? 799. Exercise. The radius of a circle is 10 ft. The area of a sector of that circle is 120 sq. ft. What is its arc in degrees ? EXERCISES 1. In a regular polygon of n sides, diagonals are drawn from one vertex. What angles do they make with each other ? 2. Show that the altitude of an inscribed equilateral triangle is | of the diameter, and that the altitude of a circumscribed equilateral triangle is 3 times the radius. 3. The radii of two circles are 4 in. and 6 in. respectively. How do their areas compare ? 4. Find the area of the ring between the circumferences of two con- centric circles the radii of which are a and h respectively. 5. The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and the circumscribed equilateral tri- angles. [See Ex. to Prop. 6. ] 6. The diagonals joining the alternate vertices of a regular hexagon form by their intersection a regular hexagon having an area one third of that of the original hexagon. 7. Eind the area of the six-pointed star in the figure of Exercise 6 in terms of the radius of the circle. 8. Erom any point within a regular polygon of n sides, perpendiculars are drawn to the sides. Prove that the sum of these perpendiculars is equal to n times the apothem of the polygon. [Join the point with the vertices and obtain an expression for the area of the polygon. Compare this with the expression for the area obtained from§ 771.] 9. Construct a circle that shall be double a given circle (§ 784). 10. Construct a circle that shall be one half a given circle. 11. Construct a circle equivalent to the sum of two given circles ; also one equivalent to their difference. [§ 646.] 244 PLANE GEOMETRY 12. If two circles are concentric, show that the area of the ring between their circumferences is equal to the area of a circle having for its diameter a chord of the larger circle that is tangent to the smaller. 13. Find the area of the sector of a circle intercepting an arc of 50°, the radius of the circle being 10 ft. [§ 776.] 14. The radius of a circle is 20 ft. What is the angle of a sector having an area of 300 sq. ft. ? 15. The radius of a circle is 20 ft., and the area of a sector of the circle is 300 sq. ft. Find the area of a similar sector in a circle having a radius 50 ft. long. 16. What is the radius of a circle having an area equal to 16 times the area of a circle with a radius 5 ft. long ? 17. Find the area of a circle circumscribed about a square having an area of 600 sq. ft. [§ 729.J 18. Show that the area of a circumscribed equilateral triangle is greater than that of a square circumscribed about the same circle. 19. Four circles, each with a radius 5 ft. long, have their centers at the vertices of a square, and are tangent. Find the area of a circle tan- gent to all of them. 20. How many degrees in the arc, the length of which is equal to the radius of the circle ? 21. A circle is circumscribed about the right- angled triangle ABC. Semicircles are described on the two legs as diameters. Prove that the sum of the crescents AD BE and BFCG is equiv- alent to the triangle ABC. 22. The radius of a regular inscribed polygon is a mean proportional between its apothem and the radius of a similar circumscribed polygon. 23. If the bisectors pf the angles of a polygon meet in a point, a circle can be inscribed in the polygon. 24. The diagonals of a regular pentagon form by their intersection a second regular pentagon. 25. Any two diagonals of a regular pentagon not drawn from a common vertex divide each other into extreme and mean ratio. [A ABC and CfD are similar.] 26. Divide an angle of an equilateral triangle into five equal parts. BOOK V 245 27. If two angles at the centers of unequal circles are subtended by arcs of equal length, the angles are inversely proportional to the radii of the circles. 28. The apothem of a regular inscribed pentagon is equal to one half the sum of the radius of the circle and a side of a regular inscribed decagon. 29. If two chords of a given circle intersect each other at right angles, and on the four segments of the chords as diameters, circles are de- scribed, the sum of the four circles is equivalent to the given circle. [Ex. 34, page 217.] 30. Divide a circle into three equivalent parts by concentric circles (§ 784). 31. The radius of a given circle ABD is 10 ft. Find the areas of the two segments BCA and BDA into which the circle is divided by a chord AB equal in length to the radius. [Sub- tract area of A from area of sector. ] 32. Find the radius of a circle that is doubled in area by increasing its radius one foot. 33. On the sides of a square as diameters, four semicircles are described within the square, forming four leaves. If the side of the square is a, find the area of the leaves. 34. In a given equilateral triangle inscribe three equal circles tangent to each other and to the sides of the triangle. 35. In a given circle inscribe three equal circles tangent to each other and to the given circle. 36. In the circle ABCD, the diameters AC and BD are at right angles to each other. With E, the middle point of 00, as a center, and BB as a radius, the arc BF is described. Prove that the radius OA i.^ divided into extreme and mean ratio at F. [Describe arc OG with E as center, and arc G^^with £ as center.] 246 PLANE GEOMETRY 37. The diameter AB of a given circle is divided into tv/o segments, AC and CB. On each segment as a diameter a semicircle is described, but on opposite sides of the diameter. Prove that the sum of the two semi-circumferences described is equal to the semi-circumference of the given circle, and that the line they form divides the given circle into parts that are to each other as the segments of the diameter. 38. If a given square is divided into four equal squares, and a circle is inscribed in each of the small squares and also in the given square, prove that the sum of the four small circles is equiva- lent to the circle inscribed in the given square. 39. If a regular polygon of n sides be circumscribed about a circle, the sum of the perpendiculars from the points of contact to any tangent to the circle is equal to n times the radius. [If J., B, C, Z), etc., are the points of contact of the polygon and P the point at which the tangent is drawn, the sum of the Js from A, B, etc., on tangent at P = sum of Js from P to tangents drawn at A, B, etc. ; and this by Ex. 8 = wP.] 40. The sum of the perpendiculars from the vertices of a regular inscribed polygon to any line without the circle is equal to n times the perpendicular from the center of the circle to the line. [Draw a tangent to the O parallel to the given line, and then use Ex. 39.] 41. The sum of the squares of the lines drawn from any point in the cir- cumference to the vertices of a regular inscribed polygon is equal to 2 nE^. [Using notation of Ex. 39, show that the square of the line from the given point P to each vertex = 2 P times the ± from the vertex to a tangent at P. Add these equations and use Ex. 39.] 42. A crescent-shaped region is bounded by a semi-circumference of radius a, and another circular arc whose center lies on tlie semi-circum- ference produced. Find the area and the perim- eter of the region. [Show that the arc is a quadrant in a O with radius = a \/2.] 43. Three points divide a circumference into equal parts. Through each pair of these points an arc of a circle is described tangent to the BOOK V 247 radii drawn to the points and lying wholly within the circle. Find the perimeter of the figure thus formed, and show that its area is 3 ( V3 — I tt) a2^ where a denotes the radius of the circle. [Show that each arc is ^ of a circumference with radius a VS.] 44. Three radii are drawn in a circle of radius 2 a, so as to divide the circumference into three equal parts ; and, with the middle of these radii as centers, arcs are drawn, each with the radius a, so as to form a closed figure (trefoil). Show that the length of the perimeter of the trefoil is equal to that of the circle, and find its area. RETURN TO the circulation desk of any University of California Library or to the NORTHERN REGIONAL LIBRARY FACILITY BIdg. 400, Richmond Field Station University of California Richmond, CA 94804-4698 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS 2-month loans may be renewed by calllna (510)642-6753 1-year loans may be recharged by bringing books Renewals and recharges may be made 4 days prior to due date DUE AS STAMPED BELOW 'mi 9 1993 YB 17302 A-/^'^ '■■J'>::.S'^r%k ■ ■ • ■;•< ' ^ ■; ,>':.:■ h-kk WMii