■ 
 
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Ray's Series, 
 
 EMBRACING 
 
 A Thorough and Progressive Course in Arithmetic, Algebra, 
 and the Higher Mathematics. 
 
 Primary Arithmetic. 
 Intellectual Arithmetic. 
 Rudiments of Arithmetic. 
 
 Higher Arithmetic. 
 
 Test Examples in Arithmetic. 
 
 New Elementary Algebra. 
 
 Practical Arithmetic. New Higher Algebra. 
 
 Plane and Solid Geometry. By Eli T. Tappan, A.M., Prcs't 
 
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 Geometry and Trigonometry. By Eli T. Tappak, A.M. 
 
 Prcs't Kenyon College. 8vo, sheep, 420 pp. 
 Analytic Geometry, By Geo. H. Howisok, A.M., Prof, in Mass. 
 Institute of Technology. Treatise on Analytic Geometry, especially 
 as applied to the Properties of Conies : including the Modern 
 Methods of Abridged Notation. 
 
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EVANS' SCHOOL GEOMETRY. 
 
 PRIMARY ELEMENTS 
 
 OF 
 
 PLANE AND SOLID GEOMETRY: 
 
 FOR SCHOOLS AND ACADEMIES. 
 
 By E. W. EVANS, M. A., 
 
 PBOrtSSOR OF MATHEMATICS IN MARIETTA COLIEOE. 
 
 WILSON, HINKLE & CO., 
 
 CIISrCINlNr^TI: 
 
 137 "Walnut St. 
 
 NEW YORK 
 28 Bond St. 
 

 Entered according to Act of Congress, in the year 1862, by 
 
 "W. B. SMITH & CO., 
 
 In the Clerk's Office of the District Court of the United States, for the 
 Southern District of Ohio. 
 
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 Cincinnati, 0. 
 
PREFACE. 
 
 The following concise treatise on Geometry has been 
 prepared chiefly for that class of students in our Pub- 
 lic Schools and elsewhere, who can not spare the requisite 
 time for mastering the larger works. It is conceived, 
 also, that it will be found useful as a course of first 
 lessons for those who intend to pursue the study more 
 at length. 
 
 The writer has aimed, in general, to select those prop- 
 ositions in both Plane and Solid Geometry which have 
 the most important direct applications ; omitting a large 
 number of those which serve chiefly as steps to some- 
 thing following them in an extended course of reasoning. 
 In order to do this without breaking the logical con- 
 nection, he has often found it necessary to present the 
 demonstrations in a considerably modified form ; though, 
 in a majority of cases, they are essentially based on 
 those found in Euclid, Legendre, and other standard 
 works. 
 
 In the treatment of parallel lines, important hints 
 have been taken from the work of Professor Peirce, 
 
6 PKEFACE. 
 
 whose method on this subject is the most simple and 
 logical. Upon the whole, the demonstrations have been 
 put in that form which was thought best suited to the 
 comprehension of beginners having no previous knowl- 
 edge of mathematics beyond arithmetic. 
 
 In the divisions of the work, the writer has aimed 
 to follow the simple and natural classification of the 
 subjects treated. The arrangement adopted has enabled 
 him to set down as simple corollaries many propositions 
 which are usually demonstrated separately. The defini- 
 tions have been distributed between the various sections, 
 instead of being crowded together at the beginning. 
 After each Section, or Book, there will be found a few 
 practical illustrations and exercises. In a Supplement, 
 some examples have been given of the application of 
 Algebra to Geometry. 
 
 The amount of Geometry contained in the work is 
 sufficient to prepare the pupil for the study of Plane 
 Trigonometry and Surveying. 
 
CONTENTS. 
 
 INTRODUCTION. 
 
 PACK 
 
 Sec. I. General Definitions 9 
 
 Sec. IT. Of Terms and Signs 10 
 
 Sea III Axioms H 
 
 BOOK I 
 PLANE GEOMETRY. — RATIOS. 
 
 Sec. IV. Straight Lines and their Angles 13 
 
 Sec. V. Triangles 13 
 
 Sec. VI. Quadrilaterals 2G 
 
 Sec. VII. Of Polygons in General 35 
 
 See. VIII. Of the Circle 38 
 
 Sec. IX. Problems in Construction 49 
 
 * 
 
 BOOK II. 
 PLANE GEOMETRY.— PROPORTIONS. 
 
 Sec. X. Of Polygons 55 
 
 Sec. XI. Of Circles 03 
 
 Sec. XII. Problems in Construction 73 
 
 (7) 
 
8 CONTENTS 
 
 BOOK III. 
 SOLID GEOMETRY. 
 
 PAGE 
 
 Sec. XIII. Planes and their Inclinations 76 
 
 Sec. XIV. Of Parallelopipeds 79 
 
 Sec. XV. The Prism and the Cylinder 83 
 
 Sec. XVI. Pyramids and Cones 86 
 
 Sec. XVII. The Sphere 92 
 
 SUPPLEMENT. 
 
 Sec. XVIII. Miscellaneous Examples 97 
 
 Sec. XIX. Applications of Algebra .'... 99 
 
INTRODUCTION. 
 
 SECTION I— GENERAL DEFINITIONS. 
 
 1. Geometry is that branch of mathematics which 
 treats of magnitudes in space. 
 
 2. A line is that magnitude which has length, with- 
 out breadth or thickness. 
 
 3. A surface is that which has length and breadth, 
 without thickness. 
 
 4. A solid is that which has length, breadth, and 
 thickness. 
 
 5. A point has position without extension. 
 
 6. A straight line is one which has the same di- 
 rection through all its consecutive parts. 
 
 7. Of bent lines, that which is composed of straight 
 lines is called a broken lirie; that of which no part is 
 straight is called a curve. 
 
 8. A plane is a surface in which if any two points 
 whatever be taken, the straight line joining them will 
 lie wholly in that surface. 
 
 9. Magnitudes of extension are lines, surfaces, and 
 solids. The relative positions of these give rise to 
 magnitudes of direction, or angles, which will be defined 
 in their proper connections. 
 
 (9) 
 
10 GEOMETRY. 
 
 SEC. II.— OF TEEMS AND SIGNS. 
 
 1. An anjom is a self-evident truth. 
 
 2. A theorem is a statement of a truth which re- 
 quires to be proved, or demonstrated. 
 
 8. A problem is a statement of something required 
 to be done, or solved. 
 
 4. The term proposition may be applied either to 
 a theorem or to a problem. 
 
 5. A corollary is an obvious consequence from 
 something that precedes. 
 
 6. A scholium is some remark relating to what 
 precedes. 
 
 7. The term hypothesis denotes the supposition 
 made, or the conditions given, in any proposition. 
 
 8. The term infinite, as used in geometry, means 
 beyond measure ; that is, either absolutely beyond 
 limits, or beyond all appreciable limits. 
 
 9. A ratio is the relation which one quantity bears 
 to another, as equal to it, greater, or less. The value 
 of the ratio is the quotient arising from dividing one 
 of the quantities by the other. If A and B represent 
 any two quantities, the ratio of A to B may be written 
 
 A 
 
 either A : B, or — • 
 
 Jj 
 
 10. A proportion is an equality of ratios. Thus, 
 if the ratio of A to B be equal to the ratio of C to 
 D, those two ratios will constitute a proportion, which 
 may be written A : B : : C : D. 
 
 11. The sign = denotes the equality of two quan- 
 tities between which it is placed. Hence, 
 
 A C 
 
 A proportion may also be written — = — • 
 
INTRODUCTION. 11 
 
 12. The sign -f" denotes addition. Thus, A-\-B 
 means the sum of A and B, and is read A plus B. 
 
 13. The sign — denotes subtraction. Thus, A — B 
 means A diminished by B, and is read A minus B. 
 
 14. The sign X denotes multiplication. Thus, AxB 
 means A multiplied by B. Sometimes, however, this 
 sign is omitted, especially if one of the factors be a 
 figure. Thus, 2B means twice B. 
 
 15. The sign -r- denotes division. Thus, A-=-B 
 
 means A divided by B. It is equivalent to — 
 
 16. The parenthesis ( ) denotes that the quantities 
 inclosed within it are to be subjected to the same 
 operation. Thus, Ax(B-f-C) means A multiplied into 
 the sum of B and C. 
 
 17. The square of any quantity, as A, is written 
 A 2 ; its cube is written A 3 . 
 
 18. The square root of any quantity, as A-|-B, is 
 written ^/A+B. 
 
 SEC. III.— AXIOMS. 
 
 There are certain axioms, relating to quantity in 
 general, which lie at the foundation of all branches 
 of mathematics. Others relate particularly to magni- 
 tudes in space. Some of those axioms, of both classes, 
 which are most frequently applied in geometry, are 
 stated here. 
 
 1. The whole is greater than any of its parts. 
 
 2. The whole is equal to the sum of all its parts. 
 
 3. Things which are equal to the same thing are 
 equal to each other. 
 
 4. If equals be added to equals, the sums will be 
 equal. 
 
12 GKOMETBY. 
 
 5. If equals be taken from equals, the remainders 
 will be equal. 
 
 6. If equals be added to unequals, the sums will be 
 unequal. + 
 
 7. If equals be taken from unequals, the remainders 
 will be unequal. 
 
 8. Doubles of the same thing are equal to each 
 other. 
 
 9. Halves of equal things are equal to each other. 
 
 10. From one point to another only one straight 
 line can be drawn ; and that is the shortest line be- 
 tween them. 
 
 11. Two magnitudes are equal if, when one is ap- 
 plied to the other, they will exactly coincide. 
 
r. 
 
 Of THE Y 
 
 U N/VE Rsrp . 
 
 
 OF 
 
 RNM 
 
 PLANE GEOMETRY. 
 
 BOOK I, 
 
 RATIOS OF MAGNITUDES LYING IN THE SAME PLANE. 
 
 SEC. IV.— STRAIGHT LINES AND THEIR ANGLES. 
 
 DEFINITIONS. 
 
 1. The divergence of two straight lines from a point 
 constitutes an angle. The quantity of the angle is 
 the difference of direction between the two lines. 
 
 The lines themselves are said to contain the angle, 
 and are called its sides. The point from which they 
 diverge is called the vertex. . C 
 
 Thus, AB and BC are the sides, and 
 B the vertex of the angle ABC. B^- A 
 
 In reading angles by three letters, the one at the 
 vertex should always occupy the middle place. 
 
 2. When the adjacent angles 
 made by one straight line with 
 another are equal, they are called 
 
 right angles. Such are the A 
 
 angles ACD and BCD. 
 
 An angle which is greater than 
 a right angle is termed OBTUSE ; an angle less than a 
 right angle, acute. 
 
 3. One straight line is said to be perpendicular to 
 another when it makes right angles with it ; oblique, 
 when it makes unequal angles. 
 
 (13) 
 
 -B 
 
14 GEOMETRY. 
 
 4. Two straight lines are said to be parallel to 
 each other when they have the 
 
 same direction. Thus, AB and 
 CD are parallel. 
 
 Corollary. Parallel lines will never meet, hoAV- 
 ever far produced; for, if they were to meet, they 
 would make an angle with each other, that is, they 
 would differ in direction, which is contrary to the 
 definition. 
 
 Scholium. Lines having the same direction may 
 also be said to have exactly opposite directions when 
 taken in different ways. Thus, the direction AB is 
 opposite to the direction DC. 
 
 5. A straight line, or any other magnitude, is said 
 to be bisected when it is divided into two equal 
 parts. 
 
 THEOREM I. 
 
 The two angles which one straight line makes with 
 another, on one side of it, are together equal to two right 
 angles. 
 
 Let the straight line DC 
 meet the straight line AB in 
 C. Then will the two angles 
 ACD, DCB, be together equal 
 to two right angles. ° 
 
 For, if DC be perpendicular to AB, each of these 
 angles will be a right angle (Definition 8, Section IV). 
 But if DC is oblique to AB, erect the perpendicular 
 CE. Now, the part ACE of the obtuse angle ACD, 
 is a right angle ; and the remaining part ECD and the 
 
BOOK I. 
 
 15 
 
 acute angle DCB together make up another right 
 angle, ECB. 
 
 Therefore, the two angles which one straight line 
 makes with another, on one side, are together equal to 
 two right angles. 
 
 Cor. 1. In a similar manner, whatever be the num- 
 ber of adjacent angles formed at C, on one side of AB, 
 it may be shown that their sum is two right angles. 
 
 Cor. 2. A right angle is the fourth part of the 
 ivhole compass in the plane about any point, as C ; 
 for, half of this compass lies on one side of AB, and 
 half on the other. 
 
 THEOREM II. 
 
 If two straight lines cut one another, the opposite or 
 vertical angles are equal. 
 
 Let AB and CD be two 
 straight lines cutting one an- 
 other in E. Then will AEC 
 be equal to its opposite or ver- 
 tical angle DEB. 
 
 For the angles AEC, AED, on one side of CD, are 
 together equal to two right angles (Theo. I) ; so, also, 
 the angles AED, DEB, on one side of AB, are together 
 equal to two right angles ; therefore, the sum of AEC 
 and AED is equal to the sum of AED and DEB 
 (Axiom 3). Now take away the common angle AED, 
 and the remaining angle AEC is equal to the remain- 
 ing angle DEB (Ax. 5). 
 
 In the same manner it may be proved that the 
 angle AED is equal to its opposite angle CEB. 
 
 Therefore, if two straight lines, etc. 
 
16 
 
 GEOMETRY. 
 
 •vl^\ 
 
 V 
 
 THEOREM III. 
 
 If a straight line intersect two parallels, the corres- 
 ponding inner and outer angles will be equal to each 
 other; also the alternate angles. 
 
 Let the straight line EF 
 intersect the two parallel 
 straight lines AB, CD, in 
 G and H. Then will any 
 two corresponding inner and 
 outer angles, as GHD and 
 EGB, be equal. ■— 
 
 For, since EF is a straight line, its part HG has 
 the same direction from H that its part GE has from 
 G (Def. G, Sec. I) ; and since HD and GB are by hy- 
 pothesis parallel, HD has the same direction from H 
 that GB has from G (Def. 4, Sec. IV). Therefore, the 
 difference in direction between HG and HD is equal 
 to the difference in direction between GE and GB ; 
 that is, the angle GHD is equal to the angle EGB 
 (Def. 1, Sec. IV). 
 
 A^ain, any two alternate angles, as AGH and 
 GHD, are equal to each other. For AGH is equal 
 to its opposite angle EGB (Theo. II). But EGB has 
 just been proved equal to GHD. Consequently, AGH 
 is equal to GHD (Ax. 3). 
 
 Therefore, if a straight line intersect, etc. 
 
 Cor. 1. It is evident that if AB is not parallel to 
 CD, but takes some other direction, as mGn, the cor- 
 responding inner and outer angles will be unequal; 
 also, the alternate angles. 
 
 Cor. 2. If two angles have their sides parallel, 
 
BOOK I. 
 
 17 
 
 each to each, and directed the same way from the ver- 
 tex, they are equal. 
 
 .F* 
 
 THEOREM IV. 
 
 If a straight line intersect two parallels, the sum of 
 the two inner angles on the same side ivill be equal to 
 two right angles. 
 
 Let EF intersect the two 
 parallels AB, CD, in G and 
 H. Then will BGH and 
 GHD be together equal to 
 two right angles. 
 
 For the sum of the an- 
 gles GHC and GHD is two 
 right angles (Theo. I). But GHC is equal to its 
 alternate angle BGH (Theo. III). Therefore, the sum 
 of BGH and GHD is equal to two right angles. 
 
 Hence, if a straight line, etc. 
 
 Cor. 1. If BGH is a right angle, GHD must also 
 be a right angle. Hence, if a straight line is perpen- 
 dicular to one of two parallels, it is perpendicular to 
 the other. 
 
 Cor. 2. If two straight lines are both perpendicular 
 to a third, they are parallel. 
 
 EXERCISES. 
 
 1. Prove that the sum of all the adjacent angles 
 made by any number of straight lines meeting in one 
 point is equal to four right angles. 
 
 2. Prove that if one of the four angles made by 
 
 Evans? Geometry. — "2 
 
18 
 
 GEOMETR Y. 
 
 two straight linos intersecting each other be a right 
 
 O O O 
 
 angle, each of the others will be a right angle. 
 
 8. Prove that the alternate outer angles EGA and 
 FIID (Figure to Theo. Ill) are equal to each other. 
 
 4. Prove that the sum of the two outer angles on 
 the same side, EGB and FHD (Fig. Theo. IV), is equal 
 to two right angles. 
 
 5. Point out all the angles equal to EGB (Fig. 
 Theo. IV) ; also, all the angles equal to EGA ; and 
 show in each case why they are equal. 
 
 SEC. V.— TRIANGLES. 
 
 DEFINITIONS. 
 
 1. A plane figure is a portion of a plane bounded 
 on all sides by lines. 
 
 2. A polygon is a plane figure bounded by straight 
 lines. 
 
 3. A triangle is a polygon of three sides. 
 
 If it has one right angle it is 
 called a right-angled triangle. The 
 side opposite the right angle is 
 called the hypotenuse. 
 
 A triangle which has no right 
 angle is called an oblique-angled 
 triangle. 
 
 An isosceles triangle is one which 
 
 has two equal sides. 
 
BOOK I. 19 
 
 4. Any side of a triangle may be considered as its 
 base, and the opposite angle as its vertex; but in an 
 isosceles triangle, that side is usually called the base 
 which is not equal to either of the others. 
 
 The altitude of a triangle is the perpendicular let 
 fall from the vertex on the base, or the base produced. 
 
 5. A triangle, or other polygon, is called equiangu- 
 lar when all its angles are equal ; equilateral, when all 
 its sides are equal. 
 
 6. Two polygons are called mutually equilateral or 
 mutually equiangular, if the sides or angles of the one 
 are equal to the sides or angles of the other, each to 
 each, taken in the same order. 
 
 THEOREM V. 
 
 Tlie three interior angles of any triangle are together 
 equal to two right angles. 
 
 Let ABC be any tri- 
 angle. It is to be proved 
 that the angles ABC, BCA, 
 
 CAB, are together equal ^/_ _ D 
 
 to two right angles. 
 
 Produce AB to any point D, and draw BE parallel 
 to AC. Now, the sum of the three angles ABC, CBE, 
 EBD, is equal to two right angles (Cor. 1, Theo. I). 
 But because CB intersects the parallels AC and BE, 
 the angle CBE is equal to its alternate angle BCA 
 (Theo. Ill)*; also, because AD intersects the same par- 
 allels, the angle EBD is equal to its corresponding 
 inner angle CAB. Therefore, the sum of ABC, BCA, 
 CAB, is equal to two right angles. 
 
 That is, the three interior angles of a triangle, etc. 
 
20 
 
 GEOMETRY. 
 
 Cor. 1. If one side of a triangle be produced, the 
 exterior angle ivill be equal to the sum of the two oppo- 
 site interior angles. For the sura of CBE and EBD, 
 that is, the whole exterior angle CBD, is equal to the 
 sum of BCA and CAB. 
 
 Cor. 2. A triangle can not have more than one rigid 
 angle; for, if it had two, the third angle would be 
 nothing. Still less can a triangle have more than one 
 obtuse angle. 
 
 THEOREM VI. 
 
 If two straight lines be drawn from the extremities 
 of one side of a triangle to a point within, their sum 
 will be less than that of the other two sides of the 
 triangle. 
 
 From the extremities of AB, let 
 straight lines be drawn to a point 
 D within the triangle ABC. It is 
 to be proved that the sum of AD 
 and DB is less than the sum of 
 AC and CB. 
 
 Produce AD to meet BC in E. Now, since AE is 
 a straight line, it is less than the sum of AC and 
 CE, which form a broken line (Ax. 10). Therefore, 
 if we add EB to both, it is evident that the sum. of 
 AE and EB is less than the sum of AC, CE, and 
 EB, that. is, less than the sum of AC and CB (Ax. 6). 
 In the same manner it may be shown that the sum 
 of AD and DB is less than the sum of AE and EB ; 
 still more, then, we may conclude it is less than the sum 
 of AC and CB. 
 
 Therefore, if two straight lines, etc. 
 
BOOK I. 21 
 
 THEOREM VII. 
 
 If two triangles have two sides and the included 
 angle of the one equal to two sides and the included 
 angle of the other, each to each, they are equal through- 
 out. 
 
 Let the triangles ABC, DEF, have the side AB 
 equal to the 
 side DE, and 
 AC equal to 
 DF, and the 
 included angle 
 A to the in- 
 
 B D " E 
 
 eluded angle D. Then will the two triangles be equal 
 throughout; that is, the third side BC will be equal 
 to the third side EF, the angle B to the angle E, the 
 angle C to the angle F, and the triangle ABC as a 
 whole to the triangle DEF as a whole. 
 
 If the triangle ABC be applied to the triangle DEF 
 so that the point B shall be upon the point E, and the 
 side BA upon the side ED, then since these sides are 
 by hypothesis equal, the point A will fall upon the 
 point D ; and since the angle A is equal to the angle 
 
 D, the side AC will fall upon the side DF ; and since 
 these sides again are equal, the point C will fall upon 
 the point F. But the point B is also upon the point 
 
 E. Therefore, the side CB must coincide with the 
 side FE (Ax. 10), and the angle C with the angle F, 
 and the angle B with the angle E, and the triangle 
 ABC as a whole with the triangle DEF as a whole ; 
 and since they coincide, each with each, they are also 
 equal (Ax. 11). , - ' ■ 
 
 Therefore, if two triangles, etc. 
 
22 
 
 GEOMETRY. 
 
 Cor. In equal triangles, equal angles are opposite 
 to equal sides. 
 
 THEOREM VIII. 
 
 The angles at the base of an isosceles triangle are 
 equal to each other. 
 
 Let ABC be an isosceles triangle 
 having the side AC equal to the side 
 BC. It is to be proved that the angle 
 A is equal to the angle B. 
 
 Draw CD bisecting the angle ACB. 
 Then in the triangles ADC, BDC, A " "^ 
 since the side AC is, by hypothesis, equal to the side 
 BC, and the side CD is common to both, and the in- 
 cluded angle ACD is, by construction, equal to the 
 included an^le BCD, it follows that the triangle ADC 
 is equal to the triangle BDC (Theo. VII), and the 
 angle A to the angle B (Cor. Theo. VII). 
 
 Therefore, the angles at the base, etc. 
 
 Cor. 1. A straight line bisecting the vertical angle 
 of an isosceles triangle is a perpendicular to the middle 
 point of the base. For AD is equal to DB ; also, the 
 angles CDA, CDB, being equal, are right angles (Def. 
 
 2, Sec. 4). 
 
 Cor. 2. An equilateral triangle is also equian- 
 gular. 
 
 THEOREM IX. 
 
 The greater of two unequal sides in a triangle has 
 the greater angle opposite to it. 
 
BOOK I 
 
 23 
 
 Let ABC be a triangle having the 
 side AB greater than the side BC. 
 Then will the angle BGA be greater 
 than the angle BAC. 
 
 Produce BC so as to make BD 
 equal to BA ; also, join DA. Now, 
 in the isosceles triangle ABD, the A 
 angles BDA, BAD, are equal (Theo. VIII). But the 
 exterior angle BCA, of the triangle ACD, is greater 
 than the opposite interior angle BDA (Cor. I, Theo. 
 V), and consequently greater than BAD ; still more, 
 then, is it greater than BAC. 
 
 Therefore, the greater of two unequal sides, etc. 
 
 Cor. Hence, if two angles of a triangle are equal, 
 the sides opposite to them can not be unequal. 
 
 THEOREM X. 
 
 If two triangles have hvo sides of the one equal to 
 two sides of the other, but the included angles unequal, 
 the base of that which has the greater angle is greater 
 than the base of the other. 
 
 Let the two tri- 
 
 angles ABC, DEF, 
 have the two sides 
 AB, AC, respect- 
 ively equal to the 
 two sides DE, DF, but the angle BAC greater than 
 the angle EDF. It is to be proved that the base BC 
 is greater than the base EF. 
 
 Because the angle EDF is, by hypothesis, less than 
 the angle BAC, the sum of the angles E and F must 
 be greater than the sum of the angles B and C (Theo. 
 
24 GEOMETRY. 
 
 V) : hence, either E is greater than B, or F is greater 
 than C. Suppose, then, that F is greater than C. 
 On the adjacent side DF, describe the triangle DGF 
 equal to the triangle ABC, and so placed that DG 
 shall be the side equal to AB. Now, the sum of DH 
 and HG is greater than DGL (Ax. 10) ; and the sum 
 of EH and HF is greater than EF. Therefore, the 
 sum of the entire lines DE and GF is greater than 
 the sum of DG and EF, Taking away from these 
 unequals the equals DE and DG, we have the remain- 
 der GF greater than the remainder EF. But GF is 
 equal to BC by construction. Therefore, BC is 
 greater than EF. 
 
 Hence, if two triangles, etc. 
 
 THEOREM XL 
 
 If two triangles have two angles and the included 
 side of the one equal to two angles and the included 
 side of the other, each to each, they are equal throughout. 
 
 C F 
 
 / 
 
 Let ABC, DEF, 
 be two triangles 
 having the angles 
 A and B respect- .X _\ B -^/_ \. £ 
 
 ively equal to the 
 
 angles D and E, and the included side AB equal to 
 the included side DE. It is to be proved that these 
 triangles are equal throughout, 
 
 If the triangle ABC be applied to the triangle DEF, 
 so that the side AB shall coincide with its equal DE, 
 then because the angle A is equal to the angle D, the 
 side AC will fall on the side DF, and the point C will 
 be found somewhere in the line DF ; also, because the 
 
BOOK I. 
 
 25 
 
 angle B is equal to the angle E, the side BC will fall 
 on the side EF, and the point C will be found some- 
 where in the line EF ; consequently the point C, being 
 in both the lines DF and EF, must be at their point 
 of intersection F. Hence, the two triangles exactly 
 coincide, and are equal throughout (Ax. 11) ; that is, 
 the side AC is equal to the side DF, the side BC to 
 the side EF, the angle C to the angle F, and the tri- 
 angle ABC as a whole to the triangle DEF as a whole. 
 
 Therefore, if two triangles, etc. 
 
 THEOKEM XII. 
 
 Two triangles which are mutually equilateral are also 
 mutually equiangular. 
 
 Let the two tri- A 
 
 angles ABC, DEF, 
 have the three sides 
 AB, BC, CA, re- 
 spectively equal to 
 the three sides DE, EF, FD. Then, also, will the 
 angles A, B, and C be respectively equal to the angles 
 D, E, and F. 
 
 If the angle A is not equal to the angle D, it must 
 be either greater or less. But it can not be greater 
 than D, for then the base BC would be greater than 
 the base EF (Theo. X), which is contrary to the hy- 
 pothesis. Neither can it be less than D ; for, in that 
 case, BC would be less than EF, which is also con- 
 trary to the hypothesis. Hence, A must be equal to 
 D : and, in the same manner, it may be proved that B 
 is equal to E, and C to F. 
 
 Therefore, two triangles, etc. 
 
 Evans Geometry. — 3 ^ 
 
26 GEOMETRY. 
 
 Cor. If two triangles are mutually equilateral they 
 are equal throughout. 
 
 EXERCISES. 
 
 1. Prove that if two angles of one triangle are 
 equal to two angles of another, the third angles arc 
 equal. 
 
 2. Prove that each angle of an equilateral triangle 
 is equal to two-thirds of a right angle. 
 
 3. Prove that if a perpendicular be erected on the 
 middle point of a straight line, any point in it will 
 be equally distant from the extremities of that line 
 (Theo. VII). 
 
 4. Prove that from a given point without a straight 
 line, only one perpendicular to that line can be drawn 
 (Cor. 2, Theo. V). 
 
 5. If the vertical angle of an isosceles triangle be 
 two-sevenths of a right angle, what will be the value 
 of each of the angles at the base ? 
 
 SEC. VI.— QUADRILATERALS. 
 DEFINITIONS. 
 
 1. A quadrilateral is a polygon of four sides. 
 
 2. A parallelogram is a quad- 
 
 rilateral having its opposite sides 
 
 parallel. L j 
 
 A parallelogram whose angles 
 are all right angles, is called a 
 rectangular parallelogram, or sim- 
 ply a RECTANGLE, 
 
BOOK I. 27 
 
 A SQUARE is a rectangle whose 
 sides are all equal. 
 
 3. A trapezoid is a quadrilat- 
 eral having only two of its op- 
 posite sides parallel. 
 
 4. A diagonal of a quadrilateral or other polygon 
 is a straight line joining two angles not adjacent to 
 each other. 
 
 5. The area of any plane figure is the amount of 
 surface which it contains. 
 
 6. Two plane figures are said to be equivalent when 
 their areas are equal. 
 
 THEOREM XIII. 
 
 The opposite, sides and angles of a parallelogram are 
 equal to each other. 
 
 Let ABCD be a parallelogram. Dj — ^C 
 
 It is to be proved that any one 
 of its sides is equal to the side 
 opposite, and any one of its an- A 
 gles to the angle opposite. 
 
 Draw the diagonal AC. Now, since AC intersects 
 the two parallels AD, BC, the alternate angles DAC, 
 BCA, are equal (Theo. Ill) ; also, since it intersects 
 the parallels AB, DC, the alternate angles BAC, DCA, 
 are equal. Hence, the two triangles ABC, ADC, have 
 two angles of the one equal to two angles of the 
 other, and the included side AC common ; they are, 
 
28 GEOMETRY. 
 
 therefore, equal (Theo. XI); and the side AB is equal 
 to the side DO (Cor., Theo. VII), and the side AD 
 to the side BC, and the angle B to the angle D ; 
 also, since the angle DAC is equal to the angle 
 BCA, and the angle BAC to the angle DCA, it fol- 
 lows that the whole angle BAD is equal to the whole 
 angle BCD. 
 
 Therefore, the opposite sides and angles, etc. 
 
 Cor. 1. A diagonal of a parallelogram divides it 
 into two equal triangles. 
 
 Cor. 2. If ABCD be a rectangular parallelogram, 
 AD and BC will both be perpendicular to each of the 
 parallels AB, DC (Def. 2, Sec. VI). Hence, the per- 
 pendicular distance between two parallels is everywhere 
 the same. 
 
 Schol. Any side, as AB, 
 of a parallelogram may be 
 taken as its base; and a per- 
 pendicular, as CE, let fall a^ 
 from any point in the oppo- 
 site parallel on the base, or the base produced, is 
 called the altitude of the parallelogram. 
 
 THEOREM XIV. 
 
 If a quadrilateral has two of its sides equal and 
 parallel., it is a parallelogram. 
 
 Let the quadrilateral ABCD j) v _£ 
 
 have the sides AB, DC, equal ,-''' \ 
 
 and parallel. Then will it be \ 
 
 a parallelogram. A^- — — ^B 
 
BOOK I. 29 
 
 Join AC ; then, since AC intersects the two paral- 
 lels AB, DC, the alternate angles BAC, DCA, are 
 equal (Theo. III). Now, in the triangles ABC, ADC, 
 we have the side AB equal to the side DC, and the 
 side AC common, and the included angle BAC equal 
 to the included angle DCA; therefore, the two tri- 
 angles are equal (Theo. VII), and the angle BCA is 
 equal to the angle CAD (Cor., Theo. VII), and these 
 being alternate angles, it follows that AD and BC are 
 parallel (Cor. 1, Theo. III). Hence, ABCD has its 
 opposite sides parallel, and is a parallelogram (Def. 
 2, Sec. VI). 
 
 Therefore, if a quadrilateral, etc. 
 
 Sehol. If a quadrilateral has its opposite sides equal 
 it is a parallelogram. For the triangles ABC, ADC, 
 will in that case be mutually equilateral, and there- 
 fore equal (Cor., Theo. XII) ; and the parallelism of 
 the opposite sides may be shown by the equality of 
 the alternate angles, as above. 
 
 THEOREM XV. 
 The diagonals of a parallelogram bisect each other. 
 
 Let ABCD be a parallelo- ^ 
 gram. It is to be proved that 
 its diagonals, AC, BD, bisect ''Br- 
 each other in E (Def. 6, Sec. A^ ^b 
 
 IV). 
 
 In the triangles AEB, DEC, since the angle ABD 
 is equal to its alternate angle BDC (Theo. Ill), and 
 the angle BAC equal to its alternate angle ACD, and 
 the included side AB equal to the included side DC 
 
30 
 
 GEOMETRY. 
 
 (Theo. XIII), it follows that the two triangles are 
 equal (Theo. XI) ; consequently, the side AE is equal 
 to the side EC, and BE to ED (Cor., Theo. VII) ; that 
 is, the two diagonals bisect each other in E. 
 Therefore, the diagonals, etc. 
 
 THEOREM XVI. 
 
 The area of a rectangle is equal to the product of 
 its base by its altitude. 
 
 Let ABCD be a rectangle. 
 It is to be proved that its 
 area is equal to the product 
 of its base AB by its altitude 
 AD. 
 
 D 
 
 9 
 
 f 
 
 i 1 
 
 e\ 
 
 Let AB be divided into a A 
 certain number of equal parts 
 
 Ab, be, etc., taken as the units of length ; also, let AD 
 be divided into a certain number of the same units, 
 A/, fg, etc. From b, c, etc., draw straight lines par- 
 allel to AD, and from /, etc., draw straight lines 
 parallel to AB. Now, it is evident that the whole 
 rectangle is divided into small squares (Cor. 2, Theo. 
 XIII), each equal to Abef; which may be taken as 
 the unit of area. Of these equal squares there are as 
 many in the tier next to AB as there are units of 
 length in AB ; and there are as many equal tiers in 
 the whole figure as there are units of length in AD. 
 Therefore, the whole number of square units in 
 ABCD is equal to the number of linear units in AB 
 multiplied by the number of linear units in AD. 
 
 Hence, the area of a rectangle, etc. 
 
BOOK I 
 
 31 
 
 Scliol. If AB and AD are incommensurable, that 
 is, if no unit can be found into which they can both 
 be divided without leaving a remainder in one of 
 them, the theorem will still hold true ; for if the unit 
 be taken smaller and smaller, the remainder can be 
 made less than any assignable quantity. 
 
 When the linear unit is one inch, the unit of area 
 is a square inch ; when the linear unit is one foot, the 
 unit of area is a square foot, etc. 
 
 Cor. Since the base and altitude of a square are 
 equal (Def. 2, Sec. VI), its area may be found by 
 multiplying one side into itself. 
 
 THEOKEM XVII. 
 
 The area of any parallelogram is equal to the area 
 of a rectangle having tlie same base and altitude. 
 
 Let ABCD 
 be a rectangle, 
 and ABFE 
 any parallelo- 
 
 gram, 
 
 on the 
 
 I) 
 
 EC F 
 
 1 / 
 
 / 
 
 / 
 
 / 
 
 / 
 
 B 
 
 same base AB, and of the same altitude, namely 
 the perpendicular distance between the parallels AB, 
 DF. Then will ABFE be equivalent to ABCD. 
 
 Since AB and DC are opposite sides of a parallelo- 
 gram they are equal (Theo. XIII); and for the same 
 reason AB and EF are equal; therefore DC is equal 
 to EF (Ax. 3). Taking away each of these in turn 
 from the whole line DF, we have the remainder DE 
 equal to the remainder CF. But DA is equal to CB 
 (Theo. XIII), and the included angle ADE is equal to 
 
32 GEOMETRY. 
 
 the included angle BCF (Theo. Ill) ; therefore, the 
 triangles ADE, BCF, are equal (Theo. VII) ; and 
 hence if each of them be taken away in turn from the 
 whole figure ABFD, the remainder ABFE will be 
 equivalent to the remainder ABCD. 
 
 Therefore, the area of any parallelogram, etc. 
 
 Cor. 1. The area of any parallelogram is equal to 
 the product of its base by its altitude. 
 
 Cor. 2. Since any triangle, as D C 
 
 ABC, is half of a parallelogram 
 
 ABCD (Cor. 1, Theo. XIII) on 
 
 the same base AB and having the 
 
 same altitude CE (Def. 4, Sec. 
 
 V), it follows that the area of a triangle is equal to 
 
 half the product of its base by its altitude. 
 
 THEOREM XVIII. 
 
 The area of a trapezoid is equal to half the product 
 of the sum of its parallel sides by its altitude. 
 
 Let ABCD be a trapezoid of D o 
 
 which AB and DC are the par- / y' \ 
 
 allel sides. Then will its area / /'' \ 
 
 be equal to half the product of Ll, \ 
 
 the sum of AB and DC into its 
 
 altitude, namely, the perpendicular distance between 
 
 AB and DC. 
 
 Draw the diagonal AC. Now, the area of the 
 triangle ABC (Cor. 2, Theo. XVII) is equal to half 
 the product of its base AB into its altitude, which is 
 the same as the altitude of the trapezoid ; again, the 
 area of the triangle ADC is equal to half the product 
 of its base DC into its altitude, which is also the same 
 
BOOK I. 
 
 33 
 
 as the altitude of the trapezoid. Therefore, adding 
 together the area of the two triangles, we have the 
 area of the whole figure equal to half the product of 
 the sum of AB and DC into the altitude. 
 Hence, the area of a trapezoid, etc. 
 
 THEOEEM XIX. 
 
 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. 
 
 Let ABC be a triangle 
 right-angled at B. It is to 
 be proved that the square 
 AEDC is equivalent to the 
 sum of the squares ABGF 
 and BHIC. 
 
 Join FC, BE, and draw 
 BK parallel to AE. Also 
 observe that since ABC and 
 
 ABG are right angles, BC and BG form one straight 
 line. 
 
 Now, in the triangles EAB, CAF, the side EA is 
 equal to the side CA, since they are sides of the same 
 square, and for the same reason the side AB is equal 
 to the side AF. But the included angles EAB, CAF, 
 are also equal ; for each of them is composed of a 
 right angle and the angle CAB. Therefore, the two 
 triangles are equal (Theo. VII). 
 
 But since the triangle CAF and the square BAFG 
 have the same base AF, and the same altitude, namely 
 the perpendicular distance between the parallels AF, 
 
34 GEOMETRY. 
 
 CG, the square must be double of the triangle (Cors. 1 
 and 2, Theo. XVII). For like reason the parallelo- 
 gram AEKL is double of the triangle EAB. But 
 doubles of equals are equals ; therefore, the paral- 
 lelogram AEKL is equivalent to the square BAFG. 
 In the same manner (by joining AI and BD), it 
 may be shown that the parallelogram CDKL is equiv- 
 alent to the square BCIH. Hence, the whole square 
 ACDE is equivalent to the sum of the squares 
 BAFG, BCIH. 
 
 Therefore, the square described, etc. 
 
 Cor. 1. The hypotenuse is equal to the square 
 root of the sum of the squares of the other two sides. 
 
 Cor. 2. The square of either of the sides contain- 
 ing the right angle is equivalent to the square of the 
 hypotenuse diminished by the square of the other 
 side. By taking the square root of the remainder the 
 side itself will be found. 
 
 Cor. 3. If two right-angled triangles have the hy- 
 potenuse and one side of the one respectively equal 
 to the hypotenuse and one side of the other, the 
 third sides will also be equal. 
 
 EXEHCISES. 
 
 1. If the side of a square be 36 inches, what is its 
 area in square inches ? What in square feet ? 
 
 2. If the base of a parallelogram be 3 feet and its 
 altitude 4 feet and 6 inches, what is its area? 
 
 3. If the base of a triangle be 50 yards and its 
 altitude 20 yards, what is its area? 
 
 4. If the parallel sides of a trapezoid be 12 rods 
 
BOOK I. 35 
 
 and 16 rods, and its altitude 8 J rods, what is its 
 area ? 
 
 5. If the sides containing the right angle of a riffht- 
 angled triangle be 3 and 4, what is the length of the 
 hypotenuse ? 
 
 6. If the hypotenuse be 10 and one of the sides 8, 
 what is the length of the other side ? 
 
 7. Prove that if a parallelogram has one right 
 angle it is a rectangle. 
 
 8. Prove that the diagonals of a rectangle are equal 
 to each other. 
 
 9. Prove that a perpendicular is the shortest line 
 that can be drawn to a straight line from a point 
 without it ; also, that of two oblique lines that which 
 is furthest from the perpendicular is the longest 
 (Theo. XIX). 
 
 SEC. VII.— OF POLYGONS IN GENERAL. 
 DEFINITIONS. 
 
 1. A polygon of five sides is called a pentagon ; 
 one of six sides, a hexagon ; of seven sides, a hepta- 
 gon ; of eight sides, an octagon ; of ten sides, a deca- 
 gon ; of twelve sides, a dodecagon. 
 
 2. A regular polygon is one which is both equilat- 
 eral and equiangular. Squares and equilateral tri- 
 angles are species of regular polygons. 
 
 8. The perimeter of a polygon is the sum of all its 
 
 sides. 
 
36 
 
 GEOMETRY. 
 
 THEOREM XX. 
 
 The sum of all the angles of any polygon is equal to 
 twice as many rigid angles, wanting four, as the figure 
 has sides. 
 
 Let ABODE be any polygon. 
 It is to be proved that the sum 
 of all its angles A, B, C, D, E, 
 is equal to twice as many right 
 angles, wanting four, as it has 
 sides. 
 
 From the vertex of each angle 
 draw straight lines AF, BF, etc., to any point F 
 within, thus dividing the polygon into as many tri- 
 angles as it has sides. Now, since the angles of each 
 triangle are together equal to two right angles (Theo. 
 V), it follows that the angles of all the triangles are 
 together equal to twice as many right angles as the 
 figure has sides. But the sum of the angles A, B, C, 
 D, E, is equal to the sum of the angles of all the 
 triangles, wanting the angles about the point F ; and 
 the angles about the point F are together equal to 
 four right angles (Cor. 2, Theo. I). Therefore, the sum 
 of the angles A, B, C, D, E, is equal to twice as 
 many right angles, wanting four, as the polygon has 
 sides. 
 
 That is, the sum of all the angles of any poly- 
 gon, etc. 
 
 Schol. The area of a polygon may be found by 
 dividing it into triangles as above (or by diagonals 
 drawn from any angle A to the opposite angles), then 
 finding the areas of these triangles separately and 
 adding them together. 
 
BOOK I. 37 
 
 THEOEEM XXI. 
 
 A regular polygon may he divided into as many 
 equal triangles as it has sides. 
 
 m E 
 
 Let ABCDEF be a regular 
 polygon. Draw AG and BG 
 bisecting the angles A and B ; 
 
 also, draw GC, GD, etc., to the A / _-V ._^ D 
 
 other angles. 
 
 Now, since the triangles 
 AGB, BGC, have the side AB B c 
 
 equal to the side BC (Def. 2, Sec. VII), and the side 
 BG common, and the included angle ABG equal to 
 the included angle CBG, the two triangles are equal 
 to each other (Theo. VII). Hence the angle GAB is 
 equal to the angle GCB (Cor., Theo. VII). But GAB 
 is by construction half of BAF ; therefore, GCB must 
 be half of the equal angle BCD, that is, BCD is bi- 
 sected by GC. Hence, it may be shown in the same 
 manner as before that the triangles BGC, CGD, are 
 equal : and so on of the other triangles in succes- 
 sion ; of which there is one on each side of the 
 polygon. 
 
 Therefore, a regular polygon, etc. 
 
 ScJwl. The center of a regular polygon is the point 
 G, found by bisecting any two of its angles. A per- 
 pendicular, as Gm, let fall from the center upon any 
 side of a regular polygon is called its apothegm. It 
 is the altitude of the triangle EFG, and consequently 
 of each of the equal triangles into which the polygon 
 is divided. 
 
 Cor. Since the area of each triangle is equal to 
 
38 
 
 GEOMETRY. 
 
 half the product of its base by its altitude, the area 
 of all the triangles taken together is equal to half 
 the product of the sum of their bases by their com- 
 mon altitude : that is, the area of a regular polygon 
 is equal to half the product of its perimeter by its 
 apothegm. 
 
 EXEECISES. 
 
 1. To how many right angles are all the angles of 
 a quadrilateral equal ? Of a pentagon ? Of a hexa- 
 gon ? Of a heptagon ? etc. 
 
 2. What is the area of a regular pentagon whose 
 side is 25, and the apothegm 17.205? 
 
 3. Prove that each angle of a regular hexagon is 
 equal to four thirds of a right angle. 
 
 SEC. VIII.— OF THE CIRCLE. 
 DEFINITIONS. 
 
 1. A circle is a plane figure bounded by a curve, 
 called the circumference, every point in "which is 
 equally distant from a point within, called the CENTER. 
 
 2. A straight line drawn from 
 the center to any point of the cir- 
 cumference is called the RADIUS. 
 
 A double radius, that is, a 
 straight line passing through the 
 center, and terminated both ways 
 by the circumference, is called 
 
 the DIAMETER. 
 
 Cor. In the same circle, all radii are equal ; also, 
 all diameters. 
 
BOOK I 
 
 39 
 
 3. A TANGENT is a straight 
 line which meets the circum- 
 ference, but being produced 
 does not cut it. 
 
 4. Any portion of the cir- 
 cumference is called an arc. 
 The chord of an arc is the 
 straight line joining its ex- 
 tremities. A chord produced one way beyond the 
 circumference is called a secant. 
 
 5. A segment of a circle 
 is the part contained by an 
 arc and its chord. 
 
 A sector is the part con- 
 tained by an arc and two 
 radii drawn to its extremities. 
 
 6. Half a circle is called a 
 
 semicircle; half a circum- 
 
 ference, a semi-circumference. A quarter of a circle 
 or of a circumference is called a quadrant. 
 
 7. An inscribed angle is one which has its vertex 
 in tire* circumference, and is contained by two chords. 
 
 A polygon is said to be inscribed in a circle when 
 all its sides are chords of that circle. 
 
 A circle is said to be inscribed in a polygon when 
 all the sides of the polygon are tangents to the circle. 
 
 When one figure is inscribed in another, the latter 
 is said' to be circumscribed about the former. 
 
 THEOREM" XXII. 
 
 A diameter divides the circle and its circumference 
 into two equal parts. 
 
40 
 
 GEOMETRY 
 
 Let ABCD be a circle of 
 which E is the center and 
 AC a diameter. It may be 
 shown that the segment ABC A 
 is equal to the segment ADC, 
 and the arc ABC to the arc 
 ADC. 
 
 Let the segment ABC be 
 applied to the segment ADC, the line AC remaining 
 common ; then the arc ABC will exactly coincide with 
 the arc ADC : for if not, suppose some point in ABC 
 to fall within or without ADC ; then we shall have 
 points in the circumference unequally distant from 
 the center, which is contrary to the definition of a 
 circle. Hence, the arc ABC is equal to the arc ADC, 
 and the segment ABC to the segment ADC (Ax. 11). 
 
 Therefore, a diameter divides, etc. 
 
 Schol. In a similar manner it may be proved that 
 two circles of equal diameters or radii are equal to 
 each other. 
 
 THEOREM XXIII. 
 
 If a straight line is perpendicular to a diameter at 
 its extremity, it is a tangent to the circle. 
 
 Let the straight line EF 
 be perpendicular to the di- 
 ameter AC at its extremity 
 C. Then will EF be a tan- 
 gent to the circle ABCD. 
 
 In EF take any two points 
 m and n, one on each side 
 of C, and connect them by 
 
BOOK I, 
 
 41 
 
 straight lines with the center G. Now, since GCm is 
 a right-angled triangle, the square of Gm is greater 
 than the square of GC (Theo. XIX) ; hence, Gm is 
 greater than GC, that is, greater than the radius. 
 Therefore, the point m, however near it may be 
 to C, is necessarily without the circle. The same 
 may be proved, in like manner, of the point n, or 
 any other point in EF on either side of C. Hence, 
 EF can meet the circumference only in the point 
 C, and it is consequently a tangent (Def. 3, Sec. 
 VIII). 
 
 Therefore, if a straight line, etc. 
 
 Cor. If a chord be drawn perpendicular to a tan- 
 gent at the point of contact, it will be a diameter. 
 
 THEOEEM XXIV. 
 
 Equal arcs in a circle have equal chords, and sub- 
 tend equal angles at the center. 
 
 Let AD and BC be equal 
 arcs, and E the center of the 
 circle. It is to be proved 
 that the chords AD, BC are 
 equal ; also, the subtended 
 angles AED, BEC. 
 
 Let the sector EBC be ap- 
 plied to the sector EAD, so that the line EB shall fall 
 upon the line EA, the point E remaining common ; 
 then, since these lines are equal, being radii (Cor., 
 Def. 2, Sec. 8), the point B will fall upon the point 
 A ; and, since all points in the circumference are 
 equally distant from the center, the arc BC will fall 
 Evans Geometry. — 4 
 
42 GEOMETRY. 
 
 upon the arc AD; and these arcs being by hypothesis 
 equal, the point C will fall upon the point D ; there- 
 fore (Ax. 10), the chord BC will coincide with the 
 chord AD, and be equal to it. 
 
 Again, in the triangles EAD, EBC, the two sides 
 EA, ED, are equal to the two sides EB, EG, since 
 they are all radii; and the base AD has just been 
 proved equal to the base BC ; therefore, the two 
 triangles are mutually equiangular (Theo. XII), and 
 the angle AED is equal to the angle BEC (Cor., 
 Theo. VII). 
 
 Hence, equal arcs in a circle, etc. 
 
 Sclwl. If the whole circumference be divided into 
 any number of equal arcs, and radii be drawn to all 
 the points of division, the whole compass about the 
 center will be divided into the same number of equal 
 angles. Hence, an arc may be taken as the measure 
 of its subtended angle. To this end the circumference 
 is conceived to be divided into 360 equal parts, called 
 degrees, each degree into 60 minutes, and each minute 
 into 60 seconds. Then whatever number of degrees, 
 etc., an arc contains, the same number will denote 
 the magnitude of the angle at the center which it 
 subtends. 
 
 Cor. A quadrant is an arc of 90 degrees ; and a 
 right angle, being measured by a quadrant, is an angle 
 of 90 degrees. 
 
 THEOREM XXV. 
 
 If a radius is perpendicular to a chord, it bisects 
 the chord, and also the arc which the chord subte?ids. 
 
BOOK I 
 
 43 
 
 Let the radius DB be perpen- 
 dicular to the chord AC. It is 
 to be proved that it bisects AC, 
 and also the arc ABC. 
 
 Draw the radii DA, DC. 
 Now, since the right-angled tri- 
 angles AED, CED, have the 
 hypotenuse AD equal to the 
 
 hypotenuse CD, and the side ED common, it follows 
 that the third side AE is equal to the third side CE 
 (Cor. 3, Theo. XIX) ; that is, AC is bisected in E. 
 
 Again, since the triangles AED, CED, are mutually 
 equilateral, the angle ADE is equal to the angle CDE 
 (Theo. XII). Hence, the arc AB which measures the 
 former angle is equal to the arc CB which measures 
 the latter (Schol., Theo. XXIV). 
 
 Therefore, if a radius is perpendicular, etc. 
 
 THEOKEM XXVI. 
 
 An inscribed angle is measured by half the arc on 
 which it stands. 
 
 Let ACB be an angle in- 
 scribed in a circle. It is meas- 
 ured by half the arc AEB. 
 
 First, 
 
 the 
 
 D 
 
 , suppose the center 
 to be within the angle. Draw 
 the diameter CE ; also, join DA 
 and DB. Now, since AD and 
 DC are radii, the triangle ADC is isosceles ; hence, 
 the angles DAC, DCA^are equal (Theo. VIII). But 
 the exterior angle ADE is equal to the sum of the 
 two opposite interior angles DAC, DCA (Cor. 1, 
 
44 
 
 GEOMETRY. 
 
 Theo. V) ; it is consequently double of DCA ; and 
 since ADE is measured by the arc AE (Scbol., Theo. 
 XXIV), it follows that DCA is measured by half of 
 AE. In the same manner it may be shown that 
 DCB is measured by half of BE. Therefore, the 
 whole angle ACB is measured by half of the whole 
 arc AEB. 
 
 Next, let the center D be 
 without the angle ACB. By 
 the above demonstration, the 
 angle ACE is measured by half 
 the arc AE, and the anfr ^ 
 
 angle 
 
 BCE is measured by half the 
 
 arc BE ; therefore, BCA, which 
 
 is the difference of these two angles, is measured by 
 
 half the difference of the two arcs, that is, by half 
 
 of AB. 
 
 Hence, an inscribed ancrle, etc. 
 
 Cor. 1. Angles inscribed in 
 the same segment, as AEC and 
 ADC, are equal; for they are 
 measured by half the same arc 
 ABC. 
 
 Cor. 2. Any angle inscribed in a semicircle is a 
 right angle; being measured by half a semi-circum- 
 ference. 
 
 THEOEEM XXVII. 
 
 An angle contained bg a tangent and a chord is 
 measured by half the intercepted arc. 
 
BOOK 1. 
 
 45 
 
 A 
 
 B 
 
 Let AC be a tangent to a 
 circle, and BF a chord drawn 
 from the point of contact B. 
 Then will the angle ABF be 
 measured by half the arc BF, 
 and the angle CBF by half the 
 arc BDF. 
 
 Draw BE perpendicular to 
 AC, and it will be a diameter of the circle (Cor., 
 Theo. XXIII). Now, since ABE is a right angle, it is 
 measured by half the semi-circumference, BFE (Cor., 
 Theo. XXIV) ; and since the part EBF is an inscribed 
 angle, it is measured by half the arc FE (Theo. 
 XXVI) ; therefore, the remaining angle ABF is meas- 
 ured by half the remaining arc BF. 
 
 Again, because CBE is measured by half the semi- 
 circumference BDE, and EBF by half the arc EF, it 
 follows that the whole angle CBF is measured by 
 half the whole arc BDF. 
 
 Therefore an angle contained, etc. 
 
 THEOREM XXVIII. 
 
 If the circumference be divided into any number of 
 equal arcs, their chords will form a regular polygon. 
 
 Let the arcs AB, BC, etc., 
 be all equal to each other. 
 Then will the inscribed polygon 
 formed by their chords AB, BC, 
 etc., be a regular polygon. 
 
 Draw the radii AG, BG, CG, 
 
 etc. Now, since two sides in 
 
46 GEOMETRY. 
 
 each of the triangles AGB, BGC, etc., are radii ; and 
 since their bases AB, BC, etc., are equal, being chords 
 of equal arcs (Theo. XXIV) ; it follows that these 
 triangles are mutually equiangular (Theo. XII). But 
 they are also isosceles ; hence (Theo. VIII), all the 
 angles at their bases, GBA, GBC, GCB, GCD, etc, 
 are equal ; hence, also, the sums, ABC, BCD, etc., 
 that is, all the angles of the polygon, are equal. And 
 its sides have already been shown to be equal. It is, 
 therefore, a regular polygon (Def. 2, Sec. VII). 
 
 Hence, if the circumference, etc. 
 
 Cor. 1. If the polygon ABCDEF be a hexagon, 
 the angle AGB, being measured by one-third of the 
 semi-circumference, will be one-third of two right 
 angles (Cor., Theo. XXIV). Hence, either of the two 
 equal angles at the base of the triangle, as GBA, will 
 also be one-third of two right angles (Theo. V). 
 Therefore, AB is equal to AG (Cor., Theo. IX) ; that 
 is, the side of a regular hexagon is equal to the radius 
 of the. circumscribed circle. 
 
 Cor. 2. A perpendicular, as Gm, let fall from the 
 center of the circumscribed circle on any side of a 
 regular polygon, will be the apothegm of the polygon 
 (Schol., Theo. XXI). 
 
 Cor. 3. If from the center G, with the apothegm 
 Gm as radius, a circle be drawn, the side FE, and 
 consequently all the other sides of the regular poly- 
 gon, will be tangents to that circle (Theo. XXIII), 
 and the circle will be inscribed in the polygon (Def. 
 7, Sec. VIII). 
 
BOOK I 
 
 47 
 
 THEOREM XXIX. 
 
 The area of a circle is equal to half the product of 
 its radius by its circumference. 
 
 Let ABC be a circle, of 
 which DB or DC is radius. 
 
 If we conceive the whole cir- 
 cumference to be divided into 
 equal arcs so small that the 
 points of division may be re- 
 garded as infinitely near to each 
 other, the perimeter of the reg- 
 ular polygon formed by their chords (Theo. XXVIII) 
 will coincide with the circumference, and the area of 
 the polygon will be equal to the area of the circle ; 
 also, the apothegm of the polygon (Cor. 2, Theo. 
 XXVIII) will be equal to the radius of the circle. 
 But the area of the polygon will be equal to half the 
 product of its apothegm by its perimeter (Cor., Theo. 
 XXI). Therefore, the area of the circle must be 
 equal to half the product of its radius by its circum- 
 ference. 
 
 Cor. 1. In like manner it may be shown that the 
 area of any sector, as BECD, is equal to half the 
 product of the radius into its arc. 
 
 Cor. 2. The area of a segment BEC, less than a 
 semicircle, may be found by subtracting the triangle 
 BDC from the sector BECD. The area of a segment 
 BAC, greater than a semicircle, may be found by 
 adding the triangle BDC to the sector BACD. 
 
48 
 
 GEOMETRY. 
 
 THEOREM XXX. 
 
 The radius, and one side of a regular inscribed 
 polygon Icing given, the side of a regular inscribed 
 polygon of twice the number of sides mag be found. 
 
 Let AB be the side of a reg- 
 ular inscribed polygon in a 
 circle whose radius is CA or 
 CB. DraAV CE perpendicular to 
 AB. This will bisect AB and 
 also the arc AEB (Theo. XXV). 
 Hence, the chords AE, EB 
 (Theo. XXVIII), will be sides 
 
 of a regular inscribed polygon having twice as many 
 sides as the first. 
 
 Now, if the lengths of CB and AB are given, we 
 shall have, in the right-angled triangle CDB, the hy- 
 potenuse CB and the side BD; hence, we may find 
 the other side DC (Cor. 2, Theo. XIX). Subtracting 
 this from the radius CE, we shall have DE. Then, in 
 the right-angled triangle BDE, we shall have the two 
 sides BD, DE, from which we can find the hypotenuse 
 EB (Cor. 1, Theo. XIX) ; which is the side required. 
 
 Schol. If the diameter be 1, the side of a regular 
 inscribed hexagon will be \ (Cor. 1, Theo. XXVIII). 
 From this we can find the side, and consequently the 
 perimeter, of a regular inscribed dodecagon; from that 
 Ave can find the perimeter of a regular inscribed polygon 
 of 24 sides, etc. By carrying this calculation on, to 
 polygons of an indefinitely large number of sides, it 
 is found that the perimeter, though it increases at 
 every step, never exceeds 3.14159, except by decimal 
 figures beyond lkjfr 9; that is, beyond the fifth place 
 of decimals. Mcnce, since the perimeter ultimately 
 
 M.cnc 
 
BOOK I. 49 
 
 coincides with the circumference, it follows that the 
 circumference can not differ from 8.14159, except by 
 decimals beyond the fifth place. Disregarding these, 
 as of trifling value, we may conclude that the circum- 
 ference of a circle whose diameter is 1, is 3.14159. 
 
 Cor. If the diameter is 1, the area is equal to 
 
 1 yQ 141 ^Q 
 
 " X " (Theo. XXIX), which by reduction is .7854. 
 
 EXEECISES. 
 
 1. How many degrees are contained in an arc cut 
 off by the side of an inscribed square ? Of a regular 
 inscribed pentagon ? Hexagon ? Heptagon ? etc. 
 
 2. What is the area of a sector whose arc is 25 
 feet, in a circle whose diameter is 16 feet? 
 
 3. Prove that in an inscribed quadrilateral the sum of 
 any two opposite angles is 180 degrees (Theo. XXYI). 
 
 4. Prove that of two unequal chords the less is the 
 further from the center (Theo. XXV, and Cor. 2, 
 Theo. XIX). 
 
 SEC. IX.— PROBLEMS IN CONSTRUCTION. 
 
 PROBLEM I. 
 
 To bisect a given straight line. 
 
 Solution. Let AB be 
 the given line. From A 
 as a center, with a radius 
 obviously greater than 
 one-half of AB, describe 
 the arc CFD. From B 
 as a center, with the 
 same radius, describe the arc CED. Join the points 
 of intersection C and D. 
 Evans Geometry. — 5 
 
50 
 
 GEOMETRY. 
 
 Now, since the opposite sides of the quadrilateral 
 ACBD are equal, it is a parallelogram (Schol., Theo. 
 XIV). But the diagonals of a parallelogram bisect 
 each other (Theo. XV). Therefore, CD bisects AB. 
 
 PROBLEM II. 
 
 At a given point in a straight line, to erect a perpen- 
 dicular to that line. 
 
 f 
 
 A D 
 
 E B 
 
 Solution. Let AB be 
 an indefinite straight line, 
 and C the given point in 
 it. Take the points D 
 and E equally distant 
 from C. From D as a center, with a radius greater 
 than DC, describe an arc; and from E as a center, 
 with the same radius, describe another arc intersecting 
 the former at some point F. Now, draw FC, and it 
 will be the perpendicular required. 
 
 Join FD and FE. Then, since the triangles DFC, 
 EFC, are by construction mutually equilateral, the 
 angles FCD, FCE, are equal (Theo. XII) ; they are, 
 therefore, right angles, and CF is perpendicular to 
 AB at the point C (Defs. 2 and 3, Sec. IV.) 
 
 PROBLEM III. 
 
 To draw a perpendicular to a straight line from, a 
 given point without. 
 
 Solution. Let C be 
 the given point without 
 the straight line, AB. 
 From C as a center, with 
 a radius greater than the 
 
 
BOOK I. 
 
 51 
 
 shortest distance to the line, draw an arc cutting AB 
 in D and E. Bisect DE in F (Prob. I). Then draw 
 CF, and it will be the perpendicular required. 
 
 Join CD and CE. Now, since the triangles DCF, 
 EOF, are mutually equilateral, the angles CFD, CFE, 
 are equal (Theo. XII) ; they are, therefore, right 
 angles, and CF is a perpendicular to AB from the 
 point C. 
 
 PEOBLEM IV. 
 
 Solution. Let BAC be the given 
 angle. From A as a center, with any 
 radius, draw an arc BDC intersecting 
 both sides of the angle. Then draw 
 AD perpendicular to the chord BC 
 (Prob. Ill), and it will bisect the angle. 
 
 For, since the perpendicular AD is radius, it bisects 
 the arc BDC (Theo. XXV). Therefore, the arcs BD, 
 DC, being equal, the subtended angles BAD, DAC, 
 must also be equal (Theo. XXIV), and the angle BAC 
 is bisected by AD. 
 
 
 PEOBLEM V. 
 
 At a given point in a straight line, to make an angle 
 equal to a given angle. 
 
 Solution. Let ACB be 
 the given angle, and F 
 the given point in a 
 straight line FE. From c sf 
 
 C, as a center, with any radius as CB, describe an arc 
 BA intersecting both the sides of the angle ; also, 
 draw the chord BA. Then, from F as a center, with 
 
52 GEOMETRY. 
 
 a radius FE equal to CB, draw an indefinite arc ; and 
 from the center E, with a radius equal to the chord 
 BA, draw an arc intersecting the other at D; also, 
 join DE and DF. The angle DFE is the angle re- 
 quired. For, by equality of the triangles DEF, ABC, 
 the angle F is equal to the angle C (Theo. XII). 
 
 PROBLEM VI. 
 
 Through a given 'point to draw a straight line par- 
 allel to a given line. 
 
 Solution. Let AB E _£_ -i 
 
 be the given straight 
 line, and C the given 
 
 point. In AB take 
 
 any point D, and join 
 
 CD. Through C draw EF, making the angle ECD 
 equal to the angle CDB (Prob. V). These being 
 alternate angles, the straight line EF must be parallel 
 to AB (Cor. 1, Theo. III). 
 
 PROBLEM VII. 
 
 Two angles of a triangle being given, to find the 
 third. 
 
 Solution. At any point C in 
 a straight line AB, make ACD 
 equal to one of the given 
 angles (Prob. V), and DCE 
 equal to the other; then will • 
 
 ECB be equal to the third angle. For the sum of 
 the three angles at C is two right angles (Cor. 1, 
 Theo. I) ; which is also the sum of the three angles of 
 a triangle (Theo. V). 
 
BOOK I. 
 
 PROBLEM VIII. 
 
 Given the hypotenuse and one side of a right-angled 
 triangle, to construct the triangle. 
 
 Solution. Draw AB equal to c 
 the given side. At B erect a 
 perpendicular BC (Prob. II). 
 From A as a center, with a 
 radius equal to " the given hy- 
 potenuse, describe an arc inter- 
 secting BC at D. Join AD. 
 It is evident that ABD is the triangle required. 
 
 PEOBLEM IX. 
 
 To draw a circle through three given points. 
 
 Solution. Let A, B, and C 
 be the given points. Draw 
 the straight lines AB and BC, 
 and from their middle points 
 erect the perpendiculars DF 
 and EF. Also, join AF, BF, 
 and CF. 
 
 Now, because the triangles 
 AFD, BFD, have the side AD equal to the side BD, 
 and the side DF common, and the included angle 
 ADF equal to the included angle BDF, the third side 
 AF is equal to the third side BF (Theo. VII). In 
 the same manner it may be shown that BF is equal 
 to CF. Therefore, the three lines AF, BF, and CF 
 are equal ; and if from F as a center, with one of 
 these equals as radius, a circle be drawn, it will pass 
 through the three points A, B, and C. 
 
54 GEOMETB Y. 
 
 Cor. Hence, the center of a given circle may be 
 found by erecting perpendiculars on the middle points 
 of two chords, and producing them till they meet. 
 
 EXERCISES. 
 
 1. Given the three sides of a triangle, to construct 
 "the triangle. 
 
 2. Given two angles and the iacluded side, to con- 
 struct the triangle. 
 
 3. Given two adjacent sides and the included angle 
 of a parallelogram, to construct the parallelogram. 
 
 4. To draw a diameter of a given circle. 
 
 5. At a given point in the circumference of a circle, 
 to draw a tangent to the circle. 
 
PLANE GEOMETBY. 
 
 BOOK II. 
 
 PROPORTIONS OF MAGNITUDES LYING IN THE SAME PLANE. 
 
 SEC. X.— OF POLYGONS. 
 
 DEFINITIONS. 
 
 1. In mutually equiangular figures, the sides which 
 are similarly situated with respect to the equal angles, 
 are called homologous 
 sides. Thus, 
 
 If the angles A, B, C, 
 are respectively equal 
 to the angles a, b, e, 
 the side AB is homol- 
 ogous to the side ab, the side BC to the side be, and 
 CA to ca. 
 
 Two triangles or other polygons are called similar 
 when they are mutually equiangular, and their homol- 
 ogous sides are proportional. 
 
 2. Magnitudes are called proportionals when the 
 first lias the same ratio to the second, that the third 
 has to the fourth, the fifth to the sixth, etc. 
 
 The first terms of the several equal ratios are called 
 the antecedents, and the second the consequents. Thus, 
 
 If A : B : : C : D : : E : F, the terms A, C, E, are 
 the antecedents, and B, D, F, the consequents. 
 
 (55) 
 
56 GEOMETRY. 
 
 If four magnitudes are in proportion, the first and 
 fourth terms are called the extremes, and the second 
 and third the means. In this case the last term is 
 called a fourth projjortional to the three others. 
 
 If the first of three magnitudes is to the second as 
 the second is to the third, the second is called a mean 
 proportional between the two others. 
 
 It does not belong to geometry to develop the prin- 
 ciples of proportion ; but, inasmuch as they are ap- 
 plied in it, some of the more important of them are 
 stated here, in explanation of the terms and phrases 
 by which they will be quoted. None will be referred 
 to that are not found in common arithmetics. 
 
 3. Multiplying extremes and means, is the phrase 
 used in citing the principle that the product of the 
 extremes of a proportion is equal to the product of 
 the means. Thus, 
 
 If A : B : : C : D, then, AxD=BxC. 
 
 4. Resolving into a projyortion, refers to the prin- 
 ciple that if the product of two magnitudes is equal 
 to the product of two others, either couple may be 
 made the extremes of a proportion and the other the 
 means. Thus, 
 
 If AxD=BxC, then A : B : : C : D. 
 
 5. Multiplying an extreme and a mean equally, 
 refers to the principle that we may multiply an 
 extreme and a mean by the same quantity without 
 destroying the proportion. Thus, 
 
 If A : B : : C : D, then A : 2B : : C : 2D. 
 
 6. Dividing an extreme and a mean equally, refers 
 to the principle that we may divide an extreme and a 
 
BOOK II. 57 
 
 mean by the same quantity without destroying the 
 proportion. Thus, 
 
 If A : B : : C : D, then A : B : : — : — . 
 
 2 2 
 
 7. By composition, implies that if any number of 
 magnitudes are proportionals, the sum of all the 
 antecedents is to the sum of all the consequents as 
 any one antecedent is to its consequent. Thus, 
 
 If A : B : : C : D : : E : F, 
 Then A-fC-f-E : B-f-D-f F : : A : B. 
 
 8. By alternation, implies that the order of the 
 means in a proportion may be inverted. Thus, 
 
 If A : B : : C : D, then A : C : : B : D. 
 
 9. By equality of ratios, implies that if two ratios 
 are respectively equal to a third, they are equal to 
 each other. Thus, 
 
 If A : B : : C : D and A : B : : E : F, 
 Then C : D : : E : F. 
 
 10. 3Iultiplying corresponding terms, is the phrase' 
 used in applying the principle that the product of the 
 first terms of two proportions is to the product of the 
 second terms, as the product of the third is to the 
 product of the fourth. Thus, 
 
 If A : B : : C : D, and E : F : : G : H, 
 Then AxE : BxF : : CxG : DxH. 
 
 11. Squaring all the terms, is a reference to tho 
 principle that if any number of magnitudes are pro- 
 portionals their squares are also proportionals. Thus, 
 
 If A : B : : C : D, then A 2 : B 2 : : C 2 : D 2 . 
 
58 
 
 GEO'METEY, 
 
 THEOREM I. 
 
 If two triangles have the same altitude, their areas 
 are to each other as their bases. 
 
 A T) 
 
 Let the triangles 
 ABC, DEF, have the 
 same altitude ; that 
 is, let the perpendic- 
 ulars AG, DH, be 
 equal. It is to be proved that the area ABC is to the 
 area DEF, as the base BC is to the base EF. 
 
 The area ABC is equal to -JAGxBC (Cor. 2, Theo. 
 XVII, Book I) ; and the area DEF is equal to 
 JDHXEF. Therefore, 
 
 ABC : DEF : : 1-AGxBC : JDHxEF. 
 But, from the hypothesis, JAG is equal to JDH. 
 Hence, dividing a mean and an extreme equally (Def. 
 6, Sec. X), we have 
 
 ABC : DEF : : BC : EF. 
 
 Therefore, if two triangles, etc. 
 
 THEOEEM II. 
 
 If a straight line be drawn parallel to the base of a 
 triangle it ivill cut the other sides proportionally. 
 
 In the triangle ABC, let DE be 
 drawn parallel to the base BC. It 
 is to be proved that it cuts AB and 
 AC proportionally. 
 
 Join BE and CD. Now, since the 
 triangles ADE, BDE, have the same B c 
 
 altitude, namely, the perpendicular distance from E 
 
BOOK II. 59 
 
 to the line of their bases AB (Def. 4, Sec. V, B. I), 
 their areas are to each other as their bases (Theo. 
 I) ; that is, 
 
 ADE : DBE : : AD : DB. 
 Also, because the triangles ADE, CDE, have the same 
 
 altitude, 
 
 ADE : CDE : : AE : EC. 
 
 But the first ratio is the same in both proportions ; 
 for ADE is common, and the triangles DBE, CDE, 
 having the same base, DE, and the same altitude, 
 namely, the perpendicular distance between the par- 
 allels DE and BC, are equivalent (Theo. I). 
 
 Therefore, by equality of ratios (Def. 9, Sec. X), 
 
 we have 
 
 AD : DB : : AE : EC; 
 
 Or, by alternation (Def. 8, Sec. X), 
 
 AD : AE : : DB : EC. 
 
 Hence, if a straight line, etc. 
 
 Cor. 1. By composition (Def. 7, Sec. X), we get 
 from the last proportion 
 
 AD+DB : AE+EC : : AD : AE, 
 that is, AB : AC : : AD : AE. 
 
 Cor. 2. If DE be not parallel to BC, but takes 
 some other direction, as DF, it is evident that it will 
 not cut the other sides proportionally. 
 
 THEOKEM III. 
 Two triangles ivliich are mutually equiangular are 
 also similar. 
 
60 
 
 GEOMETRY. 
 
 F 
 
 Let ABC, BDE, be two tri- 
 angles having the angles at A, 
 B, and C, respectively equal to 
 the angles at B, D, and E. It 
 is to be proved that these tri- 
 angles are similar. 
 
 Let them be so placed that 
 their homologous sides AB, BD (Def. 1, Sec. X), shall 
 form one straight line. Produce AC and DE till they 
 meet. Then, because the angle DBE is by hypothesis 
 equal to the corresponding inner angle BAC, the lines 
 BE and AC are parallel (Cor. 1, Theo. Ill, B. I) ; 
 and because the angle ABC is equal to the corres- 
 ponding inner angle BDE, the lines BC and DE are 
 parallel ; hence, CBEF is a parallelogram. Now, in 
 the triangle ADF we have (Theo. II) 
 
 DB : BA : : DE : EF. 
 
 But EF is equal to BC (Theo. XIII, B. I). Hence, 
 DB : BA : : DE : BC. 
 
 By a like construction it may be proved in the same 
 way that the other homologous sides are proportional. 
 The two triangles are consequently similar (Def. 1, 
 Sec. X). 
 
 Therefore, two triangles, etc. 
 
 Cor. If two triangles have two angles of the one 
 respectively equal to two angles of the other, they are 
 similar ; for in that case the third angles are neces- 
 sarily equal (Theo. V, B. I). 
 
 THEOREM IV. 
 
 Two triangles having the sides of the one successively 
 proportional to the sides of the other, are similar. 
 
BOOK II. 
 
 Gl 
 
 Let ABC, DEF, be 
 two triangles having 
 AB : DE : : BC : EF 
 : : CA : FD. It may 
 be shown that they 
 are similar. 
 
 On DE describe the 
 triangle DGE, having the angles EDG, DEG, respect- 
 ively equal to the angles A and B. Then will the 
 triangles ABC and DEG be similar (Cor., Theo. Ill) ; 
 and we shall have (Def. 1, Sec. X) 
 
 AB : DE : : BC : EG. 
 
 But by hypothesis we have 
 
 AB : DE : : BC : EF. 
 
 Hence, EG is equal to EF. In the same manner it 
 may be shown that DG is equal to DF. Therefore, 
 the two triangles DGE, DFE, are mutually equilateral 
 and consequently equal (Cor., Theo. XII, B. I). But 
 ABC is by construction similar to DGE : hence, it is 
 also similar to DFE. 
 
 Therefore, two triangles, etc. 
 
 THEOEEM V. 
 
 Two triangles having an angle of the one equal to an 
 angle of the other, and the sides about those angles pro- 
 portional, are similar. 
 
 Let the two triangles 
 ABC, DEF, have the 
 angle A equal to the 
 angle D, and the sides 
 AB, AC, proportional to 
 
62 
 
 GEOMETRY. 
 
 the sides DE, DF. Then will these triangles be 
 similar. 
 
 Take AG equal to DE, and AH to DF ; also, join 
 GH. Then the triangles AGH, DEF, having two 
 sides and the included angle of the one equal to two 
 sides and the included angle of the other, are equal 
 throughout (Theo. VII, B. I). Now, by hypothesis, 
 AB : DE : : AC : DF. 
 
 Therefore, AB : AG : : AC : AH. 
 Hence, it follows that GH is parallel to BC (Cor. 2, 
 Theo. II). Consequently the angle AGH is equal to 
 the angle ABC (Theo. Ill, B. I), and AHG to ACB. 
 Consequently, also (Cor., Theo. Ill), the triangle 
 AGH or its equal DEF is similar to ABC. 
 
 Therefore, two triangles, etc. 
 
 Cor. If a triangle, as AGH, be cut off from an- 
 other triangle, ABC, by a straight line parallel to the 
 base, the two triangles will be similar. 
 
 THEOREM VI. 
 
 Two triangles having their sides respectively parallel 
 or perpendicular to each other, are similar. 
 
 First, let the tri- 
 angles ABC, abc, have 
 their sides respect- 
 ively parallel. Then, 
 because the sides con- 
 taining the angle A 
 are parallel to the 
 sides containing the 
 angle a, these angles are equal (Cor. 2, Theo. Ill, 
 
BOOK II. 
 
 63 
 
 B. I) : and for the same reason the angles B and 
 b are equal, also C and c. Hence, the two triangles 
 are mutually equiangular, and consequently similar 
 (Theo. III). . 
 
 Secondly, let the triangles ABC, DEF, have their 
 sides respectively perpendicular to each other. Pro- 
 duce the sides of DEF to the points G, H, I. Now, 
 the sum of all the angles of the quadrilateral AIDG 
 is equal to four right angles (Theo. XX, B. I). But 
 AID and AGD are right angles by hypothesis. 
 Hence, IDG and IAG must be together equal to two 
 right angles. But IDG and EDF are also together 
 equal to two right angles (Theo. I, B. I). Therefore, 
 subtracting equals from equals, we have the angle 
 IAG equal to the angle EDF. In the same manner 
 it may be proved that the angle ACB is equal to the 
 angle DFE, and CBA to FED. Hence, the two tri- 
 angles are mutually equiangular, and consequently 
 similar. 
 
 Therefore, two triangles, etc. 
 
 THEOEEM VII. 
 
 A perpendicular let fall from the right angle iipon 
 the hypotenuse of a rigid-angled triangle divides it into 
 two triangles similar to the whole, and to each other. 
 
 Let ABC be a triangle right- 
 
 angled at B, and let BD be per- 
 pendicular to the hypotenuse vVC. 
 It is to be proved that ABD and 
 BCD are similar to ABC and to 
 
 each other. 
 
 The triangles ABD, ABC, have a right angle in 
 
64 
 
 GEOMETRY. 
 
 ^u> 
 
 each, and the angle A common ; hence, they are 
 gar (Cor., Tlieo. III). The triangles BCD, ABC, 
 e also a right angle in each, and the angle C 
 common ; and hence they are similar. Therefore, 
 also, the triangles ABD, BCD, being both similar 
 to the same triangle, arcj^Ailar to each other. 
 
 Hence, a perpendffl mu 
 
 Cor. By similaritjjB JjiLangh's ADB, BDC, we 
 
 have AD : DB : : DB :^^; that is, the perpendicular 
 from the right angle is a mean proportional between the 
 parts of the hypotenuse (Def. 2, Sec. X). 
 
 THEOREM VIII. 
 
 The areas of similar triangles are to each other as 
 the squares described on their homologous sides. 
 
 D 
 
 E 
 
 /L 
 
 H 
 
 T 
 
 Let ABC, DEF, be 
 
 two similar triangles 
 of which the angles 
 A and B are respect- 
 ively equal to the an- R 
 gles D and E. It 
 
 may be shown that their areas are to each other as 
 AB 2 is to DE 2 (Def. 1, Sec. X). 
 
 Draw the perpendiculars AG' and DH. Then, the 
 triangles ABG, DEH, having a right angle in each, 
 and the angles B and E equal, are similar (Cor., 
 Theo. Ill), and Ave have 
 
 GA : HD : : AB : DE. 
 But by similarity of ABC and DEF we have 
 BC : EF : : AB : DE. 
 
BOOK II. 65 
 
 Multiplying corresponding terms (Def. 10, Sec. X) 
 and dividing an extreme and a mean equally ilK'^^i 
 Sec. X), ^^ 
 
 BCXGA : EFXHD DE2 
 
 2 2 
 
 But the first two terms ofl| ■& proportion represent 
 the areas of the triaii^|j |£ and DEF (Cor. 2, 
 
 Theo. XVII, B. I). 
 
 Therefore, the areas, 
 
 THEOREM IX. 
 
 If between the tivo parallel sides of a trapezoid a 
 third parallel be drawn, it will cut the other sides pro- 
 portionally. 
 
 In the trapezoid ABCD, r —\ 
 
 let EG be a third parallel / „-'"' \ 
 
 to AD and BC. Then will J ^-' '' V 
 
 BE be to EA as CG is to L^- A, 
 
 GD. 
 
 Draw the diagonal BD. Now, because. EF is par- 
 allel to the base AD of the triangle BAD, we have 
 
 (Theo. II) 
 
 BE : EA : : BF : FD ; 
 
 and, because FG is parallel to the base BC of the 
 
 triangle DBC, 
 
 CG : GD : : BF : FD. 
 
 Therefore, by equality of ratios (Def. 9, Sec. X), 
 
 BE : EA : : CG : GD. 
 
 Hence, if between, etc. 
 
 Cor. 1. If the third parallel bisects the oblique sides, 
 it is equal to one-half the sum of the parallel sides. 
 Evans Geometry. — 6 
 
66 
 
 GEOMETKY. 
 
 For by similarity of the triangles BEF, BAD (Cor. 
 Theo. V), if BE be one-half of BA, EF will be one- 
 half of AD ; and in the same manner it may be 
 shown that FG will be one-half of BC : consequently, 
 the whole line EG will be one-half the sum of AD 
 and BC. 
 
 Cor. 2. The area of the trapezoid (Theo. XVIII, 
 B. I) is equal to its altitude multiplied by EG, which 
 is its mean breadth. 
 
 THEOREM X. 
 
 The perimeters of similar polygons are to each other 
 as their homologous sides. >» 
 
 Let ABCDE, 
 abcde, be two simi- 
 lar polygons, hav- 
 ing the angles A, 
 B, C, etc., respect- 
 ively equal to the 
 angles a, b, c, etc. 
 
 Then will their perimeters be to each other as any 
 two homologous sides AB and ab. 
 
 By similarity of the two polygons (Def. 1, Sec. X) 
 we have 
 
 AB : ab : : BC : be : : CD : cd : : DE : de : : EA : ea. 
 And therefore, by composition (Def. 7, Sec. X), 
 AB+BC-f-CD + DE + EA : ab+bc+cd+de+ea : : 
 
 AB : ab. 
 But the first two terms of the last proportion repre- 
 sent the perimeters of the two polygons (Def. 3, Sec. 
 VII, B. I.) 
 
 Therefore, the perimeters, etc. 
 
BOOK II. 
 
 67 
 
 THEOREM XL 
 
 Tlie areas of two regular polygons of the same num- 
 ber of sides are to each other as the squares of their 
 sides. 
 
 Let ABODE, 
 abode, be two 
 regular polygons -g 
 of the same num- 
 ber of sides, for 
 example, five. 
 The area of the 
 former is to the area of the latter as AB 2 is to air. 
 
 Since the two polygons have the same number of 
 sides and angles, it is evident (Theo. XX, B. I) that 
 the sum of all the angles of ABODE is equal to the 
 sum of all the angles of abode, and consequently that 
 each angle of ABODE is equal to each angle of 
 abode (Del 2, Sec. VII). 
 
 Bisect the angles A and B by AF and BF, and the 
 
 angles a and b by af and bf. Now, since the triangles 
 
 ABF, abf, have two angles of the one equal to two 
 
 angles of the other, they are similar (Cor., Theo. Ill) ; 
 
 hence 
 
 ABF : abf : : AB 2 : ab° (Theo. VIII). 
 
 Multiplying an extreme and a mean equally, 
 
 5 ABF : babf : : AB 2 : ab 2 (Def. 5, Sec. X). 
 But five times ABF is the area of ABODE (Theo. 
 XXI, B. I), and five times abf is the area of abode. 
 
 Therefore, the areas, etc. 
 
68 G-EOMETRY. 
 
 EXEECISES. 
 
 1. Prove that two parallelograms of the same alti- 
 tude are to each other as their bases. 
 
 2. Prove that if two isosceles triangles have their 
 vertical angles equal they are similar. 
 
 3. If the area of a regular pentagon whose side is 
 1 be 1.72, what is the area of a regular pentagon 
 whose side is 5? 
 
 4. In two similar polygons, if a side of one be 7, 
 and the homologous side of the other 13, and if the 
 perimeter of the former be 39, Avhat is the perimeter 
 of the latter? 
 
 SEC. XI.— OF CIRCLES. 
 DEFINITIONS. 
 
 1. Two ARCS of circles are called similar when they 
 are equal parts of the circumferences to which they 
 belong. 
 
 2. Two sectors are called similar when their arcs 
 are similar. 
 
 THEOREM XII. 
 
 If two chords intersect in a circle, the product of the 
 parts of the one is equal to the product of the p>arts of 
 the other. 
 
 In the circle ABCD let the D/^ "^Xn 
 
 chords AC, BD, intersect each // ^\ ^s\ 
 
 other in E. It is to be proved [ / ^-^e"\ • j 
 
 that AExEC=BExED. \L^ ^nI 
 
 Join AD and BC. Now, in \ / 
 
 the triangles ADE, BCE, the ^ ^ 
 
 vertical angles AED and BEC are equal (Theo. II, 
 
BOOK II. 
 
 69 
 
 B. I) ; also, the angles A and B are equal, being both 
 measured by half the same arc, DC (Theo. XXVI, B. 
 I) ; hence, the third angles are equal, and the two 
 triangles are similar (Theo. Ill), and we have 
 
 AE : BE : : ED : EC. 
 Then, multiplying extremes and means, we get 
 AExEC=BExED. 
 
 Therefore, if two chords, etc. 
 
 THEOREM XIII. 
 
 If from a point without a circle a tangent and a 
 secant he drawn, the tangent ivill be a mean 'propor- 
 tional between the secant and its external part. 
 
 Let AB be a tangent and AC 
 a secant to a circle. Then will 
 AC : AB : : AB : AD. 
 
 Join BC and BD. Now, the 
 angle DBA, contained by a tan- 
 gent and chord, is measured by 
 half the arc DB (Theo. XXVII, 
 B. I) ; and the angle C, being an 
 inscribed angle, is measured by half the same arc 
 (Theo. XXVI, B. I) ; therefore, these two angles are 
 equal. But the angle A is common to the two tri- 
 angles BAD and BAC. Hence, these triangles have 
 two angles of the one equal to two angles of the 
 other, and are consequently similar (Cor., Theo. III). 
 
 Therefore, AC : AB : : AB : AD. 
 That is, if from a point, etc. 
 
70 GEOMETRY. 
 
 THEOREM XIV. 
 
 Similar arcs in different circles subtend equal angles 
 at the centers. 
 
 Let AB and ah be 
 similar arcs -in two 
 circles whose centers 
 are C and c. It may- 
 be shown that the an- 
 gles at C and c are equal. 
 
 Whatever portion AB is of the circumference to 
 which it belongs, the angle C is the same portion of 
 four right angles (Schol., Theo. XXIV, B. I); and 
 whatever portion ah is of the other circumference, the 
 angle c is the same portion of four right angles. But 
 AB is the same portion of the first circumference that 
 ab is of the second (Def. 1, Sec. XI). Hence, the 
 angles C and c are equal portions of four right 
 angles ; hence, also, they are equal to each other. 
 
 Therefore, similar arcs, etc. 
 
 Cor. The sectors ABC, abc, are similar (Def. 2, 
 Sec. XI). Hence, the sides of similar sectors con- 
 tain equal angles. Hence, also, similar sectors are 
 contained an equal number of times in the circles to 
 which they belong. 
 
 Schol. Since a degree is ^^ of the circumference 
 in which it is taken (Schol., Theo. XXIV, B. I), it 
 follows that degrees in unequal circles are similar but 
 not equal arcs. 
 
BOOK II. 
 
 71 
 
 THEOEEM XV. 
 
 The circumferences of two circles are to each other 
 as their diameters. 
 
 J 
 
 C / 
 
 Let ACE 
 and ace be two 
 circles having 
 two similar 
 regular poly- 
 gons inscribed, 
 one in each. 
 
 Draw the radii AG, BG ; ag, bg. Now, the angles G and 
 g, being subtended by similar arcs, AB and ab, are 
 equal (Theo. XIV) ; and the sides containing them are 
 by construction proportional ; therefore, the two tri- 
 angles AGB, agb, are similar (Theo. V), and we have 
 
 AB : ab : : AG : ag : : 2 AG : 2 ag. 
 
 But the perimeter ABCDEF is to the perimeter 
 abcdef, as AB is to ab (Theo. X), and consequently 
 as 2 AG is to 2ag. Now, if the number of sides of 
 the similar polygons be indefinitely increased, the 
 perimeters will ultimately coincide with the circum- 
 ferences. Therefore, 
 
 Circumference ACE : circumference ace : : 2 AG : 2 ag. 
 But 2 AG and 2 ag represent the diameters of the two 
 circles (Def. 2, Sec. 8, B. I). 
 
 Therefore, the circumferences, etc. 
 
 Cor. 1. Representing the diameter of any circle by 
 
 D, and its circumference by C, and remembering that 
 
 the circumference of a circle whose diameter is 1 is 
 
 3.14159 (Schol., Theo. XXX, B. I), we have, by the 
 
 above theorem, 
 
 1 : D : : 3.14159 : C. 
 
72 
 
 GEOMETRY. 
 
 Multiplying extremes and means, we get C = 3.14159 
 XD. That is, 
 
 Tlic circumference of any circle is equal to its di- 
 ameter multiplied by 3.14159. 
 
 Cor. 2. The diameter is equal to the circumference 
 divided by 3.14159. 
 
 THEOREM XYI. 
 
 TJie areas of two circles are to each other as the 
 squares of their diameters. 
 
 The same con- 
 struction being 
 used as in the 
 last theorem, it 
 may be shown 
 as before, that 
 
 AB : ab : : 2 AG : 2 ag. 
 
 Hence, by squaring all the terms (Def. 11, Sec. X), 
 
 AB- : ab 2 : : (2 AG) 2 : (2ag) 2 . 
 But the area of the polygon ABCDEF is to the area 
 of the polygon abedef, as AB 2 is to ab 2 (Theo. XI), 
 and, consequently, as (2 AG) 2 is to (2 ag) 2 . Now, if 
 the number of sides of the two polygons be in- 
 definitely increased, their areas will ultimately coincide 
 with the areas of the circles. Hence, the area of the 
 circle ACE is to the area of the circle ace as (2 AG) 2 
 is to (2ag) 2 . 
 
 That is, the areas of two circles, etc. 
 
 Cor. Representing the diameter of any circle by 
 D and its area by A, and remembering that the area 
 
BOOK II. 73 
 
 of a circle whose diameter is 1 is .7854 (Cor., Theo. 
 XXX, B. I), we have 
 
 V : D 2 : : .7854 : A. 
 But I 2 is 1. Therefore, multiplying extremes and 
 means, we get A = .7854xD' 2 . That is. 
 
 The area of any circle is equal to the square of its 
 diameter multiplied by .7854. 
 
 EXEKCISES. 
 
 1. If a secant to a circle be 12 feet, and its external 
 part 8 feet, what will be the length of a tangent to 
 the circle, drawn from the same point ? 
 
 2. If the diameter of a circle be 24, Avhat will be 
 the length of its circumference? 
 
 3. If the circumference of a circle be 8924, what is 
 the length of its diameter ? 
 
 4. The diameter of a circle is 16: what its area? 
 
 5. Prove that the areas of similar sectors are to 
 each other as the squares of their radii. 
 
 6. Prove that the area of a circle equals the square 
 of its radius multiplied by 3.14159. 
 
 SEC. XII.— PROBLEMS IN CONSTRUCTION. 
 
 PROBLEM I. 
 To divide a straight line into any number of equal 
 parts. 
 
 Solution. Let AB be 
 the given straight line, 
 which it is required to 
 
 divide into a certain num- A-^ j^ 
 
 ber of equal parts, for example, three 
 Evans' Geometry. — 7 
 
 O B 
 
74 
 
 GEOMETRY. 
 
 From A draw any straight line AD ; and lay off 
 DE and EC in the same direction, each equal to AD. 
 Join CB, and through D draw DF parallel to CB. 
 Now, AD : AC : : AF : AB (Cor. 1, Theo. II). But 
 AD is one-third of AC, by construction ; therefore, 
 AF is one-third of AB. Now, lay off FG equal to 
 AF, and it is evident that AB will be divided into 
 three equal parts. 
 
 PROBLEM II. 
 
 To find a fourth proportional to three given straight 
 lines. 
 
 Solution. Draw two straight 
 lines containing any angle A. 
 Make AB, BC, and AD re- 
 spectively equal to the three 
 given lines, and join BD. 
 Through C draw CE parallel to BD, and produce AD 
 to meet it in E. Now, AB : BC : : AD : DE (Theo. 
 II). Hence, DE is the fourth proportional required 
 (Def. 2, Sec. X). 
 
 PROBLEM III. 
 
 To find a mean proportional between two given 
 straight lines. 
 
 Solution. Lay off, in one 
 straight line, AB and BC, 
 respectively equal to the 
 two given lines. Find E, 
 the middle point of AC. 
 Then, from E as a center, with EA or EC as radius, 
 describe a semicircle. At B erect BD perpendicular 
 to AC, and join AD and DC. Now, since the angle 
 
 A 1 -' 
 
BOOK II. 75 
 
 ADC is inscribed in a semicircle, it is a right angle 
 (Cor. 2, Theo. XXVI, B. I). Hence, BD is a mean 
 proportional between AB and BC (Cor., Theo. YII). 
 
 PROBLEM IV. 
 
 To describe a square that shall be equivalent to a 
 given parallelogram. 
 
 Solution. Let ABCD be the 
 
 given parallelogram. Between its /• 7 
 
 base AB and its altitude ED, find 
 a mean proportional, which denote 
 by M. Then, since A E E 
 
 AB : M : : M : ED, we have ABxED=M 2 . 
 
 But ABxED represents the area of the parallelogram 
 (Cor. 1, Theo. XVII, B. I). Therefore, the square 
 described on M is the square required. 
 
 Cor. It is evident that if we find a mean propor- 
 tional between the base of a triangle and half its 
 altitude (Cor. 2, Theo. XVII, B. I), we shall have the 
 side of a square equivalent to the triangle. 
 
 EXERCISES. 
 
 1. To construct a triangle that shall be similar to a 
 given triangle. 
 
 2. On a given straight line, to describe a rectangle 
 that shall be equivalent to a given square. 
 
BOOK III, 
 
 SOLID GEOMETRY 
 
 SEC. XIII.— PLANES AND THEIR INCLINATIONS. 
 
 DEFINITIONS. 
 
 1. A straight line is said to be perpendicular to a 
 plane when it is perpendicular to every straight line 
 it can meet in that plane. 
 
 2. The line in -which two planes cut one another is 
 called their intersection. 
 
 Schol. The intersection of two 
 planes is a straight line. For, 
 if A and D are any two points 
 in the intersection of the planes 
 FC and EB, the straight line 
 joining these points must be in 
 both planes (Def. 8, Sec. I) ; it is, 
 therefore, their intersection. 
 
 3. The angle of inclination of two planes is that 
 contained by any two straight lines perpendicular to 
 their intersection, one in each plane. 
 
 If the line in each plane is perpendicular to the 
 other plane, the two planes are said to be perpendicu- 
 lar to each other, the angle of inclination being a 
 right angle. 
 (76) 
 
BOOK III. 
 
 77 
 
 Thus, if AB and CB' are both perpendicular to the 
 intersection EH, then ABC is 
 the angle of inclination of the 
 planes DII and HF. If AB is 
 also perpendicular to the plane 
 HF, and CB to DII. the two D < 
 planes are perpendicular to each 
 other. 
 
 4. Two planes are said to be 
 parallel with one another when their intersections by 
 any third plane are parallel. 
 
 Cor. Two parallel planes can never meet ; for, if 
 they were to meet in any point, their intersections by 
 a third plane passing through that point would also 
 meet; which is contrary to the definition. 
 
 5. The divergence of three 
 or more planes from a single 
 point constitutes a solid an- 
 gle. Thus, A is a solid an- 
 gle contained by the three A 
 planes BAC, CAD, DAB. 
 
 THEOKEM I. 
 
 Two straight lines perpendicular to the same plane 
 are parallel to each other. 
 
 Let AB and CD be two 
 perpendiculars to the plane 
 EF. It is to be shown that 
 they are parallel. 
 
 Join BD. Now, AB and 
 CD, being perpendicular to 
 the plane, are also perpendicular to the line BD 
 
78 
 
 (tEOMETR y. 
 
 "which they meet in it (Def. 1). Also, a plane per- 
 pendicular to EF and passing through BD will contain 
 both AB and CD (Def. 3). But straight lines in the 
 same plane and perpendicular to the same straight line 
 are parallel (Cor. 2, Theo. IV, B. I). Therefore, AB 
 is parallel to CD. 
 
 That is, two straight lines, etc. 
 
 THEOREM II. 
 
 - Parallel lines intercepted between parallel planes are 
 equal. 
 
 Let FH and EG be any 
 two parallel lines intercepted 
 between two parallel planes, 
 AB and CD. It is to be 
 shown that FH is equal to 
 EG. 
 
 Through the parallels FH 
 and EG let the plane FG pass. Its intersections with 
 the parallel planes AB and CD will be parallel lines 
 (Def. 4) ; that is, FE is parallel to HG. Hence, 
 FEGH is a parallelogram, and FH is equal to EG 
 (Theo. XIII, B. I). 
 
 Therefore, parallel lines, etc. 
 
 Cor. 1. Any straight line, as FH, perpendicular to 
 
 one of two parallel planes, as CD, will be perpendicular 
 to the other. For FHG being a right angle (Def. 1), 
 HFE will also be a right angle (Cor. 1, Theo. IV, 
 B. I), whatever may be the position of the parallels 
 FE and HG in their planes. 
 
 Cor. 2. The perpendicular distance between two 
 parallel planes is everywhere the same. 
 
BOOK III. 
 
 79 
 
 SEC. XIV.— OF PARALLELOPIPEDS. 
 
 1. A PAiiALLELOPiPED is a solid bounded by six 
 plane faces, of which each one E 
 
 is parallel to its opposite. The 
 intersections of the faces are 
 called edges. r 
 
 Cor. Any face of a parallele- 
 piped is a parallelogram. For 
 the lines GH and BC, being the intersections of two 
 parallel planes, FH and AC, with a third plane GC, are 
 parallel (Def. 4, Sec. XIII); and in the same way it 
 may be shown that BG and CII are parallel ; hence, 
 BCHG is a parallelogram (Def. 2, Sec. VI, B. I). 
 
 2. A right parallelopiped is one whose edges are 
 perpendicular to the faces. Any other parallelopiped 
 is called oblique. 
 
 S. A cube is a right parallelopiped whose length, 
 breadth, and hight are equal. 
 
 4. Two parallelopipeds, or other solids, are called 
 equivalent when they contain the same amount of solid 
 space. 
 
 E 
 
 D 
 
 II 
 
 THEOREM III. 
 
 Any hvo opposite faces of a parallelopiped are equal. 
 
 Let AH be a parallelopiped. 
 It is to be proved that any 
 two opposite faces, as DEFA, 
 CHGB, are equal. Join DF 
 and CG. Now, since EFGH is 
 a parallelogram (Cor. Def. 1), EF is equal to HG 
 (Theo. XIII, B. I) ; and in the same manner it may be 
 shown that DE is equal to CH. But FG and DC, 
 
 \ 
 
 
 
 \ 
 
 F 
 
 
 / 
 
 
 
 
 / 
 
 / 
 
 
 C 
 
 / 
 
 \ 
 
 
 
 \ 
 
 c; 
 
 J: 
 
80 
 
 GEOMETRY. 
 
 being both equal and parallel to EH (Theo. XIII, 
 B. I), must be equal and parallel to each other. Hence, 
 DCGF is a parallelogram (Theo. XIV, B. I), and DF 
 is equal CG. Wherefore, the triangles DEF, CHG, are 
 mutually equilateral, and consequently equal. But the 
 former triangle is half the parallelogram DEFA (Cor. 
 1, Theo. XIII, B. I), and the latter triangle is half the 
 parallelogram CHGB : hence, these two parallelograms 
 are equal. 
 
 Therefore, any two opposite faces, etc. 
 
 Schol. Any face of a parallelopiped may be taken 
 as the base. We may designate DABC and EFGH as 
 the lower and upper bases. The perpendicular distance 
 between the bases is called the hight or altitude. 
 
 Cor. 1. If EB be a right parallelopiped, the edges 
 AF and BG, since, by definition, they are both perpen- 
 dicular to the face DB, must be perpendicular also to 
 the straight line AB, which they meet in that plane 
 (Def. 1, Sec. XIII). Hence, ABGF, or any face of a 
 rigid parallelopiped, is a rectangle. 
 
 Cor. 2. The faces of a cube are all equal squares. 
 
 THEOREM IV. 
 
 The solidity of a rigid 'parallelopiped is equal to the 
 area of its base multiplied by its hight. 
 
 Let ABC be a right paral- 
 lelopiped. It is to be proved 
 that its solidity is equal to the 
 area of its base AB, multiplied 
 by its hight DC. 
 
 Let planes pass through the 
 solid parallel to the three faces' 
 
 
 ' / / 
 
 
 / / / 
 
 / 
 
 / 
 
 / 
 
 / 
 / 
 
 / 
 / 
 
 
 
 D 
 
 
 
 
 
 
 
 
 
 
 
BOOK III. 
 
 81 
 
 AB, BC, CA, dividing the edges DA, DB, DC, into parts 
 of equal length. It is evident that the parallelopiped 
 will thus be divided into a number of equal cubes. Let 
 one of these be taken as the unit of solidity. Now, 
 in the layer next to the base, there will be as many of 
 these solid units as there are corresponding units of area 
 in the base ; and there will be as many equal layers as 
 there are corresponding linear units in the hight. 
 
 Therefore, the solidity of a right parallelopiped, etc. 
 
 Schol. If the edges are incommensurable, so that 
 no unit can be found into which they can all be divided 
 without remainder, the above theorem will still hold 
 true ; for by taking the unit smaller and smaller, the 
 remainder can be made less than anv assignable 
 quantity. 
 
 When the linear unit is one inch, the unit of solidity 
 is a cubic inch ; when the linear unit is one foot, the 
 unit of solidity is a cubic foot, etc. 
 
 Cor. 1. The solidity of a right parallelopiped is 
 equal to the product of its length, breadth, and hight. 
 
 Cor. 2. The solidity of a cube may be found by 
 raising one of its edges to the third power. 
 
 THEOKEM V. 
 
 Parallelopipeds on the same base and between the 
 same parallels are equivalent. 
 
 J) E M K 
 
 Let AGHD and AGLM 
 
 be two parallelopipeds on 
 the same base AG, and 
 between the same paral- 
 lels BG, CL; AF, DK. b o 
 It may be shown that they are equivalent. 
 
82 GEOMETRY. 
 
 The parallelograms AC, FH, being opposite faces 
 of a parallelopiped, are equal (Theo. III). Also, the 
 parallelograms DH, ML, being both equal to AG, are 
 equal to each other ; hence, if we add EN to both, 
 the parallelograms DN and EL must be equal. It is 
 also evident that the side CN is equal to the side IIL. 
 But CB is equal to HG, and the angle BCN is equal 
 to the corresponding outer angle GHL ; therefore, 
 the triangles BCN, GHL, having two sides and the 
 included angle of the one equal to two sides and 
 the included angle of the other, are equal (Theo. 
 VII, B. I). 
 
 If now the solid FGLE be applied to the solid 
 ABND, so that the face EL shall coincide with its 
 equal DN, then the faces GHL and BCN, being equal 
 and in the same plane, will coincide ; and in the same 
 manner it may be shown that the faces FEK and 
 ADM will coincide ; hence, also, the face FH will 
 coincide with the face AC, and FL with AN ; and the 
 two solids are consequently equal. If, then, from the 
 whole solid AL, we take away these equal portions in 
 turn, the remainder AGHD will be equivalent to the 
 remainder AGLM. 
 
 Therefore, parallelopipeds on the same base, etc. 
 
 Schol. Any two par- V- XM E 
 
 allelopipeds on the same x \ \ \ \ 
 
 base and between the same 
 
 D' \E 
 
 planes are equivalent. \ V 
 
 For, if their upper bases c H 
 
 are not between the same parallels, but have some 
 
 other position, as DH and MS, produce two sides of 
 
 each till they intersect as in the figure ; then UVWX 
 
BOOK III. 
 
 83 
 
 can be the upper base of a parallelopipecl on the same 
 lower base and between the same parallels with both 
 the other parallelopipeds ; hence, by the above demon- 
 stration, both the others will be equivalent to this third 
 parallelopiped ; they will, therefore, be equivalent to 
 each other. 
 
 Cor. The solidity of any parallelopiped on a rect- 
 angular base, is equal to the area of its base multi- 
 plied by its altitude ; for it is equivalent to a right 
 parallelopiped of the same base and altitude. 
 
 SEC. XV.— THE PRISM AND THE CYLINDER. 
 DEFINITIONS. 
 
 1. A prism is a solid bounded by plane faces, of 
 which two are equal and parallel polygons, and the 
 Others parallelograms. It includes the parallelopiped, 
 as one species. 
 
 The equal and parallel polygons 
 are called the bases. The other 
 faces together form the convex or 
 lateral surface. The edges join- 
 ing the corresponding angles of the 
 two bases are called the principal 
 edges. 
 
 A prism is called triangular, quadrangular, pen- 
 tagonal, etc., according as its base is a triangle, a 
 quadrilateral, a pentagon, etc. 
 
 2. A right prism is one whose principal edges are 
 perpendicular to the bases. Any other prism is called 
 oblique. 
 
84 
 
 GEOMETRY. 
 
 3. A cylinder is a solid described by the revolu- 
 tion of a rectangle about one side, which 
 remains fixed. 
 
 The fixed side is called the axis of the 
 cylinder. The opposite side describes the 
 convex surface. The circles described by 
 two other sides are called the bases. 
 
 4. A prism is said to be inscribed in 
 a cylinder when its bases are inscribed 
 
 in the bases of the cylinder, and its principal edges 
 lie in the convex surface of the cylinder. 
 
 5. The HiGHT, or altitude, of a prism or a cylin- 
 der, is the perpendicular distance between the planes 
 of its bases. 
 
 THEOREM VI. 
 
 The convex surface of a right prism is equal to the 
 perimeter of its base multiplied by its hight. 
 
 Let ABE be a right prism. Since 
 its principal edges are by defini- 
 tion perpendicular to the bases, any 
 one of them may be taken as the 
 hight of the prism. Now, AD and 
 BC, being both perpendicular to the 
 lower base, are also perpendicular to 
 AB, which they meet in the plane of 
 that base (Def. 1, Sec. XIII); hence, the parallelo- 
 gram ABCD is a rectangle, and its area is equal to 
 its base AB multiplied by its altitude BC (Cor. 1, 
 Theo. XVII, B. I). In the same manner it may be 
 shown that the area of each of the other faces com- 
 posing the convex surface is equal to its base multi- 
 
BOOK III 
 
 85 
 
 plied by the altitude of the prism. Therefore, the 
 sum of their areas is equal to the sum of their bases 
 multiplied by the common altitude. But the sum of 
 their bases constitutes the perimeter of the base of 
 the prism. 
 
 Hence, the convex surface, etc. 
 
 THEOEEM VII. 
 
 The solidity of any prism is equal to the area of its 
 base multiplied by its altitude. 
 
 Let ABCD be any prism. Now, A| 
 whatever may be the form of its 
 base, AB, it is evident that it may 
 be divided into an indefinitely large 
 number of small rectangles, so that 
 the remainder, if there be any, shall 
 be less than any assignable quantity ; and the base 
 DC may be divided into the same number of equal 
 rectangles having their sides respectively parallel to 
 the sides of the others. Now, each rectangle in the 
 lower base, and its corresponding rectangle in the 
 upper base, may be considered as the opposite bases 
 of a small parallelopiped ; and the prism will be made 
 up of such parallelopipeds. But the solidity of each 
 of these will be equal to the area of its base multi- 
 plied by its altitude (Cor., Theo. Y), which is the same 
 as the altitude of the prism : hence, the sum of their 
 solidities will be equal to the sum of their bases mul- 
 tiplied by their common altitude. 
 
 That is, the solidity of any prism is equal, etc. 
 
86 
 
 GEOMETRY. 
 
 THEOREM Till. 
 
 The convex surface of a cylinder is equal to the cir- 
 cumference of its base multiplied by its altitude; and 
 its solidity is equal to the area of its base multiplied by 
 its altitude. 
 
 Let iVBCD be a cylinder, having A^ 
 a prism inscribed in it whose base 
 is a regular polygon. Now, if the 
 number of sides of this polygon be 
 indefinitely increased, its perimeter 
 will ultimately coincide with the iC 
 circumference of the base of the 
 cylinder. Then, also, the convex surface of the prism 
 will coincide with the convex surface of the cylinder, 
 and the solidity of the prism with the solidity of the 
 cylinder. But the convex surface of the prism is 
 equal to the perimeter of the base multiplied by the 
 altitude (Theo. VI), and its solidity is equal to the 
 area of the base multiplied by the altitude (Theo. 
 VII). 
 
 Therefore, the convex surface of a cylinder, etc. 
 
 SEC. XVI.— PYRAMIDS AND CONES. 
 DEFINITIONS. 
 
 1. A pyramid is a solid bounded 
 by plane faces, of which one is any 
 polygon, and the others triangles 
 having a common vertex. The poly- 
 gon is called the base. The triangles 
 together form the convex or lateral 
 surface. 
 
 Pyramids are called triangular, quadrangular, pen- 
 
BOOK III. 
 
 87 
 
 tagonal, etc., according as their bases are triangles, 
 quadrilaterals, pentagons, etc. 
 
 2. A regular pyramid is one whose base is a reg- 
 ular polygon, and the triangular faces are equal and 
 isosceles. 
 
 3. A cone is a solid described by the revolution of 
 a right-angled triangle about one of 
 the sides containing the right angle, 
 which side remains fixed. The fixed 
 side is called the axis of the cone. 
 The hypotenuse describes the convex 
 surface. The circle described by the 
 other revolving side is called the base. 
 
 4. The altitude of a pyramid or cone is the per- 
 pendicular distance from the vertex to the plane of 
 the base. 
 
 5. The slant liiglit of a regular pyramid is the per- 
 pendicular let fall from the vertex upon the base of 
 any one of its triangular faces. The side or slant 
 hight of a cone is the straight line drawn from the 
 vertex to any point in the circumference of the base. 
 
 6. A FRUSTUM of a pyramid or a cone, is the por- 
 tion next the base cut off by a plane parallel to the 
 base. The slant hight of a frustum is that part of 
 the slant hight of the whole solid which lies on the 
 frustum. 
 
 7. A section of any solid is the surface in which 
 it is divided by a plane which passes through it. 
 
88 
 
 GEOMETKT. 
 
 THEOREM IX. 
 
 The convex surface of a regular pyramid is equal to 
 half the product of the perimeter of the base by the 
 slant hight. 
 
 Let ABODE be a regular 
 
 pyramid,* of Avhich AF is the 
 slant hight. The area of the 
 triangle ACD is equal to half 
 the product of its base CD into 
 its altitude, which is the slant 
 hight. Consequently the areas 
 of all the equal triangles com- 
 posing the convex surface are 
 
 together equal to half the product of the sum of their 
 bases by the slant hight. But the sum of their bases 
 constitutes the perimeter of the base of the pyramid. 
 
 Therefore, the convex surface of a regular pyra- 
 mid, etc. 
 
 Schol. 1. The trapezoids BF, 
 CG, etc., composing the convex 
 surface of a frustum of a regular 
 pyramid, are equal to each other; 
 for they are the differences be- 
 tween the equal triangles AEF, 
 AFGr, etc., and the equal tri- 
 angles ABC, ACD, etc. 
 
 Schol. 2. Since the area, of the trapezoid BF is 
 equal (Cor. 2, Theo. IX, B. II) to its mean breadth 
 LM multiplied by RS, which is the slant hight of the 
 frustum, and since the same is true of each of the 
 other trapezoids, it follows that the convex surface of 
 <i frustum of a regular pyramid is equal to its slant 
 
BOOK III, 
 
 89 
 
 liiglit multiplied into the perimeter of a middle section 
 between its two bases. 
 
 THEOKEM X. 
 
 Two triangular pyramids of equal bases and alti- 
 tudes arc equivalent. 
 
 Let the two pyramids 
 have their bases in the 
 same plane, and DC equal 
 to BC. Conceive a plane 
 to cut the two solids 
 parallel to the plane of 
 their bases, making the 
 triangular sections FGH 
 
 and LMN. Now, since the vertices R and S are by 
 hypothesis equidistant above the plane of the bases, a 
 third plane may pass through these two points par- 
 allel with the other two planes (Cor. 2, Theo. II). 
 Then the lines HL and RS, being the intersections 
 of two parallel planes by a third plane CRS, will 
 be parallel (Def. 4, Sec. XIII) ; hence, the triangles 
 CRS, CHL, are similar (Cor., Theo. V, B. II). In 
 the same manner it may be shown that the triangles 
 CRB, HRG, are similar ; also, CSD and LSM. 
 
 Now, CR : HR : : CB : HG. 
 
 Also, CS : LS : : CD : LM. 
 
 But by similarity of CRS and CHL, we have 
 
 CR : HR : : CS : LS. 
 Hence, by equality of ratios (Def. 0, Sec. X), 
 
 CB : HG : : CD : LM. 
 But CB is by hypothesis equal to CD ; therefore, HG 
 is equal to LM. In the same way it may be shown 
 Evans Geomelrrj. — 8 
 
90 GEOMETEY. 
 
 that the other sides of the triangle FHG are respect- 
 ively equal to the other sides of the triangle LMN ; 
 and these triangles are consequently equal throughout. 
 Hence, at equal hights, the sections parallel to the 
 bases are equal ; and the two pyramids may bo con- 
 ceived to be applied to one another so as to coincide 
 at all equal hights successivel} r , from their bases to 
 their vertices. They are, therefore, equivalent. 
 That is, two triangular pyramids, etc. 
 
 THEOKEM XI. 
 
 A triangular pyramid is one-third of a triangular 
 prism of the same base and altitude. 
 
 Let ABCDEF be a triangular a C 
 
 prism. Join AF, BF, and BD. j , 1 
 
 Now, the pyramid BDEF, cut off Umf 
 
 by the plane of the triangle BDF, h /i,l 
 
 is equivalent (Theo. X) to the / ^p'j \\ 
 pyramid FABC, cut off by the |||f ' ' W 
 plane of the triangle ABF ; for ^Hh^ 
 they have equal bases, DEF and jjj 
 
 ABC (Def. 1, Sec. XV), and the 
 same altitude, namely the altitude of the prism. But 
 the pyramid FABC is equivalent to the third pyramid 
 BADF; for they have equal bases ADF and FCA 
 (Cor. 1, Theo. XIII, B. I), and the same altitude, 
 namely the perpendicular distance of their common 
 vertex B above the plane of their bases ADFC. 
 Hence, the pyramid BDEF is one-third of the prism 
 ABCDEF. 
 
 That is, a triangular pyramid is one-third, etc. 
 
 Cor. 1. Hence, the solidity of a triangular pyramid 
 
BOOK III. 
 
 91 
 
 is equal to one-third the product of its base by its 
 altitude (Thco. VII). 
 
 Cor. 2. The solidity of any pyramid ivhatever, is 
 equal to one-third the product of its base by its alti- 
 tude; for, by dividing its base into triangles, and 
 passing planes through the lines of division and the 
 vertex, the pyramid can be divided into a number of 
 triangular pyramids ; and the sum of their solidities 
 will be equal to one-third the product of the sum of 
 their bases by their common altitude. 
 
 THEOREM XII. 
 
 Tlie convex surface of a cone is equal to half the 
 product of the circumference of the bast by the sla?it 
 hight; and its solidity is equal to one-third the product 
 of the area of the base by the altitude. 
 
 Let ABC be a cone, having a 
 regular pyramid inscribed in it. 
 If the number of sides of the poly- 
 gon constituting the base of the 
 pyramid be indefinitely increased, 
 its perimeter will ultimately coin- 
 cide with the circumference of the 
 base of the cone. Then, the slant 
 hight of the pyramid will be equal 
 to the slant hight of the cone, and the convex surface 
 and solidity of the pyramid to the convex surface and 
 solidity of the cone. But the convex surface of the 
 pyramid will be equal to half the product of the peri- 
 meter of the base by the slant hight (Theo. IX) ; and 
 the solidity of the pyramid will be equal to one-third 
 
92 
 
 GEOMETRY. 
 
 the product of the area of the base by the altitude 
 (Cor. 2, Tiieo. XI). 
 
 Therefore, the convex surfaces of a cone, etc. 
 
 Schol. In the same manner, it may be shown that 
 the convex surface of a frustum of a cone is equal to 
 the product of the slant hight into the circumference 
 of a middle section between the two bases (Schol. 2, 
 Theo. IX). And as this will hold true however small 
 the upper base may be, it will hold true of the cone 
 itself, which may be treated as a frustum whose upper 
 base is nothing. 
 
 SEC. XVII.— THE SPHERE. 
 
 * 
 
 DEFINITIONS. 
 
 1. A sphere is a solid which may be described by 
 the revolution of a semicircle around its diameter as 
 a fixed axis. 
 
 The semi-circumference de- 
 scribes the convex surface. The 
 center is the middle point of 
 the axis. The radius is a 
 straight line, from the center 
 to any point of the surface ; 
 and it is equal to the radius 
 of the semicircle. A diameter is a double radius. 
 
 Schol. As the semicircle ADB revolves about the 
 axis AB, a perpendicular, as DC, let fall on the axis 
 from any point in the circumference, will describe a 
 circle. 
 
 2. A great circle on the sphere is one whose 
 
BOOK III. 
 
 93 
 
 plane passes through the center. Its radius is the 
 same as the radius of the sphere. Its circumference 
 is also called the circumference of the sphere. Any 
 other circle on the sphere is called a small circle. 
 
 3. A segment of a sphere is any portion, as ADEF, 
 cut off from the solid by a plane passing through it. 
 
 4. A frustum of a cone is said to be inscribed in a 
 sphere, when the circumferences of its bases lie in the 
 surface of the sphere. 
 
 THEOREM XIII. 
 
 If a frustum of a cone be inscribed in a sphere, its 
 convex surface will be equal to the altitude of the frus- 
 tum multiplied by the circumference of a circle, whose 
 radius is a pcrpeyidicular from the center of the sphere 
 to the slant hight of the frustum. 
 
 Let CD and GH be both perpen- 
 dicular to the axis AB. Then, as 
 the semicircle revolves about the 
 axis, describing the sphere, the trape- 
 zoid DCGH will describe a frustum 
 of a cone, which will be inscribed in 
 the sphere. From the center K let 
 fall the perpendicular KE, on the 
 chord CG ; then E will be the middle point of CG 
 (Theo. XXV, B. I). Draw EF perpendicular to AB, 
 and CI perpendicular to GH. 
 
 Now, since the triangles GCI, KEF, have the sides 
 of the one respectively perpendicular to the sides of 
 the other, they are similar (Theo. VI, B. II). 
 Hence, GC : CI : : KE : EF. 
 
94 GEOMETKY. 
 
 Multiplying an extreme and a mean equally (Def. 5, 
 Sec. X), 
 
 GC : CI : : 2 KEx 3.14159 : 2EFX3.14159. 
 The first term of this proportion is the slant hight of 
 the frustum ; the second is its altitude ; the third is 
 the circumference of a circle whose radius is KE (Cor. 
 1, Theo. XV, B. II) ; the fourth is the circumference 
 of a circle 'whose radius is EF. Therefore, multi- 
 plying extremes and means, Ave have the product of 
 the slant hight into the circumference of a middle 
 section of the frustum, equal to the product of its 
 altitude into the circumference of a circle whose 
 radius is the perpendicular from the center to the 
 slant hight. But the former product is equal to the 
 convex surface of the frustum (Schol., Theo. XII) ; 
 consequently, the latter product is equal to the same. 
 That is, if a frustum of a cone be inscribed, etc. 
 
 Schol. If the chord CG be one side of a regular 
 inscribed polygon, it is evident that KE will be its 
 apothegm (Cor. 2, Theo. XXVIII, B. I). 
 
 THEOKEM XIV. 
 
 TJie surface of a sphere is equal to the product of 
 its diameter by its circumference. 
 
 Let ABCDEF be a semicircle, 
 having the half of a regular poly- 
 gon inscribed in it. As the semi- 
 circle revolves about the axis AF, 
 describing the sphere, each of the 
 trapezoids GBCII, IICDI, etc., will 
 describe a frustum of a cone, which 
 will be inscribed in the sphere. The convex surface 
 
BOOK III. 95 
 
 of each of these frustums will be equal to its altitude 
 multiplied by the circumference of a circle whose 
 radius is the apothegm of the polygon (Schol., Theo. 
 XIII). Therefore, the sum of their convex surfaces 
 will be equal to the circumference of such a circle 
 multiplied by the sum of the altitudes Gil, HI, etc., 
 that is, multiplied by AF, the diameter of the sphere. 
 Now if the number of sides of the semi-polygon be 
 indefinitely increased, . its perimeter will ultimately 
 coincide with the semi-circumference, and its apo- 
 thegm with the radius of the sphere. Then, the sum 
 of the convex surfaces of the frustums will be equal 
 to the surface of the sphere, and the circumference 
 of which the apothegm is radius will be the circum- 
 ference of the sphere (Def. 2). 
 
 Therefore, the surface of a sphere is equal, etc. 
 
 Cor. 1. Since the circumference is equal to 3.14159 
 XD (Cor. 1, Theo. XV, B. II), it follows that the sur- 
 face is equal to 3.141 59 XD 2 . 
 
 Cor. 2. The convex surface of any segment of a 
 sphere, as that described by the arc ABC, is equal to 
 its altitude AH multiplied by the circumference of the 
 sphere. 
 
 THEOEEM XV. 
 
 Tlie solidity of a sphere is equal to its surface mul- 
 tiplied by one-sixth of its diameter. 
 
 The sphere may be considered as made up of in- 
 finitely small pyramids, whose bases together form the 
 surface of the sphere, and whose common vertex is at 
 the center. Now, the solidity of each of these pyra- 
 mids will be equal (Cor. 2, Theo. XI) to the product 
 
96 GEOMETRY. 
 
 of its base by one-third of its altitude, that is, by 
 one third the radius of the sphere. Hence, the sum 
 of their solidities will be equal to the sum of their 
 bases multiplied by one-third of the radius, or one- 
 sixth of the diameter. 
 
 That is, the solidity of a sphere is equal, etc. 
 
 Cor. Since the surface of a sphere is equal to 
 8.141 59 XD 2 , it follows that its solidity is equal to 
 
 D 
 
 3.14159xD 2 X— , which, by reduction, becomes .5236X 
 6 
 
 D 3 . That is, the solidity of a sphere is equal to the 
 
 cube of its diameter multiplied by .5286. 
 
SUPPLEMENT. 
 
 SEC. XVIII.— MISCELLANEOUS EXAMPLES. 
 
 1. What is the side of a square inscribed in a circle 
 whose diameter is 5 feet? 
 
 2. Prove that the side of a square is to its diag- 
 onal as 1 to the square root of 2. 
 
 3. Prove that the area of a square described about 
 a circle is double the area of a square inscribed in 
 the same circle. 
 
 4. "What is the altitude of an equilateral triangle 
 whose side is 12 feet ? 
 
 5. If the altitude of a cone is 9 feet, and the di- 
 ameter of its base 5 feet, 4 inches, what is its convex 
 surface ? Its solidity ? 
 
 6. What is the surface of a sphere avIiosc diameter 
 is 7 feet? Its solidity? 
 
 7. If the altitude of a prism is 5 feet, and the area 
 of its base 18 square feet, what is its solidity? 
 
 8. What is the solidity of a cube whose edge is G 
 feet? Its surface? 
 
 9. What is the lateral surface of a regular pyramid 
 whose slant hiiiht is 15 feet, and the base 80 feet 
 square? 
 
 Evans Geometry. — 9 (27) 
 
98 GEOMETRY. ■ 
 
 10. What is the solidity of a pyramid whose alti- 
 tude is 72, d the sides of whose base are 24, 24, 
 and 40 ? 
 
 11. If the altitude of a cylinder is 7, and the 
 radius of its base 2, what is its whole surface ? Its 
 solidity ? 
 
 12. Prove that a circle described on the hypotenuse 
 of a right-angled triangle as diameter, is equivalent to 
 the sum of the circles described on the other two 
 sides. 
 
 13. Prove that if a perpendicular be let fall from 
 the right angle of a right-angled triangle on the 
 
 © © © © © 
 
 hypotenuse, either of the two sides containing the 
 right angle will be a mean proportional between the 
 hypotenuse and the adjacent part of the hypotenuse. 
 
 14. If from the top of a mountain 2.006 miles 
 high a vessel can be seen in the horizon 126 miles 
 off, what is the diameter of the earth ? (Theo. XIII, 
 B. II). 
 
 15. If the earth's diameter be 7912 miles, what is 
 its circumference ? Its surface ? Its solidity ? 
 
 16. If the circumference of the moon be 6783 miles, 
 what is its diameter? 
 
 17. Prove that the solidities, S and s, of any two 
 spheres are to each other as the cubes of their diame- 
 ters, D and d. 
 
 18. The diameter of the sun is 112 times that of the 
 earth ; what arc the relative solidities of the two bodies ? 
 
SUPPLEMENT. 99 
 
 SEC. XIX.— APPLICATIONS OF ALGEBRA. 
 
 For those acquainted with the elements of hoth branches, 
 a few examples of the application of Algebra to Geometry 
 are subjoined. The method is specially adapted to the 
 solution of that class of problems where certain parts of a 
 figure are aiven, from which to determine others ; but it 
 may also be used for abbreviating long demonstrations of 
 theorems. 
 
 1. If a perpendicular be let fall from the vertex of 
 a triazigle upon its base, the sum of the parts of thu 
 base is to the sum of the other two sides as the dif- 
 ference of the latter is to the difference of the former. 
 
 Proof. Let BD be the per- 
 pendicular. Now, BC 2 — CD- = 
 BD 2 (Cor. 2, Theo. XIX, B. I). / 
 
 Also, AB 2 — AD 2 =BD 2 . / 
 
 Therefore, A d ~~c 
 
 BC 2 — CD 2 =AB 2 — AD 2 . 
 
 Transposing, BC 2 — AB 2 =CD 2 — AD 2 . 
 
 Factoring, 
 
 (BC-fAB) (BC— AB) = (CD+AD) (CD— AD). 
 Hence, resolving into a proportion, Ave have 
 
 CD-fAD : CB+AB : : CB— AB : CD— AD. 
 
 2. The square of the side opposite to any acute 
 angle in a triangle is equal to the sum of the squares 
 of the other sides, minus twice the product of the 
 base by the distance from the acute angle to the foot 
 of the perpendicular let fall from the vertex on the 
 base, or the base produced. 
 
100 
 
 GEOMETRY. 
 
 Proof. For the case where 
 
 the perpendicular falls on the 
 
 base produced, take the annexed 
 
 figure ; for the other case, take 
 
 the preceding one. 
 
 y y a 
 
 Now, AB 2 =BD4-AD 2 (Theo. XIX, B. I). 
 
 But, AD=CD— AC (or, AC— CD). 
 
 Hence, AD- = CD--{-AC 2 — 2 CDx AC. 
 
 Substituting this in the first equation, we have 
 
 AB 2 =BD'-+ CD-+ AC 2 — 2 CD X AC. 
 
 But BD 2 +CD 2 =BC 2 ; whence, 
 
 AB-=BC-+AC 2 — 2 CDX AC. 
 
 3. The square of the side opposite an obtuse angle 
 in a triangle is equal to the sum of the squares of 
 the other sides, plus twice the product of the base by 
 the distance from the obtuse angle to the foot of the 
 perpendicular, let fall from the vertex on the base 
 produced. 
 
 For proof use the last figure. 
 
 4. Given the base and the sum of the two other 
 sides of a right-angled triangle, to find the hypote- 
 nuse and perpendicular. 
 
 Solution. Represent the base by b and the sum 
 of the other sides by s. Also, represent the perpen- 
 dicular by x, then the hypotenuse Avill be s — x. 
 
 Now, V+x 2 ={8—xf (Theo. XIX, B. I). 
 
 Hence, by reduction, we have, 
 
 O TO 
 
 8~ — 0' 
 
 X = - 
 
 2s 
 
SUPPLEMENT. 
 
 101 
 
 5. Given the base and altitude of a triangle, to find 
 the side of the inscribed square. 
 
 Solution. Represent the 
 base AC by b, the altitude 
 BH by h, and the side of the 
 inscribed square by x. Then 
 BI will be represented by 
 h — x. 
 
 Now, by similarity of triangles, 
 
 h : b : : Ji — x : x. 
 Hence, hx=bh — bx. 
 
 bh 
 
 And, 
 
 x = 
 
 b+h 
 
 6. If two parallel chords in a circle be 96 and 
 (30, and the distance between them 26, what is the 
 diameter ? 
 
 Solution. Draw the di- 
 ameter AF perpendicular to 
 the chords, and it will bisect j/ 
 them (Theo. XXV, B. I). 
 Represent AH by x and GF 
 by y. Here, since we have 
 introduced two unknown quan- 
 tities, we must have two independent equations. 
 
 Now, x(26+y)=(SQ) 2 (Theo. XII, B. II). 
 And 04-26)7/=(48) 2 . 
 
 Finding x and y from these equations, and adding 
 their sum to 26, we shall have the diameter =100. 
 
 7. Given the diagonal and perimeter of a rectangle, 
 to find the sides. 
 
102 GEOMETRY. 
 
 8. In a triangle, given the base b, the altitude h, 
 and the ratio of the other two sides, as m to n, to find 
 the sides. 
 
 Represent the sides by mx and nx. 
 
 9. Given the base and altitude of any triangle 
 and the difference of the other two sides, to find 
 the sides. 
 
 Let x stand for the half sum of the two sides, 
 and d for their half difference, then the sides will be 
 represented by x-\-d and x — d. 
 
 10. Given the base and the other two sides of a 
 triangle, to find the altitude. 
 
 11. Given the area of a rectangle inscribed in a 
 given triangle, to find the sides of the rectangle. 
 
 12. Given the radius of a circle, to find the side 
 of the inscribed equilateral triangle. 
 
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THE MOST POPULAR 
 
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 The attention of Parents, Teachers, and School-Officers 
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 Superintendents of Public Instruction of nearly all the 
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 great and abiding popularity of the Eclectic Series is 
 founded upon their rare intrinsic merit, and their admir- 
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 OHIO. 
 
 From Hon. Anson" Smtth, former Slate Supt. of Public Inst, Ohio. 
 
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 the very best works of the kind ; but the New are certainly a 
 decided improvement upon the Old. I know of no others 
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2 POPULAR SCHOOL-BOOKS. 
 
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 From Hox. Miles J. Fletcher, former Stale Supt. of Public Inst, Ind. 
 
 The public sentiment, as expressed in Indiana by the al- 
 most universal use of the Eclectic Educational Series 
 of School Books, embracing McGuffey's New Readers and 
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 cheapness. Miles J. Fletcher. 
 
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 From Hox. N. Batemax, State Supt. of Public Inst., Illinois. 
 
 No series of books has ever obtained so many voices of 
 approval from teachers as McGuffey's Eclectic Readers. 
 Certainly no other series has been so popular throughout the 
 West. We unhesitatingly say that we know of no letter books, 
 and should not take the trouble to look for any. The print- 
 ing is beautiful, the paper very fine, and the binding good, 
 and McGuffey's Readers are proverbially cheap. 
 
 Ray's Arithmetics have deservedly shared in the popu- 
 larity of the Eclectic Series. The Higher Arithmetic is better 
 than any other that wc know to be used in this country. 
 Ray's Algebras are clear, full, and comprehensive. 
 
POPULAR SCHOOL-BOOKS. 8 
 
 If I were to select from the different Series of School- 
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 Readers, and Kay's Series of Arithmetics and Algebras, as 
 combining more excellences and fewer defects than any other 
 series with which I am acquainted. The preference herein 
 expressed is the result of actual use, in my own schools, of 
 some live or six different Series of Readers, Arithmetics, and 
 Algebras, as well as a diligent comparison of all the leading 
 series which have been published in the last twenty years. 
 
 Newton Bateman. 
 
 We heartily concur in the foregoing recommendation. 
 [Signed.] C. Goudy, 
 
 Member of the Illinois State Board of Education, ami 
 Chairman of Committee on Text-Books iu the Normal University. 
 
 S. W. MOULTON, 
 
 President of the Illinois State Board of Education. 
 
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 Member of the Illinois State Board of Education, 
 
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 President of the Illinois State Normal University. 
 
 IOWA. 
 
 Fro)n Hon. Oran Faville, former State Supt. of Public Inst., Iowa. 
 
 Having recently re-examined the Eclectic Educational 
 Series of Common School Books, I am fully confirmed in 
 the opinion that they are the best Series, on the whole, now 
 in use in the West. Their remarkable popularity, and the 
 continued attachment manifested for them by practical Edu- 
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 Without specifying further, I will say that McGuffey's 
 New Eclectic Series of Readers, Speller, and Primary 
 School Charts, Pinneo's Series of Grammars, and Ray's 
 Series of Arithmetics and Algebras, are tin surpassed by 
 any similar Series with which I am acquainted. I therefore 
 recommend their continued use in our State. 
 
 Oran Faville. 
 
4 POPULAR SCHOOL-BOOKS. 
 
 From Hon. H. A. Wiltse, former State Supt. of Public Inst., Iowa. 
 
 I have long been acquainted with the Eclectic Educa- 
 tional Series, embracing McGuffey's New Readers, 
 Speakers, and Speller, Ray's Arithmetics and Alge- 
 bras, and Pinneo's Grammars. I have seen this Series 
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 through years, of comparing its results, and know that it is 
 the best Series published. The Schools that have used McGuf- 
 fey's Speller, Readers, and Speakers, are, under equal instruc- 
 tion, in advance of Schools where these books have not been 
 used. H. A. Wiltse. 
 
 WISCONSIN. 
 
 From Hon. J. L. Pickard, former State Supt. of Public Inst., Wis.; 
 at present, Supt. of Chicago Public Schools. 
 
 The books I have recommended below, [McGuffey's New 
 Readers, Ray's Arithmetics, Pinneo's Grammar, and White's 
 Class-Book of Geography,] are such as commend themselves 
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 From Edward Olney, Professor of Mathematics in the State 
 University of Michigan. 
 
 University of Michigan, July 2d, 1866. 
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 better. I have a tolerable familiarity with the various Al- 
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 I think it not only has no superior but no equal published in 
 our country ; and hence I use it till I find a better. 
 
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POPULAR SCHOOL-BOOKS. 5 
 
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 high, and in their low price and beautiful printing and bind- 
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 MIS30UEL 
 
 From Hon. James H. Robinson, late State Supt. of Public Inst., Mo. 
 After careful examination of their intrinsic and compara- 
 tive merits, I cordially recommend for use in the Public 
 Schools of Missouri, the following Text-Books : 
 
 McGuffey's Primary School Charts, 
 McGuffey's New Eclectic Speller, 
 McGuffey's New Eclectic Readers, 
 Kidd's Elocution and Vocal Culture, 
 
 Ray's Series of Arithmetics, 
 Ray's Series of Algeukas, 
 Ray's Plane and Solid Geometry, 
 Pinneo's Series of Grammars. 
 
 No other similar books have been so largely used, and are 
 so generally popular as Ray's Mathematics and McGuffey's 
 Readers, and I believe them to be better adapted to the 
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 I recommend McGuffey's New Eclectic Series of 
 Readers, Speakers, and Speller, and Ray's Series of 
 Arithmetics and Algebras to the favorable consideration 
 of the Teachers of our Public Schools. These works possess 
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 of the State generally. 
 
 I have also, after careful examination, concluded to rec- 
 ommend Pinneo's Series of Grammars. I have endeav- 
 ored to examine the most popular works on the subject of 
 
C POPULAR SCHOOL-EOOKS. 
 
 Grammar, as a teacher rather than as a critic, and, in sd 
 doing, have been compelled to give my preference to Pinneo's. 
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 tical works. Wm. R. Griffith. 
 
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 It is needless to call the attention of experienced teachers 
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 From Hon. Wm. R. White, State Supt. of Public Inst., West Va. 
 
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 2\Apr'6lG? 
 RETC'D Cd 
 
 
 ftPR 1 2 1961 
 
 
 
 LD 21-50m-8,'33 
 
QA.*53 
 
 UNIVERSITY OF CALIFORNIA LIBRARY 
 
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