AN ELEMENTARY TREATISE ON PLANE AND SOLID GEOMETRY. BY BENJAMIN PEIRCE, A. M., PERKINS PROFESSOR OF ASTRONOMP AND ATHEIMATICS IN HiARVARD UNIVERSITY. STEREOTYPE EDITION. BOSTON AND CAMBRIDGE: JAMES MUNROE AND COMPANY. 1 8 5 5. Entered accord(ing to Act of Congress, In the year 1837, by JAMES DMUNROE AND COMIPANY, in the Clerk's office of the District Court for the District of Massachusett. PREFA C E. TI-IE use of infinztely small quantities, which was" first introduced into the higher departments of Mathematics, has been gradually creeping downwards, and elementary writers are rapidly becoming reconciled to it. But at the same time, the uncompromising advocates of the ancient rigor of demonstration have, by their attacks, induced some mathematicians to waste much time in disguising the principles of the Differential Calculus under a form of words, in which the term "infinitely small" does not occur. The value of this labor may be duly estimated from the inconsistency of one, who has ostensibly discarded the infinitesimal doctrine from his theory of the Calculus, and introduced it into his treatise of Geometry. With all its boasted rigor, the ancient Geometry can indeed lead to no result more accurate, none more to be depended upon, than those of the infinitesimal theory; and I doubt if any well constituted mind, well constituted at least for mathematical investigations, ever reposes with any more confidence upon the one than upon the other. If there were iv PREFACE. any error involved in the latter theory, it must not only be infinitely small, but must remain infinitely small after all the magnifying processes to which it could possibly be subjected. But there is no error; for, if we suppose that there be an error which we may represent by.1, since the aggregate of all the quantities neglected in arriving at the result is infinitely small, that is, as small as we choose, we may choose it to be smaller than.1; and, therefore, the error./ is greater than the greatest possible error which could be obtained, a manifest absurdity, but one which cannot be avoided as long as S. is any thing. The term direction is introduced into this treatise without being defined; but it is regarded as a simple idea, and to be as incapable of definition as length, breadth, and thickness; and this innovation will probably be pardoned, when it is seen how much it contributes to the brevity and simplicity of demonstration, which I have everywhere studied. BENJAMIN PEIRCE CONTENTS. [The figures in parentheses refer to the articles.] Page. EXPLANATION OF SIGNS AND OF SOME USEFUL PROPOSITIONS IN THE DOCTRINE OF PROPORTIONS)... XV PLANE GEOMETRY. CHAPTER I. GENERAL REMARKS AND DEFINITIONS,. 3 Geometry (1), point (2), line (4), surface (5), solid (6),.. 3 CHAPTER II. THE POINT,.... 4 The position of a point (8), its direction (9), its distance (10),. 4 CHAPTER II1. THE STRAIGHT LINE, ~... 5 The direction of a line (11),... 5 Straight line (12), broken line (13), curved line (13), 5 Plane (14),.......... 5 Direction of a point of a straight line from preceding point (15), 5 Position of a straight line (16),.... 6 Shortest way from one point to another (18),... 6 CHAPTER IV. TIlE ANGLE,. a... 6 Angle, its vertex, sides (19),... 6 Right angle, perpendicular (20),. 7 Acute and obtuse angle (21),.. 7 a1 * vi CONTENTS. Complemnent and supplement of angle (22), 7 Vertical and adjacent angles (23, 24),. 7 Sum of angles about a point (25, 2(i),. 8 CHAPTER V. PARALLEL LINES,... 9 Definition of parallel lines (27),.. 9 Parallel lines cannot meet (28),. 9 Angles which have their sides parallel (29).. 9 External-internal, alternate-internal angles (30), 9 Interior angles on the same side (32),.... 10'Cases of parallel lines (31, 33, 35, 36),... 10 Parallel lines perpendicular to a third line (34), 10 Lines parallel to a third line (36),.. 11 CHAPTER VI. PERPENDICULAR AND OBLIQUE LINES,.. 11 Only one perpendicular from a point to a line (37),... 11 Oblique lines drawn from a point to a line (38, 39, 41),.. 11 Shortest distance from a point to a line (39),.... 12 Sum of two lines included by two other lines (40),... 12 Distances of a point from the extremities of a line (42),.. 13 CHAPTER VII. SIDES AND ANGLES OF POLYGONS,.. 14 Plane figure, polygon, and its perimeter (43); triangle, quadrilateral, pentagon, hexagon (44); equilateral, isosceles, and scalene triangle (45); right triangle and its hypothenuse (46); square, rectangle, parallelogram, rhombus or lozenge, trapezoid (47); diagonal (48),.14 Equilateral and equiangular polygon (49); polygons equilateral and equiangular with respect to each other, and their homologous sides and angles (50),. 15 Three cases of equal triangles (51, 53, 61),. 15 Equal sides and angles of the isosceles triangle (55, 58), 16 Equilateral and equiangular triangle (56, 59). 17 Line drawn from the vertex to the middle of the base of an isosceles triangle (57),...... 17 Greater side of a triangle opposite the greater angle (62),. 18 Case of equal right triangles (64).. 20 Sum of the angles of a triangle (65- 71),. 20 CONTENTS. vii Sum of the interior angles of a polygon (72 - 76),. 21 The diagonal of a parallelogram bisects it (77),.. 2 Opposite sides and angles of a parallelogram (78),.. 23 Parallel lines between two others (79),.. 23 Cases in which a quadrilateral is a parallelogram (80, 81), 23 Distance of parallel lines from each other (82),.. 23 Mutual bisection of the two diagonals of a parallelogram (83), of a rhombus (84),... 24 CHAPTER VIII. TIlE CIRCLE AND THE MEASURE OF ANGLES, o 24 Circumference and centre of circle (85); its radius and diameter (86),.. 24 Comparative magnitude of radii and diameters (87),.. 25 Circle bisected by the diameter (88); semicircumference and semicircle (89); arc and its chord (90),. 25 Chord less than diameter (91),....... 25 Line inscribed in a circle (92); greatest inscribed line (93),. 25 Line meets the circumference in only two points (94), 26 Angle measured by arc (95- 98, 100),. 26 Infinitely small quantities (99),.. 28 Degree, minute, second, quadrant (101- 105),... 28 Inscribed angle, triangle, and figure (106),... 29 Measure of inscribed angle (107- 111),... 30 Arcs and their chords (112-115),... 31 Radius perpendicular to middle of chord (116, 117,). 32 Secant (118); tangent, point of contact, tangent circles (118); polygon circumscribed about circle (118),... 33 Direction of tangent and curve at point of contact (119), 33 Tangent perpendicular to radius (120),. 34 Measure of angle formed by tangent and chord (121),. 34 Measure of angle formed by two secants, two tangents, or a tangent and secant (122); of angle formed by two chords (123), 34 Arcs intercepted by parallels (124),...... 35 Line joining the centres of circles which cut each other (126); or which touch each other (127), 36 CHAPTER IX. PROBLEMS RELATING TO TIHE FIRST EIGIIT CHAPTERS, 37 To find the position of a point at given distances from given points (128-131),. 37 vii Viii CONTENTS. To bisect a line (132),... 38 To draw a perpendicular to a line (133, 134),... 38 To make an arc or angle equal to a given are or angle (135, 136), 38 To bisect an arc or angle (137, 138),. 39 To draw a line parallel to a given line (139),. 40 To find the third angle of a triangle (140),. 40 To construct a triangle (141 - 144),.... 40 To construct a right triangle (145)... 41 To construct a parallelogram (146, 147),.... 41 To find the centre of an arc or circle (148, 149),. 41 To draw a tangent to a circle (150, 151),.. 42 To inscribe a circle in a triangle (152, 153.... 42 To describe a segment capable of containing a given angle (154, 155),..... 43 To find the common measure of two lines (156); or of two arcs (157)..... 44 CHAPTER X. PROPORTIONAL LINES,... 45 Lines divided into equal parts (158),. 45 To divide a line into equal parts (159),.45 A line parallel to one side of a triangle divides the other two proportionally (160 - 162); and the converse (163),. 46 To divide a line into parts proportional to given lines (164, 165), 47 To find a fourth proportional to three lines (165, 166),. 48 To divide one side of a triangle into two parts proportional to the other two sides (167),.48 To draw a line through a point in an angle so that it may be divided in a given ratio by the point (168, 169),... 49 CHAPTER XI. SIMILAR POLYGONS,... 49 Similar polygons and their homologous sides; similar arcs (170), 49 Altitude of parallelogram, of triangle, of trapezoid (171),. 50 Cases of similar triangles (172 - 180),.. 50 Intersections of lines drawn through the vertex of a triangle with lines parallel to the base (181, 182),.. 52 Division of the right triangle into similar triangles by a perpendicular to the hypothenuse (183),. 53 Leg of a right triangle is a mean proportional between hypothenuse and adjacent segment (184, 187),. 53 CONTENTS. iX Perpendicular to the diameter is a mean proportional between its adjacent segments (185, 186),. 53 To find a mean proportional (188),. 54 Chords and secants of a circle drawn through a point are divided in reciprocal proportion (189, 190),. 54 Tangent drawn to a circle through a point is a mean proportional between the parts of the secant drawn through it (191),. 55 To divide a line in extreme and mean ratio (192),... 55 Case of similar polygons (193),. 5 To construct a polygon similar to a given one (194), 56 Case in which similar polygons are equal (195),. 57 Similar polygons are composed of similar triangles (196), 57 Ratio of the perimeters of similar polygons (197),... 58 Ratio of the altitudes of similar triangles, parallelograms, and trapezoids (198, 1.99); of their perimeters (200),... 58 CHAPTER XII. REGULAR POLYGONS,.. 59 Regular polygon defined (201),.. 59 Case of regular polygon (202),.. 59 Infinitely small arc same as its chord (203),... 59 Circle a regular polygon of an infinite number of sides. (204),. 60 infinitely small quantities which are neglected are infinitely smaller than those which are retained (205),. 61 Regular polygons of the same number of'sides are similar (206), 61 Ratio of the perimeters of regular polygons (207, 233),. 62 Circles are similar regular polygons (208),... 62 Regular polygon may be inscribed (209),. 62 To inscribe a regular polygon of double the number of sides of a given inscribed regular polygon (210, 211),. 62 To inscribe a square (212); a polygon of 8, 16, 32, &c. sides (213); a hexagon (214); a polygon of 12, 24, 48, &c. sides (215); an equilateral triangle (216); a decagon (217); a polygon of 20, 40, 80, &c. sides (218); a pentagon (219); a polygon of 15 sides (220); of 30, 60, 120, &c. sides (221); 63 To circumscribe a circle about a regular polygon (222), 65 The centre of the regular polygon (223),. 66 To inscribe a regular polygon similar to a given one (227), 66 Sides of a regular polygon equally distant from its centre (228), 67 To inscribe a circle in a polygon (229),.... 67 To circumscribe about a circle a polygon similar to a given inscribed one (230, 231), 67 X CONTENTS. Ratio of the homologous sides of similar regular polygons (232,'233),.68 Ratio of the circumferences of circles (234 - 238),.. 69 Ratio of circumference to diameter or ar (237, 238). 69 CHAPTER XIII. AREAS,.. 70 Equivalent figures, area (2939); unit of surface (240), 70 Ratio of rectangles (241, 244)....... 70 Area of rectanlge (242); of' square (243), 71 Equivalent parallelogramis (245, 246), 71 Area of' parallelogram (247),....... 72 Ratio of parall!ograms (248),.. 72 Ratio of parallelogram and triangle (249),. 72 Equivalent tliangles (2950), 72 Area of triangle (251),.. 72 Ratio of triangles (252),........ 72 Area of trapezoid (253, 255),....... 73 Line joining middle points of sides of trapezoid (254),. 73 Square upon the hypothenuse of' a right triangle equivalent to the sum of squares upon its legs (256),..., 74 Square upon the diagonal of a square (25.8), 75 Ratios between the squares of hypothenuse and legs of right triangle (259, 20),.. 75 To make a square equivalent to sum or difference of squares (2G61 - 2(4),.... 76 To make a square in a given ratio to a given square (265), 77 Ratio of similar triangles (2(i6, 267),. 78 Ratio of similar polygons (268, 269); of regular polygons (270), 78 Ratio of circles (271),.. 79 To make a polygon or circle equivalent to the sum or to the difference of given similar polygons or circles (272- 274), 79 To make a polygon or circle in a given ratio to a similar polygoni or circle (275, 276),.80 Area of circumscribed polygon (277); of regular polygon (278), SO Area of circle (279 - 281,, 85).. 81 Sector, (282),.. 81 Area of sector (283, 285); of segment (284),.. 82 Similar sectors and segments (286),. 83 Ratio of similar sectors (287); of similar segments (288),. 83 To convert a polygon into an equivalent triangle (289),.. 84 To find the quadrature of a parallelogram (290); of a triangle CONTENTS. xi (291); of a circumscribed polygon (292); of a circle (293); of any polygon (294),. 85 To construct a figure similar to a given figure and equivalent to a given figure (295, 296), 86 To construct a parallelogram, equivalent to a given square, and having the sum (297), or the difference (298) of its base and altitude equal to a given line,..... 87 To find the ratio sz of circumference to diameter (301),. CHAPTER XIV. ISOPERIMETRICAL FIGURES, ~ ~. 90 Isoperiinetrical figures, maximum and minimum (302),.. 90 Maximum of isoperimetrical triangles of the same base (303),. 90 Maximum of isoperimetrical polygons of the same number of sides (304, 308),... 91 Maximum of triangles formed with two given sides (305),. 92 Maxilum of polygons formed with sides all given but one (306), 92 Maximum of polygons formed with given sides (307),.. 93 Greatest of isoperimetrical regular polygons (309),. ~. 94 Maximum of isoperimetrical figures (310),... 96 SOLID GEOMETRY. CHAPTER XV. PLANES AND SOLID ANGLES,.. 97 Three points not in the same straight line determine the position of a plane (311, 312),.97 Intersection of' planes (313),....... 97 Angle of planes (314),. 97 Measure of angle of planes (315),...... 98 Perpendicular planes (31),....... 98 Line perpendicular to plane (317),...... 98 Cases in which a line is perpendicular to a plane (318, 324, 326, 331),..........99 Distance of' point finom plane (319),.. 99 Oblique lines drawn to plane (320, 321),. 9 Parallel lines perpendicular to plane (322); and converse (323), 100 Parallel lines in the same plane (327),.... 101 Line parallel to plane, parallel planes (328),... 101 Xii CONTENTS. Cases in which a line is parallel to a plane (329, 335),. 101 Cases in which planes are parallel (:31, 332),.. 102 Lines between parallel planes (333, 336),... 103 Intersection of two parallel planes by a third plane (334),. 103 Solid angle (337).. 104 Sum of two plane angles of a solid angle greater than the third (338),..104 Sum of plane angles of a solid angle (339),.. 105 Solid angles formed by equal plane angles (340),... 106 CHAPTER XVI. SURFACE AND SOLIDITY OF SOLIDS,.. 107 Equivalent solids (341 ); lamina (341),.... 107 Case in which solids are equivalent (342),.... 107 Polyedron, and its faces and sides (343); tetraedron, hexaedroni octaedron, dodecaedron, icosaedron (344); prism, its base, its convex surface, its altitude (345); right prism (34(;); triangular, quadrangular, pentagonal, hexagonal, &c. prism (347); cylinder, its axis, right cylinder (348); generation of right cylinder (349); parallelopiped, right parallelopiped (350); cube, unit of solidity (351); volume, solidity (352),. 108 Area of convex surface of right prism or cylinder (:35:), 109 Section of prism or cylinder by plane parallel to base (354),. 110 Equivalent prisms and cylinders (355, 356, 367),. 110 Ratio of parallelopipeds (357, 363, 364),.. 111 Solidity of parallelopiped (358, 359, 361, 362;) of cube (360),. 112 Solidity of prism and cylinder (365, 366, 369),.. 113 Ratio of prisms or cylinders (368).114 Pyramid, its vertex, its base, its convex surface, its altitude (370); triangular, quadrangular, &c. pyramid (371); regular pyramid and its axis (372); cone, its axis, right cone (373); generation of right cone (374),. 114 Area of convex surface of regular pyramid (375); of right cone (376),......115 Section of' pyramid or cone by plane parallel to base (377-380), 116 Equivalent pyramids or cones(381, 388),. 17 Ratio of pyramid or cone to prism or cylinder (382, 389),.. 117 Solidity of pyramid or cone (383, 385, 390),. 118 Ratio of pyramids and cones (386, 387),..... lS Truncated prism and its base (391),..... 119 Solidily of truncated triangular prism (392),... 119 Frustum of pyramid or cone, its convex surface and bases (393); generation of frustum of right cone (304),.... 120 CONT ENT S. xiii Area of convex surface of frustum of regular pyramid or of' right cone (395- 397), ~. 120 Area of surface described by revolving line (398),..121 Solidity of frustum of pyramid or cone (400, 401), 122 Solidity of polyedron (402),...124 CHAPTER XVII. SIMILAR SOLIDS,.. 124 Similar polyedrons (403); ratio of their sides (404), of their faces (405); of the bases and of the convex surfaces of similar prisms, pyramids, cylinders or cones (406-410),...124 Ratio of similar pyramids, prisms, cylinders or cones (411- 413), 124 Ratio of similar polyedrons (414), 126 CHAPTER XVIII. THE SPHERE,... 126 Sphere, its centre (415); its generation (416); its radius and diameter (417, 418), 126 Section of sphere (419); great and small circles (420-422), 127 Pole of circle (423); distance of pole from circumference (424, 425),...27 Arcs drawn on surface of sphere (426),. 12S Case in which a point is a pole of a given circle (427),. 128 Intersection of great circles (428),... 128 Bisection of sphere (429),.. 128 Spherical triangle, its sides and angles (430); when right, isosceles, equilateral (431); spherical polygon (432); ]unary surface and spherical wedge (433); spherical pyramid (434); plane tangent to sphere (435), spherical segment and zone, their bases and altitude (436); spherical sector (437),. 129 One side of spherical triangle less than sum of other two (433), 130 Sun of sides of spherical polygon (439, 440),.. 130 Measure of angle of great circles (441, 442),.... 131 Sides and angles of polar triangles (443 - 446),.. 131 Sum of angles of spherical triangle (447),.... 133 Each angle of spherical triangle greater than difference between two right angles and sum of the other two angles (448),. 133 Case of equal or symmetrical spherical triangles (449 - 452, 458), 134 Equal sides and angles of isosceles or equilateral spherical triangle (453 - 456),. 135 Opposite sides and angles of spherical triangle (460 - 462),. 136 b Xiv CON TENTS. Degree of surface (4(63), 464),.... 139 Measure of lunary surface (465).. 139 Symmetrical triangles are equivalent (466, 467),... 140 Supplementary triangles (468),. 141 Surface of spherical triangle (469); of spherical polygon (470), 141 Area of surface generated by revolution of polygon (471, 472), 143 Surface of sphere and of zone (473-475, 484),. 144 Ratio of surfaces of spheres (476); of zones (477),... 144 Solidity of sphere (478, 482); of spherical sector or pyramid (479, 484),...145 Ratio of spherical sectors and pyramids (480, 481), 145 Ratio of spheres (482),. 145 Solidity of spherical segment (485, 46),..... 147 CHAPTER XIX. REGULAR POLYEDRONS,.. 147 Regular polyedrons, their number (487, 488),. 147 Number of faces of regular polyedrons (489); tetraedron, octaedron, icosaedron (490); hexaedron or cube (491); dodecaedron (492),. 149 EXPLANATION OF SIGNS, AND OF SOME USEFUL PROPOSITIONS IN THE DOCTRINE OF PROPORTIONS. THE sign + is plus, or added to. Thus /- + B is.J added to B. The sign -is mninus, or less. Thus, 2/ - B is A less B. The sign X is multiplied by. Thus, / X B is 2 multiplied by B; and the period (.) is also the sign of multiplication. The sign - or: is divided by. Thus, -. B or 2: B is A divided by B. The quotient of d divided by B may also be written —. B The sign = is equal to. Thus,.2 -B is 2 equal to B; and the expression in which this sign occurs is called an equation. The sign > is greatesr than. Thus,.1 > B is 2 greater than B. The sign < is less than. Thus, A1 < B is A less than B. J2 indicates the second power of 2,. 3 the third power, &c. A ratio or fraction is the quotient of one quantity divided by another, and is usually written with the sign ( ). Thus the ratio of 2A to B is A: B, or it may just as well be written in the form of a fraction, as-. t term of a ratio is called the an The first term of a ratio is called the antecedent, and xvi EXPLANATION 0' SIGNS) &C. the second the consequenzt. Thus,.3 is the antecedent of the preceding ratio, and B its consequent. The value of a ratio is not altered by multiplying or dividing both its terms by the same number. Thus, A.: B is equal to in X.3': m X B. A proportion is the equation formed by two equal ratios. Thus, if the two ratios A.: B and C: D are equal, the equation f: B- C: D is a proportion, and it may also be written B D The first and last terms of a proportion are called its extremes; and the second and third its mIeans. Thus,.3 and D are the extremes of this proportion, and tS and C its means. Theorem L The prodtuct of the imeans of a propoItion it equal to the product of its extremes. Proof. If the fractions of a proportion J9: B C: D are reduced to a common denominator, they give JX D BX C or, omitting the common denominator, A x D - B X C. This proposition is called the test of proportions. Theorem HI. If four qluantities are such that the product of the first and last ef the-m is equal to the'product of the second tand third, these four qutantities formn a proportion Proof. Let X., B, C, D, be such that q x D B x C. Dividing by B X D we have. X D _Bx C B X D BX- D EXPLANATION OF SIGNS, &C. xvii which, reduced to lower terms, and written in the form of ratios, is Ji: B= C': D. Corollary. The termns of a proportion mnay be transposed in any way, provided the product of the means is retained equal to that of the extremes, and the proportion will not be destroyed. Thus, the preceding proportion gives, by transposition, A: C -B:, B -:D: C, B: D =: C, &c. If both the means of the proportion are of the same magnitude, this mean is called the mean proportional between the extremes. Thus, if A B B B D, B is a mean proportional between d. and D.'Theoren H11. The smeanr proportional between two quantities is the square root of their product. Proof The application of the test to the preceding proportion gives B2 - — X Di the square root of which is B L (A X D). A succession of several equal ratios is called a continued proportion. Thus, d': B C:D E- E: F= &c. is a continued proportion. Theorelm I V. The sum of any number of antecedents of a continued proportion is to the sum of the corresponding consequents as one antecedent is to its consequent. Proof. Denote the common value of the ratios in the above continued proportion by.M, we have be xviii EXPLANATION OF. SIGNS, &C. J~ — J: B — C: D — &c.; whence - B X 31, C D X JM, E - F X Jf, &c. and the sum of these equations is ~ + C+ E+ S&c. =- (B + D + F — &c.) X J; whence + _C + E - &c. j C B.+ D + F +_ &c. B &c. Corollary. The sum of the antecedents of a proportion is to the sumw of its consequents as either antecedent is to its consequent; and the difference of the antecedents is to the difference. of the consequents in the same ratio. Theorem V. The sumn of the antecedents of a proportion is to their difference, as the sumi of the consequents is to their difference. Proof. The proportion:: B — C: D gives, by the preceding proposition, - +- C: B4- +D-= J - C: B - D whence, by transposing the means, +: - C C- B 2 D:B D. Theorem VI. The sumi of the first two terms of a proportion is to the sum of the last two as the first term is to the third, or as the second is to the fourth; and the difference of the frst t1wo terms is to the difference of the last tVwo in the samne ratio; also the stnz of the Sfrst twUo terms is to their difference as the sunm of the last twlo is to their difference. Proof. The proportion.:B =z C: D gives, by transposing the means, EXPLANATION OF SlGNS, &C. Xix d: C: B: D; from which we obtain, by the preceding propositions, +B: C - B C - D =: C B: CB: C-: C D J+ -j- B: - B C +-D: C - D. Two proportions, as J: B- C: D and E: F= G:, may evidently be multiplied together term by term, and the result X E: B X F C X G - D X H is a new proportion. Likewise, a proportion may be multiplied by itself any number of times in succession, and the squares, cubes, fourlth powzers, 4Sc. of the terms form a nlew proportion. Thus, the proportion.: B= — C: D gives 2: B2 - C2: D2 3: B3- C3:D3 3 4: B 4: C4 D4, &c. &c. GEOMETRY. CHAPTER I. GENERAL REMARKS AND DEFINITIONS. 1. Definition. Geometry is the Science of Position and Extension. 2. Definition. A Point has merely position, without any extension. 3. Definition. Extension has three dimensions: Length, Breadth, and Thickness. 4. Definition. A Line has only one dimension, namnely, length. 5. Definition. A Surface has two dimensions; length and breadth. 6. Definition. A Solid has the three dimensions of extension; length, breadth, and thickness. 7. Scholiunm. The boundaries of solids are surfaces, the limits of surfaces are lines, and the extremities of lines are points. The Point, then, on account of its simplicity, deserves our first consideration. 4 PLANE GEOMETRY. [CII. II. ~ tO The Position of a Point; its Direction and Distance. CHAPTER II. THE POINT. 8. The Position of a Point is determined by its Direction and Distance from any known point; in other words, the Elements of its Position are Direction and Distance. Remarks. The Direction of a Point is readily ascertained without any change in the position of the observer, whereas the determination of its distance is often more difficult, as it requires some change of place proportionate to the distance to be measured; thus, the direction of a star is seen at a glance, while the most profound science and the most accurate observations have not enabled the astonoiner to ascertain its distance. 9. The Direction of a Point from the observer may be determined by a reference to some known direction, such as that of the zenith, the pole-star, &c. The method by which one direction may thus be referred to another will be more definitely treated of in a succeeding article. 10. The Distance of a Point from the observer is the length of the shortest line drawn to the point; and it may be determined by a reference to'some known length, such as an inch, a yard, a metre, a mile, &c. CI. III. ~ 15.] THE STRAIGHT LINE. 5 The Direction of a line; the Straight and Curved Lines; the Plane. CHAPTER III. THE STRAIGHT LINE. 11. Definition. The Direction of a Line in any part is the direction of a point at that part from the next preceding point of the line. a. Thus the direction of the line J B (fig. 1) at P is the same as the direction of P from 0. b. In the same way, the direction of the line at P is the same as that of 0 from P, or the opposite direction to the preceding; and, consequently, a line has two different directions exactly opposed to each other, either of which may be assumed as the direction of the line. 12. Definition. A Straight line is one, the direction of which is the same throughout, as.B (fig. 2). 13. Definitions. A Broken or Polygonal Line is one, which is composed of straight lines, as.IBCD (fig. 3). A Curved Line is one, the direction of which is constantly changing, as AB (fig. 1). 14. Definition. A Plane is a surface in which any two points being taken, the straight line joining those points lies wholly in that plane. 15. JxIiom. The direction of any point of a straight line from any preceding point, is the same as the direction of the line itself. Thus the direction of P or B (fig. 2) from J or A. is the same as that of the line J./B. 6 PLANE GEOMIETRY. [CH. IV. ~ 19. Shortest way between two Points. The Angle. 16. Theorem. The position of a straight line is determined by means of two points. For, by the preceding axiom, these two points deter. mine its direction. 17. Theorem. All the points which lie in the same direction from a given point are in the same straight line. Proof. Thus, if P and ]M (fig. 2j) are in the same direction from Z/, the two straight lines A./P and.VJf must likewise, by ~ 15, have the same direction, and must consequently coincide in the same straight line. 18. dxiom. A straight line is the shortest way from one point to another. CHAPTER IV. THE ANGLE. 19. Definitions. An Angle is formed by two lines meeting or crossing each other. The Vertex of the angle is the point where its sides neet. The magnitude of the angle depends solely upon the difference of direction of its sides at the vertex. a. The magnitude of the angle does not depend upon the length of its sides. Thus the angle formed by the two lines.JB and SqC (fig. 4) is not changed by shortening or lengthening either or both of these lines. ci. IV. ~ 23.] TIHE ANGLE. 7 Right and Acute Angles; Complement and Siupplement of' an Angle. b. The method of denoting the angle is by the three letters BtC, the letter A which is at the vertex being placed in the middle; or the letter A may be used by itself, when this can be done without confusion. 20. Definition. When one straight line meets or crosses another, so as to make the tVwo adjacent angles equal, each of these angles is called a Right angle, and the lines are said to be perpendicular to each other. Thus the angles ABC and.BD (fig. 5), being equal, are right angles. 21. Definitioins. An lcutte angle is one less than a right angle, as 2. (fig. 4). An Obttuse angle is one greater than a right angle, as. (fig. 6)j. 22. Definitions. The Complement of an angle is the remainder, after subtracting it firom a right angle. The Sutpplement of an angle is the remainder, after subtracting it from two right angles. 23. Theorem. When one straight line meets or crosses another, the two adjacent angles are supplements of each other, and the vertical angles are equal to each other. Proof. Let /B and CD (fig. 7) be-the two lines. The adjacent angles JPC and A.PD are supplements, for, if the perpendicular P.11 be erected, we have, by insyc-e.. tion, /P C +.PD JIP C + MPD -two right angles. b. In the same way,./PC and BP-C may be proved to be supplements of each other; and therefore the vertical I 8 PLANE GEOIIETRY. [CI. IV. ~ 26. Adjacent and VTertical Angles. Sum of all the Angles about a Point. angles.JPD and BPC must be equal, since they have the same supplement.IPC. In the same way, it may be shown that the vertical angles APC and BPD are equal. c. Corollary. If either of the angles.P C,.PD, BPC, or BPD is a right angle, the other three must also be right angles. d. Scholium. As a straight line has two different directions exactly opposed to each other, it is not unfrequently considered as making an angle with itself equal to two right angles. 24. Corollary. If the two adjacent angles.JPC and.3PD (fig. 8) are supplements of each other, their exterior sides PC and PD must be in the same straight line. 25. Theorem. The sum of all the successive angles 1PB, BPC, CPD, DPE (fig. 9), formed in a plane, on the same side of a straight line SE, is equal to two right angles. Proof. For it is equal to the sum of the two right angles PJ I, JPE, formed by the perpendicular PJ.f. 26. Theorem. Tile sum of all the successive angles JPB, BPC, CPD, DPE, and EPJi (fig. 10), formed in a plane about a point, is equal to four right angles. Proof. For it is equal to the sum of the four right angles JIPN, VPJ31', MI'PP.T, JA'PJM[, formed by the two perpendiculars 3I.MM' and."JV'. CHI. v. ~ 30.] PARALLEL LINES. Parallel Lines cannot meet. Angles are equal whose Sides are lParllel. CHAPTER V. PARALLEL LINES. 2'7. Dejf iition. Parallel Lines are straight lines which have the sabme Direction, as JiB~, CD (fig. 11). 28. Theorem. Parallel lines cannot meet, however far they are produced. Proof. Thus the two lines AJB and CD (fig. 1 1) cannot meet at P; for, if two straight lines are drawn through P, in the same direction, they must coincide and form one and the same straight line. 29. Theorem. Two angles, as el and B (fig. 12)) are equal, when they have their sides parallel and directed the same way from the vertex. Proof. For, as the directions of BD and BF are respectively the same as those of J/C and JE, the difference of direction of BD and BF must be the same as that of.IE/ and RCW; that is, by ~ 19 the angle AJ is equal to the angle B. 30. Theorem. If two parallel lines AB, CD (fig. 13) are cut by a third straight line EF, the externalinternal angles, as E l(B and EhVD, or BJIIF and DXF, are equal, and the alternate-internal angles, as dQJXi and OIJND, or BJ I2 and LJjNC, are also equal. Prooff a. a The external-internal angles are equal, because their sides lhave the same direction. 10 PLANE GEOMETRY. [CH-I. V. ~ 34. Aigles made by a Line cutting l'arallel Lines. b. The alternate-internal angles are equal, as.d23L1Nand J.INjD because.JJIN is, by ~ 23, equal to its vertical angle EMIB, which has just been proved equal to XMAD. 31. Theorem. If two straight lines, lying in the same plane, as dB, CD (fig. 13), are cut by a third, EF, so that the angles EJ3IB and ENVD are equal, or /JIN.7i and.MNaD are equal, &c.; the lines AB, CD must be parallel. Proof. For the line, drawn through the point.Ji parallel to CD, must make these angles equal, and must thercfore coincide with 3B. 32. Theorem. If two parallel lines AB, CD (fig. 813) are cut by a third straight line E/, the two interior angles on the same side, as BJiL JYand lWr'D, are supplements of each other. Proof. For B.JN1 is, by ~ 23, the supplement of its adjacent angle EJIB, which is equal to M1 JJD. 33. Theorem. If two straight lines, lying in the same plane, as.1B and CD (fig. 13), are cut by a third, EF, so that the angles B dI'h? and JIND are supplements of each other, the lines./B, CD must be parallel. Proof. For the line, drawn through the point X parallel to CD, must make these angles supplements to each other, and must therefore coincide with AIB. 34. Theorem. If a straight line is perpendicular to one of two parallels, it must also be perpendicular to the other. Proof. Thus, if EJM (fig. 14), is a right angle, its equal END must also be a right angle. CIH. VI. ~ 3S8.] PERPENDICULAR AND OBLIQUE LINES. l1 Equal Oblique Lines. 35. Theorenm. Reciprocally, if two straight lines, lying in the same plane, are perpendicular to a third, they are parallel. Proof. For the line, drawn through the point U parallel to CD, must be perpendicular to EF, and must therefore coincide with 3B. 36. Theorem. If two straight lines, as diB, CD (fig. 15), are parallel to a third, EF, they are parallel to each other. Proof. For, by the definition of parallel lines, they have the same direction with this third, and are therefore parallel. CHAPTER VI. PERPENDICULAR AND OBLIQUE LINES. 37. Theorem. Only one perpendicular can be drawn from a point to a straight line. Proof. For, if two perpendiculars are erected in the same plane, at two different points, J3I and P (fig. 16) of the line AB, they are parallel, by ~ 35, and cannot meet at any point, as C. 38. Theoremn. Two oblique lines, as CE and CF,' (fig. 17), drawn fropm the point C to the line A/B, at equal distances DE and DF from the perpendicular CD, are equal. 12 PLANE GEOM5ETRY. [CH. VI. ~ 40. Shortest Distance fiomn a Line. Proof. For, if CD.B be folded over upon CBD, DB will fall upon DJ, because the right angles CDB and CDSI are equal; the point F will fall -upon E, because DF and DE are equal; and the straight lines CF and CE will coincide. 39. Theorem. A perpendicular measures the shortest distance of a point from a straight line. Proof. Let the perpendicular CD (fig. 18) and the oblique line CF be drawn from the point C to the line AB. Produce CD to DE, making DE equal to DC, and join FE, we shall, by ~ 18, have E < FC + FE. But CE - 2 CD, and FC + FE 2 FC, for FC and FE are equal, because they are oblique lines drawn from the point F to the line CE at equal distances DC and DE from the perpendicular. Therefore 2 CD < 2 I'C, or CD < F. 40. Lemma. The sum of two lines, as C,/ and CB (fig. 19), drawn to the extremities of the line J.B, is greater than that of two other lines D).f and DB, similarly drawn, but included by them. Proof. Produce DA to E. We have, by ~ 18, aC +- CE > D - + DE, and IDE + BE'> DB. ciI. VI. ~42.] PERPENDICULAR AND OBLIQUE LINES. 13 Oblique Lines unequally Distant from the Perpendicular. The sum of these inequalities is Cc + CE + DE + BE > AD + DE + DB, or, striking out the common term DE, and substituting for CE + BE, its equal BC, SIC + BC > dDW + DIB. 41. Theorem. Of two oblique lines, CF and CG (fig. iS), drawn unequally distant from the perpendicular, the more remote is the greater. Proof. For, the figure being constructed as in ~ 39, and GE being joined, we have, by the preceding proposition, GC + GE> FC+ FEi; or, as in ~ 39, 2 GC>2 FPC and GC > F> 42. Theorem. If from the point C the middle of the straight line J.IB (fig. 20), a perpendicular EC be drawn: - 1. Any point in the perpendicular EC is equally distant from the two extremities of the line AB. 2. Any point without the perpendicular, as F, is at unequal distances fiom the same extremities./ and B. Proof. 1. The distances EJ and EB are equal, since they are oblique lines drawn at equal distances (.s and CB from the perpendicular.B. 2. The distance /d is greater than FB; for FA=- FE + EA =FE + EB while FE - EB'> FB. 14 PLANE GEOMETRY. [CH. VII. ~ 47 Polygon, Triangle, Square. CHAPTER VII. SIDES AND ANGLES OF POLYGONS. 43. Definitions. A plane figure is a plane terminated on all sides by lines. If the lines are straight, the space which they contain is called a rectilinealfigure, or polygon (fig. 21), and the sum of the bounding lines is the perimeter of the polygon. 44. Definitions. The polygon of three sides is the most simple of these figures, and is called a triangle; that of four sides is called a quadrilateral; that of five sides, a pentagon; that of six, a hexagon, &c. 45. Definitions. A triangle is denominated equilateral (fig. 22), when the three sides are equal, isosceles (fig. 23), when two only of its sides are equal, and scalene (fig. 24), when no two of its sides are equal. 46. Definitions. A right-triangle is that which has a right angle. The side' opposite to the right angle is called the hypothenuse. Thus /1BC (fig. 25) is a triangle right-angled at., and the side BC is the hypothenuse. 47. Definitions. Among quadrilateral figures, we distinguish The square (fig. 26), which has its sides equal, and its angles right angles. (See ~ 73). CH. VII. ~ 51.] POLYGONS. 15 Rectangle, Parallelogram, Rhombus, trapezoid, Diagonal. The rectangle (fig. 27), which has its angles right angles, without having its sides equal. The parallelogram (fig. 28), which has its opposite sides parallel. The rhombus or lozenge (fig. 29), which has its sides equal without having its angles right angles. The trapezoid (fig. 30), which has two only of its sides parallel. 48. Definition. A diagonal is a line which joins the vertices of two angles not adjacent, as AC (fig. 30.) 49. Definitions. An equilateral polygon is one which has all its sides equal; an equiangular polygon is one which has all its angles equal. 50. Definition. Two polygons are equilateral with respect to each other, when they have their sides equal, each to each, and placed in the same order; that is, when, by proceeding round in the same direction, the first in the one is equal to the first in the other, the second in the one to the second in the other, and so on. In a similar sense are to be understood two polygons equiangular with respect to each other. The equal sides in the first case, and the equal angles in the second, are called homologous. 51. Theorem. Two triangles are equal, when two sides and the included angle of the one are respectively equal to two sides and the included angle of the other. Proof. In the two triangles AIBC, DEF, (fig. 31), let the angle A. be equal to the angle D, and the sides J.fB, SJC, respectively equal to DE, DF. to PLANE GEOMETRY. [CH. VII. ~ 55 First and Second Cases of Equal Triangles. Place the side DE upon its equal J/B. DF will take the direction A2C, because the angle D is equal to the angle Jl; the point F will fall upon C, because DFis equal to SRC; and the lines FE and BC will coincide, since their extremities are the same points. The triangles will therefore coincide, and must be equal. 52. Corollary. Hence, when two sides and the included angle of one triangle are respectively equal to those of another, the other side and angles are also equal in the two triangles. 53. Theorem. Two triangles are equal, when a side and the two adjacent angles of one triangle are respectively equal to those of the other. Proof. In the two triangles JBC, DEF (fig. 31), let the side /B be equal to the side DE, and the angles. and B respectively equal to D and E. Place the side DE upon the side rB. The side DF will take the direction JU, because the angle D is equal to A1; the side EF will take the direction BC, because the angle E is equal to B; and the point F, falling at once in each of the lines SC and BC, must fall upon their point of intersection C. The triangles will therefore coincide, and must be equal. 54. Corollary. Hence, when a side and the two adjacent angles of one triangle are respectively equal to those of another, the other sides and angle are also equal in the two triangles. 55. Theorem. In an isosceles triangle the angles opposite the equal sides are equal. Proof. In the isosceles triangle /BC (fig. 32), let the equal sides be /B and BC. CH. VII. ~ 59.] POLYGONS. 17 Equal Angles of the Isosceles Triangle. Let the line ED be drawn so as to bisect the angle.8BC. Then the two triangles JtDB and DBC will be equal, since they have two sides JB, BD, and the included angle.JBD, respectively equal to the two sides BC, BD, and the included angle DBC; and the angle 2. will be equal to C. 56. Corollary. An equilateral triangie is also equiangular. 57. Theorem. The line BD (fig. 32), which bisects the angle B, at the vertex of an isosceles triangle, is perpendicular to the base, and bisects the base. Proof. a. For, on account of the equality of the triangles ~L/BD and BCD, tAD must be equal to DC. b. Moreover, the angles BD./ and BDC are equal, and are therefore right angles by the very definition of the right angle in ~ 20. 58. Theorem. If, in a triangle, two angles are equal, the opposite sides are also equal, and the triangle is isosceles. Proof. In the triangle JIBC (fig. 32), let the angle./ be equal to the angle C. Invert the triangle, and place it in the position BCdU; and, as the two triangles JBC and CB2 have the side SC and the adjacent angles. and C of the one respectively equal to CJfl and the adjacent angles C and. of the other, their other sides must be equal, or BC must be equal to B./1. 59. Corollary. An equiangular triangle is also equilateral. 18 PLANE GEOMETRY. [CH. VI. V ~ 62. Third Case of Equal Triangiles. 60. Lemma. Two different triangles cannot be formed on a given line d B (fig. 33), of which the sides, S/D and DB, are respectively equal to CJ and CB, and terminate at the same extremities of JB. Proof. For, first, the vertex D of one triangle cannot fall within the other triangle qCB, as in fig. 19, because, by ~ 40, AD + DB must in this case be less than JC+ CB. Secondly. If D falls without ACUB, as in fig. 33, the triangles GCD and BCD are isosceles, since AC is equal to aD and BC is equal to BD. Hence / CD 2DC, and BCD== BDC; but this is impossible; for of the first members of these equations.ACD> BCD while of the second members aDC< BDC. 61. Theorem. When two triangles are equilateral with respect to each other, they must be equal, and must also be equiangular with respect to each other. P.roof. Let.BC and DEF (fig. 31) be the triangles, whose sides J/B, BC, and SC are respectively equal to DE, EF, and DF. If DE is placed upon AB, the point F must by the preceding proposition fall upon C, and the triangles must coincide. 62. Theorem. Of two sides of a triangle, that is the greater which is opposite the greater angle; and conversely, of two angles of a triangle, that is the greater which is opposite the greater side. CH. VII. ~ 63.3 POLYGONS. 19 The greatest Side of a Triangle opposite the greater Angle. Proof. 1. Suppose the angle C> B (fig. 34). Draw CD so as to make the angle BCD B. Then will BD- CD and dB = =.1D ~ DB = J.D + DC. But JD oDC >.C Flence dB > SC. 2. Conversely. Suppose JB>.dC, the angle C must be greater than B; for if Cwere equal to or less than B, J1B would by ~ 61 and the preceding demonstration, be equal to or less than SC. 63. Theorem. If two triangles have two sides of the one respectively equal to two sides of the other, and if the included angle of the first triangle is greater than the included angle of the second triangle, the third side of the first triangle, is also greater than the third side of the second triangle. Proof. Let the first triangle be LBC (figs. 19 and 33), and the second.BD, which have the sides.lB and AD, respectively equal to.QB and AC, and the included angles B.ID < BC. 1. If the point D falls within the first triangle as in fig. 19, we have by ~ 40 sC + BC> D D+ DB; whence, substracting the equals JC and JID, BC> BD. 2. If the point D falls upon the third side as at E, we nave at once BC> BE. 3. If the point D falls without the first triangle, as in fig. 33, we have in the isosceles triangle ACD, 1CD -- DC. 2 20 PLANE GEOMETRY. [CH. VIi. ~ 67. Sum of the Angles of a Triangle. But BDC > /DC, while / CD > BCD; whence B DC> BCD, so that in the triangle B CD, by ~ 62, we have BC> BD. 64. Theorem. Two right triangles are equal, when the hypothenuse and a side of the one are respectively equal to the hypothenuse and a side of the other. Proof. Let ABC and DEF (fig. 35) be the right triangles, of which the hypothenuse. C is equal to DF, and:/B equal to DE. Place DE upon siB, EF will fall upon CB produced, since the right angles tB G and DEF are equal. An isosceles triangle CA/G is thus formed, and.dB being perpendicular to its base, divides it, by ~ 57, into the two equal triangles AJBC and ABG. 65. Theorem. The sum of the three angles of any triangle is equal to two right angles. Proof. Let ABC (fig. 36) be the given triangle. Produce SC to D, and draw CE parallel to A2B. The angles.BC and.BCE, being alternate-internal angles, are equal, and BAC and ECD, being externalinternal angles, are equal. Hence the sum of the three angles of the triangle is equal to sCB -+ BCE -f- ECD, or, by ~ 25, to two right angles. 66. Corollary. Two angles of a triangle being given, or only their sum, the third will be known by subtracting the sum of these angles from two right angles. 67. Corollary. If two angles of one triangle are respectively equal to two angles of another triangle, the third of the one is also equal to the third of the other, CH. VII. ~ 73.] POLYGONS. 21 Sum of the Angles of a Polygon. and the two triangles are equiangular with respect to each other. 68. Corollary. In a triangle, there can only be one right angle, or one obtuse angle. 69. Corollary. In a right triangle, the sum of the acute angles is equal to a right angle. 70. Corollary. An equilateral triangle, being also equiangular, has each of its angles equal to a third of two right angles, or 2 of one right angle. 71. Corollary. In any triangle JBC, if we produce the side d C toward D, the exterior angle BCD is equal to the sum of the two opposite interior angles J. and B. 72. Theorem. The sum of all the interior angles of a polygon is equal to as many times two right angles as it has sides minus two. Proof. Let JBCDE, &c. (fig. 37), be the given polygon. Draw from either of the vertices, as d, the diagonals JC, SD, SE, &c. The polygon will obviously be divided into as many triangles as it has sides minus two, and the sum of the angles of these triangles is the same as that of the angles of the polygon. But the sum of the angles of each triangle is, by ~ 65, equal to two right angles; and, consequently, the sum of all their angles is equal to as many times two right angles as there are triangles, that is, as there are sides to the polygon minus two. 73. Corollary. The sum of the aingles of a quadrilateral is equal to two right angles multiplied by 4 -2; 22 PLANE GEOMETRY. [CH. VII. ~ 77. The Diagonal of a Parallelogram bisects it. which makes four right angles; therefore, if all the angles of a quadrilateral are equal, each of them will be a right angle, which justifies the definition of a square and rectangle of ~ 47. 74. Corollary. The sum of the angles of a pentagon is equal to two right angles multiplied by 5 -2, which makes 6 right angles; therefore, when a pentagon is equiangular, each angle is the fifth of six right angles, or G of one right angle. 75. Corollary. The sum of the angles of a hexagon is equal to 2 X (6 — 2), or 8 right angles; therefore, in an equiangular hexagon, each angle is the sixth of eight right angles, or 4 of one right angle. The process may be easily extended to other polygons. 76. ScholimGn. If we would apply this proposition to polygons, which have any angles whose vertices are directed inward, as CDE (fig. 38), each of these angles is to be considered as greater than two right angles. But, in order to avoid confusion, we shall confine ourselves in future to those polygons, which have angles directed outwards, and which may be called convex polygons. Every convex polygon is such, that a straight line, however drawn, cannot meet the perimeter in more than two points. 77. Theorem. The diagonal of a parallelogram divides it into two equal triangles. Proof. Let tQBCD (fig. 39) be the parallelogram and A/C its diagonal. The two triangles JB C and bJDC are equal, since they have the side J C common, the angle Bt C -- J CD, by on. viI. ~ 82.] POLYGONS. 23 Parallel Lines at Equal Distances throughout. ~ 30, on account of the parallels AB and CD, and BC.4 - CAD, on account of the parallels B C and ASD. 78. Theorem. The opposite sides of a parallelogram are equal, and the opposite angles are equal. Proof. For the triangles A CB and /CD (fig. 39) being equal, their sides CB and AB are respectively equal to A.D and DC; and the angle dBC-.ADC. In the same way it might be proved that BAD -=BCD. 79. Corollary. Two parallel lines comprehended between two other parallel lines are equal. 80. Theorem. If, in a quadrilateral ABBCD (fig. 39), the opposite sides are equal, namely,./B CD, and AIlD - BC, the equal sides are parallel, and the figure is a parallelogram. Proof. ]For the triangles JBC and JCD are equal, having their three sides respectively equal; and therefore CB: - CAD, whence B C is parallel to.QD, by ~ 31; and B C Z.d CD, whence AB is parallel to CD. 81. Theorem. If two opposite sides BC, AD (fig. 39) of a quadrilateral are equal and parallel, the two other sides are also equal and parallel, and the figure SBCD is a parallelogram. Proof For the triangles.BC and.iCD are equal, since they have the side.flC common, the side BC 3 AD, and the included angle B C.s- CAD, on account of the parallelism of B C and A/D; and therefore AB and CD must be equal and parallel. 82. Theorem. Two parallel lines are throughout at the same distance from each other. 5') 24 PLANE GEOMETRY. [CiH. VIII. ~ 86. The Circle, Radius. Proof. The two parallels JB and CD (fig. 40), being given, if through two points taken at pleasure we erect, upon JiB, the two perpendiculars EG and FH, the straight lines EG, FH will, by ~ 34, be perpendicular to CD; and they are also parallel and equal to each other, by arts. 35 and 79. 83. Theorem. The two diagonals of a parallelogram mutually bisect each other. Proof. For the triangles (fig. 41) JIDO and BOC are equal, since the side B C S = D, and the angles OCB OdD, and OBC=- ODI, on account of the parallelism of BC and SD; therefore S C)-=- C and BO=- OD. 84. Corollary. In the case of the rhombus (fig. 42), the triangles 21OB and J/OD are equal, for the sides:/B J/D, B 0 DO, and.30 is common; therefore the angles MOB and /OD are equal, and, as they are adjacent, each of them must, by definition, ~ 20, be a right angle, so that the two diagonals of a rhombus bisect each other at right angles. CHAPTER VIII. THE CIRCLE AND THE MEASURE OF ANGLES. 85. Definitions. The circumference of a circle is a curved line, all the points of which are equally distant from a point within, called the centre. The circle is the space terminated by this curved line. 86. Definitions. The radius of a circle is the straight CH. VIII. ~ 92.] TIHE-I CIRCLE AND ANGLES. 25 Diameter, Inscribed Lines. line, as AB, JtC, D (fig. 43), drawn fiom the centre to the circumference. T'he diameter of a circle is the straight line, as BD, drawn through the centre, and terminated each way by the circumference. 87. Corollary. Hence, all the radii of a circle are equal, and all its diameters are also equal, and double of ~he radius. 88. Theorem. Every diameter, as BD (fig. 43), bisects the circle and its circumference. Proof. For if the fioure B CD be folded over upon the part BED, they must coincide; otherwise there would be points in the one or the other unequally distant from the centre. 89. Definition. A semicircumference is one half of the circumference, and a senmicircle is one half of the circle itself. 90. Definition. An arc of a circle is any portion of its circumference, as BFE. The chord of an are is the straight line, as BE, Nwhich joins its extrernities. The segmnnent of a circle, is a part of a circle comprehended between an are and its chord, as EFB. 91. Theoremi. Every chord is less than the diameter. Proof. Thus BE (fig. 43) is less than DB. For, joining fE, we have BD = +B -- + E, but BE < BtQ + QE, therefore BE < BD. 92. Definition. A straight line is said to be inscribed in a circle, when its extremities are in the circumference of the circle. 26 PLANE GEOMETRY. [iCH. VIII. 5 97. Angles proportional to their Arcs. 93. Corollary. Hence the greatest straight line wnich can be inscribed in a circle is equal to its diameter. 94. Theorem. A straight line cannot meet the circumference of a circle in more than two points. Proof. For, by ~~ 38 and 41, only two equal straight lines can be drawn from the same point to the same straight line; whereas, if a straight line could meet the circumference JBD (fig. 45) in the three points, JBD, three equal straight lines C, CB, CD, would be drawn from the point C to this line. 95. Theorem. In the same circle, or in equal circles, equal angles 2CB, DCE (fig. 44), which have their vertices at the centre, intercept upon the circumference equal arcs riB, DE. Proof. Since the angles DCE and SJCB are equal, one of them may be placed upon the other; and since their sides are equal, the point D will fall upon 1, and the point E upon B. The arcs SB and DE must therefore coincide, or else there would be points in one or the other unequally distant from the centre. 96. Theorem. Reciprocally if the arcs JB, DE (fig. 44) are equal, the angles JfCB and DCE must be equal. Proof. For if the line CE be drawn, so as to make an angle DCE equal to IACB, it must pass through the extremity E of the arc DE, which is equal to A/B. 97. Theorem. Two angles, as A CB,.:CD (fig. 45), are to each other as the arcs JB,.3D intercepted CH. VIII. ~ 98.] THE CIRCLE AND ANGLES. 27 Infinitely Small Qu:Intities. between their sides, and described from their vertices as centres, with equal radii. Proof. Suppose the less angle placed in the greater, and suppose the angles to be to each other, for example, as 7 to 4; or, which amounts to- the same, suppose the angle AC a, which is their common measure, to be contained 7 times in ACD, and 4 times in ISCB; so that the angle.ACD may be divided into the 7 equal angles 2AC a, a C b, b C c, &c., while the angle.ACB is divided into the 4 equal angles AC a, &c. The arcs AB and AD are, at the same time, divided into the equal parts. a, ab, be, &c., of which AD contains 7 and AB 4; and therefore these arcs must be to each other as 7 to 4, that is, as the angles.ACD and /t CB. 98. Scholi'ton. The preceding demonstration does not strictly include the case in which the two angles are incommnensurable, that is, in which they have no common divisor. The divisor AC a, instead of being contained an exact number of times in the given angles A CB, A CD, is, in this case, contained in one or each of them a certain number of times plus a remainder less than the divisor. So that if these remainders be neglected, the angle AC a will be a common divisor of the given angles. Now the angle AC a may be taken as small as we please; and therefore the remainders, which are neglected, may be as small as we please; less, then, than any assignable quantity, less than any conceivable quantity, that is, less than any possible quantity within the limits of human knowledge. Such quantities can, undoubtedly, be neglected, without any error; and the above demonstration is thus extended to the case of incommensurable angles. 28 PLANE GEOMETRY. [CH. VIII. ~ 101 Measure of Angles. Degree, Minute, Second, &c.; Quadrant. 99. The principle, involved in the reasoning just given, is general in its application; and may be stated as follows, using the term infinitely small quantity to denote a quantity which may be taken at pleasure, as small as we please, so that it may be supposed equal to nothing whenever we please. Axionm. Infinitely small quantities may be neglected. 100. Corollary. Since the angle at the centre of a circle is proportional to the are included between its sides, either of these quantities may be assumed as the measure of the other; and we shall, accordingly, adopt, as the measure of the angle, thMe arc describedfrooml its vertex as a centre and included between its sides. But when different angles are compared with each other, the arcs, which measure them, must be described with equal radii. 101. Definitions. In order to compare together different arcs and angles, every circumference of a circle may be supposed to be divided into 360 equal arcs called degrees, and marked thus (o). For instance, 60~ is read 60 degrees. Each degree may be divided into 60 equal parts called minutes, and marked thus ('). Each minute may be divided into 60 equal parts called seconds, and marked thus ("). When extreme minuteness is required, the division is sometimes extended to thirds and fourths, &c., marked thus ("'), (""), &c. A quadrant is a fourth part of a circumference, and contains 90~. This is called the sexacgesinmal division of the circle; another which is called the centesinmal di cIi. VI1I. ~ 106.] THE CIRCLE AND ANGLES. 29 Inscribed Angle and Triangle. vision has been introduced by the French geometers. They divide the quadrant into 100 degrees, the degree into 100 minutes, &c; so that by this method of division, the whole notation is decimal. 102. Scholiurn. As all circumferences, whether great or small, are divided into the same number of parts, it follows that a degree which is thus made the unit of arcs, is not a fixed value, but varies for every different circle. It merely expresses the ratio of an arc, namely, A.t to the whole circumference of which it is a part, and not to any other. 103. Corollary. The angle may be designated by the degrees and minutes of the arc which measures it; thus the angle which is measured by the are of 17~ 28' may be called the angle of 17~0 8'. 104. Corollary. The rig'ht angle is tnen an angle of 90~, and is measured by the quadrant. 105. Corollary. The angle which is measured by the arc of one degree, that is, the angle of 10 is then -1 of a right angle, and has a fixed value, altogether independent, in its magnitude, of the radius of the arc by which it is measured. The same is the case with an angle of any other value, so that the arcs.P, J'lD', J."D", &c. (fig. 46), of the same number of degrees, all measure the same angle Cf the vertex of which is at their common centre. 106. Definitions. An inscribed angle is one, whose vertex is in the circumference of a circle, and which is formed by two chords, as B.JC (fig. 47). An inscribed triangle is a triangle whose three angles have their vertices in the circumference of the circle. 80 PLANE GEOMETRY. [CeI. VIII. ~ 109. Inscribed Angle. And, in general, an inscribed figure is one, all whose angles have their vertices in the circumference of the circle. In this case the circle is said to be circumscribed about the figure. 107. The inscribed angle B2C (figs. 47, 48, 49) has for its measure the half of the arc BC comprehended between its sides. Proof. 1. If one of the sides is a diameter, as AC (fig. 47), 0 being the centre of the circle, Join BO. Then the triangle dOB is isosceles, for the radii,/O, BO are equal. Therefore the angles' ORB and OB-d are equal, and the exterior angle BOC being equal to their sum, by ~ 71, is equal to the double of either of them, as B C. BAdC is, therefore, half of BOG and has half its measure, or half of BC. 2. If the centre 0 falls within the angle, as in (fig. 48,) Draw the diameter SOD; and, by the above, BaD has for its measure half of BD, and DA1C half of DC; so that B2D +- DSC or B, C has for its measure half of BD +-DC. or half of BC. 3. If the centre 0 falls without the angle, as in (fig. 49,) Draw the diameter s OD; and BAD — DG C, or Bd C has for its measure half of BD - DC, or half of B C. 108. Corollary. All the angles B.2C, BDC (fig. 50), &c., inscribed in the same segment are equal. Proof. For they have each for their measure the half of the same arc BE C. 109. Corollary. Every angle B,/JD (fig. 51) inscribed in a semicircle is a right angle. cH. VIII. ~ 114.] THE CIRCLE AND ANGLES. 31 Arcs and Chords. Proof. For it has for its measure the half of the semicircumference BED, or a quadrant. 110. Corollary. Every angle BdC (fig. 50) inscribed in a segment greater than a semicircle is an acute angle, for it has for its measure the half' of an arc BEG less than a semicircumference. 111. Corollary. Every angle BE C inscribed in a segment less than a semicircle is an obtuse angle; for it has for its measure the half of an are greater than a semicircumference. 112. Theorem. In the same circle, or in equal circles, equal arcs are subtended by equal chords. Proof. Let the are.AB (fig. 52) be equal to the are BC. Join AC; and, in the triangle ABBC, the angles A and C are equal, for they are measured by the halves of the equal arcs BC and dB. The triangle ABC is therefore isosceles, by ~ 58, and the chords /B and BC are equal. 113. Theorem. Conversely, in the same circle, or In equal circles, equal chords subtend equal arcs. Proof. Let the chord AiB (fig. 52) be equal to the chord B C. Join AC; and in the isosceles triangle JiB C the angles $. and C must be equal, by ~ 55, and also the arcs AB and B C, which are double their measures. 114. Theorem. In the same circle, or in equal circles, if the sum of two arcs be less than a circurmference, the greater arc is subtended by the greater chord; and, conversely, the greater chord is subtended by the greater arc. 3 32 PLANE GEOMETRY. [CH. ViII. ~ 116, Perpendicular at the Middle of a Chord. Proof. a. Let the are BC (fig. 53) be greater than the are UB. Join J.C; and the angle B,//C, being measured by half the arc B C, is greater than B C6., which is measured by half of /B; and therefore, by ~ 62, the chord BC is greater than A3B. b. Coiversely. Suppose the chord B C > AB. Join 3 C; and, by ~ 62, BL C> B CJ, and, therefore, the are BC double the measure of BSC is greater than the are AIB double the measure of B C3. 115. Corollary. If the sum of the two arcs is greater than a circumference, the greater are is subtended by the less chord, and the less arc by the greater chord. Proof. Suppose the are B CUVYJ > BAJIVC (fig. 53). Take.JVNC from each, and we have the arc BC > B.J, and consequently, by the preceding proposition, the chord BC of the less arc BJ!NC is greater than the chord B,/ of the greater are B CE J. 116. Theorem. The radius CG (fig. 54), perpendicular to a chord SB, bisects this chord and the are subtended by it. Proof. a. The radii CAl and CB are equal oblique lines drawn to the chord JB. They are, therefore, by ~ 38, at equal distances from the perpendicular, or ID DB. b. Since the line GC is a perpendicular erected at the middle of the straight line JIB, any point of it, as G, is, by ~ 42, at equal distances from its extremities, that is, the chords JIG and GB are equal; and therefore, by ~ 113, the arcs AJG and GB are equal. CH. VIII. ~ 119.] THE CIRCLE AND ANGLES. 33 Taingent to a Circle. 117. Corollary. The perpendicular erected upon the middle of a chord passes through the centre, and also through the middle of the are subtended by the chord. 118. Definitions. A secant is a line which meets the circumference of a circle in two points, as AB (fig. 55). A tangent is a line, which has only one point in common with the circumference, as CD. The common point JA_ is called the point of contact. Also two circumferences are tangents to each other (figs. 56 and 57), when they have only one point common. A polygon is said to be circumscribed about a circle, when all its sides are tangents to the circumference; and in this case the circle is said to be inscribed in the polygon. 119. Theorem. The direction of the tangent is the same as that of the circumference at the point of contact. Proof. Draw through the point M (fig. 55) the secant JME and the tangent MD. If the secant MSE is turned around the point JM so as to diminish the angle EMD, the secant ME will approach the tangent M,1D, and the point E will approach the point M. When ME? is turned so far as to pass through the point P next to M, the angle DMJE will be infinitely small, since P is at an infinitely small distance from Ai'; and the line M~E will approach infinitely near the tangent JMD, that is, it will, by ~ 99, coincide with this tangent, which has therefore, by ~ 11, the same direction with the circumference at M. 34 PLANE GEOMETRY. [CH. VIII. ~ 122. Angles tllrmedl by Secants and T'angents. 120. Theorem. The tangent to a circle is perpendicular to the radius drawn to the point of contact. Proof. The radius OJX'- O.N (fig. 58) is shorter than any other line, as OP, which can be drawn from the point O to the tangent,MP; it is therefore, by ~ 39, perpendicular to this tangent. 121. The angle BAC (fig. 59), formed by a tangent and a chord, has for its measure half the arc BJVI./ comprehended between its sides. Proof. a. Draw the diameter./D, and we have B.A C - D C - D3B. But DAC, being a right angle, has for its measure half of a semicircumference, as JIBD; also BAD has, by ~ 107, for its measure half of the arc BD. The measure of B C is therefore 2 (EdBDB- aD) X SCRUB. b. In the same way, it may be shown that BJLE has for its measure half the arc BDI. 122. Theorem. r'1he angle B.IC, formed by two secants (fig. 60), two tangents (fig. 62), or a tangent and a secant (fig. 61), and which has its vertex without the circumference of the circle, has for its measure half the concave are B3J1iC intercepted between its sides, minus half the convex arc D.N'E. Proof. Join BE; and as BEG is an exterior angle of the triangle.IBE, we have, by ~ 71 BE C - 3BE + BA C, whence BA C BE C —. JBE. Ci. VIII. ~ 125.] THE CIRCLE AND ANGLES. 35 Angles formed by Chords. Arcs intercepted by Parallels. But the measure of BEC is half of B3AC, and that of,/BE is half of DJV*'E; therefore the mneasure of' BC is V,5BJCE —- DJV2E. Scholiurn. In applying the preceding demonstration to (figs. 61 and 62), the letters B and D mrust denote the same point; and in (fig. 62) the letters C and E must also denote the same point. 123. Theorem. The angle BJiC (fig. 63), formed by two chords, and which has its vertex between the centre and the circumference, has for its measure half the are BC contained between its sides plus half the are DE contained between its sides produced. Proof. Join BE; and, as BsC is an exterior angle of the triangle.LIBE, we have, by 5 71, BdC — BESd + ABE. But the measure of BEd is, by ~ 107, half of BC; and that of ABE is half of DE; therefore the measure of BS C is BC +IDE. 124. Theorem. Two parallels JiB and DC (figs. 64, 65, 66), intercept upon the circumference equal arcs.D, B C. Proof. Join BD. The alternate-internal angles MBD and BDC are equal, by ~ 30; and therefore, the arcs AD and BC, the double of their measures, are equal. Scholiuzn. In applying this demonstration to figs. (65 and 66), the letters A and B must denote the same point; and in (fig. 66) the letters D and C must also denote the same point. 125. Corollary. The arcs lD and BC (fig. 66) being equal must be semicircumferences, and the chord BC must be a diameter. 3* 36 PLANE GEOMETRY. [CH. VIII. ~ 127. Tanogent Circumferences. 126. Theorem. When the circumferences of two circles cut each other, the line JIB (fig. 67), which joins their centres, is perpendicular to the middle of tile line CD, which joins their points of intersection. Proof: For if a perpendicular be erected upon the middle of the chord CD, it must, by ~ 117, pass through the centres d. and B of both the circles of which CD is a chord. 127. Theorenm. When two circumferences are tangents to each other, their centres and point of contact are in the same straight line perpendicular to their common tangent at the point of contact. Proof. a. If the centres of two circumferences which cut each other (fig. 67) are removed from each other, until the points C and D of intersection approach infinitely near to each other, the circles will become tangent, as in (fig. 56), the chord CD of (fig. 67) will become the tangent CD of (fig. 56); and as both the radii AMJ3 and MJB are perpendicular to their common tangent, these radii must be in the same straight line. b. In the same way, the centres of the circles (fig. 67) may be brought near to each other until the circles are tangents, as in (fig. 57), and the same reasoning may be here applied to prove that the line./BMA, perpendicular to the common tangent at MI, passes through both the centres.A and B. CO. ix. ~ 131.] PROBLEMS. 37 Position oifa Point in a plane. CHAPTER IX. PROBLEMS RELATING TO THE FIRST EIGHT CHAPTERS. 128. Problem. To find the position of a point in a plane, having given its distances from two known points in that plane. Solution. Let the known points be./ and B (fig. 68). From the point A as a centre, with a radius equal to the distance of the required point from J3, describe an are. Also, from the point B as a centre, with a radius equal to the distance of the required point from B, describe an are cutting the former are; and the point of intersection C is the required point. SchIolitun. By the same process, another point D may also be found which is at the given distances from J. and B, and either of these points therefore satisfies the conditions of the problem. 129. Corollary. If both the radii were taken of equal magnitudes, the points C and D thus found would be at equal distances from./ and B. 130. Scholizan. The problem is impossible, when the distance between the known points is greater than the sum of the given distances or less than their difference. 131. Scholimun. If the required point is to be at equal distances from the known point, its distance from either of them must be greater than half the distance between the known points. 38 PLANE GEOMETRY. [cii. ix. ~ 135. To Bisect a Line; to Erect a Perpendicular. 132. Problem. To divide a given straight line {RB (fig. 69) into two equal parts; that is, to bisect it. Solution. Find by ~ 129, a point C at equal distances from the extremities J. and B of the given line. Find also another point D, either above or below the line, at equal distances from./ and B. Through C and D draw the line CD, which bisects AB at the point E. Proof. For the perpendicular, erected at E to the line.,B, must, by ~ 42, pass through the points C and D, and must therefore, by ~ 16, coincide with the line CD. 133. Problem. At a given point Jl (fig. 70), in the line BC, to erect a perpendicular to this line. Solution. Take the points B and C at equal distances from A.; and find a point D equally distant fiom B and C. Join A.D, and it is the perpendicular required. Proof. For the point D must, by ~ 42, be a point of the perpendicular erected at./. 134. Problem. From a given point./ (fig. 71), without a straight line BC, to let fall a perpendicular upon this line. Solution. From 2. as a centre, with a radius sufficiently great, describe an arc cutting the line BC in two points B and C; find a point D equally distant from B and C, and the line ADE is the perpendicular required. Proof. For the points A/t and D, being equally distant from B and C, must, by ~ 42, be in this perpendicular. 135. Problem. To make an are equal to a given arc dB (fig. 72), the centre of which is at the given point C. Solution. Draw the chord dB. From any point D as a centre, with a radius equal to the given radius CA7, CH. IX. ~ 13.]1 PROBLEMS. 39 To make and to bisect a given Arc, or Angle. describe the indefinite are FH. From F as a centre, with a radius equal to the chord AtB, describe an are cutting the are FH in EH, and we have the arc FII[ AB. Proof. For as the chord AB the chord FH, it follows, from ~ 1 12, that the are dB the arc FIt. 136. Problem. At a given point A (fig. 73), in the line AAB, to make an angle equal to a given angle K. Solution. From the vertex K, as a centre, with any radius describe an are IL meeting the sides of the angle; and from the point AJ as a centre, by the preceding problenm, make an are BC equal to IL. Draw AC, and we have AJ K. Proof. For the angles A. and K being, by ~100, measured by the equal arcs BC and IL, are equal. 137. Problem. To bisect a given arc AB (fig. 74). Solution. Find a point D at equal distances from.A and B. Through the point D and the centre C draw the line CD, which bisects the are AB at E. Proof. Draw the chord.AB. Since the points D and C are at equal distances from A and B, the line DC is, by ~ 132o, perpendicular to the middle of the chord riB, and therefore by ~ 117, it passes through the middle E of the are RB. 138. Problem. To bisect a given angle A (fig. 75). Solution. From A as a centre, with any radius, describe an are BC, and, by the preceding problem, draw the line AE to bisect the are B C, and it also bisects the angle A. Proof. The angles BARE and EAC are equal, for they tre measured by the equal arcs BE and EC. 40 PLANE GEOIETRY. [CHI. IX. ~ 143 To construct a Triangle. 139. Problem. Through a given point a (fig. 76), to draw a straight line parallel to'a given straight line BC. Solution. Join EA, and, by the preceding problem, draw AD, making the angle ER3D —.-=iEF, and AD is parallel to B C, by ~ 31. 140. Problemn. Two angles of a triangle being given, to find the third. Solution. Draw the line A1BC (fig. 77). At any point B draw the line BD, to make the angle DBC equal to one of the given angles, and draw BE, to make EBD equal to the other given angle, and iBE is the required angle. Proof. For these three angles are, by ~ 25, together equal two right angles. 141. Problem. Two sides of a triangle and their included angle being given, to construct the triangle. Solution. Make the angle A. (fig. 78) equal to the given angle, take AJB and AC equal to the given sides, join BC, and 3BC is the triangle required. 142. PrOblem. One side and two angles of a triangle being given, to construct the triangle. Solution. If botlh the angles adjacent to the given side are not given, the third angle can be found by ~ 140. Then draw AB (fig. 78) equal to the given side, and draw AC and BC, makling the angles Al and B equal to the angles adjacent to the given side, and JIB C is the triangle required. 143. Problem. The three sides of a triangle being given, to construct the triangle. Solution. Draw ~.B (fig. 78) equal to one of the given CH. IX. ~ 148.] PROBLEMS. 41 To construct a Parallelogram. To find the Centre of a Circle. sides, and, by ~ 128, find the point C at the given distances SC and BC from the point C, join.C and BC, and B C is the triangle required. 144. Scholiunt. The problem is impossible, when one of the given sides is greater then the sum of the other two. 145. Problem. To construct a right triangle, when a leg and the hypothenuse are given. Solution. Draw,B (fig. 79) equal to the given leg. At 2i erect the perpendicular AC, from B as a centre, with a radius equal to the given hypothenuse, describe an arc cutting LAC at C. Join BC, and. ABC is the triangle required. 146. Problem. The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram. Solution. MI ake the angle A (fig. 80) equal to the given angle, take A2B and AC equal to the given sides, find the point D, by ~ 128, at a distance from B equal to,C, and at a distance from C.equal to./B. Join BD and DC, and JBCD is, by ~ 80, the parallelogram required. 147. Corollary. If the given angle is a right angle, the figure is a rectangle; and, if the adjacent sides are also equal, the figure is a square. 148. Problem. To find the centre of a given circle or of a given arc. Solution. Take at pleasure three points.d, B, C (fig. 81) cn the given circumference or arc; join the chords./B and BC, and bisect them by the perpendiculars DE and FG; the point 0 in which these perpendiculars meet is the centre required. 42 PLANE GEOMETRY. [cH. IX. ~ 152. To draw a Tanogent to a Circle. Proof. For, by ~ 117, the perpendicular DE and FG must both pass through the centre, which must therefore be at their point of meeting. 149. Scholium. By the same construction a circle may be found, the circumference of which passes through three given points not in the same straight line, or in which a given triangle is inscribed. 150. Problem. Through a given point, to draw a tangent to a given circle. Solution. a. Ifthe given point A (fig. 82) is in the circumference, draw the radius C.A, and through d draw S//D perpendicular to CA/, and A/D is, by ~ 120, the tangent required. b. If the given point A (fig. 83) is without the circle, join it to the centre by the line AC; upon S.C as a diameter describe the circumference./JJUCA,, cutting the given circumference in.71 and A"; join d2Ii and fJA', and they are the tangents required. Proof. For the angles dJ}IC and AN/C are right angles, because they are inscribed in semicircles, and therefore AM1 and AJN are perpendicular to the radii MC and.NC at their extremities, and are, consequently, tangents, by ~ 120. 151. Corollary. The two tangents ASM and AJN are equal; for the right triangles /J2MC and.AtC are equal, by ~ 64, since they have the hypothenuse AC common, and the leg JIC equal to the leg NC, and, therefore, the other legs AM2J. and NJV' are equal. 152. Problem. To inscribe a circle in a given triangle.B C (fig. 84). Solution. Bisect the angles A. and B by the lines SO CH. IS. ~ 155.] PROBLEMS. 43 To inscribe a Circle in a Triangle. and B O, and their point of intersection 0 is the centre of the required circle, and the perpendicular OD let fall from O upon the side S C is its radius. Proof. The perpendiculars OD, OE, and OFlet fall from O upon the sides of the triangle are equal to each other. For in the right triangles GOD and OGE the hypothenuse O0/ is common; the angle OD —- OGAE by construction; and the third angle ASOD = AOE, by ~ 67; the triangles are, therefore, equal, by ~ 53; and OD is equal to OE. In the same way it may be proved that OF - OD-= OE. Hence the circumference DFE passes through the points D, F, E, and the sides are'tangents to it, by ~ 120. 153. Corollary. The three lines.30, BO, and CO, which bisect the three angles of a triangle, meet at the same point. 154. Problem. Upon a given straight line,//B (figs. 85 and 86), to describe a segment capable of containing a given angle, that is, a segment such that each of the angles inscribed in it is equal to a given angle. Solution. Draw BF, making the angle.JBF equal to the given angle. Draw BO perpendicular to BF, and OC perpendicular to the middle of AIB. From 0, the point of intersection of OB and OC, with a radius OB OA, describe the circumference BMJ3JVN, and BMJLi is the segment required. Proof. Since BF is perpendicular to BO, it is a tangent to the circle, and therefore the angles J1.MB and JBF are equal, since they are each, by ~ 107 and 121, measured by half the arc /JVNB. 155. Scholium. If the given angle were a right angle, 4 44 PLANE GEOMETRY. LCH. IX. ~ 156. To find the Ratio of two Lines. the segment sought would be a semicircle described upon the diameter.JB. 156. Problem. To find a common measure of two given straight lines, JB, CD (fig. 87), in order to express their ratio in numbers. Solulion. a. The method of finding the common divisor is the same as that given in arithmetic for two numbers. Apply the smaller CD to the greater JIB, as many times as it will admit of; for example, twice with a remainder BE. Apply the remainder BE to the line CD, as many times as it will admit of; twice, for example, with a remainder DF. Apply the second remainder DF to the first BE, as many times as it will admit of; once, for example, with a remainder B G. Apply the third remainder BG to the second DF, as many times as it will admit of. Proceed thus till a remainder arises, which is exactly contained a certain number of times in the preceding. This last remainder is a common measure of the two proposed lines; and, by regarding it as unity, the values of the preceding remainders are easily found, and, at length, those of the proposed lines from which their ratio in numbers is deduced. If, for example, we find that GB is contained exactly three times in FD, GB is a common measure of the two proposed lines. b. Let GB 1; and we have FD==) GB =- 3, EB ==. FD + GB = 3 +1= 4, CD== 2.EB -FD- F 8 +3-11, AB =-. CD +- EB-2 4 — 26; CH. X. ~ 159.] PROPORTIONAL LINES. 45'To divide a Line into equal Parts. consequently, the ratio of the lines RIB, CD is as 26 to 11; that is, AdB is i{ of CD, and CD is 1 of riB. 157. Corollary. By a like process, may be found the ratio of any two quantities, which can be successively applied to each other, like straight lines, as, for instance, two arcs or two angles. CHAPTER X. PROPORTIONAL LINES. 158. Theorem. If lines a a', b b', c c', &c. (fig. 88), are drawn through two sides,.B, AJIC of a triangle -iBC, parallel to the third side BC, so as to divide one of these sides A1B into equal parts J/ a, a b, &c., the other side JC is also divided into equal parts Jfa', ab', &c. Proof. Through the points a', b', c', &c. draw the lines a' mn, b' n, c'o, &c. parallel to./B. The triangles. a a', a' In b', b' n' c, &c. are equal, by ~ 53; for the sides a' m, b" n, c' o, &c. are, by ~ 79, respectively equal to a b, be, c d, &c., and are therefore equal to each other and to A a; moreover, the angles a a', m a' b', n b' c', &c. are equal, by ~ 29, and likewise the angles A a a', a' nz b', b' n c', &c. Consequently, the sides./ a', a' b', b' c', &c. are equal. 159. Problem. To divide a given straight line JB (fig. 89) into any number of equal parts. 46 PLANE GEOMETRY. [CH. X. ~ 162. A line drawn Parallel to a Side of a Triangle. Solution. Suppose the number of parts is, for example, six. Draw the indefinite line./O0; take J2C of any convenient length, apply it six times to./SO. Join B and the last point of division D by the line BD, draw CE parallel to DB, and J/E, being applied six times to JQB, divides it into six equal parts. Proof. For if, through points of division of J.D, linetr are drawn parallel to DB, they must, by the preceding theorem, divide d/B into six equal parts, of which dIE is one. 160. Theorem. If a line DE (fig. 90) is drawn through two sides.JB, AC of a triangle./BC, parallel to the third side BC, it divides those two sides proportionally, so that we have SD: /B --./E: J/C. Proof. a. Suppose, for example, the ratio of SLD: AB to be as 4 to 7..B may then be divided into 7 equal parts.i a, a b, b c, &c., of which./D contains 4; and if lines a a', b b', c c', &Sc. are drawn parallel to B C,./ C is divided into 7 equal parts J. a', a' b', b' c', &c., of which S/E contains 4. The ratio of StE to./C is, therefore, 4 to 7, the same as that of SD/: JAB. 161. Scholizn. b. The case in which SJD and.AB are incommensurable, is included in this demonstration by the reasoning of ~ 98. 162. Corollary. In the same way./D: BD - nE: EC. and BD: B -- EC:./C. 163. Theorem. Conversely, if a line DE (fig. 90) CH. X. ~ 165.] PrROPORTIONAL LINES. 47 Division of a Line into Parts proportional to given Lines. is drawn so as to divide two sides 3B~, JC of a triangle proportionally, this line is parallel to the third side BC. Proof. For the line, which is drawn through the point D parallel to BC must, by the preceding proposition, pass through the point E, so as to divide the side SC proportionally to.sB, and must therefore coincide with the proposed line DE. 164. Problem. To divide a given straight line IB (fig. 91) into two parts, which shall be in a given ratio, as in that of the two lines ni to Xn. Solution. Draw the indefinite line 2/O. Take.1C 2.m and CD s —z. Join DB, through C draw CE parallel to DB; and E is the point of division required. Proof For, by ~ 161, IE: EB SC: CD mn: n. 165. Problem. To divide a given line JB (fig. 92) into parts proportional to any given lines, as m, n, o, &c. Solution. Draw the indefinite line {/O. Take SC -- 1, CDD —, DE o, &c. Join B to the last point E, and draw CC', DD', &c. parallel to BE. C', D', &c. are the required points of division. Proof. For, if.3E is divided into parts equal each of them to the greatest common divisor of im, n, o, &c., and if, through the points of division, lines are drawn parallel to BE; it appears, from inspection, as in ~ 160, that C"': CD' f1 C: CD m'n. and that 4* 48 PLANE GEOMETRY. [cII. x. ~ 167. To find a Fourth proportional to three given,ines. C'D': D'B CD: DE _ n: o; or, as they may be written for brevity, A/': CID': DDB at: n: o. 165. Problem. To find a line, to which a given line JiB (fig. 93) has a given ratio, as that of the lines m to n; in other words, to find the fourth proportional to the three lines m, n, and.LB. Solution. Draw the indefinite line./B, take.l C -- n, JtD == n. Join CB, draw DE parallel to BC, and S.E is the required line. Proof. For, by ~ 160, JB: E -- A C: ~JD —:'n. 166. Corollary. By making n equal to /B in the preceding solution, we find a third proportional to the two lines mn and.JB. 167. Problem. To divide one side BC (fig. 94), of a triangle.ABC into two parts proportional to the other two sides. Solution. Draw the line.D to bisect the angle BAIC, and D is the required point of division, that is, BD: DC —B B:' C. Proof. Produce BA2 to E, making JtE equal to AGC. Join CE. Then the angles.CE and ALEC are equal, by ~ 55; and the exterior angle CAB of the triangle A2CE is equal to ACE + StEC, or to 2 CEJ.1, and, as DAIB is half of B.1 C, we have DDAlB - (2 CEJ) - CEA; CH. XI. ~ 170.] SIMILAR POLYGONS. 49 To divide one Side of a Triangle into parts proportional to other Sides. and, therefore, by ~ 31, JD is parallel to CE, and, by ~ 161, BD: DC=- B': E, or, since A.1E =. C, BD: DC - B/i:'.C. 168. Problem. Through a given point P (fig. 95) in a given angle t., to draw a line so that the parts intercepted between the point and the sides of the angle may be in a given ratio. Solution. Draw PD parallel to./B. Take DC in the same ratio to S.D as the parts of the required line. Through C and P draw CPE, and this is the required line. Proof. For, by ~ 161, CP: PE — CD: DAi. 169. Corollary. When DC is taken equal to liD, PC is equal to PE. CHAPTER XI. SIMILAR POLYGONS. 170. Definitions. Two polygons are similar, which are equiangular with respect to each other, and have their homtologouts sides proportional. In different circles, similar arcs are such as correspond to equal angles at the centre. Thus the arcs,P, JI'D', &c. (fig. 46) are similar. 50 PLANE GEOMiETRY. [Cii. XI. ~ 175. Similar Polygonls and Arcls. Eqluin,11.gular triangles are similar. 171. Defin;,itio.ns.'The altitude of a parallelogram is the perpendicular, whllich measures the distance between its opposite sides considered as bases. The altitude of a triangle is the perpendicular, as S/D (fig. 96), which measures the distance of any one of its vertices, as,X2, from the opposite side BC taken as a base. The altitude of a trapezoid is the perpendicular, as EF (fig. 97), drawn between its two parallel sides. 172. Theorem. Two triangles JBC, DEF (fig. 98), which are equiangular with respect to each other, are similar. Proof. Place the angle D upon its equal A/; E must fall upon E', and F upon F'i; and F'E' is parallel to B C, because the angles dE'F' and.GCB are equal. Hence, by ~ 160, JE'.: A C S — F'.: B, that is, DE: JlC — DF: 3B. In the same way, it may be proved that DE: Jl C EF: B C ODF': J1B. 173. Corollary. Hence, and from ~ 67, it follows that two triangles are similar, when they have two angles of the one respectively equal to two angles of the other. 174. Corollary. Two right triangles are similar, when they have an acute angle of the one equal to an acute angle of the other. 175. Theorem. Two triangles are similar, when CH. XI. ~ 180.] SIMILAR POLYGONS. 51 Cases of similar'Triangles. they have the sides of the one respectively parallel to those of the other. Proof. For, in this case, the angles are equal by ~ 29. 176. Corollary. The parallel sides are homologous. 177. Theorem. Two triangles are similar, when the sides of the one are equally inclined to those of the other, each to each, as dBC, DEF (fig. 99). Proof. For if one of the triangles is turned around, by a quantity equal to the angle made by the sides of the one with those of the other, the sides of the two triangles become respectively parallel, and they are, therefore, by ~ 175, equiangular and similar. 178. Corollary. Two triangles are similar, when the sides of the one are respectively perpendicular to those of the other, and the perpendicular sides are homologous. 179. Theorem. Two triangles.BC, DEF (fig. 98) are similar, if they have an angle J/ of the one equal to an angle D of the other, and the sides including these angles proportional, that is, JB': DF=- JC: DE. Proof. Place the angle D upon.//; E falls upon E', and F upon F'; and E'F' is parallel to BC, by ~ 162, because BB F:' -- C: dE'. Hence, by ~ 30, the angle C — E'F' E, and B F2E F; that is, the triangles.JBC and DEF are equiangular, and, by ~ 172, similar. 180. Theorem. Two triangles JB C, DEF (fig. 98) are similar, if they have their homologous sides proportional, that is, 52 PLANE GEOMIETRY. [Ci. XI. ~ 181. Cases of similar Triangles. sB: DF- C: DE- BC: EF. Proof. Take SE' DE, and draw E'F parallel to BC. The triangles dE'F' and ABC are similar, by ~ 175, and we are to prove that SE'F' is equal to DEF. Now, by ~ 160, SE': AC J dF': JB, and, by hypothesis, DE or SE': A C — DF: A.B. 1Hence, on account of the common ratio SE'1: /C, F':.lB - DPF: B B; that is, dF' and DF are in the same ratio to A.lB, and are consequently equal. In the same way it may be proved that E''F and EF, being in the same ratio to BC, are equal; and as the triangle DEFE has its sides equal to those of dE'F', it is equal to &E'F', and is, therefore, similar to JB C. 181. Theorem. Lines S/F, AG, &c. (fig. 100), drawn at pleasure through the vertex of a triangle, divide proportionally the base BC and its parallel DE, so that DI: BF - IK: FG -- KL: GHI, &c. Proof. Since DIis parallel to BF, the triangles ADD,./BF are equiangular, and give the proportion, D1: BEF F I: F; also, since lK is parallel to FG, l: _ F IK: A:FG; and, therefore, on account of the common ratio Al:.A, DI: B1F — IK: FG. CHon. XI. ~ 186.] SIMILAR POLYGONS. 53 Right'Triangle divided into two similar Right Triangles. It may be shown in like way, that 1K: FG- IKL: GH, &c. 18'. Corollary. VWhen BC is divided into equal parts, the parallel DE is likewise divided into equal parts. 183. Theorem. The perpendicular SD (fig. 101) upon the hypothenuse BC of the right triangle BJC from the vertex d of the right angle, divides the triangle into two triangles B.3D, CdD, which are similar to each other and to the whole triangle B.dC. Proof. a. The right triangles BAC and BAD are similar, by ~ -174, for the acute angle B is common to them both. b. In the same way it may be shown, that DJC is similar to BdC, and, therefore, to BAD. 184. Corollary. From the similar triangles B.ID, B.dC, we have BD: B= Bd: BC, that is, the leg Bd is a mean proportional between the hypothenuse BC and the adjacent segment BD. a. In the same way AC is a mean proportional between BC and DC. 185. Corollary. From the similar triangles B.2D, CAD), we have BD: DA — D: DC, or, the perpendicular D.l is a mean proportional between the segments BD, DC of the hypothenuse. 186. Theorem. If from a point d. (fig. 102), in the circumference of a circle, a perpendicular SD is 54 PLANE GEOMETRY. [CH. XI. ~ 190. To find a Mean proportional. drawn to the diameter BC, it is a mean proportional between the segments BD, DC of the diameter. Proof. For, if the chords A/B,.1C are drawn, the triangle B11C is, by ~ 109, right-angled at t/. 187. Corollary. The chord Bsi is a mean proportional between the diameter BC and the adjacent segment BD. Likewise,,/C is a mean proportional between B C and DC. 188. Problem. To find a mean proportional between two given lines. Solution. Draw the indefinite line,/B (fig. 103). Take A.C equal to one of the given lines, and B C equal to the other. Upon AB as a diameter describe the semicircle.1DB. At C erect the perpendicular CD, and CD is, by ~ 186, the required mean proportional. 189. Theorem. The parts of two chords which cut each other in a circle are reciprocally proportional, that is (fig. 104),./0: DO=- CO: BO. Proof. Join SdD and CB. In the triangles A.OD and COB, the angles dOD and COB are equal, by ~ 23; also the angles.tDO and CBO are equal, by ~ 108, because they are each measured by half the are SC, and, therefore, the triangles iOD and COB are similar by ~ 173, and give the proportion.O0: DO== CO: BO. 190. Tlzeorem. If, from a point 0 (fig. 105), taken without a circle, secants 0., OD be drawn, the entire secants./10 and DO are reciprocally proportional to the parts BO and CO without the circle, that is, t 0: DO -- C O: BO. CHi. XI. ~ 192.] SIMILAR POLYGONS. 55 To divide a linue in extreme and mean ratio. Proof. Join`C and BD. In tle triangles AOC and BOD, the angle 0 is common, and the angles B.C and BDC are equal, by ~ 108; these triangles are, therefore, similar, by ~ 173, and give the proportion AO: DO -CO: BO. 191. Th'leorem. If, fiom a point 0 (fig. 106), taken without a circle, a tangent 0 C and a secant Od be drawn, the tangent is a mean proportional between the entire secant and the part without the circle, that is, 0': C0= C0: BO. Proof. When, in (fig. 105), the secant OC is turned about the point 0 until it becomes a tangent, as in (fig. 106), the points C and D must coincide, CO must be equal to DO, and the proportion (fig. 105) 20: DO= CO: BDO, becomes (fig. 106) 0: C00= 0: B0. 192. Problemz. To divide a given line..B (fig. 107) at the point C in extreme cantd mean ratio, that is, so that we may have the proportion B: C AC: CB3. Solulion. At B erect the perpendicular BD equal to half of B. Join MD, take DE equal to BD, and AC equal to ME, and C is the required point of division. Proof. Describe the semicircumference EBF with the radius DB to meet 2/D produced in F; and, by the preceding proposition, AF: B: -B:' A 3E; and, by the theory of proportions,.dB:'F-:B B- JE':B- iE. 5 56 rPLArNE GlEOlMETrY. L[c. xi. ~ 194. Similar Polygons composed of similar Triangles. But B Q2. BD -- EF, and // -E A C; hence s- dB - EF = E - C, and B - E B -- C - B; and the preceding proportion becomes AB: AC =- C. BC. 193. Theorem. If two polygons JBC.D, &C., A' B' C' D', &c. (fig. 108) are composed of the same number of triangles aBC, JCD, &c., CR:, C!D' &c. which are similar each to each and similarly disposed, the polygons are similar. Proof. Since the triangles ABC, &c. are similar to /'B'C', &c., their angles must be equal each to each. Hence the angle d of the first polygon, which is the sum of the angles RBC, C2D, &c. is equal to the angle A' of' the second polygon, which is the sum of B'2' C', C''D', &c. Also B= — B', C -- B CA -+.-, CD == B C/,' - A, C,'D C = C, &c.; the polygons are therefore equiangular with respect to each other. Their homologous sides are, moreover-, proportional, for the similar triangles give A2B ~.3B': B C ~ B, C, and B C B'C' A C: d C' - CD: C'D', &c. Hence, by ~ 170, the polygons are similar. 194. Problem. To construct a polygon similar to a given polygon ABCD, &c. (fig. 108) upon a given line J'B', homologous to the side./B. CH. xi. ~ 196.] SITMILAR POLYGONS 57 E:qcuilaterat si:ilar Polygons are equal. Solution. Join AdC, JD,) &c. Draw'C','D', &c., making the angles B''..'-C-=BC, C'i'D- CCJD, &c. Draw B'C', making the angle AEB'C'-.IBC, and meeting' C/ at C'. Draw C'D', making the angle t'CID' 2 JCDS, and meeting Ai'D' at D'; and so on. The polygon Jt'B'C'/D &c. thus constructed, is the required polygon. Proof For, by ~ 173, the successive triangles I2'B'C', A'C'D', &c. are similar to ABC1, iCD, &.c. each to each, and therefore, by the preceding theorem, the polygons are similar. 195. 7Theoreme. If the similar polygons ABCD &c.!'B'C'D' &c. (fig. 109) have a side dB of the one equal to the homologous side'g' of the other, the polygons are equal. Proof. The polygons are, by ~ 170, equiangular; they are also equilateral, for, by ~ 170, the ratio of BC to B' C' is the same as that of,B to A'B', or the ratio of equality; that is, BC B'C', and, in the same way CD:- C'', &c. If, then,.'B' is placed upon,B, BC''vill take the direction of BC, because the angle B'-= C'; and C' will fall upon C, because B'C' B C; and in the same way it may be shown, that B' falls upon D, E' upon E, &c.; so that the polygons coincide, and are equal. 196. Theorem. Two similar polygons JBCD &c., d'B3'C'D' &c. (fig. 108), are composed of the same number of triangles.SfBC, W CD, &c.,.1'B' C',.f C'', &c., which are similar each to each and similarly disposed. 58 PLANE GEOMiETRY. [CI. XI. ~ 198. PRatio of the Peri e tlels of sinmilar Polygons. Proof. Construct upon AM'B' homologous to JiB, by ~ 194, a polygon similar to IBCD &c., and it must also be similar to.'B'IC'D' kc., and must therefore, by the preceding proposition, coincide with it; so that J/'B'C'D' &c. must, from ~ 194, be composed of triangles similar and similarly disposed to those of AB CD &c. 197. Theorem. The perimeters of similar polygons are as their homologous sides. Proof. From the definition of ~ 170, the similar polygons JBCD &c. (fig. 108), A'BiC'D' &c. give the proportion,AB:.'B' B C: B' C' - CD: CD', &Ce. Now the sum of the antecedents of this continued proportion is JB + BC + CD + &c., or the perimeter of ABCD &e., vwhich we may denote by the letter P; and the sum of' the consequents is:!' +- B' C' + C'D'~ + &c., or the perimeter of d/'B'C'D' &c., which we may denote by P'. Hence, from the theory of proportions, P: PI'- B: A'' = BC: B'C', &c. 198. Theoremz. If two homologous sides &.B, A'B', (figs. 109, 110, 111) of two similar triangles, parallelogramns, or trapezoids, are assumed as their bases, the altitudes CE, C"E' are to each other as the homologous sides. Proof. Since the acute angles CAB, C'.'B' are, by ~ 170, equal, the right triangles dEC, 9'Bt'C' are, by ~ 174, similar, and give the proportion ~ a * CE: CIeN CH. XII. ~ 203.] REGULAR POLYGONS. 59 An inscribed Equilateral Polygon is regular. 199. Corollary. The hlomologous altitudes of two similar triangles, &c. are to each other as their homologous bases. 200. Corollary. The perimeters of two similar triangles, parallelograms, or trapezoids are to each other as their homologous altitudes. CHAPTER XII. REGULAR POLYGONS. 201. Definition. A regular polygon is one which is at the same time equiangular and equilateral. Hence the equilateral triangle is the regular polygon of three sides, and the square the one of four. 202. Theorem. Every equilateral polygon, as B CD &c. (fig. 112), which is inscribed in a circle, is regular. Proof. As the polygon.IABCD &c. is supposed to be equilateral, we have only to prove that it is also equiangular. Now the arcs.1B, B C, CD, &c. are equal, for they are subtended by the equal chords JB, BC, CD, &c.; and, therefore, twice these arcs, or the arcs JIB C, BCD, CDE, 8ac. are equal. Hence the angles.ABC, BCD, CDE, &c. are equal, since they are inscribed in equal segments. 203. Theorem. An infinitely small arc,/B (fig. 113) coincides with its chord JB. 5* 60 PLANE GEOMETRY. [CI-T. XIi. ~ 204. The Circle is a regular Polygon of an infinite number of Sides. Proof. Through C the middle of the chord AB draw the radius DO. Complete the rectangle DEA/C, by ~ 147; and, as the side DE is perpendicular to OD, it is a tangent to the circle. The are A1D is, then, less than the sum of the including lines SIE +- DE-2tQ C + DC; and 2 A2 < <2 Ac + 2 DC, or the arc JB < the chord AB + 2 DC; or the arc eB A-the chordl JB < 2 DC, that is, the diffference between the arc AB and its chord is less than 2 DC. But, by ~ 186, CU- F SC-AC: CD:: 2 C: 2 CD =the chord JsB: 2 CD; that is, 2 CD has the same ratio to the chord A2B, which the infinitely small line AIC has to CF; so that 2 CD is infinitely small in comparison with the chord AB. And, as the difference between the chord and the are is smaller than 2 CDA, it must likewise be infinitely small in comparison with either the chord or the arc, and may, by ~ 99, be neglected. The are JB is, therefore, equal to the chord AB, and must, by ~ 18 and 16, coincide with it. 204. Theorem. The circle is a regular polygon of an infinite number of sides. P3roof. Suppose the circumference J2BCD &c. (fig. 114) divided into the infinitely small and equal arcs AB, B C, CD, &c. The polygon formed by the chords of these arcs is, by ~ 202, a regular polygon of an infinite number of sides; but since, by the preceding theorem, CH. XII. ~ 206.] REGULAR POLYGONS. 61 Limitation of' axiom of' Art. 99. the arcs coincide with the chords, this polygon is the circle itself. 205. Scholiunm. The two preceding demonstrations contain the following obvious and necessary limitation of the axiom of ~ 99. The infinitely small quantities, which are neglected lby the axiom of ~ 99, must be infinitely small in comparison with those which are retained. In the present case, indeed, the difference between the infinitely small are and its chord is infinitely small, and yet it could not be neglected if it were not infinitely small in comparison with the are. For, as the sum of all these differences corresponding to all the arcs of the circle has the same ratio to the sum of all the arcs, that is, to the entire circumference, which each difference has to its are; the sum of the differences, that is, the difference between the circumference of the circle and the perimeter of the polygon of an infinite number of sides, would not be infinitely small, and, therefore, capable of being neglected, unless each difference were infinitely small in comparison with its arc. 206. Theorem. Two regular polygons OBCD, &c. R2'B'lD', &c. (fig. 115), of the same number of sides, are similar. Proof. For, they are equiangular with respect to each other, since the sum of their angles is the same, by ~ 72, and each angle of each polygon is found by dividing this common sum by the number of sides. Their homologous sides are, moreover, proportional; for since AB -B BC= CD, &c. jt~B'- _ B' C' =- CD', &c. 62 PLANE GEOMETRY. [CH. XII. ~ 211. To inscribe a Regular Polygon of twice the number of Sides, &c. we have JIB: A'B' BC: B'C' CD: C'D', &c. 207. Corollary. Hence, and by ~ 197, the perimeters of regular polygons are to each other as their homologous sides. 208. Theorem. Two circles are similar regular polygons. Proof. The number of sides of each circle is any infinite number whatever, and, if we choose, the same infinite number for all circles. 209. Theorem. A regular polygon of any number of sides may be inscribed in a given circle. Proof. Suppose the circumference.ABCD kc. (fig. 116) to be divided into any number of equal arcs AB, B C, CD, &c. Their chords SB, BC, CD, &c. are also equal, by ~ 112; and the polygon AiB CD, &c. formed by these chords is, by ~ 202, a regular polygon of a number of sides equal to that of the arcs JB, B C, CD, &c. 210. Problem. To inscribe in a given circle a regular polygon, which has double the number of sides of a given inscribed regular polygon.BCD &c. (fig. 116). Solution. Bisect the arcs AB, B C, CD, &c. at the points AX, S 0, F,, &c. Join AJM, 31, BJNB,, JC, &c., and AJiBYCOO, &c. is the required polygon. Proof. For the sides JAI, aMB, BJV', NVsC, &c. being the sides of equal arcs, are equal, and, by ~ 2-02, the polygon is regular. 211. Corollary. By bisecting the arcs.l/.f, XJB, BAN, &c., a regular inscribed polygon is obtained of coi. xii. ~ 215.] REGULAR POLYGONS. 63 To inscribe a Square and a Hexagon. 4 times the number of sides of the given polygonl; and, by continuing the process, regular inscribed p)olygons are obtained of 83, 16, 32, &c. times the number of sides of the given polygon. 212. Problem. To inscribe a square in a given circle. Solution. Draw the two diameters JB and CD (fig. 117) perpendicular to each other; join AID, DB, BC, C.I; and d.DBC is the required square. Proof The arcs AfD, BD, BC, and dC are equal, being quadrants; and therefore their chords.D, DB, BC, and CA are equal, and, by 2~ 201 and 202, J1DBC is a square. 213. Corollary. Hence, by ~~ 210 and 211, a polygon may be inscribed in a circle of 8, 16, 32, 64, &c. sides. 214. Problem. To inscribe in a given circle a regular hexagon. Solution. Take the side BC (fig. 118) of the hexagon equal to the radius /C of the circle, and, by applying it six times round the circumference, the required hexagon BCDEFG is obtained. Proof. Join AC, and we are to prove that the are BC is one sixth of the circumference, or that the angle BaC is 1 of four right angles, or I of two right angles. Now, in the equilateral triangle ABC, each angle, as BA2C, is, by ~ 70, equal to 3 of two right angles. 215. Corollary. Hence regular polygons of 12, 24, 43, &c. sides may, by ~~ 210 and 211, be inscribed in a given circle. 64 PLANE GEOMETRY. [CH. XII. ~ 217'To iscribe a Decaclon. 216. Corollcary. An equilateral triangle BDF is inscribed by joining the alternate vertices, B, D, F. 217. Problem. To inscribe in a given circle a regular decagon. Solution. Divide the radius AB (fig. 119) in extreme and mean ratio at the point C. Take BD for the side of the decagon equal to the larger part /C, and, by applying it ten times round the circumference, the required decagon BDEF &c. is obtained. Proof. Join A/D, and we are to prove that the are BD is -L of the circumference, or that the angle BAD is J- of four right angles, or - of two right angles. Join DC. The triangles BCD and.BD have the angle B common; and the sides BC and BD, which include this angle in the one triangle, are proportional to the sides BD and A/B, which include the same angle in the other triangle. For, by ~ 192, BC: C = —C: B; but, by construction, BD is equal to ASC, and, being substituted for it in this proportion, gives B C: BD B D:.B. The triangles BCD and.2BD are therefore similar, by ~ 179. Now the triangle 21BD is isosceles, and therefore B CD must also be isosceles; and the side DC is equal to BD, which is equal to.C; so that the triangle.sCD is also isosceles. We have, therefore, the angle A / the angle.D C; and, by ~ 71, the angle BCD -the angle / + the angle ADC CH. XIi. ~ 222.] REGULAR POLYGONS. 65 To inscribe a Pentagon. - twice the angle./. But, in the isosceles triangles BCD and.UCD, the angle B CD-= the angle CBD the angle JDB twice the angle J, and the sum of the three angles JBD, JiDB, and.J of the triangle ABD, or by ~ 65, two right angles, is equal to five times the arngle J.. Hence,./ is 5 of two right angles. 218. Corollary. Hence, regular polygons of 20, 40, 80, &c. sides may, ~~ 210 and 211, be inscribed in a given circle. 219. Corollary. A regular pentagon BE GIL is inscribed by joining the alternate vertices B, E, G, I, L. 220. Problem. To inscribe in a given circle a regular polygon of 15 sides. Solution. Find, by ~ 217, the are /B (fig. 120) equal to -l of the circumference, and, by ~ 214, the are SC equal to ~ of the circumference, and the chord BC, being applied 15 times round the circumference, gives the required polygon. Proof. For the are BC is — 1 — 15 of the circum ference. 221. Corollary. Hence, regular polygons of 30, 60, 120, &c. sides may, by ~~ 210 and 211, be inscribed in a given circle. 222. Problem. To circumscribe a circle about a given regular polygon A/B CD &c. (fig. 121). 66t PLANE GEOMETRY. [cii. xii. ~ 227. To circu mscribe a Circle about a Reg-ular Polygon. Solution Find, by ~ 149, the circumference of a circle which passes through the three vertices, B, C; and this circle is circumscribed about the given polygon. Proof. Suppose the circumference divided into the same number of equal arcs dB', BG', &c. as that of the sides of the given polygon. The chords B/', B'C', &c. form, by ~ 202, a regular polygon, which, by ~ 206, is similar to AB CD &%c. Hence, the angle.BC7 the angle CB'C'; and, consequently, by ~ 108, the arc JBC, which is twice the are 2iB, is equal to the are fB' C', which is twice the arc dB'. We have then, the arc kB - the are B', and the chord fB is equal to the chord RB', and coincides with it. The polygons B'C"'D' &e.,A BCD &c., must, therefore, by ~ 195, coincide; and the circle is circurnscribed about the given polygon. 223. Corollary. There is a point 0 in every regular polygon equally distant from all its vertices, and which is called the centre of tihe polyg'onm. 224. Corollary. If we join CO, BO, CO, &c., the angles SIOB, BOC, COD, &c. are all equal, and each has the same ratio to four right angles, which the are B has to the circumference. 22~5. Corollary. The isosceles triangles JIOB, BOC, COD, &c. are all equal. 26. Corollary. The angles OBW, OBJ, OBC, OCB, &c. are all equal, atd each is half of the angle dBC. 227. Probleim. To inscribe in a given circle a regu CHI. XII. ~ 230.] REGULAR POLYGONS. 67 To inscribe in a Circle any Regular Polygon. lar polygon, similar to a given regular polygon J.BCD &c. (fig. 123) Solution. From the centre O of the given polygon draw the lines U.r, BO; at the centre 0' of the given circle make the angle.d'O'B' equal to S./OB, and the chord J/I'B', being applied round the circumference as many times as ABCD &c. has sides, gives the required polygon.i'B'C'D' &c., as is evident from ~ 224. 228. Theorem. The sides of a regular polygon are all equally distant from its centre. Proof. Let fall the perpendiculars OM03, OA, OP, &c. (fig. 122), from the centre 0, upon the sides AB, BC, &c. In the right triangles 0OI/311, OBI, OB.N', O CN', OCP, &c., the hypothenuses O./, OB, OC, &c. are all equal, by ~ 223, and the legs./M, MB, JB, BNC, CP, &c. are equal, since each is, by ~ 116, half of AB, or of its equal BC, &c. The triangles O./M, OBR, OBXV, &c. are, consequently, equal, by ~ 64; and the perpenliculars OJMl, ON, OP, &c. are equal. 229. Problem. To inscribe a circle in a given regular polygon,ABCD &c. (fig. 124). Solution. From the centre 0 of the polygon, with a radius equal to OM, the distance of.iB from 0, describe a circle, and it is the required circle. Proof. The distances 0.M1, OJ, O.P, &c. are all equal, by ~ 228, and therefore the circumference passes through the points J1,, P, P&c.; and the sides AIB, BC, CD, &c. are all, by ~ 120, tangents to the circle; and the circle is, by ~ 118, inscribed in the polygon. 230. Problem. To circumscribe about a given circle a polygon similar to a given inscribed polygon JIB CD &c. (fig. 125). 6 63 PLANE GEOMETRY. [cH. [CH. ~ 232. Homologous Sides of Regular Polygons. Solution. Through the points Ji, B, C, D, &c. draw the tangents./'B', B'C', C'D', &c. and I'B'CID' is the required polygon. Proof. The triangles JB'B, BCIC, &c. are by ~ 151, isosceles; they are also equal, for the sides JIB, BC, &c. are equal, and the angles BIIB',.fBB', CB C', B CC', &c. are equal because they are measured by the halves of the equal arcs JIB, BC, &c. Hence the angles.i/, B', &c. are equal, and the sides JI'B', B'C', &c. are eqnral, and J'B' C' &c. is a regular polygon of the same number of sides with JIBC &c. 231. Corollary. A regular polygon of 4, 8, 16, &c.; 3, 6, 12, &c.; 5, 10, 20, &c.; 15, 30, 60, &c. sides; or, one similar to any given regular polygon may, therefore, be circumscribed about a circle by means of ~~ 212-221, and 228. 232. Theorem. The homologous sides of regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, and also as the radii of their inscribed circles. Proof. Let.JBCD, &c., I'B'CID', &c. (fig. 126) be regular polygons of the same number of sides, and let 0, 0' be their centres; O0, 0'/i' are the radii of their circumscribed circles, and the perpendiculars OP, OIPI are the radii of their inscribed circles. Join OB, O'B'. The triangles OAB, O'J'B' are similar, by ~ 173, for the angle OJIB OBJ — O'B'-'= O'JI'B' for each is half the angle B C = =.'B' C'. Hence, by ~ 198, OP: O'P': J-IB: A'B' 0./A: O'I'. cii. XII. ~ 237.] REGULAR POLYGONS. 69 The Ratio of a Circumf'erence to its Diameter. 233. Corollary. Hence, the perimeters of regular polygons of the same number of sides are, by ~ 207, to each other as the radii of their inscribed circles, and also as the radii of their circumscribed circles. 234. Theorem. The circumferences of circles are to each other as their radii. Proof. For circles are similar regular polygons, by ~ 208, and the radii of their inscribed and circumscribed circles are their own radii. 235. Corollary. The circumferences of circles are to each other as twice their radii, or as their diameters. 236. Cor'ollary. If we denote the circumference of a circle by C, its radius by R, and its diameter by D; also the circumference of another circle by C', its radius by R', and its diameter by D', we have C: C'R: R' —D: D'. Hence C: R C: RI'; and C: D C,: D'. HI-ence, the circumference of every circle has the same ratio to its radius; and also to its diameter. 237. Corollary. If we denote the ratio of the circumference, C, of a circle to its diameter, D, by s, we have C: D- - xlSO C= — XD 2 X R, and D-== C:, R = —- C: 2. 70 PLANE GEOMETRY. [CII. XIII. ~ 241. Unit of Surface. 238. Corollary. a is the circumference of a circle whose diameter is unity, and the semicircumference of a circle whose radius is unity. CHAPTER XIII. AREAS. 239. Definitions. Equivalent figures are those which hlave the same surface. The area of a figure is the measure of its surface. 240. Definition. The tanit of surface is the square whose side is a linear unit; so that the area of a figure denotes its ratio to the unit of surface. 241. Theorem. Two rectangles, as 8BCD, EFRG (fig. 127) are to each other as the products of their bases by their altitudes, that is, dBCD: EFG = B X C: JE X AF. Prooq a. Suppose the ratio of the bases A1B to AE to be, for example, as 4 to 7, and that of the altitudes AC: AF to be, for example, as 5 to 3. AE may be divided into 7 equal parts Aa, ab, bc, &c., of which AB contains 4; and, if perpendiculars aa', bb', &c. to JE are drawn through a, b, c, &c., the rectangle JBCD is divided into 4 equal rectangles Aaa'C, abb'a', &c., and the rectangle AEFG is divided into 7 equal rectangles Aaa"cF, abb"a", &c. Again, A/C may be divided into 5 equal parts Ain, inn, &c., of which /AF contains 3; and, if perpendiculars mm', CH. XIII. ~ 245.] AREAS. 71 Area of the Rectangle and Square. nn', &c. to.RIC are drawn through In, &e, &c., each of the partial rectangles of' B CD is divided into 5 equal rectangles; and each of the partial rectangles of /EFG is divided into 3 equal rectangles; and all these small rectangles are, evidently, equal. Hence ABBCD contains 4 x 5 of them, and AEFG' contains 3 X 7 of them; that is, iB CD: A2EFG 4 X 5: 7 X 3, which is equal to the product of the ratio 4: 7 by 5: 3, or of.B: SE by AlC: SqF, so that A B CD: JI EFG — c B X A C: 21E X JA-P. b. This demonstration is readily extended to the case where the sides are incommensurate by dividing the rectangles into infinitely small rectangles. 242. Corollary. The rectangle JB CD is, consequently, by ~ 240, to the unit of surface, as AB X AQC to unity, or as the product of its base multiplied by its altitude to unity. Hence the area of a rectangle JtBCD is the product of its base by its altitude. 243. Corollary. The area of a square is the square of one of its sides. 244. Corollary. Rectangles of the same altitude are to each other as their bases, and rectangles of the same base are to each other as their altitudes. 245. Theoremn. Any two parallelograms JBCD, ABEF (fig. 128) of the same base and altitude are equivalent. Proof. The triangles JCF and BDE are equal, by ~ 51; for the sides.iC and BD are equal, by ~ 78, being 6* 72 PLANE GEOMETRY. [CH. XIII. ~ 252. Area of the Parallelogram and'riangale. the opposite sides of B CD; also dF and BE are equal, being the opposite side of,/BFE; and the angles Cz2F and DBE are equal, by ~~ 29, since they have their sides parallel. If, now, the triangle A CF is subtracted from the whole figure,BCE, the remainder is SBFE; and if BDE is subtracted from the whole figure, the remainder is AB CD. Hence, as IB CD and./BFE are the remainders, after taking equal triangles from the same figure, they must be equivalent. 246. Corollary. A parallelogram is equivalent to a rectangle of the same base and altitude. 247. Corollary. The area of a parallelogram is the product of its base by its altitude. 248. Corollary. Parallelograms of the same base are to each other as their altitudes; and those of the same altitude are to each other as their bases. 249. Problem. Every triangle is half of a parallelogram of the same base and altitude. Proof. For the triangle.BC (fig. 39) is, by ~ 77, half of the parallelogram ABCD of the same base and altitude, and it is, therefore, by ~ 245, half of any parallelogram of the same base and altitude. 250. Corollary. All triangles of the same base and altitude are equivalent. 251. Corollary. The area of a triangle is half the product of its base by its altitude. 252. Corollary. Triangles of the same base are to each other as their altitudes, and triangles of the same altitude are to each other as their bases. CH. XIII. ~ 254.] AREAS. 73 Area of the Trapezoid. 253. Theorem. The area of a trapezoid is half the product of its altitude by the sunm of its parallel sides. Proof. Draw the diagonal./D (fig. 129); the trapezoid AB CD is divided into two triangles A CD and.tBD, the bases of which are JB and CD, and the altitude of each is, by ~ 82, EF. The area of ABD is, by ~ 251, -- EF x JIB, and the area of A CD - EFX CD; the sum of which is the trapezoid ABCD - EF X (AB ~ CD). 254. Lemma. The line, which joins the middle points of the two sides of a trapezoid which are not parallel, is parallel to the two parallel sides, aid is equal to half their sum. Proof. a. Through the middle points IH and I (fig. 129) of the sides AC and BD, draw HI, and through I draw 0' parallel to CAd. Tilhe triangles DIO and ITB have the side DI equal to iB, the angle DIO equal to the vertical angle BIT, and the angle IDO equal, by ~ 30, to IB T; and, therefore, the triangles DIO and ITB are equal, by ~ 53; and OI=HIT=- OT. But, in the parallelogram OCAT, we have, by ~ 78, CAI — OT; whence OI —- Cs= CH; so that CHIO is, by ~ 81, a parallelogram; and HI is parallel to CD, and also to /B. 74 PLANE GEOMETRY. [Ci. xIIi. ~ 256. Theorem of Pythagoras. b. Again, in the equal triangles DIO, TIB, we have DO- TB; whence HI-= CO-= CD + DO, and also HII — T -- B -B T- B -- DO; the sum of which is 2HI =-B + CD, or IIIzr 1 (.JB + CD). 255. Corollary. The area of a trapezoid is the product of its altitude by the line joining the middle points of the sides which are not parallel. 256. Theorem. The square described upon the hypothenuse of a right triangle is equivalent to the sum of the squares described upon the other two sides. Proof. Let squares be constructed upon the three sides of the right triangle ABC (fig. 130), right-angled at B. From B let fall upon SC the perpendicular BDE, and the square JJCSR is divided into the two rectangles.4DER and DCES. Now the area of LJDER is, by ~ 242, J/D X A/R -JD X.CW; and the area of the square JBXMJJI is, by ~ 243, B'B2. But, by ~ 184, AD: AB A /B': J1C; or multiplying extremes and means, AB=- == D X.a1C; that is, the square IABAN'JM is equivalent to the rectangle Q.DER. CH. XIII. ~ 260.] AREAS. 75 Ratio of the Squares of the Sides of a Right Triangle. It may be shown in the same way, that the square BCP is equivalent to the rectangle D CSE; and, therefore, the square.CSR is equivalent to the sum of the squares sBXJ.Ill and BCPO, or A C2 2B - + B C2. 257. Corollary. The square of one of the legs of a right triangle is equivalent to the difference between the square of the hypothenuse and the square of the other leg; or. B'2 -.1 C2 B 02. 258. Corollary. In the square (fig. 117),,B2 _ JD2 + DB2 _= 2 sD2 - 2 X JDB C; or the square described upon the diagonal of a square is twice as great as the square itself. Hence JLB2: lD2:1; and, extracting the square root, dB: AD — = v/ 2: 1. 259. Corollary. Since (fig. 130), AB' -=- lD X A1C, we have A C'2: /B2'== 0 C X d C: AlD X Al0 =S.iCA: lD; and, in the same way, dC2: BC2- C C: DC; or the square of the hypothenuse of a right triangle is to the square of one of its legs, as the hypothenuse is to the segment of the hypothenuse adjacent to this leg, made by the perpendicular from the vertex of the right angle. 2060. Corollary Since JB=2-D VX SC 76 PLANE GEOMIETRY. [CII. XIII. ~ 263 To make a Square equal to the Sum or Difference of given Squares. and BC2 -- DC x SC we have JB2: BC2D —D X AC: DC X AC -.2D: DC; or the squares described upon the two legs of a right triangle are to each other, as the adjacent segments of the hypothenuse made by the perpendicular from the vertex of the right angle. 261. Problem. To make a square equivalent to the sum of two given squares. Solution. Construct a right angle C (fig. 131); take CA equal to a side of one of the given squares; take CB equal to a side of the other; join //B, and./B is a side of the square sought. Proof. For, by ~ 256, A:B2 = / C2 + B C2. 262. Problem. To make a square equal to the difference of two given squares. Solution. Construct, by ~ 145, a right triangle, of which the hypothenuse BC (fig. 79) is equal to the side of the greater square, and the leg AB is equal to the side of the less square; and A.C is the side of the required square. Proof. For, by ~ 257, C2 -- B C2 - B 263. Problem. To make a square equivalent to the sum of any number of given squares. Solution. Take AB (fig. 132) equal to the side of one of the given squares. Draw BC, perpendicular to.B, and equal to the side of the second given square. Join C, and draw CD, perpendicular to J/C, and equal to the side of the third given square. cii. XIII. ~ 265.] AREAS. 77 To make a Square ill a given Ratio to a given Square. Join J]D, and draw DE, perpendicular to./D, and equal to the side of the fourth given square; and so on. The line which joins b. to the extremity of the last side is the side of the required square. Proof. For, by ~ 256,. C2 -- AB2 + B C2, A D2 =- d C2 + CD2,= 3B2 + B 2 + CD2, lE2. S.D2 + DE2 = AB2 + B C'+ CD2 + DE2; &c. 264. Scholiumn. If either of the squares BC", CD2, &c. were to have been subtracted instead of being added, the problem might still have been solved by means of ~ 262. 265. Problem. To make a square which is to a given square in a given ratio. Solution. Divide any line, as EG (fig. 133), by ~ 163, into two parts, at the point F, which are to each other in the given ratio of the given square to the required square. Upon EG describe the semicircle EJMG; draw FPM perpendicular to EG. Join FME and MJIG; take, on MIE produced if necessary, MfH —B= B the side of the given square. Draw Ill parallel to EG, meeting JMG in I, and XMI is the side of the required square. Proof. Produce MF to P; and, as the triangle H.MI is, by ~ 109, right-angled at.11, we have, by ~ 260, MH2: J112 - iP: PI. But by ~ 181, HP: P I-EF: FG; whence, on account of the commoniratio HP: PA, J1IH2: MII'= —-EF: FG. 78 PLANE GEOMETRY. [CIi. XIII. ~ 268. Ratio of Similar and Regular Polygons. 266. Theorem. Similar triangles are to each other as the squares of their homologous sides. Proof. In the similar triangles JB C, 2'_B' C' (fig. 109), we have, by ~ 199, CE: C'E' -- B: Jd'B', which, multiplied by the proportion AB 1 J'B' = — JB:'B', gives 2B X CE:~ B'X C'E' —1B 2:.IB2 and, by ~ 251, the area of dSB C: the area of./'B C' B — 2: 2J'B'B2. 267. Corollary. Hence, by ~ 197 & 198, similar triangles are to each other as the squares of their homologous altitudes, and as the squares of their perimeters. 268. Theorem. Similar polygons are to each other as the squares of their homologous sides. Proof. In the similar polygons ABBCD &c., J'IB'C'D' &c. (fig. 108), the triangles JBBC, 2'BiC', which are similar, by ~ 196, give, by ~ 266, the proportion ~AB C: AI/RB C'. C2 ~.3/ C/2; also the similar triangles ACD, JT'C')', give the proportion A CD' t' CD'-'. t2 C: A' C' 2' hence, on account of the common ratio &.2C2: t'C2, AB C:'B' C A CD: D' C'D'. In the same way may be obtained the continued proportion B C: w' B' C' - --.CD: dJ' CD' - DE: JID'E', &c. Now the sum of the antecedents,B C, A CD, ADE, &c. cn. XIII. ~ 272.] AREAS. 79 Ratio of Circles. is the polygon J3 BCD &c., and the sum of the consequents JI'B' C, JI C'D', A'D'E', &c. is the polygon J'B' C'D' &c.; so that, by ~ 266, JB CD &c.: d'B' C'D' &c. = IB C: JI'BC' C- IB2: I'B'2. 269. Corollary. Similar polygons are, therefore, to each other, by ~ 197, as the squares of their perimneters. 270. Corollary. As regular polygons of the same number of sides are, by ~ 206, similar polygons, they are to each other as the squares of their homologous sides, and, by ~ 232, as the squares of the radii of their inscribed circles, and also as the squares of the radii of their circumscribed circles. 271. Theovrem. Circles are to each other as the squares of their radii. Proof. For, by ~ 208, they are regular polygons of the same number of sides, and, as in ~ 234, the radii of their inscribed and circumscribed circles are their own radii. 272. Problem. Two similar polygons being given, to construct a similar polygon equivalent to their sum or to their difference. Solulion. Let ei and B he the homologous sides of the given polygons. Find, by ~ 261, or by ~ 262, the line X such that the square constructed upon Xis equal to the sum or the difference of the squares constructed upon JI and B. The polygon similar to the given polygons, constructed, by ~ 194, upon the side X homologous to JI or B, is the required polygon. Proof. For, by ~ 268, the, similar polygons construct7 80 PLANE GEOMETXY. [c(I. XIII. ~ 277. To make a Polygon in a given Ratio to similar Polygons. ed upon 4, B, and X, have the same ratio to each other as the squares upon d, B, and X. 273. Corollary. If d and B were the radii of two given circles, X would, by ~ 271, be tile radius of a circle equivalent to their sum or to their difference. 274. Corollary. By the process of ~ 263, a polygon might be constructed equivalent to the sum of any number of given similar polygons, and similar to them, or a circle equivalent to the sum of any number of given circles; or, if either of the given polygons or circles is to be added instead of being substracted, the resulting polygon or circle may be obtained, as in ~ 264. 275. Problemi. To construct a polygon similar to a given polygon, and having a given ratio to it. Solution. Let d1 be a side of the given polygon. Find, by ~ 265, the side Xof a square which is to the square, constructed upon 4, in the given ratio of the polygons. The polygon, constructed upon X, similar to the given polygon, is the required polygon. Proof. For, by ~ 2168, the similar polygons constructed upon./ and X, have the same ratio to each other as the squares upon 2. and X. 276. Corollary. In the same way, a circle may be constructed having a given ratio to a given circle, by taking for. and X the radii of the given and of the required circles. 277. Theorem. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle. Proof. From the centre 0 (fig. 134) of the circle draw ciH. XIII. ~ 282.] AREAS. 81 Area of a Circle. 02, OB, OC, S&c. to thle vertices of the circumscribed polygon./B CD, &c. Draw the radii GOJ, OAN, OP, &c. to the points of. contact of the sides. If, now, the sides sIB BC, CD, kc. are taken for the bases of' the triangles OJqB, OBC, OC!D, &c.; their altitudes, being the radii OGX, ON, OP, &c., are all equal. The area of each of these triangles is, then, by ~ V51, half the product of its base tB, BC, CD, &-c. by the common altitude O. The sum of the areas of the triangles, or the area of the polygon is, consequently, half the product of the sum of the sides, JlB, BC, &c. by the common altitude OGJI; that is, half the product of the perimeter.JBBCD &C. of the polygon by the radius GOM. 278. Corollary. Since a circle can, by ~ 229, be inscribed in any regular polygon, the area of the regular polygon is half the product of its perimeter by the radius of its inscribed circle. 279. Theojrem. h,'he area of a ciircle is half the product of its circumference by its radius. Proof. F'or a circle is, by ~.204, a regular polygon, and the radius of its inscribed circle is its own radius. 80. Corollary. If we use C, D, R, and T, as in ~ 237, and denote by sA the area of a circle, we have ol —. C X C R 2. o X R X RX X R2 X D2. 281. Corollary. Wvhen R=1, we lhave.3=. 282. Definition. A sector is a part of a circle comr 82 PLANE G~EOIMETRY. CIi. XIII. ~ 2S5. An infinitely simall Sector is a''riangle. prehended between an arc and the two radii drawn to its extremities, as JtQOB (fig. 135). 283. Theorem. The area of a sector is half the product of its are by its radius. Proof. Suppose the are./B (fig. 135) of the sector:OB divided into the infinitely small arcs AJLYI, JMiN, JNP, 8&c. Draw the radii O0'1, o0, OP, &c. The sector./OB is divided into the infinitely small sectors.OM1, MION, NJVOP, &c.; which may, by ~ 203, be considered as triangles, having for their bases.fAII, JI.N, NPP, &c., and for their altitudes the radii S04 0l, ON, O &c. The sum of the areas of these triangles, or the area of the sector is, then, half the product of the sum of the bases.J3f, JMN NP, &c. by the common altitude 0.ft; that is, half the product of the are JsB by the radius O. 2S4. Corollary. The area of the segment.DB is found by subtracting the area of the triangle S.OB from that of the sector ORB. 285. Scholium. In order that no doubt may exist with regard to the accuracy of the demonstrations of ~ 283, 279, and 271, it is important to show that the infinitely small quantities, whicl are neglected in considering an infinitely small sector as a triangle with a base equal to its are and an altitude equal to its radius, come within the limitation of ~ 205. Now, the difference between the infinitely small sector.1OB (fig. 113), and the triangle GOB, is the segment 1DB. But the segment.DB is less than the rectangle J1EE'B; and, by ~ 242 and 251, the rectangle JEE'B: the triangle.O GB —B X CD: I 1B X OC — CD2: OC; =-2 CD: OC; cii. xIII. ~ 288.] AREAS. 83 Ratio of Similar Sectors and Segments. and, therefore, as G2 CD is infinitely small in comparison with OU, the rectangle dJEE'B and the segment.1DB must be infinitely small in comparison with the triangle.O0B, and may be neglected by ~ 1204; so that the sector S.OB is equivalent to the triangle /OB. The base of the triangle JfOB is the chord AqB, or, by ~ 203, the arc A.B; and its altitude OC differs from the radius OD by the infinitely small quantity CD, which may be neglected. The error arising from the neglect of these infinitely small quantities is altogether insensible, and cannot be rendered sensible by any magnifying process to which the mind can submit it; it is, then, no error at all. Indeed, if there be an error, suppose it to be represented by.3. Since the aggregate of the quanties neglected is infinitely small, that is, as smaUEll as iwe choose; we can choose it to be less than the error A; a manifest absurdity, for the error cannot be greater than the aggregate of the quanties neglected, and yet we cannot escape this absurdity so long as we suppose the error.3 to be of any magnitude whatever. 286. Definition. Simnilar sectors cndl similar segnents are such as correspond to similar arcs. 287. Theorem. Similar sectors are to each other as the squares of their radii. Proof. The similar sectors JOB,./'O'B' (fig. 136) are, by the same reasoning as in ~ 97, the same parts of their respective circles, which the angle 0 01 is of four right angles; and, therefore, they are to each other as these circles, or, by ~ 271, as the squares of the radii A.o1,.'0'. 288. Theorem. Similar segments are to each other as the squares of their radii. 7* 84 PLANE GEOMETRY. [CII. XIII. ~ 289. To find a Triangle equivalent to a given Polygon. Proof. Let d./DB, 3'D'B' (fig. 136) be the similar segments. The triangles d OB and.11'O'B' are similar, by ~ 179; for 0 -= 0'; and, since./0==BO and /'OI = B' 0', we have O': d0O'1 — BO: B'l'. Hence, by ~ 266, the triangle S. OB: the triangle.' O'B' = - 102:./1' 02; also, by the preceding article, the sector. OB: the sector./' 0GB' - 0 1:.' 0'2. Hence, by the theory of proportions, the sector.1OB — the triangle A1OB: the sector./'0'B' -the triangle G' O'B'. 02 A' 0'2; that is, the segment dDB: the segment d'D'B'-= A02.:.,1012. 289. Probleem. To find a triangle equivalent to a given polygon. Solution. Let 1dBCD kc. (fig. 137) be the given polygon. Join BD; through C, draw CJi parallel to BD. Join DiM, and SIDE, &,c. is a polygon equivalent to the given polygon, and having the number of its sides less by one. In the same way, a polygon may be found equivalent to.1il/DE, and having the number of its sides less by one; and by continuing the process, the number of sides may be at last reduced to three, and a triangle is obtained equivalent to the given polygon. Proof. a. The number of sides of 1JAIDE &c. is less by one than that of dBCD &c.; forthe two sides JffM, JIiD) are substituted for the three sides d.B, BC, CD, the other sides remaining unchanged. b. The polygon J.MDE &c. is equivalent to./BCD cii. xiii. XII. ~ 294.] AR:IAs. 85 Quadrature ef tPotlygon and Circle. &c.; for the part.JBDE &c. is common to both, and the triangles DBC, DBJMI are equivalent because they have the same base BD and the same altitude, by ~ 82, their vertices C and Jl being in the line CJki parallel to this base. 290. Problem. To find a square equivalent to a given parallelogram. Solution. Let B be the base and J the altitude of the given parallelogram. Find, by ~ 188, a mean proportional Xbetween A. and B, Xis the side of the square sou ght. Proof. For we have.: X =X: B, and, therefore, X2 - x B; or, by ~~ 242 and 243, the square constructed upon Xis equivalent to the given parallelogram. 291. Corollary. A square may be found equivalent to a given triangle, by taking for its side a mean proportional between the base and half the altitude of the triangle. 292. Corollary. A square may be found equivalent to a given circumscribed polygon, by taking for its side a mean proportional between the perimeter of the polygon and half the radius of the inscribed circle. 293. Corollary. A square may be found equivalent to a given circle, by taking for its side a mean proportional between the radius and half the circumference of the circle. 294. Corollary. In general the quadrature of any given polygon may be found, that is, a square may be 86 PLANE GEOMETRY. [CH. XIII. ~ 296, To construct a Polygon of a given Area and similar to a given Polygon. found equivalent to any given polygon, by finding, by ~ 289, the triangle which is equivalent to the polygon, and, by ~ 291, the square which is equivalent to this triangle. 295. Problem. To construct a polygon equivalent to a given circle or polygon, P, and similar to a given polygon, Q. Solution. Find, by ~~ 293 or 294, Jl the side of a square equivalent to P, and.JN the side of a square equivalent to Q. Let.1 be one of the sides of Q. Find, by ~ 165, a fourth proportional X, to.N, Ml, J. The polygon constructed by ~ 194, similar to Q upon X, homologous to k3, is the required polygon. Proof. Let Ybe the polygon constructed upon X, we have only to prove that it is equivalent to P. Now we have JV: MJ Al:X, whence N.2: M, _ 2: X2 Also, by ~ 268, Q: Y= —=A2: X2, and leaving out the common ratio J2/: X2, J2 IN 2 -=Q: Y. But N2 _ Q and Ms2 P, whence Q: P= Q: Y, or Y P. 296. Problem. To construct a circle equivalent to a given polygon. Solution. Find, by ~ 294, M the side of a square equivalent to the given polygon. Find, by ~ 265, R the side of a square which is to the given square in the ratio of the diameter of a circle to its circumference. R is the radius of the requirled circle. C1H. XIII. ~ 298.] AREAS. S7 To construct a Parallelogramn equivalent to a given Square. Proof. Using Tt as in ~ 237, we have, by construction, 3P-: R 7 Cwhence 7 R2 -= Ja2 -the given polygon. That is, by ~ 280, the circle of which R is the radius is equal to the given polygon. 297. Problem. To construct a parallelogram, equivalent to a given square, and having the sum of its base and altitude equal to a given line. Solution. Upon the given line AJB (fig. 1.33) as a diameter describe a semicircle. At A1, erect the perpendicular.C equal to the side of the given square. Draw CD parallel to JIB, to meet the circumference at D. Draw DE perpendicular to AB; JE and ELB are the required base and altitude. Proof. For JiE + EB JIB, and by ~ 290, JE X EB 1 DE2 - 2 C2. 29S. Problem. To construct a parallelogram, equivalent to a given square, and having the difference of its base and altitude equal to a given line. Solution. Upon the given line.B (fig. 139) as a diameter describe a circle. At sl draw the tangent JiC equal to the side of the given square. Through the centre 0 of the circle, draw the secant COE. CD and CE are the required base and altitude. Proof. For we have CE- CD- D DE —B, and, by ~ 191, CE: AC AC: CD, whence AWC- -CE X CD. 88 PLANE GEOIMETRY. [CHI. XIII. ~ 01, Ratio of a Circu tlre rence to its i)ianetelr. 299. Lemma. If in a circle, whose radius is II, C is the chord of an arc and C' the chord of half the arc; C, C' and R will always satisfy the equation 6C2 2 R2 _ - V (4 k2 - C2). Proof. Let QB (fig. 140) be the chord C and let flq' be C'; OJFML is, by ~ 117, perpendicular to,A/B, and the triangle 03J.i gives O!2 OA - Mq: - - 12 - (1 C) 2= R2- C2 Hence, by ~ 187,'AI =/ J'o- 0oM R — /R _(R 2- C2) ~ —. J/2 _ --.2 /tM X.i'D =2pR2 R V_2(R-2 C) -2 2 - R y/(4 R2 - C2) 300. Corollary. When R 1, this equation becomes C -_ /V(4 - C2). 301. Problem. To find the ratio of the circumference of a circle to its diameter. Solution. This ratio has been denoted, in % 237, by 7; it does not admit of being expressed in numbers, and can only be obtained approximately. The principle of approximation consists in supposing the circumference to be equal to the perimeter of some one of its inscribed polygons: and the error of' this hypothesis is the less, the greater the number of sides of the polygon. First fpro.xoimallion. Let the radius J./O (fig. 140) of the circle be unity, and its circumference is, by ~ 238, 2 rr. If, now, the hexagon 1B CD Sc. is inscribed in the circle, we have, by ~ 214, for its side, JB- 1; Cn;. xIII. ~ 301.] AREAS. 89 Ratio of a Circumference to its Diameter. and for its perimeter 6 X 2B - 6; so that, by supposing this perimeter to be equal to the circumference, we have for a first approximation 2 = 6, or --- 3. Second dpproxiimaltion. Bisect the arcs.iB, BC &c by the radii O2', OB' &c. Join AA',.A'B &c., and we have an inscribed polygon of 12 sides, and, by ~ 300, JJA1,2 =_ o V/4 __ dBe, J./' == v(2-,/4 - BB2). But./B 1, whence ail' /2 - V —-V3 0'517 nearly. Hence the perimeter.Jl'BB' C Sc. - 12 x sAJd' 6'204 nearly. And, if this is assumed for the circumference, we have, for the second approximation, n7 3-102 nearly. Third.i3p)roxiimation. If now we consider YB as the side of the inscribed polygon of C12 sides,.2' is the side of the polygon of 24 sides, and we have for JAB, AB V 2 -- V3 0 517, B'2 = 2 — /3 0267, 12 -- 2 -- J4 - JB2 = 2 - / —- + 3- 0-068. Rf' 0 0261. The perimeter.I2/B &c. = 24 X.,A/' = -626; and, by assuming this perimeter for the circumference, we have v c 3l13 nearly. Further approximations might be obtained by supposing.B successively to be the side of an inscribed polygon of 24, 48, &c. sides, and by carrying the calculation to a 90 PLANE GEOSIETRY. [CIi. XIV. ~ 303. Value of vt. greater number of decimals. But it is useless to extend this process any further, as much more expeditious methods of calculating the value of u are obtained from the higher branches of mathematics, by means of which it has been calculated to 140 places of decimals. For almost all practical purposes, the value of n — 3 141G, is sufficiently accurate. CHAPTER XIV. ISOPERIMETRICAL FIGURES. 302. Definitions. Those figures which have equal perimeters are called isoperimetrical figures. Among quantities of the same kind, that which is greatest is called a. maximum; and that which is smallest a minimum,. Thus the diameter of a circle is a maximonum among all inscribed straight lines; and a perpendicular is a mirnimnzmn among all the straight lines drawn from a given point to a given straight line. 303. Theorem. The maximum of isoperimletrical triangles of the same base is that triangle in which the two undetermined sides are equal. Proof. Let the two triangles d CB and JIDB (fig. 141) have the same base dB, and the same perimeter, that is,./B +. C + B C =.B + SAD [- BD, CH. XIV. ~ 304.] ISOPERIMETRICAL FIGURES. 91 Maximum of Isoperimetrical Triangles of the samle Base. or, taking away./B, AC+B =1D+BD, and suppose 1CB isosceles, or ASC CB. We are to prove that the triangle 1GCB > the triangle 21DB. But, since these triangles have the same base AB, they are to each other as their altitudes CE and DF; so that we need only prove CE > DF. Produce 11C to -, making CH-= CB 1GC. Join BIH; and if a semicircle is described upon SEH as a diameter with the radius 1C- G —CH, it will pass through the point B; and AJB1, being inscribed in it, must be a right angle. Produce BH towards L, and take DL —DB. Join fL, and we have 1D + DL ==D + DB =C +CB mA1-G+G C=z-zH. But 1lD +- DL > 1L, or 1H> A1L. Hence, by ~ 41, BH > BL, and 1 BH> >- BL. Now, letting fall the perpendiculars C1 and D.M upon BH and BL, we have - BH-= BI-= CE, BL -BJ —F DF; whence CE > DF. 304. Theorem. The maximum of isoperimetrical polygons of the same number of sides is equilateral. Proof. Let A2BCD &c. (fig. 142) be the maximum of isoperimetrical polygons of any given number of sides. 8 92 PLANE GE OlETRY. [CII. xIV. ~ 306. Maximum of Polygons formed of sides all given but one. Join SC. The triangle.BC roust be the maximum of all the triangles which are formed upon YC, and with a perimeter equal to that of ABBC. Othervise a greater triangle./FC could be substituted fobr ABC, without changing the perimeter of the polygon, which would be inconsistent with the hypothesis that.BCD &c. is the maximum polygon. Therefore, by the preceding article, AB BC. In the same way it may be proved, that B C CD=- DE, &c. 305. Theorem. Of all triangles, formed with two given sides making any angle at pleasure with each other, the maximum is that in which the two given sides make a right angle. Proof. Let ABC, SD C (fig. 143) be triangles, formed with the side.C common and the side iB:= AJLD, and suppose BA/C to be a right angle. As these triangles have the same base.AC, they are to each other as their altitudes A1B and DE. But A.B -==-D, and, by ~ 39, dD > DE; whence hB > DE, and the triangle /B C > the triangle JD C. 306. Theorem. The maximum of polygons formed of sides, all given but one, can be inscribed in a semicircle having the undetermined side for its diameter. Proof. Let A.BCD &c. (fig. 144) be the maximum polygon formed of the given sides J/B, BC, CD &c. Draw from either vertex, as D, to the extremities.1 CH. XIV. ~ 303.] ISOPERIMIETRICAL FIGURES. 93 Maximum of Polygorns formed of given Sides. and S of the side not given, the lines D., DS. The triangle ADS must be the maximum of all triangles formed with the sides A and S; otherwise, either by increasing or else by diminishinishing the angle ADS, the triangle IADS would be enlarged, while the rest of the polygon JRBCD, DEF S&c. would be unchanged; so that the polygon would be enlarged, which is inconsistent with the hypothesis that it is the maximum polygon. The angle SJDS is, therefore, a right angle by the preceding article, and is inscribed in the semicircle which has A.iS for its diameter. 307. Theorem. The maximum of all polygons formed of given sides can be inscribed in a circle. Proof. Let ABCD &c. (fig. 145) be a polygon which can be inscribed in a circle, and.3'B'C'D' &Sc. one which cannot be inscribed in a circle, but equilateral with respect to B CD &c.:Draw the diameter MA3I. Join E Si, JF. Upon E'F', equal to EF, construct the triangle E'ij'F', equal to Ea i', and join A;'". The polygon AB DEJi, which is inscribed in the semicircle having /Ji for its diameter is, by the preceding article, greater than R'B'CI'D!E'M3 formed of the same sides but one, and which cannot be so inscribed. In the same way the polygon AJffFG kScC. > A.PI F'GIG kec. Hence, the entire polygon AIB CDEM/Fu &c. > IB'' C'D' ME'3' kSec., and, subtracting the triangle EJMIF — =bE'J'F' the polygon AB CD &e. > 2'B' C'D' &c. 308. lTheorem. The maximum of isoperimetirical polygons of the same number of sides is regular. 94 PLANE GEOMETRY. [CH. XIV. ~ 309. Greatest of' lsoperi:etrical legular Polygons. Proof. For, by ~ 304, it is equilateral; and, by the preceding article, it can be inscribed in a circle; so that, by ~ 202, it is regular. 309. Theorem. Of isoperimetrical regular polygons that is the greatest which has the greatest number of sides. Proof. Let A.BCD &c., A.'B'C'D' &c. (fig. 146) be two isoperimetrical regular polygons, of which.BCD &c. has the greater number of sides. Denote the area of JtB CD &c. by S, and the radius 0Hi of its inscribed circle by R1; and denote the area of.f'B'C'D' &c. by S', and the radius O'H' of its inscribed circle by R'; also the common perimeter of the two polygons by P. Then we have, by ~ 277, S.S'- 5 P X R': IP X RI, or, striking out the common factor~ 1 P,: SI== R:'; so that, in order to prove s> S, we have only to prove R >R,. Upon AlB', as a side, describe a polygon 3'B'C"D" &c. similar to AJBCD &c.; denote its perimeter by P', and the radius 0'q"M of its inscribed circle byv R". Join./'O' and Al'O"; describe the are JI'.N with the radius R', and the arc JMf'J' with the radius R". The half side /31'/ is, evidently, the same part of the perimeter P, which the are J3'JV is of its circumference, which circumference is, by ~ 237, equal to 2 0 X R'; that is, 3'.1M': P —' M.I': 2 or X R', CIu. xiv. ~ 309.] ISOPERIMIETRICAL FIGURES. 95 Greatest of' Isoperimetrical Regular Polygons. and in the same way, P"I': I' — = 2n X R'I: J'.7.. the product of these two proportions is, by striking out the factors common to the terms of each ratio, P":P IR" X 32IJ9: A' x M'INI. But, by ~ 233, P": P = ": R and, on account of' the common ratio P'': P, I": R Rt" X J'.MN: P' X JVXW', which, multiplied by the identical proportion R': RI/' - I: RI, gives, by striking out the common factors, Ro: R - JI'JB: JM,' so that we need only prove fJV1' > MI'J, in order to prove R > iR. Now, the angle &' O'.J1' is obtained by dividing 3600 by twice the number of sides of the polygon./'B' C'D', &c., and the angle O2!0"M12 is obtained by dividing 3600 by twice the number of sides of the polygon./'B4 C"D'" &c., but the second number of sides was supposed to be greater than the first, and, therefore, the angle'l O"JM' < the angle AltO'la'; and, therefore, as the angle O"'J'IM' is, by ~ 69, the remainder after subtracting the angle /'O"J.M' from 900, it is greater than the angle 0'./'J' which remains after subtracting ~/' 0'J' from 900; and O".1f'JI' includes O':JUlM'; so that the radius.M' 0" is greater than 1Ji 0', and the circle described with Mf'O" as a radius includes the circle described with 1'10' as a radius. 8' 96 PLANE GEONETRY. [cii. XIV. ~ 310. Maximum of' Isoperimetrical Figures. Join A"'; and upon the middle of VNI erect a perpendicular meeting the tangent JVT to the arc MJVJ' at T, which it will do, for the angle TJA'J, being less than the right angle TJ.VL, is acute. Join NJI T, and, by ~ 42,.VT A —- "'T. But since the concave broken line TAJ"I' is included by TNr'JMI', we have TJ' + JAV', > TJV + NJM, whence, omitting TJA' equal to TAN," X/Th VI, > AM/, and, therefore, R > R', and S > S'. 310. Corollary. As the circle is a polygon of an infinite number of sides, that is, of a greater number of sides than any other regular polygon, it is greater than any polygon of a finite number of sides which has a perimeter equal to the circumference of the circle. SOLID GEOMETRY. CHAPTER XV. PLANES AND SOLID ANGLES. 311. Theorem. Three points not in the same straight line determine the position of the plane in which they are situated. Proof. For if any plane, passing through two of the points, is swung around the line joining these two points, until it comes to a position in which it passes through the third point, it must remain in this position. For swinging it any further must remove it from this third point. 312. Corollary. Only one plane can be drawn through three points not in the same straight line. 313. Theorem. The common intersection of two planes, which cut each other, is a straight line. Proof. For, if any two of the points common to the two planes be joined by a straight line, this straight line must, by ~ 14, be in both of the planes; and no point out of this straight line can, by ~ 312, be in the two different planes at the same time. 314. When two planes cut each other, they form an angle, the magnitude of which does not depend b 98 SOLII GEOMETRY. [CH. XV. ~ 317. Intersection and Angle of two Planes. upon the extent, but merely upon the position of tile planes. 315. Theorem. TI'he angle of two planes, which cut each other, is measured by the angle of two lines drawn perpendicular to the common intersection of the two planes, at the same point, one in one of the planes, and one in the other. Proof. In order to show the legitimacy of this measure we have only to prove that the angle of the two lines is proportional to the angle of the two planes. Let JB (fig. 147) be the common intersection of the two planes; and let J.C and AtD be the two lines drawn in these planes perpendicular to the common intersection AB. Let a third plane be drawn having also the common intersection JfB with the two given planes, and let S.iE be drawn in this plane perpendicular to,3B. We are to prove that the angle of the planes DI3B and CJB is to that of the two planes EAB and CJAB as D3 C is to.ES C. For this purpose, suppose the angles of the planes to be to each other as any two whole numbers, and let the angle of the two planes CtB and a.B be their common divisor,./ a being perpendicular to.JB. The angle C./ a must be a common divisor of the two angles C./E and C.aD; and it is shown by precisely the reasoning so often adopted, that the angles of the planes are to each other as C-3D to CAE. 316. Corollary. When the angle C.3D is a right angle, the planes are perpendicular to each other. 317. Definitions. A straight line is perpendicular to a plane, when it is perpendicular to every straight line drawn through its foot in the plane. CII. xv. ~ 320.] PLANES AND SOLID ANGLES. 99 Iine Perpendicular to a Plane. Reciprocally, the plane, in this case, is perpendicular to the line. The foot of the perpendicular is the point in which it meets the plane. 318. Theoremn. When a straight line is perpendicular to two straight lines drawn through its foot in a plane, it is perpendicular to every other straight line drawn through its foot in the plane, and, consequently, is perpendicular to the plane. Proof. Let CAIC', DAD' (fig. 148) be the two lines to which J.B is perpendicular, and let ElIE' be any other line drawn in the plane, we are to prove that BA. is perpendicular to EWE'. Take A C equal to A.C', and ASD equal to S.D', join DC D'C'. Turn D'.3C'E' around upon the point AX, keeping AD' and A C' perpendicular to A.B until ASD' falls upon A.D, and then A/C' will fall upon./C, because the angle D'1'C2' is equal to DA C', D C' will fall upon DC, E' upon E, and A./E' upon SE. Therefore, the angle B.1E' is equal to the angle BA1E, and each is, by ~ 20, a right angle. 319. Corollary. The perpendicular B. is less than any oblique line BE, and measures the distance of the point B, from the plane. 320. Theorem. Oblique lines drawn from a point to a plane at equal distances fitom the perpendicular are equal; and of two oblique lines unequally distant the more remote is the greater. Proof. a. The oblique lines BC, BD, BE &c. (fig. 149) at the equal distances JC0,.D,.E &c. from the perpendicular BA/ are equal; for the triangles B2C, 100 SOLID GEOMETRY. [CII. XV. ~ 324. Oblique lines drawn to a Plane. BJD, BlE, &c. are equal, by ~ 51, since the angles BsC, B3D, B sE, &c. are equal, being right angles, the sides S3C, dID, S1E &c. are equal, and the side BA is common. b. Since the oblique line B C' is drawn to the line AiC' at a distance.0C' greater than SiC from the perpendicular BDd, it is, by ~ 41, greater than BC or its equal BD or BE. 321. Corollary. All the equal oblique lines BC, BD, BE &c. terminate in the circumference CDE, drawn with,t as a centre, and a radius equal to JiC. 322. Theorem. If a line is perpendicular to a plane, every line which is parallel to this perpendicular, is likewise perpendicular to the plane. Proof. Let JB (fig. 150) be the perpendicular to the plane, and let CD be parallel to JB, CD is likewise perpendicular to the plane, that is, to every straight line, as DE, drawn through its foot in that plane. For, if B.H be drawn through the foot of JB, parallel to DE, the angle DBI is, by ~ 317, a right angle; but, by ~29, the angle CDE is equal to J.BiI, and is, also, a right angle. 323. Corollary. Hence straight lines, which are perpendicular to the same plane are parallel. 324. Theorenm. If two planes are perpendicular to each other, the line, which is drawn in one of the planes perpendicular to their common intersection, lmust be perpendicular to the other plane. Proof. Let the plane MIJVN (firg. 15.1) be perpendicular to the plane PQ; and let IB be perpendicular to the common intersection l2P, we are to prove that lB is perpendicular to MJV. Draw, in the plane MAN, JlC perpendicular to JiP, B G C-I. xv. ~ 329.] PLANES AND SOLID AN~GLE S. 101 Pierpendiculars to a Plane. must, by ~ 316, be a right angle. As AB is, therefore, perpendicular to both &C and A/P, it is, by ~ 318, perpendicular to the plane JJN. 325. Corollary. If two planes are perpendicular to each other, the straight line, drawn through any point of the common intersection perpendicular to one of the planes, must be in the other plane. 326. Theorem. If two planes are perpendicular to a third plane, their common intersection is also perpendicular to this third plane. Proof. For, by the preceding article, the straight line /AB (fig. 152) drawn through the common point A/ of the three planes, perpendicular to the third plane MIJVN, must be in both of the planes./P and IQ, and must, therefore, be their common intersection. 327. Theorem. Two parallel lines are always in the same plane. Proof. Draw a plane MN.(fig. 153) perpendicular to one of the parallels J.B, it must also, by ~ 322, be perpendicular to the other parallel CD; and if a plane is drawn through the two points,8 and C, perpendicular to.MJV, AB and CD must both, by ~ 325, be in this plane. 328. Definitions../ straight line and a plane are parallel when all the points of the straight line are equally distant from the plane. Twvo planes are parallel, when all the points of one of the planes are equally distant from the other plane. 329. Theorem. A straight line and a plane are parallel, when they are perpendicular to the same straight line. 102 SOLID GEOMETRY. [CHi. XV. ~ 332. Parallel Planes and Lines. Proof. Let the straight line B C (fig. 154) and the plane.MN be perpendicular to the same straight line.BB; we are, by ~ 319, and 328, to prove that the perpendicular DC let fall from any point C of the line B C upon MJJN is equal to AB. Join AD; JAB and CD are parallel, by ~ 323, also AD is, by ~ 317, perpendicular to AB, and being in the plane of the parallels AB, CD, must, by ~ 35, be parallel to B C; so that ABCD is a parallelogram, and its opposite sides JAB and CD are equal, by ~ 78. 330. Theorem. If two planes are perpendicular to the same straight line, they are parallel. Proof. Let the planes JM', PQ (fig. 155) be perpendicular to the line AB; we are, by ~ 328, to prove that the line CD, drawn from any point of.PQ perpendicularly to JI1N, is equal to AB. Join B C, and, as BC is, by ~ 317, perpendicular to AB, it is, by ~ 329, parallel to M.U; and, therefore, CD is equal to AIB. 331. Theorem. If a straight line is perpendicular to one of two parallel planes, it must also be perpendicular to the other. Proof. Thus, if AB (fig. 155) is perpendicular to the plane JkIS, it must also be perpendicular to the plane PQ, which is parallel to 1I3N. For the plane drawn through B, perpendicular to JB, must be parallel to 3MN, and must therefore coincide with the plane PQ. 332. Theorem. If two planes are parallel to a third, they are parallel to each other. Proof. For any line perpendicular to the third plane must, by the preceding article, be perpendicular to both Ci. xv. ~ 335.] PLANES AND SOLID ANGLES. 103 Parallel Planesand Lines. the other planes; so that these other planes, being perpendicular to the same straight line, are parallel, by ~ 330. 333. Theorem. Two parallel lines, comprehended between two parallel planes, are equal. Proof. Let the two parallel lines A/B, C'D (fig. 156) be included between the two parallel planes -JIN, PQ. If the parallel lines are perpendicular to the parallel panes, they are equal, by ~ 328. Otherwise, draw firom the points 2 and C the lines SE, CF, perpendicular to M7;JVN; and join BE, DF. The triangles J.BE, CD1' are equal, by ~ 53; for the sides S1E and CF are equal, by ~ 328; the right angles JIEB and CiFD are equal; and the angles BAJK and DCF are equal, by ~ 29, because they have their sides parallel; hence JB is equal to CD. 334. Theoren. The intersections of two parallel planes by a third plane are parallel lines. Proof. Let the intersections of the plane.D (fig 156) with the parallel planes MJN, PQ be AC and BD.'rhrough A and C, in the plane JD, draw the parallel lines AB, CD; these parallels are equal by the preceding article, and, therefore, by ~ 81, 2ABCD is a parallelogram, and AC is parallel to BD. 335. Theorem. If a straight line is parallel to another straight line drawn in a plane, it is parallel to the plane. Proof. Let AC (fig. 156) be parallel to the line BD in the plane.J3i. Through any point A of the line AC, let a plane PQ be drawn parallel to IJN3. The intersection of PQ with the plane ABCD is, by the preceding article, parallel to BD; and, as it also passes through the point A, it must coincide with 20C. 9 104 SOLID GEOME0TRY. [CI. XS. ~ 338, Lines comprehended between thr.ee ]Parllel'Planes. Now, since 2tC is in the plane P Q parallel to JlNiv, all its points must, by ~ 328, be equally distant from iMJ, and it is therefore parallel to J7/IN. 336. Theoremn. Two straight lines, conprehended between three parallel planes, are divided into parts that are proportional to each other. Proof. Let the line d'C (fig. 157) meet the three parallel planes JYIN, PQ, RS at the points 2, B, C; and let the line DF meet the same planes at D, E, F. Join dF cutting the plane EPQ at I-I; join diD, BJH, HE, CF. The intersections BYH ald CF of the parallel planes P Q and RS with the plane d3CF, are parallel, and give, by ~ 160, the proportion AB13: B C -- AH: liF. In like manner, the intersections IE and diD of the parallel planes PQ and MIN with the plane Fdi are parallel, and give the proportion ai': HF - DE: EF. Hence, on account of the common ratio dil: HF, dB: BC=DE: EF. that is, the lines.dB and DF are divided proportionally at B and E. 337. Definitions. When three or a greater number of planes meet at a point, a solid angle is formed; as S (fig. 158) formed by the planes.iSB, BSC, CSD, D Sq. The point of meeting, S, of the planes, is called the vertex of the angle. 338. Theorem. If a solid angle is formed by three plane angles, the sum of either two of these angles is greater than the third. GHI. XV. ~ 339. PLANIES AND SOLID ANGLES. 105 Sum of the Plane Ai;~4es which formn a Solid A\nole. Prorf. Let S (fig. 159) be a solid angle formed by the, three plane angles.3 SB, BSC, and.JSC, and let.3SC, be the greatest of these plane angles. We need only prove that ASC < JSB +- B SC. Draw SD, making the angle CSD equal to CS]1. Draw any line t3Cy. Take SR equal to SD; join BC and Bd3. The triangles SCB and SCD are equal, by ~ 51, and CD _ C,. But, by ~ 18, Al C < s - RC. and, subtracting DC: BC, we have SD <.B. Now in the twvo triangles 2SD and $SB, the side SD is equal to SB, and AS is common; but the third side AD < riB, and therefore, by ~ 63, S2D < 2SB, and, adding CS$D- CSB.eSC CBO + OB; and, in the same way, BCS + sCD > BCO + oCD, 106 SOLTD GEOMEnTRY. [Cii. xv. ~ 340. Equal Solid Anrgles. C@S + SDE > CDo q- ODE, &c. Hence YIBS SB 6 - B CS + S CD + &c., or the suni of the angle at the bases of the triangles, which have their vertices at S, is greater than qB 0 + OB C + B C 0 + OCD + &c., or the sum of the angles at the bases of the triangles which have their vertices at 0. If, then, these two sums of the angles at the bases of the triangles are subtracted from the common suun of all the angles of each set of triangles, the remaining sum of the angles which have their vertices at S must be less than the surm of the angles which have their vertices at 0, or, by ~ 26, than four right angles. 340. Theorem. If two solid angles are respectively contained by three plane angles which are equal, each to each, the planes of any two of these angles in the one have the salme inclination to each other as the planes of the homologous angtles in the other. Proof. Let the solid angles S, $/ (fig. 161) be included by the plane angles dSB='S'.B', itSCO=1S'C/', BSC O B'S'C'. Take S =-= SA'! of any length at pleasure. Draw.//B, S.C, perpendicular to S, in the planes!SB and s2SC; and draw Ai'B', AiC'/, perpendicular to S'A4 in the planes.' S'B' and'S' C'. In the triangles ASAB, Ai'S'B, the side AS -=./', the angle ASB:= J'S'B'; and the right angle S$3B = SI'B; hence, by ~ 54, SB -= -'B' and SB - S"B'. In the same way, it may be shown that AC A —O'C', SCO-= SOC'. Join BC, B C', and, in the triangles SBC, S/B'C', the angle BSC B'S'g', the side SB - S'B', and the side S C -- S CSO; h-ence, by ~ 52, B C - B' C'. In the triangles,2B C, J'B' C' the three sides are respec cI-. xvI. ~ 342.] SOLIDS. 107 Solids of equal Heights anrd equivalenlt Sections are EIqual. tively equal, and, therefore, by } 61, the angle BAqC, which, by ~ 315, measures that of the planes JSB, J)SC is equal to B'.'C', which measures the angle of the planes d', SIB, d' S C,. In the same way, it may be shown that the angles of the other planes are equal; some changes, easily made, are, however, required in the demonstration when either of the angles MS'B, JSC is obtuse. CHAPTER XVI. SURFACE AND SOLIDITY OF SOLIDS. 341. Definitions. Equivalent solids are those which have the same bulk or magnitude. A lamina or slice is a thin portion of a solid included between two parallel planes. 342. Theorem. If two solids have equal bases and heiglhts, and if their sections, made by any plane parallel to the common plane of their bases, are equal, they are equivalent. Proof. Let ABCDEF, A'B'C'D'E-' (fig. 162) be the two solids. Let JVfO, j'i4" O' be two equal sections made by a plane parallel to the base, and let PQR, P' Q'R' be two other equal sections made by a plane infinitely near the former plane, and parallel to it. The infinitely thin laminte MJ OPQR, J]F'JV O'P'Q'R' are equal; for if i' JV'0O' be applied to its equal J'AIO, P' Q'R' must be infinitely near coincidence with its equal PQPR; and the laminio themselves can differ from coincidence only by a quantity infinitely smaller than either of them, and which may, by ~ 99 and 205, be neglected. 94 103 SOLID GEOMETRY. [Ci. XVI. ~ 346. Polyedron, Prism. But by drawing a series of parallel planes, infinitely near each other, -the given solids are divided into laminae, which are respectively equal to each other; and, therefore, their sums or the entire solids must be equivalent. 343. Definitions. Every solid bounded by planes is called a polyedroln. The bounding planes are called the faces; whereas the sides or edges are the lines of intersection of the faces. 344. Definitions. A polyedron of four faces is a tetraedronl one of six is a hexaedron, one of eight is an octaedron, one of twelve a dodecaedron, one of twenty an icosaedron, &c. The tetraedron is the most simple of polyedrons; for it requires at least three planes to form a solid angle, and these three planes leave an opening, which is to be closed by a fourth plane. 345. Definitions. A prism is a solid comprehended under several parallelograms, terminated by two equal and parallel polygons, as a1BC &c. FGHI &c. (fig. 163). The bases of the prism are the equal and uarallel polygons, as.iB C &c., and FGH &c. The convex sutface of the prismz is the sum of its parallelograms, as JQBFG + BCGH + &c. The altitude of a prism is the distance between its bases, as P Q. 346. Definitions. A right prisnm is one whose lateral faces or parallelograms are perpendicular to the bases, as JIBC &c. FGH &c. (fig. 164). In this case each of the sides 27JF, BG &c. is equal to the altitude. CH. XVI. ~ 353.] SOLIDS. 109 Cylinder, Parallelopiped, Cube, Unit of Solidity, Volume. 347. Definitions. A prism is triangular, quadrangular, pentagonal, hexagronal, &c., according as its base is a triangle, a quadrilateral, a pentagon, a hexagon, &c. 345. Definitions. The prism, whose bases are regular polygons of an infinite number of sides, that is, circles, is called a cylinder (fig. 165). The line OP, which joins the centres of its bases, is called the axis of the cylinder. In the right cylinder (fig. 166) the axis is perpendicular to the bases, and equal to the altitude. 349. Corollary. The right cylinder (fig. 166) may be considered as generated by the revolution of the right parallelogram OBBP about the axis OP. The sides O4 and PB generate, in this case, the bases of the cylinder, and the side JB generates its convex surface. 350. Definitions. A prism whose base is a parallelograrn (fig. 167) has all its faces parallelograms, and is called a parallelopiped. When all the faces of a parallelopiped are rectangles, it is called a right parallelopiped. 351. Definitions. The cube is a right parallelopiped, comprehended under six equal squares. The cube, each of whose faces is the unit of surface, is assumed as the unit of solidity. 352. Definition. The volume, solidity, or solid contents of a solid, is the measure of its bulk, or is its ratio to the unit of solidity. 353. lTheorem. The area of the convex surface of 110 SOLIID GEOMEIJ:,TRY. [CII. XVr. $ 356. Convex **uri.r tce I'l rilit Iiris1 or C, viIIt (Ie r. a right prism or cylinder is the perimeter or circumference of its base multiplied by its altitude. Proof. a. The area of each of the parallelograms JIBFG. B CGGH, &c., which compose the convex surface of the right prism (fig. 164) is, by ~ 247, the product of its base.B, BC &.c., by the common altitude JF; and the sum of their areas, or the convex surface of the prism, is the sum of these bases, or the perimeter J1JBCD &c., by the altitude SF. b. This demonstration is extended to the right cylinder by increasing the number of sides to infinity. 354. Theorem. The section of a prism or cylinder made by a plane parallel to the bases is equal to either base. Proof. a. Let LJAd.NO, &e. (fig. 163) be a section of the prism made by a plane parallel to the bases. It follows, from ~ 334, that LME is parallel to.B, MJNi. to B C, &c.; and, consequently, the angle LMZJ is equal to JAB C, by ~ 29, the angle X3d]10 to BCD, &c. Moreover, in the parallelograms d.BLJP1, BCJdJV, kc.,.B is equal to LJI, BC to MLN~, kc., and the polygons SB CD &c., LJi:/O, &c. are equiangular and equilateral with respect to each other, and are, therefore equal, by ~ 195. b. The demonstration is extended to the cylinder by increasing the number of sides to infinity. 355. Corollary. Hence, from ~ 342, two prisms or cylinders of equal bases and altitudes are equivalent. 356. Corollary. Any prism or cylinder is equivalent to a right prism or cylinder of the same base and altitudte. cIi. xvi. ~ 357.] SOLIDS. 111 Ratio of right Parallelopipeds. 357. Theoremn.' Two right parallelopipeds are to each other as the products of their bases by their altitudes. Proof. Let the two right parallelopipeds be IB CD EFGII,.JIKMI jOPQ (fig. 168), which we will denote by r G and AJP. Then, if the sides of the rectangles JBACD and JICKL1 are commensurable, the rectangles can, by ~ 241, be divided into equal rectangles; and, if, through each of the vertices of these small rectangles, perpendiculars are erected to the plane JIL, the parallelopipeds JIG and IP are divided into smaller right parallelopipeds. All the parallelopipeds of JG are equivalent, by ~ 35.5, as well as all those of JP; and the number of paralielopipeds in IAG is equal to the number of rectangles in JABCD; and the number of parallelopipeds in JP is equal to the number of rectangles in JIKL.M1. If now the altitudes JE and AJN' are commensurate, J1N can be divided into equal parts, of which JE contains a certain number; and, if', through the points of division of JIJ, planes are drawn parallel to the base JL, each of the partial parailelopipeds of AG and JP are divided into smaller equal parallelopipeds, and all these smallest parallelopipeds are equal to each other. Now, the whole number of the smallest parallelopipeds contained in JG is the product of the number of rectangles in its base JBCD by the number of divisions ~of its altitude JE, and the number contained in AJP is the product of the number of rectangles in its base JIKLJ1 by the number of divisions in its altitude AV. I-fence J G: J P — = AB D X JEI: J IL J X JrJV. This demonstration is readily extended to the case where the sides are incommensurate, by dividing the solids into infinitely small parallelopipeds. 112 SOLID GEOnIETRY. [CH. XVI. ~ 362. Solidity of the Parallelopipeds. 358. Corollary. The solidity of any ri;ght parallelopiped or its ratio to the unit of solidity is, by ~ 352, the product of its base by its altitude, that is, A1 G =- B CD X J2E. 359. Corollary. Since, by ~ 242,.B CD AWB x lID, we have AIG -/B X AID X J./E; or the solidity of a right parallelopided is the product of its three dimensions. 360. Corollary. The solidity of a cube is the cube of one of its sides. 361. Corollary. Since, by ~ 356, any parallelopiped of a rectangular base is equivalent to a iright parallelopiped of the same base and altitude, the solidity of any parallelopiped of a rectangular base is the product of its base by its altitude. 362. Theorem. TLhe solidity of any parallelopiped is the product of its base by its altitude. Proof. Any parallelopiped which has AB CD (fig. 1 69) for its base is, by ~ 356, equivalent to the parallelopiped AG, which has the same base, and its sides J/H, BE, C G, DF', perpendicular to the base BCD. But any other face may as wvell be assumed for the base of.G as JlB CD; taking, then, the rectangle WBEH for the base, the parallelopiped AG is, by ~ 361, equal to the right parallelopiped of the same base and altitude, that is, by drawing DK perpendicular to lJB, iG = —-DK: X A BEH I1-DK X MB X AI. CHI. xvI. ~ 366.] SOLIDS. 113 Solidity of the Prism and Cylinder. But, AB CD ~- DK X A/B; hence G G -- AB CBD X JII. 363. Corollary. Any two parallelopipeds of equivalent bases and the same altitude are equivalent. 364. Corollary. Parallelopipeds of the same base are to each other as their altitudes, and parallelolipeds of the same altitude are to each other as their bases. 365. Th/eorejm. The solidity of a triangular prism is the product of its base by its altitude. Proof. Let A.B C DEF (fig. 170) be a triangular prism. Draw EBG parallel to 1AC, CG parallel to.B, GH' parallel to 1AD, meeting the plane EDF in IH. Join EI-, FII; 21H is, evidently, a parallelopiped; and BCG FF11 is a triangular prism. The triangular prisms 11BC D EF and BCG EFH are equivalent, by ~.355; since their altitude is the same and their bases ABC and BCG are equal, by ~ 77. Flence each of the prisms is half of the parallelopiped S.11, and has half its measure, or the product of- -1B C G by the altitude, that is, the product of its own base by its altitude. 366. Theorem. The solidity of any prism or cylinder whatever is the product of its base by its altitude. Proof a. The prism IBGC &c. FGH S&c. (fig. 163) may be divided into the triangular prisms 1B0C FGH,;CD FrIJ &c. by the planes.tCFie, iDFI RLc., and, by the preceding section, the solidity of each of these triangular prisms is the product of its base EBC, J.CD, &c. by the altitude PQ. Hence, the sum of these prisms or the entire prism is the product of the sum of the bases 114 SOLID GEOMETRY. [CI-i. xvI. ~ 371. Pyramid. by PQ, or of the entire base./BCD &c. by the altitude PQ. b. This demonstration is extended to cylinders by increasing the number of sides to infinity. 367. Corollary. Prisms or cylinders of equivalent bases and equal altitudes are equivalent. 368. Corollary. Prisms or cylinders of equivalent bases are to each other as their altitudes; and those of the same altitude are to each other as their bases. 369. Corollary. Denoting by R the radius, and by./ the area of the base of a cylinder; and using = as in ~ 237, we have, by ~ 280, Denoting, also, by Ir the altitude, y the solidity of the cylinder, we have, by ~ 366, V-d X H — X R2 X H. 370. Definitions. A pyramid is a solid formed by several triangular planes proceeding firom the same point, and terminating in the sides of a polygon, as S&BCD &c. (fig. 171). The point S is the vertex of the pyramid. The polygon JBCD &c. is the base of the pyramid. The convex surface of the pyramid is the sum of the triangles- S.B +- S&C, &c. The altitude of the pyramid is the distance of its vertex from its base. 371. Definitions. A pyramid is triangular, quadrangular, &c., when the base is a triangle, a quadrilateral, &c. CH. XVI. ~ 375.] SOLIDS. 115 Cone. Convex Surface of the regular Pyramid. 372. Definitions. A pyramid is regular, when the base is a regular polygon, and the perpendicular let fall from the vertex upon the base, passes through the centre of the base (fig. 172). This perpendicular from the vertex is called the axis of the pyramid. 373. Definitions. When the base of a pyramid is a regular polygon of an infinite number of sides, that is, a circle, it is called a cone (fig. 173). The axis of the cone is the line drawn from the vertex to the centre of the base. A right cone is one the axis of which is perpendicular to the base (fig. 174). 374. Corollary. The right cone (fig. 174) may be considered as generated by the revolution of the right triangle S OS about the axis S O. The leg O0., in this case, generates the base, and the hypothenuse SiJ, which is called the side of the cone, generates the convex siuftce. 375. Theorem. The area of the convex surface of the regular pyramid is half the product of the perimeter of the base by the altitude of one of the triangles. Proof. The triangles S B, SBC, &c. (fig. 172) are all equal, for, by ~ 201, JIB BC CD,b &c.; and, since the oblique lines lS, SB, SC, &c., are all at equal distances Oil, OB, OC, &c., from the perpendicular SO, they are equal by ~ 320. Hence the altitudes SH, SI, SKI, &c. of these triangles are equal; and the sum of the areas of the triangles is half the product of the sum of their bases JiB, BC, CD, &c. by the common 10 116 SOLID GEOMETRY. [ciI. XVI. ~ 378. Section of Pyramid parallel to the Base. altitude S11; that is, the convex surface of the pyramid is half the product of the perimeter of its base by the altitude of one of its triangles. 376. Corollary. Wthen the base of the regular pyramid is a polygon of an infinite number of sides, the pyramid is a right cone, and the altitude of each triangle becomes the side Ss (fig. 174) of the cone. Hlence the area of the convex surface of the right cone is half the product of the circumference of the base by the side. 377. Theorem. The section of a pyramid made by a plane parallel to the base is a polygon similar to the base. Proof. Let MXJYOP &c. (fig. 171) be the section of a pyramid made by a plane parallel to its base.iBCD &c. Since AJ.IN is, by ~ 334, parallel to rB, we have SB: SAi =JB: M3A13, and since.VO is parallel to B C, we have SB: S =- B C: XNO; and, on account of the common ratio, SB: SB J, AB: MN-= B C: AWO. In the same way we might prove.B C: OIE~ B C:,NOY CD: OP, &c. whence the sides of the polygons fIBCD, c.,,CMDNOP &c. are proportional. The angles of the polygons are also equal; indeed on account of the parallel sides, we have JiAMO =- B C, JVOP - B CD, &c. The polygons are therefore similar, by ~ 170. 378. Corollary. The section of a cone made by a plane parallel to the base is a circle. CH. XVI. ~ 382.] SOLIDS. 117 Equivalent Pyramids and Cones. 379. Corollary. If the perpendicular ST is let fall firom S upon the base, meeting the section at R, we have, by ~~ 268 and 336, iJB CD &.: JMIJNOP &c. - B': JJ"2 -— _ $SJ: $SjF =ST: SR-, or, the base of a pyramid or cone is, to the section made by a plane parallel to the base, as the square of the altitude of the pyramid is to the square of the distance of the section from the vertex. 38O. Corollary. If two pyramids or cones have the same altitude and their bases in the same plane, their sections made by a plane parallel to the plane of their bases are to each other as their bases. if the bases are equivalent, the sections are equivalent. If the bases are equal, the sections are equal. 381. Theoremn. Two pyramids or two cones which have equal bases and altitudes are equivalent. Proof. For, if their bases are placed in the same plane, their sections made by a plane parallel to the plane of their bases are equal; and, therefore, by ~ 342, the pyramids are equivalent. 382. Theorent. A triangular pyramid is a third part of a triangular prism of the same base and altitude. Proof. From the vertices B, C (fig. 175) of the triangular pyramid S.BC, draw BD, CE parallel to SA. Draw SD, SE parallel to AIB, fAC, and join CE; AB C SDE is a triangular prism. The quadrangular pyramid S BCED is divided by the plane SBE into two triangular pyramids S BED, $ BEC, which are equivalent; for their bases BED, 113S SOLID GEOME.TRY. [CT. XVI. ~ 3S8. Solidity of thie Pyramid and Coiie. BEC are equal, by ~ 77; and their common altitude is the distance of their common vertex S from the plane of their bases. Again, if the plane SED is taken for the base of S BED and the point B for its vertex, the pyramid B SDE is equivalent to S dBGC; for their bases SED, S.BC are equal, and their common altitude is the altitude of the prism. But the sum of the three equal pyramids S.TBC, S BED, S BEC is the prism 3BC $DE, and, therefore, either pyramid, as S B1BC, is a third part of the prism. 383. Corollary. The solidity of a triangular pyramid is a third of the product of its base by its altitude. 384. Theorem. The solidity of any pyramid is one third of the product of its base by its altitude. Proof. The planes $SC, SAD, &c. (fig. 171) divide the pyramid S qBCD &c. into triangular pyramids, the common altitude of which is the altitude of the entire pyramid. Hence the solidity of the entire pyramid is one third of the product of the sum of their bases lJB C, A CD, &c., by the common altitude, that is, one third of the entire base by the altitude of the pyramid. 385. Corollary. The solidity of a cone is one third of the product of its base by its altitude. 386. Corollary. Pyramids or cones are to each other as the products of their bases by their altitudes. 387. Corollary. Pyramids or cones of the same altitude are to each other as their bases; and those of equivalent bases are to each other as their altitudes. S33. Corollary. Pyramids or cones of equivalent bases and equal altitudes are equivalent. ci. xVI. ~ 392.] SOLIDS. 119 Truncated Prism. 389. Corollary. Any pyramid or cone is a third part of a prism or cylinder of the same base and altitude. 390. Corollary. Denoting by Xi the radius of the base of a cone, by H its altitude, by V its solidity, and using n, as in ~ 237, we have, by ~~ 369 and 389, $V TX R2X H. 391. Definitions. A truncated prism is the portion of a prism cut off by a plane inclined to its base, as.2BC DEF (fig. 176). The base of the truncated prism is the same as the base of the prism from which it is cut. 392. Theorem. A truncated triangular prism is equivalent to the sum of three pyramids, which have for their common base the base of the prism, and for their vertices the three vertices of the inclined section. Proof. Draw the plane FaC (fig. 176), cutting off from the truncated triangular prism ABC DEF the pyramid FJBC, which has ABC for its base, and F for its vertex. There remains the quadrangular pyramid FJICDE, which the plane FE C divides into the two triangular pyramids FJEC and FCDE. Now FAsEC is equivalent to the pyramid B.EC, which has the same base AEC, and the same altitude, because the vertices F, B are in the line FB parallel to this base. But iBBC may be taken for the base of E2BC, and E for its vertex. Lastly, the pyramid FE CD - the pyramid BE CD, for they have the same base ECD, and the same altitude, 10~ 120 SOLID GEOMETRY. [CHI. XV1. ~ 395. FIrustuni of a Pvramlid or Chone. because their vertices F, B are in the line P-B lparallel to this base. Also, taking E as the vertex of' BE Cd the pyramid EB CD- =the pyramid AdB CD, for, they have the common base BDC, and their vertices J/, E are in the line SJE parallel to this base. But BMBC' may be taken for the base of. BCD, and D for its vertex. Hence the truncated prism is equivalent to the sum of three pyramids, which have the common base /B C, and for their vertices E, F, and D. 393. Definitions. If a pyramid or cone is cut by a plane parallel to its base, the portion which remains after taking away the smaller pyramid or cone, is called the frustumn of a pyramid or cone, as B CD &c. JMI2NOP &c. (fig. 171.) The convex stirface of the frustum of a pyramid is the sum of the trapezoids which compose its lateral faces. The polygons J.BC.D &c., JMIJVXOP &c. are the bases of the frustumn, and the distance between its bases is its altitude. 394. Corollary. The frustum of the right cone (fig. 174) may be considered as generated by the revolution of the trapezoid 0./l'Jt about the side 00'. The side 2AI', which is called the side of the frustznm, in this case, generates the convex sujmce. 395. 7Theorem. The area of the convex surface of the frustum of a regular pyralnid is half' the product of the sum of the perimeters of its bases by the altitude of either of its trapezoids. Proof. The trapezoids //BMNA, BCJV'O, &c. (fig. 172) CH. xvI. ~ 398.1 SOLIDS. 121 Convex Surface of a Frustumn of a Pyramid or Cone. are all equal; and the area of each is half the product of the sum of its parallel sides by their common altitude filI'. The sum of their areas, or the area of the convex surface of the frustum is, therefore, half the product of this common altitude, by the sum of all the parallel sides, that is, by the sum of the perimeters of the bases of the frustum. 396. Corollary. If a section JFM'A"O'P' &c. is made by a plane parallel to the bases, and passing through the middle point R' of the altitude, it must, by ~ 336, bisect the lines.J/M, BJ., &c.; and the area of each trapezoid is, by ~ 255, the product of its altitude by the line M'Zi/V,.V 0', ce. The area of the convex surface of the frustum is, therefore, the product of the altitude by the sum of these lines, that is, by the perimeter of the section made by the plane which bisects the lateral sides of the frustum. 397. Corollary. The area of the convex surface of the frustum of a right cone is half the product of its side by the sum of the circumferences of the bases; or it is the product of the side by the circumference of the section parallel to the bases which bisects the side. 398. Theorem. The area of the surface, described by a line revolving about another line in the same plane with it as an axis, is the product of the revolving line by the circumference described by its middle point. Proo f. a. If the revolving line is parallel to the axis, as in (fig. 166), it describes the convex surface of a right cylinder, the area of which is, by ~ 353, the product of 122 SOLID GEOMETRY. [CH. XVl. ~ 400. Surface described by a revolving Line. the circumference of the base by the altitude. But the altitude is equal to the revolving line, and the circum-. ference of the base is, by ~ 354, equal to the circumnference described by the middle point; and, therefore, in this case, the area of' the surface described is the product of the revolving line by the circumf'erence described by its middle point. b. If the revolving line is inclined to the axis without meeting it, the surface described is the convex surface of the frustum of a right cone; and its area is as, in ~ 397, the product of the revolving line by the circumference described by its middle point. c. When the revolving line meets the axis without cutting, the surface described is the convex surface of a right cone, and is included in the preceding case by considering it as a frustum whose upper base is the vertex of the cone. 399. Scholium. The case, where the revolving line cuts the axis, is not included in the preceding theorem. 400. 7Theorem. The frustusn of a pyramid or cone is equivalent to the sum of three pyramids or cones, which have for their common altitude the altitude of the firustum, and whose bases are the lower base of the frustum, its upper base, and a mean proportional between tllem. Proof. Let ABCD &c..JALOP &c. (fig. 171) be the given frustum. Denote the area of the lower base IBCD &c. by V, and that of the upper base MJVI'OP &c. by I'; and denote the altitude ST of the greater pyramid by IT, the altitude SR of the less pyramid by it', and the altitude R T of the frustum by H". Since the frustum is the difference between the pyra mids, we have for its solidity, by ~ 384, CH. xvI. ~ 335.] SOLiDS. 123 Solids of a Frustium of a Pyramid or Cone. - Vx HX - V' x H', and, for the sum of three pyramids, which have t'; for their altitude and for their bases V, VI and the mean proportional /VV' between Vand V','H" X (F+ - +,/VV) =1 Hl X V- I-11 X V I a g Hx/ /V P, and we are to prove that these solidities are equal, or that V X H V' Hx I = x H" + Y' X HI" + V V X H". Now H" IH —-H', and, by ~ 379, Y': V' = z: II' 2, whence Y/V: /V/ --- H: iH', and, multiplying extremes and means V x - -V'-/' x H. If we multiply this equation successively by V and,/ V we obtain, by transposing the members of the first product, /VV' x H- F X I', /VV'X II' VI X H; the difference between which is V/ VV X ( H —H') ==V X I' —V' X t, or./VV' x II"-= V X>< I — VX H. And if we add to this last equation the equations V X H", V X H-V X Ii' V X H" - X I — X I-I', we get, by cancelling the terms which destroy each other, VVV X H"'+V X HI' + VX H,,= VX H —V X I', which is the equation to be proved, and the solidity of the frustum is therefore equal to HI" X (VT+ V+r,/+ v). 124 SOLID GEOMETRY. [CH. XVii. ~ 405. Solidity of the Frustum of a Cone. 401. Corollary. If R is the radius of the lower base of the frustum of a cone, and R' the radius of the upper base, we have, by ~ 280, V —rt X R2 VIFfr X Rt'2, hence /YV'I v 2 X WR X RI 2 -' X R X RI, and the solidity of the frustum is 7 X I-iI X (R1? + It'12 + R X I). 402. Scholiumz. Tle solidity of any polyedron mnay be found by dividing it into pyramids. CHAPTER XVII. SIMILAR SOLIDS. 403. Definition. Similar polyedrons are those in which the homologous solid angles are equal, and the homologous faces are similar polygons. 404. Corollary. Hence, from ~ 170, the sides of similar polyedrons are plroportional to each other. 405. Corollary. From ~ 263, the faces of similar polyedrons are to each other as the square of their homologous sides; and, fi'orn the theory of proportions, the surms of the faces, or the entire surfaces of the polyedrons are also to each other as the squares of the homologous sides. CH. XVII. ~ 412.] SIMILAR SOLIDS. 125 Ratios of Similar Prisms, &c. 406. Corollary. The bases of similar prisms or pyramids are to each other as the squares of their altitudes; and the perimeters of their bases are to each other as their altitudes. 407. Corollary. The bases of similar cylinders or cones are to each other as the squares of their altitudes; and their altitudes are to each other as the circumferences of the bases, or as the radii of the bases. 403. Corollary. The convex surfaces of similar prisms, pyramids, cylinders, or cones are to each other as their bases, or as the squares of their altitudes. 409. Corollary. The convex surfaces of similar prisms or pyramids are to each other as the squares of their homologous sides. 410. Corollary. The convex surfaces of similar cylinders or cones are to each other as the squares of the radii of their bases. 411. Theorem. Similar prisms, pyramids, cylinders, or cones are to each other as the cubes of their altitudes. Proof. Prisms, pyramids, cylinders, or cones are to each other, by ~ 366 and 386, as the products of their bases by their altitudes. But where these solids are similar, their bases are to each other, by ~ 406 and 407, as the squares of their altitudes; and the products of the bases by their altitudes, or their solidities are to each other, as the products of the squares of their altitudes by their altitudes, or as the cubes of their altitudes. 412. Corollary. Similar prisms or pyramids are to each other as the cubes of their homologous sides. 12G SOLID GEOMETRY. [CII. xvIII. ~ 417. Ratio of Similar Solids. 413. Corollary. Similar cylinders or cones are to each other as the cubes of the radii of their bases. 414. Theorem. Similar polyedrons are to each other as the cubes of their homologous sides. Proof. Let a polyedron be divided into pyramids by drawing lines from one of its vertices to all its other vertices; any similar polyedron may be divided into similar pyramids by lines similarly drawn from the homologous vertex. Now these similar pyramids are to each other, by ~ 412, as the cubes of their homologous sides, or as the cubes of any two homologous sides of the polyedrons; and, from the theory of proportions, their sums, that is, the polyedrons themselves, are to each other in the same ratio, or as the cubes of their homologous sides. CHAPTER XVIII. THE SPHERE. 415. Definition. A sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within called the centre. 416. Corollary. The sphere may be conceived to be generated by the revolution of a semicircle, DsE (fig. 177) about its diameter DE. 417. Definitions. The radius of a sphere is a straight line drawn from the centre to a point in the CH. XVIII. ~ 424.1 TIIE SPHERE. 127 Great and Small Circles. Poles. surface; the diameter or axis is a line passing through the centre, and terminated each way by the surface. 418. Corollary. All the radii of a sphere are equal; and all its diameters are also equal, and double of the radius. 419. Theorem. Every section of a sphere made by a plane is a circle. Proof. From the centre C (fig. 178) of the sphere draw the perpendicular CO to the section AJ2MB and the radii C., CM3, CB, &c. Since these radii are equal, they must, by ~ 321, terminate in a circumference 4MaB, of which 0 is the centre. 420. Definitions. The section made by a plane which passes through the centre of the sphere is called a great circle. Any other section is called a small circle. 421. Corollary. The radius of a great circle is the same as that of the sphere, and therefore all the great circles of a sphere are equal to each other. 422. Corollary. The centre of a small circle and that of the sphere are in the same straight line perpendicular to the plane of the small circle. 423. Definition. The points, in which a radius of the sphere, perpendicular to the plane of a circle, meets the surface of the sphere, are called the poles of the circle; thus P, P' are the poles of.MJ1B. 424. Corollary. Since the oblique lines PR., P 1, &c. are equally distant from the perpendicular PO, they are equal; and also the arcs of great circles P2), PJ1I, 11 128 SOLID GEONJETRY. [CH. XVIII. ~ 429. Arcs traced upon a Sphere. &c. are, by ~ 113, equal; that is, the pole of a circle is equally distant fiom all the points in the circumference of the circle. 425. Corollary. Since the distance DJI (fig. 177) of a point, in the circuniference of a great circle from the pole, is measured by the right angle DC3S, it is a quadrant. 426. Scholinm.. By means of poles, arcs may be traced upon the surface of a sphere as easily as upon a plane surface. AVe see, for example, that by turning the arc DF (fig. 177) about the point D, the extremity F describes the small circle FINSG; and by turning the quadrant DF,/t about the point D, the extremity Al describes the are of a great circle./J[l. 427. Theorem. A point upon the surface of a sphere which is at the distance of a quadrant fiom each of two other points, is one of the poles of the great circle which passes through these two points. Proof. Thus, if the distances DtI, DJ (fig. 177) are quadrants, the angles DCS1 and DCAI are right angles, and, therefore, by ~ 318, DC is perpendicular to the circle J.MB, and its extremity D is, by ~ 423, a pole of the circle,BM. 428. Corollary. Since the common intersection of two great circles is, by ~ 420, a diameter, they bisect each other. 429. Theorem. Every goreat circle bisects the sphere. Proof. For if, having separated the two hemispheres from each other, we apply the base of one to that of the other, ciI. vriii. ~ 435.] THEM SPHERE. 129 Spherical Triangle, Polygon, Wedge, Pyramid. turning the convexities the same way, the two surfaces must coincide; otherwise, there would be points in these surfaces unequally distant from the centre. 430. Definitions. A spherical tricangle is a part of the sturface of a sphere comprehended by three arcs of great circles. These arcs, which are called the sides of the triangle, are always supposed to be smaller each than a sernicilrcumference. The angles, which their planes make with each other, are the angles of the triang'le. Since the sides are arcs, they may be expressed in degrees and minutes, as well as the angles. 431. Definitions. A spherical triangle takes the name of right, isosceles, and equilateral, like a plane triangle, and under the same circumstances. 432. Definition. A spherical polygon is a part of the surface of a sphere terminated by several arcs of great circles. 433. Definitions. The portion of a sphere comprehended between the halves of two great circles is called a spherical wedge, and the portion of the surface of the sphere comprehended between them is called a lunary swtface, and is the base of the wedge. 434. Definitions. A spherical pyramid is the part of a sphere comlprehended between the planes of a solid angle whose vertex is at the centre. The base of the pyramid is the spherical polygon intercepted by these planes. 435. Definition. A plane is tangent to a sphere, when it has only one point in common with the surface of the sphere. 130 SOLID GEOMETRY. [cIi. XVIII. ~ 439. Spherical Segment, Sector. 436. Definitions. When two parallel planes cut a sphere, the portion of the sphere comprehended between them is called a spherical segment, and the portion of the surface of the sphere comprehended between them is called a zone. The bases of the segment are the sections of the sphere, and the bases of the zone are the circumn-ferences of the sections. The altitude of the segment or zone is the distance between the sections. One of the cutting planes may be tangent to the sphere, in which case the zone or segment has but one base. 437. Definition. While the semicircle D.JJE (fig. 177) turning about the diameter DE describes a sphere, every circular sector, as DCF or FCHil, describes a solid, which is called a spherical sector. The base of the sector is the zone generated by the arc DF, or FIt. 438. Theorem. Either side of a spherical triangle is less than the sum of the other two. Proof. From the centre O (fig. 179) of the sphere draw the radii 0.3, OB, OC to the vertices.3, B, C of the spherical triangle ATBC. The three plane angles./OB, 2100, BOC form a solid angle at 0; and each of these angles is, by ~ 338, less than the sum of the other two. But they are measured by the arcs JB, A.C, BC; and, therefore, each of these arcs is less than the sum of the other two. 439. Theorem. The sum of the sides of a spherical polygon is less than the circumference of a great circle. CHI. XVIii. ~ 443.] THE SPHERE. 131 Sum of the Sides of a Spherical Polygon. Proof. From the centre 0 (fig. 180) of the sphere draw the radii 0A, GB, 0O, &c. to the vertices, B, C, E&c. of the spherical polygon ABC &Sc. The plane angles 4AOB, BOC, &c. form a solid angle at 0; and the sum of these angles is, by ~.339, less than four right angles. The sum of the arcs AS, BC, CD, c. is, consequently, less than a circumference of a great circle. 440. Corollary. If then, we denote the sides of a spherical triangle by a, b, c, we have a + b + c < 3600. 441. Theoreim. The angle formed by two arcs of great circles is ileasured by the arc described fiom its vertex as a pole, and included between its sides. Proof. The arc A4JM (fig. 177) measures the angle TCiM, which, by ~ 315, measures the angle of the planes DC4 and DCOJM; and therefore, by ~ 430, it measures the angle 4//D.1. 442. Corollary. The value of the are 4.M expressed in degrees, minutes, &c., is the same as that of 4DAM. 443. Theorem. If from the vertices of a given spherical triangle as poles, arcs of great circles are described, another triangle is formed, the vertices of which are the poles of tlhe sides of the given triangle. Pr1oof. Bet JBC (fig. 181) be the given triangle; let EF, DF, and DE be described, respectively, with 4., B, C as poles. Then, since E is in the are EF, the distance from E to 4 is, by ~ 42b, a quadrant; and since E is in the arc DE, the distance from E to C( is also a quadrant; and, therefore, by ~ 427, E is a pole of 4G. In the same way it may be shown, that D is a pole of B C, and' a pole of 41B. I. I" 132 SOLID GEOMETRY. [CH. XVIII. ~ 445. Sides and Angles of polar Triangle. 444. Definition. The triangle DEF is called the polar triangle of.JBC, and in the same way AJBC is the polar triangle of DEF. As several different triangles might be formed by producing the sides DE, EF, and DF, we shall limit ourselves to the one DEF, such that the pole D of B C is on the same side of BC with the vertex J; E is on the same side of flC with the vertex B; and F is on the same side of AB with the vertex C. 445.'Theore~m. If the sides and angles of a spherical triangle and of its polar triangle are expressed in degrees, minutes, &c., the sides of either triangle thus expressed are respectively supplements of the angles of the other triangle. Proof. Produce the sides AB, SC (fig. 181), if necessary, to G and II. Since Fis the pole of./lB, and E the pole of AC, we have, by ~ 425, BEH- FG - 90. Hence EF -EH- + -IF= 90~ -+ HF GH - GF- HF —= 900 - HF, and, therefore, EF + G1H 1800. But, by ~ 441 and 442, GH- the angle BF0C, whence EF - the angle B C 180~; that is, the side EF and the angle BffC are supplements of each other. In the same way it may be shown, that DF and the an cii. xvIII. ~ 448.] THIIE SPHERE. 133 Sum of the Angles of a Spherical Triangle. gle ABC, DE and the angle /CB, /lB and the angle F, BC and the angle D, 2.C and the angle E, are respectively supplements of each other. 446. Corollary. If therefore we denote the angles of a spherical triangle by J, B, C; and the sides respectively opposite by a, b, c; the angles of the polar triangle must be 1800 -a, 1800~-b, 1800 -c; and the sides of the polar triangle 180~ -,, 1800 - B, 1800 - C. 447. Theorem. The sum of the angles of a spherical triangle is greater than two right angles. Proof. Let /3, B, C be the angles of the spherical triangle. The sides of its polar triangle are 180~ —a, 180 - B, and 1800 - C. Now the sum of these sides, is, by ~ 440, less than 3600, that is, 3600 > (1800 - ) + (1800 -B) + (1800 -C) or, 3600 > 5400~ - - B-C, or, by transposition,.3 + B + C > 5400 - 3600, or, A + B + C > 1800; that is, the sum of the angles./, B, C is greater than 1800~. 448. Theorem. Each angle of a spherical triangle is greater than the difference between two right angles and the sum of the other two angles. Proof. Let /, B, C be the angles of a spherical triangle; we are to prove that either of these angles, as.3, is greater than the difference between 180~ and B + C. a. That is, if B + C is less than 1800, we are to prove.2 > 1800~-(B + C). 1 34 SOLID GEOMETRY. LCH. XV1II. ~ 449 Equilateral Spherical Triangles are equiangular. We have, from the preceding proposition,, d- B + C > 1800, whence, by transposition,. > 1800~-(B + C). b. But if B + C is greater than 180~, we are to prove > (B + C) — 80o. Now, of the three sides 1800 —-, 1800 -B, 1800~- C of the polar triangle, each is, by ~ 438, less than the sum of the other two; that is, (1800 - B) + (180-C) > 180o0 - or 3600 - B - C > 1800 -., and, by transposition, > + B C - 3600 + 1800, or, > (B + C)- 180o, as we wished to prove. 449. Theorem. If txwo spherical triangles on the same sphere, or on equal spheres, are equilateral with respect to each other, they are also equiangular with respect to eachl other. Proof. Let JIBC, DEF (fig. 182) be the spherical triangles, of which the sides.B = DE, C-== DF, and BC EF. Draw the radii 01, OB,, O, O'D, OE O'F. The angles q OB and D O'E are equal, because they are measured by the equal arcs JIB and DE; in the same way, C0 C -- D O'F, B OC — E O'F, and therefore, by ~ 340, the angle of the planes J1OB, 10C is equal to that of the planes D O'E, D O'F, that is, 11S C = EDDF. In like manner, JIB C D — DEF, and A CB D —DFE. ci-. xvIII. ~ 4:54.] TIE SPHIERE. 135 E!,qual Spherical Triangles. 450. Definition. Two spherical triangles are symmetrical, when they are equilateral and equiangular with respect to each other, but cannot be applied to each other, as J.BC, /3BC' (fig. 183). 451. Theorem. If tsvo triangles on the same sphere, or on equal spheres, have a side, and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other, they are equal5 or else they are symmetrical. Proof. If the two triangles ABC, DEF (fig. 183) have the side.B- DE, the angle BA C = EDF, and the angle ABC-DEF; the side DE caa be placed upon AB, and the sides DF, FE will fall upon AC, BC, or upon the sides.AC', BC' of the triangle J/BC', symmetrical to JB C. 452. Theorem. If two triangles on the same sphere, or on equal spheres, have two sides, and the included angle of the one respectively equal to the two sides and the included angle of the other, they are equal, or else they are symmetrical. Proof. For one of the triangles may be applied to the other, or to its symmetrical triangle. 453. T7heorem. In every isosceles spherical triangle the angles opposite the equal sides are equal. Proof. Let AB (fig. 184) be equal to AC. From A draw JD to the middle of BC. In the triangles ABD, JtCD, the side /AD is common, the side BD- DC, and the side AB AJ C; hence, by ~ 449, the angle ABC the angle JCB. 454. Corollary. Also the angle ADB --- D C, and, 136 SOLID GEOMETRY. [CuI. XVIII. ~ 458 Isosceles Triangle. therefore, each is a right angle; and also D-B - DdC, that is,.'he arlc, drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to the base, and bisects the angle at the vertex. 455. Corollary. An equilateral spherical triangle is also equiangular. 456. Theorem. If two angles of a spherical triangle are equal, the opposite sides are also equal, and the triangle is isosceles. Proof. Let the angle ABC (fig. 184) be equal to the angle A/CB. Then let Ji'BC be the symmetrical triangle, of which J'B = r.B, and.' C- d C. In the triangles JB C,.J'BC, the side B C is common; the angle d'BC JC CB, for each is equal to AB C; and the angle d' CB -= JB C, for each is equal to UCB; hence, by ~ 450 and 451, the side dC= JI-'B; and, therefore, A C-EB. 457. Corollary. An equiangular spherical triangle is also equilateral. 458. Theorem. If two spherical triangles on the same, or on equal spheres, are equiangular with respect to each other, they are also equilateral with respect to each other. Proof. Denote by A, B two spherical triangles which are equiangular with respect to each other; and by P, Q their polar triangles. Since the sides of P, Q are, by ~ 445, the supplements of the angles of d, B; P, Q must be equilateral with re.spect to each other; and, also, by ~ 449, equiangular with respect to each other. But the sides of dQ, B are, by CHI. XVIII. ~ 460.] THE SPHERE. 137 Sides compared with opposite Angles. 6 445, the supplements of the angles of P, Q, and therefore 3, B are equilateral with respect to each other. 459. Theorem. Of two sides of a spherical triangle, that is the greater which is opposite the greater angle; and, conversely, of two angles, that is the greater which s opposite the greater side. Proof. 1. Suppose the angle C> B (fig. 185). Draw CD so as to make the angle B CD -B. Then, by ~ 455, BD —DC, and DB LI - JDB C.D + D C. But, by ~ 438, /D + DC > SC, hence AB > JC. 2. Conversely. Suppose JB > S.C, the angle C must be greater than B; for if C were equal to or less than B,,/B would, by ~ 456 and the preceding demonstration, be equal to or less than t/C. 460. Theorem. If, of two sides of a spherical triangle, that which differs most from 900 is acute the opposite angle is acute, and if it is obtuse the opposite angle is obtuse. Proof. Of the two sides./B,.C (fig. 186) of the spherical triangle.IBC, let.3C be the one which differs the most from 900. Produce J/B, BC to B'. Since JsB, JB' are, by ~ 428, supplements of each other, one of them is acute and the other obtuse. Suppose either of them, as J.B' to be acute. Take Bl! -B'H= 900, and take ITC' -the difference between./C and 900. HF/ is the difference between.B and 900~; therefore HC' > H./. 138 SOLID GEOMETRY. [CII. XVIii. ~ 461. Sides compared with opposite Angles. a. If, then, SC is acute, we have B' Ct = C, that is AC < B'aJ. Hence, by ~ 459, in the triangle AB' C, the angle B' < the angle s/CB'. But since B' and B are each equal to the angle of the planes BAB', BGB', they are equal; and, therefore, the angle B < the angle a CB'. Again, since AlC is acute'and dB obtuse, AC < AB; and, in the triangle aBC, by ~ 459, the angle B < the angle A CB. That is, the angle B is less than either the angle.iCB or its, supplement S2CB'; but one of these angles must be acute, and therefore the angle B is acute. b. If d C is obtuse, we have BC' -- d C, that is, AC > BA; and, therefore, by ~ 459, the angle B > the angle B CAJ. Also, as B'd is acute, A/C > B'a, and, therefore, by ~ 459, the angle B' > the angle B' C; that is, the angle B is greater than either the angle ACB or its supplement A CB'; but one of these angles must be obtuse, and therefore the angle B is obtuse. 461. Corollary. Of two sides of a spherical triangle, the one which differs most from 900 is opposite the angle which differs most from 900; and, conversely, of two angles of a spherical triangle, the one which dif CH. XVIII. ~ 465.] THE SPHERE. 139 Degrees of Surface, Area of lunary Surface. fers most from 90~ is opposite the side which differs most from 900. 462. Corollary. If, of two angles of a spherical triangle, that which differs most from 900 is acute, the opposite side is acute; and if it is obtuse, the opposite side is obtuse. 463. Definition. If we suppose the surface of the hemisphere to be divided into 360 equal parts, each of these may be called a degree of spherical surface; and the degree may be subdivided into 60 minutes, and the minute into 60 seconds. 464. Corollary. Any spherical surface may, then, be expressed by that number of degrees, minutes, &c. which has the same ratio to 3603, that the given surface has to the hemisphere; it is also measured by an angle of the same number of degrbes, minutes, &c. 465. Theorem. A lunary surface is measured by double the angle of its bounding circles. Proof. Let double the angle M.3/JV' (fig. 187), expressed in degrees and minutes, be to 3600, in any ratio as 5 to 48, that is, 2 J: 5360~0.~: 1800 =- 5: 48. Suppose the arcs of great circles.t a./,.J b J1/, &c. to be drawn, so that the angles JMA/ a, a./ b, &c. may be all equal to each other, and each 418 part of 180~. The hemisphere JkIJP.i' is divided into 48 equal lunary surfaces.lasaJl',. a b.3', &c., of which the lunary surface AJ1.MYTJ' contains 5. Hence, the lunary surface.JJ.VfI': the hemisphere =5: 48 2 J2J.N": 360~, 12 140 SOLID GEOMETRY. [CH. XV1II. ~ 467. Symmetrical Triangles are equivalent. or 2.MdAJV is, by ~ 463, the measure of the lunary surface.J1J.1/.'. The demonstration is extended to the case in which the angle JWMU is incommensurate with 1800, by the principles of ~ 98. 466. Theorem. Two symmetrical spherical triangles are equivalent. Proof. Let dBC, DEF (fig. 188) be two symmetrical triangles, of which JiB=-DE, SC _- DF, and BC EF. Let P be the pole of a small circle passing through the three points d., B, C; then the distances P.1, PB, PC must be equal. Draw D Q making the angle QDE equal to P3B, and draw QE making the angle DE Q equal to ABP. Join QF. In the triangles dBP and QDE the side DE-.B, the angle QDE: — PB, and QED- PBI; and, therefore, by ~ 451, the side QD PP and QE — PB; and since these triangles are isosceles, they can be applied to each other, and are equal. In the triangles PA C, QDF, the side PJ — QD, the side S C — DF, and the angle PIC, being the sum of PJQB and B.3C, is equal to QDF, which is the sum of QDE and EDF; and, therefore, by ~ 452, the side QF PC; and since these triangles are isoceles, they are equal. In the same way, it may be proved that the isosceles triangle PBC is equal to QEE. But the triangle dB C = PA C + PB C- P.3B, and the triangle DEF — QDF + QEF- QDE; whence the triangle A-JBC the triangle DEF. 467. Corollary. Hence all spherical triangles, which CH. XVIII. ~ 469.] THE SPHERE. 141 Area of a Spherical Triangle. are equilateral or equiangular with respect to each other, are equivalent. 468. Lemma. If two spherical triangles have all angle of the one equal to an angle of the other; and the sides which include the angle in one triangle are supplements of those which include it in the other triangle; the sum of the surfaces of the two triangles is measured by double the included angle. Proof. Let the triangles be JBC and DEF (fig. 189), in which.2 and D are equal; and 2B and A1C are respectively supplements of DE and DF. Produce SiB and JSC till they meet in SI'. SB3/ and ACJSI are, by ~ 428, semicircumferences. In the triangles./'BC and DEF, the angles I' and D are equal, being both equal to Ji; J'IB and DE are equal, being supplements of SB; and S;'C and DF are equal, being supplements of SIC. It follows, therefore, from ~ 467, that they are equal in surface. But AIB C and IB C compose the lunary surface SIB CSI' which is measured by 2 S. Therefore the sum of JSBC and DEF is also measured by 2 S. 469. Theorem. The surface of a spherical triangle is measured by the excess of the sum of its three angles over two right angles, or 180~. Proof. Iet GIBC (fig. 190) be the given triangle. Produce S1C to form the circumference dSCSI'C', also produce SIB and BC to form the semicircuinferences ASBSI/ and CB C'. Then, by ~ 465, the lunary surface CSB C' 2 C, the lunary surface ASB CS' = 2 S, 142 SOLID GEOMIETRY. [C-I. XVIII. ~ 470. Area of a Spherical Polygon. or the surface J/B C + the surface JIB C' 2 C, the surface JB C -4- the surface JI'B C 2./1; and, by ~ 468, the surface J/B C + the surface JE'B C'= 2 B, for the sides BC and ASB are supplements of BC' and JIB; and the angle tIBC is equal to the angle J'BC'. The sum of these three equations is 3 X the surface JB C + the surface JI'BC + the surface /B C' + the surface Jl'B C' -2J +2 B +2 C. But the surface of the hemisphere is, by ~ 463, the surface AB C + the surface J/'BC + the surface AIB C' - the surface L'B C' = 3600; which, subtracted from the previous one, gives 2 X surface AB C - 2 Ji + 2 B -- 2 C — 360~, or the surface JIB C -J + B + C - 1800~. 470. Theorem. The surface of a spherical polygon is equal to the excess of the sum of its angles over as many times two right angles, as it has sides minus two. Proof. Let ABCCDEK (fig. 191) be the given polygon. Draw from the vertex J the arcs SJC, CJD, &c., which divide it into as many triangles as it has sides minus two. By the preceding theorem, the sum of the surfaces of all these triangles, or the surface of the polygon, is equal to the sum of all their angles diminished by as many times two right angles as there are triangles; that is, the surface of the polygon is equal to the sum of all its angles diminished by as many times two right angles, as it has sides minus two. CH. XVIII. ~ 471.] THE SPHERE. 143 Surface described by the revolution of a regular portion of a Polygon. 471. Theorenm. If a portion JIBCD (fig. 192) of a regular polygon, situated entirely upon the same side of a line PG drawn through the centre 0 of the polygon, revolve about FG as an axis, the surface generated by JBCD has for its measure the product of the circumference inscribed in the polygon by JI/Q, which is the altitude of this surface, or the part of the axis conlprehended between the extreme perpendiculars Jd.I, D Q. Proof. Let I be the middle of AiB, 01 is the radius of the inscribed circle. Draw 1K, BA", CP, perpendicular to F G, and A/X perpendicular to BN. The measure of the surface described by AB is, by ~ 398, i/B X circumference of which KI is radius, which circumference we will denote by circurnf. KI. The triangles 01K, A.BX are similar, since their sides are perpendicular to each other; whence, by ~ 178 and 234, JIB: X -- OI: IK — circumf. OI: circumf. 1K, or, since /X — JjN, iB:.-N'z circumf. 01: circumf. 1K; and, multiplying extremes and means, /B> X circumf. IK=.M7NJ X circumf. OL. Whence the area of the surface described by AiB is the product of the circumference of the inscribed circle by the altitude JI.N. In like manner the area of the surface described by BC is the product of the circumference of the inscribed circle by the altitude /NP; and that described by CD is the product of this circumference by P Q. Hence the area of the entire surface described by ABCD is the product of the circumference of the in-. 12* 144 SOLID GEOMETRY. [CIH. XVIII. ~ 476. Area of the Surface of the Sphere. scribed circle by the surn of the altitudes MN, N', PQ; that is, by the entire altitude JIQ. 472. Corollary. If the axis FG passes through the opposite vertices F, G, the area of the surface described bly the semipolygon Fd.CG is the product of the circumference of the inscribed circle by the axis FG. 473. Corollary. If the sides of the polygon are infinitely small, the polygon becomes a circle, the entire surface generated is that of a sphere, of which the generating circle is a great circle; and the surface generated by the circular segment AB CD is a zone. Hence the area of the surface of a sphere is the product of its diameter by the circumference of a great circle. And, the area of a zone is the product of its altitude by the circumference of a great circle. 474. Corollary. Since the. area of the great circle is, by ~ 279, half the product of its radius by its circumference; or one fourth of the product of its diameter by its circumference, it is one fourth of the surface of the sphere; that is The surface of a sphere is equivalent to four great circles. 475. Corollary. If we denote by R the radius of the sphere, by C the circumference of a great circle, by $ the surface of the sphere, and by 7r the ratio of the circumference to the diameter, as in q 237; we have C = 2 7 X R S=2 —- X R X 2 R — 4 r X R2. 476. Corollary. If we denote in the same way, by R and S' the radius and surface of.a second sphere, we have S' a 4 7r X R'2, CH. XVIII. ~ 481.] THE SPHERE. 145 Solidity of the Sphere. whence S: S'-47 X R2: 4 X R 2'2 n:R'2, that is, the surfaces of spheres are to each other as the squares of their radii. 477. Corollary. Zones upon the same sphere are to each other as their altitudes; and a zone is to the surface of its sphere as its altitude is to the diameter of the sphere. 478. Theorem. The solidity of a sphere is one third of the product of its surface by its radius. Proof. For the surface of the sphere may be considered as composed of infinitely small planes; and each of these planes may be considered to be the base of a pyramid, which has its vertex at the centre of the sphere, and, consequently, an altitude equal to the radius of the sphere. The sum of the solidities of these pyramids is, then, one third of the product of the sum of their bases by their common altitude, that is, the solidity of the sphere is one third of the product of its surface by its radius. 479. Corollary. In the same way, the base of a spherical pyramid or sector may be considered as composed of planes, and, therefore, the solidity of a spherical pyramid or sector is one third of the product of the polygon or zone, which serves as its base, by its radius. 480. Corollary. Spherical pyramids or sectors of the samre sphere are to each other as their bases; and a spherical pyramid or sector is to the sphere of which it is a part, as its base to the surface of the sphere. 481. Corollary. Hence, by ~ 477, spherical sec. 146 SOLID GEOMETRY. [Cel. XVIII. ~ 485. Area of a Zone and Solidity of a Sector. tors upon the same sphere are to each other as the altitudes of the zones, which serve as their bases; and a spherical sector is to the sphere of which it is a part, as the altitude of its base to the diameter of the sphere. 482. Corollary. Denoting by R the radius of the sphere, by S its surface, by Vits solidity, and by r the ratio of a circumference to its diameter, we have, by ~ 475, and 478, =-4-74 X Ri2 V — R X S-r4 7X X 3. 483. Corollary. Denoting, in like manner, by R' and AV the radius of another sphere, we have VI =4r x f13 whence V: VYI —-4 rX: 4 X RI 3 — R X 3: RX 3 that is, spheres are to each other as the cubes of their radii. 484. Corollary. Denoting by R the radius of a sphere, by C the circumference of a great circle, by H the altitude of a zone, by Z the surface of the zone, by Vthe solidity of its corresponding sector, and using n as before, we have, C 2 7r X R, from ~ 473, Z- - C X HIf 2 7 X X HI-I and, from ~ 479, VJ= Z x R:= 7 X R2 X H. 455. Corollary. The solidity of the spherical segment of one base less than a hemisphere, generated by the revolution of the portion 2tBC (fig. 193) of CH. XIX. ~ 487.] REGULAR POLYEDRONS. 147 Solidity of the Spherical Segment. a circle about the radius OP0, may be found by subtracting that of the right cone generated by OBC, from that of the spherical sector generated by t OB. In like manner, the solidity of the spherical segment of one base, greater than a hemisphere generated by the revolution of lB'C', may be found by adding that of the right cone generated by OB'C', to that of the sector generated by.fOB'. 486. Corollary. The solidity of the spherical seglnent of two bases generated by CBB'C' (fig. 193), about the diameter.J20O', may be found by subtracting that of the segment of one base generated by JIBC from that of the segment of one base generated by /B'C'. CHAPTER XIX. REGULAR POLYEDRONS. 487. Definitions. A regular polyedron is one, all whose faces are equal regular polygons, and all whose solid angles are equal to each other. These conditions can be fulfilled in only a small number of cases. a. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of these triangles, by four, or by five. No other polyedron can be formed with equilateral triangles, for six angles of such a triangle are equal to four right angles, and cannot, by ~ 339, form a solid angle. 148 sOLID GEOMETRY. [Ci. XIX. ~ 489. Five Regular Polyedrons. b. If the faces are squares, their angles may be arranged by threes. But four angles of a square are equal to four right angles, and cannot form a solid angle. c. If the faces are regular pentagons, their angles may likewise be arranged by threes. d. We can proceed no further; for three angles of a regular hexagon are equal to four right angles; three of a heptagon are greater. 488. Corollary. There can be only five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons; and in three of these polyedrons each solid angle is formed by three plane angles, and in one of them by four, and in one by five plane angles. 489. Problem. To find the number of faces of the regular polyedrons. Solutlion. Denote the number of plane angles by which each solid angle is formed by m1, and the number of sides of each face by n. Now it is evident from the symmetrical character of the regular polyedron, that a sphere can be circumscribed about it; and, if the adjacent vertices of the polyedron are joined by arcs of great circles, the surface of the sphere is divided into as many equal regular spherical polygons as the polyedron has faces, and the number of sides of each spherical polygon is n, or the same as that of the face of the polyedron. Moreover, the number of spherical angles which are formed at each vertex is mn; but their sum is equal to that of four right angles, and, since they are equal to each other, each must be represented by 360~ divided by m; that is, denoting each spherical angle by.3, CH. XIX. ~ 491.1 REGULAR POLYEDRONS. 149 Number of Faces of the Regular Polyedrons. -- 360~. nm. Again, the sum of the angles of each spherical polygon is n >X A; and therefore the surface of the polygon, which we shall denote by S, is, by ~ 470, S-n X. —(n —2) X 1800, or S —-n X 360~ n — (n —2) X 180~. Htence the number of faces is easily found, and is equal to the number of times which S is contained in the surface of the sphere, or, by ~ 464 in 7200~. 490. Corollary. When the polyedron is composed of equilateral triangles, we have n -3, whence S -— 1080~0 - 2a — 180~. a. If, then, the number of plane angles at each vertex is 3, we have m = 3, whence S = 360~ - 1800~ 1800~, which is contained 4 times in 7200, and therefore this polyedron is a tetraedron. b. If the number of plane angles at each vertex is 4, we have m<, = 4, whence S 2700~ -- 1800 - 900, which is contained 8 times in 7200, and, therefore, this polyedron is an octaedron. c. If the niunber of plane angles at each vertex is 5, we have nm -5, whence S - 02160 - 180~ 36~, which is contained 20 times in 720O, and therefore this polyedron is an icosaedron. 491. Corollary. When the polyedron is conmposed of squares, we have n -4, and, by ~ 486, m - 3, whence S -- 4800 - 360~ 120~, 150 SOLID GEOMETRY. [CH. XIX. ~ 492. Number of Faces of the Regular Polyedrons. which is contained 6 times in 720~, and therefore this polyedron is a hezaedron or cube. 492. Corollary. When the polyedron is conmposed of regular pentagons, we have n 5, and, by ~ 486, n -3, whence S == 600 -- 540~0 600, which is contained 12 times in 720~, and therefore this polyedron is a dodecaedron. THE END. iI - \'N,. ~. ~ ~.. ~~~3-~k I~S t~',l ~ I ~ I.~\ ic ~' K,- - —' —- - c t ~ ~~~ I, ~. I ~: ~ ~ C~ - ~'1IC~,r N / I - % ~ ~.~' -',,. ~. ~ ~. ~... N ~ z ~ \ f L N "~.;; ~I —--.\ /......... \ ) N) e \)N \ N-. N OjZZt c c () N b-~ ~ I \ \ /l'>.,Y().IrA I'J?['.0 l':.,5..'1.-/. 177 />'.; /.' II /,5. 51 */,// /(J7 /Kx,/ 55-v ii., l/1t.,,1 //'. t19 B 13 n 7 / /, Ii h" D h,,^ 1I'/: 71~/ ~J/z/ —v'9 lot2///./:iY -Di./w A7 A. 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