552. Cor. The angle B A D is equal to DA C, and the angle B D A is equal to A DC; the last two are therefore right angles; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to the base, and bisects the vertical angle. PROPOSITION XVI.-THEOREM. 553. In a spherical triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side. In the triangle ABC, let the angle A be greater than B; then will the side BC, opposite to A, be greater than A C, opposite to B. D B A с Take the angle BAD equal to the angle B; then, in the triangle ABD, we shall have the side AD equal to DB (Prop. XV.). But the sum of AD plus DC is greater than A C; hence, putting DB in the place of AD, we shall have the sum of D B plus D C, or B C, greater than A C. Conversely. Let the side BC be greater than AC; then the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC equal to AC; and if B A C were less than A B C, we should then have, as has just been shown, B C less than A C. Both of these results are contrary to the hypothesis; hence the angle BAC is greater than A B C. PROPOSITION XVII.-THEOREM. 554. If two triangles on the same sphere, or on equal spheres, are mutually equilateral, they are equivalent. D A 44 F P C E B Let ABC, DEF be two triangles, having the three sides of the one equal to the three sides of the other, each to each, namely, AB to DE, AC to DF, and CB to EF; then their triangles will be equivalent. Let O be the pole of the small circle passing through the three points A, B, C; draw the arcs O A, O B, O C, and they will all be equal (Prop. V. Sch.). At the point F make the angle DFP equal to ACO; make the arc FP equal to CO; and draw DP, EP. The sides D F, FP are equal to the sides A C, CO, and the angle D F P is equal to the angle A CO; hence the two triangles DFP, A CO are equal in all their parts (Prop. XIV.); hence the side DP is equal to AO, and the angle D P F is equal to AO C. In the triangles DFE, A B C, the angles D FE, ACB, opposite to the equal sides DE, A B, are equal (Prop. XII.). Taking away the equal angles DFP, ACO, there will remain the angle PFE, equal to OCB. The sides PF, FE are equal to the sides OC, CB; hence the two triangles FPE, COB are equal in all their parts (Prop. XIV.); hence the side PE is equal to OB, and the angle FPE is equal to COB. Now, the triangles D F P, A CO, which have the sides. equal, each to each, are at the same time isosceles, and may be applied the one to the other. For, having placed OA upon its equal PD, the side O C will fall on its equal PF, and thus the two triangles will coincide; consequently they are equal, and the surface DPF is equal to A OC. For a like reason, the surface FPE is equal to CO B, and the surface D PE is equal to A OB; hence we have AOC — СОВ АОВ COB-AOB DPF+FPE-DPE, = or, ABC=DEF. Hence the two triangles A B C, D E F are equivalent. 555. Cor. 1. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are equivalent. For in that case the triangles will be mutually equilateral. 556. Cor. 2. Hence, also, if two triangles on the same sphere, or on equal spheres, have two sides and the included angle, or have two angles and the included side, in the one equal to those in the other, the two triangles are equivalent. 557. Scholium. The poles O and P might lie within the triangles ABC, DEF; in which case it would be requisite to add the three triangles DPF, FPE, DPE together, to form the triangle DEF; and in like manner to add the three triangles AO C, COB, AOB together, to form the triangle ABC; in all other respects the demonstration would be the same. PROPOSITION XVIII.-THEOREM. 558. The area of a lune is to the surface of the sphere as the angle of the lune is to four right angles, or as the arc which measures that angle is to the circumference. A Let A CBD be a lune upon a sphere whose diameter is A B; then will the area of the lune be to the surface of the sphere as the angle DOC to four right angles, or as the arc DC to the circumference of a great circle. C D B F E For, suppose the arc CD to be to the circumference CDEF in the ratio of two whole numbers, as 5 to 48, for example. 20* Then, if the circumference CDEF be divided into 48 equal parts, CD will contain 5 of them; and if the pole A be joined with the several points of division by as many quadrants, we shall have 48 triangles on the surface of the hemisphere ACDEF, all equal, since all their parts are equal. Hence, the whole sphere must contain 96 of these triangles, and the lune A CBD 10 of them; consequently, the lune is to the sphere as 10 is to 96, or as 5 to 48; that is, as the arc CD is to the circumference. B If the arc CD is not commensurable with the circumference, it may still be shown, by a mode of reasoning exemplified in Prop. XVI. Bk. III., that the lune is to the sphere as CD is to the circumference. D A E 559. Cor. 1. Two lunes on the same sphere, or on equal spheres, are to each other as the angles included between their planes. 560. Cor. 2. It has been shown that the whole surface of the sphere is equal to eight quadrantal triangles (Prop. X. Sch.). Hence, if the area of a quadrantal triangle be represented by T, the surface of the sphere will be represented by 8T. Now, if the right angle be assumed as unity, and the angle of the lune be represented by A, we have, Area of the lune: 8T:: A: 4, which gives the area of lune equal to 2 A × T. 561. Cor. 3. The spherical ungula included by the planes AC B, AD B, is to the whole sphere as the angle DOC is to four right angles. For, the lunes being equal, the spherical ungulas will also be equal; hence, two spherical ungulas on the same sphere, or on equal spheres, are to each other as the angles included between their planes. PROPOSITION XIX.-THEOREM. 562. If two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed is equivalent to a lune, whose angle is equal to the angle formed by the circles. Let the great circles BAD, CA E intersect on the surface of a hemisphere, ABCDE; then will the sum of the opposite triangles, BAC, DAE, B be equal to a lune whose angle is DAE. A E D F For, produce the arcs AD, AE till they meet in F; and the arcs BAD, ADF will each be a semi-circumference. Now, if we take away AD from both, we shall have D F equal to BA. For a like reason, we have EF equal to CA. DE is equal to BC. Hence, the two triangles BAC, DEF are mutually equilateral; therefore they are equivalent (Prop. XVII.). But the sum of the triangles D EF, DA E is equivalent to the lune ADFE, whose angle is DA E. PROPOSITION XX.-THEOREM. 563. The area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles, multiplied by the quadrantal triangle. Let A B C be a spherical triangle; its area is equal to the excess of the sum of its angles, A, B, C, above two right angles multiplied by the quadrantal triangle. For produce the sides of the triangle A B C till they |