PROPOSITION XII.-THEOREM. 547. If two triangles on the same sphere, or on equal spheres, are mutually equilateral, they are mutually equiangular; and their equal angles are opposite to equal sides. Let ABC, ABD be two triangles on the same sphere, or on equal spheres, having the sides of the one respectively equal to those of the other; then the angles opposite to the equal sides, in the two triangles, are equal. For, with three given sides, A B, A C, BC, there can be constructed only two triangles, A CB, ABD, and these triangles will be equal, each to D A B C each, in the magnitude of all their parts (Prop. XI.). Hence, these two triangles, which are mutually equilateral, must be either absolutely equal, or equal by symmetry; in either case they are mutually equiangular, and the equal angles lie opposite to equal sides. PROPOSITION XIII.-THEOREM. 548. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are mutually equilateral. Let A and B be the two given triangles; P and Q, their polar triangles. Since the angles are equal in the triangles A and B, the sides will be equal in the polar triangles P and Q (Prop. IX.). But since the triangles P and Q are mutually equilateral, they must also be mutually equiangular (Prop. XII.); and, the angles being equal in the triangles P and Q, it follows that the sides are equal in their polar triangles A and B. Hence, the triangles A and B, which are mutually equiangular, are at the same time mutually equilateral. PROPOSITION XIV. A THEOREM. D 549. If two triangles on the same sphere, or on equal spheres, have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles are equal in all their parts. In the two triangles ABC, DEF, let the side A B be equal to the side DE, the side A C to the side DF; and the angle BAC to the angle EDF; then the triangles will be equal in all their parts. B C F G E Let the triangle DEG be symmetrical with the triangle DEF (Prop. XI. Sch.), having the side E G equal to E F, the side G D equal to F D, and the side ED common, and consequently the angles of the one equal to those of the other (Prop. XII.). Now, the triangle ABC may be applied to the triangle DEF, or to D E G symmetrical with D E F, just as two rectilineal triangles are applied to each other, when they have an equal angle included between equal sides. Hence, all the parts of the triangle A B C will be equal to all the parts of the triangle D E F, each to each; that is, besides the three parts equal by hypothesis, we shall have the side BC equal to EF, the angle ABC equal to DE F, and the angle A C B equal to D F E. 550. Cor. If two triangles, A B C, DEF, on the same sphere, or on equal spheres, have two angles and the included side in the one equal to two angles and the included side in the other, each to each, the two triangles are equal in all their parts. For one of these triangles, or the triangle symmetrical with it, may be applied to the other, as is done in the corresponding case of rectilineal triangles. PROPOSITION XV.-THEOREM. 551. In every isosceles spherical triangle, the angles opposite the equal sides are equal; and, conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. Let ABC be an isosceles spherical triangle, in which the side A B is equal to the side A C; then will the angle B be equal to the angle C. B A D For, if the arc AD be drawn from the vertex A to the middle point, D, of the base, the two triangles ABD, ACD will have all the sides of the one respectively equal to the corresponding sides of the other, namely, AD common, B D equal to D C, and A B equal to AC; hence their angles must be equal; consequently, the angles B and C are equal. Conversely. Let the angles B and C be equal; then will the side A C be equal to A B. For, if A C and A B are not equal, let A B be the greater of the two; take BO equal to A C, and draw O C. The two sides BO, BC in the triangle B OC are equal to the two sides AB, BC in the triangle B AC; the angle O B C, contained by the first two, is equal to AC B, contained by the second two. Hence, the two triangles BOC, BAC have all their other parts equal (Prop. XIV. Cor.); hence the angle OCB is equal to ABC. But, by hypothesis, the angle A B C is equal to A CB; hence we have OCB equal to AC B, which is impossible; therefore A B cannot be unequal to A C; consequently the sides A B, A C, opposite the equal angles B and C, are equal. 552. Cor. The angle B A D is equal to D A C, and the angle BDA 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. 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. Take the angle BAD equal to the angle B; then, B D Α C in the triangle ABD, we shall have the side AD cqual 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 A C; 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. small circle passing through the three points A, B, C; draw the arcs OA, OB, 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 ACO; 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 D F E, 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 |