1 to the same radius AO; therefore the angle FAG is equal to the angle of the planes OAB, OAC (349), which is that of the arcs AB, AC, and which is designated by BAC. In like manner, if the arc AD is equal to a quadrant, and also AE, the lines OD, OE, will be perpendicular to AO, and the angle DOE will be equal to the angle of the planes AOD, AOE; therefore the arc DE is the measure of the angle of these planes, or the measure of the angle CAB. 472. Corollary. The angles of spherical triangles may be compared with each other by means of the arcs of great circles, described from their vertices as poles and comprehended between their sides; thus it is easy to make an angle equal to a given angle. 473. Scholium. The angles opposite to each other at the Fig. 238. vertex, as ACO, BCN (fig. 238), are equal; for each is equal to the angle formed by the two planes ACB, OCN (350). Fig. 227. It will be perceived also that in the meeting of two arcs ACB, OCN, the two adjacent angles ACO, OCB, taken together, are equal to two right angles. THEOREM. 474. The triangle ABC (fig. 227) being given, if, from the points A, B, C, as poles, the arcs EF, FD, DE, be described, forming the triangle DEF, reciprocally the points D, E, F, will be the poles of the sides BC, AC, AB. Demonstration. The point A being the pole of the arc EF, the distance AE is a quadrant; the point C being the pole of the arc DE, the distance CE is likewise a quadrant; consequently the point E is distant a quadrant from each of the points A, C; therefore it is the pole of the arc AC (467). It may be shown, in the same manner, that D is the pole of the arc BC, and F that of the arc AB. 475. Corollary. Hence the triangle ABC may be described by means of DEF, as DEF is described by means of ABC. THEOREM. 476. The same things being supposed as in the preceding theorem, Fig. 227. each angle of one of the triangles ABC, DEF (fig. 227), will have for its measure a semicircumference minus the side opposite in the other triangle. Demonstration. Let the sides AB, AC, be produced, if necessary, till they meet EF in G and H; since the point A is the pole of the arc GH, the angle A will have for its measure the arc GH. But the arc EH is a quadrant, as also GF, since E is the pole of AH, and F the pole of AG (465); consequently EH+GF is equal to a semicircumference. But EH + GF is the same as EF + GH; therefore the arc GH, which measures the angle A, is equal to a semicircumference minus the side EF; likewise the angle В has for its measure circ. DF, and the angle C, circ. - DE. This property must be reciprocal between the two triangles, since they are described in the same manner, the one by means of the other. Thus we shall find that the angles D, E, F, of the triangle DEF have for their measure respectively circ. - BC, circ. AC, circ. AB. Indeed, the angle D, for example, has for its measure the arc MI; but MI+ BC= MC + BI = 1⁄2 circ.; therefore the arc MI, the measure of the angle D, = 1 circ. — BC, and so of the others. 477. Scholium. It may be remarked that, beside the triangle DEF, three others may be formed by the intersection of the three arcs DE, EF, DF. But the proposition applies only to the central triangle, which is distinguished from the three others by this, that the two angles A, D, are situated on the same side of BC, the two B, E, on the same side of AC, and the two C, F, on the same side of AB. Different names are given to the triangles ABC, DEF; we shall call them polar triangles. LEMMA. 478. The triangle ABC (fig. 229) being given, if, from the pole Fig. 229 A and with the distance AC, an arc of a small circle DEC be described, if, also from the pole B and with the distance BC, the arc DFC be described, and from the point D where the arcs DEC, DFC, cut each other, the arcs of great circles AD, DB, be drawn; we say that of the triangle ADB thus formed the parts will be equal to those of the triangle ACB. AC, Demonstration. For, by construction the side AD = Fig. 230. have the sides equal, each to each. We say, moreover, that the angles opposite to the equal sides are equal. Indeed, if the centre of the sphere be supposed in O, we can suppose a solid angle formed at the point O by the three plane angles AOB, AOC, BOC; we can suppose likewise a second solid angle formed by the three plane angles AOB, AOD, BOD. And since the sides of the triangle ABC are equal to those of the triangle ADB, it follows that the plane angles, which form one of the solid angles, are equal to the plane angles which form the other solid angle, each to each. But the planes of any two angles in the one solid have the same inclination to each other as the planes of the homologous angles in the other (359); consequently the angles of the spherical triangle DAB are equal to those of the triangle CAB, namely, DAB = BAC, DBA CBA, and ᎯᎠᏴ = ᏁᏟᏴ ; therefore the sides and the angles of the triangle ADB are equal to the sides and angles of the triangle ACB. = 479. Scholium. The equality of these triangles does not depend upon an absolute equality, or equality by superposition, for it would be impossible to make them coincide by applying the one to the other, at least except they should happen to be isosceles. The equality then under consideration is that which we have already called equality by symmetry, and for this reason, we shall call the triangles ACB, ADB, symmetrical triangles. THEOREM. 480. Two triangles situated on the same sphere, or on equal spheres, are equal in all their parts, when two sides and the included angle of the one are equal to two sides and the included angle of the other, each to each. Demonstration. Let the side AB=EF (fig. 230), the side AC = EG, and the angle BAC = FEG, the triangle EFG can be placed upon the triangle ABC, or upon the triangle symmetrical with it ABD, in the same manner as two plane triangles are applied, when they have two sides and the included angle of the one respectively equal to two sides and the included angle of the other (36). Therefore all the parts of the triangle EFG will be equal to those of the triangle ABC, that is, beside the three parts which were supposed equal we shall have the side BC= FG, the angle ABC = EFG, and the angle ACB= EGF. THEOREM. 481. Two triangles situated on the same sphere, or on equal spheres, are equal in all their parts, when a side and the two adjacent angles of the one are equal to a side and the two adjacent angles of the other, each to each. Demonstration. For one of these triangles may be applied to the other as has been done in the analogous case of plane triangles (38). THEOREM. 482. If two triangles situated on the same sphere, or on equal spheres, are equilateral with respect to each other, they will also be equiangular with respect to each other, and the equal angles will be opposite to equal sides. Demonstration. This proposition is manifest from the reasoning pursued in art. 478, by which it is shown that with three given sides AB, AC, BC, only two triangles can be constructed, differing as to the position of their parts, but equal as to the magnitude of these parts. Therefore two triangles, which are equilateral with respect to each other, are either absolutely equal, or at least equal by symmetry; in either case they are equiangular with respect to each other and the equal angles are opposite to equal sides. THEOREM. 483. In every isosceles spherical triangle the angles opposite to the equal sides are equal; and conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. Demonstration. 1. Let AB be equal to AC (fig. 231); we say Fig. 231. that the angle C will be equal to the angle B. For, if from the vertex A the arc AD be drawn to the middle of the base, the two triangles ABD, ADC, will have the three sides of the one equal to the three sides of the other, each to each, namely, AD common, BD= DC, AB=AC; consequently, by the preceding theorem, the two triangles will have their homologous angles equal, therefore B= C. 2. Let the angle B be equal to C; we say that AB will be equal to AC. For, if the side AB is not equal to AC, let AB be Fig. 232. Fig. 233. = the greater; take BO AC, and join OC. The two sides BO, BC, are equal to the two AC, BC; and the angle OBC contained by the first is equal to the angle ACB contained by the second. Consequently the two triangles have their other parts equal (480), namely OCB = ABC; but the angle ABC is, by hypothesis, equal to ACB; whence OCB is equal to ACB, which is impossible; AB then cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles B, C, are equal. 484. Scholium. It is evident, from the same demonstration, that the angle BAD = DAC, and the angle BDA = ADC. Consequently the two last are right angles; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts. THEOREM. 485. In any spherical triangle ABC (fig. 232), if the angle A is greater than the angle B, the side BC opposite to the angle A will be greater than the side AC opposite to the angle B; conversely, if the side BC is greater than AC, the angle A will be greater than the angle B. Demonstration. 1. Let the angle A> B; make the angle BAD = B, and we shall have AD = DB (483); but AD + DC > AC; in the place of AD substitute DB, and we shall have DB + DC or BC AC. 2. If we suppose BC > AC, we say that the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC = AC; and if BAC were less than ABC, it would follow, according to what has just been demonstrated, that BC AC, which is contrary to the supposition, therefore the angle BAC is greater than ABC. THEOREM. 486. If the two sides AB, AC (fig. 233), of the spherical triangle ABC are equal to the two sides DE, DF, of the triangle DEF described upon an equal sphere, if at the same time the angle A is greater than the angle D, we say that the third side BC of the first triangle will be greater than the third side EF of the second. |