PROPOSITION IX. THEOREM. If around the vertices of the two angles of a given spherical triangle, as poles, the circumferences of two circles be described which shall pass through the third angle of the triangle; if then, through the other point in which these circumferences intersect and the two first angles of the triangle, the arcs of great circles be drawn, the triangle thus formed will have all its parts equal to those of the given triangle. Let ABC be the given triangle, CED, DFC, the arcs described about A and B as poles; then will the triangle ADB have all its parts equal to those of ABC. For, by construction, the side ADAC, DB=BC, and AB is common; hence these two triangles have their sides equal, each to each. We are now to show, that. the angles opposite these equal sides are also equal. D T E If the centre of the sphere is supposed to be at O, a solid angle may be conceived as formed at O by the three plane angles AOB, AOC, BOC; likewise another solid angle may be conceived as formed by the three plane angles AOB, AOD, BOD. And because the sides of the triangle ABC are equal to those of the triangle ADB, the plane angles forming the one of these solid angles, must be equal to the plane angles forming the other, each to each. But in that case we have shown that the planes, in which the equal angles lie, are equally inclined to each other (Book VI. Prop. XXI.); hence all the angles of the spherical triangle DAB are respectively equal to those of the triangle CAB, namely, DAB=BAC, DBA=ABC, and ADB ACB; hence the sides and the angles of the triangle ADB are equal to the sides and the angles of the triangle ACB. = Scholium. The equality of these triangles is not, however, an absolute equality, or one of superposition; for it would be impossible to apply them to each other exactly, unless they were isosceles. The equality meant here is what we have already named an equality by symmetry; therefore we shall call the triangles ACB, ADB, symmetrical triangles. PROPOSITION X. THEOREM. Two triangles 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. Suppose the side AB-EF, the side AC=EG, and the angle BAC-FEG; then will the two triangles be equal in all their parts. B C G G For, the triangle EFG may be placed on the triangle ABC, or on D ABD symmetrical with ABC, just as two rectilineal triangles are placed upon each other, when they have an equal angle included between equal sides. Hence all the parts of the triangle EFG will be equal to all the parts of the triangle ABC; that is, besides the three parts equal by hypothesis, we shall have the side BC=FG, the angle ABC=EFG, and the angle ACB=EGF. PROPOSITION XI. THEOREM. Two triangles on the same sphere, or on equal spheres, are equal in all their parts, when two angles and the included side of the one are equal to two angles and the included side of the other, each to each. For, one of these triangles, or the triangle symmetrical with it, may be placed on the other, as is done in the corresponding case of rectilineal triangles (Book I. Prop. VI.). PROPOSITION XII. THEOREM. If two triangles on the same sphere, or on equal spheres, have all their sides equal, each to each, their angles will likewise be equal, each to each, the equal angles lying opposite the equal sides. This truth is evident from Prop. IX, where it was shown, that with three given sides AB, AC, BC, there can only be two triangles ACB, ABD, differing as to the position of their parts, and equal as to the magnitude of those parts. Hence those two triangles, having all their sides re- D spectively equal in both, must either be absolutely equal, or at least symmetrically so; in either of which cases, their corres ponding angles must be equal, and lie opposite to equal sides. PROPOSITION XIII. THEOREM. 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. First. Suppose the side AB=AC; we shall have the angle C=B. 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, BD=DC, and AB= AC: hence by the last Proposition, their angles will be equal; therefore, B=C. B D Secondly. Suppose the angle B⇒C; we shall have the side AC-AB. For, if not, let AB be the greater of the two; take BO=AC, and draw OC. The two sides BO, BC, are equal to the two AC, BC; the angle OBC, contained by the first two is equal to ACB contained by the second two. Hence the two triangles BOC, ACB, have all their other parts equal (Prop. X.); hence the angle OCB-ABC: but by hypothesis, the angle ABC-ACB; hence we have OCB ACB, which is absurd; hence it is absurd to suppose AB different from AC; hence the sides AB, AC, opposite to the equal angles B and C, are equal. Scholium. The same demonstration proves the angle BAD= DAC, and the angle BDA=ADC. Hence the two last are right angles; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to that base, and bisects the vertical angle. PROPOSITION XIV. THEOREM. In any spherical triangle, the greater side is opposite the greater angle; and conversely, the greater angle is opposite the greater side. Let the angle A be greater than the angle B, then will BC be greater than AC; and conversely, if BC is greater than AC, then will the angle A be B greater than B. First. Suppose the angle A>B; make the angle BAD=B; then we shall have AD=DB (Prop. XIII.): but AD+DC is greater than AC; hence, putting DB in place of AD, we shall have DB+DC, or BC>AC. Secondly. If we suppose BC>AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC=AC; if BAC were less than ABC, we should then, as has just been shown, find BC<AC. Both these conclusions are false: hence the angle BAC is greater than ABC. PROPOSITION XV. THEOREM. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they will also be 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 their polar triangles P and Q (Prop. VIII.): but since the triangles P and Q are mutually evuilateral, they must also be mutually equiangular (Prop. XII.); and lastly, 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 mutually equiangular triangles A and B are at the same time mutually equilateral. Scholium. This proposition is not applicable to rectilineal triangles; in which equality among the angles indicates only proportionality among the sides. Nor is it difficult to account for the difference observable, in this respect, between spherical and rectilineal triangles. In the Proposition now before us, R* as well as in the preceding ones, which treat of the comparison of triangles, it is expressly required that the arcs be traced on the same sphere, or on equal spheres. Now similar arcs are to each other as their radii; hence, on equal spheres, two triangles cannot be similar without being equal. Therefore it is not strange that equality among the angles should produce equality among the sides. The case would be different, if the triangles were drawn upon unequal spheres; there, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as the radii of their spheres. PROPOSITION XVI. THEOREM. The sum of all the angles in any spherical triangle is less than six right angles, and greater than two. For, in the first place, every angle of a spherical triangle is less than two right angles: hence the sum of all the three is less than six right angles, Secondly, the measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (Prop. VIII.); hence the sum of all the three, is measured by the three semicircumferences minus the sum of all the sides of the polar triangle. Now this latter sum is less than a circumference (Prop. III.); therefore, taking it away from three semicircumferences, the remainder will be greater than one semicircumference, which is the measure of two right angles; hence, in the second place, the sum of all the angles of a spherical triangle is greater than two right angles. Cor. 1. The sum of all the angles of a spherical triangle is not constant, like that of all the angles of a rectilineal triangle; it varies between two right angles and six, without ever arriving at either of these limits. Two given angles therefore do not serve to determine the third. Cor. 2. A spherical triangle may have two, or even three of its angles right angles; also two, or even three of its angles obtuse. |