Page:Spherical Trigonometry (1914).djvu/34

33. Identical and symmetrical equality of triangles. If $ABC$, $A''B'C''''$ be antipodal triangles, the plane of the arc $BC$ is the same as the plane of the arc $B'O''$, and similarly for $CA$, $C'A''$ and $AB$, $A''B'$. Hence the angles of the one triangle are respectively equal to those of the other; and as the distance between two points is equal to the distance between the diametrically opposite points, the sides of one triangle are equal to the corresponding sides of the other. Thus the triangles have all their corresponding elements equal.

There is however this difference between them, that if we go round the two triangles in such a manner as to take corresponding elements in the same order, we shall go round one triangle in the clockwise, the other in the counter-clockwise sense. And so if the triangle $A''B'C''$ be shifted bodily in the surface of the sphere until $B'$ coincides with $B$, and $C'$ with $C$, then the remaining vertices $A''$ and $A$ will not coincide, but will be on opposite sides of the common arc $BC$. The triangles therefore are not superposable. If however the triangle $A''B'C'$, regarded as a material film, were lifted off the sphere and, as it were, turned inside out, so that the formerly convex side of its surface would become concave, the altered triangle could then be exactly superposed on the triangle $ABC$.

Antipodal triangles are accordingly equal to one another in every respect, and yet not superposable in the ordinary meaning of the term. Triangles having this sort of equality are said to be symmetrically^[1] equal, as distinguished from triangles which are superposable and which are said to be identically equal, or congruent.

34. The proof, by the method of superposition, of the equality of plane triangles under certain circumstances, as used for example in Euclid, I, 4, 8, and 26, may be applied equally well to spherical triangles on the same sphere; the

1. This term is due to Legendre (Géométrie, VI, Def. 16.)