# Isomorphic triangles?

From my previous post I have learnt that spherical triangles can have different interior angle sums. Is this enough to argue that the triangles are not isomorphic? I am not sure how isomorphism works on a sphere. Thanks in advance.

(I only know isomorphism in the context of algebra where it is a bijective homomorphism...)

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The term isomorphic is (in English) not used for triangles. The standard term for what I think you are trying to convey is congruent. Certainly different angle sum implies that the triangles are not congruent. More interesting is that if we are looking at two triangles on the same sphere, and the angles match, then the triangles are actually congruent. This is very different from the ordinary plane, where you can have two triangles that are similar but not congruent, like a small equilateral triangle and a big one. – André Nicolas Feb 9 '12 at 21:27
@AndréNicolas: Thank you! And that is a very interesting fact :) – sphere Feb 9 '12 at 21:32
In the plane, two triangles are congruent if there is a sequence of rotations, translations, and reflections (aka isometries) that brings one triangle to the other. The triangles are similar if you allow for dilations too. These things are transformations on the plane. With the sphere, dilation isn't exactly an option. We say triangles are congruent if isometries (rotations) can bring one triangle to the other. If you allowed for some other transformation on the sphere that preserved straight lines (great circles), you might get a different kind of "similarity" between spherical triangles. – alex.jordan Feb 9 '12 at 21:47
@alex.jordan: Under general great-circle-preserving transformations, all nondegenerate spherical triangles are equivalent. – Henning Makholm Feb 10 '12 at 0:58
@Henning Yes, just like under general affine transformations of $\mathbb{R}^2$, all triangles are equivalent. But the subgroup of translations, rotations, reflections, and dilations give the usual notion of symmetry. So what I meant was, maybe some subgroup of general great-circle-preserving transformations would provide some kind of interesting equivalence. – alex.jordan Feb 10 '12 at 1:30