Triangle inequality without parallel postulate Without the parallel postulate, how can one prove that in any triangle $ABC$, $AB+AC>BC$? Most often, we introduce the triangle inequality _a priori_ as part of the definition of a metric, but here everything we have is the four basic postulates (save the parallel postulate) and the five principles in Euclid's _Elements_. The proof Euclid himself made is actually pretty close, only that he invoked the proposition that the exterior angle is greater than an interior angle, which should be false without the parallel postulate (e.g. in spherical geometry). Is it possible to completely avoid the parallel postulate and prove the triangle inequality?

I already discussed this point several times (see this answer, for instance): Euclid's exterior angle theorem is perfectly valid in absolute geometry (i.e. euclidean geometry without parallel postulate), as its proof does not rely on the parallel postulate (and if you do not believe that, you should point out where this postulate is used in the proof).

Elliptic geometry IS NOT absolute geometry, because it also does away with Euclid's second postulate.