Triangle area on surfaces of constant curvature I am looking for an elementary derivation of the formula for the area of a geodesic triangle lying in a surface of constant curvature $\kappa$, depending on the angles and side length.
Of course, the formula can easily be derived from the Gauss–Bonnet formula to be
$$A = \frac{1}{\kappa}(\alpha + \beta + \gamma - \pi)$$
for $\kappa \neq 0$. However, I would like to have an elementary geometric proof.
Does anybody know a reference?
 A: All the "elementary derivations of that type" are cheating (it may look nice but it proves nothing). 
The only elementary way to introduce area is adding it as an axiom (which is already kind of cheating).
You have to say that there is a additive area-functional on the set of all polygons.
Then you probably want to prove that this functional is unique (or include it in the same axiom).
It remains to notice that your functional $A$ satisfies the same properties and nothing left to prove.
A: M. Berger, Geometrie, vol. V. MR0536874
Edit. Let me sketch a proof for the spherical triangle. Let the sphere have area $4\pi$.
First you derive the area of digon. It is $2\alpha$, where $\alpha$ is the angle,
by completely elementary reasons. Now consider a triangle. Extend its sides to three
full great circles. These three circles make several digons and two equal triangles
(the second one is centrally symmetric to the original one).
Make a picture showing how these three circles partition the sphere. As the areas of all
digons are known the area of a triangle is simply derived by the exclusion-inclusion formula!
Notice: this proof is truly elementary in the sense that it only uses the existence of the area
for a digon and triangle, its invariance with respect to rotations, and finite additivity. Euclid COULD give a rigorous
proof of this. As rigorous as his investigation of areas of Euclidean triangles.
