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?



Triangle area on surfaces of constant curvature
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.


