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?
Best Answer
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.