Girard's theorem

Remarks

The quantity $E = \alpha + \beta + \gamma - \pi$ is the spherical excess; the theorem shows that it is positive. Proof idea: a lune, the region between two half great circles meeting at angle $\theta$, has area $2\theta R^2$, since it is the fraction $\theta / (2\pi)$ of the sphere of area $4\pi R^2$. The three great circles through the sides of $T$ cut the sphere into eight triangles. Let $T'$ be the antipodal triangle; the six lunes with angles $\alpha$, $\beta$, $\gamma$ at the vertices of $T$ and of $T'$ cover the sphere. They cover $T$ and $T'$ three times each and every other point once, so $2(2\alpha + 2\beta + 2\gamma) R^2 = 4\pi R^2 + 4 \operatorname{Area}(T)$, which gives the formula. Example: the octant triangle with vertices $(R, 0, 0)$, $(0, R, 0)$, $(0, 0, R)$ has three right angles, so its area is $R^2 (3\pi/2 - \pi) = \pi R^2 / 2$, one eighth of $4\pi R^2$.

Going deeper

For a student

The theorem is the spherical case of the Gauss–Bonnet theorem: the integral of the Gaussian curvature $1/R^2$ over the triangle equals the angle excess. The hyperbolic counterpart says that a hyperbolic triangle in curvature $-1$ has area $\pi - (\alpha + \beta + \gamma)$, the defect. Since the excess is positive and tends to $0$ for small triangles, it explains why the angle sum of a small spherical triangle is close to $\pi$. It also shows that two spherical triangles with the same angles have the same area. In fact such triangles are congruent, by the law of cosines for angles.