Theorem: In any hyperbolic triangle, the sum of interior angles is strictly less than $\pi$.

Proof (via the hyperboloid model):

Consider hyperbolic space as the upper sheet of $-x^2 - y^2 + z^2 = -1$ in Minkowski space with metric $ds^2 = dx^2 + dy^2 - dz^2$.

Step 1: Let $\triangle ABC$ be a hyperbolic triangle with vertices $A, B, C$ on the hyperboloid. The sides are geodesics—intersections of the hyperboloid with planes through the origin.

Step 2: Project the triangle vertically onto the $xy$-plane. This projection is area-increasing: the hyperbolic area element $dA_{\text{hyp}} = dA_{\text{flat}}/z^2$ satisfies $dA_{\text{hyp}} < dA_{\text{flat}}$ since $z > 1$ on the hyperboloid.

Step 3: The Euclidean angle sum for the projected triangle equals $\pi$. The hyperbolic angles are smaller than or equal to the projected angles (as the metric has negative curvature, it "bends inward").

Step 4: More rigorously, use the Gauss-Bonnet formula:

For a geodesic triangle, geodesic curvature $\kappa_g = 0$ on sides, and $K = -1$ (constant negative curvature):

Since area $A_{\triangle} > 0$, we have $\alpha + \beta + \gamma < \pi$. ∎