Hyperbolic triangle

Remarks

Example: in the Poincaré disk model, take $A = 0$, $B = 1/2$ and $C = i/2$. The only geodesic through $0$ and $1/2$ is the real diameter, which does not contain $i/2$, so $A, B, C$ span a hyperbolic triangle. Its angle at $A$ is $\pi/2$, since the model is conformal and the sides at $A$ are perpendicular diameters. Non-example: the points $0$, $1/4$, $1/2$ lie on the real diameter, which is one geodesic line, so they do not form a hyperbolic triangle. In curvature $-1$, the angles satisfy $\alpha + \beta + \gamma < \pi$, and the area is $\pi - (\alpha + \beta + \gamma)$. One also allows vertices at infinity, on the ideal boundary; a triangle with all three vertices ideal is an ideal triangle, with all angles $0$ and area $\pi$. The notion is the hyperbolic counterpart of the Euclidean triangle.