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A subset of a Riemannian manifold is geodesically convex when any two of its points can be joined by a minimizing geodesic segment lying in the subset. In the hyperbolic plane the joining geodesic segment is unique. Thus in the Beltrami-Klein model, geodesically convex subsets correspond to Euclidean convex sets.

A six-sided polygon in the hyperbolic plane whose sides are geodesic segments. The especially useful right-angled case is a right-angled hyperbolic hexagon.

At an interior point $w$ of the geodesic segment from $z_1$ to $z_2$, the two angles made with the segment toward $w^{\prime }$ add to $\pi$. At least one is at least $\pi /2$. In the corresponding hyperbolic triangle, part (a) says that its opposite side $w^{\prime }{z}_i$ is at least the adjacent side $w^{\prime }w$. ThereforeIf $w$ is an endpoint the assertion is immediate. The same inequality also holds when $w^{\prime }$ lies on the complete geodesic through $z_1,z_2$: parameterize that geodesic by arclength and use the elementary inequality $|t-s|\le \max (|t-a|,|t-b|)$ for $a\le s\le b$. This extension will allow repeated use of the result in part (c).

Consider the filled hyperbolic triangle with vertices $A,B,C$. Every point $P$ is on some geodesic segment $Aw$ with $w\in BC$: for instance, in the Beltrami-Klein model these are the ordinary straight segments filling a convex triangle. For any other point $Q$ of the triangle, apply part (b) first on $Aw$ and then on $BC$:Applying the same argument to $Q$, with each vertex held fixed, bounds every term on the right by the maximum of the three side lengths. The endpoints of a longest side attain that bound. Thus the diameter is exactly the longest side length: