Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 / 4 / a

Let $[u,v]$ be one side of the geodesic quadrilateral, and draw a diagonal from $u$ to the opposite vertex. A point $p\in [u,v]$ lies, by $\delta$-thinness of the first geodesic triangle, within $\delta$ either of an adjacent side or of the diagonal. In the latter case, $\delta$-thinness of the second triangle places the nearby point of the diagonal within another $\delta$ of one of the other two sides. The triangle inequality then places $p$ within $2\delta$ of the remaining three sides. The argument applies to every side, proving the geodesic quadrilateral in a hyperbolic metric space bound.