Let be the midpoint of a geodesic and put . Apply the geodesic quadrilateral in a hyperbolic metric space bound to the quadrilateral with consecutive vertices . The point is within of one of the other three sides. It cannot be within of , because every point of has distance greater than from every point of .
By symmetry there is therefore a point with . The triangle inequality gives
and hence
Since and is a closest point of to , we also have . Therefore , as required.

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