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For curvature $-1$, the Poincare disc model has metric and area elementThe given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal toRadial symmetry therefore identifies the whole hyperbolic circle in the Poincare disc. Integrating its enclosed hyperbolic area gives

For a geodesic hyperbolic triangle with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem states $A=\pi -(\alpha +\beta +\gamma )$, in particular $A\le \pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. ConsequentlyThis uses hyperbolic area bounds an inscribed disc and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.