Problem

Source: RMO Mumbai 2016, P1

Tags: geometry, incenter



Let $ABC$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $I$ be the incenter of $ABC$. Draw a line perpendicular to $AI$ at $I$. Let it intersect the line $CB$ at $D$. Prove that $CI$ is perpendicular to $AD$ and prove that $ID=\sqrt{b(b-a)}$ where $BC=a$ and $CA=b$.