Problem

Source: Mexico National Olympiad 2020 P2

Tags: geometry, incenter, circumcircle



Let $ABC$ be a triangle with incenter $I$. The line $BI$ meets $AC$ at $D$. Let $P$ be a point on $CI$ such that $DI=DP$ ($P\ne I$), $E$ the second intersection point of segment $BC$ with the circumcircle of $ABD$ and $Q$ the second intersection point of line $EP$ with the circumcircle of $AEC$. Prove that $\angle PDQ=90^\circ$. Proposed by Ariel GarcĂ­a