Question 170
In a triangle $ABC$ with incenter $I$, let $P$ be the intersection point of the bisector of the angle $A$ with the circumcircle other than $A$, $D$ the point of tangency of the incircle to the side $BC$, and $Q$ the intersection point of $PD$ with the circumcircle other than $P$. Show that $PI = QI$ if $PD$ is equal to the inradius.