Question 74
Let $ABC$ be an acute triangle, and $P$ a point on the interior of side $BC$. Let $I$ be the incenter of triangle $ABC$, and denote by $D$ the foot of the altitude from $I$ to $BC$. Line $BI$ meets the internal angle bisector of $\angle APC$ at $X$, while line $CI$ meets the internal angle bisector of $\angle APB$ at $Y$. Show that the points $D, P, X, Y$ lie on a circle.