In a triangle $ABC$ a point $I$ is a centre of inscribed circle. A line $AI$ meets a side $BC$ in a point $D$ . A bisector of $AD$ meets lines $BI$ and $CI$ respectively in a points $P$ and $Q$. Prove that heights of triangle $PQD$ meet in the point $I$.
I've tried to show that sides of triangle $PQD$ are parallel to sides of triangle $ABC$ but it didn't work out. That's why I ask you for help.