Problem

Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2003 Seniors p1

Tags: geometry, concurrent, concurrency, perpendicular, angle bisector, Champions Tournament



Consider the triangle $ABC$, in which $AB > AC$. Let $P$ and $Q$ be the feet of the perpendiculars dropped from the vertices $B$ and $C$ on the bisector of the angle $BAC$, respectively. On the line $BC$ note point $B$ such that $AD \perp AP.$ Prove that the lines $BQ, PC$ and $AD$ intersect at one point.