Problem

Source: Bosnia and Herzegovina 2018 TST

Tags: geometry, perpendicular, orthocenter



In acute triangle $ABC$ $(AB < AC)$ let $D$, $E$ and $F$ be foots of perpedicular from $A$, $B$ and $C$ to $BC$, $CA$ and $AB$, respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$. Prove that $\angle ADP = \angle PBQ$