Problem

Source: Ukraine TST 2011 p10

Tags: concurrent, concurrency, equal angles, geometry, orthocenter



Let $ H $ be the point of intersection of the altitudes $ AP $ and $ CQ $ of the acute-angled triangle $ABC$. The points $ E $ and $ F $ are marked on the median $ BM $ such that $ \angle APE = \angle BAC $, $ \angle CQF = \angle BCA $, with point $ E $ lying inside the triangle $APB$ and point $ F $ is inside the triangle $CQB$. Prove that the lines $AE, CF$, and $BH$ intersect at one point.