Let $ H $ be the intersection point of the altitudes $ AP $ and $ CQ $ of the acute-angled triangle $ ABC $. On its median $ BM $ marked points $ E $ and $ F $ so that $ \angle APE = \angle BAC $ and $ \angle CQF = \angle BCA $, and the point $ E $ lies inside the triangle $ APB $, and the point $ F $ lies inside the triangle $ CQB $. Prove that the lines $ AE $, $ CF $ and $ BH $ intersect at one point. (Vyacheslav Yasinsky)