RU-ALLRUSSIAN 2015 Problem 107

In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$ , respectively, so that the $QM \perp AC$ and $PM \perp AB$ . The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$ M. Didin