BALKAN-SL 2012 G6

Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$ . Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$ , and $F$ be the foot of perpendicular from $Q$ to the line $AB$ . Let $M$ be intersection of the lines $DE$ and $AB$ . Prove that $MP \perp CF$