Question 26
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 the intersection of the lines $DE$ and $AB$. Prove that $MP \perp CF$.