Balkan 2012 shortlist

Let $G$ be the foot of the perpendicular from $P$ to the line $AB$, and $H$ and $I$ be the feet of the perpendiculars from $Q$ to the lines $CB$ and $CA$, respectively. Observe that we also have $\angle PCA=\angle QCB$ by the trigonometric form of Ceva's Theorem.

The quadrilaterals $AEPG$ and $AFQI$ are similar, so $\angle AEG=\angle AFI$, and therefore points $E,F,G,I$ lie on a circle $k_1$. Similarly, points $D,E,I,H$ lie on a circle $k_2$, and points $D,H,F,G$ on a circle $k_3$.

If the circles $k_1,k_2$ and $k_3$ are all different, the radical axes of pairs of these circles are the lines $AB,BC$ and $CA$, a contradiction. Therefore the points $D,E,F,G,H,I$ are cyclic.

Let $K$ and $L$ be the centers of the circles $CDPE$ and $PFG$. Since $MD\cdot ME=MF\cdot MG$, the line $MP$ is the radical axis of these two circles, and therefore perpendicular to $KL$. Since $K$ and $L$ are the midpoints of segments $PC$ and $PF$, the lines $KL$ and $CF$ are parallel, and therefore $MP\perp CF$.