, 2012

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 $MP \perp CF$.