Let $X$ be the second intersection of the circles $CMEP$ and $CFNQ$. Then, $\angle CXP= \angle CXQ=90^{\circ}$, so $P,X,Q$ collinear. Also, $$\angle EXC = \angle EMC=\angle DMC =\angle DNB=\angle FNB=\angle FNC=\angle FXC $$which means that $E,X,F $ collinear. Lines $EF$ and $AB$ are homothetic with center $C$ and ratio $2/3$, so the cetroid of $ABC$ is on the line $EF$, and $$\angle ACX =\angle MCX= \angle MEX=\angle MDB=\angle MDA=\angle CAD$$$X$ is the centroid, which finishes the proof since $X \in PQ$