The $C$-altitude is perpendicular to $AB$. So, the isogonal conjugate of the $C$-altitude, $CM$ is perpendicular to $PQ$ since it is antiparallel to $AB$.

Dear Mathlinkers, 1. Note Tc the tangent to the circumcir cle of ABC at C 2. According to Reim's theorem, Tc // PQ and we are done... Sincerely Jean-Louis

Both the solutions above are overkill. $\angle PQC=B$ as $BAQP$ is cyclic. And in triangle $MAC$, $\angle AMC=2B, MA=MC\Longrightarrow \angle MCQ=90^{\circ}-B$. So, $CM$ is perpendicular to $PQ$.