Saudi Arabia Mathematical Competitions

Let $K$, $L$, $M$, $B^{\prime }$, $C^{\prime }$ be the midpoints of $BP$, $CQ$, $PQ$, $CA$, and $AB$, respectively. Since $CA\parallel LM$, we have $\widehat{LMP}=\widehat{QPA}$. Since $\mathcal{C}$ touches the segment $PQ$ at $M$, we find $\widehat{LMP}=\widehat{LKM}$. It follows $$\widehat{QPA}=\widehat{LKM}\text{\hspace{1em}(1)}$$ Similarly, from $AB\parallel MK$ we get $$\widehat{PQA}=\widehat{KLM}\text{\hspace{1em}(2)}$$ From (1) and (2) we obtain that triangles $APQ$ and $MKL$ are similar, hence $$\frac{AP}{AQ}=\frac{MK}{ML}=\frac{\frac{QB}{2}}{\frac{PC}{2}}=\frac{QB}{PC}\text{\hspace{1em}(3)}$$ Now (3) is equivalent to $AP\cdot PC=AQ\cdot QB$ which means that the power of points $P$ and $Q$ with respect to the circumcircle of $△ABC$ are equal, hence $OP=OQ$.