Mongolian Mathematical Olympiad

Let $K$, $N$, and $L$ be the midpoints of segments $BP$, $BC$, and $PC$, respectively. Since $AB$ is parallel to $MN$ and $AP$ is parallel to $ML$, we have $\angle BAP=\angle NML$.

Considering the parallelogram $KNCL$, $\angle BCP=\angle BCL=\angle NKL$. Moreover, since $\angle NML=\angle NKL$, we have $K$, $N$, $L$, and $M$ lie on the same circle.

Since $\angle BQP=90^{\circ }$, we can conclude that $QK=\frac{BP}{2}$. $Q$, $K$, $N$, $L$, and $M$ all lie on the same circle, because $QK=NL$.

Since $QM=\frac{CP}{2}=KN$, $QK$ is parallel to $AB$. Consequently, we can deduce that $RQ=QP$.