Team selection tests for JBMO 2018

Let $K$ be the circumcenter of triangle $PEF$. Note that $$\angle APB=\angle ADB=45^{\circ }$$ so triangle $KEF$ is right isosceles at $K$. Thus, quadrilateral $OEKF$ is cyclic. Hence, $\angle FOK=\angle FEK=45^{\circ }$, this means $OK$ is bisector of angle $DOC$, it is also perpendicular bisector of $CD$. Since $KP=KQ$, $OP=OQ$, we have $OK\perp PQ$. Therefore, $PQ$ and $CD$ are parallel. $\square$