Autumn tournament

Let $T$ be the midpoint of $CQ$ (it is anyway the center of the important circle with diameter $CQ$). Then $OC$ is the perpendicular bisector of $AB$ (as $AC=BC$), triangle $ONT$ is isosceles by symmetry (as $P$ and $Q$ are symmetric with respect to the midpoint $M$ of $AB$ and hence with respect to $CO$, by the problem condition) and $OT$ is the perpendicular bisector of $CE$, so $\angle OCF=90^{\circ }-\angle OCT=90^{\circ }-\angle OCD=\angle ODM=\angle ODF$, done! $\square$