Trivial angle chasing, but still a decent problem! To not leave out the midpoint $N$ alone, 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!