Ireland_2017

Because $P$, $A$, $Q$, $C$ are on a circle, $\angle PCA=\angle PQA$ and $\angle APC+\angle CQA=180^{\circ }$. Hence $\angle EPA+\angle AQF=180^{\circ }$. Because $A$, $B$, $F$, $Q$ are on a circle, $\angle AQF+\angle FBA=180^{\circ }$. Finally, as $F$, $E$, $B$, $A$ are concyclic, $\angle EPA+\angle ABE=180^{\circ }$.



The last three equations imply $\angle ABE=\angle AQF$, $\angle EPA=\angle FBA$ and $\angle ABE+\angle FBA=180^{\circ }$ which means that $E$, $B$ and $F$ are collinear. Hence $$\begin{gathered} \angle FEP+\angle PQF=\angle FEP+\angle PQA+\angle AQF \\ =\angle FEP+\angle PCA+\angle ABE=180^{\circ } \end{gathered}$$ because the angle sum in triangle $EBC$ is $180^{\circ }$. Therefore, the quadrilateral $EFQP$ is cyclic.

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Alternative solution.

The point $C$ is on the radical axis of the two original circles, hence the power of the point $C$ is the same for both circles: $$|CQ|\cdot |CF|=|CA|\cdot |CB|=|CP|\cdot |CE|$$ hence $E$, $F$, $Q$ and $P$ are on a circle.