Problem 947

As shown in Figure 4, draw $B K / / P Q$, intersecting line $C D$ at point $J$ and circle $\odot O$ at point $K$. Connect $C K$, intersecting circle $\odot O$ at point $L$.

Since $C D$ and $P Q$ are the internal and external angle bisectors of $\angle C$, respectively, and $C D \perp P Q, O J \perp B K$, it follows that $B J=K J$.
Thus, $C B=C K$.
By the power of a point theorem, $C L \cdot C K=C F \cdot C B$.
Therefore, $C L=C F$.
Also, since $\angle K C J=\angle B C J$ and $C D \perp P Q$, we have $\angle L C P=\angle F C Q$.
Notice that,
$$\begin{array}{l} \angle C P E=\angle E B K=\angle E L K \\ \Rightarrow P, E, L, C \text{ are concyclic} \\ \Rightarrow \angle C E P=\angle C L P . \end{array}$$
$$\begin{array}{l} \text{ and } \angle C F Q=\angle A F B=\angle A E B=\angle C E P \text{, hence } \\ \angle C L P=\angle C F Q . \end{array}$$

From equations (1), (2), and (3), we know
$$\begin{array}{l} \triangle C L P \cong \triangle C F Q \Rightarrow C P=C Q \\ \Rightarrow \angle P O C=\angle Q O C \end{array}$$