As shown in Figure 2, connect $O A, O D, O E, O F, O P,$ and $O Q$. To prove $Q R / / A B$, we need to prove that
$$\frac{E R}{R F}=\frac{A Q}{Q F}.$$

By Menelaus' theorem, we have
$$\frac{E R}{R F}=\frac{E P}{P A} \cdot \frac{A C}{C F}.$$

It is also easy to see that $\triangle A O C \sim \triangle O F C$, hence
$$\frac{A C}{C F}=\frac{S_{\triangle A O C}}{S_{\triangle O F C}}=\frac{O A^{2}}{O F^{2}}.$$

Therefore, $\frac{E R}{R F}=\frac{E P}{P A} \cdot \frac{O A^{2}}{O F^{2}}$.
Thus, we need to prove that $\frac{A Q}{Q F}=\frac{E P}{P A} \cdot \frac{O A^{2}}{O F^{2}}$, which is equivalent to
$$\frac{A Q}{Q F} \cdot \frac{A P}{P E}=\frac{O A^{2}}{O F^{2}}.$$

On the other hand, by the tangent length theorem, we have
$$\begin{array}{l} \angle A O Q=\angle A O F - \angle Q O F \\ =\frac{1}{2} \angle E O F - \frac{1}{2} \angle D O F = \frac{1}{2} \angle E O D = \angle P O E. \end{array}$$

Similarly, $\angle A O P = \angle Q O F$.
Thus, $\frac{A Q}{Q F} \cdot \frac{A P}{P E}=\frac{S_{\triangle A O Q}}{S_{\triangle Q O F}} \cdot \frac{S_{\triangle A O P}}{S_{\triangle A O E}}$
$$=\frac{A O \sin \angle A O Q}{O F \sin \angle Q O F} \cdot \frac{A O \sin \angle A O P}{O E \sin \angle P O E}=\frac{O A^{2}}{O F^{2}}.$$

Therefore, the conclusion holds.