Given the circle $\Gamma$ with center $O'$, let $H$ be the antipode of point $A$ with respect to $\odot O'$, and the extension of $HO$ intersects $\odot O'$ at point $P$.
$$\begin{array}{l} \text { Then } \angle B F O=\frac{1}{2}\left(180^{\circ}-\angle B O F\right) \\ =\frac{1}{2}\left(180^{\circ}-2 \angle C\right)=90^{\circ}-\angle C \\ =\angle B A H=\angle B P H=\angle B P O \end{array}$$
$\Rightarrow B, P, F, O$ are concyclic
$$\begin{aligned} \Rightarrow & \angle A P F=\angle A P H+\angle F P H \\ & =90^{\circ}+\angle F B O=90^{\circ}+90^{\circ}-\angle C \\ & =180^{\circ}-\angle A E F \end{aligned}$$
$\Rightarrow A, E, F, P$ are concyclic
$$\begin{aligned} \Rightarrow & \angle P C F=\angle P B E, \\ & \angle P F C=180^{\circ}-\angle A F P \\ & =180^{\circ}-\angle A E P=\angle P E B \\ \Rightarrow & \triangle P F C \sim \triangle P E B \\ \Rightarrow & \frac{P F}{P E}=\frac{F C}{E B}=\frac{P C}{P B} . \end{aligned}$$

Since points $B$ and $C$, $P$ and $Q$ are symmetric with respect to the perpendicular bisector of $BC$, we have,
$$\begin{aligned} P C & =Q B, P B=Q C \\ \Rightarrow & Q B \cdot E B=Q C \cdot F C \\ \Rightarrow & \frac{1}{2} Q B \cdot E B \sin \angle A B Q \\ & =\frac{1}{2} Q C \cdot F C \sin \angle A C Q \\ \Rightarrow & S_{\triangle Q B E}=S_{\triangle Q C F} . \end{aligned}$$