Proof: From Figure 3, we have
$$\begin{array}{l} \angle B E C + \angle B F C \\ = \angle A + \angle A C E + \angle A + \angle A B F \\ = \angle A + \angle A C E + \angle B C E + \angle A B F + \angle C B F \\ = \angle A + \angle A C B + \angle A B C \\ = 180^{\circ}. \end{array}$$

From the area relationship, we get
$$\frac{S_{\triangle P B E}}{S_{\triangle P C F}} = \frac{P B \cdot P E}{P C \cdot P F} = \frac{E B \cdot E P}{F C \cdot F P} \Rightarrow \frac{P B}{P C} = \frac{B E}{C F}.$$

Since $P B = P C$, it follows that $B E = C F$.
Connecting $B S$, $B N$, $E S$, and $F N$, we have
$$S_{\triangle C N S} = S_{\triangle E N S}, \quad S_{\triangle B S N} = S_{\triangle F S N}.$$

Therefore, $\frac{C S}{S F} = \frac{S_{\triangle C N S}}{S_{\triangle A S N}} = \frac{S_{\triangle E N S}}{S_{\triangle B S N}} = \frac{E Q}{Q B}$.
Thus, $\frac{C S}{S F} = \frac{E Q}{Q B}$, which implies $\frac{C F}{S F} = \frac{E B}{Q B}$.
Since $B E = C F$, it follows that $B Q = F S$.
(Yuan Anquan, Hechuan Taihe Middle School, Chongqing, 401555)