Problem 928

2. As shown in Figure 2, connect $B C$, $R P$, and $D F$.
From $A B=A C$, we know $\angle A D C=\angle A F B$.
Therefore, $S$, $D$, $F$, and $T$ are concyclic.
Thus, $\angle Q S K=\angle T D F=\angle R A C$.
Combining $\angle S K Q=\angle A C E$, we get
$\triangle Q S K \backsim \triangle R A C$.
Similarly, $\triangle Q T K \backsim \triangle P A B$.
Hence, $\frac{S K}{K Q}=\frac{A C}{C R}$, $\frac{K Q}{K T}=\frac{B P}{B A}$.
Therefore, $\frac{S K}{K T}=\frac{S K}{K Q} \cdot \frac{K Q}{K T}=\frac{A C}{C R} \cdot \frac{B P}{B A}=\frac{B P}{C R}$.
By Pascal's theorem, $P$, $Q$, and $R$ are collinear.
Let point $J$ be on the ray $C D$ such that $\triangle B C J \backsim \triangle B A P$, and connect $P J$.
$$\begin{array}{l} \text { From } \frac{B P}{B J}=\frac{A B}{C B}, \text { and } \\ \angle A B C=\angle P B A-\angle P B C \\ =\angle J B C-\angle P B C=\angle J B P, \end{array}$$

we get $\triangle B P J \backsim \triangle B A C$.
Combining $A B=A C$, we know $P B=P J$.
Also, $\angle D P E=\angle B P A=\angle B J C$, so $B$, $J$, $D$, and $P$ are concyclic.
Thus, $\angle P J Q=\angle D B E=\angle D C E$.
Therefore, $P J \parallel C R$.
Hence, $\frac{B P}{C R}=\frac{P J}{C R}=\frac{P Q}{Q R}$.
From equations (1) and (2), the proposition is established.