Connect $P D$, $A S$,
$R C$, $B R$, $A P$,
$S D$. From
$\triangle E B R \sim \triangle E P A$,
$\triangle F D S \sim \triangle F P A$,
we have $\frac{B R}{P A}=\frac{E B}{E P}$,
$$\frac{P A}{D S}=\frac{F P}{F D} \text {. }$$

Multiplying the two equations, we get $\frac{B R}{D S}=\frac{E B \cdot F P}{E P \cdot F D}$.
Furthermore, from $\triangle E C R \sim \triangle E P D$, $\triangle F P D \sim \triangle F A S$, we have $\frac{C R}{P D}=\frac{E C}{E P}$, $\frac{P D}{A S}=\frac{F P}{F A}$.
Multiplying these two equations, we get $\frac{C R}{A S}=\frac{E C \cdot F P}{E P \cdot F A}$.
From (1) and (2), we have $\frac{B R \cdot A S}{D S \cdot C R}=\frac{E B \cdot F A}{E C \cdot F D}$.
Thus, $\frac{B R}{R C} \cdot \frac{C D}{D S} \cdot \frac{S A}{A B}=\frac{E B}{B A} \cdot \frac{A F}{F D} \cdot \frac{D C}{C E}$.
By Menelaus' theorem, we have
$$\frac{E B}{B A} \cdot \frac{A F}{F D} \cdot \frac{D C}{C E}=1 \text {. }$$

From (3) and (4), we get $\frac{B R}{R C} \cdot \frac{C D}{D S} \cdot \frac{S A}{A B}=1$.
Therefore, $B D$, $R S$, and $A C$ intersect at one point.
That is, $R$, $T$, and $S$ are collinear.