Let $O_{1}$, $O_{2}, O_{3}$ be the circumcenters of $\triangle A P S$, $\triangle B Q P$, $\triangle C S Q$. After constructing the hexagon $\mathrm{O}_{1} \mathrm{PO}_{2} \mathrm{QO}_{3} \mathrm{~S}$, by the properties of circumcenters, we have
$$\begin{array}{l} \angle P O_{1} S=2 \angle A, \\ \angle Q O_{2} P=2 \angle B, \\ \angle S O_{3} Q=2 \angle C . \\ \therefore \angle P O_{1} S+\angle Q O_{2} P+\angle S O_{3} Q=360^{\circ} . \text { Thus, } \end{array}$$

we know $\angle O_{1} P O_{2}+\angle O_{2} Q O_{3}+\angle O_{3} S O_{1}=360^{\circ}$.
Rotating $\triangle O_{2} Q O_{3}$ around point $O_{3}$ to $\triangle K S O_{3}$, it is easy to determine that $\triangle K S O_{1} \cong \triangle O_{2} P O_{1}$, and we can also get $\triangle O_{1} O_{2} O_{3} \cong \triangle O_{1} K O_{3}$.
$$\begin{array}{c} \therefore \angle O_{2} O_{1} O_{3}=\angle K O_{1} O_{3}=\frac{1}{2} \angle O_{2} O_{1} K \\ =\frac{1}{2}\left(\angle O_{2} O_{1} S+\angle S O_{1} K\right) \\ =\frac{1}{2}\left(\angle O_{2} O_{1} S+\angle P O_{1} O_{2}\right) \\ =\frac{1}{2} \angle P O_{1} S=\angle A ; \end{array}$$

Similarly, we have $\angle O_{1} O_{2} O_{3}=\angle B$. Therefore, $\triangle O_{1} O_{2} O_{3} \sim \triangle A B C$.