In $\triangle ABC$ inscribed in $\odot O$, $AB > AC > BC$. Point $D$ is on $\overparen{BC}$, and perpendiculars from $O$ to $AB$ and $AC$ intersect $AD$ at $E$ and $F$, respectively. The rays $BE$ and $CF$ intersect at point $P$. When $PB = PC + PO$, what is $\angle BAC$?
Solution: Connect $O B, O C$. It is easy to know that $\triangle E A B$ and $\triangle F A C$ are both isosceles triangles, and the exterior angle of the vertex angle in an isosceles triangle is equal to twice the base angle. Therefore,
$$
\begin{array}{l}
\angle B P C=\angle A E P+ \\
\angle C F D=2(\angle B A D+\angle C A D)=2 \angle B A C \\
=\angle B O C \Rightarrow B, C, P, O \text { are concyclic. }
\end{array}
$$

Considering quadrilateral $O B C P$ as a cyclic quadrilateral, by Ptolemy's theorem we get
$$
\begin{array}{l}
P E \cdot O C=P C \cdot O F \div P O \cdot B C \\
\Rightarrow(P B-P C) \cdot O C=P O \cdot B C .
\end{array}
$$

Given $P B=P C+P O \Rightarrow P B-P C=P O$, hence $O C=B C$, $\triangle O B C$ must be an equilateral triangle.
$$
\therefore \angle B A C=\frac{1}{2} \angle B O C=30^{\circ} \text {. }
$$
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)