(12) $\triangle A B C$ is an inscribed triangle in circle $O$, $A B>A C>B C$, point $D$ is on arc $B C$, perpendiculars from point $O$ to $A B$ and $A C$ intersect $A D$ at $E$ and $F$, respectively, and rays $B E$ and $C F$ intersect at point $P$. When $B P=P C+P O$, then $\angle B A C=$ $\qquad$
(12) $30^{\circ}$ Hint: Connect $O B, O C$, it is easy to know that $\triangle E A B, \triangle F A C$ are both isosceles triangles, and the supplement of the vertex angle of an isosceles triangle is equal to twice the base angle, so
$$
\begin{aligned}
\angle B P C= & \angle A E P+\angle C F D \\
& =2(\angle B A D+\angle C A D) \\
& =2 \angle B A C=\angle B O C,
\end{aligned}
$$

Therefore, $B, C, P, O$ are concyclic. By Ptolemy's theorem, we have:
$$
B P \cdot O C=P C \cdot O B+P O \cdot B C,
$$

which means
$$
(B P-P C) \cdot O C=P O \cdot B C,
$$

Given that $B P=P C+P O$, we can get $O C=B C$, so $\triangle O B C$ is an equilateral triangle, hence
$$
\angle B A C=\frac{1}{2} \angle B O C=30^{\circ} \text {. }
$$