12. $\triangle A B C$ is an inscribed triangle in $\odot O$, $A B>A C$ $>B C$, point $D$ is on arc $\overparen{B C}$, perpendiculars from $O$ to $A B$ and $A C$ intersect $A D$ at points $E$ and $F$, respectively, and rays $B E$ and $C F$ intersect at point $P$. If $B P=P C+P O$, then $\angle B A C=$ $\qquad$ .
12. $30^{\circ}$.

Connect $O B, O C$. It is easy to know that $\triangle E A B$ and $\triangle F A C$ are both isosceles triangles.

From the property that the supplement of the vertex angle of an isosceles triangle is twice the base angle, we have
$$
\begin{array}{l}
\angle B P C=\angle A E P+\angle C F D \\
=2(\angle B A D+\angle C A D) \\
=2 \angle B A C=\angle B O C .
\end{array}
$$

Therefore, points $B, C, P, O$ are concyclic.
By Ptolemy's theorem, we get
$$
B P \cdot O C=P C \cdot O B+P O \cdot B C \text {, }
$$

which simplifies to $(B P-P C) O C=P O \cdot B C$.
Since $B P=P C+P O$, we have
$$
O C=B C \text {. }
$$

Thus, $\triangle O B C$ is an equilateral triangle.
Hence, $\angle B A C=\frac{1}{2} \angle B O C=30^{\circ}$.