3. In $\triangle A B C$, $A B=A C, A D, B E$ are the angle bisectors of $\angle A, \angle B$ respectively, and $B E=2 A D$. Then the degree measure of $\angle B A C$ is $\qquad$ .
3. $108^{\circ}$.

As shown in Figure 5, extend $A D$ to point $P$ such that
$$
A P=2 A D=B E \text {. }
$$

It is easy to see that $A C / / B P, \angle A E B=\angle E B P$,
$$
\angle E A P=\angle A P B \text {. }
$$

Draw a line through point $E$ parallel to $A P$, intersecting the extension of $B P$ at point $N$.
Thus, quadrilateral $A E N P$ is a parallelogram, and $E N=A P=B E$.
Therefore, $\angle A E B=\angle E B N=\angle E N B=\angle E A P$.
Let $\angle E B C=\alpha$. Then
$$
\begin{array}{l}
\angle A C B=\angle A B C=2 \alpha, \\
\angle C A D=\angle A E B=\angle E B C+\angle E C B \\
=\alpha+2 \alpha=3 \alpha .
\end{array}
$$

In the right triangle $\triangle A D C$,
$$
\begin{array}{l}
5 \alpha=\angle C A D+\angle D C A=90^{\circ} \Rightarrow \alpha=18^{\circ} \\
\Rightarrow \angle C A D=3 \alpha=54^{\circ} \\
\Rightarrow \angle B A C=2 \angle C A D=2 \times 54^{\circ}=108^{\circ} .
\end{array}
$$