4. In an acute triangle $\triangle A B C$, $A B=A C>B C$, point $D$ is on side $A B$, and $A D=B C$. If $\angle B D C=30^{\circ}$, then $\angle A=(\quad)$.
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $15^{\circ}$
(D) $22.5^{\circ}$
4.B.

As shown in Figure 5, construct an equilateral $\triangle A D E$ outside $\triangle A B C$ with $A D$ as a side, and connect $E C$. It is easy to see that
$$
\begin{array}{l} 
\triangle A D C \cong \triangle E D C \\
\Rightarrow \angle A C D=\angle E C D, A C=E C \\
\Rightarrow \triangle A B C \cong \triangle C A E \\
\Rightarrow \angle B A C=\angle A C E \\
=2 \angle A C D .
\end{array}
$$

Also, $\angle B D C=\angle B A C+\angle A C D=30^{\circ}$, so,
$$
\angle A C D=10^{\circ}, \angle B A C=20^{\circ} \text {. }
$$