4. In $\triangle A B C$, it is known that $A D$ is the angle bisector of $\angle B A C$, and $A D=2$. Then $A B+A C(\quad$.
(A) greater than 4
(B) less than 4
(C) not greater than 4
(D) not less than 4
4. A.

As shown in Figure 1, since $\angle A D C > \angle B$, we can construct $\angle A D E$ $=\angle B$ within $\angle A D C$, where $D E$ intersects $A C$ at point $E$. It is easy to see that
$$
\begin{array}{l} 
\triangle A B D \backsim \triangle A D E \\
\Rightarrow \frac{A B}{A D}=\frac{A D}{A E} \\
\Rightarrow A D^{2}=A B \cdot A E2 A D=4$.
$$