4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$
4.9.

From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$.
By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin 60^{\circ} \\
\Rightarrow \frac{1}{a}+\frac{1}{c}=1 .
\end{array}
$$

Therefore, $4 a+c=(4 a+c)\left(\frac{1}{a}+\frac{1}{c}\right)$
$$
=5+\frac{c}{a}+\frac{4 a}{c} \geqslant 5+2 \sqrt{\frac{c}{a} \cdot \frac{4 a}{c}}=9 \text {. }
$$

Equality holds if and only if $c=2 a=3$. Therefore, the minimum value of $4 a+c$ is 9.