4. Given in $\triangle A B C$, side lengths $A C=6, B C=9, \angle C=120^{\circ}$, find the length of the angle bisector $C D$ of $\angle C$.
4. Let $C D=x$. By the triangle area formula $S=\frac{1}{2} a b \sin C$, we get:
$$
S_{\triangle A C D}=\frac{\sqrt{3}}{4} \cdot 6 x, S_{\triangle D C B}=\frac{\sqrt{3}}{4} \cdot 9 x, S_{\triangle A B C}=\frac{\sqrt{3}}{4} \times 6 \times 9
$$

Then $\frac{\sqrt{3}}{4} \cdot 6 x+\frac{\sqrt{3}}{4} \cdot 9 x=\frac{\sqrt{3}}{4} \times 54 \Rightarrow x=\frac{18}{5}$