9. As shown in Figure 4, in $\triangle A B C$, $C D$ is the altitude, and $C E$ is the angle bisector of $\angle A C B$. If $A C$ $=15, B C=20, C D$ $=12$, then the length of $C E$ is
9. $\frac{60 \sqrt{2}}{7}$.

By the Pythagorean theorem, we know $A D=9, B D=16$.
Therefore, $A B=A D+B D=25$.
Hence, by the converse of the Pythagorean theorem, $\triangle A C B$ is a right triangle, and $\angle A C B=90^{\circ}$.
As shown in Figure 4, draw $E F \perp B C$, with the foot of the perpendicular at $F$. Let $E F=x$.

Since $\angle E C F=\frac{1}{2} \angle A C B=45^{\circ}$, we have $C F=x$.
Thus, $B F=20-x$.
Because $E F \parallel A C$, we have $\frac{E F}{A C}=\frac{B F}{B C}$, which means $\frac{x}{15}=\frac{20-x}{20} \Rightarrow x=\frac{60}{7}$.
Therefore, $C E=\sqrt{2} x=\frac{60 \sqrt{2}}{7}$.