3. In Rt $\triangle A B C$, it is known that $\angle C=90^{\circ}, B C$ $=6, C A=3, C D$ is the angle bisector of $\angle C$. Then $C D=$ $\qquad$ .
$3.2 \sqrt{2}$.
Let $CD=t$. Then, by the relationship of the areas of triangles, we have
$$
\begin{array}{l}
\frac{1}{2} \times 3 t \sin 45^{\circ}+\frac{1}{2} \times 6 t \sin 45^{\circ} \\
=S_{\triangle M C B}=\frac{1}{2} \times 3 \times 6 .
\end{array}
$$

Solving this, we get $t=2 \sqrt{2}$.