In $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the angle bisector of $\angle C$, and $C A=3$, $C B=4$. Find the length of $C D$.
Three, Proof As shown in the figure, it is obvious that $AB=5$.
$\because CD$ bisects
$\angle C$,
$$
\therefore \frac{AD}{BD}=\frac{3}{4} \text{. }
$$

Then $AD=\frac{15}{7}$,
$$
BD=\frac{20}{7} \text{. }
$$

Draw $DE \perp BC$ at $E$, thus $DE \parallel AC$,
$$
\therefore \frac{BE}{CE}=\frac{BD}{AD}=\frac{4}{3} \text{. }
$$

Then $CE=\frac{12}{7}$.
In $\mathrm{Rt} \triangle CDE$, from $\angle DCE=45^{\circ}$ we get
$$
CD=\frac{12}{7} \sqrt{2} \text{. }
$$