3. Given Rt $\triangle A B C$ with the two legs $A C=5$, $B C=12, D$ is a point on $B C$. When $A D$ is the angle bisector of $\angle A$, then $C D=$ $\qquad$
3. $\frac{10}{3}$.

As shown in Figure 5, construct Rt $\triangle ABC$,
on side $AB$, mark $AE = AC$, draw $ED \perp AB$, intersecting $BC$ at point $D$. Then $AD$ is the angle bisector of $\angle A$.

Let $CD = x$. By $BE = 13 - 5 = 8$, $DE = x$, $BD = 12 - x$ and the Pythagorean theorem, we get $x^2 + 8^2 = (12 - x)^2$. Solving for $x$ yields $x = \frac{10}{3}$.