Problem 397

Solve As shown in Figure 2, rotate $C D$ clockwise around point $C$ by $60^{\circ}$ to get $C E$, and connect $D E$ and $A E$. Then $\triangle C D E$ is an equilateral triangle.
Since $A C=B C$,
$$\begin{aligned} & \angle B C D \\ = & \angle B C A+\angle A C D \\ = & \angle D C E+\angle A C D=\angle A C E, \end{aligned}$$

Therefore, $\triangle B C D \cong \triangle A C E \Rightarrow B D=A E$. Also, $\angle A D C=30^{\circ}$, so $\angle A D E=90^{\circ}$. In the right triangle $\triangle A D E$, given $A E=5, A D=3$, we get $D E=\sqrt{A E^{2}-A D^{2}}=4 \Rightarrow C D=D E=4$. Hence, the answer is B.