Example 1 As shown in Figure 1, in quadrilateral $A B C D$, $A C$ and $B D$ are diagonals, $\triangle A B C$ is an equilateral triangle, $\angle A D C=30^{\circ}, A D=3, B D=$ 5. Then the length of $C D$ is ( ). ${ }^{[1]}$
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5
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.