Example 2. In Rt $\triangle A B C$, $\angle C=90^{\circ}, \angle A B C$ $=66^{\circ}, \triangle A B C$ is rotated around $C$ to the position of $\triangle A^{\prime} B^{\prime} C^{\prime}$, with vertex $B$ on the hypotenuse $A^{\prime} B^{\prime}$, and $A^{\prime} C$ intersects $A B$ at $D$. Find $\angle B D C$. (1993, Huanggang Region, Hubei Province Junior High School Mathematics Competition)
Solve $\because \triangle A B C$ is a right triangle, $\angle A B C=66^{\circ}$,
$$
\therefore \angle A=90^{\circ}-66^{\circ}=24^{\circ} \text {. }
$$

Also, $\triangle A B C$ is rotated around $C$ to $\triangle A^{\prime} B^{\prime} C^{\prime}$, so
$$
\begin{array}{l}
\angle A C D=\angle A B A^{\prime} \\
=\angle B C B^{\prime} . \\
\because C B=C B^{\prime}, \\
\therefore \angle B^{\prime} \\
=\angle C B B^{\circ}=66^{\circ} .
\end{array}
$$

Thus, $\angle B^{\prime} C B=180^{\circ}-66^{\circ} \times 2=48^{\circ}$.
Therefore, $\angle A C D=48^{\circ}$.
Hence, $\angle B D C=48^{\circ}+24^{\circ}=72^{\circ}$.