Example 2 As shown in Figure 2, in $\triangle A B C$, it is known that $\angle A B C=50^{\circ}$. Rotate $\triangle A B C$ counterclockwise around point $B$ to $\triangle A^{\prime} B C^{\prime}$, such that $A A^{\prime} / / B C$. Find the size of $\angle C B C^{\prime}$ at this time.
【Analysis】Obviously, $\angle C B C^{\prime}=\angle A B A^{\prime}$.
Furthermore, we find that $\triangle A B A^{\prime}$ is an isosceles triangle, thus it is easy to find $\angle A B A^{\prime}$, which can be done using $\angle 2=\angle 1=50^{\circ}$.
In fact, by rotation we know
$$
\begin{array}{l}
B A=B A^{\prime} \\
\Rightarrow \angle 2=\angle 1 .
\end{array}
$$

By $A A^{\prime} / / B C$
$$
\Rightarrow \angle 1=\angle A B C=50^{\circ} \text {. }
$$

Therefore, $\angle A B A^{\prime}=180^{\circ}-(\angle 2+\angle 1)=80^{\circ}$.
Thus, $\angle C B C^{\prime}=\angle A B A^{\prime}=80^{\circ}$.