Example 2 As shown in Figure 2, in $\triangle A B C$, $\angle A B C$ $=46^{\circ}, D$ is a point on side $B C$, $D C=A B$, 
$\angle D A B=21^{\circ}$. Try to find 
the degree measure of $\angle C A D$.
Solve: Fold $\triangle A B D$ along the line $A D$ to get $\triangle A E D$. Then
$$
\begin{array}{l}
\angle E A D=21^{\circ}, A E=A B, D E=B D . \\
\text { It is easy to see that, } \angle A D C=21^{\circ}+46^{\circ}=67^{\circ} \text {. } \\
\text { Therefore, } \angle A D E=\angle A D B \\
=180^{\circ}-67^{\circ}=113^{\circ} \text {, } \\
\angle C D E=113^{\circ}-67^{\circ}=46^{\circ} .
\end{array}
$$

Connect $C E$.
Since $D C=A B$, we have,
$$
\triangle C D E \cong \triangle A B D \cong \triangle A E D .
$$

Thus, quadrilateral $A D E C$ is an isosceles trapezoid.
Therefore, $\angle A C O=46^{\circ}=\angle O A C$.
Hence, $\angle C A D=180^{\circ}-\angle A D E=67^{\circ}$.