Example 3 As shown in Figure 3, in $\triangle A B C$, $\angle B=46^{\circ}$, point $D$ is on side $B C$, and satisfies $\angle B A D=21^{\circ}$. If $A B$ $=C D$, find the degree measure of $\angle C A D$. ${ }^{[3]}$
Solve: $\triangle D E C \cong \triangle B D A$, as shown in Figure 3.
Then $\angle E D C=\angle B=46^{\circ}$,
$E C=A D, D E=B D$.
Thus, $\angle A D E=\angle A D C+\angle E D C$
$=113^{\circ}=\angle A D B$.
Also, $A D=A D$, hence, $\triangle B D A \cong \triangle E D A$.
Therefore, $\triangle D E C \cong \triangle E D A$.
Then $\angle D A E=\angle D C E$.
Thus, points $A, D, E, C$ are concyclic.
Hence $\angle D A C=180^{\circ}-\angle D E C$
$=180^{\circ}-\angle A D B=67^{\circ}$.