Four, (10 points) As shown in Figure 5, in $\triangle A B C$, $\angle A B C=46^{\circ}, D$ is a point on side $B C$, $D C=$ $A B, \angle D A B=21^{\circ}$. Determine the degree measure of $\angle C A D$.


As shown in Figure 8, construct the axisymmetric figure $\triangle A E D$ of $\triangle A B D$ with respect to $A D$, then
$$
\begin{array}{l}
\angle E A D=21^{\circ}, \\
A E=A B .
\end{array}
$$

Therefore, $D E=B D$.
It is easy to see that
$$
\angle A D C=21^{\circ}+46^{\circ}=67^{\circ} \text {. }
$$

Thus, $\angle A D E=\angle A D B=180^{\circ}-67^{\circ}=113^{\circ}$,
$$
\angle C D E=113^{\circ}-67^{\circ}=46^{\circ} \text {. }
$$

Connect $C E$.
Since $D C=A B$, we have
$\triangle C D E \cong \triangle A B D \cong \triangle A E D$.
Let $O$ be the intersection of $A E$ and $D C$.
Since $\angle O D E=\angle O E D=46^{\circ}$, thus,
$$
O D=O E \text {. }
$$

Also, $D C=A E$, so
$$
\begin{array}{l}
A O=C O \Rightarrow \angle O C A=\angle O A C \\
\Rightarrow \angle C O E=2 \angle A C O .
\end{array}
$$

It is easy to see that $\angle C O E=2 \times 46^{\circ}=92^{\circ}$.
Therefore, $2 \angle A C O=\angle C O E=92^{\circ}$.
Thus, $\angle A C O=46^{\circ}=\angle O A C$.
Hence, $\angle D A C=\angle D A E+\angle E A C=67^{\circ}$.