4. (10 points) As shown in the figure, fold a triangular paper piece $A B C$ so that point $C$ lands on the plane of triangle $A B C$, with the fold line being $D E$. Given $\angle A B E=74^{\circ}, \angle D A B=70^{\circ}, \angle C E B=20^{\circ}$, then $\angle C D A$ equals $\qquad$ .
【Analysis】Before folding, the degree of $\angle C$ can be obtained by using the sum of the interior angles of a triangle. After folding, the degree of $\angle CDA$ can be obtained by using the sum of the exterior angles of a triangle and the sum of the interior angles of a quadrilateral.

【Solution】According to the analysis, before folding, by the sum of the interior angles of a triangle, $\angle C=180^{\circ}-74^{\circ}-70^{\circ}=36^{\circ}$, after folding,
$$
\begin{array}{l}
\angle E O D=\angle C+\angle C E O=36^{\circ}+20^{\circ}=56^{\circ} ; \\
\angle B O D=180^{\circ}-\angle D O E=180^{\circ}-56^{\circ}=124^{\circ}, \\
\angle C D A=360^{\circ}-\angle A B E-\angle B A E-\angle B O D=360^{\circ}-70^{\circ}-74^{\circ}-124^{\circ}=92^{\circ} .
\end{array}
$$

Therefore, the answer is: $92^{\circ}$.