10. (5 points) As shown in the figure, it is a plane figure after folding a triangular paper piece, with the crease being $DE$. It is known that: $\angle B=74^{\circ}, \angle$ $A=70^{\circ}, \angle CEB=20^{\circ}$. What is the measure of $\angle ADC$?
【Analysis】We can use the sum of the exterior angles of a triangle and the sum of the interior angles of a quadrilateral. $\angle C=180^{\circ}-\angle A-\angle B=180^{\circ}$
$-74^{\circ}-70^{\circ}=36^{\circ}, \angle C O E=180^{\circ}-\angle C-\angle C E O=180^{\circ}-36^{\circ}-20^{\circ}=124^{\circ}$, and
$\because \angle O E D+\angle O D E=\angle C O E=124^{\circ}$, according to the sum of the interior angles of a quadrilateral: $\angle O E D+\angle O D E+\angle C D A+$ $\angle A+\angle B=360^{\circ}$, finally we can find the degree of $\angle A D C$.

【Solution】Solution: According to the analysis, as shown in the figure, $\angle C=180^{\circ}-\angle A-\angle B=180^{\circ}-74^{\circ}-70^{\circ}=36^{\circ}$, $\angle C O E=180^{\circ}-\angle C-\angle C E O=180^{\circ}-36^{\circ}-20^{\circ}=124^{\circ}$,

and $\because \angle O E D+\angle O D E=\angle C O E=124^{\circ}$, according to the sum of the interior angles of a quadrilateral:
$$
\begin{array}{l}
\angle O E D+\angle O D E+\angle C D A+\angle A+\angle B=360^{\circ} \\
\therefore \angle C D A=360^{\circ}-\angle C O E-\angle A-\angle B=360^{\circ}-124^{\circ}-74^{\circ}-70=92^{\circ}
\end{array}
$$