【Question 10】As shown in the figure, in $\triangle A B C$, it is known that $A B=A C, A E \perp B C, C D=C A, A D=B D$, then $\angle D A E=$ $\qquad$ degrees.
【Analysis and Solution】
Let $\angle B=x^{\circ}$;
Since $A B=A C$; therefore $\angle B=\angle C=x^{\circ}$;
Since $A D=B D$; therefore $\angle B=\angle B A D=x^{\circ}$;
Thus, $\angle A D C=\angle B+\angle B A D=2 x^{\circ}$;
Since $C D=C A$; therefore $\angle A D C=\angle D A C=2 x^{\circ}$;
Thus, $\angle B A C=\angle B A D+\angle D A C=3 x^{\circ}$;
The sum of the interior angles of a triangle is $180^{\circ}$, so $\angle B A C+\angle B+\angle C=5 x^{\circ}=180^{\circ}$;
Solving for $x$ gives $x=36$;
$$
\angle A D C=2 x^{\circ}=72^{\circ} \text {; }
$$

Since $A E \perp B C$; therefore $\angle A D C+\angle D A E=90^{\circ}$;
Thus, $\angle D A E=90^{\circ}-\angle A D C=90^{\circ}-72^{\circ}=18^{\circ}$.