[Sol] As shown in the figure, by the exterior angle property of a triangle, we have
\[
\begin{aligned}
\angle x=\angle A D C- & \angle A D E=\angle A D C-\angle A E D \\
= & \angle A D C-(\angle x+\angle C),
\end{aligned}
\]

Therefore, \(\angle x=\frac{1}{2}(\angle A D C-\angle C)\)
\[
\begin{array}{l}
=\frac{1}{2}[(\angle B+\angle B A D)-\angle C](\because \angle B=\angle C) \\
=\frac{1}{2} \angle B A D=15^{\circ} .
\end{array}
\]

Thus, the answer is \((D)\).
17・14 As shown in the figure, $A B=A C, \angle B A D=30^{\circ}$, and $A E=A D$, then $x$ equals
(A) $7.5^{\circ}$.
(B) $10^{\circ}$.
(C) $12.5^{\circ}$.
(D) $15^{\circ}$.
(E) $20^{\circ}$.
(7th American High School Mathematics Examination, 1956)