17・12 In $\triangle A B C$, $A C=C D$, and $\angle C A B-\angle A B C=30^{\circ}$, then the degree measure of $\angle B A D$ is
(A) $30^{\circ}$.
(B) $20^{\circ}$.
(C) $22.5^{\circ}$.
(D) $10^{\circ}$.
(E) $15^{\circ}$.
(8th American High School Mathematics Examination, 1957)
［Solution］By the exterior angle property of a triangle and the given conditions, we have
$$
\begin{aligned}
\angle B A D & =\angle C D A-\angle B=\angle C A D-\angle B \\
& =(\angle C A B-\angle B A D)-\angle B .
\end{aligned}
$$

Therefore, $2 \angle B A D=\angle C A B-\angle B=30^{\circ}$, which means $\angle B A D=15^{\circ}$. Hence, the answer is $(E)$.