$17 \cdot 4$ As shown in the figure, outside the equilateral triangle $A B C$, $A D=$ $A B$ is constructed. Connect $D C, B D$. Then $\angle C D B$ equals
(A) $60^{\circ}$.
(B) $45^{\circ}$.
(C) $30^{\circ}$.
(D) Cannot be determined.
[Solution] Given $A B=A C=A D$, then points $B, C, D$ are all on the circle with $A$ as the center and $A B$ as the radius.
Therefore, $\angle C D B \stackrel{m}{=} \frac{1}{2} \overparen{B C}$,
and $\overparen{B C} \stackrel{\mathrm{m}}{=} \angle A B C=60^{\circ}$,

Thus, $\angle C D B=\frac{1}{2} \angle A B C=30^{\circ}$.
Hence, the answer is $(C)$.