$20 \cdot 3$ As shown in the figure, $\angle E=40^{\circ}$, and the lengths of arcs $\overparen{A B}, \overparen{B C}$, and $\overparen{C D}$ are all equal. The degree measure of $\angle A C D$ is
(A) $10^{\circ}$.
(B) $15^{\circ}$.
(C) $20^{\circ}$.
(D) $\left(\frac{45}{2}\right)^{\circ}$.
(E) $30^{\circ}$.
(28th American High School Mathematics Examination, 1977)
[Solution] Let $\widehat{A B} \stackrel{\mathrm{m}}{=} x^{\circ}, \widetilde{A D} \stackrel{\mathrm{m}}{=} y^{\circ}$, then from the problem we have
$$
3 x+y=360, \quad \frac{1}{2}(x-y)=40 \text {. }
$$

Solving these equations, we get
$$
\begin{array}{c}
x=110, \quad y=30 . \\
\angle C=15^{\circ} .
\end{array}
$$
$$
\therefore \quad \angle C=15^{\circ} \text {. }
$$

Therefore, the correct choice is $(B)$.