$20 \cdot 2$ As shown in the figure, chords $AC$ and $BD$ intersect at $E$, and $AB = BC = CD, \angle BEC = 130^{\circ}$. Then the degree measure of $\angle ACD$ is 
(A) $15^{\circ}$.
(B) $30^{\circ}$.
(C) $84^{\circ}$.
(D) $105^{\circ}$.
(China Guangzhou, Wuhan, Fuzhou, etc. Junior High School Mathematics League;
China National Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1984)
[Solution 1] Let $\overparen{A \mathrm{mD}} \stackrel{\mathrm{m}}{=} x^{\circ}$, minor arc $\overparen{A B}=\overparen{B C}=\overparen{C D}$ $\stackrel{m}{=} y^{\circ}$,
By the given conditions $\left\{\begin{array}{l}x+3 y=360, \\ \frac{1}{2}(x+y)=130,\end{array}\right.$ solving these equations yields $x=210$,
Thus, $\angle A C D \stackrel{\mathrm{m}}{=} \frac{1}{2} \widehat{A \mathrm{~m} D} \stackrel{\mathrm{m}}{=} 105^{\circ}$.
Therefore, the answer is (D).
[Solution 2] Let minor arc $\overparen{A B} \stackrel{\mathrm{m}}{=} x^{\circ}$, then $\overparen{A \mathrm{mD}} \stackrel{\mathrm{m}}{=} 360^{\circ}-3 x^{\circ}$.
Also, $\angle B E C=130^{\circ}$, so $130^{\circ}=\frac{1}{2}\left(360^{\circ}-3 x^{\circ}+x^{\circ}\right)$.
Thus, $x=50$, and $\angle A C D=\frac{1}{2}\left(360^{\circ}-3 \cdot 50^{\circ}\right)=105^{\circ}$. Therefore, the answer is $(D)$.