$17 \cdot 143$ In $\triangle A B C$, $E$ is the midpoint of side $B C$, and $D$ is on side $A C$. If the length of $A C$ is $1, \angle B A C=60^{\circ}, \angle A B C=100^{\circ}, \angle A C B=20^{\circ}, \angle D E C=80^{\circ}$, then the area of $\triangle A B C$ plus twice the area of $\triangle C D E$ equals
(A) $\frac{1}{4} \cos 10^{\circ}$.
(B) $\frac{\sqrt{3}}{8}$.
(C) $\frac{1}{4} \cos 40^{\circ}$.
(D) $\frac{1}{4} \cos 50^{\circ}$.
(E) $\frac{1}{8}$.
[Solution] Let $F$ be a point on the extension of side $AB$ through $B$, and $AF=1$. Since $AF=AC$ and $\angle FAC=60^{\circ}$, $\triangle ACF$ is an equilateral triangle. Let $G$ be a point on segment $BF$ such that $\angle BCG=20^{\circ}$. Then $\triangle BCG \sim \triangle DCE$, and $BC=2EC$. Also, $\triangle FGC \cong \triangle ABC$,
so $S_{\triangle ACF}=S_{\triangle ABC}+S_{\triangle GCF}+S_{\triangle BCG}$,
$$
\frac{\sqrt{3}}{4}=2 S_{\triangle ABC}+4 S_{\triangle CDE}
$$

Therefore, $S_{\triangle ABC}+2 S_{\triangle CDE}=\frac{\sqrt{3}}{8}$.
Hence, the answer is $(B)$.