smc

Let $F$ be the point on the extension of side $AB$ past $B$ for which $AF=1$. Since $AF=AC$ and $\measuredangle FAC=60^{\circ }$,$△ACF$ is equilateral. Let $G$ be the point on line segment $BF$ for which $\measuredangle BCG=20^{\circ }$. Then $△BCG$ is similar to $△DCE$ and $BC=2(EC)$. Also $△FGC$ is congruent to $△ABC$. Therefore, $[△ACF]=([△ABC]+[△GCF])+[△BCG]$. Plugging in the values that we know and then dividing by 2 results in an answer of $\boxed{\frac{\sqrt{3}}{8}}$ This solution is from the solution manual but was typed here by alpha_2.