1979 AMC 12 Problem 30

Extend $AB$ to $F$ so that $AF=AC$, creating an equilateral triangle

Choose $G$ on $BF$ with $\angle BCG=20^\circ$; compare $\triangle BCG$ with $\triangle DCE$

Extend $AB$ through $B$ to $F$ with $AF=AC=1.$ Then $\triangle ACF$ is equilateral. Choose $G$ on $BF$ so that $\angle BCG=20^\circ.$ The angle conditions give $\triangle FGC\cong\triangle ABC,$ while $\triangle BCG\sim\triangle DCE.$ Since $E$ is the midpoint of $BC,$ the similarity scale is $2,$ so $$[BCG]=4[CDE].$$ Decomposing the equilateral triangle gives $$\frac{\sqrt3}{4} =2[ABC]+4[CDE].$$ Dividing by $2$ yields $[ABC]+2[CDE]=\frac{\sqrt3}{8}.$

Therefore, the correct answer is B.