AHSME 1979 Problem 30

In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals $\textbf{(A) }\frac{1}{4}\cos 10^\circ \textbf{(B) }\frac{\sqrt{3}}{8} \textbf{(C) }\frac{1}{4}\cos 40^\circ \textbf{(D) }\frac{1}{4}\cos 50^\circ \textbf{(E) }\frac{1}{8}$