AoPS Wiki talk:Problem of the Day/November 5, 2011

We see that $AC=AB=FC=FD$ , and $\angle ACB=\angle ABC=\angle FCD=\angle FDC=30$ . Thus, $\angle BCF=60$ and $\angle ACE$ is right, which means that $\triangle ACE$ is a 45-45-90 triangle.

So $AC=AB=FC=FD=10\sqrt{2}$ , and by the Law of Cosines on $\triangle ABC$ , $\begin{align*}[BCDE]=BC^2=AB^2+AC^2-2(AB)(AC)\cos\theta=(10\sqrt{2})^2+(10\sqrt{2})^2-2(10\sqrt{2})(10\sqrt{2})\cos 120=2(200)+200=\boxed{600}\end{align*}$

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AoPS Wiki talk:Problem of the Day/November 5, 2011