2010 AMC 12A Problem 8

Let $\angle BAE=\angle ACD=x.$ Because $\triangle CFE$ is equilateral, $\angle CFE=60^\circ.$ Rays $FA$ and $FE$ are opposite, so $\angle AFC=120^\circ.$ In $\triangle AFC,$ we get $\angle FAC=180^\circ-120^\circ-x$ $=60^\circ-x.$ Since $F$ lies on $AE,$ this is $\angle EAC.$ Hence $\angle BAC=x+(60^\circ-x)=60^\circ.$ Put $AC=s,$ so $AB=2s.$ The Law of Cosines gives $$\begin{aligned} BC^2 &= s^2+(2s)^2 \\ &\quad-2(s)(2s)\cos60^\circ \\ &=3s^2. \end{aligned}$$ Thus the side lengths are in the ratio $1:\sqrt3:2,$ with $AB$ the hypotenuse, so $\angle ACB=90^\circ.$ Thus, C is the correct answer.