2010 AMC 10A Problem 14

Let $\angle BAE = \angle ACD = x.$ Note that $\angle CFE = 60^{\circ}$ since $\triangle CFE$ is equilateral. We then have that $$\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}.$$ Then: $$\begin{aligned} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{aligned}$$ We then get that $$\begin{aligned} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{aligned}$$ Since $AB = 2 \cdot AC$ and $\angle BAC = 60^{\circ},$ we have that $\triangle ABC$ is a $30-60-90$ triangle. Thus, C is the correct answer.