2004 AMC 10A Problem 20

Let the square have side $1,$ and by symmetry let $ED = DF = x,$ so $AE = 1 - x.$ Since $\triangle BEF$ is equilateral, $EF^2 = EB^2,$ giving $$2x^2 = 1 + (1 - x)^2,$$ which simplifies to $x^2 = 2(1 - x).$ The right triangles have areas $[DEF] = \tfrac12 x^2$ and $[ABE] = \tfrac12(1 - x),$ so $$\begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned}$$ Thus, the correct answer is D.