jmc

First, assume that $AB=1$, and let $ED=DF=x$. Then, we have $[DEF]=\frac{x^2}{2}$ and $[ABE]=\frac{(AE)(AB)}{2}=\frac{(1-x)(1)}{2}$, so $$\frac{[DEF]}{[ABE]}=\frac{x^2}{1-x}.$$ By the Pythagorean Theorem applied to $△DEF$, we have $$E{F}^2=D{E}^2+D{F}^2=2x^2.$$ Applying the Pythagorean Theorem to $△AEB$, we have $$E{B}^2=A{B}^2+A{E}^2=1+(1-x{)}^2=2-2x+x^2.$$ Since $△EFB$ is equilateral, we have $EF=EB$, so $$2x^2=2-2x+x^2,$$ or $x^2=2-2x=2(1-x)$. Hence the desired ratio of the areas is $$\frac{[DEF]}{[ABE]}=\frac{x^2}{1-x}=\boxed{2}.$$