imc

Let the length of $AD$ be $x$, so that the length of $AB$ is $2x$ and $\text{[}ABCD\text{]}=2x^2$. Because $ABCD$ is a rectangle, $\angle ADC=90^{\circ }$, and so $\angle ADE=\angle EDF=\angle FDC=30^{\circ }$. Thus $△DAE$ is a $30-60-90$ right triangle; this implies that $\angle DEF=180^{\circ }-60^{\circ }=120^{\circ }$, so $\angle EFD=180^{\circ }-(120^{\circ }+30^{\circ })=30^{\circ }$. Now drop the altitude from $E$ of $△DEF$, forming two $30-60-90$ triangles. Because the length of $AD$ is $x$, from the properties of a $30-60-90$ triangle the length of $AE$ is $\frac{x\sqrt{3}}{3}$ and the length of $DE$ is thus $\frac{2x\sqrt{3}}{3}$. Thus the altitude of $△DEF$ is $\frac{x\sqrt{3}}{3}$, and its base is $2x$, so its area is $\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}$. To finish, $\frac{\text{[}△DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\frac{\sqrt{3}}{6}}$.