In rectangle $ABCD$, $DC=4\cdot CB$. Points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$?

1. If $AD=h$ and $DC=kh$, then the trisecting rays give $AE=\frac{h}{\sqrt3}$ and $AF=h\sqrt3$. Thus $EF=\frac{2h\sqrt3}{3}$, $[DEF]=\frac{h^2\sqrt3}{3}$, and $[ABCD]=kh^2$.
2. With $k=4$, the ratio $\dfrac{[DEF]}{[ABCD]}=\boxed{\dfrac{\sqrt3}{12}}$.