In rectangle $ABCD$, $DC=2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$? $\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad$ $\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad$ $\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad$ $\textbf{(D)}\ \frac{1}{3}\qquad$ $\textbf{(E)}\ \frac{\sqrt{2}}{4}$
Solution
The ratio is unchanged by scaling, so assume $AD=6$, $CD=12$; then $[ABCD]=72$.
$DE$ and $DF$ trisect the right angle $ADC$, so $\angle ADE=30^\circ$ and $\angle ADF=60^\circ$. In the right triangles $DAE$ and $DAF$, $AE=6\tan30^\circ=2\sqrt3$ and $AF=6\tan60^\circ=6\sqrt3$ (both less than $12$, so $E,F$ are on $\overline{AB}$).
So $EF=4\sqrt3$, and since the height from $D$ to $AB$ is $AD=6$, $[DEF]=\frac12\cdot4\sqrt3\cdot6=12\sqrt3$.
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 $ABCD$?
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad$ $\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad$ $\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad$ $\textbf{(D)}\ \frac{1}{3}\qquad$ $\textbf{(E)}\ \frac{\sqrt{2}}{4}$ 
