In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$?

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20}$
The area of $\triangle BFD$ is the area of square $ABCE$ subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be $6$.

The ratio of the area of $\triangle BFD$ to the area of $ABCE$ is
\[\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}\]