In triangle $\triangle ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line  $\overline{BC}$ and line $\overline{AE}$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?


$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
We use the line-segment ratios to infer area ratios and height ratios.
Areas:
$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$.
$BE:BD = 1:2 \text{ (midpoint)}  \implies  [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$.
Heights:
Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$.
$AD:DC = 1:2  \implies CD:CA = 2:3  \implies  \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$.

$BE:BD = 1:2 \text{ (midpoint)}  \implies  \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$. 

Conclusion:
$\frac{[EBF]} {[ABF]} =  \frac{[EBF]} {[EBF] + [ABE]} =   \frac{[EBF]} {[EBF]+60}$,  and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$.

So, $\frac{[EBF]} {[EBF] + 60} =  \frac{1}{3}$, and thus, $[EBF] = \boxed{30}$.