AHSME 1980 Problem 21

In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$, and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$. The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is $\text{(A)} \ \frac 15 \text{(B)} \ \frac 14 \text{(C)} \ \frac 13 \text{(D)}\ \frac{2}{5} \text{(E)}\ \text{none of these}$