Find the ratio $\frac{AF}{FC}$

Question

In the figure below, $AD$ is the median on $BC$. The point $E$ divides $A$ and $D$ in the ratio $1:2$. $BE$ produced meets $AC$ at $F$. Find the value of $AF:FC$

My try: I joined $E,C$.Let area of $\Delta ABC=x$
Then we get $$ar(BED):ar(ABE)=2:1$$
Also
$$ar(ABD)=\frac{x}{2}$$
$\implies$
$$ar(BED)=\frac{2x}{6}$$
$$ar(AEB)=\frac{x}{6}$$
So $$ar(ECD)=\frac{2x}{6}$$
$\implies$
$$ar(AEC)=\frac{x}{6}$$
Any way from here?

Answer 1

Let $G$ halves $FC$, then $FG= GC = 2b$ and $DG||BF$ (by Thales theorem).

Then, again by Thales: $AF: FG = AE: ED = 1:2$ so $AF = b$ and thus $AF:FC = 1:4$