Show that $Area(\triangle BFE)=2Area(\triangle FED)$.

Question

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the midpoint of $BC$. If $AE$ intersects $BC$ at $F$, show that $Area(\triangle BFE)=2Area(\triangle FED)$

Solution given :

I didn't understand in the solution that $FB=2FD$ .

Any hint is appreciated.

Answer 1

You have $\triangle ABF \sim \triangle EDF \Rightarrow \dfrac{AB}{ED} = \dfrac{BF}{DF} = \dfrac{FA}{FE}$, but $AB = 2BD=2ED \Rightarrow BF = 2FD$

Answer 2

Let M be the midpoint of BD. Do you see EM//AD and AD = 2EM? Then it is easy to see FD = 2MF. If you come through to this point then you can deduce that BF = 2FD. Because BF = BM + MF and MD = MF + FD and BM = MD.