Question

In Fig., $DE||AC$ and $DF||AE$. Prove that $\frac{BF}{FE}=\frac{BE}{EC}$.

Answer 1

In $△ABC,$

$DE\parallel AC$

$\therefore$ By proportionality theorem,

$\frac{BD}{DA}=\frac{BE}{EC}............\left(1\right)$

Similarly,

In $△ABE,$

$DF\parallel AE$

$\therefore$ By proportionality theorem,

$\frac{BD}{DA}=\frac{BF}{FE}............\left(2\right)$

$\therefore$ from $\left(1\right)$ and $\left(2\right)$,

$\frac{BE}{EC}=\frac{BF}{FE}$.

Hence Proved.