Question

In given figure, If $EF||DC||AB$. prove that $\frac{AE}{ED}=\frac{BF}{FC}$.

Answer 1

Given $EF||DC||AB$ in the given figure.

PROVE $\frac{AE}{ED}=\frac{BF}{FC}$

CONSTRUCTION Produce DA and CB to meet at p (say).

PROOF In $△PEF$, we have
$AB||EF$

$\therefore$ $\frac{PA}{AE}=\frac{PB}{BF}$ [By Thale's Theorem]

$\Rightarrow$ $\frac{PA}{AE}+1=\frac{PB}{BF}+1$ [Adding 1 on both sides]

$\Rightarrow$ $\frac{PA+AE}{AE}=\frac{PB=BF}{BF}$

$\Rightarrow$ $\frac{PE}{AE}=\frac{PF}{BF}$............(i)

In $△PDC$, we have

$EF||DC$

$\therefore$ $\frac{PE}{ED}=\frac{PF}{FC}$ [By Basic Proportionality Theorem]........(ii)

On dividing equation (i) by equation (ii),, we get

$\frac{\frac{PE}{AE}}{\frac{PE}{ED}}=\frac{\frac{PF}{BF}}{\frac{PF}{FC}}$

$\Rightarrow$ $\frac{ED}{AE}=\frac{FC}{BF}$

$\Rightarrow$ $\frac{AE}{ED}=\frac{BF}{FC}$