Question

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

Answer 1

We have

$EF||AB||DC$ and $EP||DC$

Thus, in $△ADC$, We have

$EP||DC$

Therefore, by basic proportionality theorem, we have

$\frac{AE}{ED}=\frac{AP}{PC}$.............(i)

Again, $EF||AB||DC$

$\Rightarrow$ $FP||AB$

Thus, in $△CAB$, we have

$FP||BA$

Therefore, by basic proportionality theorem, we have

$\frac{BF}{FC}=\frac{AP}{PC}$

From (i) and (ii), we have

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