Question

In fig., DE $\parallel$ BC and CD $\parallel$ EF. Prove that AD²= AB x AF.

Answer 1

In $△$ABC, we have

DE$\parallel$BC

$\Rightarrow$ $\frac{AB}{AD}$ = $\frac{AC}{AE}$ ........ (i)

In $△$ADC, we have

FE $\parallel$ DC

$\Rightarrow$ $\frac{AD}{AF}$ = $\frac{AC}{AE}$ ........ (ii)

From (i) and (ii), we get

$\frac{AB}{AD}$ = $\frac{AD}{AF}$

$A{D}^2$ = AB $\times$ AF