Question

In given figure, $DE||BC$ and $CD||EF$. Prove that ${AD}^2=AB\times AF$

Answer 1

In $△ABC$, we have

$DE||BC$

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

In $△ADC$, we have

$FE||DC$

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

From (i) and (ii), we get

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

$\Rightarrow$ ${AD}^2=AB\times AF$ $\left[Henceproved\right]$