Question

In figure $DE\parallel BC$ and $CD\parallel EF.$ Prove that $A{D}^2=AB\times AF.$

Answer 1

$\Delta AEF$ & $\Delta ACD$ are similar triangle. Hence their sides are in proportion.

$\frac{AE}{AF}=\frac{AC}{AD}\Rightarrow AF=\frac{AE.AD}{AC}$ __(1)

$\Delta ADE$ & $\Delta ABC$ are also similar

$\frac{AD}{AB}=\frac{DE}{BC}\Rightarrow AD=\frac{DE.AB}{BC}$ __(2)

$AD\times AD=\frac{AF\times AC}{AE}\times \frac{DE\times AB}{BC}$

$A{D}^2=(AF\times AB)\times \frac{AC.DE}{AE.BC}$

But, $\frac{AC}{BC}=\frac{AE}{DE}$ (similarity proportional)

$\therefore A{D}^2=AF\times AB\times 1$

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