Question

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

Answer 1

Given that, In $\Delta ABC$ $DE\parallel BC$ and $CD\parallel EF$

$In\Delta ABC,DE\parallel BC$

By basic proportionally theorem (BPT)

$\frac{AD}{DB}=\frac{AE}{EC}........\left(1\right)$

$In\Delta ABC,EF\parallel CD$

By basic proportionally theorem (BPT)

$\frac{AF}{FD}=\frac{AE}{EC}........\left(2\right)$

By equation (1) and (2) to we get,

$\frac{AD}{DB}=\frac{AF}{FD}........\left(3\right)$

Taking reciprocal and we get,

$\frac{DB}{AD}=\frac{FD}{AF}$

Adding $1$ both side and we get,

$\frac{DB}{AD}+1=\frac{FD}{AF}+1$

$\frac{DB+AD}{AD}=\frac{FD+AF}{AF}$

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

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

Hence proved.