Question

In this figure $\angle DEF={90}^{\circ }$ seg $EP\perp DE$
Prove that
$\frac{EF}{FP}=\frac{DF}{DE}$

Answer 1

Given $\angle DEF={90}^o$ To prove $\frac{DF}{DE}=\frac{EF}{EP}$

$seqEP\perp DE$

As $segEP\perp DE\Rightarrow \angle EPF={90}^o$

Now in $△DFE$ & $△EFP$

$\angle DEF=\angle EPE\left[Botharea{90}^o\right]$

$\angle DFE=\angle EFB\left[commonangle\right]$

$EF=EF\left[commonside\right]$

$△DFE\cong △EFP\left[ByAASRule\right]$

Now, By congurancy we know

$\frac{DF}{DE}=\frac{EF}{EP}$

Hence proved