Question

In fig. considering triangle $BEP$ and $CPD$, prove that $BP\times PD=EP\times PC$

Answer 1

To prove:- $BP\times PD=EP\times PC$

Proof:-

In $△BEP$ and $△CDP$

$\angle BEP=\angle CDP\left[\text{each}90°\right]$

$\angle BPE=\angle CPD\left[\text{Vertically opposite angles}\right]$

$△BEP\sim △CDP\left[\text{By AA criteria}\right]$

$\because$ Corresponding sides of similar triangles are proportional.

$\therefore \frac{BP}{CP}=\frac{EP}{DP}$

$\Rightarrow BP\times DP=CP\times EP$

Hence proved that $BP\times PD=EP\times PC$.