Question

$D$ is a point on the side $QR$ of triangle $PQR$ such that angles $PDR$ and $QPR$ are equal. Prove that $QR.DR={PR}^2.$

Answer 1

Given,

$\angle PDR=\angle QPR$

To prove :

$QR.DR={PR}^2$

Proof :

Consider the triangle $\Delta PQR$ and $\Delta PDR$

$\angle PDR=\angle QPR$ (given)

$\angle PRD=\angle PRQ$ (given)

$\therefore$ $\Delta PQR\sim \Delta PDR$ by $AA$ congruence criterion.

$\therefore$ the corresponding sides will be in proportion

$\Rightarrow \frac{QR}{PR}=\frac{PR}{DR}$

$\Rightarrow QR.DR={PR}^2$

Hence proved.