Question

In $△PQR,N$ is a point on $PR$ such that $QN\perp PR$. If $PN\times NR={QN}^2$, prove that $\angle PQR={90}^{\circ }$

Answer 1

$PN\times NR=Q{N}^2$

Given: $PN=QN,QN=NR$

In $\Delta PNQ$ and $\Delta QNR$

$\angle PNQ=\angle QNR={90}^o$

$PN=QN,QN=NR$

$\Delta PNQ\sim \Delta QNR$ by S.A.S

$\therefore \angle PQN=\angle QRN$

In $\Delta QNR$,

$\angle QNR+\angle QRN+\angle NQR={180}^o$ by angle sum property

$\Rightarrow {90}^o+\angle PQN+\angle NQR={180}^o$

$\Rightarrow \angle PQN+\angle NQR={180}^o-{90}^o={90}^o$

$\Rightarrow \angle PQR={90}^o$

Hence proved.