Question

Question 11
In a $\Delta$PQR, N is a point on PR such that $QN\perp PR$. If $PN.NR=Q{N}^2$, then prove that $\angle PQR={90}^{\circ }$.

Answer 1

Given $\Delta PQR$ N is a point on PR, such that $QN\perp PR$
and $PN.NR=Q{N}^2$

To prove that $\angle PQR={90}^{\circ }$
$\Rightarrow$ PN.NR = QN.QN
$\Rightarrow \frac{PN}{QN}=\frac{QN}{NR}$ …..(i)

In $\Delta QNPand\Delta RNQ,\frac{PN}{QN}=\frac{QN}{NR}$
and $\angle PNQ=\angle RNQ$ [each equal to ${90}^{\circ }$]
$\therefore$ $\Delta QNP\sim \Delta RNQ$ [by SAS similarity criterion]
Then, $\Delta$QNP and $\Delta$ RNQ are equiangulars.
i.e., $\angle PQN=\angle QRN$
$\angle RQN=\angle QPN$
On adding both sides, we get
$\angle PQN+\angle RQN=\angle QRN+\angle QPN$
$\Rightarrow \angle PQR=\angle QRN+\angle QPN$ ….(ii)
We know that, sum of angles of a triangle = ${180}^{\circ }$

In $\Delta PQR,\angle PQR+\angle QPR+\angle QRP={180}^{\circ }$
$\Rightarrow \angle PQR+\angle QPN+\angle QRN={180}^{\circ }$ $[\because \angle QPR=\angle QPNand\angle QRP=\angle QRN]$
$\Rightarrow \angle PQR+\angle PQR={180}^{\circ }$
$\Rightarrow 2\angle PQR={180}^{\circ }$
$\Rightarrow \angle PQR=\frac{{180}^2}{2}={90}^{\circ }$
$\therefore$ $\angle PQR={90}^{\circ }$