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

(N/A) Given: In $\Delta PQR$,$N$ is a point on $PR$ such that $QN \perp PR$ and $PN \cdot NR = QN^2$.
To prove: $\angle PQR = 90^{\circ}$.
Proof: We have $PN \cdot NR = QN^2$.
This can be written as $\frac{PN}{QN} = \frac{QN}{NR}$.
In $\Delta QNP$ and $\Delta RNQ$:
$1$. $\frac{PN}{QN} = \frac{QN}{NR}$ (Given)
$2$. $\angle PNQ = \angle RNQ = 90^{\circ}$ (Given $QN \perp PR$)
By $SAS$ similarity criterion,$\Delta QNP \sim \Delta RNQ$.
Since the triangles are similar,their corresponding angles are equal:
$\angle PQN = \angle QRN$ (let this be $\alpha$)
$\angle RQN = \angle QPN$ (let this be $\beta$)
In $\Delta PQR$,the sum of angles is $180^{\circ}$:
$\angle P + \angle R + \angle PQR = 180^{\circ}$
$\angle QPN + \angle QRN + (\angle PQN + \angle RQN) = 180^{\circ}$
Substituting the equal angles:
$\beta + \alpha + (\alpha + \beta) = 180^{\circ}$
$2(\alpha + \beta) = 180^{\circ}$
$\alpha + \beta = 90^{\circ}$
Since $\angle PQR = \alpha + \beta$,we have $\angle PQR = 90^{\circ}$.
Hence proved.