(A) In a PQR, N is a point on PR, such that QN ⊥ PR. If PN × NR = QN^2, prove that ∠ PQR =90^°.

\(\displaystyle \mathrm{PN} \times \mathrm{NR}=\mathrm{QN}^{2}\)

\(\displaystyle \frac{\mathrm{PN}}{\mathrm{QN}}=\frac{\mathrm{QN}}{\mathrm{NR}}\)

\[\begin{aligned} & \angle \mathrm{PNQ}=\angle \mathrm{QNR} \\ & \Delta \mathrm{PNQ} \sim \Delta \mathrm{QNR} \\ & \Rightarrow \angle 2=\angle \mathrm{P} \text { and } \angle 1=\angle \mathrm{R} \\ & \Rightarrow \angle 1+\angle 2=\angle \mathrm{P}+\angle \mathrm{R} \\ & \Rightarrow \angle \mathrm{PQR}=\angle \mathrm{P}+\angle \mathrm{R} \\ & \text { In } \triangle \mathrm{PQR}, \angle \mathrm{P}+\angle \mathrm{PQR}+\angle \mathrm{R}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{PQR}=180^{\circ} \Rightarrow \angle \mathrm{PQR}=90^{\circ} \end{aligned} \]

Draw \(\displaystyle \mathrm{AL} \perp \mathrm{BC}\) and \(\displaystyle \mathrm{DM} \perp \mathrm{BC}\)

In \(\displaystyle \triangle A O L\) and \(\displaystyle \triangle D O M\),

\[\begin{aligned} & \angle A O L=\angle D O M \\ & \angle A L O=\angle D M O \end{aligned} \]

\[\begin{aligned} \frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DBC})} & =\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AL}}{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{DM}} \\ & =\frac{A L}{D M}=\frac{A O}{D O}[\text { using (i) }] \end{aligned} \]