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In a $\Delta PQR$, N is a point on PR, such that $QN \perp PR$. If $PN \times NR = QN^2$, prove that $\angle PQR = 90^\circ$.
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$$\begin{aligned}& PN \times NR = QN^2 \\ & frac{PN}{QN} = \frac{QN}{NR} \\ & \angle PNQ = \angle QNR\end{aligned}$$ (each $90^\circ$)
$\Delta PNQ \sim \Delta QNR$ (SAS similarity)
$\Rightarrow \angle 2 = \angle P$ and $$\begin{aligned}& \angle 1 = \angle R \\ & Rightarrow \angle 1 + \angle 2 = \angle P + \angle R \\ & Rightarrow \angle PQR = \angle P + \angle R\end{aligned}$$In $\Delta PQR$, $$\begin{aligned}& \angle P + \angle PQR + \angle R = 180^\circ \\ & Rightarrow 2 \angle PQR = 180^\circ \Rightarrow \angle PQR = 90^\circ\end{aligned}$$_a_1.png)
$\Delta PNQ \sim \Delta QNR$ (SAS similarity)
$\Rightarrow \angle 2 = \angle P$ and $$\begin{aligned}& \angle 1 = \angle R \\ & Rightarrow \angle 1 + \angle 2 = \angle P + \angle R \\ & Rightarrow \angle PQR = \angle P + \angle R\end{aligned}$$In $\Delta PQR$, $$\begin{aligned}& \angle P + \angle PQR + \angle R = 180^\circ \\ & Rightarrow 2 \angle PQR = 180^\circ \Rightarrow \angle PQR = 90^\circ\end{aligned}$$