Question

In the following figure angle $PQR={90}^{\circ }$ and $\overline{SQ}$ is perpendicular to PR. Show that $Q{R}^2\times P{S}^2=P{Q}^2\times Q{R}^2$

Answer 1

Given that PQR is a right angled triangle and QS is perpendicular to PR. $\Delta PQR$ and $\Delta PSQ$ are similar
$\frac{PQ}{PS}=\frac{PR}{PQ}\Rightarrow P{Q}^2=PR\times PS$
$\Rightarrow PS=\frac{P{Q}^2}{PR}.....\left(1\right)$
In $\Delta PQR$ and $\Delta QSR,\frac{QR}{RS}=\frac{PR}{QR}$
$\Rightarrow Q{R}^2=PR\times RS\Rightarrow RS=\frac{Q{R}^2}{PR}.....\left(2\right)$
and in $\Delta PQR,O{S}^2=PS\times SR.....\left(3\right)$
Substitute (1) and (2) in (3), $Q{S}^2=\frac{P{Q}^2}{PR}\times \frac{Q{R}^2}{PR}$
$Q{S}^2\times P{R}^2=P{Q}^2\times Q{R}^2$. Hence proved