Question

Given: $PQRS$is a rectangle.

$\angle A$ is a right angle.

Prove: $BS·RC=PS·QR={\left(PS\right)}^2$

Short answer

The result is,

$BS·RC=PS·QR=P{S}^2$.

Step-by-step solution

Step 1. Given information.

The triangle is shown below:

$PQRS$is rectangle.

Step 2. Consider triangles BSP and PAQ.

$\begin{array}{l}\angle BSP=\angle PAQ=90°\\ \angle PBS=\angle QPA\left(\text{Correspondingangle}\right)\\ \therefore \Delta BSP~\Delta PAQ\mathrm{......}\left(1\right)\end{array}$

Step 3. Consider triangles PAQ and QRC.

$\begin{array}{l}\angle QRC=\angle PAQ=90°\\ \angle PQA=\angle QCR\left(\text{Correspondingangle}\right)\\ \therefore \Delta QRC~\Delta PAQ\mathrm{......}\left(2\right)\end{array}$

From equation $\left(1\right)$ and $\left(2\right)$,

$\Delta QRC~\Delta BSP$

Hence,

$\begin{array}{l}\frac{BS}{QR}=\frac{PS}{RC}\\ BS\cdot RC=PS\cdot QR\mathrm{......}\left(3\right)\end{array}$

Also, $PQRS$ is rectangle.

Hence, side $PS$ is equal to $QR$.

$\begin{array}{c}BS\cdot RC=PS\cdot PS\\ BS\cdot RC=P{S}^2\mathrm{......}\left(4\right)\end{array}$

Thus, from equations $\left(3\right)$ and $\left(4\right)$,

$BS·RC=PS·QR=P{S}^2$

Therefore, it is proved that,$BS·RC=PS·QR=P{S}^2$.