Chapter 8.1, Problem 44WE

To determine

To prove: $BS\cdot RC=PS\cdot QR=P{S}^2$ .

Explanation of Solution

Given:

Thetriangleis shown below:

  

PQRS is rectangle.

Calculation:

Consider trianglesBSP and PAQ,

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

Consider triangles PAQ and QRC,

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

From equation (1) and (2),

  $\Delta QRC\sim \Delta BSP$

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 (3) and (4),

  $BS\cdot RC=PS\cdot QR=P{S}^2$