In the given figure, RQ and TP are perpendicular to PQ, also $TS\perp PR$ prove that ST. RQ = PS. PQ.

In $\Delta RPQ$
$\angle 1+\angle 2+\angle 4={180}^{\circ }$
$\angle 1+\angle 2+{90}^{\circ }={180}^{\circ }$
$\angle 1+\angle 2={180}^{\circ }-{90}^{\circ }$
$\angle 1={90}^{\circ }-\angle 2$ ...(i)
$\because$ $TP\perp PQ$
$\therefore$ $\angle TPQ={90}^{\circ }$
$\Rightarrow \angle 2+\angle 3={90}^{\circ }$
$\angle 3={90}^{\circ }-\angle 2$ ...(ii)
From eq.(i) and eq. (ii)
$\angle 1=\angle 3$
Now in $\Delta RQP$ and $\Delta PST$
$\angle 1=\angle 3$ [Proved above]
$\angle 4=\angle 5$ [Each ${90}^{\circ }$]
So by AA similarity
$\Delta RQP$ ~ $\Delta PST$
$\frac{RQ}{PS}=\frac{PQ}{ST}$
$\Rightarrow ST.RQ=PS.PQ$ Hence Proved.