Question

In the figure, $\angle PQR=\angle TPQ=\angle TSR={90}^0.$ Hence $:ST\times RQ=PS\times PQ$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given $\angle TPQ=\angle RQP={90}^{\circ }$
Sum of angles, $\angle TPQ+\angle RQP=90+90={180}^{\circ }$
Since, the sum of angles is 180, hence, $PT\parallel RQ$
In $△PST$ and $△PQR$,
$\angle PST=\angle PQR$ (each ${90}^{\circ }$)
$\angle TPS=\angle PRQ$ (Alternate angles for parallel lines PT and RQ)
$\angle STP=\angle RPQ$ (Third angle)
Thus, $△PST\sim △RQP$
Hence, $\frac{PS}{QR}=\frac{ST}{PQ}$
$PS\times PQ=ST\times QR$