In the given figure, ST is a straight line and ray QP stands on it.
∴ \(∠\)PQS + \(∠\)PQR = 180º (Linear Pair)
\(∠\)PQR = 180º − \(∠\)PQS............ (1)
\(∠\)PRT + \(∠\)PRQ = 180º (Linear Pair)
\(∠\)PRQ = 180º − \(∠\)PRT............ (2)
It is given that \(∠\)PQR =\(∠\) PRQ.
Equating equations (1) and (2), we obtain
180º − \(∠\)PQS = 180º− \(∠\)PRT
\(∠\)PQS = \(∠\)PRT