Solution :
Given, \(QR\over QS\) = \(QT\over PR\)
So, \(QT\over QR\) = \(PR\over QS\) ……….(1)
Also given, \(\angle\) 1 = \(\angle\) 2
Since, sides opposite to equal \(\angle\)s are equal,
So , PR = PQ ………(2)
From (1) and (2), we get
\(QT\over QR\) = \(PQ\over QS\) or \(PQ\over QT\) = \(QS\over QR\)
In \(\triangle\)s PQS and TQR, we have :
\(PQ\over QT\) = \(QS\over QR\) and \(\angle\) PQS = \(\angle\) TQR = \(\angle\) Q
Hence, By SAS criterion of similarity, \(\triangle\) PQS ~ \(\triangle\) TQR