In the fig.6.36, \(\frac{QR}{QS} \) = \(\frac{QT}{PR} \) and \(\angle\) 1 = \(\angle\) 2. Show that \(\triangle\) PQS ~ \(\triangle\) TQR.

In \(\triangle\) PQR,
\(\angle\) PQR = \(\angle\) PRQ
PQ = PR ………………………(i)
Given,
\(\frac{QR}{QS} \) = \(\frac{QT}{PR} \)

Using equation (i), we get
\(\frac{QR}{QS} \) = \(\frac{QT}{QP} \) ……………….(ii)

In \(\triangle\) PQS and \(\triangle\) TQR, by equation (ii),
\(\frac{QR}{QS} = \frac{QT}{QP} \)
\(\angle\) Q = \(\angle\) Q

\(\triangle\) PQS ~ \(\triangle\) TQR [By SAS similarity criterion]