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# 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]