# In the figure, $$QR\over QS$$ = $$QT\over PR$$ and $$\angle$$ 1 = $$\angle$$ 2. Show that $$\triangle$$ PQS ~ $$\triangle$$ TQR.

## 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