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In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

3.  In Fig. 6.15, \angle PQR = \angle PRQ, then prove that \angle PQS = \angle PRT.

                

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Given that,
ABC is a triangle such that \angle PQR = \angle PRQ  and ST is a straight line.
Now, \angle PQR + \angle PQS = 180^0     {Linear pair}............(i)
Similarly, \angle PRQ + \angle PRT = 180^0..................(ii)

equating the eq (i) and eq (ii), we get

\angle PQR +\angle PQS =\angle PRT + \angle PRQ   {but \angle PQR = \angle PRQ }
Therefore, \angle PQS = \angle PRT
Hence proved.

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