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Question

In figure 1=2 and ΔNSQΔMTR, then prove that ΔPTSΔPRQ.
1319768_05b8ac64c06946a9bb619c4c0457ebb0.png

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Solution

Since,1=2i,e,PS=PT(i)andΔNSQΔMTRSQ=TR(ii)InΔPTS&ΔPRSP=P(common)(iii)PSPT=SQTRi,e.PSSQ=PTTRPS+SQSQ=PT+TRTRPSSQ=PRPT(iv)fromequation(iii)&(iv),wegetΔPTSΔPRS

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