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Question

In adjacent figure, PR>PQ and PS bisects QPR. Prove that PSR>PSQ.

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Solution

In the given triangle PQR,PR>PQ and PS bisects QPR
In ΔPQR
PR>PQ ....(Given )
PQR>PRQ (Angle opposite to longer side is larger) .... (1)
In ΔPQS and ΔRPS
PQR=PQS=180°(PSQ+QPS)
PRQ=PRS=180°(PSR+RPS)
Given PS is bisector of QPR
Then QPS=RPS .... (2)
PQR>PRQ (As per (1))
180°(PSQ+QPS)>180°(PSR+RPS)
PSR>PSQ (QPS=RPS)

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