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Question 5
In the given figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ
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Solution

PR>PQ                           (Given)
PQR > PRQ.........(1) (In any triangle, the angle opposite to the longer side is larger.)
We also have PQR+QPS+PSQ=180 (Angle sum property of triangle)
PQR=180QPSPSQ.....(2)
And, PRQ+RPS+PSR=180   (Angle sum property of triangle)
PRQ=180PSRRPS .........(3)
Substituting (2) and (3) in (1), we get
180QPSPSQ > 180PSRRPS
We also have QPS=RPS, using this in the above equation, we get
PSQ > PSR
PSQ < PSR

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