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Question

In given fig, PR>PQ and PS bisects QPR. Prove that PSR>PSQ.
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Solution

In ΔPQR,
PR>PQ (Given)
PQR>PRQ ...(1)
(Angle opposite to side of greater length is greater

PS is the bisector of P, sox=y
Adding x in (1)
PQR+x>PRQ+x
PQR+x>PRQ+y .... (2)

In ΔPQS,
PQS+x+PSQ=180
(Angle sum property of triangle)

PQS+x=180PSQ ..... (3)

In ΔPSR ,
PRS+y+PSR=180
(Angle sum property of triangle)

PRS+y=180PSR

Using equation (1), (2), (3) we get

180PSQ>180PSR ...(4)

PSQ>PSR

PSQ<PSR

So,
PSR>PSQ [henceproved]

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