In Figure, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.
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12/07/2021 12:31 pm
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In Figure, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.
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12/07/2021 12:32 pm
It is given that PR > PQ and PS bisects QPR
Now we will have to prove that angle PSR is smaller than PSQ i.e. PSR > PSQ
Proof:
QPS = RPS — (ii) (As PS bisects ∠QPR)
PQR > PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)
PSR = PQR + QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)
PSQ = PRQ + RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)
By adding (i) and (ii)
PQR + QPS > PRQ + RPS
Thus, from (i), (ii), (iii) and (iv), we get
PSR > PSQ
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