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Ex 7.4, 5 - In figure, PR > PQ and PS bisects ∠QPR - Side inequality

  1. Chapter 7 Class 9th Triangles
  2. Serial order wise
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Ex7.4, 5 In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ. Given PR > PQ, ∴ ∠PQR > ∠PRQ PS is the bisector of ∠QPR. ∴ ∠QPS = ∠RPS Let ∠QPS = ∠RPS = x In Δ PQS, ∠PSR is the exterior angle ∠PSR = ∠PQR + x Now, ∠ PQR > ∠ PRQ Adding x both sides ∠PQR + x > ∠PRQ + x ∠PSR > ∠PSQ Hence proved

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