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Question 5 - Side inequality - Chapter 7 Class 9 Triangles
Last updated at August 5, 2026 by Teachoo
Chapter 7 Class 9 Triangles
Concept wise
- Congruency - Introduction
- SAS
- ASA/AAS
- Opposite Angles of equal sides
- SSS
- RHS
-
Side inequality
Transcript
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
