<img height="1" width="1" style="display:none" src="https://www.facebook.com/tr?id=539359806247306&ev=PageView&noscript=1"/>

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

  1. Chapter 7 Class 9 Triangles
  2. Serial order wise
Ask Download


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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.