Last updated at March 16, 2017 by Teachoo

Transcript

Ex 6.3 ,6 In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR= 1/2 ∠QPR Given TQ is the bisector of ∠ PQR. So, ∠ PQT = ∠ TQR = 1/2 ∠ PQR Also, TR is the bisector of ∠ PRS So, ∠ PRT = ∠ TRS = 1/2 ∠ PRS In Δ PQR, ∠ PRS is the external angle ∠ PRS = ∠ QPR + ∠ PQR In Δ TQR, ∠ TRS is the external angle ∠ TRS = ∠ TQR + ∠ QTR Putting ∠ TRS = 1/2 ∠ PRS & ∠ TQR = 1/2 ∠ PQR 1/2 ∠PRS = 1/2 ∠ PQR + ∠ QTR 1/2 ∠PRS = 1/2 ∠ PQR + ∠ QTR Putting ∠ PRS = ∠ QPR + ∠ PQR from (1) 1/2 (∠ QPR + ∠ PQR) = 1/2 ∠ PQR + ∠ QTR 1/2 ∠ QPR + 1/2∠ PQR = 1/2 ∠ PQR + ∠ QTR 1/2 ∠ QPR + 1/2∠ PQR – 1/2 ∠ PQR = ∠ QTR 1/2 ∠ QPR = ∠ QTR ∠ QTR = 1/2 ∠ QPR Hence proved

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.