Last updated at March 16, 2017 by Teachoo

Transcript

Example 8 In figure, the sides AB and AC of ∆ABC are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90° − 1/2∠ BAC. BO is the bisector of ∠ CBE So, ∠ CBO = ∠ EBO = 1/2 ∠ CBE Similarly, CO is the bisector of ∠ BCD So, ∠ BCO = ∠ DCO = 1/2 ∠ BCD ∠ CBE is the exterior angle of Δ ABC Hence, ∠ CBE = x + z ∠ CBE = x + z 1/2(∠ CBE) = 1/2 (x + z) ∠ CBO = 1/2 (x + z) Similarly, ∠ BCD is the exterior angle of Δ ABC Hence, ∠ BCD = x + y 1/2(∠ BCD) = 1/2 (x + y) ∠ BCO = 1/2 (x + y) In Δ OBC ∠ BOC + ∠ BCO + ∠ CBO = 180° ∠ BOC + 1/2 (x + y) + 1/2 (x + z) = 180° ∠ BOC + 1/2 (x + y) + 1/2 (x + z) = 180° ∠ BOC + 1/2 (x + y + x + z) = 180° In Δ ABC x + y + z = 180° Putting (2) in (1) ∠ BOC + 1/2 (x + y + x + z) = 180° ∠ BOC + 1/2 (x + 180° ) = 180° ∠ BOC + 𝑥/2 + 1/2 × 180° = 180° ∠ BOC + 𝑥/2 + 90° = 180° ∠ BOC = 180° – 90° – 𝑥/2 ∠ BOC = 90° – 𝑥/2 ∠ BOC = 90° – 𝑥/2 ∠ BOC = 90° – 1/2 × ∠ BAC 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

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .