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Example 8 - Sides AB & AC of ∆ABC are produced to points E & D - Triangle - Problems

Example 8 - Chapter 6 Class 9 Lines and Angles - Part 2
Example 8 - Chapter 6 Class 9 Lines and Angles - Part 3
Example 8 - Chapter 6 Class 9 Lines and Angles - Part 4

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.