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Example 3 - In figure, AB is a diameter of circle, CD - Examples

Example 3 - Chapter 10 Class 9 Circles - Part 2
Example 3 - Chapter 10 Class 9 Circles - Part 3

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Example 3 In figure, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that ∠ AEB = 60°. Given: AB is diameter of circle Chord CD, where CD = Radius of circle To prove: ∠AEB = 60° Construction: Join OC , OD & BC Proof: In Δ OCD OC = OD = CD = Radius of circle Since all sides are equal, Δ OCD is equilateral triangle ∠ COD = 60° Now, For arc CD subtends ∠ COD at centre & ∠ CBD at point B ∴ ∠ COD = 2 ∠ CBD 60° = 2∠ CBD 2∠ CBD = 60° ∠ CBD = (60°)/2 = 30° Now, Since AB is a diameter So, ∠ ACB = 90° Since AE is a line ∠ ACB + ∠ ECB = 180° 90° + ∠ ECB = 180° ∠ ECB = 180° – 90° = 90° In Δ ECB ∠ CEB + ∠ ECB + ∠ CBE = 180° ∠ CEB + 90° + 30° = 180° ∠ CEB + 120° = 180° ∠ CEB = 180° – 120° ∠ CEB = 60°

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Davneet Singh

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.