Slide3.JPG

Slide4.JPG

Slide5.JPG

 

Go Ad-free

Transcript

Example 2 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°

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo