Question

In the following fig., AB is a diameter of the circle. CD is a chord equal to ths radius of the circle. AC and BD when extended intersect at a point E. Prove that, ∠AEB = 60°.

Answer 1

Solution:

Join OC, OD and BC.

△ODC is equilateral.

∴ ∠COD = 60°

Now, ∠CBD = $\frac{1}{2}$ ∠COD (The angle substended by an arc at the centre is double the angle substended by it at any poiny on the remaining part of the circle.)

This gives, ∠CBD = 30°

∠ACB = 90° ((The angle substended by an arc at the centre is double the angle substended by it at any poiny on the remaining part of the circle.)

∠BCE = 180° - ∠ACB = 90°

Which gives, ∠CEB = 90° - 30° = 60°

i.e., ∠AEB = 60°

Hence, the proof.

Answer 2

Step-by-step explanation:

Join OC, OD and BC.

△ODC is equilateral.

∴ ∠COD = 60°

Now, ∠CBD = \frac{1}{2}

2

1

∠COD (The angle substended by an arc at the centre is double the angle substended by it at any poiny on the remaining part of the circle.)

This gives, ∠CBD = 30°

∠ACB = 90° ((The angle substended by an arc at the centre is double the angle substended by it at any poiny on the remaining part of the circle.)

∠BCE = 180° - ∠ACB = 90°

Which gives, ∠CEB = 90° - 30° = 60°

i.e., ∠AEB = 60°

Hence, the proof