AB ia a diameter of the circle,CD is a chord equal to the radius of the circle.AC and BD when extended intersect t a point E. Prove that angle AEB=60 DEGREE.

We have the following situation-

Construction: Join OC, OD and BC.

Consider ∆COD,

OC= OD = CD = r (Radius of circle)

∴ ∆COD is equilateral.

Thus, ∠COD = 60°

 [Angle subtended by an arc on the circle is half the angle subtended by the same arc on the centre]

∠ACB = 90° [Angle in a semi circle]

∴ ∠ECB = 180° – ∠ACB

= 180° – 90°

= 90°

Consider ∠BCE,

∠ECB + ∠CBE + ∠BEC = 180°

⇒ 90° + 30° + ∠BEC = 180°

⇒ 120° + ∠BEC = 180°

⇒ ∠BEC = 180° – 120°

⇒ ∠BEC = 60°

⇒ ∠BEA = 60°.

 Hence proved..........

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