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..........