# Example 3 : AB is a diameter of the circle, CD is a chord equal to theradius of the circle. AC and BD when extended intersect at a point E. Prove that AEB = 60Â°?

$CD=radius,\;r\\*OC=OD=radius\\*\triangle OCD\;is\;equilateral\;triangle.\\* \Rightarrow \angle DCO=\angle COD=\angle ODC=60^{\circ}\\* \angle ACB=90^{\circ}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(Angle\;in\;semicircle)\\* \angle DOC=2\angle DBC\;\;\;\;\;\;\;\;\;\;\;\;\;\;(half\;angle)\\* \angle DBC=30^{\circ}\\* \angle ECB+\angle BCA=180^{\circ}\;\;\;\;\;\;\;\;\;\;\;\;\;\(linear\;pair)\\*\angle ECB=180-90=90^{\circ}\\* In,\;\triangle ECB\\* \angle CEB+\angle ECB+\angle CBE=180^{\circ}\\* \angle CEB+90^{\circ}+30^{\circ}=180^{\circ}\\* \angle CEB=60^{\circ}\\*Hence,\;proved\;that\;\angle CEB=\angle AEB=60^{\circ}$

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