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

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Question

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
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Prakhar Bindal · Accepted Answer

Join OC, OD and BC
Triangle ODC is equilateral
Therefore, ∠ COD = 60°
Now, ∠ CBD = half of ∠ COD
This gives ∠ CBD = 30°
Again, ∠ ACB = 90°
So, ∠ BCE = 180° – ∠ ACB = 90°
Which gives ∠ CEB = 90° – 30° = 60°, i e
∠ AEB = 60
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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 =

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

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