AB is a diameter of the circle CD is chord equal to the radius of the circle.AC and bd when extended intersect at a pnt E. Prove that angle AEB =60
Join OC, OD and BC.
Triangle ODC is equilateral
Ã COD = 60Â°
Now, This gives Again, So, Which gives
This givesÃCBD = 30Â°
So,Ã BCE = 180Â° â Ã ACB = 90Â°
Ã CEB = 90Â° â 30Â° = 60Â°, i.e., Ã AEB = 60Â°