Question
Sun February 27, 2011 By: Baskaran Ramanan

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

Expert Reply
Mon February 28, 2011
Dear Student,
 
Here is the solution:
 

Join OC, OD and BC.

Triangle ODC is equilateral

Therefore,

Ð COD = 60°

Now,

ÐCBD =1/2ÐCOD 

This gives

ÐCBD = 30°

Again,

 

ÐACB = 90°

So,

Ð BCE = 180° – Ð ACB = 90°

Which gives

 
 

Ð CEB = 90° – 30° = 60°, i.e., Ð AEB = 60°

 Regards

Team Topperlearning

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