Question

bisector AD of angle BAC of ABC passes through the centre O of the circumcircle of the triangle prove that AB=AC

Answer 1

Given that ∠OAB = ∠OAC (AD bisects ∠BAC) -- (1) 
Since O is the center of the circumcircle of triangle ABC, OA = OB = OC
=> OA = OB, OA = OC and OB = OC
=> ∠OAB = ∠OBA, ∠OAC = ∠OCA  and ∠OBC = ∠OCB -- (2)

From (1) and (2)

∠OAB = ∠ OAC = ∠ OBA = ∠OCA -- (3)

Now ∠ABD = ∠OBA + ∠OBC
= ∠OCA + ∠OCB [Using (2) and (3)]
= ∠ACD

=> ABC  = ∠ACB

So AB = AC

Answer 2

Hope it helps you...... I have written the answer and not typed it....