In Figure, ` hat A B~= hat A C` and `O` is the centre of the circle. Prove that `O A` is the perpendicular bisector of `B C`

Explanation

Join `OB and OC.`
`AB~= AC`(Given)
chord AB = chord AC.(If two arcs of a circle are congruent, then their corresponding chords are equal)
`/_AOB = /_AOC` ....(i) (Equal chords of a circle subtend equal angles at the centre)
In `/_\OBC and /_\OCD, /_DOB = /_DOC` (From (1))
`OB = OC`(Radii of the same circle)
`OD = OD`(Common)
`/_\OBD ~= /_\OCD` (By SAS)
`/_ODB =/_ODC` ....(ii) (By cpctc)
`BD = CD` ...(ii) (By cpctc)
But `/_BDC = 180^0`
`/_ODB + /_ODC = 180^0`
`=>/_ODB + /_ODB = 180^0`From equation (ii)
`=> 2/_ODB = 180^0 = /_ODB = 90^0`
`/_ODB = /_ODC = 90^0` ....(iv)
(From (ii)) So, by (iii) and (iv),
OA is the perpendicular bisector of BC.