If `O` is the circumcentre of a ` A B C` and `O D_|_B C ,` prove that `/_B O D=/_Adot`

Explanation

∴ `O` is the point of intersection of perpendicular bisectors of the sides of `triangle ABC`.
`⇒ D` is the midpoint of `BC`
`⇒ BD = DC`
In `triangle OBD` and `triangle OCD`,
`OB = OC `(Radius of circle)
`OD = OD `(Common)
`BD = DC `(Proved)
`triangle OBD ≅ triangle OCD `(`SSS` congruency criterion)
`⇒ angle BOD = angle COD` (By `CPCT` )
`⇒ angle BOD = angle COD=1/2 angle BOC`
We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
`⇒ angle BOC = 2angle BAC`
`⇒ 2angle BOD = 2angle BAC` (From equation (1))
`⇒ angle BOD = angle BAC`
Hence Proved.