Two chords `A B` and `A C` of a circle are equal. Prove that the centre of the circle lies on the angle bisector of `/_B A Cdot`

Explanation

Giventhat
`AB` and` AC` are two equal chords of `C (O, r)`.
To prove: Centre, `O` lies on the bisector of `angleBAC`.
Construction: Join `BC`. Let the bisector of `angleBAC` intersects `BC` in `P`.
Proof:-
In `triangleAPB `and `triangleAPC`,
`AB = AC` (Given)
`angleBAP = angleCAP `(Given)
`AP = AP` (Common)
`triangleAPB cong triangleAPC` (SAS congruence cririon)
`BP=CP` and `angleAPB=angleAPB` (CPCT)
`angleAPB+angleAPC=180^0` (linear pair)
`2angleAPB=180^0 (angleAPB=angleAPC)`
`angleAPB=90^0`
Now, `BP=CP` and `angleAPB=90^0`
Therefore `AP` is the perpendicular bisector of chord `BC`
`AP` passes through the center, `O` of the circle