Given:
AB and AC are equal chords of a circle
Let O be the centre of the circle.
In ∆OAB and ∆OAC,
AB = AC (given)
OA = OA (common)
OB = OC (radius of the circle)
By SSS property,
∆OAB ≅ ∆OAC
Therefore, ∠OAB = ∠OAC (by C.P.C.T.)
Thus, ∠BAC = 2∠OAB.
Hence, the bisector of ∠BAC passes through the centre.