Let AB and BC be equal chords of circle with center O.

Draw the angle bisector AD of ∠BAC.

Join BC, meeting AD at M.

In triangle DAM and CAM:

AB = BC (given that the chords are equal)

∠BAM = ∠CAM (∵ AD is angle bisector of A)

AM = AM (common side)

∴ ΔABD = ΔACD (by SAS congruence rule)

∴ BM = CM and ∠AMB = ∠AMC = x (say) (by CPCT) ………(1)

But ∠AMB + ∠AMC = 180° ………(2)

From equation (1) and (2), we have:

x + x = 180°

⇒ 2x = 1280°

⇒ x = 90°

⇒ ∠AMB = ∠AMC = 90°

Therefore, AD is the perpendicular bisector of chord BC, but the perpendicular bisector of any chord always passes from the center of the circle.

∴ AD passes through the center O of the circle.

Thus, O lies on the angle bisector of the angle BAC.

Hence, proved.