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.
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