AB and AC are two

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.

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