Q. 43.5( 13 Votes )

AB and AC are two

Answer :

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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