Q. 4

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the center of the circle.

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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