Question

The two chords AB and AC of a circle are equal. I prove that, the bisector of angle BAC passes through the centre.​

Answer 1

Answer:

Given: A circle C(O, r) in which AB and AC are two equal chords. AD is the bisector of ∠BAC.  To prove: O lies on AD.  Construction: O lies on AD.  Proof: In ∆BAM and ∆CAM,  AB = AC(Given)  ∠BAM = ∠CAM (Given)  AM = AM (common)  ∴ ∆BAM ≅ ∆CAM (By SAS congruency Rule) and  BM = CM (By CPCT)  ∠BMA = ∠CMA (By CPCT)  ∠BMA = ∠CMA = 90° [∵∠BMA + ∠CMA = 180°]  => AM is a perpendicular bisector of chord BC.  => AD is the perpendicular bisect of the chord BC. But the perpendicular bisector of a chord always passes through the centre of the circle.  => AD passes through the centre O of the circle.  => O lies on AD

Answer 2

Your Answer is in the Attachment mate :)