Question

Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the bisector of angle BAC.

Show that the bisector of angle BAC is a perpendicular bisector of chord BC

Answer 1

In ΔBAM and ΔCAM,

AB = AC and ∠BAM = ∠CAM [given]

and AM = AM [common sides]

ΔBAM ≅ ΔCAM [by SAS congruency]

∴ BM = CM

and ∠BMA = ∠CMA [by CPCT] ---- i)

But ∠BMA + ∠CMA = 180° ----- (ii)

From Eqs. (i) and (ii), we get

∠BMA = ∠CMA = 90°

⇒ AD is the perpendicular bisector of chord BC, but the perpendicular bisector of a chord always passes through the centre of circle.

∴ AD passes through the centre O of circle.

O lies on AD. Hence proved.