In the Fig. AB and AC are two equal chords of a circle. Prove that bisector of ∠BAC passes through the centre of the circle.
Let AB and AC be two equal chords of a circle and ∠BAD = ∠CAD in ΔAMB and ΔAMC, AB = AC (given), AM = AM (Common) and
∠BAD = ∠CAD.
... ΔAMB @ DAMC
... BM = CM and ∠CMA = ∠BMA=90°
Hence, AD is a perpendicular bisector of chord BC. But perpendicular bisector of a chord passes through the centre.
Hence AD passes through the centre O.
∠BAD = ∠CAD.
... ΔAMB @ DAMC
... BM = CM and ∠CMA = ∠BMA=90°
Hence, AD is a perpendicular bisector of chord BC. But perpendicular bisector of a chord passes through the centre.
Hence AD passes through the centre O.
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