Question

two chords AB and AC of a circle are equal . prove that the centre of the circle lies on the bisectorvof angle BAC

Answer 1

Join BC , meeting AD at M.

In ΔDAM and ΔCAM, 

AB = AC and ∠RAM = ∠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. 

Answer 2

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