two chord AB and AC of a circle are equal.prove that the bisector of angle BAC passes through the centre O of the circle

Dear Student!

Given: AB and AC are two equal chords of C (O, r).

To prove: Centre, O lies on the bisector of ∠BAC.

Construction: Join BC. Let the bisector of ∠BAC intersects BC in P.


In  ΔAPB and ΔAPC,

AB = AC  (Given)

∠BAP = ∠CAP (Given)

AP = AP (Common)

 (SAS congruence criterion)

⇒ BP = CP and ∠APB = ∠APC (CPCT)

∠APB + ∠APC = 180° (Linear pair)

∴ 2∠APB = 180° (∠APB = ∠APC)

⇒ ∠APB = 90°

Now, BP = CP and ∠APB = 90°

∴ AP is the perpendicular bisector of chord BC.

⇒ AP passes through the centre, O of the circle.


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