Question

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

Answer 1

Given AB=AC

The two triangle formed

∆AOB , ∆AOC

We have to make them congruence

AO = AO( Common)

AB = AC ( Given)

Angle BAO = Angle CAO ( Because the AO bisects Angle BAC)

∆ AOB (=congruence) ∆AOC

Answer 2

Answer:

Required to prove: Centre, O lies on the bisector of /BAC.

Construction : Join BC, Let the bisector of ZBAC intersects BC in P. Proof:

In A and AAP C, AB = AC [Given]

ZBAP = ZCAP [Given]

AP - AP [Common] Hence, AAPB ≈ AAPC by SAA congruence criterion SO, by CPCT we have BP - CP and ZAPB = ZAPC

And

ZAP B+ ZAPC = 180° [Linear pair] 2/APB = 180° [ZAPB = ZAPC]

ZAP B = 90°

Now, BP

CP and ZAPB = 90°

Therefore, AP is the perpendicular bisector of chord BC.

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