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.

Answer 1

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.

Proof:

In ΔAPB and ΔAPC,

AB = AC   ...(Given)

∠BAP = ∠CAP  ...(Given)

AP = AP  ...(Common)

∴ ΔAPB ≅ ΔAPC  ...(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.