#### 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.

#### Solution

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)

∴ Δ APB ≅ 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.

Is there an error in this question or solution?

Solution Two Chords Ab and Ac of a Circle Are Equal. Prove that the Centre of the Circle Lies on the Bisector of Angle Bac. Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).