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

Given: Two chords AB and AC of a circle are equal.

The bisector of angle ∠BAC intersects chord BC at point P.

To Prove: The center of the circle lies on the bisector of ∠BAC.

Construction: Join BC.

Proof:

In Δ APB and Δ APC,

AB = AC (Given)

∠BAP = ∠CAP (Given)

AP = AP (Common Side)

By using ASA congruency criterion,

Δ APB ≅ Δ APC

By corresponding parts of congruent triangles,

BP = CP and ∠APB = ∠APC

∠APB and ∠APC form a linear pair(they lie on the straight line BC).

⇒ ∠APB + ∠APC = 180°

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = $\dfrac{180°}{2}$

⇒ ∠APB = 90°

Therefore, AP is the perpendicular to chord BC.

The perpendicular bisector of a chord passes through the center of the circle.

Hence, the centre of the circle lies on the bisector of ∠BAC.