In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: 'The centre of the circle lies on the angle bisector of $\angle BAC$.'

In triangles OAB and OAC:

$$OA = OA \quad [\text{Common}]$$

$$OB = OC \quad [\text{Radii of circle}]$$

$$AB = AC \quad [\text{Given}]$$

By SSS congruence, $\triangle OAB \cong \triangle OAC$.

Hence, corresponding angles are equal

$$i.e., \quad \angle OAB = \angle OAC$$

This means the line AO bisects $\angle BAC$, i.e., the centre O lies on the angle bisector of $\angle BAC$.