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 ∠BAC”.

Step-by-step solution

Idea: Show that the line AO makes equal angles with AB and AC. Congruent triangles OAB and OAC give exactly that.

1. Join OA, OB and OC. In ΔOAB and ΔOAC: OB = OC (radii), AB = AC (given), OA = OA (common).1 mark
2. So ΔOAB ≅ ΔOAC (SSS congruence rule).1 mark
3. Hence ∠OAB = ∠OAC (CPCT): the line AO divides ∠BAC into two equal parts. So AO is the bisector of ∠BAC and the centre O lies on it.1 mark

Check: Another view: A and O are both equidistant from B and C, so both lie on the perpendicular bisector of BC; in isosceles ΔABC this line is also the bisector of ∠BAC.

Answer to write in the exam

Common mistakes that cost marks

• Using SAS with an angle at A, which is what we are trying to prove equal.
• Writing OA = OB = OC and then claiming the triangles are equilateral.
• Concluding “AO bisects BC” instead of “AO bisects ∠BAC” (both are true, but the question asks about the angle).

How this can come in the exam

Try one yourself

In a circle with centre O, AB = AC and ∠OAB = 28°. Find ∠BAC and ∠OBA.

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