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Chords and their perpendicular bisectors · 3 marks

Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Answer: The centre O is equidistant from the ends of every chord AB (OA = OB, radii). Every point equidistant from A and B lies on the perpendicular bisector of AB, so O lies on it.

Step-by-step solution

Given: A circle with centre O; A chord AB and its perpendicular bisector l
To find: Prove that O lies on l

Idea: Use the locus of points equidistant from two points: it is exactly the perpendicular bisector of the segment joining them. Or use the earlier result that the line from the centre to the midpoint of a chord is perpendicular to it.

ABOM
  1. Let M be the midpoint of AB. (If AB is a diameter, M = O and there is nothing more to prove.) Join OM, OA and OB.½ mark
  2. In ΔOMA and ΔOMB: OA = OB (radii), AM = BM (M is the midpoint), OM is common. So ΔOMA ≅ ΔOMB (SSS) and ∠OMA = ∠OMB.1 mark
  3. ∠OMA + ∠OMB = 180° (linear pair), so each is 90°: OM ⟂ AB.1 mark
  4. Through M there is only one line perpendicular to AB, the perpendicular bisector. OM is such a line, so the perpendicular bisector is the line OM and passes through O.½ mark
OA = OB, so O is equidistant from A and B; the line OM through the midpoint is perpendicular to AB, so the perpendicular bisector of AB is the line OM, which passes through the centre.

Answer to write in the exam

Given: chord AB of a circle with centre O; M the midpoint of AB. To prove: the perpendicular bisector of AB passes through O.

In ΔOMA and ΔOMB: OA = OB (radii), AM = BM, OM = OM (common)

∴ ΔOMA ≅ ΔOMB (SSS) ⇒ ∠OMA = ∠OMB (CPCT)

∠OMA + ∠OMB = 180° (linear pair) ⇒ ∠OMA = 90° ⇒ OM ⟂ AB

Only one line through M is ⟂ AB ⇒ OM is the perpendicular bisector of AB

∴ The perpendicular bisector of AB passes through O.

Common mistakes that cost marks

  • Assuming the perpendicular bisector passes through O and then proving OM ⟂ AB: the argument must end with O on the bisector.
  • Forgetting the uniqueness step: only one line through M is perpendicular to AB.
  • Using RHS before the right angle has been proved.

How this can come in the exam

MCQ (1 mark)

The perpendicular bisectors of two non-parallel chords of a circle meet at

  1. a point on the circle
  2. the centre of the circle
  3. the midpoint of a chord
  4. no point
Show answer

(B) the centre of the circle
Each perpendicular bisector passes through the centre, and two non-parallel lines meet in one point, so they meet at the centre.

Try one yourself

Explain how the result of this question lets you find the centre of a circular plate using a ruler and set square.

Show answer

Draw two non-parallel chords on the plate and the perpendicular bisector of each. Both bisectors pass through the centre, so the point where they cross is the centre.

More questions like this

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