A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.

Question

Philoid · Accepted Answer

A set of lines or curves are said to be concurrent if they all intersect at the same point.

Let A, B, C be points on a circle. Draw perpendicular bisector of AB and AC which meet at point O.

Join OA, OB and OC.

We need to prove that the perpendicular bisector of BC also passes through O (if so, then perpendicular bisectors of AB, BA and CA are concurrent as they all will intersect at the same point O).

So, in ΔOEB and ΔOEA:

AE = BE (∵ E is the perpendicular bisector of AB)

∠AEO = ∠BEO = 90°

OE = OE (common)

∴ ΔOEB ≅ ΔOEA (by SAS congruence rule)

∴ OA = OB (By CPCT)

Similarly, ΔOFA ≅ ΔOFB (by SAS congruence rule)

∴ OA = OC (By CPCT)

So, OA = OB = OC = x (say)

Construct a perpendicular line from O to the line BC which intersect line BC at M and join them.

So, in ΔOMB and ΔOMC:

OB = OC (proved above)

OM = OM (common)

∠OMB = ∠OMC = 90°

∴ ΔOEB ≅ ΔOEA (by RHS congruence rule)

⇒ BM = MC (by CPCT)

∴ M is the perpendicular bisector of BC and hence OL, ON and OM are concurrent.