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.
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