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.


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