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Question 3
A, B and C are three points on a circle, prove that perpendicular bisectors of AB, BC and CA are concurrent.

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Solution

Given a circle passing through three points A, B and C.
Construction : Draw the perpendicular bisectors of AB and AC and let they meet at a point O. Join OA, OB and OC.
Meet at a point O. Join OA, OB and OC
To prove that perpendicular bisector of BC also passes through O i.e., LO, ON and OM are concurrent.

Proof
In Δ OEA and Δ OEB
AE = BE [OL is the perpendicular bisector of AB]
AEO=BEO [each angle is 90]
And OE = OE [common side]
ΔOEAOEB [by SAS congruence rule]
OA = OB [by CPCT]

Similarly, ΔOFAΔOFC [by SAS congruence rule]
OA = OC
OA = OB = OC = r [by CPCT]

Now, we draw a perpendicular from O to the BC and join them.
In ΔOMB and ΔOMC
OB = OC [proved above]
OM = OM [Common side]
And OMB=OMC [each angle is 90]
ΔOMBΔOMC [by RHS congruence rule]
BM = MC [by CPCT]
Hence, OM is the perpendicular bisector of BC.
Hence, OL, ON and OM are concurrent.


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