Given A circle passing through three points, A, B and C.
Construction Draw perpendicular bisectors of AB and AC and they meet at a point O. Join OA, OB and OC.
To prove Perpendicular bisector of BC, also passes through O i.e, LO,ON and OM are concurrent.
Proof In `DeltaOEA and DeltaOEB`,
AE=BE
[OL is the perpendicular bisector of AB]
`angleAEO=angleBEO ["each "90^(@)]`
and OE=OE [common side]
`:. Delta OEA cong OEB` [by SAS congruence rule]
`:. OA=OB`
Similarly, `DeltaOFA cong DeltaOFC` [by SAS congruence rule]
`:. OA=OC` [by CPCT]
`:. OA=OB=OC=r` [say]
Now, we draw a perpendicular from O to the BC and join them.
In `DeltaOMB and DeltaOMC`,
OB=OC [proved above]
OM=OM [common side]
and `angleOMB = angleOMC ["each "90^(@)]`
` :. DeltaOMB cong DeltaOMC` [by RHS congruence rule]
`rArr BM=MC` [by CPCT]
Hence, OM is the perpendicular bisector of BC.
Hence, OL, ON and OM are concurrent.
