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

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

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Given: A, B, C are 3 points on a circle

Construction \Rightarrow Draw perpendicular dissector of AB and AC and they meet at a point O. Then join OA, OB and OC.

To prove: perpendicular bisector of AB, BC and CA are concurrent.

Proof: In \triangleOEA and \triangleOEB

AE = BE [OL is ^ar bisector of AB]

OE = OE [Common side]

\angleAEO = \angleBEO [each 90°]

\therefore \triangleOEA \cong \triangleDEB [By SAS congruence Rule]

OA = OC

Similarly, \triangleOFA \cong \triangleOFC [by SAS congruence]

\therefore OB = OC = OA = t [Say]

SO, we draw a perpendicular from O to the BC on join

In \triangleOMB and \triangleOMC

OB = OC [Proved Above]

OM = OM [Common side]

\angleOMB = \angleOMC [90° each]

\therefore \triangleOMB \cong \angleOMC [by RHS congruence rule]

\Rightarrow BN = MC [by corresponding parts of congruent triangle]

Hence, OM is the perpendicular bisector of side BC.

Hence OL, ON, OM are concernment.

Hence proved. 

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