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

AB, BC and CA are concurrent.

Hi!

Here is the proof of your question.

A, B and C are three non-collinear points on the circle. Let O be the centre of the circle.

Now, the sides AB, BC and AC of ∆ABC are the chords of the circle.

It is known that the perpendicular bisector of a chord of a circle passes through the centre of that circle.

Thus, the perpendicular bisector of each chords AB, BC and AC passes through O.

This shows that the perpendicular bisectors of AB, BC and AC intersect at O.

Thus, the perpendicular bisectors of AB, BC and AC are concurrent .

Hope! You got the proof.

Cheers!

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