Prove that any th

We draw a circle of any radius and take any three points A, B and C on the circle.

We join A to B and B to C.

We draw perpendicular bisectors of AB and BC.

We know that perpendicular from the center bisects the chord.

Hence the center lies on both of the perpendicular bisectors.

The point where they intersect is the center of the circle.

The perpendiculars of the line segments drawn by joining collinear points is always parallel whereas in circle any three point’s perpendicular bisector will always intersect at the center.

Hence, any three points on the circle cannot be collinear.