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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.

Answer:

According to the question,

Three non-collinear points A, B and C are on a circle.

To prove: Perpendicular bisectors of AB, BC and CA are concurrent.

Construction: Join AB, BC and CA.

Draw:

ST, perpendicular bisector of AB,

PM, perpendicular bisector of BC

And, QR perpendicular bisector of CA

As point A, B and C are not collinear, ST, PM and QR are not parallel and will intersect.

Proof:

O lies on ST, the ⊥ bisector of AB

OA = OB … (1)

Similarly, O lies on PM, the ⊥ bisector of BC

OB = OC … (2)

And, O lies on QR, the ⊥ bisector of CA

OC = OA … (3)

From (1), (2) and (3),

OA = OB = OC

Let OA = OB = OC = r

Draw circle, with centre O and radius r, passing through A, B and C.

Hence, O is the only point equidistance from A, B and C.

Therefore, the perpendicular bisectors of AB, BC and CA are concurrent.

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