NCERT Exemplar Solutions Class 9 Mathematics Solutions for Circles - Exercise 10.3 in Chapter 10 - Circles
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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.
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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