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Question
A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.
Solution
Given: A, B and C are three points on a circle.
To prove: That perpendicular bisector of AB, BC and CA.
Construction: Draw perpendicular bisectors ST of AB, PM of BC and QR of CA. Join AB, BC and CA.
Proof: OA = OB ...(i) [O lies on ST, the perpendicular bisector of AB]
Again, OB = OC ...(ii) [O lies on PM the perpendicular bisector of BC]
And OC = OA ...(iii) [O lies on QR, the perpendicular bisector of CA]
Now, from equations (i), (ii) and (iii),
OA = OB = OC = r
So, draw a circle with centre O and radius r, that will pass through A, B and C.
That means a circle passing through the point A, B and C. Since, ST, PM or QR can cut each other at one and only one point O.
Therefore, O is the only one point which is equidistance from A, B and C.
Hence, the perpendicular bisector of AB, BC and CA are concurrent.
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