prove that the perpendicular bisectors of two chords of a circle intersect at the centre.

Here consider ACD and CDB
So CD = CD
Angle ADC = Angle BDC (90 degree)
AD = DB (perpendicular bisector)
So ACD and CDB are congruent (SAS)
So CA = CB

As only from centre the length is same to the circumference, so C is center.
Similarly do for other chord. You will get the same result for any chord.

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