Let AB and CD be two equal chords of a circle with centre O. It is to be proved that the chords AB and CD are equidistant from the centre.


Draw from O, the perpendiculars OE and OF to the chords AB and CD respectively. Join O,A and O,C.

Step-1: OE⊥AB and OF⊥CD (Perpendicular from the centre bisects the chord)
Therefore, AE = BE and CF = DF.
∴ AE = 1/2 AB
and CF = 1/2 CD

Step-2: But AB = DC (supposition)
∴ AE = CF.

Step-3: Now b etween the right-angled ΔOAE and ΔOCF (radius of same circle)
hypotenuse OA = hypotenuse OC and AE = CF.
∴ OE = OF.

Step-4: But OE and OF are the distances from O to the chords AB and CD respectively.
Therefore, the chords AB and CD are equidistant from the centre of the circle. (Proved)

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