Proposition: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.
Construction: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.
∴ ΔOAE ≅ ΔOCF
∴ 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)