# Equal chords of a circle are equidistant from the centre.

## 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) Digital STUDY Center

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