Draw a circle on a tracing paper and draw two equal chords and drop a perpendicular from center to the chord. Fold the paper such that the two chords coincide. You will find that the two perpendiculars also coincide.

True

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Solution

The correct option is A
True

Draw perpendiculars from centre to chord AB and chord CD.

We know the property that perpendicular drawn from centre to the chord bisects the chord. Therefore,

AE = BE and DF = CF.

In $Δ$ OEB and $Δ$ OFD

OB = OD (radii of the circle)

$∠$ OEB = $∠$ OFD (right angle)

Therefore, by RHS congruence rule, $Δ$ OEB and $Δ$ OFD are congruent.

Hence, OE = OF.

Hence equal chords are equidistant from the centre.

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