If I draw a circle on a tracing paper and draw two equal chords and drop perpendicular from centre to the chord. Fold the paper such that the two chords coincide. Then, the two perpendiculars are also coinciding.
True
Given two equal chords AB and CD. Draw perpendiculars from the centre of the circle to these chords.
We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴AE=BE=AB2=CD2=DF=CF.
In ΔOEB and ΔOFD
OB = OD (radii of the circle)
∠OEB = ∠OFD (right angle)
BE=DF
Therefore, ΔOEB≅ΔOFD by RHS Congruency.
Hence, OE = OF.
So, we see that the distances of equal chords from the centre of the circle are equal.