Given: A circle with centre O. in which chords AB=CD.
To Prove: OM=ON
Construction: Draw OM⊥AB and ON⊥CD
Proof: it is given that
⇒ AB=CD
So,
⇒ ½ × AB=½ × CD
And, since the perpendicular drawn from the centre of a circle bisects a chord:
⇒BM=CN
In ΔOMB and ΔONC
⇒ BM=CN (proved above)
⇒ OB=OC (radii of same circle)
⇒∠OMB=∠ONC (each 90°)
∴ ΔOMB≅ΔONC (by RHS)
⇒ OM=ON (by CPCT)
Hence proved that equal chords of a circle are equidistant from the centre.