**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.

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