Theorem 10.6 : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres) Class 9

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.

Leave a Reply

%d bloggers like this: