Theorem 10.7 : Chords equidistant from the centre of a circle are equal in length Class 9

Given : A circle with centre ‘O’.

AB and CD are two chords of the circle. OM=ON and ∠OMB=∠ONC=90°

To prove: AB=CD

Proof: In ΔOMB and ΔONC

OB=OC (radii of the circle)

OM=ON (given)

∠OMB=∠ONC=90° (given)

∴ ΔOMB ≅ ΔONC (by R.H.S.)

MB=CN (CPCT). ……(i)

as we know, perpendiculars from centre to the chord bisect the chord

∴ AB = 2MB and CD = 2CN ….. (ii)

MB=CN

2MB=2CN (from eq(i))

AB=CD (proved)

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