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)