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Here AB and CB is the chord. And as we know that Perpendicular line from centre of circle bisect the cord. So in chord AB, AL = LB Similarly in chord CB, CM = MB Now in triangle AOL and COM, we have ::Angle CLA = Angle AMC = 90⁰ ::OA = OC (BOTH ARE RADIUS) ::Angle LOA = Angle MOC (Vertically opposite angles)

Therefore, triangle AOL is congruent to triangle COM by AAS criterion of congruency => AL = MC (BY C.P.C.T) =>2AL = 2MC (Multiplying both the sides by 2) =>AB = BC (2AL = AB and 2MC = BC)

Hence both the chords are equal....

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given :o is the center :ol and om are perpendicular to ab and cb respectively :<oab=<ocb to prove :ab=cb PROOF consider tri.AOL and COM <l=<m=90 <loa=<moc (vert.opp.ang.) oa=oc (radii) hence,triangle AOL is congruent to tri. COM by aas congruence criteria so, AL=CM (CPCT) but, L and M are midpoints of ab and cb as OL and OM are the perpendiculars from the center and the perpendiculars from the center to the chord bisect the chord. Thus, 2AL=2CM AL=LB & CM=MB so, AB=CB HENCE THE PROOF ...