General enunciation: We have to show that between the two chords the larger one is nearer to the center than the smaller one.
Particular enunciation: Let ABCD is a circle with centre O. AB and CD are two chords and AB>CD. OE and OF are respectively perpendiculars from the centre O to the chords AB and CD. We have to prove that OE<OF.
Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
So, AE = BE = ½AB ————-(1)[∵ O be the centre and OE⊥AB ]
Since OE⊥AB and OF⊥CD
So, ∆OAE and ∆OCF are right angled.
⟹ OE2+AE2= OF2+CF2
⟹ AE2– CF2= OF2– OE2——————————-(3)
⟹AE>CF [from (1)]
From (2) and (3)
∴ OE>OE (Proved)