**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 ]

Again, OF⊥CD

∴ CF=DF=½CD——————(2)

Since OE⊥AB and OF⊥CD

So, ∆OAE and ∆OCF are right angled.

∴ OA^{2}=OE^{2}+AE^{2}

and OC^{2}=OF^{2}+CF^{2}

But, OA=OC

⟹ OA^{2}=OC^{2}

⟹ OE^{2}+AE^{2}= OF^{2}+CF^{2}

⟹ AE^{2}– CF^{2}= OF^{2}– OE^{2}——————————-(3)

Now,

AB>CD

⟹½AB>½CD

⟹AE>CF [from (1)]

⟹ AE^{2}>CF^{2}

⟹ AE^{2}-CF^{2}>0

From (2) and (3)

OF^{2}-OE^{2}>0

⟹ OF^{2}>OE^{2}

⟹ OF>OE

∴ OE>OE **(Proved)**