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Home / Geometry / Prove that between the two chords the larger one is nearer to the center than the smaller one.

# Prove that between the two chords the larger one is nearer to the center than the smaller one.

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.

∴ OA2=OE2+AE2

and OC2=OF2+CF2

But, OA=OC

⟹ OA2=OC2

⟹ OE2+AE2= OF2+CF2

⟹ AE2– CF2= OF2– OE2——————————-(3)

Now,

AB>CD

⟹½AB>½CD

⟹AE>CF         [from (1)]

⟹ AE2>CF2

⟹ AE2-CF2>0

From (2) and (3)

OF2-OE2>0

⟹ OF2>OE2

⟹ OF>OE

∴ OE>OE (Proved)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...