Question

Of any two chords of circle , show that one which is greater is nearer to center geozebra

Answer 1

Diagram is given in image,
To prove: Chord AB is nearer to the centre of the circle I.e. OL<OM, where OL and OM are perpendiculars from O to AB and CD respectively.
Construction: Join OA and OC
Proof: Since the perpendicular from the centre of a circle to a chord. Therefore,
OL is perpendicular to AB
AL = 1/2 AB
OM is perpendicular to CD
CD =1/2 CD

In right triangles OAL and OCM, we have
OA^2 = OL^2 + AL^2
And,
OC^2 = OM^2 + CM^2
By equating both, we get
(as OA = OC)
OL^2 + AL^2 = OM^2 + CM^2 ( Equation I )

Now,
AB > CD
1/2 AB > 1/2 CD
AL > CM
AL^2 > CM^2
OL^2 + AL^2 > OL^2 + CM^2 ( Adding OL^2 both sides)

OM^2 + CM^2 > OL^2 + CM^2 ( Using equation I )

OM^2 > OL^2
OM > OL
OL < OM
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Answer 2

Given two chords ab and CD of a circle C (O, r) OE perpendicular AB and OF perpendicular CD such that
OE
To prove:AB>CD

Construction :join OA and OC

PROOF we know that the perpendicular from the centre of a circle to a chord bisects the chord

AE=1/2 AB
And, CF=1/2 CD
Also, OA=OC=r. [2 nd]
And, OEOF>OE
OF SQUARE >OE SQUARE[3 rd]
now from the right triangle OEAand OFC we have;
OA square =OE square +AE square and OC square =OF square +CF square
OE square +AE square =OF square +OF square [OA=OC=r]
OA square +AE square =OF SQUARE +CF SQUARE >OE SQUARE +CF SQUARE [USING 3]
AE SQUARE >CD SQUARE
AB>CD
Hence, AB>CD