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Of any two chords of a circle show that ...

Of any two chords of a circle show that the one which is nearer to the centre is larger.

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Given : Two chords AB and CD of a circle with centre O such that `ABgtCD`
To Prove : `OMlt ON`.
Construction : Join OA and OC
Proof: Since `ABgtCD`
`rArr(1)/(2) ABgt(1)/(2)CD`
`rArrAMgt CN` (because Perpendicular drawn from centre, to the chord bisects the chord)
`rArrAM^2gtCN^2` (on squaring both sides)
`rArr-AM62larrCN^2` (if we change the sign both sides then the sign of inequality also changes)
`rArr r^2-AM^2ltr^2-CN^2 (adding r^2 on both sides)`
`rArrOA^2-AM^2ltOC^2-CN^2 (because OA=OC=r)`
`rArrOM^2ltON^2` (by Pythagoras theorem)
`rArrOMltON` (taking square root on both sides).
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