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Prove that all the chords of a circle th...

Prove that all the chords of a circle through a given point within it, the least is one which is bisected at that point.

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Given A circle C(O,r) and a point M within it, AB is a chord with midpoint M and CD is another chrod through M.
To Prov `AB lt CD`.
Construction Join OM and draw `ON _|_ CD`.
Proof In the right `Delta ONM ,` OM is the hypotenus.
`:. ON lt OM`
`implies` chord CD is nearer to O than chord AB.
We know that of any two chords of a circle, the one which is nearer to the centre is longer.
`:. CD gt AB`
Hence, `AB lt CD`
Thus, of all chords through M, the shortest is the one which is bisected at M.
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