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Prove that the perpendicular bisector of...

Prove that the perpendicular bisector of a chord of a circle always passes through the centre.

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Let DE be the perpendicular bisector of the given chord AB of a circle C(O,r ) .
Then , AD `=` DB and `/_ ADE =90^(@)` ….(i)
Now,we have to show that DE passes through O.
If possible, suppose DE does not pass through O. Join OD
We know that the line joining the centre of a circle to the midpoint of a chord is always perpendicular to the chord.
`:. OD _|_ AB implies /_ADO =/__ 90^(@)` ...(ii)
From (i) and (ii) , we get `/_ADE =/_ADO`.
This is a contradiction , since `/_ADO` is a part of `/_ ADE`.
The contradiction arises by assuming that DE not passthrough O.
Hence, DE must pass through O.
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