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Prove that the perpendicular bisector \(ON\) of a chord \(LM\) passes through the centre of the circle \(O\).
Proof:
Consider a circle with centre \(O\) and a chord \(LM\).
Let \(ON\) be the perpendicular bisector from the centre of a circle \(O\) meeting the chord \(LM\) at the midpoint \(N\).
Join the points \(OL\) and \(OM\).
Consider the \(\Delta OLN\) and \(\Delta OMN\).
Here, \(OL\) \(=\) [].
Also, \(N\) is the midpoint of the chord, \(NL =\) .
And, the side \(ON\) is to the triangle \(ONL\) and \(ONM\).
Therefore, \(\Delta OLN\) \(≅\) \(\Delta OMN\) [].
This implies, \(\angle ONL = \angle ONM\). []
It is observed that \(\angle ONL\) and \(\angle ONM\) are .
So, \(\angle ONL\) \(+\) \(\angle ONM\) \(=\) \(^{\circ}\)
This implies, \(\angle ONL = \angle ONM =\) \(^{\circ}\).
We say that , \(ON\) \(LM\).
Here, the line through \(N\) perpendicular to the chord \(LM\) contains the centre \(O\).
Therefore, perpendicular bisector of a chord passes through the centre of the circle.
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