Deduce that the perpendicular from the centre of a circle to a chord of the circle bisect the chord.
Proof:
Consider a circle with centre \(O\) and a chord \(LM\).
Let \(ON\) be drawn perpendicular from the centre of a circle to the chord \(LM\).
Join the points \(OL\) and \(OM\).
Consider the \(\Delta OLN\) and \(\Delta OMN\).
Here, \(OL\) \(=\) [].
And, the side \(ON\) is to the triangle \(ONL\) and \(ONM\).
Also, \(ON \perp LM\) [].
So, \(\angle ONL = \angle ONM =\) \(^{\circ}\).
Therefore, \(\Delta OLN\) \(≅\) \(\Delta OMN\) []
This implies, \(NL =\) .
Therefore, perpendicular from the centre of a circle to a chord of the circle bisects the chord.
Hence, proved.