Prove that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point.
Proof:
Let \(P\) be the given point inside a circle with centre \(O\).
Draw the chord \(AB\) which is perpendicular to the diameter \(XY\) through \(P\).
Let \(CD\) be any other chord through \(P\).
Draw \(ON\) perpendicular to \(CD\) from \(O\).
Then \(∆ONP\) is a .
Therefore, its \(OP\) is than \(ON\).
We know that, .
Thus, \(CD\) \(AB\).
In other words, \(AB\) is the of all chords passing through \(P\).
Hence, proved.