Prove that among all the chords of a circle passing through a given point inside the circle,the one that is perpendicular to the diameter passing through the point is the smallest.

(N/A) Let $P$ be the given point inside a circle with centre $O$. Draw the chord $AB$ which is perpendicular to the diameter $XY$ passing through $P$. Let $CD$ be any other chord passing through $P$. Draw $ON$ perpendicular to $CD$ from $O$.
In $\Delta ONP$,$\angle ONP = 90^{\circ}$. Since $OP$ is the hypotenuse,$OP > ON$.
We know that the chord nearer to the centre is larger than the chord which is farther from the centre.
Since $ON < OP$ (where $OP$ is the distance of chord $AB$ from the centre),the chord $CD$ is farther from the centre than the chord $AB$.
Therefore,$CD > AB$.
In other words,$AB$ is the smallest of all chords passing through $P$.