Question

Prove that out of all the chords which passing through any point of circle, that chord will be smallest which is perpendicular on diameter which passes through that point.

Answer 1

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 right triangle.

Therefore, its hypotenuse $OP$ is larger than $ON$.

We know that the chord nearer to the centre is larger than the chord which is farther from the centre.

Therefore, $CD>AB$

In other words, $AB$ is the smallest of all chords passing through $P$ and perpendicular to the diameter $XY$.
Hence, proved.