Prove that all the chords of a circle through a given point within it, the least is one which is bisected at that point.

Explanation

Let P be the given point inside a circle with center 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.

`therefore` `triangleONP` is a right triangle.

Now, its hypotenuse OP is larger than ON.

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

Hence, CD>AB

In other words, AB is the smallest of all chords passing through P.