Correct Option: D

Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
⇒ chord CD is nearer to O in comparison to AB.
⇒ CD > AB
⇒ AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .