Which is the smallest of all the chords of a circle passing through a given point in it?
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
⇒ Smallest one is the one bisected at the point.