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

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.

Draw ON ⊥ CD from O.

Then ΔONP is a right angled triangle.

⇒Its hypotenuse OP is larger than ON.

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

⇒ CD > AB

∴ AB is the smallest of all chords passing through P.