Question

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

Answer 1

Solution :

Given -

A circle C(O, r) and a point M within it; AB is a chord with midpoint M and CD is another chord through M.

TO PROVE : AB < CD

CONSTRUCTION : Join OM and draw ON perpendicular to CD.

PROOF : In the right ∆ONM , OM is the hypotenuse.

Hence , ON < OM

➨ Chord CD is the nearer to O than Chord AB.

We know that , of any two chords of a circle , the one which is nearer to the centre is longer.

Hence , CD > AB.

And , AB > CD.

Thus ,

Of all the chords through M , the shortest of two equal chords of a circle subtends equal angles with the chords.

$\boxed{Hence\: proved\:!}$