Question

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Answer 1

Given: A circle with centre O and radius r. OM ⊥ AB and ON ⊥ CD Also AB > CD

To prove: OM < ON

Proof: Join OA and OC.

In right ΔAOM,

AO² = AM² + OM²

`=> r^2 = (1/2 AB)^2 + OM^2`

`=> r^2 = 1/4 AB^2 + OM^2`    ...(i)

Again in right ΔONC,

OC² = NC² + ON²

`=> r^2 = (1/2 CD )^2 + ON^2`

`=> r^2 = 1/4 CD^2 + ON^2`    ...(ii)

From (i) and (ii)

`1/4 AB^2 + OM^2 = 1/4  CD^2 + ON^2`

But, AB > CD   ...(Given)

∴ ON > OM

`=>` OM < ON

Hence, AB is nearer to the centre than CD.