#### Question

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

#### Solution

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 Rt. ΔAOM,

`AO^2 = AM^2 + OM^2`

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

`r^2 = 1/4 AB^2 + OM^2` ………..(i)

Again in Rt. ΔONC,

`OC^2 = NC^2 + ON^2`

`⇒ 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.

Is there an error in this question or solution?

Solution Prove That, of Any Two Chords of a Circle, the Greater Chord is Nearer to the Centre. Concept: Arc and Chord Properties - If Two Chords Intersect Internally Or Externally Then the Product of the Lengths of the Segments Are Equal.