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