Of any two chords of a circle show that one which is larger is nearer to the centre.

Explanation

Given: `AB` and `CD` are two chords of a circle `C(O,r)`. `OPTo prove: `CD>AB`
Proof:
`AQ=1/2AB` (Perpendicular from the centre of a circle to a chord bisects the cord)
`CP=1/2CD` (Perpendicular from the centre of a circle to a chord bisects the cord)
In right `ΔAOQ`,
`⇒ AO^2=OQ^2+AQ^2`
`⇒AQ^2=AO^2-OQ^2` ....(1)
In right `ΔCOP`,
`⇒CO^2=CP^2+OP^2`
`⇒CP^2=CO^2-OP^2` ....(2)
`⇒OP`⇒OP^2`⇒-OP^2<-OQ^2`
`⇒CO^2-OP^2`⇒CP^2>AQ^2` (from (1) and (2))
`⇒1/4CD^2> 1/4AB^2`
`⇒CD^2>AB^2`
`⇒CD>AB`
Hence Proved.