If (x+y) `prop`(x-y), then prove that `(x^(2)+y^(2)) prop` xy .

Explanation

(x+y) `prop` (x-y) `rArr x+y =k (x-y)`
( where k `ne ` 0= variation constant )
`rArr (x+y)^(2) =k^(2) (x-y)^(2) ` (squaring both the sides )
`rArrx^(2) +2xy +y^(2) =k^(2) (x^(2)-2xy+y^(2))`
`rArrx^(2) +y^(2) =k^(2) (x^(2)+y^(2)) -2k^(2)xy-2xy`
`rArr(x^(2)+y^(2)) -k^(2)(x^(2)+y^(2))=-2k^(2)xy-2xy`
`rArrx^(2)+y^(2) -k^(2)(x^(2)+y^(2))=-xy(2k^(2)+2)`
`rArr (x^(2) +y^(2))(1-k^(2))=-(2k^(2)+2) xy `
`rArr (x^(2) +Y^(2)) (k^(2)-1)=(2k^(2)+2)xy`
`rArr x^(2) + y^(2) =((2k^(2)+2)xy)/(k^(2)-1)`
`rArr x^(2) + y^(2) =(2k^(2)+2)/(k^(2)-1).xy`
`rArr x^(2)+y^(2) =mxy.(where m=(2k^(2)+2)/(k^(2)-1)ne0)`
`rArrx^(2)+y^(2) propxy [because mne0="variation constant"]`
`therefore(x^(2)+y^(2)) propxy. (proved )`