If x + y ∝ x – y, let us show that

x² + y²∝ xy

Given: x + y ∝ x – y

It can be written as,

x + y = k(x – y)

Square on both sides,

⇒ x² + y² + 2xy = k(x² + y² + 2xy)

⇒ x² + y² + 2xy = kx² + ky² + k2xy

⇒ x² - kx² + y²- ky² = k2xy - 2xy

⇒ x² (1-k) + y² (1-k) = 2xy (k-1)

⇒ (1-k) {x² + y²} = -2xy(1-k)

⇒ x² + y² = -2xy

As -2 is also a constant term.

Hence,

⇒ x² + y²∝ xy

Hence proved.