If \[ a + b \propto a - b \] Then, \[ a + b = k(a - b) \quad \text{[where \(k\) is a non-zero constant]} \] So, \[ \frac{a + b}{a - b} = k \] Squaring both sides: \[ \frac{(a + b)^2}{(a - b)^2} = k^2 \] Now, using the sum and difference of squares: \[ \frac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} = \frac{k^2 + 1}{k^2 - 1} \] [Using addition and subtraction process] \[ = \frac{2(a^2 + b^2)}{4ab} = \frac{k^2 + 1}{k^2 - 1} \] \[ \Rightarrow \frac{a^2 + b^2}{ab} = \frac{2(k^2 + 1)}{k^2 - 1} = \text{constant} \] \[ \therefore a^2 + b^2 \propto ab \quad \text{(Proved)} \]