Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$ 
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$


*

*I managed to get $ab=cd$. Don't know how to proceed further.


Alternative approach:
From the constraints, 
$(a + b)^2 = (c + d)^2 \implies$ 
[since $a^2 + b^2 = c^2 + d^2$] 
$2ab = 2cd \implies ab = cd \implies$ 
$a(c + d - a) = cd$.
This allows a quadratic equation in $a$ to be formed, that may be solved in terms of $c,d$.
$a^2 - (c+d)a + cd = 0.$
Therefore, 
$\displaystyle a = \frac{1}{2}\left[
(c + d) \pm \sqrt{(c + d)^2 - 4cd}\right].$
This equals 
$$\frac{1}{2}\left[
(c + d) \pm \sqrt{(c - d)^2}\right].\tag 1$$
Due to the symmetry in the expression in (1) above, you have that without loss of generality, $c \geq d$.
Therefore $\displaystyle a = \frac{1}{2}\left[
(c + d) \pm (c - d)\right].$
The two solutions generated will be $a = c$ or $a = d$.
