I would like to prove this identity :
$$(a^2 + b^2 + c^2 + d^2)^2 = (a^2 + b^2 - c^2 - d^2)^2 + (2ac + 2bd)^2 + (2ad - 2bc)^2$$
Without developing each term but using some elements of geometry .
I explain :
The inequality is equivalent to :
$$\frac{(a^2 + b^2 + c^2 + d^2)^2}{(2ab + 2cd)^2} = \frac{(a^2 + b^2 - c^2 - d^2)^2}{(2ab + 2cd)^2} + (\frac{2ac+2bd}{2ab+2cd})^2 + \frac{(2ad - 2bc)^2}{(2ab + 2cd)^2}$$
But now there is an interesting fact :
For a cyclic quadrilateral with successive sides $a, b, c, d,$ semiperimeter s, and angle $A$ between sides $a$ and $b$, the trigonometric functions of $A$ are given by :
$$cos(A)=\frac{(a^2 + b^2 - c^2 - d^2)^2}{(2ab + 2cd)^2}$$
So we have :
$$\frac{(a^2 + b^2 + c^2 + d^2)^2}{(2ab + 2cd)^2} = cos(A)^2 + (\frac{2ac+2bd}{2ab+2cd})^2+ \frac{(2ab - 2dc)^2}{(2ab + 2dc)^2}$$
Furthermore the second term in the RHS is just the second theorem of Ptolemy.
But I can't explain what is the the geometric meaning of the others terms like :
$$\frac{(a^2 + b^2 + c^2 + d^2)^2}{(2ab + 2dc)^2}$$
And
$$ \frac{(2ab - 2dc)^2}{(2ab + 2dc)^2} $$
In this context so can someone help me ?
Thanks