Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions. 
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$

Solutions in the answers.

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Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.

The identity is trivially satisfied if $\,a=b=0\,$ or $\,c=d=0\,$. Otherwise $\,a^2+b^2 \ne 0\,$ and $\,\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1\,$, so there exists an angle $\,\alpha\,$ such that $\,\frac{a}{\sqrt{a^2+b^2}} = \sin \alpha\,$, $\,\frac{b}{\sqrt{a^2+b^2}} = \cos \alpha\,$. Similarly, define $\,\gamma\,$ such that $\,\frac{c}{\sqrt{c^2+d^2}} = \sin \gamma\,$ and $\,\frac{d}{\sqrt{c^2+d^2}} = \cos \gamma\,$.
After dividing by $\,(a^2+b^2)(c^2+d^2) \ne 0\,$, the identity can then be written as:
$$
(\sin\alpha\sin\gamma \pm \cos\alpha\cos\gamma)^2 + (\sin\alpha\cos\gamma \mp \cos\alpha\sin\gamma)^2 = 1
\\ \iff\quad \cos^2 (\alpha \mp \gamma) + \sin^2 (\alpha \mp \gamma) = 1 \quad\quad
$$
