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.

Simpler Solution.
\begin{align} & (ac+bd)^2+(ad-bc)^2 \\ = \; & (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2 \\ = \; & a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\ = \; & (a^2+b^2)(c^2+d^2) \end{align}
