Prove $(a^2+b^2)(c^2+d^2)= 2ac + 2bd$

Question

I arrived at this question on reading about another issue, here.

I am unable to prove the above, even though attempt geometrical proof below:
Take $a,b,c,d$ as magnitudes of quantities (lengths). As no product term of $ab, cd$ is given so assuming right angle triangles (with right angle between sides of lengths $a,b$ & $c,d$), then $(a^2+b^2)$ leads to square of hypotenuse, let $h_1 (=\sqrt{a^2+b^2})$, similarly $(c^2+d^2)$ too, i.e. $h_2^2$. Their multiplication is $h_1^2h_2^2$.

The r.h.s. gives twice the product of sides of different triangles, i.e. $(a,c)$ & $(b,d)$. This algebraically means that the two triangles' sides should be multiplied. This in the simplest case is of a rectangle with sides of lengths $a,b,c,d$ with length $a=c, b=d$. This leads to r.h.s. as $2(a^2+b^2)$, so it leads to twice the sum of two hypotenuse's (let, $h_a, h_b$) square, i.e. $2(h_a^2+h_b^2)$. It satisfies for rectangle with side lengths $a=c, b=d$ & also for square with $a=b=c=d$. But, except for this simplest case, it makes difficult to interpret something useful from multiplying the different triangles sides.

I request proofs using any / all of the 3 criteria:
(i) geometrical,
(ii) algebraic,
(iii) complex number based approach

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Update : Sorry, for faulty question, as one comment to post has shown.

Answer 1

This

$$(a^2+b^2)(c^2+d^2)= 2ac + 2bd$$

is not true for all values of $a,b,c,d.$

As a counter example let $a=2,b=3,c=4,d=5$

We get $(13)(41)=16+30$ which is not true.

Answer 2

I'm sorry, but the claim is not true. Let $a = c = 1$ and $b = d = 2$. Then $(1+4)^2 \neq 2(1+4)$.

On the bright side, this explains why you were having trouble proving the claim.