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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


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

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    $\begingroup$ Try $a=1,b=0,c=0,d=1$ $\endgroup$ – saulspatz Jul 4 '18 at 21:53
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    $\begingroup$ What are $a,b,c,$ and $d$? Are they related in some way that you haven't told us? Why would you assume there are right triangles involved, and why should we? $\endgroup$ – Cameron Buie Jul 4 '18 at 21:53
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    $\begingroup$ Maybe the objective is to find a criterion that makes this true? $\endgroup$ – Mason Jul 4 '18 at 22:03
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    $\begingroup$ Well on the linked page it has written $a^2+b^2=c^2+d^2=2ac+2bd$ in a few spots... But you have written a product above. For the product we use this Fermat's Identity: $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. $\endgroup$ – Mason Jul 4 '18 at 22:09
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    $\begingroup$ Let's delete this question and we'll meet you on math.stackexchange.com/questions/105330/… If you post a question in the comments about how the argument goes through someone will surely pick up this discussion. $\endgroup$ – Mason Jul 4 '18 at 22:15
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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.

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    $\begingroup$ In mathematics an identity should be true for all values not just for some. Of course your equation is satisfied for lots of values but it does not make it an identity. $\endgroup$ – Mohammad Riazi-Kermani Jul 4 '18 at 22:13
  • $\begingroup$ @GuusPalmer I said it is not true for all values. I did not say it is not true for any value. $\endgroup$ – Mohammad Riazi-Kermani Jul 4 '18 at 22:15
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    $\begingroup$ @GuusPalmer "Prove" is used for identities and "Solve" is used for equations. You have used the word " Prove" instead of the intended "Solve." $\endgroup$ – Mohammad Riazi-Kermani Jul 4 '18 at 22:34
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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.

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    $\begingroup$ Your answer has typos: $(a^2+b^2) = 1 +4 =5 = (c^2+d^2)$, so $25 \ne 2(1+4)$. But, how it will aid in understanding my fault is unclear still. $\endgroup$ – jiten Jul 4 '18 at 22:01
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    $\begingroup$ Good catch. I've fixed my answer. Good luck! $\endgroup$ – HermitianCrustacean Jul 4 '18 at 22:39

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