The answer is $$\boxed{(a,b,c,d)=(1,k,1,k)\quad k\in \mathbb{N}^+}$$

Probably something with $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 - (ad-bc)^2$ Thus we also need $ac+bd|ad-bc$ then maybe something with $|ac+bd| \le |ad - bc|$

Note that $(ac+bd)^2 + (ad-bc)^2 = (a^2+b^2)^2$ So $\exists r,s$ with $gcd(r,s)=1$ so that $ac+bd = g(r^2-s^2)$ ; $ad-bc = 2grs $ ; $a^2 +b^2 = g(r^2+s^2)$ .Now note that $r^2-s^2|r^2+s^2$ and as $r^2-s^2$ is odd so $r^2-s^2 | s^2$ giving $r^2-s^2 = 1$ so $r =1 ; s = 0 $ Now $ac+bd = g = a^2+b^2 \implies (a-c)^2 + (b-d)^2 = 0 $ $a=c , b=d$ Hence done !! @2below fixed

$a^2 + b^2 = c^2 + d^2=: m^2$ so we build a circumference with diameter $m$, such that quadrilateral abcd is inscribed in it. By strong Ptolemy $$m^2 = (ac+bd) (ab+cd)/(ad+bc) = a^2+b^2 $$Since $ ac+bd | a^2+b^2$, $(ab+cd)/(ad+bc)$ is an integer. WLOG, $ a\leq b,c,d$, then $ a\leq b,c\leq d$, so that the integer in consideration is between 0 and 1. Therefore after some manipulation $a=c, b=d$.

Another useful approach below. One can WLOG assume $a\geq c$ together with $d\geq b$, so the next equality is true: $$(a-c)(a+c)=(d-b)(d+b).$$That means there are nonnegative integers $p,q,r,s$ such that $$a-c=pq,\,\,a+c=rs,\,\,d-b=pr,\,\,d+b=qs,$$or $$2a=pq+rs,\,\,2c=rs-pq,\,\,2d=pr+qs,\,\,2b=qs-pr.$$One can rewrite divisibility condition as $$r^2s^2-p^2q^2+q^2s^2-p^2r^2|r^2s^2+p^2q^2+q^2s^2+p^2r^2,$$or $$(s^2-p^2)(q^2+r^2)|(p^2+s^2)(q^2+r^2).$$Hence positive integer $s^2-p^2$ divides $p^2+s^2$. Due to @chakrabortyahan's solution it is possible iff $s=0$, so $a=b$ and $b=d$.

chakrabortyahan wrote: Note that $(ac+bd)^2 + (ad-bc)^2 = (a^2+b^2)^2$ So $\exists r,s$ with $gcd(r,s)=1$ so that $ac+bd = g(r^2-s^2)$ ; $ad-bc = 2rs $ ; $a^2 +b^2 = g(r^2+s^2)$ .Now note that $r^2-s^2|r^2+s^2$ and as $r^2-s^2$ is odd so $r^2-s^2 | s^2$ giving $r^2-s^2 = 1$ so $r =1 ; s = 0 $ Now $ac+bd = g = a^2+b^2 \implies (a-c)^2 + (b-d)^2 = 0 $ $a=c , b=d$ Hence done !! It seems that $ad-bc=2grs$ .

A solution using vectors: Consider vectors $(a, b)$ and $(c, d)$. The first constraint is that they have equal modulus, call it $m$. The second is that the dot product divides $m^2$, or equivalently that the cosine of the angle they form is the inverse of a natural number. By taking the cross product (adding a third dimension) we get that the sine is rational similarly. Now since the sum of the squares of the sine and the cosine is $1$, we can easily get that the only possibility is that the cosine is $1$ and the sine is $0$. Using the fact that both vectors lie on the first quadrant, the only possibility is that both are equal. That gives only solutions of the form $(a, b, a, b)$.