Can one find explicitly $a_i,b_i,c_i\in\Bbb N,i=1,2,3,4$ so that $$ a_i<b_i, \qquad \text{ and } \qquad a_i^2+b_i^2=c^2_i \qquad\text{for } i=1,2,3,4$$ and $$a_1b_1=a_2b_2=a_3b_3=a_4b_4, \qquad c_1<c_2<c_3<c_4.$$
Some context:
A Pythagorean triple is a triple $(a,b,c)\in\Bbb N$ so that $a^2+b^2=c^2$. We say that $(a,b,c)$ is primitive if $a,b$ and $c$ are coprime. In the dedicated wikipedia article the following is written:
$\big((20, 21, 29), (12, 35, 37)\big)$ is the first pair of primitive Pythagorean triples such that the induced triangles have same area $=210$.
$\big( (4485, 5852, 7373), (3059, 8580, 9109), (1380, 19019, 19069)\big)$ is a triple of primitive Pythagorean triples such that the induced triangles have same area $=13123110$.
For each natural number $n$, there exist $n$ Pythagorean triples with different hypotenuses and the same area.
An equivalent formulation of the question above is: Is there any explicitly known quadruple of Pythagorean triples such that the induced triangles have same area?
Note: A093536 claims that for such a quadruple, the area in question will be $\geq 10^{17}$.
Note: These problem are equivalent for the following reasons: It follows from the basic fact that the area of a triangle associated to a Pythagorean triple $(a,b,c)$ is given by $ab/2$ and $2n^2=k^2$ has only $(0,0)$ as integer solution. BTW note that $a_i,b_i,c_i$ are coprime if and only if $\gcd(a_i,\gcd(b_i,c_i))=1$ for $i=1,\ldots,4$.