Given two pythagorean triples, generate another I don't know if this has been asked before, but I could not find any existing answer.
I noticed that for any pair of primitive pythagorean triples (not necessarily distinct), let's say:
a² + b² = c²
d² + e² = f²

Then there is at least another primitive triple where:
g² + h² = (cf)²

And there are 2 if the starting triples are distinct.
So, for example:
(3,4,5) and (5,12,13) -> (16, 63, 65) and (33, 56, 65)
(5,12,13) and (8,15,17) -> (21, 220, 221) and (140, 171, 221)
(3,4,5) (5,12,13) (8,15,17) -> (817,744,1105) (943,576,1105) (1073,264,1105) (1104,47,1105)
(3,4,5) and (3,4,5) -> (7,24,25)

I think there is an explanation for that, a property of pythagorean triples, or in general of diophantine equations.
Is it true in every case? Is there a way to calculate the two legs of the resulting triple(s)?

You can always find new triplets this way, although they are not necessarily primitive. This is just the Brahmagupta–Fibonacci identity in action.
For example, for $(5,12,13)$ and $(8,15,17)$, the identity states that:
$$(5\cdot8 - 12\cdot15)^2 + (5 \cdot 15 + 12 \cdot 8)^2 = (-140)^2 + (171)^2$$
$$= (5\cdot8 + 12\cdot15)^2 + (5 \cdot 15 - 12 \cdot 8)^2 = (220)^2 + (-21)^2$$
$$= (5^2 + 12^2) (8^2 + 15^2) = 13^2 \cdot 17^2 = 221^2$$
When the starting triplets are the same, then $a=d, b=e$. Using the given identity, $(ad - be)^2 + (ae + bd)^2 = (a^2-b^2)^2 + (ab + ba)^2 = a^4+b^4-2a^2b^2+4a^2b^2$ $ = a^4 + b^4 + 2a^2b^2 = (a^2+b^2)^2 = (a^2+b^2)(d^2+e^2)$.
The other possibility is $(ad + be)^2 + (ae - bd)^2 = (a^2 + b^2)^2 + (ab - ba)^2 = (a^2 + b^2)^2$, where one of the numbers of the new triplet is $0$. This explains why there is only one non-trivial triplet in this case.
Here is a counterexample that they are not always primitive. With $(13,84,85)$ and $(16,63,65)$, $ad+be, ae-bd$ gives $(5500,525,5525)$ which are all divisible by $25$. A further conjecture could be made on if $c,f$ are both divisible by $k$, then whether the lowest common divisor would be $k^2$.
