Starting with $(165,52,173)$, we attempt to find the $p,q$ pair which generates this triple, i.e., $p^2-q^2=165,2pq=52,p^2+q^2=173$. Clearly, we have $2q^2=173-165=8\implies q=2$ and thus $p=13$.
The ancestor of this triple arises from $(|p-2q|,q)\text{ or }(q,|p-2q|)$, whichever places $p',q'$ in largest-to-smallest order. In this case, we have $p-2q=9$ and thus the ancestor pair is $(p',q')=(9,2)$ and therefore the ancestor triple is $(p'^2-q'^2,2p'q',p'^2+q'^2)=(77,36,85)$.
The views of these people only apply to a small number of triples and sometimes depend on which of $A or B$ is odd or even. Below are original functions gleaned from 8 million spreadsheet formulas.
These generate all triples where GCD(A,B,C) is the square of an odd number. $\mathbf {\text{This includes all primitives}}$ and excludes all non-odd-square multiples of primitives. In the following sample of sets of triplets ($Set_1, Set_2, Set_3, \text{ and }Set_{25}$), we can see a great variation in the differences between $A,B,C$. We can also see that $\mathbf {(C-B) \text{ is the }n^{th} \text{odd square}}$. In the example: $C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$.
$$\begin{array}{c|c|c|c|c|}
\text{$Set_n$}& \text{$Triplet_1$} & \text{$Triplet_2$} & \text{$Triplet_3$} & \text{$Triplet_4$}\\ \hline
\text{$Set_1$} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline
\text{$Set_2$} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline
\text{$Set_3$} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
\text{$Set_{25}$} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline
\end{array}$$
Note: These triples can be generated by a variation of Euclid's formula where $A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad$ but we will use a formula developed from observations of these sets because it is easier to see that $n$ is a set number and $k$ is a member number.
Theorem:
$$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k$$
Proof: Let $$A=(2n-1)^2+2(2n-1)k$$
Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively, and substituting $A$, we find that
$$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$
We can then show that
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$
The Plato family $C-B=1$ applies only to members of $Set_1$ as we can see in the sample above. The Pythagoras family $C-A=2$ applies only to the first members of all sets. I haven't heard of the Fermat family $| A-B |=1$ and it seems to apply at random as in these examples of $f(n,k)=A,B,C$.
$$f(1,1)=3,4,5$$
$$f(2,2)=21,20,29$$
$$f(4,5)=119,120,169$$
$$f(9,12)=697,696,985$$
$$f(21,29)=4059,4060,5741$$
$$f(50,70)=23661,23660,33461$$
I stopped checking at $Set_{50}$ and I don't see a pattern, do you? Do note that the first two families are nothing more than the members of $Set_1$: f(1,k) or the first members of all sets: f(n,1). I hope I have shown you there is much more to explore than the families mentioned.
If you wanted to explore other configurations, I have another set of equations for side $A$ even:$$A=2n^2+2{N}n+4kn-4n\text{, }B=2n(2k-1)+(2k-1)^2\text{, }C=2n^2+2n(2k-1)+(2k-1)^2$$ and it generates the same triplets rotated $90$ degrees in the example with $A$ and $B$ reversed.
Update: I have recently inferred, but not proven, that triples where $GCD(A,B,C)\ne 2$ and $GCD(A,B,C)\ne x^2,x\in\mathbb{N}$ cannot be generated by Euclid's formula $(A=m^2-n^2\quad B=2mn\quad C=M^2+n^2)$ alone without a multiplier, as shown here.
Best Answer
I'd say the following is one of the oldest and real-lifiest applicatios: In old Egypt, it was necessary after each Nile flood to reestablish land measurements. For this a quick construction of right angles was needed. You can construct a right angle by partitioning a rope into twelve equal parts and then spanning a triangle with side lengths 3parts, 4 parts, 5 parts.