A proposer remark: In fact this is a lemma of a much harder and much weirder problem. One version is the following: Let $A,B,C$ be three lattice points such that the incenter and three excenters of $\triangle ABC$ are also lattice points. If the lengths of $AB,AC,BC$ are $x+y,x+z,y+z$, respectively, for some positive integers $x,y,z$ such that \[\gcd(x+y, x+z, y+z)=\gcd(y,z)=1\]and \[x\mid yz,\]show that $\angle BAC = 90^{\circ}$. I tried to find necessary and sufficient conditions for $x,y,z$ so that there is a lattice triangle with the prescribed side length and its incenter also a lattice point. I didn't find any good description of those $x,y,z$, but I think it is certainly an interesting problem to think about.

Let $p=prime$ such that $p^a||x$ then $p^a|yz(y+z)$ using that $(x,y,z)=1$ we have that $p|$ only one of the numbers $y,z,y+z$.And so $p^a|$ only one of the numbers $y,z,y+z$ Let $p=prime$ such that $p^a||x+y+z$ then $p^a|xyz$ using that $(x,y,z)=1$ we have that $p|$ only one of the numbers $x,y,z$.And so $p^a|$ only one of the numbers $x,y,z$ Now if we take a prime number $p$ such that $p|xyz(x+y+z)$we consider two cases: If $U_p(x)=max(U_p(x),U_p(y),U_p(z),U_p(x+y+z))$ and let$U_p(x)=a$ then from the first we have that $p^a|$ only one of the numbers $y,z,y+z$ while $p$ doesn't dievise the other two so we have that $U_p(xyz(x+y+z))=2a$ If $U_p(x+y+z)=max(U_p(x),U_p(y),U_p(z),U_p(x+y+z))$ as before we have that $U_p(xyz(x+y+z))=2U_p(x+y+z)$. As every prime how devise $xyz(x+y+z)$ is at an even power we have that $xyz(x+y+z)$ is a perfect square.

Let $x=x_1d_{xy}d_{xz}$ where $d_{xy}=(x,y)$ and $d(x,z)=(x,z)$. Write $y, z$ similarly. $xyz=(d_{xy}d_{yz}d_{zx})^2x_1y_1z_1(x+y+z)$. Let $p$ be a prime number. Note that $p|x+y+z \implies p$ cannot divide $d_{xy}d_{yz}d_{zx} \implies p|x_1y_1z_1$. Thus if we prove that for any prime $p|x_1, v_p(x_1)=v_p(x+y+z)$ we are done (this will be also true for $y_1$ and $z_1$ by symmetry). $v_p(d_{xy})v_p(d_{xz})=0$. Assume that $v_p(d_{xz})=0$. Then $p$ cannot divide $y_1z_1d_{yz}d_{zx}$. Observe that first condition gives $v_p(x+y+z) \ge v_p(x_1) \implies v_p(d_{xy})=0$ otherwise $p|z$. Also last one gives $v_p(x+y+z) \le v_p(x_1)$. Therefore $v_p(x_1)=v_p(x+y+z)$ as desired.

By the fourth condition, we only need to prove that all the prime divisors of $x,y,z$ appear an even number of times in $xyz(x+y+z)$. By the gcd assumption, there are only types of such primes: those dividing exactly one of $x,y,z$ and those dividing exactly two of them. First consider $p \mid x,y$ (w.l.o.g). So $p^a \| x$ and $p^b \| y$ and $p$ divides neither $z$ nor $x+y+z$. But then the first condition implies $a \le b$ and the second implies $b \le a$ and hence $a=b$ so that indeed $p$ appears an even number of times in $xyz(x+y+z)$. Now consider $p \mid x$ (w.l.o.g), but $p \nmid y,z$. So $p^a \| x$ and $p^b \| x+y+z$. But now the first condition implies $a \le b$ and the fourth implies $b \le a$ and hence again $a=b$ so that indeed $p$ appears an even number of times in $xyz(x+y+z)$. The End.

Let's assume gcd(x,y)=a , gcd(x,z)=b , gcd(y,z)=c , Because gcd(x,y,z)=1 It can be seen that a,b,c are pairwise coprime too. Let' say x=a.b.g , y=a.c.h and z=b.c.j Lemma1 : gcd(g,c) , gcd(h,b) and gcd(j,a) are all 1. Proof: Because these are all symetric we are going to prove only gcd (g,c)=1 and rest is same . Assume not if g and c have a common prime divisor p, hence (x,z)=b and (b,c)=1 , this divisor p doesn't divide b, But divides both x and z .And b was greatest common divisors but is not divisible by p, contradiction. Lemma2: g,h and j are pairwise coprime. If not then one of the a,b or c must be higher but they were the greatest common divisor. From question statement x=a.b.g divides y.z(x+y+z) , y.z=a.b.(c^2).h.j , since (g,c)=(g,j)=1 , g divides x+y+z , by doing similar applications you can find h divides x+y+z and j divides x+y+z. Which makes g.h.j divides x+y+z while x+y+z divides xyz. Say x+y+z=g.h.j.k , if k is not 1.Then k is divisible a common divisor of abc. But since x+y+z is coprime with a,b and c ,contradiction. Which makes k=1.And now xyz(x+y+z)= (abc)^2.(ghj)^2.

Let $p$ be an arbitrary prime divisor of $xyz$, and let $v_p(x)$ be the $p$-adic valuation of $x$. Note that $p$ cannot divide all three of $x,y,z$ since $\gcd(x,y,z)=1$. Now it suffices to show that $v_p(xyz(x+y+z))$ is even. Case 1: Suppose $p$ divides exactly one of $x,y,z$. WLOG let it be $x$. Then: $$v_p(x)\le v_p(yz(x+y+z))=v_p(yz)+v_p(x+y+z)=v_p(x+y+z)\le v_p(xyz)=v_p(yz)+v_p(x)=v_p(x)$$so equality holds. Therefore, $v_p(x+y+z)=v_p(xyz)=v_p(x)$, so: $$v_p(xyz(x+y+z))=v_p(x)+v_p(x+y+z)+v_p(yz)=v_p(x)+v_p(xyz)+v_p(yz)=2v_p(x)+2v_p(yz)$$which is even. Case 2: Suppose $p$ divides exactly two of $x,y,z$. WLOG let they be $x,y$ with $v_p(x)\le v_p(y)$. Then since $p$ divides $x+y$ and $p$ doesn't divide $z$: $$v_p(y)\le v_p(xz(x+y+z))=v_p(x)+v_p(z(x+y+z))=v_p(x)$$which implies $v_p(x)=v_p(y)$. Then: $$v_p(xyz(x+y+z))=v_p(x)+v_p(y)+v_p(z(x+y+z))=v_p(x)+v_p(y)=2v_p(x)$$which is even.

Suppose, $p, q, r$ are primes which divide $x, y$ and $z$ respectively. Then $pqr|x+y+z,$ and so $(pqr)^2 |xyz(x+y+z).$ Suppose, $m$ is a prime factor of $x+y+z,$ then $m|xyz$ and so $m^2|xyz(x+y+z)$

We want to show that $\nu_p(xyz(x+y+z))$ is even for all primes $p$. If $p\nmid xyz$, the value is $0$, so assume $p$ divides $xyz$. Suppose $\nu_p(xyz(x+y+z))$ was odd for some prime $p\mid xyz$. Let\[\nu_p(x) = a, \nu_p(y) = b, \nu_p(z) = c, \nu_p(x+y+z) = N\] Then $a+b+c + N$ is odd. The divisibilities become\begin{align*} a \le b + c + N \\ b \le a + c + N \\ c\le a + b + N \\ N \le a + b + c \\ \end{align*} We may WLOG that $a\ge b\ge c$, in which case we have $c = 0$ because $\gcd(x,y,z) = 1$. Therefore, we get $ a\le b + N$, $N\le a+b$, and $a+b+N$ is odd. If $b=0$, then $a\le N$ and $N\le a$, so $a=N$, which implies $a + b + N = 2N$ is even, contradiction. If $b>0$, the idea is that $p$ divides $x+y$, and not $z$, so $N = \nu_p(x+y+z) = 0$. In this case, we have $a\le b$, so $a = b$, which means $a+b+N = 2a$ is even, contradiction. Therefore $xyz(x+y+z)$ must be a perfect square.

you can just show the the sum of 2 of them is equal to the third then evaluate $\nu_p(x+y+z)xyz$ which is clearly even

$ $