find an equation satisfied by the three bisectors of a right triangle let $a,b,c$ be the sides of a generic triangle and $\alpha, \beta, \gamma$ the bisectors of the respective opposite angle.
It is well know that, for example, $\alpha$ divides the side $a$ in two segments, $x$ and $a-x$, and that we have $(a-x)c=bx$, in addition $\alpha$ satisfies the equation $\alpha^2=cb-x(a-x)$, thus we can eliminate $x$ obtaining $(b^2+2bc+c^2)\alpha^2=-a^2bc+b^3c+2b^2c^2+bc^3$.
Hence I have three equations involving only the sides and the bisectors, assuming that such triangle is right-angled (let $a$ be the hypotenuse), can we find other relations in order to (computationally) eliminate $a, b, c$ and get an equation in the only $\alpha, \beta, \gamma$? My attempts have not been successful so far.
 A: COMMENT.-Basically this problem can be posted as follows: Let $$f(x,y,z)=\dfrac{yz[(y+z)^2-x^2]}{(y+z)^2}$$ and given
$$\alpha^2=f(a,b,c)\\\beta^2=f(b,a,c)\\\gamma^2=f(c,a,b)$$ find out a function $F(\alpha,\beta,\gamma)=0$ in which $a,b,c$ does not appear having the condition $a^2=b^2+c^2$. In other words it is a problem of elimination for which it is maybe useful using parameterization $$a=t^2+s^2\\b=t^2-s^2\\c=2ts$$ so we have to get rid only two variables.
NOTE.-Pythagorean triples come from an identity so it is valid also for irrationals (and even complex numbers and in a commutative ring).
A: (Too long for a comment.)
In the case of a right triangle with $\,b^2+c^2=a^2\,$ the bisector formulas simplify to:

*

*$\require{cancel}(b+c)^2\alpha^2 = bc\left((b+c)^2-a^2\right)=bc\left(\cancel{b^2+c^2}+2bc-\cancel{a^2}\right)=2b^2c^2\,$;


*$(c+a)^2\beta^2=ca\left(c^2+2ac+\underbrace{a^2-b^2}_{=\,c^2}\right)=2ac^2(c+a) \implies (c+a)\beta^2=2ac^2\,$.
This gives the system of equations:
$$
\begin{cases}
\begin{align}
(b+c)^2\alpha^2 - 2b^2c^2 &= 0
\\ (c+a)\beta^2 - 2ac^2 &= 0
\\ (a+b)\gamma^2 - 2ab^2 &= 0
\end{align}
\end{cases}
$$
Together with $\,b^2 + c^2 = a^2\,$, these are $\,4\,$ algebraic equations, between which it is technically possible to eliminate $\,a,b,c\,$ using polynomial resultants in order to get a relation in  $\,\alpha,\beta,\gamma\,$ alone.
Variable $\,a\,$ can be eliminated by hand from the last two equations, by isolating the terms in $\,a\,$, squaring, then substituting $\,a^2=b^2+c^2\,$:
$$
\begin{align}
(c+a)\beta^2 - 2ac^2 = 0 \;\;\implies\;\; a(2c^2-\beta^2)=c\beta^2 \;\;\implies\;\; (b^2+c^2)(2c^2-\beta^2)^2 = c^2\beta^4
\end{align}
$$
This leaves three equations with $\,b,c\,$ remaining to be eliminated between them:
$$
\begin{cases}
\begin{align}
p(b,c) &= (b+c)^2\alpha^2 - 2b^2c^2 &= 0
\\ q(b,c) &= (b^2+c^2)(2c^2-\beta^2)^2 - c^2\beta^4 &= 0
\\ r(b,c) &=(b^2+c^2)(2b^2-\gamma^2)^2 - b^2\gamma^4 &= 0
\end{align}
\end{cases}
$$
The final relation between $\,\alpha,\beta,\gamma\,$ can be derived as $\,\text{res}_b\big(\text{res}_c(p,q), \text{res}_c(p,r)\big)=0\,$, which WA calculates to be a polynomial of total degree $\,120\,$.
