<s>[INST] The distance from the incenter to an acute vertex of a right triangle I'm seeking an alternative proof of this result:

Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that 
  $$|CI|^2=\frac12\left(\left(\;|BC|-|AB|\;\right)^2+|AC|^2\right)$$

My Proof. Draw $ID \perp AB$, $IE\perp BC$, and $IF\perp CE$.
We have $|ID|=|IE|=|IF|=x$. Since $\triangle ADI$ is right isosceles triangle, we also have that $|AD|=|ID|=x$. Respectively, we have: $$|ID|=|IF|=|IE|=|AD|=|AF|=x$$ 
$\triangle BDI=\triangle BEI \Rightarrow |BD|=|BE|=y$. And $|CE|=|CF|=z$
We have:
$$|CI|^2=|CE|^2+|IE|^2=x^2+z^2 \tag{1}$$
And 
$$\begin{align}\frac12\left(\left(|BC|-|AB|\right)^2+|AC|^2\right) &=\frac12\left(\left(\;\left(y+z\right)-\left(x+y\right)\;\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac12\left(\left(x-z\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac22\left(x^2+z^2\right) \\[4pt]
&=x^2+z^2
\tag{2}\end{align}$$
From $(1);(2)$ we are done. $\square$
 [/INST] Alternative proof: by Stewart's theorem the length $\ell_c$ of the angle bisector through $C$ is given by
$$ \ell_c^2 = \frac{ab}{(a+b)^2}\left[(a+b)^2-c^2\right] $$
and by Van Obel's theorem and the bisector theorem $\frac{CI}{\ell_c}=\frac{a+b}{a+b+c}$. It follows that, in general:
$$ CI^2 = \frac{ab}{(a+b+c)^2}\left[(a+b)^2-c^2\right]=\frac{ab(a+b-c)}{(a+b+c)} $$
and it is enough to prove that the given hypothesis ($a^2=b^2+c^2$) ensure
$$ \frac{ab(a+b-c)}{(a+b+c)} = \frac{(a-c)^2+b^2}{2} $$
or 
$$ 2ab(a+b-c) = (a+b+c)(a^2+b^2+c^2-2ac). $$
That is trivial since the difference of the RHS and the LHS of the last expression is
$$ (b+c-a)(a^2-b^2-c^2). $$
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