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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. $\square$

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Let $AB = c, AC = b, BC = a$, and let $IT\perp AC $ at $I$. Then, $CI$ can be written as $$\begin{split}CI^2 &= CT^2 + IT^2 \\&= \left(\frac{a+b-c}{2}\right)^2+\left(\frac{b+c-a}{2}\right)^2\\&= \frac{2a^2+2b^2+2c^2-4ac}{4}\\&=\frac{(a-c)^2+b^2}{2} \\&=\frac{(BC-AB)^2+AC^2}{2}\end{split}$$

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  • $\begingroup$ i think $CI^2=CT^2+IT^2$ if you right, why $AT=IT$ ? $\endgroup$ – Word Shallow May 22 '17 at 3:53
  • $\begingroup$ please the second question : why $CT^2=(\frac{a+b-c}{2})^2$ $\endgroup$ – Word Shallow May 22 '17 at 4:03
  • $\begingroup$ Also for right triangles AT=IT is true $\endgroup$ – Lazy Lee May 22 '17 at 4:03
  • $\begingroup$ This is an important property of incenters: Let $D,E$ be the point on $AB,AC$ where $ID\perp AB, IE \perp AC$, then $AD = AE = \frac{b+c-a}{2}$. $\endgroup$ – Lazy Lee May 22 '17 at 4:11
  • $\begingroup$ IT_|_AC; IE | AC -> IT coincident IE $\endgroup$ – Word Shallow May 22 '17 at 4:20
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Let the circle about $B$ through $C$ meet the extension of $\overline{AB}$ at the point $C^\prime$. By symmetry and a little angle chasing in isosceles $\triangle CBC^\prime$, we find that $\triangle CIC^\prime$ is an isosceles right triangle. Consequently, $$|IC|^2 + |IC|^2 = |IC|^2 + |IC^\prime|^2 = |CC^\prime|^2 = |AC|^2 + |AC^\prime|^2 = |AC|^2 + ( |BC| - |AB| )^2$$

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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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