Isosceles Triangle Incenter Problem Solving Question The triangle $\triangle ABC$ is an isosceles triangle where $AB = 4\sqrt{2}$ and $\angle B$ is a right angle. If $I$ is the incenter of $\triangle ABC,$ then what is $BI$?
Express your answer in the form $a + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect squares integers other than $1.$
Below is a picture of what I have done. I have found the answer, which I believe is correct, but i did not find it in the square root version. Could someone please help me find the square root version and how to get to the answer. Thanks a lot!
Edit: the property of an incenter is that it is equidistant from all of the sides.

 A: Note that the inradius of the triangle is $r=\dfrac{(2-\sqrt2)}{2}(4\sqrt2).$ 
The proof is simple: use the fact that $$\text{the area of the whole triangle}=\text{sum of 3 individual triangles}.$$
Then the line from point $I$ to $AB$ is equal in length to the line from point $I$ to $BC$. This is a square. So by Pythagorean theorem, $$r^2+r^2=(BI)^2\implies BI=\sqrt{2}\left[\dfrac{(2-\sqrt2)}{2}(4\sqrt2)\right]=8-4\sqrt2.$$
Addendum: I add the proof of the inradius here. Let the inradius be $r$. Observe $$\begin{align}\text{the area of the whole triangle}&=\text{sum of 3 individual triangles}\\\dfrac{(4\sqrt2)^2}{2}&=\dfrac{1}{2}(4\sqrt2r+4\sqrt2r+8r) \\r&=\dfrac{(4\sqrt2)^2}{2(4+4\sqrt2)}=4(\sqrt2-1)=\dfrac{(2-\sqrt2)}{2}(4\sqrt2).\end{align}$$
A: 
\begin{align} 
|AC|=|BB'|&=\sqrt2|AB|=8
,\\
|BI|=x&=\sqrt2 r
,\\
|BB'|&=2x+2r=2x+\sqrt2\sqrt2 r
=x(2+\sqrt2)
,\\
x&=\frac{8}{2+\sqrt2}
\\
&=\frac{8(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}
\\
&=\frac{8(2-\sqrt2)}{(2^2-(\sqrt2)^2)}
\\
&=8-4\sqrt2
.
\end{align}  
