jmc

We might try sketching a diagram: Since $△ABC$ is isosceles, we might try extending $BI$ to meet $AC$ at $D.$ That is advantageous to us since it will also be the perpendicular bisector and median to side $AC.$ In addition, let us draw a radius from $I$ that meets $AB$ at $E.$ Given $r$ as the inradius, we can see that $DI=r$ and $IB=r\sqrt{2},$ since $△IEB$ is also a little isosceles right triangle on its own. Therefore, $BD=r\sqrt{2}+r=r(\sqrt{2}+1).$

However, we have a nice way of finding $BD,$ from $△ABD,$ which is also an isosceles right triangle, thus $DB=\frac{AB}{\sqrt{2}}=\frac{4\sqrt{2}}{\sqrt{2}}=4.$

Setting the two expressions for $DB$ equal, we have: $$\begin{array}{rl}r(\sqrt{2}+1)& =4\\ r& =\frac{4}{\sqrt{2}+1}=\frac{4}{\sqrt{2}+1}\cdot \frac{\sqrt{2}-1}{\sqrt{2}-1}\\ & =\frac{4(\sqrt{2}-1)}{1}=4\sqrt{2}-4.\end{array}$$ Our answer is $BI=r\sqrt{2}=(4\sqrt{2}-4)\cdot \sqrt{2}=\boxed{8-4\sqrt{2}}.$