Familiar Triangles

First observe that $\angle B=90^\circ$ because $\angle CIA=90+\frac{\angle B}{2}\Rightarrow 135=90+\frac{\angle B}{2}\Rightarrow \angle B=90^\circ.$
Let $D$ be a point on $AC$ such that $ID$ is $\perp$ to $AC.$ Let $R$ be the circum-radius and $r$ be the in-radius.
$AO=R, IO=\sqrt{R^2-2Rr}$ (Well known fact), $ID=r, AI=\sqrt{2Rr}.$
Now $\triangle ADI$ is similar $\triangle AIO.$
So, $\dfrac{ID}{IO}=\dfrac{AI}{AO}\Rightarrow \dfrac{r}{\sqrt{R^2-2Rr}}=\dfrac{\sqrt{2Rr}}{R}.$ After some algebraic manipulation, we get $r=\frac{2R}{5}.$
Now let $E$ be a point on $AB$ such that $IE$ is $\perp$ to $A.$
We get $AI=\frac{2R}{\sqrt{5}}, IE=r,$ so $AE=\frac{4R}{5}$ by applying Pythagoras Theorem in $\triangle AIE.$ Then $BE=r=\frac{2R}{5}.$ So, $AB=\frac{6R}{5}.$
We also know that $AC=2R.$ So we get $BC=\frac{8R}{5}$ by again applying Pythagoras Theorem in $\triangle ABC.$
Therefore $\dfrac{AB}{BC}=\boxed{\dfrac{3}{4}}$ $.$

$SOLUTION :$

Observe that $\angle AIC=90^\circ+\frac{\angle B}{2}\Rightarrow 135=90+\frac{\angle B}{2} \Rightarrow \angle B=90^\circ.$

Then draw the circumcircle of $\triangle ABC$ . Because $\angle B\ = 90^\circ$ , AC will be the diameter.

And in a right angled triangle The circumcentre will be the midpoint of the hypotenuse.

Extend AI to meet the circumcircle at D. Then ID=DC. Moreover, $\triangle ADC$ is similar to $\triangle AIO$ (Common angle and $90^\circ$ )

Thus by similarity we find that

$DC$ = $2IO$ and $AD$ = $2AI$

$AD$ = $2AI$ $\Rightarrow$ $DI$ = $AI$ .

$DC$ = $DI$ $=$ $AI$

$tan \frac{\angle A}{2} = \frac{DC}{AD} = \frac{AI}{2AI} = \frac{1}{2}$

$tan \angle A = \frac{2tan \angle A/2}{1- tan ^2 \angle A/2}$ = $\frac{ 2 . \frac{1}{2}}{1-\frac{1}{4}}$ = $\frac{4}{3}$

So, $\frac{AB}{BC}$ = $\frac{1}{tan \angle A}$ = $\frac{3}{4}$ = $\boxed{0.75}$