Geometry fraction

Let $\angle IBO = \theta$ . Then, $\angle IOB = 90^\circ - \theta$ . And since incenter $I$ is the meeting point of angle bisectors of $\triangle ABC$ , then $\angle ABI = \angle IBO = \theta$ and $\angle ICO = \frac 12 \angle ACO = \frac 12 (90^\circ - \angle ABO) = 45^\circ - \theta$ .

Let $BC$ be $a$ and the altitude of $\triangle BIC$ and $\triangle BIO$ , $ID=h$ . We note that:

$\begin{aligned} \triangle BIO: \quad \frac h{\tan \theta} + \frac h{\tan (90^\circ - \theta)} & = OB \\ \frac h{\tan \theta} + \frac h{\cot \theta} & = \frac a2 \\ \frac 1{\tan \theta} + \tan \theta & = \frac a{2h} & ...(1) \end{aligned}$

$\begin{aligned} \triangle BIC: \quad \frac h{\tan \theta} + \frac h{\tan (45^\circ - \theta)} & = BC \\ \frac h{\tan \theta} + \frac {h(1+\tan \theta)}{1- \tan \theta} & = a \\ \frac 1{\tan \theta} + \frac {1+\tan \theta}{1- \tan \theta} & = \frac ah & ...(2) \end{aligned}$

$\begin{aligned} 2(1) = (2): \quad \frac 2{\tan \theta} + 2\tan \theta & = \frac 1{\tan \theta} + \frac {1+\tan \theta}{1- \tan \theta} \\ \frac 1{\tan \theta} + 2\tan \theta & = \frac {1+\tan \theta}{1- \tan \theta} \\ (1- \tan \theta)(1+ 2 \tan^2 \theta) & = \tan \theta (1+\tan \theta) \\ 1 - \tan \theta + 2 \tan^2 \theta - 2 \tan^3 \theta & = \tan \theta + \tan^2 \theta \\ 2\tan^3 \theta - \tan^2 \theta + 2\tan \theta - 1 & = 0 \\ \implies \tan \theta & = \frac 12 \end{aligned}$

Now we have $\dfrac {AB}{AC} = \dfrac 1{\tan 2\theta} = \dfrac {1-\tan^2 \theta}{2\tan \theta} = \frac {1-\frac 14}{2 \times \frac 12} = \boxed{\dfrac 34}$

I attempted to solve it using trigonometry but resorted to geometry.

$c = 2(x+y)$ - given

$a=c-x+r$ - inspection

$b=x+r$ - inspection

$\frac{y}{r}=\frac{r}{x}$ implies $r=\sqrt{x y}$

Therefore

$a=x+2 y+\sqrt{x y}$

$b=x+\sqrt{x y}$

Since $c^2=a^2+b^2$ we can solve for $y$ in terms of $x$ , obtaining (after a lot of algebra) $y=\frac{x}{4}$ , so that $\frac{b}{a}=\frac{3}{4}$ .

I am curious if anyone else did it this simple way, if they found an elegant way to do the algebra. Mine was ugly.