Find the relation of the sides. How can I find the following relation $\frac{AO}{r}$?

$BO = OC, AB = AC, AD = BD$
I tried to do the following $r = \frac{AO + OC + \sqrt{{OC}^{2}+{AO}^{2}}}{2} $ but I was not able to infer the relation.
$CD$ is touched by the inner circle.

Let's define, as in the diagram below: $a=AG$, $b=GC$, $c=CH$. From Pythagoras' theorem applied to triangle $AOC$ we have:
$(a+r)^2+(b+r)^2=(a+b)^2$, that is:
$$
\tag{1}
ar+br+r^2=ab.
$$
As $AD=BD$, we also have $AH=BL=BJ=2OC-JC$, that is: $a+b+c=2b+2r-c$, or: $2c=2r+b-a$. On the other hand, triangles $EGC$ and $FCH$ are similar (because angle bisectors $CE$ and $CF$ are perpendicular), which gives: 
$b:r=r:c$, or: $c=r^2/b$. Combining these two expressions for $c$ we get:
$$
\tag{2}
2r^2=2rb+b^2-ab.
$$
From equations $(1)$ and $(2)$ we can obtain $a$ and $b$ as a function of $r$. If we set $x=a/r$ and $y=b/r$ these equations can be rewritten as
$$
\cases{
x+y+1=xy\\
2+xy=2y+y^2
}.
$$
By eliminating $xy$ we get $x=y^2+y-3$ and plugging this into the first equation yields $y^3-5y+2=0$, which can be factored as $(y-2)(y^2+2y-1)=0$. The solutions are then $y=2$ and $y=-1\pm\sqrt2$, but only the first one leads to positive values for both $x$ and $y$. We obtain then $y=2$ and $x=3$, so that:
$$
{AO\over r}={a+r\over r}=x+1=4.
$$

