Germanian Geometranian!

First we can draw all the diagonals to the cyclic cuadrilaterals: $\overline{OC}$ , $\overline{OB}$ , $\overline{BD}$ and $\overline{AC}$ . Let $\angle{ADE}= \alpha$ , therefore we can use circle properties, from which:

$\angle{ODB}=\angle{OCB}=\angle{OBC}=\angle{OAF}=\alpha$

And

$\angle{OAC}=\angle{DBC}=\angle{OCA}=\beta$ $\Rightarrow \angle{DOC} = \angle{DEC}=2\beta \Rightarrow \angle{BDE}=\beta$

Therefore:

$\overline{BF}=\overline{ED}=10 \Rightarrow \bigtriangleup{AOC} \sim \bigtriangleup{BED}$

And

$\angle{CAF}=\angle{ACF}=\angle{OBD} = \angle{ODB}=\alpha-\beta$ $\Rightarrow \bigtriangleup{ODB} \sim \bigtriangleup{FAC}$

As those triangles are proportional to each other, we have:

$\frac{r}{\overline{AC}}=\frac{8}{\overline{BD}} \wedge \frac{r}{\overline{BD}}=\frac{10}{\overline{AC}}$ $\Rightarrow r=\sqrt{80}$

Using circle laws we get 1.angle ABO=angle AFO=angle COD and 2.angle DCO=DEO=FOA .Then using triangle laws of trigonometry we get Sin(FOA)/Sin(AFO)=(AD/8)=(10/AD). This means AD²=8x10 .AD=square root of 80 ,which is 8.944