Find $ED$ in the triangle $ABC$ 
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$

HInt:$$AB = 6,BC = 8,AC = 10\\AC^2=BC^2+AB^2\\10^2=6^2+8^2$$ SO $$\triangle ABC$$ Is right angle, $\widehat{B}=90$
IF you draw a line $OF$ then $BGOF$ is a rectangle so,$ BF =FG$
then for $OG=OF=OD$ they are the radius of the circle .
after that 
$GA,HA$ are tangent to the circle, so
$$OH=OG=R\\H=G=90\\OA=OA \\\to \triangle AOG=\triangle AOH \Longrightarrow AG=AH$$
