<s>[INST] 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 $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$
 [/INST] 
Let
$BC=8=a$,
$AC=10=b$,
$|AB|=6=c$,
$\angle BAC=\angle GEC=\alpha$.
Since $b^2=a^2+c^2=100$,
$\angle CBA=\angle CGE=90^\circ$.
Inradius of the right triangle
$r=\tfrac12(a+c-b)=2$.
\begin{align} 
|ED|&=
r\,\cot\alpha
=r\cdot\frac ca
=\frac32
.
\end{align}
</s>