<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] The radius $r$ of circle inscribed in right $\Delta ABC$ with legs $AB=6$, $BC=8$ and hypotenuse $AC=10$ is given by $$r=\frac{\text{Area}}{\text{semi-perimeter}}=\frac{\frac12(6)(8)}{\frac{1}{2}(6+8+10)}=2\implies OG=OF=GB=r=2$$
$$CG=CB-GB=8-2=6\implies CD=CG=6$$
In similar right triangles $\Delta EGC\sim \Delta ABC$,
$$\frac{CE}{CA}=\frac{CG}{CB}\implies \frac{CD+DE}{10}=\frac{6}{8}$$
$$\frac{6+DE}{10}=\frac68\implies DE=\frac32$$
$$\boxed{\color{blue}{ED=\frac32}}$$
</s>