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 $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$
 A: 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 $\Delta EGC\sim \Delta ABC$,
$$\frac{CE}{CA}=\frac{CG}{CB}\implies \frac{CD+DE}{10}=\frac{6}{8}$$
$$\frac{6+DE}{10}=\frac68\implies DE=\frac32$$
$$\boxed{\color{blue}{ED=\frac32}}$$
A: 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
$$OH=OG=R\\H=G=90\\OA=OA \\\to \triangle AOG=\triangle AOH \Longrightarrow AG=AH$$
A: 
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}
