Find $ED$ in the triangle $ABC$

Question

$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$

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 $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$

Answer 1

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 $$OH=OG=R\\H=G=90\\OA=OA \\\to \triangle AOG=\triangle AOH \Longrightarrow AG=AH$$

Answer 2

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 $\Delta EGC\sim \Delta ABC$, $$\frac{CE}{CA}=\frac{CG}{CB}\implies \frac{CD+DE}{10}=\frac{6}{8}$$ $$\frac{6+DE}{10}=\frac68\implies DE=\frac32$$ $$\boxed{\color{blue}{ED=\frac32}}$$

Answer 3

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}