Finding $\frac{FC}{EG}$ from $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ If $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ .How to find $\frac{FC}{EG}$

 A: This is really more an algebraical than a geometrical exercise. Essentially we are given 4 equations in 6 variables. Then we add one more variable $FC$ and exclude $EG$ by introducing the ratio. So the total number of variables remains 6. 
Therefore, we must infer the two missing relations from the given geometrical construction. Adding line segments $CG$ and $AG$ and considering pairs of right-angled triangles $\triangle CFG$ and $\triangle GEC$ sharing hipotenusa $CG$ we obtain
$$FC^{2}+FG^{2}=EC^{2}+EG^{2}$$
 $$\left(\frac{FC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{EC}{EG}\right)^{2}+1$$
 $$\left(\frac{FC}{EG}\right)^{2}=\left(\frac{EC}{EG}\right)^{2}-\left(\frac{FG}{EG}\right)^{2}+1 \tag{1}$$
Now consider $\triangle GFA$ and $\triangle ADG$ sharing $AG$
 $$AF^{2}+FG^{2}=DG^{2}+AD^{2}\tag{2}$$
This is basically where geometry ends and all that follows is a tedious algebraic manipulation. Obviously we want to find $\frac{FG}{EG}$ and $\frac{EC}{EG}$ to use them in $(1)$ so let's divide $(2)$ by $EG$
 $$\left(\frac{AF}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{DG}{EG}\right)^{2}+\left(\frac{AD}{EG}\right)^{2}\tag{3}$$
Now it is time to use the given relations to start eliminating the unknowns. Take
$$AD=2EC$$
and substitute in $(3)$
 $$\left(\frac{AF}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{DG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}\tag{4}$$
Now combine
$$AF-FG=DG+EG$$
and
$$DG=FG+EG$$
to obtain
$$AF=2FG+2EG$$
Plug this in $(4)$
$$\left(\frac{2FG+2EG}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG+EG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}$$
 $$3\left(\frac{FG+EG}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=4\left(\frac{EC}{EG}\right)^{2}$$
 $$4\left(\frac{FG}{EG}\right)^{2}+6\frac{FG}{EG}+3=4\left(\frac{EC}{EG}\right)^{2}$$
 $$\left(\frac{EC}{EG}\right)^{2}-\left(\frac{FG}{EG}\right)^{2}=\frac{3}{2}\frac{FG}{EG}+\frac{3}{4}\tag{5}$$
We have not used one more relation, namely
 $$AF=FG+2EC$$
Insert this in $(4)$ together with $DF=FG+EG$ to obtain
 $$\left(\frac{FG+2EC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG+EG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}$$
 $$\left(\frac{FG}{EG}\right)^{2}+4\frac{FG}{EG}\cdot\frac{EC}{EG}+4\left(\frac{EC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG}{EG}\right)^{2}+2\frac{FG}{EG}+1+4\left(\frac{EC}{EG}\right)^{2}$$
 $$\left(\frac{FG}{EG}\right)^{2}+2\frac{FG}{EG}\left(2\frac{EC}{EG}-1\right)-1=0\tag{6}$$
Now $(5)$ and $(6)$ give the system to determine $\frac{FG}{EG}$ and $\frac{EC}{EG}$. I trust WolframAlpha that the only solution that makes sense is
 $$\frac{FG}{EG}=\frac{1}{3} \qquad \frac{EC}{EG}=\frac{7}{6}$$
Hence, substituting these values in $(1)$ we obtain
 $$\left(\frac{FC}{EG}\right)^{2}=\frac{9}{4}$$
Finally
 $$\frac{FC}{EG}=\frac{3}{2}$$
