Quadrilateral geometry 
It's given distance between


*

*$AB = 27$

*$BC = 752$

*$CD = 26.75$

*$AD = 758$

*$CE = 1$

*$0 < FC < 752$


How do I find $FG = x$ for point $F$ on line $BC$? Is it even possible?
EDIT:
As André mentioned in comments, just by defining lengths doesn't make the structure "rigid" to calculate x.
What if we define $90 < \angle C < 120$, does this makes approximate result possible?

Here is my thought on the question;
If one has a straight line of length $758$, points $A(0,0)$ and $D(758,0)$ apart. If one constructs a circle  of radius $27$ centered at point $A$ and a circle of radius $26.75$ centered at point $D$. Are there two unique points $B(a_1,b_1)$ and $C(a_2,b_2)$, $B$ on the first circle and $C$ on the second circle such that $BC=752$.
We have the following in order, $B$ on the first circle, $C$ on the second circle, the length of $BC$, 
$$(a_1-0)^2+(b_1-0)^2=27^2 \\ (a_2-758)^2+(b_2-0)^2=26.75^2 \\ (a_1-a_2)^2+(b_1-b_2)^2=752^2 $$
Three equations are clearly insufficient for four unknowns, I don't know how  WolframAlpha got real solutions.
Even if one does find unique solution to the above system, unless $G=A=E$, line $FG$ is not unique.
