Trapezoids in a square Good day

As part of a problem I need to show that AB is parallel to CD, with the given info on the image. All the segments marked red are equal, all 1-stripe grey equal etc.
I'd like to prove EA=AD, then it should prove AB||DC
Suggestions/tips? I'd appreciate pointers rather than a complete answer so as to try it self first!
Can we do the following?

We the right-angles on AC. Now we draw in AI parallel to EF. We then have parallelogram AIFH - thus $\angle AHF$ and $\angle AIF$ are both 90 degrees. Thus $\angle HAI$ = $\angle IFH$ = 90 degrees. Thus AB||DC ?
 A: Place the square on a Cartesian coordinate grid. We can choose the units so the square is a unit square. The coordinates of the vertices $B,F,D,E$ are then obvious. (I limited this diagram to only what is necessary for my solution: your diagram has unneeded line segments and not enough labels for the points.)

Point $C$, the midpoint of $\overline{BF}$, has coordinates $(0.5,1)$. The line $\overleftrightarrow{CD}$ then has the equation $y=2x-1$.
According to your diagram, the point I have labeled $G$ in my diagram is a unit distance from point $B$, which is the origin. Point $G$ therefore lies on the unit circle $x^2+y^2=1$. We can solve those two simultaneous equations for line $\overleftrightarrow{CD}$ and the unit circle to get the only intersection point in the first quadrant (inside the square), and we get $G=(0.8,0.6)$.
The point I have labeled $H$, the midpoint of $\overline{EG}$, then has coordinates $(0.4,0.8)$. The line $\overleftrightarrow{BH}$ then has the equation $y=2x$.
Point $A$ is defined as the intersection of line $\overleftrightarrow{BH}$ with the top of the square, $y=1$, so the coordinates of $A$ are $(0.5,1)$.
We then easily see that $\overline{AB}\parallel\overline{DC}$: for one thing, the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{DC}$ have the same slope, namely $2$.
Now you can translate that argument into a purely Euclidean geometric argument, if you like.
A: Borrowing @Rory Daulton's notation, construct a second square directly beneath the original, and extend the line $\overleftrightarrow{DC}$ to meet the lower square at vertex $K$.

The circle centered at $B$ with radius $BE$ passes through points $E$, $G$ and $K$. Therefore the inscribed angle $\angle EKG$ is half the central angle $\angle EBG$, since both angles subtend the same arc $\stackrel\frown{EG}$. But $\overline {BA}$ bisects the central angle $\angle EBG$, so angle $\angle EBA$ equals $\angle EKG$. This implies that $\overline {AB}$ is parallel to $\overline{KG}$, which coincides with $\overline {CD}$.
