What is the measurement of the angle x in the figure below? For reference: In the figure, ABCD is a parallelogram and $ \measuredangle ABE = 100^o$ , $AE = AD$ and $FC = CD$.
Calculate x.

My progress:
In this one I couldn't develop much because $FC$ and $EB$ are not parallel...I think the "output" is by some auxiliary construction

by geogebra

 A: In $\Delta EAB$ and $\Delta BCF$
$\begin{align} 
EA&=BC &(\because AD=BC, \text{ opposite sides of parallelogram})\\
AB&=CF & (\because CD=AB, \text{ opposite sides of parallelogram)}\\
\angle EAB&=\angle BCF=y \text{  (let)} & (\because 90°-\angle BAC=90°-\angle BCD)\\
\therefore \Delta EAB &\cong \Delta BCF &\text{  (by SAS congruency criterion)}\\
\implies \angle EBA&=\angle BFC= 100°&\text{ (corresponding angles of congruent triangles)}\\
\text{and}\qquad EB&=BF &\text{ (corresponding sides of congruent triangles)}\\
\end{align}$
In parallelogram $ABCD$,
$\begin{align}&\angle BAD =90°-y\\
 \implies &\angle ABC=180°-(90°-y)=90°+y \end{align}$
In triangle $BCF$, $\angle FBC= 180°-(100°+y)=80°-y$
In triangle $EBF$, $\angle BEF=\angle BFE=x-100° \; (\because EB=BF)$
$\implies \angle EBF=180°-2(x-100°)=380°-2x$
Sum of all angles around point $B$ $= 100°+(90°+y)+(80°-y)+(380°-2x)=360°$
$\therefore x=145°$
A: 
All you need is to show points F, B, D and C are cyclic, then we have:
$\angle ADB=45$
$\angle FBD=90$
$\rightarrow \angle FAB=35\rightarrow \angle BAD=\angle BCD=55=\angle BFD$
$\Rightarrow \angle ABD=80$
$\angle BFC=55+45=100$
$\angle BEF=45$
$\angle FBE=360-(100+80+90)=90$
$\Rightarrow \angle EFB=45$
Finally:
$x=45+100=145$
A: Another solution.

Perform a $90^{\circ}$ clock-wise rotation around $D$ and let the rotated image of $A$ be denoted by $H$. Then $$AD = DH = AE$$ which means that $ADHE$ is a square. If you rotate $90^{\circ}$ counter-clock-wise around $D$ the segment $CD$, you get the segment $GD$ such that $GD = CD$ and $GD \, \perp \, CD$. Hence, $CDGF$ is a square. Because of these new extra constructions, during the $90^{\circ}$ clock-wise rotation around $D$, The points $A$ and $G$ are rotated to the points $H$ and $C$ and consequently, the triangle $ADG$ is rotated to the triangle $HDC$ which means that
$$\angle\, HCD = \angle\, AGD$$
$$HC = AG$$
$$HC \, \perp \, AG$$
Now, if you translate triangle $HDC$ along $DA$  you get as a parallel translated image the triangle $EAB$. Furthermore, if you translate triangle $ADG$ along $AB$  you get as a parallel translated image the triangle $BCF$. Therefore,
$$\angle\, EBA = \angle \, BFC = 100^{\circ}$$
$$EB = FB$$
$$EB \, \perp \, FB$$
Hence, triangle $EBF$ is right-angled and isosceles, which means that $$\angle \, EFB = 45^{\circ}$$
Now putting all these angles together
$$x = \angle \, EFC = \angle\, EFB + \angle \, BFC = 45^{\circ} + 100^{\circ} = 145^{\circ}$$
