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
Best Answer
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 $B$ $= 100°+(90°+y)+(80°-y)+(380°-2x)=360°$
$\therefore x=145°$