Quadrilateral with given angles 
We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!

A proof by Sine Law goes as follows:
Considering triangles $AED, BED, ABD, ABE$:
\begin{align}
\frac {\sin y}{AD} &= \frac {\sin 48^\circ}{ED}\\
\frac {\sin x}{EB} &= \frac {\sin 38^\circ}{ED}\\
\frac {\sin 46^\circ}{AD} &= \frac {\sin 64^\circ}{BA}\\
\frac {\sin 22^\circ}{EB} &= \frac {\sin 74^\circ}{BA}\\
\end{align}
Equating $ED$ in the first two equations we have:
$$\frac {EB \sin 38^\circ}{\sin x} = \frac {AD \sin 48^\circ}{\sin y}$$
$EB$ and $AD$ can be expressed in terms of $BA$:
$$\frac {AB \sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin x} = \frac {AB \sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$
Noting that $x = 68^\circ - y$,
$$\frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ (\sin 68^\circ \cos y - \cos 68^\circ \sin y)} = \frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin (68^\circ - y)} = \frac {\sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$
Rearranging:
$$(\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ) \sin y = \sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ \cos y$$
Giving the expression for $\tan y$:
$$\frac {\sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ} {\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ}$$
WolframAlpha says it is $50^\circ$. One could probably reduce the expression above, but right now I don't see how.
