Problem:
Here is what I've tried:
I drew symmetry of $BC$ and $AB$ along $AC$. Now $AE$ is angle bisector of $\angle DAC$. It seems it is enough to only consider $\triangle ACD$ and segments in it, but I still can't see a connection between orange segment and green segment.
On
Construct a circle that has centre $D$ and radius $DA$ as in the picture below and create a circumscribed angle $\angle APC$ on the large arc $AC$.
Then $\angle APC=50$. When observing the quadrilateral $ABCP$. You can easily say that it is cyclic which means points $A, B, C, P$ lies on the circumference of the same circle. This leads that points B and P are lying on the circumference of the circle that has radius $y$
Finally $BD=y$ as $\triangle ABD$ is equilateral $x=y$
@Math Lover's solution is more logical than mine.
As in comments, you already solved using law of sines. Here is a construction that shows $x = y$.
Extend $CB$ and drop a perp from $A$. Also drop a perp from $D$ to $AC$.
As $\angle ACE = 30^\circ, AE = AC/2$ and $\angle ABE = 50^\circ$. Also, as $\triangle ADC$ is isosceles triangle, $AH = CH = AC/2$.
That leads to $\triangle ABE \cong \triangle ADH$ (by A-S-A) and hence $x = y$.