How to find the angle in a triangle inside of a quadrilateral? The problem is as follows:

Find the angle $x$ as indicated in the figure from below:


The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&36^{\circ}\\
2.&40^{\circ}\\
3.&20^{\circ}\\
4.&30^{\circ}\\
5.&32^{\circ}\\
\end{array}$
This problem has left me go in circles. I don't know exactly if there's an isosceles or what?. The only thing which I could find was that the angle opposing $50^{\circ}$ is $50^{\circ}$ hence making the upper triangle an isosceles. But that's how far I went. How exactly can this information be used to find the requested angle. Can someone help me with this?.
 A: *

*We have $\angle ABC=50^{\circ}$ and $\angle BAC=80^{\circ}$, which implies that $\angle ACB=50^{\circ}$. Therefore $\overline{AB}=\overline{AC}$.

*We have $\angle ACE=80^{\circ}$ and $\angle CAE=20^{\circ}$, which implies that $\angle AEC=80^{\circ}$. Therefore $\overline{AC}=\overline{AE}$.

*From 1. and 2. we have $\overline{AB}=\overline{AE}$, and note that $\angle BAE=60^{\circ}$. Therefore $\triangle ABE$ is an equilateral triangle. Hence $\angle AEB=60^{\circ}$.

*Take a look at $\triangle ADE$. We now have $\angle AEC=80^{\circ}$ and $\angle DAE=40^{\circ}$, which implies that $\angle ADE=40^{\circ}$. Therefore $\overline{AE}=\overline{DE}$. And since $\overline{AE}=\overline{BE}$, we have $\overline{BE}=\overline{DE}$.

*$\angle BED=180^{\circ}-\angle AEC-\angle AEB=40^{\circ}$. Therefore $\angle BDE=\frac{180^{\circ}-\angle BED}{2}=70^{\circ}=\angle ADE+x\Longrightarrow \color{red}{x=30^{\circ}}$.

The five blue segments in the image below have the same length, and $\triangle ABE$ is an equilateral triangle.
