How to solve for missing angle? (Triangles) I was looking around at geometry questions and came across this 
Seemed simple but I got stuck after filling in everything I knew. This is what I couldn’t get passed

Only thing I could think of doing is drawing auxiliary lines to fill values but I couldn’t think of useful ones.
Help
 A: Start by noticing that $ABE$ is a particular triangle. After that, draw the line parallel to $AB$ going through $E$, and the one parallel to $AE$ going through $B$. Once you are done with that, calculate the new angles with the newly formed triangles (very easy given what you already have). Then use what you know about the triangle $ABE$ to draw a very nice line going through $A$ and cutting $[EB]$ in half.
Once you are done with that, observe that the triangle $OED$ (where $O$ is the point formed by the intersection of $DB$ and the line parallel to $AB$ going through $E$) is VERY SIMILAR to a newly formed triangle that you know the angles of...
A: This is a fairly simple problem, here's my approach:

1.) Locate point $C$ on $BD$ and join it with $A$ and $E$ such that $\angle BAC=20$. Now that we know that $\angle BAC=20$ and $\angle ABC=80$, we can infer that $\angle ACB=80$, therefore $AB=AC=AE$.
2.) Since $\angle EAC=60$ and $AE=AC$, we can say that $\triangle AEC$ is equilateral, therefore $AB=AC=AE=EC$. Notice that $\angle DAC=\angle ADC=40$, therefore, $AB=AC=AE=EC=DC$. This proves that $C$ is the circumcenter of $\triangle AED$, therefore via the inscribed angle theorem, $x=30$
