Question : 
I have received this question as a challenge from my teacher in Year 8 and I seem to make no progress.
All I know is that $DC$ is parallel to $AB$, and that $DA$ is parallel to $AB$ because angle $DCA$ is equal to angle $BAC$, which means that both angles, which only occurs when $DC$ and $AB$ are parallel. I assume the converse of the alternate angle rule is true? This also applies to angles $CBD$ and $ADB$ and using my assumption above I think that $DA$ and $CB$ is also parallel.
Assuming that line $DB$ and $AC$ are a straight line, I worked out a number of angles in terms of $x$, shown below.
Angle $CDB$ and $ABD$ = $x – 30$ degrees
Angle $DEC$ and $AEB$ = $180 – x$ degrees
Angle $BCA$ and $DAC$ = $150 – x$ degrees
If I add these together and try to equal it to $360$ degrees (the sum of the angles in any quadrilateral) I get a trivial $360 = 360$ because the $x$'s cancel out. The same occurs on each triangle, such as the small ones (e.g. triangle $AEB$) and the larger ones (e.g. triangle $ABC$) and I yet again get a trivial $180 = 180$ whenever I try to equal the algebra to $180$ to try and find the value of $x$.
Can anyone see where I have went wrong in my method as I see nothing wrong with what I am doing and yet I am getting trivial results and not making much progress?
Note: My notation might be rather bad, as I was taught to notate an angle you select the labelled vertex where the angle is located at and put it in the middle of other two letter. For example, if there were two line connected together, let's say, $AB$ and $BC$, then I would notate the angle at $B$ as angle $ABC$, however, there might be other ways to notate the same angle, so you might be a little confused, sorry about that. The same goes for shape notation. If anyone would like to let me know how to notate shapes and angles correctly, then please do let me know.
Best Answer
$$\measuredangle ACB=\measuredangle CAD=150^{\circ}-x$$ and $$\measuredangle BDC=\measuredangle ABD=x-30^{\circ}.$$ Thus, since $$\frac{AE}{DE}=\frac{CE}{DE},$$ we obtain: $$\frac{\sin30^{\circ}}{\sin(150^{\circ}-x)}=\frac{\sin(x-30^{\circ})}{\sin30^{\circ}}$$ or $$\cos(180^{\circ}-2x)-\cos120^{\circ}=\frac{1}{2},$$ which gives $x=45^{\circ}.$