Geometry problem, finding missing angles One of my students showed me a problem that she says is similar to what they would do in high school in her home country (which i am attaching here.
. The goal of the problem is to find the measure of $\angle DEC$ using the given angle measures provided. I've tried working on this to see what other angle measures I could deduce, and I'm including that here  This is where I'm stuck. I've tried:

*

*Labeled one unknown angle as x and determined all other unknown angles in terms of $x$, but it's fully consistent and nothing seems to simplify to indicate what $x$ is.

*Drawn in lines parallel to the sides through various points, and use what I know about parallel lines cut by transversals, but it doesn't seem to get me any closer to the target.

I suspect I might need to draw in some additional line or extend the diagram in some way but I can't figure out what. Any help would be greatly appreciated. Thanks!

Here's the proof with sine rule. This will be the shortest solution, but it is inelegant, as is the case with all "World's Hardest Easy Geometry Problem" type problems. And as with all those questions, there will be a nice solution, (hopefully.) Consider the "simplified" diagram below.

I messed up the naming so we are looking for $\theta = \angle EDC$.
In $\triangle CDE: \dfrac {EC}{\sin \theta} = \dfrac {DC}{\sin (180^\circ - 80^\circ - \theta)} = \dfrac {DC}{\sin (80^\circ + \theta)}$
In $\triangle ACE: \dfrac {EC}{\sin 40^\circ} = \dfrac {AC}{\sin 80^\circ}$
In $\triangle ACD: \dfrac {AC}{\sin 150^\circ} = \dfrac {DC}{\sin 10^\circ}$
Hence:
$$\frac {\sin (80^\circ + \theta)}{\sin \theta} = \frac {DC}{EC} = \frac {AC \sin 10^\circ}{\sin 150^\circ} \cdot \frac {\sin 80^\circ}{AC \sin 40^\circ} = \frac {\sin 10^\circ \sin 80^\circ}{\sin 150^\circ \sin 40^\circ}$$
$$\frac{\sin (80^\circ + \theta)}{\sin \theta} = \frac {\sin 80^\circ \cos \theta + \cos 80^\circ \sin \theta}{\sin \theta} = \sin 80^\circ \cot \theta + \cos 80^\circ$$
Thus we have:
\begin{align}\theta &= \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 150^\circ \sin 40^\circ}-\cot 80^\circ\right)\\
& = \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 30^\circ \sin 40^\circ}-\tan 10^\circ\right)\\
& = \cot^{-1} \left(\frac {2\sin 10^\circ}{4\sin10^\circ \cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ}{\cos 10^\circ}\right)\\
& = \cot^{-1} \left(\frac {1}{2\cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \left(\frac {\sin 30^\circ - \sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \left(\frac {\cos 10^\circ \sin 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \tan 20^\circ\\
& = 70^\circ
\end{align}
The last steps are only possible if we know in advance the solution is nice.
