Mysterious Angle Relation

I will take you through how this problem was created in detail. First, I will reveal the hidden configuration:

Construct the circumcenters of $\triangle AED, \triangle BEC$ and denote them $O_1,O_2$ respectively.

First, we claim that $EO$ bisects $O_1O_2$ .

To prove this, I will construct one more point for convenience, the circumcenter $O_3$ of $ABE$ . Clearly $OO_1O_3O_2$ is a parallelogram, so it now suffices to prove $F\in OO_3$ , which is also immediate from the fact that $AEB, CED$ are homothetic through $E$ .

Next, we show that $EO_1O_2, EFG$ are directly similar (This means the angle formed by rotating counterclockwise $\overline {EO_1}$ to $\overline {EO_2}$ and $\overline {EF}$ to $\overline {EG}$ is the same, or $\angle (EF,GE)=\angle EO_1,EO_2$ in directed angle notation), which is equivalent to the two triangles can be mapped to each other by a spiral similarity transformation.

Note that $EG,EO_1$ are isogonals wrt $\angle AED$ and likewise $EF,EO_2$ are isogonals wrt $\angle BEC$ . Therefore $\angle GEF=\angle O_1EO_2$ .

We now show that $\frac {EF}{EG}=\frac {AD}{BC}=\frac {EO_1}{EO_2}$ . The front half of this equality can be proven using the fact taht $AED,BEC$ have the same area. The second half can be done by proving either $ADO_1\sim BCO_2$ or extended sine rule.

Hence $EO_1O_2\sim EFG$ . To avoid introducing new concepts, we can conclude without rigor their directness visually with a diagram. Since $N,M$ are two corresponding points in their spiral similarity, by the properties of the transformation:

$\angle MEN=\angle FEO_1=180-|\angle DEO_1-\angle BEF|=180-|\angle DAC-\angle DBC|=\boxed {170^{\circ}}$