How will this be solved via the use of complex numbers (angle bisector)? In the quadrilateral ABCD, side AD is equal to side BC, and lines AD and BC intersect at point E. Points M and N are the midpoints of sides AB and CD, respectively. Prove that the segment MN is parallel to the bisector of $\angle{AEB}$
This has a very "easy" synthetic solution which involves constructing the midpoint of lets say diagonal AC and doing further reasoning which will not be included in this post.
I personally don't like this solution as it is very hard to find without previously encountering such problems and not quite intuitive.
I have tried using polar coordinates to encode the angle bisector condition but I am not sure how to do so for the equal segments and further finish the problem.
I will be very grateful if someone provides me a hint or even a solution :D (any analytic approach would do I just figured out complex was the most suitable)
 A: We will use complex numbers.
Let the angle bisector divide the angle in two equal angles of $\theta$ degrees and intersect $\overline{DC}$ at point T. Therefore:
$\frac{t-e}{|t-e|} = e^{i\theta}\frac{d-e}{|d-e|}$
$\frac{c-e}{|c-e|} = e^{i\theta}\frac{t-e}{|t-e|}$
Multiplying those two yields:
$\frac{(t-e)^2}{|t-e|^2} = \frac{(d-e)(c-e)}{|d-e||c-e|} \Longleftrightarrow \frac{t-e}{\bar{t} -\bar{e}} = e^{i2\theta}$
Now:
WLOG: AD = BC =1
Therefore
$\frac{n-m}{\bar{n}-\bar{m}} = \frac{a+b-c-d}{\bar{a} + \bar{b} - \bar{c} - \bar{d}} \Longleftrightarrow \frac{n-m}{\bar{n}-\bar{m}} = \frac{e^{i2\theta}+1}{e^{-i2\theta}+1}$
$\overline{MN} \parallel \overline{ET}\Longleftrightarrow \frac{n-m}{\bar{n}-\bar{m}} = \frac{t-e}{\bar{t} -\bar{e}}$
WTS:
$\frac{n-m}{\bar{n}-\bar{m}} = \frac{t-e}{\bar{t} -\bar{e}}$
$\therefore \frac{e^{i2\theta}+1}{e^{-i2\theta}+1} = e^{i2\theta}$
$\therefore e^{i2\theta}+1 =(e^{-i2\theta}+1) e^{i2\theta}$
$\therefore e^{i2\theta}+1 = 1+ e^{i2\theta}$
Which is true hence the condition is established.
A: Here is a "natural" solution using analytic geometry.
Take $E$ as the origin, and the line bissector of angle $E$ as the $x$-axis of coordinates (and, of course the second axis directly orthogonal to the first one in $E$).

Lines $EDA$ and $ECB$ have equations $y=+ax, \ y=-ax$ resp. for a certain constant $a$.
Let $A(x_A,y_A=ax_A)$, etc...
Remark : we can assume WLOG that :
$$x_A>x_D>0 \ \ \text{and} \ \ x_B>x_C>0 \tag{0}$$
Our parallelism issue boils down to check that $M$ and $N$ have the same ordinates.
As $$\vec{AD}\binom{x_D-x_A}{a(x_D-x_A)}, \ \vec{BC}\binom{x_C-x_B}{-a(x_C-x_B)}$$
Constraint $length(AD)^2=length(BC)^2$ gives :
$$(1+a^2)(x_D-x_A)^2=(1+a^2)(x_C-x_D)^2 \ \iff \ x_A-x_D=x_B-x_C\tag{1}$$
(as a consequence of Remark (0) above).
Please note that (1) is equivalent to :
$$x_A-x_B=x_D-x_C\tag{2}$$
We can now compute :
$$\begin{cases}y_M=\tfrac12(y_A+y_B)=\tfrac12(ax_A-ax_B)=\tfrac{a}{2}(x_A-x_B)\\ y_N=\tfrac12(y_D+y_C)=\tfrac12(ax_D-ax_C)=\tfrac{a}{2}(x_D-x_C)\end{cases}$$
which, indeed, are equal, due to (2), ending the proof.
