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 😀 (any analytic approach would do I just figured out complex was the most suitable)
Best Answer
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.