# 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)

• Helixglich, are you allowed to use analytic geometry ? Feb 5 at 19:21
• @Angelo isn't that obvious from the post. Sorry if I didn't make myself clear. Yes I am indeed allowed to use analytic geo. Feb 5 at 19:38
• Helixglich, if you are allowed to use analytic geometry, why did you ask how your problem could be solved via complex numbers ? Feb 5 at 19:46
• @Angelo I just think that complex numbers are best here. Should I edit the question? Feb 5 at 19:52
• Bruh I found the solution. It was polar coordinates all along. Just WLOG AD = BC =1 and it is easy from there. Use complex to finish. I will write the solution when I have time Feb 5 at 20:56

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.

• I'm impressed by this solution compared to mine; thank you for submitting it. I hadn't thought of using the angle bisector with Cartesian coordinates, but setting it as the x axis is a great simplification. Feb 6 at 5:38

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:
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. It is convenient to place the intersection point $$\ E \$$ of the extensions of sides $$\ AD \$$ and $$\ BC \$$ of the quadrilateral at the origin in the Argand plane and to arrange that side $$\ AD \$$ of the quadrilateral be along the $$\ \mathfrak{Re}-$$axis, with $$\ A \$$ and $$\ D \$$ at distances $$\ r \$$ and $$\ s \$$ from the origin, respectively. We take side $$\ BC \$$ to lie on a ray from the origin in the direction $$\ \theta \ \ , \$$ with the distance between $$\ B \$$ and $$\ C \$$ being the same as that between $$\ A \$$ and $$\ D \$$ but displaced along the ray by an amount $$\ \Delta \$$ relative to $$\ AD \ \ ; \$$ we then have $$\ B \$$ and $$\ C \$$ at $$\ (r + \Delta) \ cis \ \theta \$$ and $$\ (s + \Delta) \ cis \ \theta \ \ . \$$ (The diagram uses $$\ \Delta \ > \ 0 \ \ , \$$ but the sign of $$\ \Delta \$$ is immaterial to this argument.)

We now determine the position of the midpoint $$\ M \$$ of $$\ AB \$$ as $$\left[ \ \frac{r \ + \ (r + \Delta) · \cos \ \theta}{2} \ \right] \ + \ i·\left[ \ \frac{0 \ + \ (r + \Delta) \ \sin \ \theta}{2} \ \right]$$ $$= \ \ \left[ \ \frac{r \ + \ (r + \Delta) · \cos \ \theta}{2} \ \right] \ + \ i· \frac{ r + \Delta }{2} · \sin \ \theta \ \ ,$$ and similarly, midpoint $$\ N \$$ of $$\ CD \$$ is $$\ \left[ \ \frac{s \ + \ (s + \Delta) · \cos \ \theta}{2} \ \right] \ + \ i· \frac{ s + \Delta }{2} · \sin \ \theta \ \ . \$$ The angle $$\ \alpha \$$ that the segment $$\ \overline{MN} \$$ makes to the direction of the positive $$\ \mathfrak{Re}-$$axis is thus given by $$\tan \alpha \ \ = \ \ \frac{\frac{ s + \Delta }{2} · \sin \ \theta \ - \ \frac{ r + \Delta }{2} · \sin \ \theta}{ \left[ \ \frac{s \ + \ (s + \Delta) · \cos \ \theta}{2} \ \right] \ - \ \left[ \ \frac{r \ + \ (r + \Delta) · \cos \ \theta}{2} \ \right] } \ \ = \ \ \frac{( s \ - \ r) · \sin \ \theta }{ ( s \ - \ r) \ + \ ( s \ - \ r)· \cos \ \theta }$$ $$= \ \ \frac{ \sin \ \theta }{ 1 \ + \ \cos \ \theta } \ \ = \ \ \tan \ \frac{\theta}{2} \ \ ,$$ applying one form of the "tangent half-angle" formula.

In a sense, we are done: we have shown that the slope of $$\overline{MN} \$$ is the same as that of the angle bisector of $$\ \angle CED \ \ , \$$ represented by $$\ \overline{EG} \ \ . \$$ While we're at it, though, we might look at the converse of the "angle bisector theorem" by marking the point $$\ G \$$ so as to satisfy the ratio $$\ \large{ \frac{EC}{ED} = \frac{CG}{DG} } \ \ . \$$ The length of the sides $$\ EC \$$ and $$\ ED \$$ of triangle $$\ \triangle ECD \$$ are the moduli of the numbers at $$\ C \$$ and $$\ D \ \ , \$$ so $$\ \frac{EC}{ED} = \frac{s \ + \ \Delta}{s} \ \ . \$$ If we parameterize the equation of the line on which $$\ DC \$$ lies as $$\ \sigma \ = \ s \ + \ t·[ \ (s + \Delta) \ cis \ \theta \ - \ s \ ] \ \ , \ \ 0 \ \le \ t \ \le \ 1 \ \ ,$$ point $$\ G \$$ is located at $$\ t \ = \ \frac{s}{s + (s + \Delta)} \ = \ \frac{s}{2s + \Delta}$$ $$\Rightarrow \ \ \sigma_G \ = \ s \ + \ \frac{s}{2s + \Delta}·[ \ (s + \Delta) \ cis \ \theta \ - \ s \ ]$$ $$= \ \left[ \ \frac{s·(2s + \Delta) \ - \ s·s}{2s + \Delta} \ + \ \frac{s·(s + \Delta)}{2s + \Delta} · \cos \theta \ \right] \ + \ i·\frac{s·(s + \Delta)}{2s + \Delta} · \sin \theta$$ $$= \ \left[ \ \frac{s^2 \ + \ s· \Delta}{2s + \Delta} \ + \ \frac{s·(s + \Delta)}{2s + \Delta} · \cos \theta \ \right] \ + \ i·\frac{s·(s + \Delta)}{2s + \Delta} · \sin \theta \ \ .$$

Hence, the angle $$\ \beta \$$ which the line $$\ EG \$$ makes to the direction of the positive $$\ \mathfrak{Re}-$$axis is

$$\tan \beta \ \ = \ \ \frac{\frac{s·(s + \Delta)}{2s + \Delta} · \sin \theta \ - \ 0}{ \left[ \ \frac{s·(s + \Delta)}{2s + \Delta} \ + \ \frac{s·(s + \Delta)}{2s + \Delta} · \cos \theta \ \right] \ - \ 0 } \ \ = \ \ \frac{s·(s + \Delta) · \sin \ \theta }{ s·(s + \Delta) \ + \ s·(s + \Delta)· \cos \ \theta }$$ $$= \ \ \frac{ \sin \ \theta }{ 1 \ + \ \cos \ \theta } \ \ = \ \ \tan \ \frac{\theta}{2} \ \ ,$$ indicating that $$\ EG \$$ indeed bisects $$\ \angle CED \ \ . \$$ (An analogous argument can be made for $$\ EF \ \ . \ )$$

The application of complex numbers in the proofs of these propositions is marginal, since they can be managed with or without the use of trigonometry (as Jean Marie does).