# Prove that the points M, A, N are collinear.

Let $$ABCD$$ be a parallelogram, the symmetrical point $$M$$ point $$B$$ to point $$D$$, and $$N$$ a point located on the right $$BC$$ like this so that $$B \in (CN)$$ and $$BN = 2 \cdot BC$$. Prove that the points $$M, A, N$$ are collinear. Whether point $$E$$ is half of $$BN$$. At the bottom of the picture, you can see my ideas. Also, I thought of the propriety of the centre of gravity and that we can put a point that equals BD and then make a triangle with medians. Then we can also apply the theorem of Ceva. Hope one of you can help me! Any idea is welcome! Feel free to comment with your idea!

• Working in the complex plane $\,m=2d-b, n=3b-2c, a=b+d-c\,$ so $\,m+n=\dots$
– dxiv
Apr 1 at 19:03
• But MA∥BD∥AN cant happen. We must show that MAN are colinear which means that they cant be parallel.
– user1104319
Apr 1 at 19:10
• @dxiv I dont understand who is m,d,b,n and so on... what is a complex plane
– user1104319
Apr 1 at 19:12
• Those are the complex numbers associated with the namesake points in the complex plane, sometimes called the affix of the point.
– dxiv
Apr 1 at 19:15
• Nah, i dont really understand, because i didn't learn them. Do you have any ideas? Thank anyway!
– user1104319
Apr 1 at 19:16

As suggested by Calvin Lin's comment, join $$E$$ to $$D$$, as shown below We then have $$EN\parallel BC\parallel AD \;\to\; EN\parallel AD$$. Also, $$\lvert EN\rvert = \lvert AD\rvert$$. Since one of the sufficient conditions for a quadrilateral to be a parallelogram, as stated in Parallelogram characterizations, is

One pair of opposite sides is parallel and equal in length

this means $$ADEN$$ is a parallelogram. Thus, we have

$$ED\parallel AN \tag{1}\label{eq1A}$$

Also, as you've already determined, $$AEBD$$ is a parallelogram, so $$AE\parallel DB$$ and $$\lvert AE\rvert=\lvert DB\rvert$$. Since $$\lvert MD\rvert=\lvert DB\rvert$$ and $$MDB$$ is a straight line, we then also get that $$MD \parallel AE$$ and $$\lvert MD\rvert=\lvert AE\rvert$$, so $$AMDE$$ is a parallelogram as well. Thus,

$$MA\parallel ED \tag{2}\label{eq2A}$$

Therefore, \eqref{eq1A} and \eqref{eq2A} together show that $$MA \parallel AN$$, i.e., the points $$M$$, $$A$$ and $$N$$ are collinear.

• Now it seems more simple.
– user1104319
Apr 2 at 13:34
• $$ADBE$$ is a parallelogram by construction.
• The diagonals of a parallelogram bisect each other.
• Let $$AB$$ intersect $$ED$$ at $$X$$, so $$BX = XA$$ (and also $$DX = XE$$.)
• Notice that the expansion/homothety centered at $$B$$ with a scale factor of 2 sends $$D$$ to $$M$$, $$X$$ to $$A$$ and $$E$$ to $$N$$.
• Since $$DXE$$ is a straight line, hence so is $$MAN$$.
• In fact, $$MA = AN$$.

Construct a translated copy of $$ABCD$$, $$A'B'C'D'$$, such that $$B'$$ is at $$D$$, which puts $$D'$$ at $$M$$. Then $$\triangle MAC' \cong \triangle ANB$$, again a translated copy, so $$MAN$$ is a straight line.

Expanding on my comment for the benefit of those acquainted with the complex plane or position vectors. The following uses the notation $$\,\textbf{x}\,$$ for the complex number or position vector associated with point $$X$$.

By construction $$\,\textbf{b} - \textbf{a} = \textbf{c} - \textbf{d} \;\;\iff\;\; \textbf{a} = \textbf{b} + \textbf{d} - \textbf{c}\,$$ since $$\,ABCD\,$$ is a parallelogram, and:

• $$\textbf{m}-\textbf{d} = \textbf{d} - \textbf{b} \;\;\iff\;\; \textbf{m}=2\,\textbf{d}−\textbf{b}$$

• $$\textbf{n} - \textbf{b} = 2\,(\textbf{b} - \textbf{c}) \;\;\iff\;\; \textbf{n}=3\,\textbf{b}−2\,\textbf{c}$$

It follows that $$\,\textbf{m}+\textbf{n} = 2\,\textbf{b} + 2\,\textbf{d} - 2\,\textbf{c} = 2\,\textbf{a}\,$$ so $$\,A\,$$ is the midpoint of segment $$\,MN\,$$.