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

Question

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!

Answer 1

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.

Answer 2

• $ADBE$ 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, $MA = AN$.

Answer 3

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.

Answer 4

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 $\,MN\,$.