3

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. enter image description here

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!

Bumblebee
  • 1,209
  • Working in the complex plane $,m=2d-b, n=3b-2c, a=b+d-c,$ so $,m+n=\dots$ – dxiv Apr 01 '23 at 19:03
  • But MA∥BD∥AN cant happen. We must show that MAN are colinear which means that they cant be parallel. –  Apr 01 '23 at 19:10
  • @dxiv I dont understand who is m,d,b,n and so on... what is a complex plane –  Apr 01 '23 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 01 '23 at 19:15
  • Nah, i dont really understand, because i didn't learn them. Do you have any ideas? Thank anyway! –  Apr 01 '23 at 19:16
  • @CalvinLin FYI, your hint is good, and I mention it in my answer, but you have a typo. I believe your $MA\parallel \color{red}{B}D\parallel AN$ should be $MA\parallel \color{red}{E}D\parallel AN$ instead. – John Omielan Apr 01 '23 at 19:30
  • @IonelaBuciu You can also write it in vector notation as $\overrightarrow{OM},=,2, \overrightarrow{OD} - \overrightarrow{OB}$ etc. – dxiv Apr 01 '23 at 20:04

4 Answers4

2

As suggested by Calvin Lin's comment, join $E$ to $D$, as shown below

Diagram of OP, with line ED added

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.

John Omielan
  • 47,976
1
  • $ADBE$ 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, $MA = AN$.
Calvin Lin
  • 68,864
0

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 $\,MN\,$.

dxiv
  • 76,497
0

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.

Joffan
  • 39,627