I have found this unusual yet very neat definition of a vector from an old math lesson of mine, it is quite clever, so that I decided to prove some theorems and share it here, so that they be verified !
Definition 1 : we call bipoint any couple of point of a plane $P$. Let $A$ be a point of $P$, the bipoint $(A,A)$ is called a null bipoint
Definition 2 : We say that two bipoints, say $(A,B) \& (C,D)$ are equipollents iff both segments [A;D] and [C;B] share the same midpoint. We denote this $(A,B) \sim (C,D)$
It is clear that for all points A & B, $(A,B) \sim (A,B)$
Lemma 1 : $\sim$ is an equivalence relation*, moreover we have $(A,A) \sim (B,B)$
Please note : ABBA is considered as a flat parallellogram, as $(AB)\parallel(AB)$ Let us now define collinearity, length, and directions of bipoints.
Definition 3&4 : We call length of a bipoint the length of the segments that joins the two points. two non-null bipoints say $(A,B) \& (C,D)$ are called collinear if $(AB)\parallel(CD)$.
Definition 5 : two collinear bipoints $(A,B) \& (C,D)$ with A,B,C,D not being aligned, are said to have the same direction if $ABDC$ is a trapezoid.
Let A,B,C,D be aligned points on a line E, let P be a point that doesn't lie on E, and let $\delta$ be the line parallel to E through P. Let $\epsilon_1$ be the line perpendicular to E through A and $\epsilon_2$ the line perpendicular to E through B, they cross line $\delta$ at point $A_2 \& B_2$, respectively. Then it is clear that $(A;B)$ has the same direction as $(A_2;B_2)$ (this forms a rectangle, which is a parallellogram and a trapezoid).
We say that (A,B) has the same direction as (C,D) iff ($A_2$,$B_2$) has the same direction as $(C,D)$
Fundamental theorem: Two non-null bipoints are equipollents iff they are collinear & have both the same length and direction.
Proof :
$(\Rightarrow)$ : ABDC is a trapezoid as both bipoints have the same direction. Moreover they both have the same length AB=CD, so ABDC is a parallellogram. So $(A,B)\sim(C,D)$
$(\Leftarrow)$ : $(A,B) \sim (C,D) $ so ABDC is a parallellogram, so it is a trapezoid so same direction is proven, moreover AB=CD so they have the same length.
Definition 6 : The set of all bipoints equipolents to $(A;B)$ (which is not empty as it contains $(A,B)$ ), is denoted by $\overrightarrow{AB}$. the vector $\overrightarrow{AA}$ is noted by $\vec{0}$. Vectors may also be denoted by a letter, in which case a rightarrow sits atop of it .
(*) I did not know what equivalence relation was back when I was taught this, so,$ \overrightarrow{AB}$ is the equivalence class of $(A,B)$ , right ?
So that finally :
Fundamental theorem, reloaded : Two vectors are equal iff they have the same direction and length.
which leads to
Theorem 3 : let $\vec u$ be a vector and A a point of P, then there exist a unique point B of P such that $\vec u=\overrightarrow{AB}$
Please feel free to correct anything in this post ! Thanks for reading this quite long question,
T.D