I teach mathematics in a Parisian school. My students are twelve years old and are sometimes very bright children. That is why I am looking for a demonstration that is both mathematically correct and adapted to their knowledge of the fact that the diagonals of a parallelogram ABCD have the same middle. This is an affine result and I don't want to use metric results to do it.
I use as definition of parallelogram: "quadrilateral whose opposite sides are parallel".
I propose in particular, in addition to the usual axioms such as that through a point there passes a single line parallel to a given line, the following axiom:
the image of a line by a central symmetry is a parallel line.
Here is what I propose: consider a parallelogram $ABCD$; call $O$ the midpoint of the diagonal $[AC]$ and then "work in $ABC$": $B\in (AB)$ and $A\to C$ by the symmetry $s_O$ of centre $O$. So $s_O(B)$ belongs to the image of $(AB)$ by $s_O$ which is a line that passes through $C$ and is parallel to $(AB)$ according to the axiom: "the image of a line by a central symmetry is a parallel line." Thus, since there is a single line parallel to $(AB)$ through $C$ according to the other axiom recalled, it must be $(CD)$ (1) [note: the definition of the parallelogram $ABCD$, "a quadrilateral whose opposite sides are parallel", was used here]. Similarly, $B\in(BC)$ and $C\to A$ so $s_O(B)$ belongs to $(AD)$ (2). Therefore, from (1) and (2), $s_O(B)\in (CD)\cap (AD)=\{D\}$. Therefore $s_O(B)=D$. Therefore the midpoint of $[BD]$ is $O$, by definition of central symmetry of centre $O$. Therefore the common midpoint of $[BD]$ and $[AC]$ is $O$. qed∎
Is this demonstration mathematically correct? Can we simplify it to make it as accessible as possible to as many 12 year olds as possible?
Best Answer
According to the comments of @Insipidintegrator, @Arthur, @Anne Bauval, the only thing left to settle is to define the midpoint of a segment to the pupils without using a distance in the plane. If I have understood correctly, the support of the segment, in other words the straight line, is identified with the field $\mathbb{R}$ associated with the affine plane and it suffices to have defined the middle of a segment there: what do the pupils ? They place the zero of their ruler on one end of the segment, associate a number with the segment, which they divide in two. How do we translate this mathematically up to a translation? Let $x \in \mathbb{R}, y \in \mathbb{R}$; the midpoint of $[xy]$ is $x+\frac{y-x}2=\frac{x+y}2=\frac12x+\frac12y$.
That suits me. As I wrote in a comment above, I am trying to betray as little as possible the fact that in a real vector space, $(a,b,c,d)$ is a parallelogram $\iff b-a=c-d \iff b+d=a+c \iff \frac12(b+d)=\frac12(a+c)$