Using circles and symmetry , show that if a convex (non-crossed) quadrilateral has two pairs of opposite sides of same length, then it is a parallelogram.
I try to show this fondamental property about parallelogram using circles, symmetry and "relative position principles".
It is the same thing to show that the quadrilateral $ABCD$ has its pairs of opposite sides parallels.
My idea was to start with non-aligned three points $A, B$ and $C$ such that $AB \neq BC$. We also denote by $D$, the last point of the (non-crossed) quadrilateral.
In a first time, let us draw the circle $\mathcal{C}_1$ of center $A$ and radius $BC$ and then a second circle $\mathcal{C}_2$ of center C and radius $AB$. We notice that : $\mathcal{C}_1 \cap \mathcal{C}_2 \neq \emptyset$ and this intersection contains exactly two points. Let us call them $C'$ and $A'$ where $D$ is closer to $C'$ than $A'$.
In fact we need to prove that $D = C'$.
Let us eliminate the case $D = A'$. In that case, it would mean that $ABCD$ is a crossed quadrilateral and it is not possible by assumption.
Then, why $D = C'$ ? In fact, by assumptions, $BC = AD$ and $AC'$ is also a radius for $\mathcal{C}_1$. Hence $AC'=AD$. In the same way using $\mathcal{C}_2$, we have that $CC'= CD$. Hence $D = C'$.
Let us introduce $[AC]$, the diagonal of the quadrilateral $ABCD$ and consider $O$ the midpoint of $[AC]$. Now, why $(BC)$ is parallel to $(DA)$ ? By defintion $A$ is obtained from $C$ after doing an half-turn around $O$. And let us denote $C''$ the symmetric of $B$ around $O$. Using properties of the half-turn, we can deduce that $(BC)$ is parallel to $(AC'')$. It remains to show that $D = C''$.
We notice that in particular, $C'' \in \mathcal{C}(O,OB)$. Also by property of the half-turn, we have that $AC'' = AD$, hence $C'' \in \mathcal{C}_1$ and also as $CC''=CD$ we deduce that $C'' \in \mathcal{C}_2$.
Hence, $C''\in \mathcal{C}(O,OB) \cap \mathcal{C}_1 \cap \mathcal{C}_2 = \mathcal{C}(O,OB) \cap \{D\}$. It remains to show that $\mathcal{C}(O,OB) \cap \{D\} = \{D\}$. Why is this last intersection could not be empty ?
Are the ideas of this proof correct ? Thank you in advance !
