If $(A,B,C,D)$ and $(A,B',C,D')$ are parallelograms, then $(B,B',D,D')$ is a parallelogram

Question

I'm stuck with the following problem:

In an affine plane, suppose that $(A,B,C,D)$ and $(A,B',C,D')$ are parallelograms. Prove that $(B,B',D,D')$ is a parallelogram.

If $(A,B,C,D)$ is a parallelogram then $\overline {AB} \| \overline{CD}$ and $\overline {AC} \| \overline{BD}$

$(A,B',C,D')$ parallelogram means $\overline {AB'} \| \overline{CD'}$ and $\overline {AC} \| \overline{B'D'}$

We have to prove:

1. $\overline {BB'} \| \overline{DD'}$
2. $\overline {BD} \| \overline{B'D'}$

$\overline {BD} \| \overline{AC} \| \overline{B'D'}$, so condition 2 is proven, but how do I prove condition 1?

Answer 1

The essence of the problem is the commutativity and associativity of addition of vectors in the vector space associated with the affine plane. In the affine plane there exists a unique translation $T$ which maps point $A$ to point $C$ given by addition of a vector $AC$. This translation also maps line segment $AB$ to $CD$ because they are parallel to each other by assumption that $ABCD$ is a parallelogram, and also maps point $C$ to point $D$ by commutativity. Any translation preserves the parallel line relationship and carries a line to a line parallel to it. If now $AB'CD'$ is another parallelogram, then by transitivity of the parallel relation $BD$ and $B'D'$ are parallel because they are both parallel to $AC$. Since by assumption $AB'CD'$ is a parallelogram, line segment $AB'$ is parallel to $CD'$ and thus the translation $T$ maps $B'$ to $D'$. But now the line segment $BB'$ is mapped to $DD'$ by translation $T$ by using associativity and thus are parallel to each other. This proves that $BB'DD'$ is a parallelogram.

In more detail, given point $B'$, there are vectors $v_1=B'A$, $v_2=AB$, and $v_3=B'D'$ which uniquely determine the six points $ABCDB'D'$ by translations using commutativity and associativity of addition of vectors. Thus, stating that $AB'CD'$ is a parallelogram is equivalent to $v_1+v_3=v_3+v_1$ since $v_1$ maps $B'$ to $A$ and then $v_3$ maps $A$ to $C$ while, also, $v_3$ maps $B'$ to $D'$ and $v_1$ maps $D'$ to $C$ by commutativity. Similarly, stating that $ABCD$ is a parallelogram is equivalent to $v_2+v_3=v_3+v_2$ since $v_2$ maps $A$ to $B$ and $v_3$ maps $B$ to $D$ while, also, $v_3$ maps $A$ to $C$ and $v_2$ maps $C$ to $D$. Finally, stating that $BB'DD'$ is a parallelogram is equivalent to $(v_1+v_2)+v_3=v_3+(v_1+v_2)$ but $v_1+v_2$ maps $B'$ to $B$ and $v_3$ maps $B$ to $D$ while $v_3$ maps $B'$ to $D'$ and $v_1+v_2$ maps $D'$ to $D$.

Answer 2

Hint: $ABCD$ is parallelogram if and only if midpoint of $AC$ coincides with midpoint of $BD$.