Yesterday I asked a question concerning an argumentation, and keep on thinking about the problem I realized that what I was missing is probably a basic result in linear algebra. Actually I am not sure this is a theorem or not (probably it is), but it should be as follow.
Pseudo theorem in $\Re^2$:
Let $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ four vectors in $\Re^2$ such that it is not possible that three of them lie on the same line. Construct a quadrilateral with the four vectors as vertices. Then, if the difference between two adjacent vertices (say $\mathbf{a}$ and $\mathbf{b}$) is equal to the difference of the opposite vertices (say $\mathbf{c}$ and $\mathbf{d}$), the line upon which $\mathbf{a}$ and $\mathbf{b}$ lie is parallel to the line upon which $\mathbf{c}$ and $\mathbf{d}$ lie.
Now, I never found anything like this anywhere (I am self-taught), and I do not really know how to come up with the proof, if it exists. Moreover I am wondering if - assuming it is true in $\Re^2$ - the result can be extended in higher dimensions.
Any feedback concerning references or a proof is more than welcome.
Thanks in advance.
Your question is a little ambiguous. There are several things which I could interpret as "the difference between two adjacent vertices (say $\mathbf{a}$ and $\mathbf{b}$) is equal to the difference of the opposite vertices (say $\mathbf{c}$ and $\mathbf{d}$)".
One interpretation would be that side $AB$ has the same length as side $CD$. This condition is not sufficient for $AC$ to be parallel to $CD$. You can draw any two arbitrary line segments of the same length, oriented however you want, and still construct a quadrilateral. I'm sure it will not be difficult for you to construct a counter-example with pen and paper.
A second interpretation would be that the vector $\mathbf{b-a}$ is equal to the vector $\mathbf{d-c}$. Then indeed, the two sides are parallel and of the same length. This is a trivial equality because the direction vector of the lines $AB$ and $CD$ are determined by $\mathbf{b-a}$ and $\mathbf{d-c}$ respectively, and these two vectors are equal. This condition is also strong enough to ensure that the other pair of opposite sides are also parallel.
A third interpretation, related and stronger than the first, would be to require both pairs of opposite sides to be equal in length. Let us name our quadrilateral $ABCD$ with vertices labelled in clockwise order. Then we require, $|AB|=|CD|$ and $|BC|=|DA|$. In this case, the quadrilateral (assuming it is non-self intersecting) is in fact a parallelogram. This is an elementary result which you can find in any book on Euclidean geometry. For completeness, I will reproduce a proof here.
Proof: Consider the diagonal $AC$. Triangles $\triangle ABC$ and $\triangle CDA$ have all three sides equal since by assumption $|AB|=|CD|$ and $|BC|=|DA|$. This means that the two triangles are congruent. Repeating the argument with the other diagonal, this proves that the two pairs of opposite angles are congruent: $$\alpha = \angle ABC = \angle CDA,\ \ \ \ \text{and}\ \ \ \ \beta = \angle BCD = \angle DAB$$ Notice that since internal sums of quadrilaterals always sum to $360^\circ$, we must have $$\alpha + \beta = 180^\circ$$ But this is precisely the condition needed for the opposite sides to be parallel. $\square$
As for the generalization into higher dimensions, you will have to clearly state what exactly you wish to generalize.