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This is a sketch for the case of squares: The vertices $A_1,B_1,C_1,D_1$ of the first square are of the form $x_1,y_1,y_1+D(y_1-x_1),x_1+D(y_1-x_1)$ for $2$-vectors $x_1,y_1$ and so on for the other $i$ ($D$ is a rotation through a right angle). If you use the wedge product to compute the areas in your formulae then you see that the statement is equivalent to the vanishing of a bilinear form in the variables $x_1,y_1,\dots$. It suffices to check this on the standard basis elements (i.e., $(1,0)$ and $(0,1)$). There are $2^8$ such combinations, but by symmetry you only need to check one of these.

The general case can be shown analogously— just a question of notation.