I am looking for a proof that:

if $A_{11}A_{12}...A_{1n}$; $A_{21}A_{22}...A_{2n}$; $\cdots$; $A_{i1}A_{i2}...A_{in}$; $\cdots$; $A_{m1}A_{m2}...A_{mn}$ are $m$ oriented regular polygons ($n$-gons), where $n=2k$, then$\DeclareMathOperator\Area{Area}$
$$
\begin{align*}
& \Area(A_{11}A_{21}...A_{m1})+\Area(A_{1\;k+1}A_{2\;k+1}...A_{m\;k+1})\\ =\ & \Area(A_{12}A_{22}...A_{m2})+\Area(A_{1\;k+2}A_{2\;k+2}...A_{m\;k+2})\\=\ & \cdots\\ =\ & \Area(A_{1i}A_{2i}...A_{mi})+\Area(A_{1\;k+i}A_{2\;k+i}...A_{m\;k+i})\\ =\ & \cdots
\end{align*}
$$

**Reference:**

* [Areas que suman lo mismo](http://www.xente.mundo-r.com/ilarrosa/GeoGebra/AreasIg_Npoligonos.html)

* A case $m=4$ and $n=4$, I posed in [here](https://math.stackexchange.com/questions/1447773/two-conjectures-of-four-squares) is near six year ago. But no have a proof.