Prove that in a $4n$-gon, every other diagonal passes through a common point 
Suppose two regular $2n$-gons in the plane, which interesect one another to form a  $4n$-gon. Prove that every other diagonal of this $4n$-gon, i.e. $P_{1}P_{2n+1},P_{3}P_{2n+3},...,P_{2n-1}P_{4n-1}$ passes through a common point with every other.

A $4n$-gon has an even number of sides, so the diagonals of a regular $4n$-gon all meet in a single point. We can obtain a regular polygon easily by using e.g. two congruent polygons rotated by $ 45^{\circ}$ and this fulfills the condition. Even if the resulting polygon is not regular, it should hold (for every other diagonal). In all examples I attempted this is the case, but I had trouble proving it.
I'll be glad to hear your ideas.
 A: This is not an answer.
It is a writeup with an image to explain what is confusing about the question (but couldn't be fit into a comment).

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*Based on my interpretation of OP's comment, I read the question as


Suppose two regular $2n$-gons in the plane, which interesect one
another to form a  $4n$-gon. Prove that every other diagonal of this
$4n$-gon, i.e. $P_{1}P_{2n+1},P_{3}P_{2n+3},...,P_{2n-1}P_{4n-1}$
passes through a common point with every other.

This edit was not made by the OP, other than saying "the first sentence is simply the title".
2. Clarity on how the intersection happens is needed.

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*When two regular 2n-gons intersect, with each edge cutting an edge of the other polygon, it results in a 8n-gon, not a 4n-gon.

*If the 2n-gons intersect with vertices contained without another, then we will not end up with 4n vertices.

*It's not clear to me how they want to get a polygon with 4n vertices (based on the labelling) and 4n edges (based on the name of the 4n-gon).



*Relatedly, clarity on how to label the vertices would be helpful.

*If the two regular 2n-gons do not have the same center, then the diagonals will not pass through a common point.


