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Consider a parallelogram $ABCD$ and a plane, $P$. Let $A'B'C'D'$ be the orthogonal projective image of $ABCD$ onto $P$. If $P$ cuts the segments $AC$ and $CD$, then $BB'=AA'+CC'+DD'$. If $P$ cuts the segments $AB$ and $CD$, then $CC'+ DD'=AA'+BB'$. I have an idea of a proof using the distance formula from a point to a plane, so my questions are: 1) is this known? 2) Does it generalizes further?

Edit: I have constructed a regular parallelogram in the image, so I do not know if this works for a non-regular.

enter image description here

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If one uses signed distances, then the result can be stated in the unified form $$AA'-BB'+CC'-DD'=0$$ regardless of the configuration. The result is sharp in the sense that if a skew-quadrilateral satisfies the condition for any plane, then it is a (planar) parallelogram. I am unaware of any reference.

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  • $\begingroup$ Thank you for your comment! What if we consider an spherical parallelogram and project onto a plane? $\endgroup$ Dec 25, 2021 at 15:06
  • $\begingroup$ I would have to think about that $\endgroup$ Dec 25, 2021 at 15:32

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