# Distances from a parallelogram to a plane

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.

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.