This is an analytical proof of Varignon's theorem (of course it is a well-known proof)
I came up with an idea, which is to take advantage of the two-dimensional situation to prove the three-dimensionality
(Of course, Varignon's theorem remains true even if the vertices of the quadrilateral are not located in the same plane, and its geometric proof is easily possible through the theorem of the line passing through the middle of two sides of a triangle being parallel to the third side.)
But I want to exploit the two-dimensional analytical proof to prove the three-dimensional case, and this is my proof:
It can be noted that parallel projection is nothing but neglecting one of the coordinates, and therefore when we neglect the coordinates of $X$, we obtain that the first parallel projection of the quadrilateral is a parallelogram, and when we neglect the coordinates of $Y$, we obtain that the second parallel projection of the quadrilateral is also a parallelogram, and we can apply the same matter. Neglecting the coordinates of $Z$, thus we have four points in space connected by four lines, and we have the parallel projections of this quadrilateral on three non-parallel planes that are parallelograms, so the quadrilateral is a parallelogram, which is what is required.
My question is, if we are in the general case, is it necessary for three parallel projections on non-parallel planes of a quadrilateral to be parallelograms in order to prove that the quadrilateral is parallelograms, or is it sufficient for two parallel projections on two non-parallel planes to be parallelograms in order to prove that the quadrilateral is parallelograms?