Let e, F, G and H be the midpoint of BC, CD, DA and AB in the tetrahedron ABCD, respectively

Let e, F, G and H be the midpoint of BC, CD, DA and AB in the tetrahedron ABCD, respectively

According to the theorem of triangle median line, there are: AC ‖ GF, AC ‖ he, AC ‖ plane efgh. Similarly, there are: BD ‖ EF, BD ‖ Hg, BD ‖ plane efgh. That is to say, the plane efgh is parallel to AC and BD, and the plane efgh passes the midpoint of AD, CD, BC and ab. the distances from AC and BD to the plane efgh are equal, and ∵ d-efgh and c-efgh are of the same base and equal height

In tetrahedral a-bcd, efgh is ab BC CD Da midpoint (1) if AC = BD, efgh is rhombic (2) AC is parallel to plane efgh

The efgh in tetrahedral a-bcd is ab BC CD da
(1) If AC = BD, efgh is diamond
The efgh in tetrahedral a-bcd is the midpoint of AB BC CD da
EH / / BD / / FG, and eh = 0.5bd = FG
EF / / AC / / Hg, and EF = 0.5ac = Hg
Because AC = BD
So eh = FG = EF = Hg
So efgh is rhombic
(2) AC parallel to plane efgh
Because AC and BD do not intersect, and EF / / AC / / Hg, eh / / BD / / FG,
So AC is parallel to the plane efgh