If $ABCD$ is a regular tetrahedron where $AB=b$.
$I,J,K,$ and $L$ are the midpoints of $[AB],[BC],[CD],$ and $[AD]$ respectively. I proved $(AB)$ orthogonal to $(CD)$. What would be the nature of $IJKL$?
I proved it to be a rhombus where I applied midpoint theorem to triangles $ABC, ADB, BCD,$ and $ACD$ to show that $IJ=JK=KL=IL=\frac {AC} 2=\frac {BD}2=\frac b 2 $
Could it possibly be a square? How ?
Hint: If you take the midpoints of all the six edges, you will obtain the vertices of a smaller regular octahedron. $I$, $J$, $K$, $L$ are four of the vertices. You may draw the octahedron to visualize the $4$ vertices.