Prove that arbitrary tetrahedron can be intersected with a plane so that in the intersection will be a rhombus Prove that arbitrary tetrahedron can be intersected with a plane so that in the intersection will be a rhombus.
My idea is to find $4$ coplanar points $A,B,C,D$ (vertices of rhombus) on the pyramid sides and then to prove that $ABCD$ is rhombus, but I have no idea what points to choose. How it can be proved in this way, or, if you have another simplier way, how to prove it?
 A: Say $P,Q,R,S$ are the vertices of the tetrahedron. Consider the family $F$ of planes parallel to both edges $PQ$ and $RS$ (and parallel to each other as a consequence). Start with a plane in this family that contains $PQ$ and gradually move it to the opposite side $RS$ (keeping it parallel to both $PQ$ and $RS$). Note that the intersection of the plane with the tetrahedron at any moment of time is a parallelogram, and this parallelogram starts "tall and narrow" near $PQ$ (assuming $PQ$ is oriented "about vertical") and ends "shallow and wide" near $RS$, so there must be a moment of time in between then the parallelogram is just as "tall" as "wide", that is, a rhombus.  
A: Recall that in $\triangle XYZ$, if $P$ and $Q$ are taken on edges $\overline{XY}$ and $\overline{XZ}$ such that 
$$\frac{|\overline{XP}|}{|\overline{XY}|} = k = \frac{|\overline{XQ}|}{|\overline{XZ}|}$$
for some $0 < k < 1$, then, by similar triangles,
$$\overline{PQ}\parallel\overline{YZ} \qquad\text{and}\qquad |\overline{PQ}| = k\;|\overline{XY}|$$
(When $k=1/2$, this is "The Midpoint Theorem".)

Now, given tetrahedron $ABCD$ with opposite edges $\overline{AB}$ and $\overline{CD}$, consider points $P$, $Q$, $R$, $S$ positioned at "equal proportions" along edges $\overline{AC}$, $\overline{AD}$, $\overline{BC}$, $\overline{BD}$; that is, assume
$$\frac{|\overline{AP}|}{|\overline{AC}|} = \frac{|\overline{AQ}|}{|\overline{AD}|} = k = \frac{|\overline{BR}|}{|\overline{BC}|} = \frac{|\overline{BS}|}{|\overline{BD}|} \tag{$\star$}$$
for some $k$. Applying the preceding to segments $\overline{PQ}$ and $\overline{RS}$ in faces $\triangle ACD$ and $\triangle BCD$, we have 
$$\overline{PQ} \parallel \overline{CD} \parallel \overline{RS} \qquad\text{and}\qquad |\overline{PQ}| = k \; |\overline{CD}| = |\overline{RS}|$$
Therefore, for having parallel and congruent opposite sides, $\square PQRS$ is at least a parallelogram. To determine when it might be specifically a rhombus, consider what happens on the other two faces of the tetrahedron.
Note that $(\star)$ can be re-written as
$$\frac{|\overline{CP}|}{|\overline{CA}|} = \frac{|\overline{CR}|}{|\overline{CB}|}= 1-k = \frac{|\overline{DQ}|}{|\overline{DA}|} = \frac{|\overline{DS}|}{|\overline{DB}|}$$
so that, for segments $\overline{PR}$ and $\overline{QS}$ in faces $\triangle CAB$ and $\triangle DAB$, we have
$$\overline{PR} \parallel \overline{AB} \parallel \overline{QS} \qquad\text{and}\qquad |\overline{PR}| = (1-k)\;|\overline{AB}| = |\overline{QS}|$$
Parallelogram $\square PQRS$ will be a rhombus when adjacent sides are congruent. So, we solve:
$$k\;|\overline{CD}| = (1-k)\;|\overline{AB}| \qquad\to\qquad k = \frac{|\overline{AB}|}{|\overline{AB}| + |\overline{CD}|}$$
(Note that this $k$ is between $0$ and $1$. The plane of the rhombus definitely cuts through the interiors of those four tetrahedral edges.)
