A problem with similar tetrahedrons Consider an $ABCD$ tetrahedron. For each $XY$ edge $X(\neq) Y\in\{A,B,C,D\}$ consider two points on the edge $P_{XY}, P_{YX}$, such that $P_{XY}<P_{YX}$. Suppose that $\dfrac{AP_{AB}}{AB}=\dfrac{AP_{AC}}{AC}=\dfrac{AP_{AD}}{AD}$ and this cyclic inequality is also true for $B,C,D$.
Prove that if $\{X\neq Y\neq Z\neq W\}=\{A,B,C,D\}$, then the volume of the $P_{XY}P_{YZ}P_{ZW}P_{WX}$ is always the same!
I only tried to use the facts about a tetrahedron (volume, similarity, so on), but it didn’t help. Please help! The problem is based on a problem from Viktor v. Prasolov - Problems in Plane and Solid Geometry Vol 2.

We use the affine-linearity of the volume of a tetrahedron in each of its vertices: if $X$ is a point on $AB$ with $AX = x \cdot AB$, then the volume of $XCDE$ is given by
$$
   [XCDE] = (1-x) [ACDE] + x [BCDE]
$$
that is, the volume changes at a constant rate as $X$ moves from $A$ to $B$ at a constant rate. This is always true if we use signed volume, and otherwise it is true only if $X$ does not cross the plane $CDE$. Checking that is troublesome, so we'll stick with signed volume.

Let $\alpha = \frac{AP_{AB}}{AB}$ and define $\beta, \gamma, \delta$ similarly for the ratios at $B,C,D$.
If we begin computation by
$$
  [P_{AB} P_{BC} P_{CD} P_{DA}] = (1-\alpha) [A P_{BC} P_{CD} P_{DA}] + \alpha [B P_{BC} P_{CD} P_{DA}]
$$
and similarly eliminate $P_{BC}, P_{CD}, P_{DA}$, then the only nonzero terms left at the end will be
$$
  [P_{AB}P_{BC}P_{CD}P_{DA}] = (1-\alpha)(1-\beta)(1-\gamma)(1-\delta) [ABCD] + \alpha\beta\gamma\delta [BCDA].
$$
Since $BCDA$ is an odd permutation of $ABCD$,  this means that $[BCDA] = -[ABCD]$, so the volume of this tetrahedron is a $(1-\alpha)(1-\beta)(1-\gamma)(1-\delta) - \alpha\beta\gamma\delta$ factor of the volume of $ABCD$.
This is a symmetric expression in $A,B,C,D$, so we'll get the same answer for any ordering of the vertices.
