5. The edge of a regular tetrahedron $A B C D$ is 1. Through a point $M$, lying on the face $A B C$ (but not on the edge), planes parallel to the other three faces are drawn. These planes divide the tetrahedron into parts. Find the sum of the lengths of the edges of the part that contains point $D$.
5. See the figure at the end of the file.

Note that the part of the tetrahedron we are interested in is bounded by three of its faces containing point $D$, and by three planes parallel to the faces. Therefore, this is a parallelepiped.

Consider the faces $A B C$ and $A D C$. Align these faces so that vertex $B$ coincides with $D$, and edge $A C$ remains in place. Note that the lines of intersection of these faces with the plane passing through $M$ parallel to $(B C D)$ will coincide, since they are at the same distance from $C D$ (from $B C$). The same is true for the plane parallel to $(A B D)$. As a result, one of the faces of our parallelepiped transforms into a parallelogram with diagonal $BM$.

Reasoning similarly for the other two faces, we see that three faces of our parallelepiped are equal to three parallelograms with vertex $M$ lying in the face $A B C$. If we denote the edges of the parallelepiped as $a, b$, and $c$, then the result shown in Figure 2 is obtained. Since all triangles in this figure are equilateral, $a+b+c=1$. Therefore, the sum of the lengths of the edges of the parallelepiped is $4 a+4 b+4 c=4$.

Answer: 4.