When we draw the 2D net of the tetrahedron we get a triangle splited by its medial triangle(or the midlines) and all of those triangles are indeed congruent.

We open the tetrahedron from the vertex $A$ and put it in the 2D plane. We name vertices $A_1, A_2, A_3$ such that they make a triangle that has been a face of tetrahedron before with $BC, CD, DB$ respectively. Since $\angle ACB + \angle BCD + \angle ACD = \pi$, then $\angle A_1CB + \angle BCD + \angle A_2CD = \pi$ as well and we get $A_1, C, A_2$ lie on one line. In the same way we get to collinearity of sets $A_2, D, A_3$ and $A_3, B, A_1$. Having that with the fact that $A_1A_2A_3$ must make a tetrahedron with base $BCD$ we get that four sides must be congruent and $BCD$ must be the medial triangle.