All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$.
a.) If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length;
b.) If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.
Rotate the triangle BCD around the edge BC until ABCD are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting X and Z. Therefore, XYB=ZYC.
Summing the four equations like this, we get exactly <ABC+<ADC=<BCD+<BAD.
Now, draw all four faces in the plane, so that BCD is constructed on the exterior of the edge BC of ABC and so on with edges CD and AD.
The final new edge AB (or rather A'B') is parallel to the original one (because of the angle equation). Call the direction on AB towards B "right" and towards A "left". If we choose a vertex X on AB and connect it to the corresponding vertex X' on A'B'. This works for a whole interval of vertices X if C lies to the left of B and D and D lies to the right of A. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
Finally, regard the sine in half the isosceles triangle ACA' which gives the result with the angles around C instead of A, but the role of the vertices is symmetric.