Let $ ABCD$ be a regular tetrahedron and $ M,N$ distinct points in the planes $ ABC$ and $ ADC$ respectively. Show that the segments $ MN,BN,MD$ are the sides of a triangle.
WLOG $AB=1$. Then the height is $h=\sqrt{\frac2 3}$ and $BN+MD>2h>1>MN$. Now it remains to show that $|BN-MD|<MN$. Let the points $M', N'$ be symmetric to $M, N$ respectively wrt the bisector plane of dihedron $AC$. If $M'=N$ then $BN=MD$ and everything is obvious. Otherwise define $O=MN\cap M'N'$. We have $|BN-MD|=|DN'-MD|<MN'<MO+ON'=MN$ as desired. The result follows.