The midpoints of an isosceles tetrahedron I recently came across this geometric property:

If the tetrahedron $ABCD$ is isosceles (all faces are congruent), and $F,E$ are the midpoints of $AD,BC$ respectively, then $FE\perp BC$ (the same can be said for all opposite edges).

Can someone help explain why this is the case?
Because $BF$ and $CF$ they are similar medians of congruent triangles, which gives $$BF=CF$$ and
$$FE\perp BC.$$
The line FE is contained in the plane of symmetry DAE and therefore FE is orthogonal to BC.