The midpoints of an isosceles tetrahedron

Question

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?

Answer 1

Because $BF$ and $CF$ they are similar medians of congruent triangles, which gives $$BF=CF$$ and $$FE\perp BC.$$

Answer 2

The line FE is contained in the plane of symmetry DAE and therefore FE is orthogonal to BC.

Answer 3

Picture the tetrahedron as a physical body, with AD held horizontally, then BC will also sit horizontally. Note that due to the congruence of faces you cannot actually differentiate between AD and BC. FE now is an orthogonal line on both, hanging (or standing) upright.