Can you prove that in every tetrahedron there is a vertex where the three edges meeting at that vertex have lengths which could be the sides of a triangle?
Show that the edges $AD$ and $BC$ of a tetrahedron $ABCD$ are mutually perpendicular if and only if $AB^2 +CD^2 = AC^2+BD^2$. This problem uses the scalar product of two vectors.
In a right-angled tetrahedron prove that the sum of the squares of the areas of the 3 faces in mutually perpendicular planes equals the square of the area of the sloping face. A generalisation of Pythagoras' Theorem.