In tetrahedron ABCD heights AE, BF, CG, DH intersect in one point. Prove that plane ABC is parallel to EFG, BCD is parallel to HEG, ACD is parallel to EHF and BAD is parallel to FEH.
Note that by symmetry, the 4 statements are equivalent.
I think the easiest way to go is to show that the normal vectors of both planes are parallel.
1. Define AB and AC.
2. Cross-product to get the normal vector.
3. Define EF and EG.
4. Cross-product to get the normal vector .
5. Cross-product both normal vectors and get 0 (meaning they are parallel and hence that the planes are parallel).