Proving that the angle between the main diagonal of a cube and a skew diagonal of a face of the cube is 90 degrees

Question

I need to prove that the angle between the main diagonal of a cube and a skew diagonal of the face of the cube is 90 degrees. I can do this with vectors, but I have to use applications from projective geometry to prove this. Note that the skew diagonal is on the bottom face of the cube. I am a little unsure how to do this with projective geometry.

Answer 1

Make two cubes, one behind the other, so that the main diagonal of the back cube starts at its front lower left point $P$, the same point as where the skew diagonal of the bottom face of the front cube starts. Since the main diagonals of the two cubes are parallel, a copy of the angle you seek is now in a single triangle with one vertex at $P$, one at the lower right front of the front cube, and one at the upper right back of the back cube. The side lengths will then be $\sqrt{2},\ \sqrt{3},\ \sqrt{5}$ which satisfy $a^2+b^2=c^2$, making the triangle a right triangle by the converse of Pythagoras' theorem.