How to proof that two lines in cube are perpendicular, without use of vectors Given: Cube $ABCDA_1B_1C_1D_1$
Prove that $BD$ is perpendicular to $AC_1$

I don't have any idea how to proof this. Also I can't use vectors(we didn't study them in school). I can use all theorems from the stereometry(I think another name for this is solid geometry, but basically we deal with 3d figures(finding their volume, area, angles between different sides etc..), planes, and lines in the space)

As long as we're answering old questions ...
Let $B'$ and $D'$ be the respective midpoints of $\overline{BB_1}$ and $\overline{DD_1}$, so that $\overline{B'D'}\parallel\overline{BD}$. Both $\overline{B'D'}$ and $\overline{AC_1}$ contain the center of the cube, making $\square AB'C_1D'$ a planar quadrilateral; this quadrilateral's sides are clearly congruent, making it a rhombus, so its diagonals, $\overline{AC_1}$ and $\overline{B'C'}$, are perpendicular. $\square$
