How to proof that two lines in cube are perpendicular, without use of vectors

Question

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)

Answer 1

Perpendicular means if you translate $BD$ so that it begins at $A$ instead, the resulting lines are perpendicular. So translate $ABCD$ over to the left to get a square in the same plane, say $A'ADD'.$ Note that $C_1 D' = \sqrt{5}, AC_1 = \sqrt{3},$ and $AD' = \sqrt{2},$ so this is a right triangle.

Answer 2

Let $M$ be the midpoint of $BD$ and $O$ be the midpoint of $AC_1$. Then $OM$ is perpendicular to $DB$, and $AC$ is perpendicular to $DB$. It follows that $DB$ is perpendicular to two different lines in the plane $A\vee C\vee C_1$, hence to all lines in this plane, in particular to $A\vee C_1$.