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)
 A: 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 $A\vee C_1$.
A: 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.
A: Without calculation. Just use axioms.

*

*$AC$ is perpendicular to $BD$.

*$A'C'$ is perpendicular to $B'D'$.

There is an axiom saying that

There is a single plane passing through two parallel lines.


*

*The plane $ACC'A'$ is perpendicular to $BDD'B'$.

Another axioms:

If a line $L$ is perpendicular to a plane then every line on the plane is perpendicular to $L$.


*

*$BD$ is perpendicular to plane $ACC'A'$ so the line $AC'$ on the plane is also perpendicular to $BD$.


*Done!
A: 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. $\square$
