# 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)

• Hint: Project AC1 onto the plane ABCD. Commented Jan 13, 2016 at 18:13
• So the point C is the projection of C1 on the plane ABCD => AC is the projection of AC1 in (ABCD) and since BD and AC are diagonals of the square ABCD BD is perpendicular to AC. Am I right? Commented Jan 13, 2016 at 18:16
• These lines are not coplanar, they are skew! Commented Jan 13, 2016 at 18:24
• It might be not so easy. Being perpendicular in 3D space is not the same as being perpendicular in 2D projection. Think of projecting in the D C C1 D1 plane you get an angle of 45°. The other way round you can think of non perpendicular lines appearing perpendicular in one special projection. Commented Jan 13, 2016 at 18:25
• Kellerspeicher could you be more specific, about the special projection? Commented Jan 13, 2016 at 18:35

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 $A\vee C_1$.

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.

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!

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. $$\square$$