Basic geometry proof about tetrahedron Suppose tetrahedron ABCD such that AB is orthogonal to CD and AC is orgthongal to BD. Show that AD is orthogonal to BC.
So i made a picture of a tetrahedron in 3 space and sort of look down at it from the z axis, sort of the jist i get from this is that the vectors coming out of the vertices of the base must be perpendicular with the opposite side if the base is a triangle, which it must be for a tetrahedron. Ive sort of come to a road block because i cant show specifically why this must be true.
I sort of attempted to go about it with proof by contradiction ~(P->Q)= P and ~Q. I thought that i would arrive at a contradiction that one of the faces would not have a property of being a triangle. I sort of was also questioning this because if i wrote something like suppose tetrahedron ABCD such that AB is orthogonal to CD and AC is orthogonal to BD and AD not orthogonal to BC wouldnt this be a fundamental problem because i suppose the existence of a tetrahedron and gave it properties which makes it not a tetrahedron, seems troublesome to me. If someone could give me insight on this proof and sort of give me an idea of showing it with and without contradiction that would be great. My area of mathematics is partial differential equations so please use more advanced notation if needed. Im doing differential geometry and just trying to do the basics, which as you can see is not going to well.
 A: For this kind of thing, I usually coordinatize the tetrahedron by aligning $\overline{AB}$ with the $z$-axis, and having $\overline{CD}$ in the $xy$-plane, parallel to the $y$ axis and at distance $h$ from the origin:
$$A = (0,0,a) \qquad B = ( 0, 0, b ) \qquad C = (h, c, 0 ) \qquad D = ( h, d, 0 )$$
By design, $\overline{AB}\perp\overline{CD}$, as we can verify by computing $(A-B)\cdot(C-D) = 0$.
Also, $$(A-C)\cdot(B-D) = h^2 + a b + c d = (A-D)\cdot(B-C)$$
so that the dot products of the remaining pairs of opposing vectors vanish simultaneously, or not at all; thus,
$$\overline{AC}\perp\overline{BD} \quad \Leftrightarrow \quad \overline{AD}\perp\overline{BC}$$
A: Use vectors, and take the origin to be $A$, and let the position vectors of $B, C, D$ be $b, c, d$.
You are given that $AB$ is perpendicular to $CD$, so that, in term of scalar products:
$$b.(c-d) = 0$$
i.e. $$b.c = b.d$$
Similarly, 
$$c.(b-d) =0$$
i.e. 
$$c.b = c.d$$
You can then deduce that $b.d = c.d$, which gives $d.(b-c) = 0$ which gives the required conclusion.
