Concurrency of the heights of a tetrahedron with opposite edges perpendicular. Can anyone give me a vectorial solution to the following problem:

Prove that if each pair of opposite edges of the tetrahedron $ABCD$ is perpendicular (that is, $AB \perp CD$ and $AC \perp BD$ and $AD \perp BC$), then the heights of the tetrahedron are concurrent.

Here, the heights (also known as the altitudes) of tetrahedron $ABCD$ are the perpendicular from $A$ to the plane $BCD$, and three other similarly defined perpendiculars.
Tetrahedra satisfying the condition of this problem are called orthocentric, and this appears to be a known result.
 A: That is a simple exercise in visualization. Imagine that $A,B,C$ are embedded in the $xy$ plane (the screen) and $D$ lies on the $z$-axis (orthogonal to the screen), so that the origin $O$ is the projection of $D$ on the plane through $A,B,C$. Since $DB\perp AC$ (in $3$D) we have $OB\perp AC$ (in $2$D). Similarly we get $OA\perp BC$ and $OC\perp AB$, hence $O$ is the orthocenter of $ABC$.

The orthocenter $H_A$ of the $BCD$ face lies on on the line joining $D$ with its projection on $BC$, hence the projection of $H_A$ on the $ABC$ plane lies on the $AO$ line. In particular the lines $AH_A,BH_B,CH_C,DH_D$ are concurrent when projected on the $ABC$ plane. The same holds by replacing $ABC$ with any face of the tetrahedron, hence the lines $AH_A,BH_B,CH_C,DH_D$ are concurrent in the $3$D space.
A: Here is a proof using vectors (since I don't trust pictures too much):
We let $\left\langle \mathbf{x},\mathbf{y}\right\rangle $ denote the scalar
product of two vectors $\mathbf{x}$ and $\mathbf{y}$. If $X$, $Y$ and $Z$ are
three non-collinear points in space, then $XYZ$ shall denote the plane through $X$, $Y$ and $Z$.
We first notice the following simple fact:

Lemma 1. Let $\mathcal{P}$ be a plane, and let $x$ and $y$ be two
  non-parallel lines on $\mathcal{P}$. Let $v$ be a line such that $v\perp x$
  and $v\perp y$. Then, $v\perp\mathcal{P}$.

This is just saying that a line $v$ is orthogonal to a plane $\mathcal{P}$ if
it is orthogonal to some two non-parallel lines on $\mathcal{P}$. If you
define your stereometry synthetically by some axioms (not that I know what
these axioms would be), then Lemma 1 is probably one of the fundamental
results or even one of the axioms. If you define your stereometry as the study
of vectors in $\mathbb{R}^{3}$, then it easily follows from the bilinearity of
the scalar product (indeed, if we pick nonzero vectors $\mathbf{v}$,
$\mathbf{x}$ and $\mathbf{y}$ running along the lines $v$, $x$ and $y$,
respectively, then $v\perp x$ and $v\perp y$ lead to $\left\langle
\mathbf{v},\mathbf{x}\right\rangle =0$ and $\left\langle \mathbf{v}
,\mathbf{y}\right\rangle =0$, which entails that $\left\langle \mathbf{v}
,\lambda\mathbf{x}+\mu\mathbf{y}\right\rangle =0$ for all $\lambda,\mu
\in\mathbb{R}$; but this yields that $v$ is orthogonal to any line on
$\mathcal{P}$).
Now, let $H$ be a common point of


*

*the plane through $B$ orthogonal to the line $AC$;

*the plane through $C$ orthogonal to the line $AD$;

*the plane through $D$ orthogonal to the line $AB$.
(Such an $H$ exists, since the lines $AB$, $AC$ and $AD$ are pairwise non-parallel.)
The point $H$ lies on the plane through $B$ orthogonal to the line $AC$ (by
its definition). Thus, $HB\perp AC$. Similarly, $HC\perp AD$ and $HD\perp BC$.
We have $\overrightarrow{HD}=\overrightarrow{HB}+\overrightarrow{BD}$, and
thus
\begin{align*}
\left\langle \overrightarrow{HD},\overrightarrow{AC}\right\rangle  &
=\left\langle \overrightarrow{HB}+\overrightarrow{BD},\overrightarrow{AC}
\right\rangle =\underbrace{\left\langle \overrightarrow{HB}
,\overrightarrow{AC}\right\rangle }_{\substack{=0\\\text{(since }HB\perp
AC\text{)}}}+\underbrace{\left\langle \overrightarrow{BD},\overrightarrow{AC}
\right\rangle }_{\substack{=0\\\text{(since }BD\perp AC\text{)}}}\\
& =0+0=0.
\end{align*}
In other words, $HD\perp AC$. Now, the lines $BC$ and $AC$ are two
non-parallel lines on the plane $ABC$. Hence, from $HD\perp BC$ and $HD\perp
AC$, we obtain $HD\perp ABC$ (by Lemma 1, applied to $\mathcal{P}=ABC$,
$v=HD$, $x=BC$ and $y=AC$). Thus, $H$ belongs to the height of the tetrahedron
$ABCD$ from the vertex $D$.
We have shown that $HD\perp ABC$. The same argument (but with $C$, $D$ and $B$
playing the roles of the points $B$, $C$ and $D$) yields $HB\perp ACD$
(because our definition of $H$ does not change if we cyclically rotate the
variables $B$, $C$ and $D$). Thus, $H$ belongs to the height of the
tetrahedron $ABCD$ from the vertex $B$.
We have shown that $HB\perp ACD$. The same argument (but with $C$, $D$ and $B$
playing the roles of the points $B$, $C$ and $D$) yields $HC\perp ADB$
(because our definition of $H$ does not change if we cyclically rotate the
variables $B$, $C$ and $D$). Thus, $H$ belongs to the height of the
tetrahedron $ABCD$ from the vertex $C$.
We have $HD\perp ABC$ and thus $HD\perp BC$ (since the line $BC$ lies on the
plane $ABC$). Now, $\overrightarrow{HA}=\overrightarrow{HD}
-\overrightarrow{AD}$, so that
\begin{align*}
\left\langle \overrightarrow{HA},\overrightarrow{BC}\right\rangle  &
=\left\langle \overrightarrow{HD}-\overrightarrow{AD},\overrightarrow{BC}
\right\rangle =\underbrace{\left\langle \overrightarrow{HD}
,\overrightarrow{BC}\right\rangle }_{\substack{=0\\\text{(since }HD\perp
BC\text{)}}}-\underbrace{\left\langle \overrightarrow{AD},\overrightarrow{BC}
\right\rangle }_{\substack{=0\\\text{(since }AD\perp BC\text{)}}}\\
& =0-0=0.
\end{align*}
In other words, $HA\perp BC$. The same argument (but with $C$, $D$ and $B$
playing the roles of the points $B$, $C$ and $D$) yields $HA\perp CD$ (because
our definition of $H$ does not change if we cyclically rotate the variables
$B$, $C$ and $D$). Now, the lines $BC$ and $CD$ are two non-parallel lines on
the plane $BCD$. Hence, from $HA\perp BC$ and $HA\perp CD$, we obtain $HA\perp
BCD$ (by Lemma 1, applied to $\mathcal{P}=BCD$, $v=HA$, $x=BC$ and $y=CD$).
Thus, $H$ belongs to the height of the tetrahedron $ABCD$ from the vertex $A$.
We have now proven that the point $H$ belongs to all four heights of the
tetrahedron $ABCD$. Hence, these four heights are concurrent. $\blacksquare$
A: Let $\vec{DA}=\vec{a},$ $\vec{DB}=\vec{b}$ and $\vec{DC}=\vec{c}.$
Thus, $$AB^2+CD^2=AC^2+BD^2$$ it's
$$(\vec{b}-\vec{a})^2+\vec{c}^2=(\vec{c}-\vec{a})^2+\vec{b}^2$$ or
$$\vec{a}\vec{b}=\vec{a}\vec{c}$$ or
$$\vec{a}(\vec{b}-\vec{c})=0$$ or $$AD\perp BC.$$
Similarly, we obtain that $BD\perp AC$ and $CD\perp AB.$
Now, let $DK$ be an altitude of the tetrahedron and $AK\cap BC=\{E\}.$
Also, let $AF$ be an altitude of $\Delta ADE$.
We see that $BC\perp AE$ and $BC\perp DK$.
Thus, $BC\perp(ADE)$ and since $AF\subset(ADC),$ we obtain $BC\perp AF$.
Id est, $AF\perp BC$ and $AF\perp DF,$ which says $AF\perp(DBC)$ and we got that $AF$ is an altitude of the  tetrahedron.
But $AF\subset(ADE)$ and $DK\subset(ADE),$ which says that $AF$ and $DK$ intersect. 
Can you end it now?
