If the four altitudes of a tetrahedron $ABCD$ intersect in a point $H$, $ \overrightarrow{HA} \cdot ( \overrightarrow{HB}-\overrightarrow{HC} ) = \overrightarrow{HA} \cdot \overrightarrow{CB} = 0 $ so $ \overrightarrow{HA} \cdot \overrightarrow{HB} = \overrightarrow{HA} \cdot \overrightarrow{HC} $ and so on then $ AB^2 + CD^2 = |\overrightarrow{HA} - \overrightarrow{HB}|^2 + |\overrightarrow{HC} - \overrightarrow{HD}|^2 $ $ = HA^2 + HB^2 + HC^2 + HD^2 - 2\overrightarrow{HA} \cdot \overrightarrow{HB} - 2\overrightarrow{HC} \cdot \overrightarrow{HD} $ $ = HA^2 + HB^2 + HC^2 + HD^2 - 2\overrightarrow{HB} \cdot \overrightarrow{HC} - 2\overrightarrow{HA} \cdot \overrightarrow{HD} $ $ = |\overrightarrow{HB} - \overrightarrow{HC}|^2 + |\overrightarrow{HA} - \overrightarrow{HD}|^2 = BC^2 + AD^2 $ similarly $ BC^2 + AD^2 = CA^2 + BD^2 $, hence \[AB^2 + CD^2 = BC^2 + AD^2 = CA^2 + BD^2.\]

Conversely, if $ AB^2 + CD^2 = BC^2 + AD^2 = CA^2 + BD^2 $, Since $ BC^2+AD^2 = AB^2+AC^2-2\overrightarrow{AB}\cdot\overrightarrow{AC}+AD^2 $ and $ CA^2+BD^2 = CA^2+AB^2+AD^2-2\overrightarrow{AB}\cdot\overrightarrow{AD} $, from $ BC^2 + AD^2 = CA^2 + BD^2 $ we have $ \overrightarrow{AB}\cdot\overrightarrow{AC} = \overrightarrow{AB}\cdot\overrightarrow{AD} $, so $ \overrightarrow{AB}\cdot\overrightarrow{CD} = \overrightarrow{AB}\cdot\overrightarrow{AD} - \overrightarrow{AB}\cdot\overrightarrow{AC} = 0 $, that is, $AB$ is orthogonal to $CD$ Similarly $AC$ is orthogonal to $BD$, $AD$ is orthogonal to $BC$ Let $AA'$, $BB'$, $CC'$, $DD'$ be altitudes of tetrahedron $ABCD$. Since $AA'$ and $AB$ are both orthogonal to $CD$, $A'$ and $B$ are on the plane orthogonal to $CD$ and containing $A$. Since $BB'$ is also orthogonal to $CD$, $B'$ is also on the plane. Hence $AA'$ and $BB'$ are on the same plane and they are not parallel (otherwise $ABCD$ will degenerate), so they intersect at a point, called $H$ We now have $AH$ orthogonal to plane $BCD$ and $BH$ orthogonal to plane $ACD$. Since $AD$ is orthogonal to $BC$ and $BH$, it is also orthogonal to $CH$, and similarly $BD$ is orthogonal to $CH$. Hence $CH$ is orthogonal to plane $ABD$, so $H$ is on the altitude $CC'$. Similarly $DH$ is orthogonal to plane $ABC$, $H$ is on the altitude $DD'$. Finally we have the four altitudes of tetrahedron $ABCD$ intersect in point $H$