Solution:

Let $P$ be the point of intersection of the lines $AA'$, $BB'$, $CC'$, $DD'$, and consider the plane $\Pi$ passing through $A$, $P$, $B$. Since $\Pi$ contains the lines $AP$ and $BP$, it contains the points $A'$, $B'$, which lie on such lines; thus, denoting by $AK$ the bisector of the angle at $A$ in the triangle $ADC$ and by $BL$ the bisector of the angle at $B$ in the triangle $BCD$, we have that the points $K$ and $L$ also belong to $\Pi$, since the lines $AA'$ and $BB'$ (which contain these points) do. On the other hand, the line $CD$ is not contained in the plane $\Pi$ (otherwise the tetrahedron would be degenerate); it contains $K$ and $L$ and has at most one point of intersection with the plane, so $K$ and $L$ coincide. From this it obviously follows that $DK : KC = DL : LC$; but, by the angle bisector theorem, $DK : KC = AD : AC$ and $DL : LC = BD : BC$, whence $AD : AC = BD : BC$, that is $AD \cdot BC = AC \cdot BD$. The same reasoning applied to the plane through $B$, $P$, $C$ gives $AB \cdot CD = AC \cdot BD$, whence the thesis.