I think I have the proof and I just need a second check (and some advice to make it less messy).

Given four concurrent planes, for any straight line interesecting the four planes in the points $a,b,c,d$ the cross product defined as ${\frac {ac} {bc}} / {\frac {ad} {bd}}$ is costant, regardless of the position of the straight line.

I will use a double projection.


Let $A,B,C,D$ be four planes concurrent in the straight line $l$. Let $r,r'$ be two straight lines intersecting the planes in $a,b,c,d$ and $a',b',c',d'$, respectively, and laying ont the planes $P$ and $P'$ which in turn intersect the $l$ in $O$ (we can choose, without loss of generality, a single point on $l$ in which the two planes $P$ and $P'$ and the straight line $l$ intersect).

Choose a third straight line $r''$ such that it is parallel to $r'$ and intersecting with $r$. Then $r''$ will lay on $P$. By means of a projection from $O$, we find the points $a'',b'',c'',d''$ on $r''$. Then, for previous results (invariance of the cross product after a central projection) we will have $${\frac {a'c'} {b'c'}} / {\frac {a'd'} {b'd'}} = {\frac {a''c''} {b''c''}} / {\frac {a''d''} {b''d''}}$$ We can now prove the proposition. By means of a parallel projection of $r''$ into $r$, we find that $a'',b'',c'',d''$ will correspond to $a,b,c,d$, respectively. In fact, having chosen $r''$ on as concurrent to $r$, these two straight lines will lay on the same plane $Q$ that intersects $A,B,C,D$ in four parallel straight lines $p,q,r,s$ where $a,b,c,d$ and $a'',b'',c'',d''$ lay. But then, for the fact that a parallel projection holds the cross product of any four points, we have $${\frac {ac} {bc}} / {\frac {ad} {bd}} = {\frac {a''c''} {b''c''}} / {\frac {a''d''} {b''d''}}$$ and therefore, for transitivity, the conclusion. $${\frac {a'c'} {b'c'}} / {\frac {a'd'} {b'd'}} = {\frac {a'c'} {b'c'}} / {\frac {a'd'} {b'd'}}$$ Q.E.D.


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