Complete quadrilateral: projective version of Newton-Gauss line

Question

There is a well known theorem that the midpoints of the three diagonals of a complete quadrilateral are collinear (on the Newton-Gauss line). It appears that if you intersect the diagonals with a line, the harmonic conjugates of those intersection points will also be collinear. My question is: how do you prove it?

As shown in the diagram, a complete quadrilateral has vertices $A,B,C,D,E,F$ and diagonals $AC,BD,EF$. A line $\mathscr l$ intersects the diagonals at points $I,J,K$. The harmonic conjugates of these points with respect to the segments $[AC],[BD],[EF]$ are $I',J',K'$. I'd like to show that $I',J',K'$ are collinear.

The motivation is that this is a generalization of the collinearity of midpoints, because midpoints are the harmonic conjugates of the intersections that result when $\mathscr l$ is the line at infinity.

Answer 1

Apply a central projection that takes $l$ to the line at infinity. Since the cross-ratio is preserved, the projections of $I', J', K'$ are the midpoints of the projections of $AC$, $BD$, $EF$ and are collinear by the quoted theorem, therefore $I', J', K'$ are collinear.

Answer 2

I'm learning the ropes of projective geometry and thought that proving the proposition using ratio chasing would be a good exercise. It took a while, but I learned a lot. The proof here is adapted from the Euclidean proof given at https://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml. It is the Proof #2 on that page, and uses Menelaus' theorem and the following diagram.

The basic idea is to show that the Menelaus proportions for the triangle $EDC$ with respect to the line $AFB$ are the same as the proportions for the triangle $PQR$ with respect to the putative line $KLM$. Because the proportions are the same, $KLM$ is indeed a line. We have to do a little work to translate this argument to the projective case, which is shown in the following diagram.

The diagram is drawn to correspond to the Euclidean diagram, so is labeled, arranged, and colored differently from the OP diagram. Line $\mathscr l$ has been renamed to $\Omega$. $K,L,M$ (corresponding to OP $I',J',K'$) are the harmonic conjugates of the intersections of the diagonals with $\Omega$.

We want to prove that $K,L,M$ are collinear.

1) Let $PQR$ be the Cevian triangle of $CDE$ with perspectrix $\Omega$ (the green line). Triangle $PQR$ is constructed by intersecting the side lines of $CDE$ with $\Omega$ and then taking harmonic conjugates of the intersections. For example, $P'=DE\cap\Omega$ and $P$ is the harmonic conjugate of $P'$ with respect to the segment $[DE]$. Similarly, $Q$ and $R$ are the harmonic conjugates of $Q'=CD\cap\Omega$ and $R'=EC\cap\Omega$ (not shown).

2) The lines $P'E,P'R,P'C,\Omega$, coincident at $P'$, are a harmonic set $H$ because they intersect a harmonic set of points $E,R,C,R'$ (where $R'=EC\cap\Omega$). Thus the intersection of any line with $H$ yields a harmonic set or points. In particular, intersecting the diagonal $AC$ with $H$ yields the points $M',A,X,C$. But since $M$ is the harmonic conjugate of $M'$, $M$ must be on the line $P'R$.

As in the Euclidean proof, we are ready to compare the ratios $RM/MQ$ and $EA/AD$. Unlike the Euclidean proof, they are not equal. Fortunately we can chase ratios to establish how they are related.

3) The cross ratios $(RQ;MP')$ and $(ED;AP')$ are equal via the perspective center $C$. Thus $$\frac{RM}{MQ}:\frac{RP'}{P'Q}=\frac{EA}{AD}:\frac{AP'}{P'D}$$ and $$\frac{RM}{MQ}=\frac{EA}{AD}\cdot\frac{RP'}{P'Q}:\frac{AP'}{P'D}.$$

By symmetric arguments we can show that $K$ is on line $QP$ and $L$ is on $PR$ and derive formulae for $QK/KP$ and $PL/LR$.

4) Define a Menelaus function $\operatorname{Mn}$ on triangles and points: $$\operatorname{Mn}(\triangle ABC;XYZ)=\frac{AX}{XB}\cdot\frac{BY}{YC}\cdot\frac{CZ}{ZA}.$$ Putting together the results from 3) we get $$\operatorname{Mn}(\triangle PQR;KLM)=\operatorname{Mn}(\triangle CDE;AFB)\cdot \operatorname{Mn}(\triangle PQR;P'Q'R') : \operatorname{Mn}(\triangle CDE;P'Q'R')$$

Because $P',Q',R'$ and $A,F,B$ are both collinear, the three terms on the right hand side are all $-1$. Thus $\operatorname{Mn}(\triangle PQR;KLM)=-1$, showing that $K,L,M$ are collinear.