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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?

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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.

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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.

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  • $\begingroup$ Nice answer, especially because it was more or less in front of my nose. Do you have any idea of whether the proposition could be proven by drawing a couple of extra lines with some ratio chasing and a dash of Menelaus? It seems hard enough to prove it for the Newton-Gauss (midpoint) case. $\endgroup$ – brainjam Jun 2 '18 at 21:39
  • $\begingroup$ You can try to generalize one of the proofs from here or here. But it seems like doing extra work. $\endgroup$ – Maxim Jun 3 '18 at 13:19
  • $\begingroup$ Another candidate for generalization: math.stackexchange.com/a/1618522/1257 $\endgroup$ – brainjam Jun 26 '18 at 3:12
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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. enter image description here

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.

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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.

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