Show that the harmonic conjugates are collinear - Menelaus and Ceva's Theorems with homogeneous coordinates

Question

Suppose we have a triangle $\triangle ABC$ such that $D$ is an arbitrary point on $BC$, $E$ is an arbitrary point on $AB$ and $F$ is an arbitrary point on $AC$. Let $G$ be the harmonic conjugate of $E$ with respect to $AB$ (the cross ratio $(A,B,E,G)=-1$), $I$ the harmonic conjugate of $D$ with respect to $BC$ (the cross ratio $(B,C,D,I)=-1$) and $H$ the harmonic conjugate of $F$ with respect to $AC$ (the cross ratio $(A,C,F,H)=-1$). Show that $G,I,H$ are collinear if and only if $AD$, $CE$ and $BF$ have a point in common.

The picture would look like this (I amended the pic as pointed out in the comments).

My idea is to convert $A,B,C$ to homogeneous coordinates, for example, $A=(1:0:0), B=(0:0:1)$ and $C=(0:1:0)$, but I am not sure where to go from there. I read that these are equivalent statements from Menelaus and Ceva's theorems but I cannot find anything on the internet with homogeneous coordinates, so any help is appreciated.

Answer 1

First have a look here showing that this property, called "trilinear polarity" is explained using trilinear coordinates (cousin of barycentric coordinates) with the very nice correspondence:

$$\text{Common point's trilinear coord.} \ (p,q,r) \ \leftrightarrow \ \text{line GIH trilinear equ. } \ \tfrac{x}{p}+\tfrac{y}{q}+\tfrac{z}{r}=0$$

Answer 2

I finally worked out the solution and I'd like to share it below. Let $A=(1:0:0)$, $B=(0:0:1)$, $C=(0:1:0)$, and denote $M:=AD\cap CE$. Then let $M=(1:1:1)$ since $A,B,C,M$ are in general position so we can choose this labelling. Since $F\in AC$ we can write it as $F=(1:a:0)$ for some $a\neq 0$ by homogeneity. We use harmonic conjugacy as follows.

We know that two points $F'$ and $G'$ are harmonic conjugates with respect to $D',E'$ if $(D',E',F',G')=-1$ so if $F=\lambda_1D'+ \mu_1E'$ then $G=-\lambda_1D'+\mu_1E\sim \lambda_1D'-\mu_1E'$ since the cross-ratio is then $\frac{\lambda_1}{\mu_1}:\frac{-\lambda_1}{\mu_1}=-1$.

Then $(1:a:0)=F=1\cdot A + a\cdot C$ so $H=-1\cdot A +a \cdot C$ because they are harmonic conjugates. So $H=(-1:a:0)$. Similarly $I=(0:-1:1)$ and $G=(-1:0:1)$.

The three lines are concurrent if and only if $M\in BF$. The equation for $BF$ is $X_1=aX_0$ so $M\in BF$ if and only if $a=1$.

$G,H,I$ are collinear if and only if the determinant

$$\begin{vmatrix}-1 & a & 0 \\ 0 & -1 & -1 \\ -1 & 0 & 1\end{vmatrix}=0,$$ which happens if and only if $a=1$.

So we have equivalent statements.