Let $X=\{p_1,p_2,p_3,p_4\}$ be four points in $\mathbb{R}^2$, not three of them on a line. We use the square bracket notation $[i,j,k]=\det \begin{pmatrix} p_i & p_j & p_k \\ 1 & 1 & 1 \end{pmatrix},$ which denotes twice the signed area of the triangle spanned by $p_i,p_j,p_k$. For every pair $i,j$ from $\{1,2,3,4\}$ we set
$A_{ij}:= \frac{[i,j,a][i,j,b]}{[i,j,k][i,j,l]},\quad \text{with $\{i,j,k,l\}=\{1,2,3,4\}$}.$
Note that $\{i,j,k,l\}=\{1,2,3,4\}$ implies that $i,j,k,l$ are distinct.
Then for any two points $p_a$ and $p_b$ (not necessarily from $X$) the following holds $ \sum_{1\le i\lt j\le4} A_{ij}=1. $ Or if you prefer it to write out the definition of the 6 $A_{ij}$s $\frac{[1,2,a][1,2,b]}{[1,2,3][1,2,4]}+\frac{[1,3,a][1,3,b]}{[1,3,2][1,3,4]} +\frac{[1,4,a][1,4,b]}{[1,4,3][1,4,2]}+\cdots +\frac{[3,4,a][3,4,b]}{[3,4,1][3,4,2]} =1.$ This can be shown algebraically, see here on page 717.
I am interested in a geometric interpretation of this invariant. Can this be proven by elementary theorems from (projective?) geometry?
I want to understand the invariant, because I am interested in higher dimensional analogues.