Let ABCD be a plane quadrilateral (convex, concave or crossed), let E be the intersection of lines AB and CD, let F be the intersection of lines BC and DA, let G be the intersection of diagonals AC and BD and let the equations (oblique/orthogonal cartesian, trilinear or barycentric ) of its sides AB, BC, CD, DA be respectively
$$L_1=p_1x+q_1y+r_1z=0$$ $$L_2=p_2x+q_2y+r_2z=0$$ $$L_3=p_3x+q_3y+r_3z=0$$ $$L_4=p_4x+q_4y+r_4z=0$$
(in a oblique/orthogonal cartesian coordinate system we assume z=1)
Then we may say:
I. the diagonal AC can be represented by the following identical equations $$D_1=\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4=0$$ $$D_1=\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3=0$$
II. the diagonal BD can be represented by the following identical equations $$D_2=\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1-\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2=0$$ $$D_2=\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3=0$$
III. the line EF can be represented by the following identical equations $$D_3=\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1+\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3=0$$ $$D_3=\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2+\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4=0$$
IV. the line EG can be represented by the following equation $$D_4=\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3=0$$
V. the line FG can be represented by the following equation: $$D_5=\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4=0$$
Proof
First, we note that all these determinants are different from zero, because three sides of a quadrilateral cannot be concurrent at a point.
Second, it's obvious that $$\begin{vmatrix}p_1 & q_1 & r_1 & 0\\p_2 & q_2 & r_2 & 0\\p_3 & q_3 & r_3 & 0\\p_4 & q_4 & r_4 & 0\end{vmatrix} \equiv 0$$
Now multiply the first columm of the determinant by $x$, the second columm by $y$, the third columm by $z$ and add them to the fourth columm. As doing this the determinant value doesn't change, we may state that $$\begin{vmatrix}p_1 & q_1 & r_1 & L_1 \\p_2 & q_2 & r_2 & L_2 \\p_3 & q_3 & r_3 & L_3 \\p_4 & q_4 & r_4 & L_4\end{vmatrix} \equiv 0$$
Therefore
$$-\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1+\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3+\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4 \equiv 0$$
Or: $$\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3 \equiv \begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4$$
Equating separately each member of this identity to zero, we get two identical equations of the same line:
$$\begin{vmatrix}p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_1-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_3 & q_3 & r_3\\\end{vmatrix}L_4=0$$ $$\begin{vmatrix}p_1 & q_1 & r_1\\p_3 & q_3 & r_3\\p_4 & q_4 & r_4\\\end{vmatrix}L_2-\begin{vmatrix}p_1 & q_1 & r_1\\p_2 & q_2 & r_2\\p_4 & q_4 & r_4\\\end{vmatrix}L_3=0$$
From the first equation we see that this line passes through A, and from the second we note that it passes through C. Therefore both equations represent the diagonal AC.
Likewise we arrive at the equations of BD and EF, $D_2$ and $D_3$.
As for $D_4$, it's easy to see that, besides being an equation of a line passing through E, $D_4=D_1+D_2$, so that this line passes also through G, concurrency point of diagonals AC and BD. Therefore $D_4$ represents the line EG.
Finally, as $D_5=D_1-D_2$, in like manner we prove that $D_5$ represents the line FG.
Is this proof correct?