Intersections of lines formed by the pentagon are collinear. Given a convex pentagon $ABCDE$ consider 5 points $$I=\overleftrightarrow{AB}\cap \overleftrightarrow{CE},G=\overleftrightarrow{BC}\cap \overleftrightarrow{DA}, H=\overleftrightarrow{CD}\cap \overleftrightarrow{EB},J=\overleftrightarrow{DE}\cap \overleftrightarrow{AC}, F=\overleftrightarrow{EA}\cap \overleftrightarrow{BD}$$
(We assume that all of these lines intersect.)

I want to prove the following:

If four of the points $F,G,H,I,J$ are collinear then all of them are collinear.

I tried to apply Pappus and Desargues theorems and I will be satisfied with the solution from these theorems.
This problem can be also considered as incidence geometry property (replace the "convex" assumption with "no three of the vertices are collinear"). I wonder if this is true in any Pappian/Desarguesian projective plane.
Any help appreciated!
PS. The figure is not fully precise.
 A: This is not yet the kind of proof based on Pappus and Desargues that you're after. But it might shed some light so I'm posting it now and may add more later in an edit or a separate answer.
This is a coordinate-based approach. As an incidence configuration, the whole setup is invariant under projective transformations. Therefore without loss of generality, assume
$$
A=\begin{pmatrix}1\\0\\0\end{pmatrix}\quad
B=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad
C=\begin{pmatrix}0\\0\\1\end{pmatrix}\quad
D=\begin{pmatrix}1\\1\\1\end{pmatrix}\quad
E=\begin{pmatrix}x\\y\\1\end{pmatrix}\;.
$$
Then you get the intersections as
$$
F=\begin{pmatrix}1\\y\\1\end{pmatrix}\quad
G=\begin{pmatrix}0\\1\\1\end{pmatrix}\quad
H=\begin{pmatrix}x\\x\\1\end{pmatrix}\quad
I=\begin{pmatrix}x\\y\\0\end{pmatrix}\quad
J=\begin{pmatrix}x-y\\0\\1-y\end{pmatrix}\;.
$$
Now the collinearity $FGH$ becomes a vanishing determinant
$$[FGH]=\begin{vmatrix}
1&0&x\\
y&1&x\\
1&1&1
\end{vmatrix}
= 1 + xy - 2x \overset!=0
$$
and likewise $FGI$ gives you
$$[FGI]=\begin{vmatrix}
1&0&x\\
y&1&y\\
1&1&0
\end{vmatrix}
= xy - x - y \overset!=0 \;.
$$
You can combine them to $[FGH]-[FGI]=1-x+y \overset!=0$ to conclude $y=x-1$. Substituting that into the collinearity $FGJ$ you get
$$[FGJ]=\begin{vmatrix}
1&0&1\\
x-1&1&0\\
1&1&2-x
\end{vmatrix}
= 0\;.
$$
So you see, whenever the first two collinearities hold, the last one will hold as well.
You can also substitute $y$ into one of the first two collinearities to determine
$$x_{1,2}=\frac{3\pm\sqrt5}2$$
so up to projective transformations there are only two distinct pentagons which satisfy your constraints. One of them is surely the regular pentagon (with all its projectively transformed variants), since for that all the intersections lie on the line at infinity. I don't have a good intuition yet about the second solution, and I don't have a good dynamic geometry application at hand just now to better explore this.
So when you try to prove this in a more generic setting, for arbitrary planes, one possible approach might be to first show that four collinear points already imply one of the two special cases, and then demonstrate that the fifth collinear point is satisfied for both of these.
Note that some Desarguesian planes might not have any such pentagon at all. The rational projective plane would be one such example. But that doesn't mean a proof using Desargues is bound to fail: if the prerequisite of four collinear points can never be met, we don't have to worry about the conclusion.
