Suppose you're given a quadrilateral $WXYZ$ and a line $l$ which divides the quadrilateral into two parts.
I'm having some trouble proving the following statement:
The line $l$ intersects either $WY$ or $XZ$ or both of them.
At the very beginning, it seems very obvious. You're given a quadrilateral and a line through it, so $l$ has to intersect somewhere at least one diagonal, right?
Despite being intuitive, I have no idea on how to prove rigourusly the statement.
I tried to consider the diagonals as line dividing the plane into four fourth-planes, but had no succeed...
Any hints?
This seems to call for Pasch's axiom.
We are given that $l$ "divides" the quadrilateral. I take this to mean that it intersects at least one of its edges, wlog. it intersects the line segment $XY$. Then by Pasch, it intersects either line segment $XZ$ or $YZ$. In the first case we are done, hence assume $l$ intersects line segment $YZ$. By repeating the argument, $l$ also intersects line segment $ZW$, and by repeating again, also line segment $WX$. Such a constellation is in fact possible:
As you see, $l$ intersects none of the diagonal line segments. Now we need to argue that the number of parts the quadrilateral is divided into does not equal two. "Obviously" there are three parts. For a formal proof, however, we'd need a sufficiently rigorous definition of what a "part" is.