Question about Pasch's Postulate, line going through all three sides of a triangle I've been reading the textbook Elementary Geometry from an Advanced Standpoint by Edwin E. Moise (3rd ed.).  My problem with his wording of Pasch's Postulate, and then a subsequent problem which aggravates my confusion.
To start, this is his exact wording for Pasch's Postulate:
"The Posutulate of Pasch: Given a triangle $\triangle ABC$, and a line $L$ in the same plane.  If $L$ contains a point $E$, between $A$ and $C$, then $L$ intersects either $\overline{AB}$ or $\overline{BC}$."
This could mean one of two things:  Either that the line $L$ must intersect exactly one of either $\overline{AB}$ or $\overline{BC}$; or that the line $L$ must intersect at least one of either $\overline{AB}$ or $\overline{BC}$ (it can't turn around and leave).
When I look at the other resources (mostly wikipedia), I lean towards the former.  However, in the same section, the reader is asked to prove that: "If $L$ contains no vertex of the triangle, then $L$ cannot intersect all of the three sides" from Pasch's Postulate.
If the former were true, the proof of this statement would be "Pasch's Postulate QED," which seems a little bit too easy.  However, I've spent hours trying to figure out how to do it with the latter defintion (it could intersect both), but I just get so confused because my lines aren't straight.  I'm completely at a loss.
So my questions is:
Which of my two interpretations is correct?  Can this problem be proved without using the former definition?
EDIT:
I've realized (with the help of one Andre Nicolas in the comments), that clearly the former definition doesn't work because the line could pass through $B$.  However, that leaves me with the same problem:  How does one go about proving that the line does not go through all three sides of the triangle without passing through a vertex?
 A: See Hilbert's statement of Pasch's axiom :

David Hilbert uses Pasch's axiom in his book Foundations of Geometry (1902 – English translation by E. J. Townsend, 1950) .  It is numbered II.5 and is stated as:

Let $A, B, C$ be three points that do not lie on a line and let $\mathit a$ be a line in the plane $ABC$ which does not meet any of the points $A, B,C$. If the line a passes through a point of the segment $\overline {AB}$, it also passes through a point of the segment $\overline {AC}$, or through a point of segment $\overline {BC}$.
The fact that both segments $\overline {AC}$ and $\overline {BC}$ are not intersected by the line $\mathit a$  is proved in Supplement I,1, written by P. Bernays.


We stay with a “standard” reading of Pasch’s axiom, considering the “or” as inclusive, and we have to prove that :

If $A, B, C$ are noncollinear points and $\mathit l \cap \{ A,B,C \}  = \emptyset$, then $\mathit l$ cannot intersect all three sides of $\Delta ABC$.

Note. I will use the relation $between(x,z,y)$, $x, y, z$ points, to say that $z$ is between $x$ and $y$.
Proof
Suppose $\{ D \} = \mathit l \cap \overline {AB}$ (thus : $between(A, D, B)$), $\{ E \} = \mathit l \cap \overline {AC}$ (thus : $between(A, E, C)$) and $\{ F \} = \mathit l \cap \overline {BC}$ (thus : $between(B, F, C)$), and suppose $between(D, E, F)$.
Now $\overleftrightarrow{BD} = \overleftrightarrow{AB}$ and $\overleftrightarrow{BF} = \overleftrightarrow{BC}$, so $B, D$, and $F$ are not collinear.
Now $\overline {AC} \cap \overleftrightarrow{DF} =  \{ E \}$, because $E \in \overline {AC} \cap \mathit l$ and if there was an $X \in \overline {AC} \cap \overleftrightarrow{DF}$ with $X \ne E$, then $\overleftrightarrow{AC}$ and $\mathit l$ would have two points in common and they would coincide (Axiom I.1).
If we apply Pasch’s axiom to $\Delta DBF$ and line $\overleftrightarrow{AC}$, we must have either :

$\overleftrightarrow{AC} \cap \overline {BD} \ne \emptyset$ or $\overleftrightarrow{AC} \cap \overline {BF} \ne \emptyset$.

But $\overleftrightarrow{AC} \cap \overline {BD} \subset \overleftrightarrow{AC} \cap \overline {BA} = \{ A \}$, and $A \notin \overline {BD}$ (since $between(A, D, B)$); so $\overleftrightarrow{AC} \cap \overline {BD} = \emptyset$.
Also, $\overleftrightarrow{AC} \cap \overline {BF} \subset \overleftrightarrow{AC} \cap \overline {BC} = \{ C \}$, and $C \notin \overline {BF}$ (since $between(B, F, C)$); so $\overleftrightarrow{AC} \cap \overline {BF} = \emptyset$.
Hence our assumptions contradict Pasch’s axiom.
