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?


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?

  • $\begingroup$ The line could intersect both by going through $B$. So it is "at least one." But usually only one. $\endgroup$ – André Nicolas May 15 '14 at 20:58
  • $\begingroup$ @Andre That's a good point. Which suggests the latter interpretation is correct, but then I'm back to this problem I cannot for the life of me solve. How do I prove it doesn't go through both of the others, and not pass through a vertex? $\endgroup$ – weirdesky May 15 '14 at 21:04
  • $\begingroup$ In the instructor's manual for the fourth edition of Greenberg's Euclidean and Non-Euclidean Geometry, there is a list of about 73 statements that are equivalent to the parallel postulate, in the presence of the rest of the axioms presented. There may be several things called Pasch's postulate, not sure. Meanwhile, I do not have Moise's book, so I cannot immediately help. $\endgroup$ – Will Jagy May 15 '14 at 21:22
  • $\begingroup$ I do not know what axioms of betweeness Moise uses. The proof will lie there. $\endgroup$ – André Nicolas May 15 '14 at 23:10

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 $y$.


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 $\overline {BD} = \overline {AB}$ and $\overline {BF} = \overline {BC}$, so $B, D$, and $F$ are not collinear.

Now $\overline {AC} \cap \overline {DF} = \{ E \}$, because $E \in \overline {AC} \cap \mathit l$ and if there was an $X \in \overline {AC} \cap \overline {DF}$ with $X \ne E$, then $\overline {AC}$ 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 $\overline {AC}$, we must have either :

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

But $\overline {AC} \cap \overline {BD} \subset \overline {AC} \cap \overline {BA} = \{ A \}$, and $A \notin \overline {BD}$ (since $between(A, D, B)$); so $\overline {AC} \cap \overline {BD} = \emptyset$.

Also, $\overline {AC} \cap \overline {BF} \subset \overline {AC} \cap \overline {BC} = \{ C \}$, and $C \notin \overline {BF}$ (since $between(B, F, C)$); so $\overline {AC} \cap \overline {BF} = \emptyset$.

Hence our assumptions contradict Pasch’s axiom.

  • $\begingroup$ Yep, thank you. I found exactly this proof in "Hilbert's Axioms of Plane Order" by Wylie when the site was down. It took me hours (literally) to find it. Very clear proof, though, thank you. I'd vote you up, but I can't. $\endgroup$ – weirdesky May 16 '14 at 13:31

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