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Background

Many geometry books used to teach high-schoolers these days try to transfer Hilbert's reworking of Euclid's axioms into a (somewhat) palatable form for students. They don't usually seem to go into as much detail, say, about betweenness, but they include an SAS "Postulate" rather than an SAS "Theorem."

In particular, this book by Jurgensen, Brown, and Jurgensen has the following eleven postulates in its first three chapters (I could have them a little off):

  1. Ruler
  2. Segment addition
  3. Protractor
  4. Angle addition

  5. A line contains two points; a plane contains three noncollinear points; space contains four noncoplanar points

  6. Two points define a line
  7. Given three points, there exists a plane containing them; three noncollinear points determine a plane
  8. If a plane contains two points, then it contains the line connecting them
  9. If two planes intersect, their intersection is a line

  10. If a transversal crosses parallel lines, then corresponding angles are equal

  11. If a transversal crosses two lines and corresponding angles are equal, then the lines are parallel

and goes on to state the following theorem without proof:

3-11. If two lines are parallel to a third line, then the lines are parallel.

The Question

In my tutoring sessions with geometry students, we always use this book because I feel that it's pretty good. But when we get to this point, we go, "Okay, we're pretty competent at geometry; let's try to prove it." And we always breeze through the case where the lines are in a plane together, and come up dry in the general case.

Looking into Euclid XI tonight, I see that this is the ninth proposition. Its proof uses the sixth:

"If lines make right angles to the same plane, then they are parallel."

This, in turn, uses I.4 (SAS) and I.8 (SSS), which in Jurgensen/Brown/Jurgensen are postulates in the next chapter.

So is 3.11 provable from the eleven postulates? Or are the authors just careless--is there a model of postulates 1 through 11 in which there are two noncoplanar lines parallel to a third line?

Edit:

Some definitions for this context.

Two lines are parallel if they are coplanar and nonintersecting.

A transversal is a line in a plane intersecting two other lines in that plane at different points.

A line and a plane are perpendicular if the line is perpendicular to all lines in the plane through the point of intersection.

I'm also not sure if the following (from Chapter 4) can be proved without any triangle postulates: "If a line is perpendicular to two lines in a plane, then it is perpendicular to all lines in that plane through the point of intersection."

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If lines A and B are parallel to line C, and D is a transversal crossing all three lines, then, by 10., corresponding angles for A and C are equal, and corresponding angles for B and C are equal. So corresponding angles for A and B are equal, and by 11., A and B are parallel. Am I missing something? –  Dilip Sarwate Oct 27 '11 at 4:15
    
This was my thought, but do we have a guarantee that a transversal of two lines must cross the third? Edit: We don't now that I remember these are not coplanar. –  AMPerrine Oct 27 '11 at 4:20
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"If lines make right angles to the same plane, then they are parallel." Don't you have this from postulate 11 and the fact that the intersection of a plane with two lines (not in that plane) defines a transversal? Which should let you use a plane orthogonal to all three lines to prove the theorem. –  AMPerrine Oct 27 '11 at 4:47
    
@AMP, I think that's an answer. Maybe you would post it as such. –  Gerry Myerson Oct 27 '11 at 5:14
    
@AMP I notice that there is no definition of what is meant by parallel lines in 10, and suspect that 11 and 10 have been reversed by the OP since 11 can be read as a definition of sorts: if a transversal crosses two lines (in the same plane) and the corresponding angles are equal, then the lines are said to be parallel. I have added the four italicized words in parentheses; the OP did say he might not have the postulates exactly. –  Dilip Sarwate Oct 27 '11 at 13:23
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3 Answers 3

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+100

The eleven postulates are sufficient to prove 3.11.

Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.

Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.

Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.

By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).

Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.

If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.

If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.

$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.

$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.

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Comment $\Rightarrow$ answer:

"If lines make right angles to the same plane, then they are parallel." Don't you have this from postulate 11 and the fact that the intersection of a plane with two lines (not in that plane) defines a transversal? Which should let you use a plane orthogonal to all three lines to prove the theorem.

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Sorry, I should have mentioned that a transversal in this context is a line intersecting two lines at different points where all three lines are in the same plane. Also, what it means for "a line and a plane to be perpendicular" is defined in the next chapter to be that that line is perpendicular to all lines in that plane through the point of intersection. They mention that it can be proved that if a line is perpendicular to two lines in the plane through that point, then it is perpendicular to the plane. But I'm not sure if this can be proved without triangle postulates. –  Daniel Briggs Oct 27 '11 at 15:24
    
I think AMPerrine's solution accomodates all that. Let lines L and M be perpendicular to plane P. Let L meet P at A, let M meet P at B. The line through A and B is a transversal to L and M and meets them both at right angles. So L and M are parallel. –  Gerry Myerson Oct 27 '11 at 21:34
    
@GerryMyerson I am not sure that AMPerrine's solution accommodates all that. Is something similar in spirit to Euclid's fifth postulate needed: Through any point in a line, there is exactly one plane that is perpendicular to the line? Since OP Daniel Briggs is giving out information in dribs and drabs, this question may never get resolved without going through the book oneself. –  Dilip Sarwate Oct 28 '11 at 11:05
    
@GerryMyerson: Sorry I'm giving out information in "dribs and drabs," as Dilip put it. I thought there would be a lot of people on here who went through the standard (US, at least) geometry education in high school, and all five textbooks or so I've seen so far present it essentially the same way. If I have more time soon, I will write up the definitions and the theorems proved, and maybe start a bounty. The line through A and B may not be a traversal because there may be no plane containing L and M. –  Daniel Briggs Oct 31 '11 at 21:15
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I think that 3.11 is provable from the postulates.

Let $a$, $b$, $c$ straightlines, $\alpha$ and $\beta$ planes such that: $$a \parallel b, \quad (\mathrm{I}) $$ $$b \parallel c, \quad (\mathrm{II})$$ $$a \subset \alpha, \quad (\mathrm{III})$$ $$c \subset \beta, \quad (\mathrm{IV})$$ and $$b = \alpha \cap \beta. \quad (\mathrm{V})$$ If $a$ is not parallel to $c$ then we have two cases: 1. $a$ and $c$ intersect each other; 2. $a$ and $c$ are skew lines.

Let's begin by the first case:

(1). $a$ and $c$ intersect each other;

Let $\{P \}=a \cap c$ then $\alpha$ and $\beta$ have at least three common points (two different point in $b$ and point $P$). Hence $\alpha =\beta$. Let $\theta$ ($\theta \neq 0$) an angle formed between $a$ and $c$, then by postulate (10) $a$ (the transversal line) will intersect $b$ and there will be a corresponding angle $\theta$ between $a$ and $b$. That's a contradiction since $a \parallel b.$

Now the second case:

(2). $a$ and $c$ are skew lines.

Since $b \parallel a$, $b \subset \beta$ and $a$ and $c$ are skew, there is a point $Q$ such that $\{Q \}= b \cap c$. But that's a contradiction since $b \parallel c$.

Therefore $a \parallel c$.

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I'll modify the prove of the second case. –  RicardoCruz Jul 10 '13 at 2:38
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