# In neutral geometry, can a family of parallel lines leave holes in the plane?

In neutral plane geometry, Euclidean geometry without the parallel postulate, I want to show that the family of parallel lines all perpendicular to a given line pass through all of the plane, leaving no holes.

My present formulation of this idea is as follows: Given two lines $l$ and $m$ and a transversal $t$ perpendicular to $l$ at point $A$ and passing through line $m$ at $B$, and given another point $C$ on line $m$ forming a triangle $\triangle ABC$. Then there exists a line $n$ perpendicular to $l$ (and therefore parallel to $t$) which passes through the interior of side $\overline{AC}$ (and therefore the interior of the triangle).

This claim is easy to show in Euclidean geometry where all parallel lines are equidistant. In neutral geometry, parallel lines may bend away from each other, and this claim is much less obvious. Since it appears to be true in models of hyperbolic geometry, I'm guessing it is true in neutral geometry, but I'm at a loss for how to show it.

Update: It occurs to me that this question seems a bit more esoteric than it really is. So let me point out that neutral geometry is really very familiar territory for many of you. As I mentioned it is just ordinary Euclidean geometry without the axiom that asserts the uniqueness of a line parallel to a given line and through a given point and any consequences that follow from that.

So the usual suspects (theorems) are present:

Isosceles triangle theorems, SAS, ASA, SSS, AAS.

The exterior angle theorem, triangle inequality, scalene inequality, hinge theorem.

Actually, many parallel line theorems hold, namely. those that say things like "If two lines are cut with a transversal are pair of congruent angles, then the lines are parallel." or " If two line share a common perpendicular, then they are parallel."

What is not true are theorems that assert the converse: "If two lines are parallel, then ..." and also not true "If lines $l$ and $m$ are parallel and lines $m$ and $n$ are parallel, then $l$ and $n$ are parallel," and that parallel lines are equidistant (but equidistant lines are parallel).

Others theorems that are missing include:

The Pythagorean theorem

The angle sum theorem (instead the sum of the angles in a triangle must be 180 or less)

Rectangles may not exist.

And the weird one: there may be no such thing as similarity.

Update: To make the answer below understandable, you need (also in the comments but hidden):

Proof that two lines perpendicular to a given line are parallel: If they were not parallel, they would intersect forming a isosceles triangle with two 90 degree angles. the complimentary angle to either angle would also be 90 degrees. But by the exterior angle theorem, that exterior angle must be strictly greater than 90 degrees, which is a contradiction.

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What is "neutral geometry"? E.g. what are the geodesics in your geometry, what axioms are you using? If you assume the axiom that for each line and each point not on the line, there exists a line (not necessarily unique) passing through the point and not intersecting the given line, then you are done, since any point is then covered. –  Alex B. Jan 28 '11 at 7:06
I imagine there are quite a number of axiom systems for Euclidean geometry, but one of the axioms will usually be equivalent to "Given a line $l$ and point $P$, the line parallel to $l$ passing through $P$ is unique." Clairaut's axiom and the angle-sum postulate are examples. An axiom system for Euclidean geometry without that axiom is an axiom system for neutral geometry. Adding the axiom "Given a line $l$ and point $P$, there at least two lines parallel to $l$ passing through $P$" to neutral geometry gives you hyperbolic geometry. –  Eric Nitardy Jan 28 '11 at 7:32
Also, it is true in neutral geometry that for each line and each point not on the line, there exists a line passing through the point and not intersecting the given line, but that does not address my question, which is about the family of parallel lines perpendicular to a given line. Does one of those parallel lines pass through each point on the plane? –  Eric Nitardy Jan 28 '11 at 7:43
The Wikipedia article on absolute geometry (aka "neutral geometry") has a footnote that describes it as the intersection of the theories of Euclidean geometry and hyperbolic geometry. (en.wikipedia.org/wiki/Absolute_geometry) What seems open to inquiry is whether the description of all the lines perpendicular to a given line can be shown to be a "family of parallel lines". The question implies that it is so, and in Euclidean geometry it is, but in hyperbolic geometry the relation of parallelism between lines is not transitive (though it holds for parallel rays in the same direction). –  hardmath Jan 28 '11 at 10:03
"... in hyperbolic geometry the relation of parallelism between lines is not transitive" -- true indeed; nonetheless, the set of lines perpendicular to a given line will all be parallel to each other. –  Eric Nitardy Jan 28 '11 at 10:10
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According to the Wikipedia article absolute geometry the first 28 propositions of Euclid's Elements do not require the parallel postulate and hence may be proven for neutral or "absolute" geometry.

Prop. 11 is the construction of a perpendicular to a line from a point on the line. Prop. 12 is the construction of a line perpendicular to a given line through a point not on the line. So, combined with OP's proof that all lines perpendicular to a given line are parallel, the result is established: The family of parallel lines perpendicular to a given line covers all points.

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There are parallel lines in neutral geometry as mentioned in the comments to the question. In the geometry of the sphere, two points do not determine a unique line (using the poles as the points). So it does violate a very basic axiom. In elliptical geometry, a line does not divide the plane into two disjoint sets (the two sides of the line). So that also does not satisfy all the axioms of neutral geometry. Euclidean and hyperbolic geometry are the only two models of neutral geometry that I am aware of. –  Eric Nitardy Jan 28 '11 at 9:05
@hardmath Prop. 12 is true in neutral geometry and this implies that given a line $l$ and a point I could construct a parallel line through that point. Hence, I could construct a line parallel to $l$ through every point on the plane, but such lines might not be all perpendicular to one line. Indeed, they might not be parallel to each other. –  Eric Nitardy Jan 28 '11 at 16:15
@Eric: You're misreading his answer. You have a line $l$. Prop 12 says that for any point $P$ not on $l$, there is a line perpendicular to $l$ passing through $P$. In other words, every point $P$ not on $l$ is on some line in the (mutually parallel) family $\text{Perp}(l) = \{ m \vert m \perp l\}$. –  mjqxxxx Jan 28 '11 at 16:27
@Eric Nitardy: Prop. 12 is about constructing a line through point I perpendicular to line $\ell$, provided point I is not on line $\ell$. "To draw a straight line perpendicular to a given infinite straight line from a given point not on it." So the family of lines perpendicular to $\ell$ cover all points not on line $\ell$, leaving admittedly a gap in my proof about covering the points which are on line $\ell$. To fill that lacuna we need Prop. 11 just preceding: "To draw a straight line at right angles to a given straight line from a given point on it." I'll add that to the Answer. –  hardmath Jan 28 '11 at 16:36
Thank you all. I've been over-thinking this one. @hardmath After you clarify your answer, I'll accept it. –  Eric Nitardy Jan 28 '11 at 16:55
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There might be a similar proof in an excellent textbook: "Euclidean and Non-Euclidean Geometries" by Marvin J. Greenberg. The first several chapters are devoted to neutral geometry.

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Thank you. I'll try to find that. I'm working with an unpublished manuscript by Jack Lee. It's very clear and readable. It should published someday. –  Eric Nitardy Jan 28 '11 at 20:09