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Consider an absolute plane (i.e. it satisfies all axioms except the parallel axiom). Let g be a straight line and P a point not in g. Then there is a unique straight line going through P which is perpendicular on g.

I asked a question regarding the existence in the same problem here. Now I have a question regarding Uniqueness.

Suppose $A$ and $B$ are points on $g$ such that the lines $PA$ and $PB$ are both perpendicular to $g$. Suppose toward contradiction that $A\neq B$.

Now consider the unique point $A'$ on $AP^+$ such that d(A',P)=d(A,P). Similarly consider the unique point $B'$ on $BP^+$ such that $d(B',P) = d(B,P)$.

Then the triangles $\triangle PA'B'$ and $\triangle PAB$ are congruent.

How do I generate a contradiction from this? Am I on the right track?

Community
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iwriteonbananas
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  • should you not be looking for contradiction in that the angles of a triangle make up $180^\circ?$ how do you know the prime points are different from the non prime points? – abel Jan 27 '15 at 16:09
  • @abel, without the parallel axiom, the angles of a triangle don't make 180 degrees. – Gerry Myerson Jan 27 '15 at 16:17
  • We've proved this only for a Euclidean planes (sum of triangles is 180 deg). In this case we're considering a plane which does not satisfy the parallel axiom. $A'$ lies on $PA^-$ whereas $A$ lies on $PA^+$, so they must be distinct – iwriteonbananas Jan 27 '15 at 16:17
  • What do your notations $AP^{*}$ and $PA^{-}$ mean? – Gerry Myerson Jan 27 '15 at 16:19
  • Keep in mind that on a sphere you can have two perpendiculars from a given point to a given line, so you must use whichever property it is that the sphere is missing. – Gerry Myerson Jan 27 '15 at 16:20
  • By $PA^+$ I mean the ray starting at $P$, going through $A$ and off to infinity. By $PA^-$ I mean the ray starting in $P$ going off in the opposite direction – iwriteonbananas Jan 27 '15 at 16:20
  • @GerryMyerson: On the sphere there are two perpendicular rays, but only a single perpendicular line, which intersects $g$ in two antipodal points. Right? So if you concentrate on lines instead of rays, the sphere might work as well. Except if $P$ is the pole of $g$… – MvG Jan 27 '15 at 22:13
  • @MvG, I was thinking of the case where $P$ is a pole. – Gerry Myerson Jan 27 '15 at 22:52
  • Gerry, I'm not sure which property you're referring to – iwriteonbananas Jan 28 '15 at 07:02
  • If you want to be sure I see a comment directed to me, you have to put an at-sign before my name. If you keep all the postulates of Euclidean geometry except the 5th postulate, you get Lobachevskian geometry. There is another geometry, spherical geometry, that resembles both Euclidean and Lobachevskian, but it differs in more than just the 5th postulate. So go look up spherical geometry, find out where exactly its postulates differ from Euclidean and Lobachevskian, and then you know what's going to be important in your proof. – Gerry Myerson Jan 29 '15 at 03:02

2 Answers2

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There are two ways to prove this. Use one of these theorems:

  1. If alternate angles are congruent, then the lines are parallel.

  2. An exterior angle of a triangle is greater than interior angle not adjacent to it.

Kulisty
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Saccheri-Legendre Theorem (https://www.math.unl.edu/~bharbourne1/M812TSummer2014/Day6/Saccheri-LegendreTheorem.pdf) states that "If one assumes Euclid's postulates other than the parallel postulate, then the sum pf the interior angles of any triangle is at most 180 degrees." This Theorem is true in Your case. Now, look at the triangle PAB that appears in Your case. If A and B are different points, then the interior angle APB is greater than 0, interior angles PAB and PBA are both 90 degrees. Hence the sum of the interior angles of the triangle PAB is greater than 180 degrees. A contradiction. Hence, there could be at most one perpendicular to g from P.

Mart Abel
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