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Isoceles triangle in absolute geometry

  1. Jul 29, 2006 #1
    Im working to prove that an interior point on a chord is also interior to the circle containing the chord. Ive got the entire proof almost finished and laid out, however im stuck on one part. My proof would be complete if I could show this:

    Given an isoceles triangle, in absolute geometry (no parallel postulate, sum of triangle <= 180)

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    For an arbitrary point P, interior to segment AC. Prove that segment PB is always less then AB=CB.



    I establish that since P is interior to AC, P will never equal A or C. From there Ive been trying to establish that angle A or C will always be smaller then the angle created by P, so that by the scalene inequality side AC or BC will always be larger (since they correspond to angles created by P).

    However I cannot figure out how to make a general proof for this in absolute geometry, can anyone offer any insight?
     
  2. jcsd
  3. Jul 30, 2006 #2

    HallsofIvy

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    You can't prove it- it's not true. Consider an is triangle on the surface of a sphere: One vertex at the "north pole", the other two on the "equator". The length of any line segment through the "north pole" to the "equator" interior to the triangle has length equal to the two sides.
     
  4. Jul 30, 2006 #3
    I see where the lengths would be equal for a sphere but not this case for the circle.
     
    Last edited: Jul 30, 2006
  5. Jul 31, 2006 #4
    Here's what I'm thinking: In your original triangle, you start with a 60 degree angle for both A and C. Whenever you draw the line PB, in essence you're making a new triangle. Whenever you do that, you take some of the degrees away from what was angle B in your original triangle. Angle P must make up for that loss in degrees. So, angle A=60; B=60-x; P=60+x. I don't think that's really a proof, but I don't know what else to say!
     
  6. Jul 31, 2006 #5

    HallsofIvy

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    The question was about proving that an line segment passing through the vertex of an isosceles triangle must be shorter than the two sides, in absolute geometry. My point is that there is a counterexample to that in spherical geometry so it cannot be true for general absolute geometry.
     
  7. Jul 31, 2006 #6

    Office_Shredder

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    Doesn't spherical geometry have sum of triangle >= 180 degrees?
     
  8. Jul 31, 2006 #7

    HallsofIvy

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    Thanks, office shredder, I didn't notice that "sum of triangle <= 180". Okay, so his version of "absolute geometry" does includes hyperolic but not elliptic. That's often done since elliptic geometry does not satisfy the "two points determine a line" axiom. My "counter-example" doesn't work in this case.
     
  9. Aug 1, 2006 #8
    If I Understood The Problem...
     

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