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If two alternate angles are same, two poincare lines are parallel.

(i.e. If two poincare lines cut by a transversal have a pair of congruent alternate interior angles, then the two poincare lines are parallel.)

I want to show this statement by using poincare disc model.

I think the converse is false.

Is there someone to help?

The following figure is just a supplementary figure from 'Points, Lines, and Triangles in Hyperbolic Geometry'.

enter image description here

jakeoung
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  • What is your definition of “parallel”? Some people define lines which intersect on the unit circle as “limit parallel”, those which do not even intersect there as “hyperparallel”. Do you want to encompass both these cases, or only one of them? – MvG Oct 03 '13 at 16:52
  • parallel : Two lines do not intersect (i.e. do not meet) – jakeoung Oct 03 '13 at 18:59
  • What is true of parallel lines is not true of ultra parallels. – Narasimham Aug 28 '15 at 17:40

1 Answers1

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Assume that in your graphic, $\measuredangle WVS = \measuredangle WVP = \alpha$. Then $\measuredangle QWV = \pi-\alpha$ since $\measuredangle QWP=\pi$. If the rays $VS$ and $WQ$ were to meet, they would form a triangle. Two of its interior angles would be $\measuredangle WVS=\alpha$ and $\measuredangle QWV=\pi-\alpha$, which already add up to $\pi$. Since the angle sum in a hyperbolic triangle is less than $\pi$ (as you know), and an interior angle cannot be negative, there can be no such triangle. The same argument holds for the rays $VR$ and $WP$, so the lines can't meet on that side of $VW$ either. Since the lines can't meet on either side, they have to be parallel.

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MvG
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