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'.
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.