Ultraparallel theorem: Difference between revisions

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The four hyperbolic motions that produced <math>z</math> above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius <math>\sqrt{z}</math> to yield the unique hyperbolic line perpendicular to both ultraparallels <math>p</math> and <math>q</math>.

==Proof in the Beltrami-Klein model==

In the [[Beltrami-Klein model]] of the hyperbolic geometry:

In the [[Klein model]] of the hyperbolic plane, two ultraparallel lines correspond to two non-intersecting [[chord (geometry)|chord]]s. The [[Pole and polar|poles]] of these two lines are the respective intersections of the [[tangent line]]s to the [[unit circle]] at the endpoints of the chords. Lines ''perpendicular'' to line A are modeled by chords whose extension passes through the pole of A. Hence we draw the unique line between the poles of the two given lines, and intersect it with the unit disk; the chord of intersection will be the desired common perpendicular of the ultraparallel lines. If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter is perpendicular as well in the hyperbolic plane, and so we draw a line through the pole of the other line intersecting the diameter at right angles to get the common perpendicular.

* two ultraparallel lines correspond to two non-intersecting [[chord (geometry)|chord]]s.

* The [[Pole and polar|poles]] of these two lines are the respective intersections of the [[tangent line]]s to the boundary [[circle]] at the endpoints of the chords.

* Lines ''perpendicular'' to line ''l'' are modeled by chords whose extension passes through the pole of ''l''.

* Hence we draw the unique line between the poles of the two given lines, and intersect it with the boundary circle ; the chord of intersection will be the desired common perpendicular of the ultraparallel lines.

If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter it is also perpendicular in the Beltrami-Klein model, and so we draw a line through the pole of the other line intersecting the diameter at right angles to get the common perpendicular.

The proof is completed by showing this construction is always possible.:

* If both chords are diameters, they intersect.(at the center of the boundary circle)
* If only one of the chords is a diameter, the other chord projects orthogonally down to a section of the first chord contained in its interior, and a line from the pole orthogonal to the diameter intersects both the diameter and the chord.
* If both lines are not diameters, thethen we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the unit circle inscribed within it.{{how}} The poles are opposite vertices of this quadrilateral, and the chords are lines drawn between adjacent sides of the vertex, across opposite corners. Since the quadrilateral is convex, the line between the poles intersects both of the chords drawn across the corners, and the segment of the line between the chords defines the required chord perpendicular to the two other chords.{{how}}

<!-- ??? "then we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the unit circle inscribed within it " this is not always the case , they not always form a quadrilateral -->

==References==