Ultraparallel theorem: Difference between revisions

Revision as of 17:04, 20 July 2016

Revision as of 21:34, 21 July 2016

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Let r and s be two ultraparallel lines.

From any two distinct points A and C on s draw AB and CB' perpendicular to r with B and B' on r.

If it happens that AB = CB', then the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the [[Saccheri quadrilateral]] ACB'B).

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Then D' ≠ D. They are the same distance from r and both lie on s. So the perpendicular bisector of D'D (a segment of s) is also perpendicular to r.<ref>{{cite book|last1=coxeter|title=non euclidean geometry|isbn=978-0-88385-522-5|pages=190-192}}</ref>

(If r and s were asymptotically parallel rather than ultraparallel, this construction would fail because s' would not meet s. Rather s' would be asymptotically parallel to both s and r.)

==Proof in the Poincaré half-plane model ==