Ultraparallel theorem: Difference between revisions

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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).

If not, we may suppose AB < CB' without loss of [[generality]]. Let E be a point on the line s on the opposite side of A from C. Take A' on CB' so that A'B' = AB. Through A' draw a line s' (A'E') on the side closer to E, so that the angle B'A'E' is the same as angle BAE. Then s' meets s in an ordinary point D'. Construct a point D on ray AE so that AD = A'D'.

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=H. S. M. Coxeter|author1-link=H. S. M. Coxeter|title=Non-euclidean Geometry|isbn=978-0-88385-522-5|pages=190–192}}</ref>