Ultraparallel theorem: Difference between revisions

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==HilbertsHilbert's construction==

Let r and s be two non-intersectingultraparallel lines.

From any two 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' the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the isocleses[[Saccheri birectanglequadrilateral]] ACB'B ).

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If not, suppose AB < CB'.

Take A' on CB' so that A'B' = AB.

Through A' draw a line s', making the same angle with A'B' that s makes with AB.

Then s meets s' in an ordinary point D.

Take a point D' on ray AC so that AD' = A'D.

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 the perpendicular bisector of DD' is also perpendicular to r.<ref>{{cite book|last1=coxeter|title=non euclidean geometry|isbn=978-0-88385-522-5|pages=190-192}}</ref>

Then the perpendicular bisector of DDD'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>

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