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→Hilberts construction: fix punctuation, spelling, wording, and most importantly the fact that the congruent structures are pointed the same way rather than being towards each other. |
→Hilbert's construction: more detail |
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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', 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=coxeter|title=non euclidean geometry|isbn=978-0-88385-522-5|pages=190-192}}</ref>
==Proof in the Poincaré half-plane model ==
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