Plane Geometry - Bruce E. Shapiro

264 SECTION 46. HYPERBOLIC GEOMETRYFigure 46.1: Illustration of the hyperbolic parallel postulate. Lines m, n,p, and q all pass through the point P and all are parallel to the line l butnone of them are parallel to each other.parallel postulate is true.Proof. Let l be a line and let P be a point not on l. Assume that theEuclidean parallel postulate is false. Then no rectangle can exist (theorem33.3).We construct a line m that is parallel to l as follows: drop a perpendiculart from P to its foot Q on l. Then construct a perpendicular to t throughP . By the alternatie interior angles theorem, m ‖ l.Pick R ∈ l such that R ≠ Q and construct a perpendicular line u throughR.Drop a perpendicular line from P to u and call the foot S. By the alternateinterior angle theorem applied to transversal u, ←→ P S ‖ l.□P QRS is a Lambert Quadrilateral. It cannot be a rectangle because wehave already stated that no rectangle can exist.Hence ∠QP S is not a right angle. Therefore ←→ P S ≠ m.Hence there are two distinct lines, ←→ P S and m, both through P , both parallelto l. Hence the Hyperbolic parallel postulate is true.In this section we assume that the hyperbolic parallel postulate holds. Asconsequences of the Universal Hyperbolic Theorem, we can then immediatelyaccept as true any state that is equivalent to the negation of theEuclidean parallel postulate.Theorem 46.4 For every triangle △ABC, σ(△ABC) < 180.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012