Plane Geometry - Bruce E. Shapiro

266 SECTION 46. HYPERBOLIC GEOMETRYFigure 46.3: A Saccheri quadrilateral (top) and Lambert quadrilateral (bottom)in Hyperbolic geometry.Proof. In neutral geometry we know that the fourth angle is either acuteor right. It it were right, the quadrilateral would be a rectangle, which isnot possible in Hyperbolic geometry. Hence the angle is acute.Theorem 46.10 In a Lambert quadrilateral, the length of a side betweentwo right angles is less than the length of the opposite sides.Proof. Let A, B, and C be the vertices with the right angles, as in figure46.3 (bottom).By corollary 31.25 we already know that BC ≤ AD.Suppose that BC = AD (RAA).This would make the □ABCD a Saccheri quadrilateral with summit CD.Then by theorem 31.16 the summit angles of a Saccheri Quadrilateral arecongruent, meaning ∠D = ∠C = 90. Thus □ABCD is a rectangle, whichviolates theorem 46.6.Hence BC < AD.Definition 46.11 The altitude of a Saccheri quadrilateral is the length ofthe segment joining midpoint of its base and the midpoint of its summit.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012