Plane Geometry - Bruce E. Shapiro

272SECTION 47.PERPENDICULAR LINES IN HYPERBOLICGEOMETRYFigure 47.2: Lines m and n admit a common perpendicular because thetwo poitns D and F satisfy d(D, m) = d(F, m).Proof. Suppose that there are, in fact, three distinct points P , Q, and Ron m that are a fixed distance D from l,D = d(P, l) = d(Q, l) = d(R, l) > 0Let P ′ , Q ′ , and R ′ be the feet of perpendiculars dropped from P , Q, andR to l from m.None of the points P , Q or R lie on l because they are each a distance Daway. Hence at least two of them must lie on the same side of l. Supposethese points are P and Q (else, relable the points accordingly).Then □P P ′ Q ′ Q is a Saccheri Quadrilateral. Hence m ‖ l (because allSaccheri Quadrilaterals are parallelograms by theorem 31.18).Since m ‖ l, then all points on m are on the same side of l. Hence P , Q,and R are all on the same side of l.Relabel the points, if necessary, so that P ∗ Q ∗ R. Then both □P P ′ Q ′ Qand □QQ ′ R ′ R are Saccheri quadrilaterals. Since the summit angles of aSaccheri quadrilateral are acute in (theorem 31.20), ∠P QQ ′ and ∠RQQ ′are acute.Since ∠P QQ ′ and ∠RQQ ′ are supplements, this is a contradiction.Hence the RAA hypothesis is false, and we conclude that there are at mosttwo points P and Q a on m that are a fixed distance from l.Theorem 47.2 (Existence of Common Perpendiculars) If l ‖ m areparallel lines and there are two points on m that are equidistant from lthen l and m admit a common perpendicular.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012