Plane Geometry - Bruce E. Shapiro

SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRY 281Figure 48.7: If two lines admit a common perpendicular then thay are notasymptotically parallel (theorem 48.13).Let D ∈ m such that R ∗ P ∗ D.Since □SAP R is a Lambert quadrilateral, then ∠AP R is acute; hence,∠AP D is obtuse.Hence −−→ P D is not a limiting parallel because the angle is not acute.Let B ∈ l on the same side of P A as D.Let C ∈ l such that C ∗ S ∗ BLet E ∈ m such that E ∗ R ∗ D.Then the ray −→−→P E cannot be a limiting parallel ray of AC, because if itwere, that would mean that −→−→RE was a limiting parallel ray of SC, and wehave already ruled that out.Hence there can be no limiting parallel ray to l in m originating at P . SinceP was chosen arbitrarily, that means that no subray of m can be a limitingparallel ray to l.Theorem 48.14 If l and m are asymptotically parallel lines, then l andm do not admit a common perpendicular.Proof. This is the contrapositive of theorem 48.13.Theorem 48.15 Suppose that l ‖ m; P, Q, R are points on m with P ∗Q∗R;and A, B, C are the feet of the perpendiculars from P, Q, R to l, respectively.Then1. ←→ P A ⊥ m =⇒ P Q < QB < RC2. −→ P A| −→ AB =⇒ P A > QB > RCProof. (Exercise.)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.