Plane Geometry - Bruce E. Shapiro

SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRY 279Figure 48.5: The angle of parallelism is a decreasing function (Theorem48.6).By the corresponding angles theorem ←→ QE ‖ ←→ P D.Hence ←→ QE does not intersect ←→←→P D.AB (because otherwise it would have to crossHence θ is not in the intersecting set of Q and −→ AB. Hence κ(b) ≤ θ. Henceκ(b) ≤ κ(a). which means κ is non-increasing.Theorem 48.7 Every angle of parallelism is acute, and every critical numberis less than 90.Proof. Let l be a line P ∉ l a point.Drop the perpendicular from P to its foot A ∈ l.Let m be a line through P such that m ⊥ ←→ P A. Hence m ‖ l.By the hyperbolic parallel postulate there is at least one additional line nthorugh P such that n ‖ l.Since n is distinct from m it is not perpendicular to ←→ P AHence it must make an acute angle with ←→ P A on one side of P A. Call themeasure of this angle θ.Since n ‖ l, θ ∉ K. Hence the critical angle must be less than θ, and henceit must be less than 90.Theorem 48.8 Suppose −−→ P D| −→ AB, Q ∈−−→ P D, and let C be the foot of theperpendicular from Q to ←→ AB. If B ′ ∈ −→ AB such that A∗C ∗B ′ and D ′ ∈ −→ P Qsuch that P ∗ Q ∗ D ′ , then −−→ QD ′ | −−→ CB ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.