Plane Geometry - Bruce E. Shapiro

286 SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRYDrop a perpendicular from B to −−→ P D and call its foot Q. Sinceσ(△BP Q) = µ(∠P BQ) + 90 + 90 − ɛ=⇒ δ(△BP Q) < ɛ> 180 − ɛDefined = min[d(A, ←→ P Q), d(A, ←→ BQ)]Let C be any point in the interior of △BP Q not on ←→ AB such that AC < d.By the crossbar theorem there is a point E such that −→ AC intersects P Q.By repeated applications of additivity of defect,δ(△P QB) = δ(△BQE) + δ(△ABC)+ δ(△AP C) + δ(△P EC)> δ(△ABC)Since δ(△P QB) < ɛ then δ(△ABC) < ɛ.Figure 49.3: Proof of theorem 49.4.Theorem 49.5 (Continuity of Defect) Let △ABC be a triangle and letc = AB. For every number x ∈ [0, c] define P (x) to be the unique pointP (x) ∈ AB such that AP (x) = x. Defineand define f(0) = 0. Thenis continuous.f(x) = δ(△AP (x)C)f(x) : [0, c] ↦→ R« CC BY-NC-ND 3.0. Revised: 18 Nov 2012