Plane Geometry - Bruce E. Shapiro

290 SECTION 50. AREA IN HYPERBOLIC GEOMETRYDefinition 50.5 Let △ABC be a triangle. The Associated SaccheriQuadrilateral (see figure 50.1) □ABDE is constructed as follows: Let Mand N be the midpoints of AC and BC. Drop perpendiculars from A andB to ←−→ MN and call their feet D and E.Figure 50.1: The associated Saccheri Quadrilateral.Theorem 50.6 The Associated Saccheri Quadrilateral is a Saccheri Quadrilateral.Proof. We need to show that BE = AD.Drop a perpendicular from C to MN and call the foot F .By AAS △BNE ∼ = △CNF (∠BNE = ∠CNF , ∠BEN = ∠CF N andNC = BN).By congruency, BE = F C. By a similar argument, AD = F C. HenceAD = BE.Theorem 50.7 Let △ABC have an associated Saccheri Quadrilateral □ABDE.Then in neutral geometry they are equivalent by dissection, i.e., △ABC ≡□ABDE.Proof. (See Venema.)Lemma 50.8 Let △ABC have an associated Saccheri Quadrilateral □ABDE,then δ(□ABDE) = δ(△ABC).Proof. (Exercise.)Lemma 50.9 Suppose that δ(△ABC) = δ(△DEF ) and that AB = DE.Then △ABC ≡ △DEF .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012