Plane Geometry - Bruce E. Shapiro

268 SECTION 46. HYPERBOLIC GEOMETRYThen we can choose a point B ′ ∈ AB and C ′ ∈ AC such that AC ′ = DFand AB ′ = DE.By SAS △AB ′ C ′ ∼ = △DEF . Hence∠AB ′ C ′ = ∠E = ∠B∠AC ′ B ′ = ∠F = ∠CSince ∠BB ′ C ′ + ∠AB ′ C ′ = 180 then∠BB ′ C ′ + ∠B = 180Similarly, since ∠B ′ C ′ A + ∠B ′ C ′ C = 180 then∠B ′ C ′ C + ∠C = 180Hence σ(□BB ′ C ′ C) = 360, which contradicts theorem 46.5 which says thatthe angle sum must be strictly less than 360.Hence our RAA hypothesis, which is that hte triangles are not congruent,is false.Theorem 46.15 Let □ABCD and □A ′ B ′ C ′ D ′ be Saccheri quadrilateralswith equal defect and congruent summits. Then □ABCD ∼ = □A ′ B ′ C ′ D ′ .Proof. Let □ABCD and □A ′ B ′ C ′ D ′ be Saccheri quadrilaterals such thatδ(□ABCD) = δ(□A ′ B ′ C ′ D ′ )CD ∼ = C ′ D ′Since σ(□) = 360 − δ(□), for any quadrilateral,σ(□ABCD) = σ(□ ′ B ′ C ′ D ′ )Since the summit angles in any Saccheri Quadrilateral are congruent∠C = ∠D, ∠C ′ = ∠D ′Since both quadrilaterals have the same measure and their base angles areall right angles, then the sum of the summit angles are equal.∠C + ∠D = ∠C ′ + ∠D ′=⇒ 2∠C = 2∠C ′=⇒ ∠C = ∠D = ∠C ′ = ∠D ′« CC BY-NC-ND 3.0. Revised: 18 Nov 2012