Plane Geometry - Bruce E. Shapiro

SECTION 50. AREA IN HYPERBOLIC GEOMETRY 291Figure 50.2: Proof that the associated quadrilateral is a Saccheri Quadrilateral.Proof. Suppose that δ(△ABC) = δ(△DEF ) and that AB = DE (hypothesis).Let □ABB ′ A ′ and DEE ′ D ′ be their corresponding Saccheri quadrilaterals.By lemma 50.8δ(□ABB ′ A ′ ) = δ(△ABC)= δ(△DEF )= δ(□DEE ′ D ′ )By hypothesis the two quadrilaterals have congruent summits. Then bytheorem 46.15□(ABB ′ A ′ ) ∼ = □DEE ′ D ′Hence by transitivity of equivalence,△ABC ≡ □ABB ′ A ′≡ □DEE ′ D ′≡ △DEFProof. (Boylai’s theorem, theorem 50.4.)Suppose that δ(△ABC) = δ(△DEF ).Assume that DF ≥ AC (if not, relabel).Let M be the midpoint of AC and let N be the midpoint of BC.Choose G ∈ ←−→ MN such that AG = (1/2)DF .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.