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Plane Geometry - Bruce E. Shapiro

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292 SECTION 50. AREA IN HYPERBOLIC GEOMETRYFigure 50.3: Proof of Boylai’s theorem.Choose H ∈ ←→ AG such that A ∗ G ∗ H and GH = AG.Then △ABC and △ABH share the same Saccheri Quadrilateral □AA ′ B ′ B.Hence by transitivity of equivalence and lemma 50.8 thenSince△ABC ≡ △ABHδ(△ABH) = δ(△ABC) = δ(△DEF )by hypothesis, and DF = AH, then by lemma 50.9,. By transitivity of equivalence,△ABH ≡ △DEF△ABC ≡ △DEFTheorem 50.10 For any two triangles △ABC and △DEF ,α(△ABC)δ(△ABC) = α(△DEF )δ(△DEF )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

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