Plane Geometry - Bruce E. Shapiro

140 SECTION 30. TRIANGLES IN NEUTRAL GEOMETRYSubstituting equation 30.1 into equation 30.2 givesσ(△ABC) = σ(△ABD)which is the first result promised by theorem 30.4.To get the second result, observe that by the angle addition postulateζ = ɛ + αHence eitherɛ < 1 2 ζ or α < 1 2 ζBut because △AEC ∼ = △DEB (from above), β = α. Hence eitherɛ < 1 2 ζ or β < 1 2 ζwhich is the second conclusion of the theorem.Corollary 30.5 Let T be a triangle, and let one of its interior angles beα. Then there exists a triangle T ′ with an interior angle α ′ such thatσ(T ) = σ(T ′ ) and α ′ ≤ 1 2 α.Proof. Let T = △ABC and let α = ∠CAB. Then by theorem 30.4 thereexists a point D ∉ ←→ AB such that one of the interior angles of T ′ = △ABDhas measure less than or equal to 1 2 α. Call this angle α′ .Theorem 30.6 (Saccheri-Legendre Theorem) For any triangle△ABC,σ(△ABC) ≤ 180Proof. Let T = △ABC be given and let α = µ(∠CAB).Suppose σ(ABC) > 180 (RAA hypothesis).Then there is some number ɛ > 0 such thatσ(ABC) = 180 + ɛBy the Archimedian ordering property there is some integer n such that2 n ɛ > α. (Choose any n > log 2 ( α ɛ ).)By corollary 30.5 there is a triangle T 1 with an interior angle α 1 ≤ 1 2 α suchthat σ(T ) = σ(T 1 )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012