Plane Geometry - Bruce E. Shapiro

SECTION 50. AREA IN HYPERBOLIC GEOMETRY 293Figure 50.4: Proof that area and defect are proportional.Proof. If δ(△ABC) = δ(△DEF ), by Boylai’s theorem △ABC ≡ △DEF ,hence they have equal area, and α(△ABC) = α(△DEF ).Now suppose that δ(△ABC)/δ(△DEF ) is rational. Then there exist positiveintegers a and b such thatδ(△ABC)δ(△DEF ) = a bAssume that a < b (if not, relabel the triangles).By the additivity of defect and continuity of defect we can define pointsP 0 , P 1 , . . . , P b on DE such thatD = P 0 , P i−1 ∗ P i ∗ P i+1 , P b = Eandδ(△F P i P i+1 ) = 1 b δ(△DEF )(By additivity of defect the sum of the defects of the smaller triangles mustadd up to the defect of the larger triangle. Hence by choosing n sufficientlylarge we can ensure that each triangle has a defect that is sufficiently small.)By the first part of the proof, all the smaller triangles have the same area.δ(△ABC) = a b δ(△DEF )= aδ(△F P i P i+1 )= δ(△F P 0 P a )Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.