Plane Geometry - Bruce E. Shapiro

294 SECTION 50. AREA IN HYPERBOLIC GEOMETRYwhere the last step follows by additivity of defect. Henceby Boylai’s theorem and thererforehence△ABC ≡ △F P 0 P aα(△ABC) = α(△F P 0 P a )= aα(△F P i P i+1 )= a b α(△DEF )α(△ABC)α(△DEF ) = a b = δ(△ABC)δ(△DEF )which proves the theorem when the ratio is rational.Now suppose that x,δ(△ABC)δ(△DEF ) = xα(△ABC)α(△DEF ) = ywhere x and y are real numbers. Let r be any rational number such that0 < r < x. Choose C ′ ∈ BC such thatThenand by the first part of the theorem,r = δ(△ABC′ )δ(△DEF )α(△ABC ′ ) < α(△ABC)r = α(△ABC′ )α(△DEF )< α(△ABC)α(△DEF ) = ySince r was chosen as an arbitrary rational such that r < x, then for everyrational r < x, we have r < y.By a similar argument, for every rational number r < y, we can show thatr < x.Hence by the comparison theorem for real numbers (theorem 6.5), x =y.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012