Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro 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
SECTION 50. AREA IN HYPERBOLIC GEOMETRY 295Theorem 50.11 Area is proportional to defect: there exists a constant ksuch thatα(△ABC) = kδ(△ABC)for every triangle △ABC (same k for every triangle).Corollary 50.12 There is an upper bound to the area of triangles.Proof. Since δ(△ABC) < 180 for every triangle, thenα(△ABC) < 180kCorollary 50.13 If α(△ABC) = α(△DEF ) then △ABC ≡ △DEF .Proof.δ(△ABC) = 1 k α(△ABC)By Boylai’s theorem △ABC ≡ △DEF .= 1 k α(△DEF )= δ(△DEF )Theorem 50.14 Let R 1 and R 2 be polygonal regions such that α(R 1 ) =α(R 2 ). Then R 1 ≡ R 2 .Theorem 50.15 There exist triangles with the same base and height butdifferent areas.Proof. Let l be a line and P a point with d(P, l) = 1.Let Q 0 be the foot of the perpendicular from P to l.Define Q 1 , Q 2 , . . . in l such that Q i Q i+1 = 1.Define T i = △P Q i−1 Q i .equal to 1.Then the base and height of each T i are eachSuppose that all the T i have the same area, α(T i ) = a for all i.Then all the T i have the same defect, δ(T i ) = d for all i. By additivity ofdefectδ(△P Q 0 Q n ) = nd < 180Since we can choose n arbitrarily large, this violates the Archimedian orderingproperty. Hence the defects cannot all be equal, and hece the areascannot all be equal.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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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
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