Plane Geometry - Bruce E. Shapiro

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 161Theorem 32.10 (Angle Sum Postulate.) σ(△ABC) = 180 ⇐⇒ EPPis true.The proof requires the following lemma.Lemma 32.11 Let P Q be a segment and R a point such that ∠P QR = 90.Then for every ɛ > 0 there exists a point T ∈ ←→ QR such that µ(∠P T Q) < ɛ.Proof. (of lemma) Let P, Q, R be given as described and let ɛ > 0 be given.Choose any S ∈ H ←→ R, P Qsuch that ←→ P S ⊥ ←→ P Q.Then every point T ∈ −→ QR is in the interior of ∠QP S.Define T 1 ∈ −→ QR such that P Q = QT1 (see figure 32.5).Chose T 2 such that Q ∗ T 1 ∗ T 2 and T 1 T 2 = P T 1 .Chose T 3 such that Q ∗ T 2 ∗ T 3 and T 2 T 3 = P T 2 .Inductively chose T n such that Q ∗ T n−1 ∗ T n and P T n−1 = T n−1 T n .Constructed in this way, by the isoceles triangle theorem,By the angle addition postulate,∠QT n P ∼ = ∠T n−1 P T n (32.2)∠QP T n = ∠QP T 1 + ∠T 1 P T 2 +∠T 2 P T 3 + · · · + ∠T n−1 P T n< ∠QP S = 90Suppose that ∠T i−1 P T i ≥ ɛ for all ɛ (RAA).By the Archimedian ordering property there is some number n such thatnɛ > 90.Hence∠QP T 1 +∠T 1 P T 2 +∠T 2 P T 3 + · · · + ∠T n−1 P T n≥ nɛ > 90(The first inequality follows from the RAA hypothesis; the second forArchimedes.) This is a contradiction; hence we conclude that for somei,∠T i−1 P T i < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.