Plane Geometry - Bruce E. Shapiro

126 SECTION 27. SCALENE AND TRIANGLE INEQUALITYBy the isosceles triangle theorem α = β.Angle β is an exterior angle of triangle △ADC. Hence β > δ.Therefore γ > α = β > δ.( ⇐ ) Assume γ > δ.By trichotomy one of the following must hold: AB > BC, AB = BC, orAB < BC.First, suppose that AB = BC. Then α = δ, which contradicts the hypothesis.Hence AB ≠ BC.Next, suppose that AB < BC. Then γ = ∠ACB < ∠BAC = δ by the firstpart of the theorem. But this contradicts the hypothesis, so AB > BC.Theorem 27.2 (The Triangle Inequality) If A, B, C are non-collinearpoints thenAC < AB + BC (27.2)Proof. Let A, B and C be non-collinear (see figure 27.2).Figure 27.2: Proof of the Triangle Inequality (theorem 27.2.)Let D ∈ ←→ AD such that BD = BC and A ∗ B ∗ D (ruler postulate).Since A ∗ B ∗ D, β > α (protractor postulate).By the isosceles triangle theorem, α = γ (because BC = BD).Hence β > γ. By the scalene inequality,AD > ACSince A ∗ B ∗ D, AD = AB + BD,AB + BD > AC« CC BY-NC-ND 3.0. Revised: 18 Nov 2012