Plane Geometry - Bruce E. Shapiro

84 SECTION 16. ANGLESFigure 16.6: Illustration of Theorem 16.9.Proof. Assume that D is not in the interior of ∠BAC ( figure 16.6) anduse this to prove that C is in the interior of ∠BAD. The proof of thealternative case is analogous.C and D are on the same side of ←→ AB by hypothesis. HenceD ∈ H C,←→ AB(16.2)Since D is not in the interior of ∠BAC by assumption, by the definition ofinterior,D ∉ H ←→ C, AB∩ H ←→ B, AChence either D ∉ H C,←→ ABor D ∉ H B,←→ AC.But by equation 16.2, D ∈ H C,←→ ABso we concludeD ∉ H B,←→ ACHence by the plane separation postulate, B and D must be on oppositesides of ←→ AC, andBD ∩ ←→ AC ≠ ∅−→ −→ −→Let E be the unique point of intersection. Since B ∗ E ∗ D, AB ∗ AE ∗ AD(theorem 16.8) and therefore E is in the interior of ∠BAD (definition ofbetweenness for rays), andE ∈ H ←→ D, ABHence E and D lie on the same side of ←→ AB. By hypothesis we alreadyassumed that C, D are on the same side of ←→ AB. Hence C, D, and E are onthe same side of line ←→ AB.Therefore −→ AC and −→ AE can not be opposite rays.Therefore −→ AC = −→ AE.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012