Plane Geometry - Bruce E. Shapiro

SECTION 16. ANGLES 83Figure 16.5: Illustration of theorem 16.8, illustrating B ∗ D ∗ C ⇐⇒ −→ AB ∗−→AD ∗ −→ ACProof. Suppose that B ∗ D ∗ C (hypothesis).Then by theorem 16.4, C and D are on the same side of ←→ AB, and B and Dare on the same of line ←→ ACHence by the definnition of interior of an angle (definition 15.8), D is inthe interior of angle ∠BAC.Hence by the definition of betweenness for rays (definition 16.1),To prove the converse, assume equation 16.1.−→AB ∗ −→ −→ AD ∗ AC (16.1)Then by corollary 16.7 point D is in the interior of angle ∠BAC.Hence B and D are on the same side of ←→ AC by the defition of interior ofan angle.Therefore C ∉ BD and hence B ∗ C ∗ D is not possible.By a similar argument, C and D are on the same side of ←→ AB, B ∉ CD, andC ∗ B ∗ D is not possible.Since D ∈ ←→ BC then one of the following must hold:(B ∗ C ∗ D) ∨ (C ∗ B ∗ D) ∨ (B ∗ D ∗ C)Since the first two possibilities have been eliminated then we must haveB ∗ D ∗ C by corollary 14.26.Theorem 16.9 Let A, B, C, D be four distinct points such that C and Dare on the same side of ←→ AB and D ∉ ←→ AC. Then either C is in the interiorof ∠BAD or D is in the interior of ∠BAC.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.