Plane Geometry - Bruce E. Shapiro

SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRY 149(⇐) Let □ABCD be a quadrilateral with E = AC ∩ BD with A ∗ E ∗ Cand B ∗ E ∗ D.Since A ∗ E ∗ C, A and E are on the same side of the line ←→ CD (theorem16.4.Similarly, since B ∗ E ∗ D, B and E are on the same side of ←→ CD.Hence A and B are on the same side of ←→ CD (Plane separation postulate,axiom 15.2), i.e, A ∈ H B,←→ CD.By a similar argument A and D are on the same side of ←→ BC, i.e, A ∈ H ←→ D, BC.Hence A ∈ H ←→ B, CD∩ H ←→ D, BC, and thus A is in the interior of ∠BCD.By a similar argument, each of the other vertices is in the interior of itsopposite angle. Hence by definition of convexity, the quadrilateral is convex.Theorem 31.12 If □ABCD and □ACBD are both quadrilaterals then□ABCD is not convex.Proof. LetP = “□ABCD is quadrilateral”Q = “□ACBD is quadrilateral”R = “□ABCD is not a convex quadrilateral”To prove the theorem we must show thatP ∧ Q ⇒ Ror equivalently, its contrapositive,¬R ⇒ ((¬P ) ∨ (¬Q))Assume □ABCD is a convex quadrilateral (i.e, assume that R is false).Then AC ∩ BD ≠ ∅, i.e., the diagonals of □ABCD share an internal point.Hence □ACBD is not a quadrilateral, i.e, ¬P is true. Hence ¬R ⇒ ¬P .By the rules of logic, ¬P ⇒ (¬P ) ∨ (¬Q).Hence ¬R ⇒ ((¬P ) ∨ (¬Q)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.