Plane Geometry - Bruce E. Shapiro

88 SECTION 17. THE CROSSBAR THEOREMFigure 17.2: Illustration for proof of the crossbar theorem (theorem 17.1).possibilities:−→AD ∩ BC ≠ ∅ (17.1)−→AD ∩ BQ ≠ ∅ (17.2)−→AP ∩ BC ≠ ∅ (17.3)−→AP ∩ BQ ≠ ∅ (17.4)Since P ∗ A ∗ D, P and D lie on opposite sides of ←→ AC (Plane Separation).But B and D are on the same side ←→ AC (because D is in the interior of∠CAB), so B and P are on opposite sides of ←→ AC.−→ −→By the Z-theorem (see figure 17.3), QB ∩ AP = ∅.Hence QB ∩ −→ AP = ∅. This eliminates equation 17.4.Figure 17.3: Elimination of equation 17.4 by the Z-theorem (heavy lines).Since B and P are on opposite sides of ←→−→AC, as we have just argued, CB ∩−→AP = ∅ (which which also follows from the Z-theorem).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012