Plane Geometry - Bruce E. Shapiro

70 SECTION 14. BETWEENNESSExample 14.6 Let P = (0, 0), Q = (2, 0). The circles of radius 2 centeredat P and Q do not interset in Q 2 . Their intersection in R 2 is (1, ± √ 3)(figure 14.3).Theorem 14.25 (Betweenness Theorem for Points) Let A, B, C bedistinct points on the line l, and let f : l ↦→ R be a coordinate function forl. Then the A ∗ C ∗ B if and only if eitherf(A) < f(C) < f(B) (14.5)orf(A) > f(C) > f(B) (14.6)Proof. Suppose that f(A) < f(C) < f(B). ThenAC + CB = |f(C) − f(A)| + |f(B) − f(C)|= f(C) − f(A) + f(B) − f(C)= f(B) − f(A)= |f(B) − f(A)|= ABso that A ∗ C ∗ B by definition 14.12.A similar argument holds if f(A) > f(C) > f(B).Now consider the converse. Assume that A ∗ C ∗ B so thati.e.,AC + CB = AB|f(C) − f(A)| + |f(B) − f(C)| = |f(B) − f(A)|But by algebra, we also havehencef(C) − f(A) + f(B) − f(C) = f(B) − f(A)|f(C) − f(A) + f(B) − f(C)| = |f(B) − f(A)|Let u = f(C) − f(A) and v = f(B) − f(C). Then|u| + |v| = |u + v|From algebra we know that this means either u and v are both positive orboth negative. [Assume the converse. If u > 0 and v < 0 then this gives« CC BY-NC-ND 3.0. Revised: 18 Nov 2012