Plane Geometry - Bruce E. Shapiro

SECTION 14. BETWEENNESS 71u − v = u + v which implies v = 0 or f(B) = f(C); but C and B aredistinct points so f(B) ≠ f(C). If u < 0 and v > 0 then −u + v = u + vwhich implies u = 0 or f(C) = f(A) which is impossible because A and Care distinct points.]Since u and v have the same sign, then both f(C) − f(A) and f(B) − f(C)have the same sign.If both f(C) − f(A) > 0 and f(B) − f(C) > 0 then f(C) > f(A) andf(B) > f(C) so that f(B) > f(C) > f(A).If both f(C) − f(A) < 0 and f(B) − f(C) < 0 then f(C) < f(A) andf(B) < f(C) so that f(B) < f(C) < f(A).Corollary 14.26 If A, B, C are distinct collinear points then exactly oneof them lies between the other two.Proof. Since A, B, C are distinct then they correspond to real numbersx, y, z. Then this is a property of the real numbers, exactly one of x,y, andz lies between the other two.Corollary 14.27 Let A, B, C be points such that B ∈ −→ AC. ThenA ∗ B ∗ C ⇐⇒ AB < ACProof. By theorem 14.25 one the following must hold:If 14.7 holds thenIf 14.8 holds thenf(A) < f(B) < f(C) (14.7)f(A) > f(B) > f(C) (14.8)AB = f(B) − f(A) < f(C) − f(A) = ACAB = f(A) − f(B) < f(A) − f(C) = ACTo prove the converse, suppose that AB < AC.By the corollary one of A,B, C lies between the other two. We have threepossibilities: A ∗ B ∗ C, B ∗ A ∗ C or A ∗ C ∗ B.But B ∗ A ∗ C is not possible B ∈ −→ AC and B is distinct from A.So suppose A ∗ C ∗ B. Then eitherf(A) < f(C) < f(B)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.