Plane Geometry - Bruce E. Shapiro

SECTION 28. CHARACTERIZATION OF BISECTORS 131Since △P AM and △P BM share a side, are right triangles, and have congruenthypotenuses, they are congruent by the Hypotenuse Leg Theorem.Hence AM = BM, meaning that ←−→ P M is a bisector of AB. Since by construction←−→ P M ⊥ AB it is a perpendicular bisector.Figure 28.3: Theorem 28.3.xyf(x)f(y)Theorem 28.3 (Continuity of Distance) Let A, B, and C be noncollinearand let d = AB. For each x ∈ [0, d], there exists a unique pointD x ∈ AB such that AD x = x. Define the function f : [0, d] ↦→ [0, ∞) byf(x) = CD x . Then the function f is continuous.Proof. Let A, B, C, f be as defined in the statement of theorem 28.3 (seefigure 28.3), and let ɛ > 0 be given. Choose δ = ɛ.Suppose y ∈ [0, d] such thatBy the triangle inequalityD x D y = |x − y| < δ = ɛCD x < CD y + D x D y < CD y + ɛandHenceCD y < CD x + D x D y < CD x + ɛCD x − ɛ < CD y < CD x + ɛf(x) − ɛ < f(y) < f(x) + ɛ−ɛ < f(y) − f(x) < ɛ|f(y) − f(x)| < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.