Plane Geometry - Bruce E. Shapiro

SECTION 39. CIRCLES 207Theorem 39.14 Let △OCQ be a right triangle with right angle at C. IfP is a point such that O ∗ P ∗ Q and OP = OC, then CP < CQ.Proof. (Exercise; see figure 39.6.)Theorem 39.15 Let C be the point of tangency of line l to Γ = C(O, r).Pick any point B ∈ l and let β = µ(∠BOC). Define P (α) : (0, β) ↦→ Γsuch that µ(∠P (α)OC) = α. Thenlim CP (α) = 0α→0 +Figure 39.6: See theorem 39.15.Proof. See figure 39.6.For any x, by the crossbar theorem ←−−−→ OP (x) intersects BC at some pointQ(x).By theorem 39.14, 0 < CP (x) < CQ(x)By the continuity axiom (theorem 20.2), the function mapping the distanceCQ(x) to α is continuous. Hencelim CQ(α) = 0α→0 +The result of the theorem follows by the Sandwich theorem for limits fromCalculus.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.