Plane Geometry - Bruce E. Shapiro

SECTION 39. CIRCLES 205Proof. Since P, Q ∈ Γ, OP = OQ = r. By the point-wise characterizationof perpendicular bisectors (theorem 28.2), O is on the perpendicularbisector of P Q.Figure 39.4: Proof of the Secant Line Theorem. Since points P and Q areequidistant from O, O must lie on their perpendicular bisector.Theorem 39.10 Let l be a secant line that intersects circle Γ = C(O, r)at distinct points P and Q. Then every point on the interior of segmentP Q is inside Γ, and every point on ←→ P Q − P Q lies outside of Γ.Proof. (Exercise)Theorem 39.11 (Weak Circular Continuity a ) Let Γ = C(O, r) be acircle and let l be a line with points A, B ∈ l such that A is inside Γ andB is outside Γ. Then l is a secant line of Γ.a We call this result weak circular continuity to distinguish it from the the similar result(theorem 39.17) that deals with the intersection of two circles, which we will call strongcircular continuity.Proof. Since A ∈ l is inside Γ, then l cannot be a tangent line (corollary39.8). Hence the number of intersections is either 0 or 2, but not 1 (theorem39.6). If we can show that any intersection exists, then since a singleintersection is not possible, we know that there must be two intersections.Let d = AB.By the ruler postulate, ∀x ∈ [0, d], ∃ a unique point D(x) ∈ AB such thatAD(x) = x.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.