Plane Geometry - Bruce E. Shapiro

SECTION 40. CIRCLES AND TRIANGLES 215let n = ←→ AC and let l be the perpendicular bisector of ACIf k ‖ l the either m = n or m ‖ n. Since the lines through any two sidesof a triangle are neither equal nor parallel we conclude that k ̸‖ l.Hence k and l intersect. Call this point O.Let r = OA. By the point-wise characterization of perpendicular bisectors,since O ∈ k, it is equidistant from A and B, and since O ∈ l, it is equidistantfrom A and C.Hence OA = OB = OC. Therefore the circle C(O, r) passes through allthree vertices.Corollary 40.5 If the Euclidean Parallel Postulate holds then all threeperpendicular bisectors of the sides of any triangle are concurrent and meetat the circumcenter of the triangle.Theorem 40.6 If the Euclidean Parallel Postulate fails then there exists atriangle that cannot be circumscribed.Figure 40.2: The triangle △CDE cannot be circumbscribed if the EuclideanParallel Postulate fails, because the lines l and m are parallel; if the EuclideanParallel Postulate held, they would intersect at the center of thecircumscribing circle.Proof. See figure 40.2.Assume that the Euclidean Parallel Postulate fails.This is equivalaent to assuming that the hyperbolic parallel postulate holds.Let □ABCD be any Saccheri quadrilateral with base AB and summit CD,and define M and N as the midpoints of AB and CD.By a result from Hyperbolic Geometry (that we have not yet proven),MN < AD.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.