Plane Geometry - Bruce E. Shapiro

220 SECTION 41. EUCLIDEAN CIRCLESFigure 41.1: Proof of theorem 41.1.Theorem 41.3 Let △ABC be a triangle and let M be the midpoint ofAB. If ∠ACB is a right angle, then AM = MC.Proof. Construct a point D on −−→ MC such that MD = MA. By theorem41.1, ∠ADB = 90.Suppose that C ≠ D. Then either M ∗ D ∗ C or M ∗ C ∗ D. Suppose thatM ∗ D ∗ C.Then (see figure 41.2), γ is an exterior angle of △ADC, hence γ > ɛ(exterior angles theorem).Similarly, β is an exterior angle of △BDC, hence β > δ (also by the exteriorangles theorem).Hence,which is a contradiction (90 ¿90).A similar result holds if M ∗ C ∗ D.90 = ∠BDA = β + γ > δ + ɛ = 90Hence C = D and therefore MC = MD = MA.Corollary 41.4 If ∠ACB is a right angle, then AB is a diameter of thecircle that circumscribes △ABC.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012