Plane Geometry - Bruce E. Shapiro

174 SECTION 34. THE PARALLEL PROJECTION THEOREMBy Proclus’ Axiom (Theorem 32.6), since m and n are parallel to l, linest ′′ and t ′′′ intersect m and n at points B ′′ and C ′′ . Define the angles α β,γ, δ and ɛ, as shown in figure 34.1.If A = A ′ then B ′′ = B and t = t ′′ hence AB = A ′ B ′′ .If B ′ = B then C = C ′′ , t ′′′ = t, and hence BC = B ′ C ′′ .If both A ≠ A ′ and B ′ ≠ B then □AA ′ B ′′ B and □BB ′ C ′′ C are parallelograms.Since the opposite sides of parallelograms are congruent (corollary32.7), AB = A ′ B ′′ and BC = B ′ C ′′ .By hypothesis, AB = BC, hence A ′ B ′′ = B ′ C ′′ .By transitivity of parallelism, t ′′ = t ′′′ or t ′′ ‖ t ′′′ . Hence γ and β arecorresponding angles. By the corresponding angles theorem (theorem 29.7),γ = δ.Similarly, by repeated application of the corresponding angles theorem,α = ɛ = β.Hence by ASA, △A ′ B ′′ B ′ ∼ = △B ′ C ′′ C ′ .Hence by the properties of congruent triangles A ′ B ′ = B ′ C ′ .Theorem 34.2 (The Parallel Projection Theorem) Let l, m, and nbe distinct parallel lines; let t and t ′ be transversals that cut these threelines at points A, B, C and A ′ , B ′ , C ′ , respectively, such that A ∗ B ∗ C.ThenABAC = A′ B ′A ′ C ′Proof. Either AB/AC is rational, or it is irrational.First suppose that AB/AC is rational. Then there exist positive integers pand q such thatABAC = p qAccording to the ruler postulate we can choose equally spaced points A 0 , A 1 , ..., A qsuch thatA i A i+1 = ACqThen A p = B (see figure 34.3). Define the lines l i ‖ l through each A i anddefine the intersection of each line with t ′ as A ′ i .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012