§22.1 Back to the Foundations 1 Remember there are three Cases: 1. There exists no line through P parallel to l. (Elliptic, spherical or Riemannian Geometry.) 2. There exists one line through P parallel to l. (Parabolic or Euclidean geometry.) 3. There exists more than one line through P parallel to l. (Hyperbolic or Lobachevskian geometry.)

Our Direction of Study 2 Let us then take up with absolute geometry without the Parallel Postulate of Euclid which will take us to the third case. 3. There exists more than one line through P parallel to l. (Hyperbolic or Lobachevskian geometry.)

3 Hyperbolic Parallelism Definition. Two lines are parallel iff they meet only at infinity. 3 Axiom: Hyperbolic Parallel Postulate. If l is any line and P any point not on l, there exist more than one line passing through P parallel to l.

4 Hyperbolic Parallelism What are some ways in which to deny the possibility of two or more parallel lines? 4 Try to make a drawing or find a real life example of two parallel lines. Perhaps use the equal distance between parallel lines as an argument. If you look around it seems that it would be easy to refute such and assumption.

Theorem 1: The sum of the angles of any right triangle is less than Given:  ABC with ∡ C = 90 Prove: Σ ΔABC < 180 (1) Line AP = m perpendicular to AC Construction. (2) m is parallel to BC Alt Interior  s. (3) n = AQ is parallel to CB Previous Axiom. (4) Pick W on CB so that  θ <  t Construction. What is given? What will we prove? Why? QED (5) Σ ∡ ’s  ACW = 90 + ∡ θ + ∡ CAW Why? Definition. (6) < 90 + ∡ t + ∡ CAQ Why? ∡ t > ∡ θ, ∡ CAQ > ∡ CAW DRAWING ON NEXT SLIDE. (7) = 90 + ∡ CAP Why? ∡ t + ∡ CAQ = ∡ CAP = 90 (8) Σ ∡ ’s  ACW < 180 Why? Substitute 7 into 5 (9) Σ ∡ ’s  ABC < 180 Why? Previous Theorem

Drawing for Theorem 1 Place on board. A B C Q PW n m Back t θ

Corollary: The sum of the angles of any triangle is less than Given:  ABC Prove: Σ ΔABC < 180 (1) Altitude to largest sideConstruction. (2) Σ ∡ ’s  ABD < 180 Theorem 1. (3) Σ ∡ ’s  BDC < 180 Theorem 1. (4) Σ ∡ ’s  ABC < 180 Algebra. What is given? What will we prove? Why? QED A B C D

Defect 8 Since the sum of the angles of a triangle is less than 180 we can define the defect of the triangle as that difference, or δ(  ABC) = ∡ A - ∡ B - ∡ C The defect of a convex polygon P 1 P 2... P n is the number δ (P 1 P 2... P n ) = 180(n -2) – P 1 – P P n

Defect 9 The defect of a triangle has the following properties: 1. Existence of area. 2. Relative size. 4. Congruency 3. Additively.

10 Definition of Area 10 The area of a convex polygon P 1 P 2... P n is defined by the number: K = k [180 (n – 2) – m  P 1 – m  P 2... – m  P n ] Where k is some predetermined constant for the entire plane (not depending on each given polygon). The value for k is frequently taken as π/180, which converts degree measure to radian measure. That is Area = k δ, where δ is the defect of the polygon.

11 Definition of Area 11 That is the area is some constant k times the sum of the deficits of each of the triangles the polygon is broken down to by drawing all the diagonals from one vertex of from some interior point. Area = k [180 (7 – 2) – m  P 1 – m  P 2... – m  P 7 ] P 4 δ 1 δ 2 δ 3 δ 4 δ 5 P 1 P 2 P 3 P 5 P 6 P 7 Area = k (δ 1 + δ 2 + δ 3 + δ 4 + δ 5 ) Area = k [(180 – m  P 1 ) + (180 -– m  P 2 ) + (180– m  P 3 ) + (180 -– m  P2) + (180 -– m  P2) + (180 -– m  P2) + (180 -– m  P2) ]

Lemma 12 Let AD be a Cevian of  ABC and δ 1 and δ 2 denote the defects of the subtriangles  ABD and  ADC, then δ (  ABC ) = δ 1 + δ 2

Lemma: Let  ABC be if AD is a Cevian of  ABC and δ 1 and δ 2 denote the defects of the subtriangles  ABD and  ABC, then δ(  ABC ) = δ 1 + δ 2 13 Given:  ABC with Cevian ADProve: δ(  ABC ) = δ 1 + δ 2 (1) δ 1 = ∡ B - ∡ 1 - ∡ 3 Definition. (2) δ 2 = ∡ C - ∡ 2 - ∡ 4 Definnition. (3) δ 1 + δ 2 = ∡ B - ∡ 1 - ∡ 3 - ∡ C - ∡ 2 - ∡ 4 Addition. (4) δ 1 + δ 2 = ∡ A - ∡ B - ∡ C = δ(  ABC ) Substitution. What is given? What will we prove? Why? QED A DC B δ1δ1 δ2δ

Theorem 2: AAA Congruency. 14 Given:  ABC and  PQR with  A =  P,  B =  Q, and  C =  R Prove: ΔABC  ΔPQR (1) If AB = PQ then ΔABC  ΔPQR SAS. (2) D on AB and E on AC AD = PQ and AE = PR Construction. (3) ΔADE  ΔPQR ASA. What is given? What will we prove? Why? QED (4) δ(  ABC ) = δ (  ADE ) + δ (  BDC ) + δ (  CDE ) Why?Lemma. (5) δ(  ADE ) = δ(  PQR) Why?Given. DRAWING ON NEXT SLIDE (6) δ (  ABC ) = δ (  PQR) + δ (  BDC ) + δ (  CDE ) Why? Substitution. (7) δ(  ABC ) > δ(  PQR) Why?  have same  s By Contradiction! Let AB > PQ and AC > PR.  AB = PQ and ΔABC  ΔPQR

Drawing for Theorem 2 P Q δ4δ4 R C A D B δ1δ1 δ2δ2 E δ3δ3 Back

Corollary 16 There do not exist any similar, noncongruent triangles in hyperbolic geometry.

Assignment: §22.1