235 V .4 : Angle defects and related phenomena In the previous section we showed that the angle sums of triangles in a neutral plane can behave in one of two very distinct ways. In fact, it turns out that there are essentially only two possible neutral planes, one of which is given by Euclidean geometry and the other of which does not satisfy any of the 24 properties listed in Section 2. The purpose of this section is to study some of these properties for a non – Euclidean plane. Definition. A neutral plane (P, L, d, µµµµ) is said to be hyperbolic if Playfair’s Parallel Postulate does not hold. In other words, there is some pair (L, X), where L is a line in P and X is a point not on L, for which there are at least two lines through X which are parallel to L. The study of hyperbolic planes is usually called hyperbolic geometry. The name “hyperbolic geometry” was given to the subject by F. Klein (1849 – 1925), and it refers to some relationships between the subject and other branches of geometry which cannot be easily summarized here. Detailed descriptions may be found in the references listed below: C. F. Adler, Modern Geometry: An Integrated First Course (2nd Ed.). McGraw – Hill, New York, 1967. ISBN: 0–070–00421–8. [see Section 8.5.3, pp. 219 – 226] A. F. Horadam, Undergraduate Projective Geometry. Pergamon Press, New York, 1970. ISBN: 0–080–17479–5. [see pp. 271 – 272] H. Levy, Projective and Related Geometries. Macmillan, New York, 1964. ISBN: 0–000–03704–4. [see Chapter V, Section 7] A full development of hyperbolic geometry is long and ultimately highly nonelementary, and it requires a significant amount of differential and integral calculus. We shall discuss one aspect of the subject with close ties to calculus at the end of this section, but we shall only give proofs that involve “elementary” concepts and techniques. In the previous section we showed that the angle sum of a triangle in a neutral plane is either always equal to 180° or always strictly less than 180°. We shall begin by showing that the second alternative holds in a hyperbolic plane. Theorem 1. In a hyperbolic plane there is a triangle ABC such that |∠∠∠∠CAB | + |∠∠∠∠ABC | + | ∠∠∠∠ACB | < 180°. By the results of the preceding section, we immediately have numerous consequences. Theorem 2. In a hyperbolic plane, given an arbitrary triangle ABC we have |∠∠∠∠CAB | + |∠∠∠∠ABC | + | ∠∠∠∠ACB | < 180°. Corollary 3. In a hyperbolic plane, suppose that we have a convex quadrilateral ABCD such that AB is perpendicular to both AD and BC. 1. If ABCD is a Saccheri quadrilateral with base AB such that d(A, D) = d(B, C), then |∠∠∠∠ADC | = |∠∠∠∠BCD | < 90°. 2. If ABCD is a Lambert quadrilateral such that |∠∠∠∠ABC | = |∠∠∠∠BCD | = |∠∠∠∠DAB | = 90°, then |∠∠∠∠ADC | < 90°.236 Proof of Corollary 3. If we split the convex quadrilateral into two triangles along the diagonal [AC], then by Theorem 2 we have the following: |∠∠∠∠CAB | + |∠∠∠∠ABC | + | ∠∠∠∠ACB | < 180° |∠∠∠∠CAD | + |∠∠∠∠ADC | + | ∠∠∠∠ACD | < 180° Since is a convex quadrilateral we know that C lies in the interior or ∠∠∠∠DAB and A lies in the interior of ∠∠∠∠BCD. Therefore we have |∠∠∠∠DAB | = |∠∠∠∠DAC | + | ∠∠∠∠CAB | and |∠∠∠∠BCD | = |∠∠∠∠ACD | + | ∠∠∠∠ACB |; if we combine these with the previous inequalities we obtain the following basic inequality, which is valid for an arbitrary convex quadrilateral in a hyperbolic plane: |∠∠∠∠ABC | + |∠∠∠∠BCD | + | ∠∠∠∠CDA | + | ∠∠∠∠DAB | = |∠∠∠∠CAB | + |∠∠∠∠ABC | + | ∠∠∠∠ACB | + |∠∠∠∠CAD | + |∠∠∠∠ADC | + | ∠∠∠∠ACD | < 360° To prove the first statement, suppose that ABCD is a Saccheri quadrilateral, so that |∠∠∠∠ADC | = |∠∠∠∠BCD | by the results of the previous section. Since |∠∠∠∠DAB | = |∠∠∠∠ABC | = 90°, the preceding inequality reduces to 180 + |∠∠∠∠BCD | + | ∠∠∠∠CDA | = 180° + 2 |∠∠∠∠BCD | = 180° + 2 | ∠∠∠∠CDA | < 360° which implies |∠∠∠∠ADC | = |∠∠∠∠BCD | < 90°. To prove the first statement, suppose that ABCD is a Lambert quadrilateral, so that |∠∠∠∠BCD | = 90°. Since |∠∠∠∠DAB | = |∠∠∠∠ABC | = 90°, the general inequality specializes in this case to 270° + | ∠∠∠∠CDA | < 360°, which implies |∠∠∠∠ADC | < 90°. Proof of Theorem 1. In a hyperbolic plane, we know that there is some line L and some point A not on L such that there are at least two parallel lines to L which contain A. Let C be the foot of the unique perpendicular from A to L, and let M be the unique line through A which is perpendicular to AC in the plane of L and A. Then we know that L and M have no points in common (otherwise there would be two perpendiculars to AC through some external point). By the choice of A and L we know that there is a second line N through A which is disjoint from L .237 The line N contains points U and V on each side of AC, and they must satisfy U∗A∗V. Since N is not perpendicular to AC and |∠∠∠∠CAU | + |∠∠∠∠CAV | = 180°, it follows that one of |∠∠∠∠CAU |, |∠∠∠∠CAV | must be less than 90°. Choose W to be either U or V so that we have θθθθ = |∠∠∠∠CAW | < 90°. The line L also contains points on both sides of AC, so let X be a point of L which is on the same side of AC as W. CLAIM: If G is a point of (CX, then there is a point H on (CX such that C∗G∗H and |∠∠∠∠CHA | ≤ ½ |∠∠∠∠CGA |. To prove the claim, let H be the point on (CX such that d(C, H) = d(C, G) + d(G, A) ; it follows that C∗G∗H holds and also that d(G, H) = d(A, G) . The Isosceles Triangle Theorem then implies that |∠∠∠∠GHA | = |∠∠∠∠GAH |, and by a corollary to the Saccheri – Legendre Theorem we also have |∠∠∠∠CGA | ≥ |∠∠∠∠GHA | + |∠∠∠∠GAH | = 2|∠∠∠∠GHA | = 2|∠∠∠∠CHA |, where the final equation holds because ∠∠∠∠GHA = ∠∠∠∠CHA. This proves the claim.