Hyperbolic Plane Geometry - Study Guide

Chapter 4 Hyperbolic Plane Geometry “It is quite simple, .... , but the way things come out of one another is quite lovely.” — Clifford 36. Introduction. There is much to be said in favor of carefully formulating, at this point, a set of explicitly stated assumptions as a foundation for the study of the geometry of the Hyperbolic Plane.1 For very mature students and for those who have been over the ground before, this is doubtless the best procedure. But for others, such a precise, rigorously logical treatment would only prove confusing. For this reason, keeping in mind our objectives, we propose to follow the path of the pioneers, to avail ourselves of the familiar foundations of Euclidean Geometry, replacing the Fifth Postulate by a contradictory one, and making such other changes as may consequently be forced upon us. Thus all of the Euclidean propositions which do not depend upon the Fifth Postulate, in particular the first twenty-eight, are immediately available. Nor are simplicity and economy the only advantages of such an approach. This presentation of the material, in the way it was acquired, has, we believe, the sanction of sound pedagogical and psychological principles. Refinement and extreme rigor may very well come later. 37. The Characteristic Postulate of Hyperbolic Geometry and Its Immediate Consequences. In Euclidean Plane Geometry, the Fifth Postulate is essentially equivalent to the statement that through a given point, not on a given line, one and only one line can be drawn which does not intersect the given line. In its place we introduce the following as the Characteristic Postulate of Hyperbolic Plane Geometry. Postulate. Through a given point, not on a given line, more than one line can be drawn not intersecting the given line. We have already observed that, if there is more than one line through the given point not intersecting the given line, there are an infinite number of them. If P is the given point, ` the given line, and AB and CD two lines through P which do not intersect `, then no line, such as EF , lying within the vertical angles AP C and DP B, which do not contain the perpendicular P Q from P to `, will cut `. A D E P F C B nR So ` Q 41 G H For if EF , produced for example to the right, were to cut `, then, since P F and the perpendicular P Q would both intersect `, P B would also have to intersect ` by the Axiom of Pasch. If one starts with the perpendicular P Q from P to ` and allows P Q to rotate about P in either direction, say counterclockwise, it will continue to intersect ` awhile and then cease to intersect it. Thus a situation is reached in which the lines through P are divided into two sets, those which cut ` and those which do not, each line of the first set preceding each line of the second. Under these circumstances, the Postulate of Dedekind asserts that there exists a line through P which brings about this division of the lines into the two sets. Since this line itself either cuts ` or does not cut it, it must either be the last of the lines which intersect ` or the first of those which do not. But there is no last cutting line. For, if one assumes that P G is the last of the cutting lines, and measures off any distance GH on the side of G opposite Q, then P H is a cutting line and a contradiction has been reached. Hence the dividing line is the first of those which do not intersect `. A similar situation is encountered if P Q is rotated clockwise. Thus there are two lines P R and P S, through P , which do not cut `, and which are such that every line through P , lying within the angle RP S, does cut `. Furthermore, the angles RP Q and SP Q are equal. If they are not, one of them is the greater, say RP Q. Measure angle IP Q equal to angle SP Q [by I.23]. P nR So I ` J Q K Then P I will cut ` in a point J. Measure off on `, on the side of Q opposite J, QK equal to QJ. Draw P K. It follows from congruent right triangles [SAS] that angle QP K equals angle QP J and hence angle QP S. But P S does not intersect ` and a contradiction has been reached. One concludes that angles RP Q and SP Q are equal. It can easily be shown that these two angles are acute. If they were right angles, P R and P S would lie on the same straight line and this line would be the line through P perpendicular to P Q. But that perpendicular does not intersect ` (Euclid I.28), and furthermore it is not the only line through P which does not intersect `. Consequently there will be lines through P within the angle RP S which do not cut ` under these circumstances and again a contradiction is encountered. These results can be summarized in the following theorem: Theorem 1: If ` is any line and P is any point not on `, then there are always two lines through P which do not intersect `, which make equal acute angles with the perpendicular from P to `, and which are such that every line through P lying within the angle containing that perpendicular intersects `, while every other line through P does not. All of the lines through P which do not meet ` are parallel to ` from the viewpoint of Euclid. Here, however, we wish to recognize the peculiar character of the two described in the theorem above. These two 42 lines we call the two parallels to ` through P , and designate the others as non-intersecting with reference to `. We shall discover shortly that the size of the angle which each of the two parallels makes with the perpendicular from P to ` depends upon the length h of this perpendicular. The angle is called the angle of parallelism for the distance h and will be denoted by Π(h) in order to emphasize the functional relationship between the angle and the distance. On occasion it will be found possible and convenient to distinguish between the two parallels by describing one as the right-hand, the other as the left-hand, parallel. Exercises 1. If two lines BA and BC are both parallel to line `, show that the bisector of angle ABC is perpendicular to `. 38. Elementary Properties of Parallels. Certain properties of Euclidean parallels hold also for parallels in Hyperbolic Geometry. Three of these are described in the following theorems. Theorem 2: If a straight line is the parallel through a given point in a given sense to a given line, it is, at each of its points, the parallel in the given sense to the given line. Statement Reason 1. AB k ` toward the right from P 1. given Case I 2. R is on AB to the right of P A 2. added premise P R N U B o T ` Q S M 3. P Q ⊥ `, RS ⊥ ` 4. line RT lies within 6 SRB 3. constructed (I.12) 4. added premise 5. U is a point on line RT 5. lines contain points 6. P U intersects line ` in point M 6. P U lies within 6 QP B 7. P U intersects QR in point N 7. Axiom of Pasch for 4QP R 8. RU intersects QM 8. Axiom of Pasch for 4QN M 9. RT intersects ` 9. follows from (8) 43 Case II 10. R is on AB to the left of P A 10. added premise R P T B I o ` S Q M 11. P Q ⊥ `, RS ⊥ ` 11. constructed (I.12) 12. RT lies within 6 QRP (within 6 SRQ is trivial) 12. added premise 13. RT intersects P Q 13. Axiom of Pasch for 4QRP 6 14. 6 T RP < QP B 15. 6 6 14. I.16 BP I = T RP 15. constructed (I.23) 16. P I intersects ` at M 16. Theorem 1 17. RT does not intersects P M 17. I.28 18. RT intersects QM 18. Axiom of Pasch for 4QP M 19. RT intersects ` 19. follows from (18) 44 Theorem 3: If one line is parallel to a second line, then the second line is parallel to the first line. A P R L I B E o F K G Q C H J Statement D Reason 1. AB k CD toward the right from P 1. given 2. P Q ⊥ CD, QR ⊥ AB, R is to the right of P 2. constructed (I.12), I.16 3. E is an arbitrary point inside 6 RQD 3. assumed 4. P F ⊥ QE 4. constructed (I.12) 5. F and E are on the same side of P Q 5. I-16 6. P G = P F , note P G < P Q 6. constructed (I.3), I.19 for 4P QF 7. GH ⊥ P Q 7. constructed (I.11) 8. 6 6 GP I = F P B 8. constructed (I.23) 9. P I intersects CD at J 9. Theorem 1 (6 QP I < 6 QP B) 10. GH cuts P Q, GH does not cut CD 10. by construction, I.28 11. GH cuts P J at K 11. Axiom of Pasch for 4P QJ 12. P L = P K, join F and L 13. 4GP K ∼ = 4F P L 12. constructed (I.3) 14. 6 13. SAS P F L is a right angle 14. CPCTE 15. QE and QL are the same line 15. both are perpendicular to P F 16. QE intersects AB at L 16. by statement 15 17. QD k AB toward the right from Q 17. point E was arbitrary 18. CD k AB toward the right from C 18. Theorem 2 (note that Q was not arbitrary) 45 Theorem 4: If two lines are both parallel to a third line in the same direction, then they are parallel to one another. A G E H I B F D C Statement Reason 1. AB k EF , CD k EF toward the right 1. given, WLOG AB is “above” CD Case I (EF is between AB and CD) 2. AC cuts EF at G 2. A and C on opposite sides of EF 3. AH is a line within 6 CAB 3. added premise 4. AH cuts EF in a point I 4. AH falls within 6 GAB 5. EF k CD toward the right 5. Theorem 3 6. IF k CD toward the right 6. Theorem 2 7. AH intersects CD 7. AH = AI falls within 6 CIF 8. AB does not cut CD 8. AB does not cut EF 9. AB k CD toward the right 9. AH arbitrary line in 6 CAB A G C H B D E F Case II (EF is “below” CD; Case III with EF “above” AB is similar) 10. assume AB 6k CD toward the right 10. added premise 11. take point G on AB 11. lines contain points 12. GB 6k CD toward the right 12. Theorem 2 13. through G draw GH k CD toward the right 13. such a parallel exists 14. GH 6= GB 14. obvious from (12) and (13) 15. EF k CD toward the right 15. Theorem 3 given CD k EF 16. GH k EF toward the right 16. Case I 17. GB k EF toward the right 17. Theorem 2 given AB k EF 18. GH = GB 18. parallels are unique 19. AB k CD toward the right 19. proof by contradiction ((14) and (18)) 46 39. Ideal Points. We wish to introduce at this point an important concept in connection with parallel lines. Two intersecting lines have a point in common, but two parallel lines do not, since they do not intersect. However, two parallel lines do have something in common. It is convenient to recognize this relationship by saying that two parallel lines have in common, or intersect in, an ideal point. (Also called, more frequently than not, a point at infinity or infinitely distant point.) Thus all of the lines parallel in the same sense to any line, and consequently parallel to one another, will be thought of as being concurrent in an ideal point and constituting a sheaf of lines with an ideal vertex. Every line contains thus, in addition to its ordinary or actual points, two ideal points through which pass all of the lines parallel to it in the two directions. While ideal points are concepts, so also, for that matter, are ordinary points. The introduction of these ideal elements is primarily a matter of convenient terminology. To say that two lines intersect in an ideal point is merely another way of saying that they are parallel; to refer to the line joining an ordinary point to one of the ideal points of a certain line amounts to referring to the line through the ordinary point parallel to that line in the sense designated. But we shall not be surprised to find these new entities assuming more and more significance as we go on. In the history of mathematics can be found more than one example of an idea introduced for convenience developing into a fundamental concept. As a matter of fact, the use of such ideal elements has been an important factor in the development of geometry and in the interpretation of space. We shall return to this later. It will gradually be recognized that, in so far as we are concerned with purely descriptive properties, we need not discriminate between ordinary and ideal points. Two distinct points, for example, determine a line, regardless of whether both points are ordinary, both ideal, or one ordinary and the other ideal. In no case is this more strikingly illustrated than in that of the triangle with two vertices ordinary points and the third ideal. We study this figure next. 40. Some Properties of an Important Figure. The figure formed by two parallel lines and the segment joining a point of one to a point of the other plays an important role in what is to follow. Let AΩ and BΩ be any two parallel lines. Here we follow the convention of using the large Greek letters (generally Ω) to designate ideal points. Let A, any point of the first, and B, any point of the second, be joined. The resulting figure is in the nature of a triangle with one of its vertices an ideal point; it has many properties in common with ordinary triangles. We prove first that Pasch’s Axiom holds for such a triangular figure. 47 Theorem 5: If a line passes within the figure ABΩ through one of the vertices, it will intersect the opposite side. A o P B Ω Q Statement Reason 1. AΩ k BΩ toward the right 1. given 2. point P is inside figure ABΩ 2. given 3. AP intersects BΩ in Q 4. BP intersects AΩ 3. 4. 5. P Ω k AΩ toward the right 5. parallels exist 6 6 BAP lies within 6 BAΩ ABP lies within 6 ABΩ 6. P Ω k BΩ toward the right 6. Theorem 4 7. P Ω does not intersect BQ 7. follows from 6 8. P Ω intersects AB 8. Axiom of Pasch for 4ABQ 48 Theorem 6: If a straight line intersects one of the sides of ABΩ, but does not pass through a vertex, it will intersect one and only one of the other two sides. A G H I o Ω B Statement Reason 1. AΩ k BΩ toward the right 1. given Case I (Case II is similar for BΩ) 2. line ` intersects AΩ in point G 2. given 3. line ` = GH intersects AB 3. Axiom of Pasch for 4ABG 4. GΩ k BΩ toward the right 4. Theorem 2 5. line ` = GI intersects BΩ 5. Theorem 5 applied to .BGΩ A S R o Ω T B Case III 6. line ` intersects AB in point R 6. given 7. RΩ k AΩ toward the right 7. parallels exist 8. RΩ k BΩ toward the right 8. Theorem 4 9. line ` = RS intersects AΩ 9. Theorem 5 applied to .RAΩ 10. line ` = RT intersects BΩ 10. Theorem 5 applied to .BRΩ 49 Theorem 7: The exterior angles of ABΩ at A and B, made by producing AB, are greater than their respective opposite interior angles. (In the figure below, γ > α and δ > β.) L A δ α M B o Ω β γ N C Statement Reason 1. AΩ k BΩ toward the right 1. given 2. assume γ = α 2. added premise 3. M is the midpoint of AB 3. constructed (I.10) 4. M N ⊥ BΩ 4. constructed (I.12) 5. AL = BN , join M to L 6. 4M BN ∼ = 4M AL 5. constructed (I.3) 7. 6. SAS LM A + 6 AM N = 6 BM N + 6 AM N = π 6 7. CPCTE, I.13 8. LM N is a straight line 8. I.14 9. LΩ k BΩ toward the right 9. Theorem 2 10. 6 N LΩ = π/2 10. CPCTE 11. γ 6= α 11. (9) and (10) contradict Theorem 1 A α o B β α D C 12. 6 CBD = α 12. constructed (I.23) 13. BD does not intersect AΩ 13. I.28 6 14. BD lies within CBΩ 6 14. Theorem 1 6 15. γ = CBΩ > CBD = α 15. follows from (14) 50 Ω Next we describe the conditions under which two such figures, ABΩ and A0 B 0 Ω0 , are congruent. Theorem 8: If AB and A0 B 0 are equal, and angle BAΩ is equal to angle B 0 A0 Ω0 , then angle ABΩ is equal to angle A0 B 0 Ω0 and the figures are congruent. (SA criterion) A A0 D α C o o Reason 1. AB = A0 B 0 , 6 BAΩ = 6 B 0 A0 Ω0 = α 0 0 0 1. given 0 0 ABΩ 6= A B Ω , WLOG ABΩ > A B Ω 6 6 6 0 2. added premise ABC = 6 A0 B 0 Ω0 6 3. 3. constructed (I.23) 4. BC intersects AΩ at D 0 0 0 4. Theorem 1 0 0 5. select D on A Ω so that A D = AD 6. 4ABD ∼ = 4A0 B 0 D0 8. 9. 5. constructed (I.3) 6. SAS ABD = 6 A0 B 0 D0 6 7. 7. CPCTE A 0 B 0 D 0 = 6 A 0 B 0 Ω0 6 0 6 10. 0 0 6 0 0 ABD < ABΩ 6 Ω0 B0 Statement 2. D0 Ω B 6 α 8. since both equal 6 ABC 0 9. B 0 D0 is a cutting line ABΩ = 6 A0 B 0 Ω0 10. proof by contradiction It is perhaps superfluous to remark that this theorem still holds if one of the figures is reversed, either by drawing the parallels in the opposite direction or by interchanging the two angles. 51 Theorem 9: If angles BAΩ and B 0 A0 Ω0 are equal and also angles ABΩ and A0 B 0 Ω0 , then segments AB and A0 B 0 are equal and the figures are congruent. (AA criterion) A α A0 o C Ω o β B B0 Reason BAΩ = 6 B 0 A0 Ω0 = α, 6 ABΩ = 6 A0 B 0 Ω0 = β, 6 0 0 0 0 0 2. AB 6= A B , WLOG AB > A B 1. given 0 2. added premise 3. AC = A B 4. ACΩ ∼ = A 0 B 0 Ω0 6 0 3. constructed (I.3) 4. Theorem 8 0 ACΩ = 6 A B Ω 5. Ω0 β Statement 1. α 0 5. follows from (4) 6 6 6. exterior ACΩ equals interior CBΩ for .BCΩ 0 6. both equal β 0 7. AB = A B ∼ .A0 B 0 Ω0 , 8. .ABΩ = 7. (6) contradicts Theorem 7 8. Theorem 8 Theorem 10: If segments AB and A0 B 0 , angles ABΩ and BAΩ, and angles A0 B 0 Ω0 and B 0 A0 Ω0 , are equal, then all four angles are equal to one another and the figures are congruent. A0 A β D o C β β Ω o E B0 B Reason 1. AB = A0 B 0 , 6 ABΩ = 6 BAΩ = α, 6 A0 B 0 Ω0 = 6 B 0 A0 Ω0 = β 1. given 2. α 6= β, WLOG α > β 2. added premise 3. 6 BAD = β = ABC Ω0 β Statement 6 E0 3. constructed (I.23) 4. BC intersects AΩ at T (not shown) 4. Theorem 1 5. AD intersects BC at E 5. Axiom of Pasch for 4ABT 6. choose E 0 on A0 Ω0 so that A0 E 0 = AE, draw B 0 E 0 7. 4EAB ∼ = 4E 0 A0 B 0 6. constructed (I.3) 8. 9. 6 6 0 0 ABE = A B E 6 0 7. SAS 8. CPCTE A0 B 0 E 0 = 6 A0 B 0 Ω 0 9. both equal β 10. α = β 10. (9) is a contradiction 11. .ABΩ ∼ = .A0 B 0 Ω0 11. Theorem 8 52 Exercises 2. In the figure ABΩ, the sum of angles ABΩ and BAΩ is always less than two right angles. 3. If a transversal meets two lines, making the sum of the interior angles on the same side equal to two right angles, then the two lines cannot meet and are not parallel, they are non-intersecting lines 4. Given two parallel lines, AΩ and BΩ, and two other lines, A0 C 0 and B 0 D0 , prove that if segments AB and A0 B 0 , angles BAΩ and B 0 A0 C 0 , and angles ABΩ and A0 B 0 D0 , are equal, then A0 C 0 and B 0 D0 are parallel. 5. If angles ABΩ and BAΩ are equal, the figure is in the nature of an isosceles triangle with vertex an ideal point. Prove that, if M is the midpoint of AB, M Ω is perpendicular to AB. Show also that the perpendicular to AB at M is parallel to AΩ and BΩ and that all points on it are equally distant from those two lines. 6. Prove that if, in figure ABΩ, the perpendicular to AB at its midpoint is parallel to AΩ and BΩ, then the angles at A and B are equal. 7. If, for two figures ABΩ and A0 B 0 Ω0 , angles ABΩ and A0 B 0 Ω0 are equal but segment AB is greater than segment A0 B 0 , then angle BAΩ is smaller than angle B 0 A0 Ω0 . 41. The Angle of Parallelism. From Theorem 8 of the preceding section, it is evident that the angle of parallelism Π(h) for any given distance h is constant. Furthermore, as a consequence of Theorem 7, it follows that 0 < h1 < h2 ⇒ Π(h1 ) > Π(h2 ), that is, Π is a strictly decreasing function. We know, from the theorem of Section 37 [Theorem 1], that every distance has a corresponding angle of parallelism. It has just been pointed out that this angle is always the same for any given distance, that the angle increases as the distance decreases and decreases as the distance increases. Presently it will be shown that to every acute angle there corresponds a distance for which the angle is the angle of parallelism. In this case equal angles must, of course, have equal corresponding distances. Putting these results together, we conclude that
lim Π(h + δ) − Π(h) = 0, δ→0 and consequently that Π(h) varies continuously if h does. [The function Π is monotone and satisfies the intermediate value property and is thus continuous.] It should perhaps be noted here that, so far, no particular unit has been specified for measuring either distances or angles. The functional relationship implied has been purely geometric. Later, when definite units have been adopted, the analytic form of Π(h) [which turns out to be 2 arctan(e−h )] will be obtained. However, as h approaches zero, Π(h) approaches a right angle, and we may write Π(0) = π/2, where π is, for the present, used merely as a symbol to denote a straight angle. As h becomes infinite, Π(h) approaches zero, or, in the conventional notation, Π(∞) = 0. Moreover, there is no reason why we should not attach a meaning to Π(h) for h negative. There is nothing which compels us to do this; we do it solely because it will prove convenient. Such a generalization will enable us to avoid certain exceptions later on. The definition of Π(h), when h is negative, is a matter of choice, but we shall choose methodically. As h, being positive, decreases, Π(h) increases; when h is zero, Π(h) is a right angle. If we think of h as continuing to decrease, becoming negative, we naturally choose to regard Π(h) as continuing to increase and becoming obtuse. Briefly, Π(−h) is defined by the relation Π(h) + Π(−h) = π. 53 42. The Saccheri Quadrilateral. It will be recalled that, as a basis for his investigations, Saccheri made systematic use of a quadrilateral formed by drawing equal perpendiculars at the ends of a line segment on the same side of it and connecting their extremities. This birectangular, isosceles quadrilateral is commonly called a Saccheri Quadrilateral. We shall study some of its properties. The side adjacent to the two right angles is known as the base, the opposite side as the summit and the angles adjacent to the summit as the summit angles. Theorem 11: The line joining the midpoints of the base and summit of a Saccheri Quadrilateral is perpendicular to both of them; the summit angles are equal and acute. D ................................. ...........................................H ... ..................................................................................C... ... .. .... ... ... .. ... . . ... .. .... ... ... .. ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... ... .. ... ... .. ... . . ... .. .... . . ... .. .... . . ............ ..................................................................................................................................................................... A B M Statement Reason 1. tABCD is a Saccheri Quadrilateral 1. given 2. M , H are the midpoints of AB, CD 3. tDAM H ∼ = tCBM H 2. constructed (I.10) 4. 6 6 6 3. SASSS 6 AM H = BM H, DHM = CHM 4. CPCQE 5. M H is perpendicular to both AB and CD 6. 6 ADC = 6 BCD 5. follows from (5) 6. CPCQE ..............D ........................................................................................................................................................C............................... E . ... α ...................θ.........− α ...... θ ...α .......β ...... ............ .. ...... ............ ...... ............ . ...... ............ ... ...... ............ . ....... ............ . ....... ............ ... ....... ............ . ............ ....... . ............ ....... ... ............ . ....... ............ ....... . ............. ....... ... ...... ............. ....... ............. ....... . ............. ....... ... . .. ............. ............. . ............. ... ............. .. ............. .............. ... .... .. ... .. ... .. ........... . . . . . . . . ................................................................................................................................................................................................................................................ A 8. 9. 6 ADΩ = BCΩ (labeled α) 6 10. 7. assumed to exist 8. equal angles of parallelism (Thm 8) 6 ADC = BCD (labeled θ) 6 Ω B 7. DΩ, CΩ parallel toward the right to AB extended 6  9. equal summit angles (from (6)) ECΩ > 6 CDΩ (β > θ − α as labeled) 10. Theorem 7 for .CDΩ 6 11. π = β + α + θ > 2θ 11. DCE = π, (10) 12. θ is an acute angle 12. simple consequence of (11) Corollary 12: The base and summit of a Saccheri Quadrilateral are non-intersecting lines. 54 43. The Lambert Quadrilateral. The reader will recollect that Lambert used, as a fundamental figure in his researches, a quadrilateral with three of its angles right angles. This trirectangular quadrilateral, which we shall call a Lambert Quadrilateral, has an important part to play in later developments. Theorem 13: In a Lambert Quadrilateral the fourth angle is acute. F D C E A B Statement Reason 1. tABCD is a Lambert Quadrilateral with right angles indicated 1. given 2. extend BA to E so that AE = BA 2. constructed (I.3) 3. erect EF ⊥ BE so that EF = BC 3. constructed (I.11, I.3) 4. join F to D 5. tDABC ∼ = tDAEF 4. postulate 6. 6 5. SASAS F DA is a right angle 6. CPCQE 7. F DC is a straight line 7. I.14 8. tF EBC is a Saccheri Quadrilateral 8. by definition 9. 6 DCB is acute 9. Theorem 11 A useful theorem in regard to a more general quadrilateral, with only two right angles, may very well be inserted here. From it come immediately some important properties of the Saccheri and Lambert Quadrilaterals. 55 Theorem 14: If, in the quadrilateral ABCD, the angles at two consecutive vertices A and B are right angles, then the angle at C is larger than or smaller than the angle at D according as AD is larger than or smaller than BC, and conversely. D C E A B Statement Reason 1. tABCD with right angles at A and B 1. given 2. BCD = 6 C, 6 3. AD = BC ⇒ 6 6 ADC = 6 D 2. notation 3. tABCD is a Saccheri Quadrilateral 6 C= D 4. AD > BC 4. added premise 5. choose E on AD so that AE = BC 5. constructed (I.3) 6 7. 6 6. tABCE is a Saccheri Quadrilateral AEC = 6 BCE 6. 6 6 BCE = AEC > D 7. I.16 8. AD > BC ⇒ 6 C> D 8. statements (4)-(7) 9. AD < BC ⇒ 6 C<6 D 9. similar reasoning 6 6 C=6 D 10. 10. added premise 11. AD 6> BC, AD 6< BC 11. by (8) and (9) 12. AD = BC 12. law of trichotomy 13. 14. 15. 6 6 6 C = 6 D ⇒ AD = BC 13. statements (10)-(12) C < 6 D ⇒ AD < BC C > 6 D ⇒ AD > BC 14. similar reasoning 15. similar reasoning Corollary 15: In a Lambert Quadrilateral, the sides adjacent to the acute angle are greater than their respective opposite sides. Corollary 16: The summit of a Saccheri Quadrilateral is larger than its base. 56 Exercises 8. 9. 10. 11. 12. 13. Prove Corollary 12. Prove Corollary 15. Prove Corollary 16. If, in quadrilateral ABCD, the angles at A and B are right angles and the angles at C and D are equal, prove that the figure is a Saccheri Quadrilateral. Prove that the diagonals of a Saccheri Quadrilateral are equal but that these diagonals do not bisect each other. Prove that, if perpendiculars are drawn from the extremities of one side of a triangle to the line passing through the midpoints of the other two sides, a Saccheri Quadrilateral is formed. As a consequence, prove that the perpendicular bisector of any side of a triangle is perpendicular to the line joining the midpoints of the other two sides. Prove that the segment joining the midpoints of two sides of a triangle is less than one-half the third side. Show that a line through the midpoint of one side of a triangle perpendicular to the line which bisects a second side at right angles bisects the third side. Prove that the line joining the midpoints of the equal sides of a Saccheri Quadrilateral is perpendicular to the line joining the midpoints of the base and summit and that it bisects the diagonals. 14. 15. 16. 44. The Sum of the Angles of a Triangle. Theorem 17: The sum of the angles of every right triangle is less than two right angles. Q A α D β M β C B P Statement Reason 1. 4ABC is a right triangle with 6 C = π/2 1. given 2. 6 A = α, 6 B = β 2. notation 3. 6 BAD = 6 B = β 3. constructed (I.23) 4. M is midpoint of AB 4. constructed (I.10) 5. M P ⊥ CB 5. constructed (I.12) 6. P is on CB between C and B 6. Axiom of Pasch and I.28 7. AQ = BP , join M and Q 8. 4M BP ∼ = 4M AQ 7. constructed (I.3) 9. 6 6 8. SAS 6 AM Q = BM P , M QA = π/2 9. CPCTE 10. π = 6 BM P + 6 AM P = 6 AM Q + 6 AM P 10. AM B is a straight line, (9) 11. QM P is a straight line 12. tCP QA is a Lambert Quadrilateral 11. I.14 12. definition 13. α + β < π/2 13. Theorem 13 14. 6 A+6 B+6 C <π 14. follows from (1) and (13) 57 Theorem 18: The sum of the angles of every triangle is less than two right angles. A α1 B α2 γ β C D Statement Reason 1. 4ABC is a triangle, 6 A = α = α1 + α2 , 6 B = β, 6 C = γ 1. given, notation 2. WLOG angles β and γ are acute 2. I.17 3. AD ⊥ BC 3. constructed (I.12) 4. D is on BC between B and C

5. α1 + β + π/2 + α2 + γ + π/2 < π + π 4. I.16 6. α + β + γ < π 6. follows from (5) 5. Theorem 17 The difference between two right angles and the angle-sum of a triangle is called the defect of the triangle. Corollary 19: The sum of the angles of every quadrilateral is less than four right angles. 58 Theorem 20: If the three angles of one triangle are equal, respectively, to the three angles of a second, then the two triangles are congruent. A A0 γ β D E γ β B C B Statement 6 1. γ β 0 C0 Reason A = 6 A0 , 6 B = 6 B 0 , 6 C = 6 C 0 , 0 0 0 2. AB 6= A B , WLOG AB > A B 1. given 0 2. added premise 3. AD = A0 B 0 , AE = A0 C 0 3. constructed (I.3) 4. AC > A0 C 0 so E between A and C 5. 4DAE ∼ = 4B 0 A0 C 0 4. see below (10–11) 6. 6 0 6 6 6 5. SAS 6 0 6 EDA = B = B = β, AED = C = C = γ, 6. CPCTE, hypothesis, notation 7. angle sum for quadrilateral tBCED is 2π 0 7. ADB and AEC are straight angles 0 8. AB = A B 9. 4ABC ∼ = 4A0 B 0 C 0 8. (7) contradicts Corollary 19 9. ASA A A D D B C=E B C F E 10. AC = A0 C 0 = AE ⇒ 6 ACD = 6 ACB, a contradiction 10. D is between A and B 0 6 CEF = 6 ACF , a contradiction 11. I.16 0 11. AC < A C = AE ⇒ Thus we reach the remarkable conclusion that in Hyperbolic Geometry similar triangles, or indeed similar polygons, of different sizes do not exist. We shall show in Section 54 how to construct a triangle, given the three angles. 59 Exercises 17. Prove that two Saccheri Quadrilaterals with equal summits and equal summit angles, or with equal bases and equal summit angles, are congruent. 18. A segment joining a vertex of a triangle to a point on the opposite side is called a Cevian. A Cevian divides a triangle into two subtriangles, and one or both of these can be subdivided by Cevians, and so on. Prove that, if a triangle is subdivided by arbitrary Cevians into a finite number of triangles, the defect of the triangle is equal to the sum of the defects of the triangles in the partition. What does this suggest about the defect of a triangle as compared to its size? 60