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1/7 Archimedean Neutral Geometry (Geometry: Euclid and Beyond - Chapter 35 ) Adding Archimedes' axiom (A) to neutral geometry (geometry with Hilbert's axioms of incidence, betweenness, and congruence - without parallel axiom), we get the fact that the angle sum of a triangle is less than or equal to two right angles (2RA), i.e. the semielliptic case is impossible. Preliminary reminders (A) Archimedes axiom Given a line segment AB and CD, there is a natural number n such that n copies of AB added together will be greater than CD. (P) Playfair's or parallel axiom For each point P and each line l, there exists at most one line through P parallel to l. Lemma 35.1 In a Hilbert plane with (A), let α, β be given angles. Then there exists an integer n > 0 such that nα > β, or else nα becomes undefined by exceeding 2RA. Proof First reduce β to β/2 to get a triangle with a RA: Construct two equal segments OA, OB to get AB Construct the perpendicular OC to get two congruent triangles ΔACD, and ΔBDC by (SAS). Prove the lemma for β/2 in ΔACO by contradiction. Rename points A, C and angle β/2 as in figure Assume nα ≤ β for all n (*) Let α cut off segements AA1, A1A2, ... on AB to get sequence A1, A2, A3, ... Claim: AA1 < A1A2 < A2A3 < ... Proof claim: Let Ai = C, Ai+1 = D, Ai+2 = E Construct DF such that DF = CD Then ΔODF = ΔOCD by (ASA) Using (I.16) we have γ < δ (δ is extension to ΔODF) Using (I.16) we have γ > ε (γ is extension to ΔODE) Conclude δ > ε Larger angle subtends larger side (I.19), thus DE > DF = CD By (A), there exists n such that n · AA1 > AB But AiAi+1 < Ai+1Ai+2, hence AAn > AB contradicting the assumption (*) □ 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 2/7 Theorem 35.2 In a Hilbert plane with Archimedes' axiom (A), the sum of the angles of a tringle is less than or equal to two right angles. Proof (by contradiction) Assume there exists a triangle ABC with sum of angles greater than 2RA. sum of angles = 2RA + ε where ε > 0 Start with any triangle ABC Construct mid-point between BC to get D Extend AD with AD to get E By (I.15) the opposite angles at D are equal By (SAS) we have ΔABD = ΔDEC The sum of angles in ΔABC is α + β + γ + δ, the sum of angles in ΔAEC α + β + γ + δ, therefore the sum of angles is unchanged. It is either α ≤ 1/2 (α + β) or β ≤ 1/2 (α + β). Repeating the construction above to the smallest angle of the preceding triangle, we get a sequence of triangles T0 = ΔABC, T1= ΔAEC, T2, T3, ... where Tn has one angle equal or less than (1/2)^n (α + β). By previous lemma (35.1) there exists n such that T n has an angle less than ε, and the sum of the two other angles will be more than 2RA, contradicting (I.17) □ Corollary 35.3 In any triangle, the exterior angle is greater than or equal to the sum of the opposite interior angles. Proof The exterior angle plus the adjoint interior angle is equal to 2RA. The sum of all angles in ΔABC is less or equal to 2RA. The sum of the opposite angles is therefore less or equal to the exterior angle. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 3/7 Proposition 35.4 In a Hilbert plane with (A), if every triangle is Euclidean, then (P) holds. Proof By proving the contraposition, i.e. if (P) does not hold, then there exists a non Euclidean triangle If (P) does not hold, then there exists a line l and point P not on l with at least two distinct lines parallel to l Construct a perpendiculer line to l through P. Add a perpendicular at P to get m, which must be parallel to l by (I.27). Let m' be another parallel to l since there must be at least two distinct parallel lines to l. Choose B and let C be such that PB = BC. Then α is exterior angle to ΔPCB. Therefore α ≥ 2γ by previous corollary (35.3). Repeating this procedure n times to find F with angle PFA ≤ (1/2)^n α. By previous proposition (35.1) there exists n such that angle at PFA is less than β. Consider ΔAPF. The angle at A is a RA, at F the angle is less than β. The angle at P is less than (RA - β) because m' cannot lie inside ΔAPF. Thus, the sum of the angles is less than 2RA. Therefore the trianlge ΔAPF is not Euclidean. □ Legendre's Axiom Given an angle α and given a point P in the interior of the angle α, there exists a line through P that meets both sides of the angle. Remark The parallel axiom can be proven with Legendre's axiom. See next proposition (35.5). It can be further shown with Legendre's axiom, that for any angle α, there exists a line l entirely contained in the angle α (see exercise 35.4). Legendre then says, that this is a contradiction to the nature of a straight line, i.e. that such a line should be entirely contained within an angle. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 4/7 Proposition 35.5 In a Hilbert plane with (A), if Legendre's axiom holds for a single angle α, then also (P) holds. Proof Starting from a triangle ΔABC with angle α at A we construct triangle ΔBDC by repeating the angle ACB at point B and the side BD = AC. Then these two triangles are congruent, i.e. ΔABC = ΔBDC by (SAS). By (I.27) AC is parallel to BD and AB is parallel to CD. Let l be a line through D meeting the sides of the angle α at E and F Therefore E and F lie beyond B, C, i.e. further from A than B, C. Therefore ΔAEF contains ΔABC and ΔBDC. Assume (P) does not hold, then by contraposition of previous proposition (35.4), there exists a not Euclidean triangle and by theorem from previous section (34.7) we see that, every triangle has the sum of angles not equal to 2RA. Therefore triangle ΔABC has a defect δ > 0 and by lemma from previous section (34.8) ΔAEF has a defect greater than 2δ. Repeating the procedure of constructing a larger triangle we have eventually a triangle with defect greater than 2RA, which is absurd. Therefore (P) has to hold. □ Theorem 34.7 (without proof) In any Hilbert plane: (a) If there exists a triangle whose sum is less than 2RA, then every triangle has angle sum less than 2RA. (b) The following are equivalent: (i) There exists a triangle with sum = 2RA. (ii) There exists a rectangle. (iii) Every triangle has angle sum = 2RA. (c) If there exists a triangle whose sum is greater than 2RA, then every triangle has angle sum greater than 2RA. Lemma 34.8 (without proof) If a triangle ABC is cut into two triangles by single transversal BD, the defect of the big triangle is equal to the sum of the defects of the two small triangles. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 5/7 Proposition 35.6 In a Hilbert plane with (A), Aristotle's axiom holds, namely, given any acute angle, the perpendicular from a point on one arm of the angle to the other arm can be made to exceed any given segment. Proof Let the given acute angle α be at A. Construct any triangle ΔABC with RA at C. Extend AB by BD = AB to get D. Construct a perpendicular to the extension of CB to get F. Then the triangles ΔABC and ΔBFD are congruent, i.e. ΔABC = ΔBFD, by (AAS). Therefore CF = 2 BC By (A), previous theorem (35.2) and theorem from previous section (34.7) the angle FDE must be acute. By proposition from previous section (34.2), DE ≥ FC = 2 BC □ Proposition 34.2 (without proof) Let ABCD be a quadrilateral with right angles at A and B, and unequal sides AC, BD, Then the angles at C is greater than the angle at D if and only if AC < BD. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 6/7 Exercise 35.3 Given an angle aAb, show that there exists a point B in the ray Ab such that the perpendicular to b at B does not meet a. Solution: By contradiction. Assume that for all points B on b, the perpendicular meets a. Therefore, Legendre's axiom holds. Consider proposition 35.5: If Legendre's axiom and (A) hold, then (P) holds. In our case Legendre's axiom and (A) hold. Then (P) holds, which is a contradiction to the hypothesis that (P) does not hold. Exercise 35.4 Show that for any angle α, however small, there exists a line l entirely contained in the inside of the angle. Hint: Apply Exercise 35.3 to the angle bisector of α. Solution (Sketch): Taking the picture of 35.3 and its mirror image and combining them, we get something like the following picture. This is already a line l entirely contained in the inside of the angle. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli 7/7 Appendix: Used propositions from Euclids Elements (without proofs) (I.15) For let the two straight-lines AB and CD cut one another at the point E. I say that angle AEC is equal to (angle) DEB, and (angle) CEB to (angle) AED. (I.16) Let ABC be a triangle, and let one of its sides BC have been proceeded to D. I say that the external angle ACD is greater than each of the internal and opposite angles, CBA and BAC. (I.17) Let ABC be a triangle. I say that (the sum of) two angles of triangle ABC taken together in any (possible way) is less than two right-angles. (I.19) Let ABC be a triangle having the angle ABC greater than BCA. I say that side AC is also greater than side AB. (I.27) For let the straight-line EF, falling across the two straightlines AB and CD, make the alternate angles and EFD equal to one another. I say that AB and CD are parallel. 23.04.15 Seminar Euclidean Geometry Pierre Haeberli