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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 (*)
□
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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.
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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.
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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.
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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.
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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.
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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