Side-Angle-Side, Side-Angle-Angle, and the Non-Continuous Absolute Plane

Abstract

We prove that if we start with a non-continuous absolute plane, remove Side-Angle-Side as an axiom, and replace it with the three new axioms, Side-Angle-Angle, angle addition, and the existence of angle bisectors, then the result is also an absolute plane.

1. Introduction

Plane geometry is a classical subject that dates back to the time of the ancient Greeks. One of the first known texts on the subject is Euclid’s Elements. In Elements, Euclid developed an axiom system for plane geometry [1]. One of the axioms that Euclid assumed in his axiomatic development was the famous Euclidean Parallel Postulate. Over the centuries, many mathematicians have worked on the question of whether one could prove the Euclidean Parallel Postulate as a theorem of the other axioms of plane geometry. It was finally shown separately by Bolyai, Lobachevsky, and Gauss that the Euclidean Parallel Postulate is independent of the other axioms of plane geometry. While showing that the Euclidean Parallel Postulate cannot be proven as a theorem, Bolyai developed the notion of absolute geometry [2,3,4]. There has been much interest in the foundations of geometry, with many interesting new results during the past several years (see [1,5,6,7,8,9,10,11]).

Absolute geometry is plane geometry in which we assume no parallel postulate. Absolute geometry can be thought of as being a common ground between Euclidean geometry and hyperbolic geometry, and the axioms of absolute geometry are satisfied by both Euclidean and hyperbolic geometry [2,3,4]. For more on absolute geometry, see [5,6,12,13,14].

An axiom system for absolute geometry was given by David Hilbert in Grundlagen der Geometrie [15]. Hilbert first developed plane geometry in a non-continuous, abstract way, and only toward the end introduced the concept of continuity using the axioms of continuity. This approach to developing an absolute plane was also used by M. J. Greenberg in Euclidean and Non-Euclidean Geometries: Development and History, and by R. Hartshorne in Geometry: Euclid and Beyond [2,3]. Once continuity is introduced using the axioms of continuity, one can define the notions of distance and angle measure (see pages 122–124 in [2]), at which point we get a continuous absolute plane.

In these developments, the Side-Angle-Side criterion for the congruence of triangles is assumed as an axiom. After assuming Side-Angle-Side, one can prove that the Angle-Side-Angle, Side-Angle-Angle, and Side-Side-Side criteria for the congruence of triangles also hold [2,3,4]. Although it is not strictly a criterion for the congruence of triangles, one can also prove various statements in continuous absolute geometry involving Side-Side-Angle [16]. Triangle congruence has played an important role in many aspects of plane geometry, not only in a foundational role (as in this paper), but in deriving many facts about geometry, and in a multitude of applications (see [14,17]).

It is well known that if one starts with an absolute plane, removes Side-Angle-Side, and replaces it with Angle-Side-Angle as a new axiom, then the result is also an absolute plane. In particular, one can prove that Side-Angle-Side still holds as a theorem in the new axiom system [18].

In [16,18,19,20], the author showed that if one starts with a continuous absolute plane, removes Side-Angle-Side, and then replaces it with exactly one of Side-Angle-Angle, Side-Side-Side, or a statement involving Side-Side-Angle, then in each of these cases, the result is a continuous absolute plane. Again, one can show that Side-Angle-Side still holds.

In [21], a model was constructed showing that if we use the approach of Hilbert to define a non-continuous absolute plane (using the axioms given in [3]) and replace Side-Angle-Side with Angle-Angle-Angle, then, in general, we cannot prove Side-Angle-Side as a theorem.

In this paper, we show that if we start with a non-continuous absolute plane (using the axioms given in [3]), remove Side-Angle-Side as an axiom, and replace it with the three new axioms, Side-Angle-Angle, angle addition, and the existence of angle bisectors, then the result is also an absolute plane. In particular, without using any assumptions about continuity, we prove that Side-Angle-Side holds as a theorem in the new axiom system. In the final section of this paper, we further investigate the assumption about the existence of angle bisectors. The approach that we use to prove Side-Angle-Side is similar to the approach used in [18]. However, since we do not use continuity in any of our arguments, we first prove several preliminary results in Section 3 and Section 4, which are necessary to complete these arguments. These preliminary results allow us to avoid the need for continuity throughout the paper.

The models of the axiom system used in this paper are often referred to as Hilbert planes. An algebraic characterization of Hilbert planes is given by Pejas in [13].

The author would like to express a very belated thank you to Matthew Brin, Robin Hartshorne, and Patricia McCauley for their many helpful comments, without which this paper could not have been written.

2. The Axioms

In this section, we state the axioms of absolute geometry as given in [3], with the exception of Side-Angle-Side. We note that, as in [3], betweenness and congruence for segments and angles are taken here to be undefined notions.

• The Incidence Axioms:

(I1)

Given any two distinct points A and B, there exists a unique line l containing A and B.

(I2)

Every line contains at least two distinct points.

(I3)

There exist three distinct noncollinear points.

• The Betweenness Axioms:

(B1)

If B is between A and C (written $A-B-C$), then A, B, and C are three distinct collinear points, and also $C-B-A$.

(B2)

Given any two distinct points A and B, there exists a point C such that $A-B-C$.

(B3)

Given three distinct points on a line, one and only one of them is between the other two.

(B4)

(PASCH) Let A, B, and C be three distinct noncollinear points, and let l be a line not containing any of A, B, or C. If l contains a point D such that $A-D-B$, then l also contains a point lying between A and C, or else a point lying between B and C, but not both.

• It is shown in [2,3,4] that PASCH is equivalent to the Plane Separation Postulate:

• Let $\mathcal{P}$ denote the set of all points. For each line l, there exist two nonempty convex sets ${\mathcal{H}}_1$ and ${\mathcal{H}}_2$ such that the following statements hold:

(1)

$\mathcal{P}\setminus l$ = ${\mathcal{H}}_1\cup {\mathcal{H}}_2$.

(2)

If $P\in {\mathcal{H}}_1$, $Q\in {\mathcal{H}}_2$, and $P\ne Q$, then $\overline{PQ}\cap l\ne ⌀$.

The sets ${\mathcal{H}}_1$ and ${\mathcal{H}}_2$ are called half-planes (or $sides$) of l. If $A,B\in {\mathcal{H}}_i$ (i = 1,2), then we say that A and B are on the same side of l. If $A\in {\mathcal{H}}_1$ and $B\in {\mathcal{H}}_2$, then we say that A and B are on opposite sides of l. Also, l is called an edge of each of the half-planes $H_1$ and $H_2$. It can be proven that $H_1$ and $H_2$ are disjoint. When referring to the Plane Separation Postulate, we abbreviate it as $PSP$.

Given two distinct points A and B, we define the line segment $\overline{AB}$ to be the set consisting of A and B, together with all points P such that $A-P-B$. Given two line segments $\overline{AB}$ and $\overline{CD}$, we use the notation $\overline{AB}\cong \overline{CD}$ to denote when $\overline{AB}$ and $\overline{CD}$ are congruent to each other.

Given a segment $\overline{AB}$, we define the interior of $\overline{AB}$ to be the set of all points C such that $A-C-B$. We denote the interior of $\overline{AB}$ by $int\left(\overline{AB}\right)$.

Given two distinct points A and B, we define the ray $\overrightarrow{AB}$ to be the set consisting of A and B, together with all points P such that exactly one of the following is true: $A-P-B$ or $A-B-P$. We refer to the point A as the vertex of ray $\overrightarrow{AB}$.

Given three distinct noncollinear points A, V, and B, we define $\angle AVB$ = $\overrightarrow{VA}$ ∪ $\overrightarrow{VB}$. We call $\angle AVB$ an angle with vertex V and sides $\overrightarrow{VA}$ and $\overrightarrow{VB}$. We note that since A, V, and B are not collinear, we do not consider “zero angles” or “straight angles” in this paper. Given two angles $\angle ABC$ and $\angle DEF$, we use the notation $\angle ABC\cong \angle DEF$ to denote when $\angle ABC$ and $\angle DEF$ are congruent to each other.

Given an angle $\angle ABC$, we define the interior of $\angle ABC$ to be the intersection of the half-plane of line $\overleftrightarrow{AB}$ containing C, together with the half-plane of line $\overleftrightarrow{AC}$ containing B. We denote the interior of $\angle ABC$ by $int(\angle ABC)$. If $G\in int(\angle ABC)$, we say that ray $\overrightarrow{BG}$ is in the interior of $\angle ABC$. (We note that this is a slight abuse of terminology since the vertex point B is not in the interior of $\angle ABC$, but that all other points on ray $\overrightarrow{BG}$ are in the interior of $\angle ABC$).

Given three noncollinear points A, B, and C, we define $▵ABC$ = $\overline{AB}$ ∪ $\overline{BC}$ ∪ $\overline{CA}$. We call $▵ABC$ the triangle with vertices A, B, and C. We say that two triangles $▵ABC$ and $▵DEF$ are congruent, and we write $▵ABC$ ≅ $▵DEF$ if $\overline{AB}$ ≅ $\overline{DE}$, $\overline{BC}$ ≅ $\overline{EF}$, $\overline{CA}$ ≅ $\overline{FD}$, $\angle ABC$ ≅ $\angle DEF$, $\angle BCA$ ≅ $\angle EFD$, and $\angle CAB$ ≅ $\angle FDE$.

Note that when using the notation $▵ABC$ ≅ $▵DEF$, the order of the vertices for triangles $▵ABC$ and $▵DEF$ is important. In other words, triangles $▵ABC$ and $▵DEF$ are congruent if there is a one-to-one correspondence between their vertices such that corresponding sides are congruent and the corresponding angles are congruent (see [2], page 83).

Next, we state the congruence axioms.

• The Congruence Axioms:

(C1)

Given a line segment $\overline{AB}$, and given a ray r originating at a point C, then there exists a unique point D on the ray r such that $\overline{AB}\cong \overline{CD}$.

(C2)

If $\overline{AB}\cong \overline{CD}$ and $\overline{AB}\cong \overline{EF}$, then $\overline{CD}\cong \overline{EF}$. Every line segment is congruent to itself.

(C3)

(Addition of Segments) Given three points A, B, and C satisfying $A-B-C$, and three other points D, E, and F satisfying $D-E-F$, if $\overline{AB}\cong \overline{DE}$ and $\overline{BC}\cong \overline{EF}$, then $\overline{AC}\cong \overline{DF}$.

(C4)

Given an angle $\angle BAC$ and given a ray $\overrightarrow{DF}$, there exists a unique ray $\overrightarrow{DE}$ on a given side of line $\overleftrightarrow{DF}$ such that $\angle BAC\cong \angle EDF$.

(C5)

Given any three angles $\alpha$, $\beta$, and $\gamma$, if $\alpha \cong \beta$ and $\alpha \cong \gamma$, then $\beta \cong \gamma$. Every angle is congruent to itself.

We do not include the sixth congruence axiom Side-Angle-Side given in [3] in our list of axioms. Instead, we assume the three new axioms listed below and prove Side-Angle-Side as a theorem in our system.

• New Axioms:

(N1)

(Side-Angle-Angle) Given triangles $▵ABC$ and $▵DEF$, if $\overline{AB}\cong \overline{DE}$, $\angle ABC\cong \angle DEF$, and $\angle BCA\cong \angle EFD$, then $▵ABC$ ≅ $▵DEF$. When referring to Side-Angle-Angle, we abbreviate it as $SAA$.

(N2)

(Addition of Angles) Assume that $\angle BAC$ is an angle and that ray $\overrightarrow{AD}$ is in the interior of angle $\angle BAC$. Suppose that $\angle D^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle DAC$, $\angle B^{\prime }{A}^{\prime }{D}^{\prime }\cong \angle BAD$, and that rays $\overrightarrow{A^{\prime }{B}^{\prime }}$ and $\overrightarrow{A^{\prime }{C}^{\prime }}$ are on opposite sides of line $\overleftrightarrow{A^{\prime }{D}^{\prime }}$. Then, rays $\overrightarrow{A^{\prime }{B}^{\prime }}$ and $\overrightarrow{A^{\prime }{C}^{\prime }}$ form an angle $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$ such that $\angle B^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle BAC$ and such that ray $\overrightarrow{A^{\prime }{D}^{\prime }}$ is in the interior of $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$.

(N3)

(Existence of Angle Bisectors) Given an angle $\angle BAC$, there exists a ray $\overrightarrow{AD}$ in the interior of angle $\angle BAC$ such that $\angle BAD\cong \angle CAD$.

As a consequence of Axiom (C4) and angle addition, we have that the following is true: given angles $\angle BAC$ and $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$, if ray $\overrightarrow{AD}$ is in the interior of angle $\angle BAC$, ray $\overrightarrow{A^{\prime }{D}^{\prime }}$ is in the interior of $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$, $\angle D^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle DAC$, and $\angle B^{\prime }{A}^{\prime }{D}^{\prime }\cong \angle BAD$, then $\angle B^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle BAC$.

We also have the following lemma on angle subtraction.

Lemma 1.

(Angle Subtraction) Given angles $\angle BAC$ and $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$, if ray $\overrightarrow{AD}$ is in the interior of angle $\angle BAC$, ray $\overrightarrow{A^{\prime }{D}^{\prime }}$ is in the interior of $\angle B^{\prime }{A}^{\prime }{C}^{\prime }$, $\angle B^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle BAC$, and $\angle B^{\prime }{A}^{\prime }{D}^{\prime }\cong \angle BAD$, then $\angle D^{\prime }{A}^{\prime }{C}^{\prime }\cong \angle DAC$.

It can be proven that the angle bisector of any angle is unique.

We now state Side-Angle-Side for the sake of completeness and for reference.

(Side-Angle-Side) Given triangles $▵BAC$ and $▵EDF$, if $\overline{AB}\cong \overline{DE}$, $\angle BAC\cong \angle EDF$, and $\overline{AC}\cong \overline{DF}$, then $▵BAC\cong ▵EDF$.

When referring to Side-Angle-Side, we abbreviate it as $SAS$.

We define a Hilbert plane to be a given set $\mathcal{P}$, together with certain subsets l of $\mathcal{P}$, which have undefined notions of betweenness and congruence for line segments, as well as the undefined notion of congruence for angles, such that Axioms (I1)–(I3), (B1)–(B4), and (C1)–(C6) are satisfied. The elements of $\mathcal{P}$ are called points, and the subsets l of $\mathcal{P}$ are called lines (see [3], pp. 96–102).

An algebraic characterization of Hilbert planes is given by Pejas in [13] (also see [3], pages 423–427).

We show in this paper that even though we are not assuming (C6) as an axiom, but instead assuming Axioms (N1)–(N3), we still end up with a Hilbert plane (after proving $SAS$ as a theorem).

3. Basic Results

In this section, we prove some basic results that we will use throughout the rest of this paper. Several of the lemmas given in this paper are standard and well-known in the literature on the subject of absolute geometry. We include the proofs of many of these lemmas here, not only for the sake of completeness, but also to show that the results of this paper do not depend on $SAS$ or any assumption of continuity. However, many of the lemmas in this paper require new proofs in the context of the current study. The new proofs of these lemmas do not depend on $SAS$, but instead depend only on the axioms given in Section 2. The following is a consequence of Axiom (B4) (PASCH) and is proven in [2,3,4].

Lemma 2.

(Crossbar) Given angle $\angle ABC$, assume that G is a point in the interior of $\angle ABC$. Then, ray $\overrightarrow{BG}$ intersects segment $\overline{AC}$ at a point K such that $B\ne K$ and $A-K-C$.

Given angles $\angle ABC$ and $\angle DEF$, we say that angle $\angle DEF$ is less than $\angle ABC$ if there exists a ray $\overrightarrow{BG}$ in the interior of $\angle ABC$ such that $\angle ABG\cong \angle DEF$. We denote this by $\angle DEF<\angle ABC$ (or, equivalently, by $\angle ABC>\angle DEF$).

The following lemmas do not depend on Side-Angle-Side. We include the proofs here for the sake of completeness.

Lemma 3.

Let A and D be points on the same side of line $\overleftrightarrow{VB}$ such that $\angle AVB<\angle DVB$. Let $\angle FWG$ be an angle such that $\angle FWG\cong \angle DVB$. Then $\angle AVB<\angle FWG$.

Proof. 

Since $\angle AVB<\angle DVB$, it follows from the uniqueness of Axiom (C4) that $\overrightarrow{VA}$ is in the interior of $\angle DVB$. Suppose that $\angle AVB>\angle FWG$. In this case, there exists a unique ray $\overrightarrow{VK}$ in the interior of $\angle AVB$ such that $\angle FWG\cong \angle KVB$. Since $\overrightarrow{VA}$ is in the interior of $\angle DVB$, the points B and D are on opposite sides of $\overrightarrow{VA}$. In particular, $\overrightarrow{VD}$ is not in the interior of $\angle AVB$, which implies that $\overrightarrow{VD}\ne \overrightarrow{VK}$. Thus, we have distinct rays $\overrightarrow{VD}$ and $\overrightarrow{VK}$ on the same side of $\overleftrightarrow{VB}$ such that $\angle KVB\cong \angle DVB$, which is a contradiction.

A similar argument shows that it cannot be the case that $\angle AVB\cong \angle FWG$. □

Lemma 4.

Let A and D be points on the same side of line $\overleftrightarrow{VB}$ such that $\angle AVB<\angle DVB$. Let K and F be points on the same side of line $\overleftrightarrow{WG}$ such that $\angle KWG\cong \angle AVB$ and $\angle FWG\cong \angle DVB$. Then $\angle KWG<\angle FWG$.

Proof. 

Since $\angle AVB<\angle DVB$ and $\angle DVB\cong \angle FWG$, it follows from Lemma 3 that there exists a unique ray $\overrightarrow{WT}$ in the interior of $\angle FWG$ such that $\angle TWG\cong \angle AVB$. Suppose that $\angle KWG>\angle FWG$. Then it follows that the ray $\overrightarrow{WK}$ is not in the interior of $\angle FWG$, which implies that $\overrightarrow{WK}\ne \overrightarrow{WT}$. Thus, we have distinct rays $\overrightarrow{WK}$ and $\overrightarrow{WT}$ on the same side of $\overleftrightarrow{WG}$ such that $\angle KWG\cong \angle TWG$, a contradiction.

A similar argument shows that it cannot be the case that $\angle KWG\cong \angle FWG$. □

Lemma 5.

Let A and D be points on the same side of line $\overleftrightarrow{VB}$ such that $\angle AVB<\angle DVB$. Let $\angle KWG$ be such that $\angle KWG\cong \angle AVB$. Then $\angle KWG<\angle DVB$.

Proof. 

Since $\angle AVB<\angle DVB$, it follows from the uniqueness of Axiom (C4) that $\overrightarrow{VA}$ is in the interior of $\angle DVB$. Suppose that $\angle KWG>\angle DVB$. In this case, there exists a unique ray $\overrightarrow{WT}$ in the interior of $\angle KWG$ such that $\angle TWG\cong \angle DVB$. Since $\angle TWG\cong \angle DVB$ and $\angle AVB<\angle DVB$, it follows from Lemma 3 that $\angle AVB<\angle TWG$. Thus, there exists a unique ray $\overrightarrow{WF}$ in the interior of $\angle TWG$ such that $\angle FWG\cong \angle AVB$. Therefore, we have distinct rays $\overrightarrow{WF}$ and $\overrightarrow{WK}$ on the same side of $\overleftrightarrow{WG}$ such that $\angle KWG\cong \angle FWG$, which is a contradiction.

A similar argument shows that it cannot be the case that $\angle KWG\cong \angle DVB$. □

Lemma 6.

Assume that $\angle ABC\cong \angle KWH$, $\angle DEF\cong \angle QVG$, and $\angle ABC<\angle DEF$. Then $\angle KWH<\angle QVG$.

Proof. 

Since $\angle ABC<\angle DEF$, there exists a unique ray $\overrightarrow{EJ}$ in the interior of $\angle DEF$ such that $\angle DEJ\cong \angle ABC$. Since $\angle DEJ<\angle DEF$ and $\angle DEF\cong \angle QVG$, it follows from Lemma 3 that $\angle DEJ<\angle QVG$. Thus, there exists a unique ray $\overrightarrow{VT}$ in the interior of $\angle QVG$ such that $\angle TVG\cong \angle DEJ$. From Axiom (C5), we have that $\angle TVG\cong \angle DEJ\cong \angle ABC\cong \angle KWH$. Since $\angle KWH\cong \angle TVG$ and $\angle TVG<\angle QVG$, it follows from Lemma 5 that $\angle HWK<\angle QVG$. □

Lemma 7.

(Trichotomy of Angles) Given angles $\angle ABC$ and $\angle DEF$, exactly one of the following is true:

(i) 

$\angle ABC\cong \angle DEF$;

(ii) 

$\angle ABC<\angle DEF$;

(iii) 

$\angle DEF<\angle ABC$.

Proof. 

Let $\overrightarrow{EK}$ be the unique ray on the same side of line $\overleftrightarrow{DE}$ as F such that $\angle ABC\cong \angle DEK$. Then exactly one of the following is true: $\overrightarrow{EK}$ = $\overrightarrow{EF}$, K and D are on the same side of line $\overleftrightarrow{EF}$, or K and D are on opposite sides of line $\overleftrightarrow{EF}$ (in which case F and D are on the same side of line $\overleftrightarrow{EK}$). In the first case, we have that $\angle ABC\cong \angle DEF$; in the second case, we have that $\angle ABC<\angle DEF$; and in the third case, we have that $\angle DEF<\angle ABC$. □

Let A, B, C, D, and E denote five distinct noncollinear points such that $A-B-C$ and $D-B-E$. We say that angles $\angle ABD$ and $\angle DBC$ are a linear pair of angles. We also refer to angles $\angle ABD$ and $\angle DBC$ as supplementary angles. We call angles $\angle ABD$ and $\angle CBE$ vertical angles. If $\angle ABD\cong \angle DBC$, then we say that angles $\angle ABD$ and $\angle DBC$ are right angles. We denote this by $\overleftrightarrow{AB}$⊥$\overleftrightarrow{BD}$.

Given segments $\overline{AB}$ and $\overline{CD}$, we say that $\overline{AB}$ is less than $\overline{CD}$ if there exists a point K such that $C-K-D$ and $\overline{CK}\cong \overline{AB}$. We denote this by $\overline{AB}<\overline{CD}$ (or, equivalently, $\overline{CD}>\overline{AB}$).

Let $\overline{AB}$, $\overline{CD}$, and the point E be such that $A-B-E$ and $\overline{BE}\cong \overline{CD}$. Then we say that $\overline{AE}$ is the sum of segments $\overline{AB}$ and $\overline{CD}$, and we write $\overline{AE}$ ≅ $\overline{AB}$ + $\overline{CD}$. We say that a segment $\overline{TQ}$ is less than $\overline{AB}$ + $\overline{CD}$ if there exists a point K such that $A-K-E$ and $\overline{AK}\cong \overline{TQ}$. That is, $\overline{TQ}$ is less than $\overline{AB}$ + $\overline{CD}$ precisely when $\overline{TQ}$ is less than $\overline{AE}$. We denote this by $\overline{TQ}<\overline{AB}+\overline{CD}$ (or, equivalently, $\overline{AB}+\overline{CD}>\overline{TQ}$).

The following is proven in [2,3].

Lemma 8.

(Properties of Segment Inequality)

(i) 

Let segments $\overline{AB}$, $\overline{A^{\prime }{B}^{\prime }}$, $\overline{CD}$, and $\overline{C^{\prime }{D}^{\prime }}$ be such that $\overline{AB}\cong \overline{A^{\prime }{B}^{\prime }}$ and $\overline{CD}\cong \overline{C^{\prime }{D}^{\prime }}$. Then $\overline{AB}<\overline{CD}$ if and only if $\overline{A^{\prime }{B}^{\prime }}<\overline{C^{\prime }{D}^{\prime }}$.

(ii) 

If $\overline{AB}<\overline{CD}$ and $\overline{CD}<\overline{EF}$, then $\overline{AB}<\overline{EF}$.

(iii) 

Given two line segments $\overline{AB}$ and $\overline{CD}$, exactly one of the following three conditions is true: $\overline{AB}<\overline{CD}$, $\overline{AB}\cong \overline{CD}$, or $\overline{CD}<\overline{AB}$.

Given segment $\overline{AC}$, we say that a point M is the midpoint of $\overline{AC}$ if $A-M-C$ and $\overline{AM}$ ≅ $\overline{CM}$.

Lemma 9.

(Existence of Midpoints) Given segment $\overline{AC}$, then there exists a unique point M such that M is the midpoint of $\overline{AC}$.

Proof. 

Let B be a point not on line $\overleftrightarrow{AC}$. First, assume that $\angle BAC\cong \angle BCA$. Let $\overrightarrow{BD}$ denote the angle bisector of $\angle ABC$. By Crossbar, there exists a unique point of intersection M of ray $\overrightarrow{BD}$ and $\overline{AC}$ such that $A-M-C$. Applying $SAA$ to triangles $▵MBA$ and $▵MBC$, we have that $▵MBA\cong ▵MBC$. Thus, $\overline{AM}\cong \overline{CM}$.

Next, assume that $\angle BAC \ncong \angle BCA$. We may assume, without loss of generality, that $\angle BCA<\angle BAC$. Thus, there exists a unique ray $\overrightarrow{AT}$ in the interior of $\angle BAC$ such that $\angle TAC\cong \angle BCA$. By Crossbar, there exists a unique point of intersection P of ray $\overrightarrow{AT}$ and $\overline{BC}$ such that $B-P-C$. Let $\overrightarrow{PK}$ denote the angle bisector of $\angle APC$. By Crossbar, there exists a unique point of intersection N of ray $\overrightarrow{PK}$ and $\overline{AC}$ such that $A-N-C$. Applying $SAA$ to triangles $▵NPA$ and $▵NPC$, we have that $▵NPA\cong ▵NPC$. Thus, $\overline{AN}\cong \overline{CN}$ (see Figure 1).

Thus, in either case, we see that there exists a midpoint of segment $\overline{AC}$.

Suppose that there exist distinct midpoints $M_1$ and $M_2$ of segment $\overline{AC}$. We may assume, without loss of generality, that $A-M_1-M_2-C$. Thus, we see that $\overline{A{M}_1}<\overline{A{M}_2}\cong \overline{C{M}_2}<\overline{C{M}_1}\cong \overline{A{M}_1}$, which is a contradiction. Hence, the midpoint of $\overline{AC}$ is unique. □

4. Properties of Angles and Triangles

Lemma 10.

(Supplements of Congruent Angles) Given angle $\angle ABC$, let point P be such that $P-B-C$, so that $\angle ABC$ and $\angle ABP$ are supplementary angles. Similarly, given angle $\angle DEF$, let point Q be such that $Q-E-F$, so that $\angle DEF$ and $\angle DEQ$ are supplementary angles. If $\angle ABC\cong \angle DEF$, then $\angle ABP\cong \angle DEQ$. That is, supplementary angles of congruent angles are congruent.

Proof. 

Suppose that $\angle ABP \ncong \angle DEQ$. We may assume, without loss of generality, that $\angle DEQ<\angle ABP$. Thus, there exists a unique ray $\overrightarrow{BG}$ in the interior of angle $\angle ABP$ such that $\angle ABG\cong \angle DEQ$. Thus, we have angle $\angle CBG$, where point G is on the same side of line $\overleftrightarrow{BC}$ as A. By Axiom (C4), there exists a unique ray $\overrightarrow{EK}$ on the same side of line $\overleftrightarrow{EF}$ as the point D such that $\angle CBG\cong \angle FEK$. (Note that since $Q-E-F$, the points Q, E, and F are collinear, and therefore do not form an angle). Since ray $\overrightarrow{BG}$ is in the interior of angle $\angle ABP$, then ray $\overrightarrow{BA}$ is in the interior of angle $\angle GBC$. This implies that $\angle ABC<\angle GBC$. Since $\angle CBG\cong \angle FEK$, $\angle ABC\cong \angle DEF$, and $\angle ABC<\angle GBC$, it follows that $\angle DEF<\angle KEF$. Therefore, we have that ray $\overrightarrow{ED}$ is in the interior of $\angle KEF$. Since $\angle CBG\cong \angle FEK$, $\angle ABC\cong \angle DEF$, $\overrightarrow{BA}$ is in the interior of angle $\angle GBC$, and $\overrightarrow{ED}$ is in the interior of angle $\angle KEF$, it follows from angle subtraction that $\angle ABG\cong \angle DEK$. Thus, by Axiom (C5), we have that rays $\overrightarrow{EK}$ and $\overrightarrow{EQ}$ are two distinct rays on the same side of line $\overleftrightarrow{ED}$ such that $\angle DEK\cong \angle ABG\cong \angle DEQ$, which contradicts Axiom (C4). Hence, $\angle ABP\cong \angle DEQ$. □

The proof of the following lemma does not require the use of continuity or Side-Angle-Side, and is given in [18].

Lemma 11.

(Vertical Angles) Let A, B, C, D, and E be five distinct noncollinear points such that $A-E-B$ and $D-E-C$. In particular, $\angle AEC$ and $\angle DEB$ are vertical angles. Then $\angle AEC\cong \angle DEB$. That is, vertical angles are congruent.

The following lemma follows from the existence of angle bisectors, together with Axioms (N2) and (C4).

Lemma 12.

Let angles $\angle ABC$ and $\angle DEF$ be such that $\angle ABC\cong \angle DEF$. Let rays $\overrightarrow{BG}$ and $\overrightarrow{EK}$ denote the angle bisectors of angles $\angle ABC$ and $\angle DEF$, respectively. Then $\angle ABG\cong \angle DEK$.

Proof. 

Since $\overrightarrow{BG}$ is the angle bisector of angle $\angle ABC$, it follows that $\angle ABG\cong \angle CBG$. Similarly, since $\overrightarrow{EK}$ is the angle bisector of angle $\angle DEF$, it follows that $\angle DEK\cong \angle FEK$. Suppose that $\angle ABG \ncong \angle DEK$. We may assume, without loss of generality, that $\angle DEK<\angle ABG$. Thus, there exists a unique ray $\overrightarrow{BT}$ in the interior of angle $\angle ABG$ such that $\angle ABT\cong \angle DEK$. Since ray $\overrightarrow{BT}$ is in the interior of angle $\angle ABG$, ray $\overrightarrow{BG}$ is in the interior of angle $\angle TBC$. Thus, $\angle TBC>\angle GBC\cong \angle ABG>\angle DEK$. Therefore, there exists a unique ray $\overrightarrow{BH}$ in the interior of angle $\angle TBC$ such that $\angle TBH\cong \angle DEK$. In particular, rays $\overrightarrow{BH}$ and $\overrightarrow{BC}$ are distinct. By Axiom (N2), we have that $\angle ABH\cong \angle DEF\cong \angle ABC$, which is a contradiction. Hence, $\angle ABG\cong \angle DEK$. □

The proofs of Lemmas 13–15 below do not require the use of continuity or Side-Angle-Side. The proofs of these lemmas are similar to the proofs given in [18].

Lemma 13.

Let angles $\angle ABC$ and $\angle DEF$ be such that $\angle DEF<\angle ABC$. Then there exists a unique point $K\in int\left(\overline{AC}\right)$ such that $\angle ABK\cong \angle DEF$.

Lemma 14.

(The Exterior Angle Theorem) An exterior angle of a triangle is larger than either of its remote interior angles.

Let $\overleftrightarrow{AB}$, $\overleftrightarrow{CD}$, and $\overleftrightarrow{FG}$ be three distinct lines such that $A-F-B$ and $C-G-D$, with A and C on the same side of $\overleftrightarrow{FG}$. We call $\overleftrightarrow{FG}$ a transversal of $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$, and we say that lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ are cut by the transversal $\overleftrightarrow{FG}$. We call $\angle AFG$ and $\angle DGF$ alternate interior angles. We say that two lines l and q are parallel if they have no points in common. We denote this by the standard notation l∥q.

Lemma 15.

If two lines are cut by a transversal such that a pair of alternate interior angles are congruent, then the two lines are parallel.

The following lemma will be used in the proof of the Pons Asinorum below. It also provides an alternate proof for the existence of a midpoint in any segment, without the use of angle bisectors.

Lemma 16.

Given $\overleftrightarrow{AB}$, let C and D be points on opposite sides of $\overleftrightarrow{AB}$ such that $\angle ABC \cong \angle BAD$. Let G = $\overleftrightarrow{AB}$ ∩ $\overline{CD}$. Then $A-G-B$. Furthermore, if $\overline{AD}$ ≅ $\overline{BC}$, then $▵DAG$ ≅ $▵CBG$ and G is the midpoint of both $\overline{AB}$ and $\overline{CD}$.

Proof. 

Since C and D are on opposite sides of $\overleftrightarrow{AB}$, it follows that $\overline{CD}$ intersects $\overleftrightarrow{AB}$ at some point G. Suppose that G = A. In this case, we have exterior angle $\angle BAD$ of $▵ABC$, with remote interior angle $\angle ABC$, such that $\angle ABC$≅$\angle BAD$, which is a contradiction. Thus, A ≠ G. A similar argument shows that B ≠ G. Thus, we have exactly one of $A-B-G$, $A-G-B$, or $G-A-B$.

Suppose that $A-B-G$. Since $B\in int\left(\overline{AG}\right)$, it follows by PASCH that $\overleftrightarrow{CB}$ intersects another side of $▵AGD$ at some point N. Since $\angle ABC$ ≅ $\angle BAD$, it follows from Lemma 15 that $\overleftrightarrow{CB}$ and $\overleftrightarrow{AD}$ are parallel. Thus, $N\in int\left(\overline{GD}\right)$. Since $C-G-D$, it follows that C ≠ N. But then we have that $\overleftrightarrow{BC}$ intersects $\overleftrightarrow{GD}$ at both C and N, with C ≠ N. Thus, $\overleftrightarrow{BC}$ = $\overleftrightarrow{GD}$. This implies that D is a point on $\overleftrightarrow{BC}$, which is a contradiction since $\overleftrightarrow{BC}$ and $\overleftrightarrow{AD}$ are parallel. Hence, it cannot be the case that $A-B-G$. A similar argument shows that we cannot have $G-A-B$. Therefore, it must be the case that $A-G-B$.

Assume that $\overline{AD}$ ≅ $\overline{BC}$. Since $A-G-B$ and $C-G-D$, it follows that $\angle AGD$ and $\angle CGB$ are vertical angles. Thus, $\angle AGD$ ≅ $\angle CGB$. Therefore, it follows from $SAA$ that $▵DAG$ ≅ $▵CBG$. This implies that $\overline{AG}$ ≅ $\overline{BG}$ and $\overline{CG}$ ≅ $\overline{DG}$. Since $A-G-B$ and $\overline{AG}$ ≅ $\overline{BG}$, it follows that G is the midpoint of $\overline{AB}$. Likewise, since $C-G-D$ and $\overline{CG}$ ≅ $\overline{DG}$, it follows that G is the midpoint of $\overline{CD}$. □

The proofs of Lemmas 17 and 18 below do not require the use of continuity or Side-Angle-Side. The proofs of these lemmas are similar to the proofs given in [18] (see Figure 2).

Lemma 17.

Given $▵ABC$, if $\angle BAC\cong \angle BCA$, then $\overline{BA}\cong \overline{BC}$.

Lemma 18.

(Pons Asinorum) Given $▵ABC$, if $\overline{BA}\cong \overline{BC}$, then $\angle BAC\cong \angle BCA$.

Even though we have not yet proven the existence of right angles, we use the following lemma to prove the uniqueness of perpendicular lines.

Lemma 19.

Given any two right angles $\angle ABC$ and $\angle KVW$, $\angle ABC\cong \angle KVW$. That is, all right angles are congruent to each other.

Proof. 

Let D be a point such that $D-B-A$, and let X be a point such that $K-V-X$. Suppose that $\angle ABC<\angle KVW$. Then there exists a unique ray $\overrightarrow{VT}$ in the interior of $\angle KVW$ such that $\angle ABC\cong \angle KVT$. It follows from Lemma 10, together with Axiom (C5), that $\angle TVX\cong \angle DBC\cong \angle ABC$. Since $\overrightarrow{VT}$ is in the interior of $\angle KVW$, it follows that $\overrightarrow{VW}$ is in the interior of $\angle TVX$. This implies that $\angle WVX<\angle TVX$. Thus, we see that $\angle KVW\cong \angle WVX<\angle TVX\cong \angle ABC$, which is a contradiction. □

Lemma 20.

Assume that lines $\overleftrightarrow{DV}$ and $\overleftrightarrow{PK}$ intersect at a point M such that $D-M-V$ and $P-M-K$. If $\angle PMV\cong \angle KMV$, then $\angle PMV\cong \angle DMP$. That is, the supplementary angles of right angles are also right angles.

Proof. 

Since $\angle PMV\cong \angle KMV$, $\angle KMV$ and $\angle DMP$ are vertical angles, and by Lemma 11 vertical angles are congruent, it follows that $\angle KMV\cong \angle DMP$. Thus, by Axiom (C5), $\angle PMV\cong \angle KMV\cong \angle DMP.$ □

Let M denote the midpoint of segment $\overline{PQ}$. The line l passing through M that is perpendicular to line $\overleftrightarrow{PQ}$ is called the perpendicular bisector of $\overline{PQ}$. We say that a point T is equidistant from P and Q if $\overline{PT}$≅$\overline{QT}$.

Let $\mathcal{P}$ denote the set of points in the plane. Given a line l, then the reflection in l is a function ${\rho }_l:\mathcal{P}\to \mathcal{P}$ defined by ${\rho }_l\left(P\right)$ = P if $P\in l$, and ${\rho }_l\left(P\right)$ = Q, where l is the perpendicular bisector of $\overline{PQ}$, if $P\notin l$.

We can now use arguments similar to those given in [18] without the use of continuity to prove that the reflection in a line l preserves lines, betweenness, segments, rays, angles, congruence of segments, congruence of angles, fixes l pointwise, and interchanges the halfplanes of l. We use this fact in the following section to prove (without continuity) that Side-Angle-Side holds as a theorem in our axiom system.

5. A Proof of Side-Angle-Side

In this section, we prove our main theorem. In the proof of this theorem, we denote the identity function on the set of points $\mathcal{P}$ in the plane by $\iota$.

Theorem 1.

(Side-Angle-Side) Given triangles $▵BAC$ and $▵EDF$, if $\overline{AB}\cong \overline{DE}$, $\angle BAC\cong \angle EDF$, and $\overline{AC}\cong \overline{DF}$, then $▵BAC\cong ▵EDF$.

Proof. 

Let triangles $▵BAC$ and $▵EDF$ be such that $\overline{AB}\cong \overline{DE}$, $\angle BAC\cong \angle EDF$, and $\overline{AC}\cong \overline{DF}$.

If $A=D$, we let ${\sigma }_1=\iota$. If $A\ne D$, we let ${\sigma }_1={\rho }_{t_1}$, where ${\rho }_{t_1}$ is the reflection in the line $t_1$, and $t_1$ is the perpendicular bisector of segment $\overline{AD}$. In either case, we have that ${\sigma }_1\left(A\right)=D$. Let ${\sigma }_1\left(B\right)=B_1$ and ${\sigma }_1\left(C\right)=C_1$.

Since ${\sigma }_1$ preserves the congruence of segments, we have that $\overline{D{B}_1}\cong \overline{AB}\cong \overline{DE}$. If $B_1$ is on ray $\overrightarrow{DE}$, it follows that $B_1=E$. In this case, let ${\sigma }_2=\iota$. If $E-D-B_1$, then D is the midpoint of segment $\overline{E{B}_1}$. In this case, D is on the perpendicular bisector of segment $\overline{E{B}_1}$. If E, D, and $B_1$ are not collinear, it follows from the fact that $\overline{D{B}_1}\cong \overline{DE}$, that D is on the perpendicular bisector of segment $\overline{E{B}_1}$. In either of the previous two cases, we let ${\sigma }_2={\rho }_{t_2}$, where ${\rho }_{t_2}$ is the reflection in the line $t_2$, and where $t_2$ is the perpendicular bisector of segment $\overline{E{B}_1}$. In any of the above three cases, we have that $D={\sigma }_2\left(D\right)={\sigma }_2\left({\sigma }_1\left(A\right)\right)$ and $E={\sigma }_2\left(B_1\right)={\sigma }_2\left({\sigma }_1\left(B\right)\right)$. Let $C_2={\sigma }_2\left(C_1\right)={\sigma }_2\left({\sigma }_1\left(C\right)\right)$.

Since ${\sigma }_2\left({\sigma }_1\left(x\right)\right)$ preserves congruence of segments and the congruence of angles, we have that $\overline{D{C}_2} \cong \overline{D{C}_1} \cong \overline{AC} \cong \overline{DF}$, and $\angle ED{C}_2 \cong \angle B_{1}D{C}_1 \cong \angle BAC \cong \angle EDF$.

If $C_2$ and F are on the same side of $\overleftrightarrow{DE}$, it follows that $C_2=F$. In this case, let ${\sigma }_3=\iota$.

Assume that $C_2$ and F are on opposite sides of $\overleftrightarrow{DE}$. Let T denote the point of intersection of segment $\overline{F{C}_2}$ and $\overleftrightarrow{DE}$. Since $\angle ED{C}_2\cong \angle EDF$, it follows (by Lemma 10, if necessary) that $\angle TD{C}_2\cong \angle TDF$. By the Pons Asinorum, we have that $\angle T{C}_{2}D\cong \angle TFD$. It follows from $SAA$ that $▵TDF\cong ▵TD{C}_2$. Thus, $\angle DTF$ is a right angle, and T is the midpoint of $\overline{F{C}_2}$. Since $\overleftrightarrow{DE}=\overleftrightarrow{DT}$, it follows that $\overleftrightarrow{DE}$ is the perpendicular bisector of $\overline{F{C}_2}$. In this case, we let ${\sigma }_3={\rho }_{t_3}$, where $t_3=\overleftrightarrow{DE}$.

We have that $D={\sigma }_3\left({\sigma }_2\left(D\right)\right)={\sigma }_3\left({\sigma }_2\left({\sigma }_1\left(A\right)\right)\right)$ and $E={\sigma }_3\left({\sigma }_2\left(B_1\right)\right)={\sigma }_3\left({\sigma }_2\left({\sigma }_1\left(B\right)\right)\right)$, and let $F={\sigma }_3\left(C_2\right)={\sigma }_3\left({\sigma }_2\left(C_1\right)\right)={\sigma }_3\left({\sigma }_2\left({\sigma }_1\left(C\right)\right)\right)$.

Since each of ${\sigma }_1$, ${\sigma }_2$, and ${\sigma }_3$ preserve the congruence of segments and the congruence of angles, we have that $\angle ABC\cong \angle DEF$, $\angle ACB\cong \angle DFE$, and $\overline{BC}\cong \overline{EF}$. Hence, $▵BAC\cong ▵EDF$. □

6. The Existence of Angle Bisectors

In this section, we further examine the assumption made in Axiom (N3) concerning the existence of angle bisectors. In particular, we first give sufficient conditions for the existence of an angle bisector for a given angle, and then present two models, $\mathbb{A}$ and $\mathbb{B}$, of non-continuous plane geometry. Both models satisfy Axioms (I1)–(I3), (B1)–(B4), (C1)–(C5), and (N2), but neither satisfies $SAS$ or $SAA$. In the model $\mathbb{A}$, there exist angles that do not have an angle bisector. This shows that we cannot prove Axiom (N3) using only Axioms (I1)–(I3), (B1)–(B4), (C1)–(C5), and (N2). In the model $\mathbb{B}$, every angle has an angle bisector. This shows that when assuming Axiom (N3), we are not tacitly assuming $SAS$, $SAA$, or continuity. Since $SAS$ does not hold in the model $\mathbb{B}$, we are not giving a circular argument when proving $SAS$ as a theorem in the arguments above in the preceding sections. Thus, the models $\mathbb{A}$ and $\mathbb{B}$ together show that Axiom (N3) is independent of Axioms (I1)–(I3), (B1)–(B4), (C1)–(C5), and (N2). Therefore, to prove the existence of angle bisectors, we would need to use $SAS$, $SAA$, or the assumption of continuity.

For the proofs of the next two lemmas, we assume Axioms (I1)–(I3), (B1)–(B4), (C1)–(C5), (N1) and (N2).

Lemma 21.

Let $\angle ABC$ and $\angle DEF$ be such that $\angle ABC\cong \angle DEF$ and such that $\angle ABC$ has an angle bisector. Then $\angle DEF$ also has an angle bisector. That is, if an angle β has an angle bisector, every angle congruent to β also has an angle bisector.

Proof. 

Let $\angle ABC$ and $\angle DEF$ be such that $\angle ABC\cong \angle DEF$ and such that $\angle ABC$ has an angle bisector $\overrightarrow{AG}$. Since $\angle ABG<\angle ABC$ and $\angle ABC\cong \angle DEF$, there exists a unique ray $\overrightarrow{EK}$ in the interior of $\angle DEF$ such that $\angle DEK\cong \angle ABG$. Thus, by angle subtraction, we have that $\angle DEK\cong \angle ABG\cong \angle CBG\cong \angle FEK$. □

Lemma 22.

Given $\angle ABC$, if every angle $\angle DEF$ such that $\angle DEF<\angle ABC$ has an angle bisector, then $\angle ABC$ also has an angle bisector.

Proof. 

Let $\angle DEF$ be an angle such that $\angle DEF<\angle ABC$. Then there exists a ray $\overrightarrow{BQ}$ in the interior of $\angle ABC$ such that $\angle ABQ\cong \angle DEF$. Since ray $\overrightarrow{BQ}$ is in the interior of $\angle ABC$, it follows that $\angle QBC<\angle ABC$. By hypothesis, there exists a ray $\overrightarrow{BT}$ in the interior of $\angle QBC$ such that $\overrightarrow{BT}$ is the angle bisector of $\angle QBC$; in particular, $\angle QBT\cong \angle CBT$. Similarly, there exists a ray $\overrightarrow{EK}$ in the interior of $\angle DEF$ such that $\overrightarrow{EK}$ is the angle bisector of $\angle DEF$; in particular, $\angle DEK\cong \angle FEK$. Let $\overrightarrow{EM}$ be the unique ray on the opposite side of line $\overleftrightarrow{ED}$ from K such that $\angle DEM\cong \angle QBT$. Let $\overrightarrow{EN}$ be the unique ray on the opposite side of line $\overleftrightarrow{EF}$ from K such that $\angle FEN\cong \angle QBT$. Since $\angle DEK\cong \angle FEK$ and $\angle DEM\cong \angle QBT\cong \angle FEN$, it follows from angle addition that $\angle KEM\cong \angle KEN$, and also that $\angle NEM\cong \angle ABC$. Thus, ray $\overrightarrow{EK}$ is the angle bisector of $\angle NEM$. It now follows from Lemma 21 that $\angle ABC$ also has an angle bisector. □

• The Model $\mathbb{A}$:

In this section, we define the model $\mathbb{A}$.

The set of points in the model $\mathbb{A}$ is the set $\mathbb{Q}\times \mathbb{Q}$ of points in ${\mathbb{R}}^2$, both of whose coordinates are rational numbers. Since ${\mathbb{Q}}^2$ is countably infinite, we can enumerate the points $P_1,P_2,\dots ,P_j,\dots$ in ${\mathbb{Q}}^2$. To define lines in the model $\mathbb{A}$, we let $\mathcal{L}$ be the set of all lines in ${\mathbb{R}}^2$ whose equations are $y=mx+b$ or $x=k$, where $m,b,k\in \mathbb{Q}$. We define a line in $\mathbb{A}$ to be the intersection of a line $l\in \mathcal{L}$ with $\mathbb{Q}\times \mathbb{Q}$. We define the distance and angle measure in the model $\mathbb{A}$ in a non-continuous way and then use this distance and angle measure to determine when two segments or two angles are congruent. Since there is an enumeration of the points $P_1,P_2,\dots ,P_j,\dots$ in ${\mathbb{Q}}^2$, when defining the angle measure below, we treat the points in ${\mathbb{Q}}^2$ as a sequence $\left(P_n\right)$. We denote the distance in $\mathbb{A}$ between two points P and Q by $t(P,Q)$. We denote the angle measure in $\mathbb{A}$ of an angle $\angle ABC$ by $m\angle ABC$.

We define the distance in $\mathbb{A}$ using the taxicab metric t. More specifically, given two points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$, then we define $t(P_1,P_2)=|x_2-x_1|+|y_2-y_1|$. Since $x_1,x_2,y_1,y_2\in \mathbb{Q}$, it follows that $t(P_1,P_2)\in \mathbb{Q}$ as well.

It is proven in [21,22] that Axioms (I1)–(I3), (B1)–(B4), and (C1)–(C3) all hold in the model $\mathbb{A}$. It is left to the reader to check that every segment in $\mathbb{A}$ has a midpoint.

Next, we define the angle measure in $\mathbb{A}$. Given a point $P_j$ = $(a,b)$, where $a,b\in \mathbb{Q}$, let $E_j$= $(a+1,b)$, $W_j$= $(a-1,b)$, $N_j$ = $(a,b+1)$, and $S_j$ = $(a,b-1)$. If we are given a point $P_j$ = $(a,b)$ in $\mathbb{A}$, where $a,b\in \mathbb{Q}$, then the lines $x=a$ and $y=b$ divide the model $\mathbb{A}$ into four "quadrants" similar to the way that the x-axis and y-axis divide the $xy$-plane into four quadrants. We use a variation of this terminology when defining and altering the angle measure to construct the model $\mathbb{A}$. We refer to the set $\left\{\right(x,y)\in \mathbb{A}\mid x>a\mathrm{and}y>b\}$ as $P_j$-quadrant I. We refer to the set $\left\{\right(x,y)\in \mathbb{A}\mid x<a\mathrm{and}y>b\}$ as $P_j$-quadrant $II$. We refer to the set $\left\{\right(x,y)\in \mathbb{A}\mid x<a\mathrm{and}y<b\}$ as $P_j$-quadrant $III$. Finally, we refer to the set $\left\{\right(x,y)\in \mathbb{A}\mid x>a\mathrm{and}y<b\}$ as $P_j$-quadrant $IV$.

Given a point $P_j\in {\mathbb{Q}}^2$, we define ${\displaystyle m\angle E_j{P}_j{N}_j=m\angle E_j{P}_j{S}_j=m\angle W_j{P}_j{N}_j=m\angle W_j{P}_j{S}_j=\frac{\pi }{2}}$.

Let $\left(D_m\right)$ denote the subsequence of $\left(P_n\right)$ consisting of points that lie in $P_j$-quadrant I. Given a ray $\overrightarrow{P_j{D}_m}$ in $P_j$-quadrant I, we let ${\widehat{D}}_m$ denote the point on $\overrightarrow{P_j{D}_m}$ that precedes all other points on $\overrightarrow{P_j{D}_m}$ with respect to the enumeration of $\left(P_n\right)$. We note that $\overrightarrow{P_j{D}_m}$ = $\overrightarrow{P_j{\widehat{D}}_m}$. Let $T_m$ be a point in ${\mathbb{Q}}^2$ such that $T_m-P_j-{\widehat{D}}_m$. Thus, $T_m$ lies on the opposite ray of $\overrightarrow{P_j{\widehat{D}}_m}$, and $T_m$ is in $P_j$-quadrant $III$. We note that every point $P_i$ in $P_j$-quadrant $III$ is of the form $T_m$ for some point ${\widehat{D}}_m$ in $P_j$-quadrant I.

Let $K_1$ denote the first point in $\left(D_m\right)$, and let $K_2$ denote the first point from $\left(D_m\right)$ that is in the interior of angle $\angle K_1{P}_j{N}_j$. We define ${\displaystyle m\angle E_j{P}_j{K}_1=m\angle K_1{P}_j{K}_2=m\angle K_2{P}_j{N}_j=\frac{1}{3}m\angle E_j{P}_j{N}_j=\left(\frac{1}{3}\right)\left(\frac{\pi }{2}\right)}$.

Let $K_3$ denote the first point in $\left(D_m\right)$ in the interior of angle $\angle E_j{P}_j{K}_1$, and let $K_4$ denote the first point from $\left(D_m\right)$ that is in the interior of angle $\angle K_3{P}_j{K}_1$. We define ${\displaystyle m\angle E_j{P}_j{K}_3=m\angle K_3{P}_j{K}_4=m\angle K_4{P}_j{K}_1=\frac{1}{3}m\angle E_j{P}_j{K}_1=\frac{1}{9}m\angle E_j{P}_j{N}_j=\left(\frac{1}{9}\right)\left(\frac{\pi }{2}\right)}$.

Continuing with this process, let $K_5$ denote the first point in $\left(D_m\right)$ in the interior of angle $\angle E_j{P}_j{K}_3$, and let $K_6$ denote the first point from $\left(D_m\right)$ that is in the interior of angle $\angle K_5{P}_j{K}_3$. We define ${\displaystyle m\angle E_j{P}_j{K}_5=m\angle K_5{P}_j{K}_6=m\angle K_6{P}_j{K}_1=\frac{1}{3}m\angle E_j{P}_j{K}_1=\frac{1}{27}m\angle E_j{P}_j{N}_j=\left(\frac{1}{27}\right)\left(\frac{\pi }{2}\right)}$.

We continue this process indefinitely. In particular, we continue it for every ray in $P_j$-quadrant I and for all sub-angles of $\angle E_j{P}_j{N}_j$. In this way, we construct infinitely many nested sequences of sub-angles. Each sub-angle in a sequence has an angle measure exactly ${\displaystyle \frac{1}{3}}$ the angle measure of the preceding angle in the sequence. Thus, starting with the initial angle measure ${\displaystyle m\angle E_j{P}_j{N}_j=\frac{\pi }{2}}$ (and regarding of ${\displaystyle \angle E_j{P}_j{N}_j}$ as the first angle in a sequence), the angle measure of the ${\displaystyle m^{th}}$ angle in the sequence is ${\displaystyle \frac{1}{3^{m-1}}m\angle E_j{P}_j{N}_j=\left(\frac{1}{3^{m-1}}\right)\left(\frac{\pi }{2}\right)}$.

Let $P_{n_1}$ and $P_{n_2}$ be two points in $P_j$-quadrant I such that $P_j$, $P_{n_1}$, and $P_{n_2}$ are not collinear. To compute the measure of the angle ${\displaystyle \angle P_{n_1}{P}_j{P}_{n_2}}$, we first use the method just described to compute the measures of all angles of the form ${\displaystyle \angle K_t{P}_j{K}_s}$ for all points $K_{i_1},K_{i_2},\dots ,K_{i_b}$ that precede and include both $P_{n_1}$ and $P_{n_2}$ in $\left(P_n\right)$. We then compute the measure of the angle ${\displaystyle \angle P_{n_1}{P}_j{P}_{n_2}}$ by angle addition, summing the measures of those sub-angles of ${\displaystyle \angle P_{n_1}{P}_j{P}_{n_2}}$ whose measures have already been computed. Since only finitely many points precede $\left(P_n\right)$ that come before $P_{n_1}$ and $P_{n_2}$ in $\left(P_n\right)$, this is precede is finite process and well-defined.

Let $T_{m_1}$ and $T_{m_2}$ be two points in $P_j$-quadrant $III$ such that $P_j$, $T_{m_1}$, and $T_{m_2}$ are not collinear. It follows from the definitions of $T_{m_1}$ and $T_{m_2}$ that $P_j$, $D_{m_1}$, and $D_{m_2}$ are also not collinear. We define ${\displaystyle m\angle T_{m_1}{P}_j{T}_{m_2}=m\angle D_{m_1}{P}_j{D}_{m_1}}$. Since every point $P_i$ in $P_j$-quadrant $III$ is of the form $T_m$ for some point ${\widehat{D}}_m$ in $P_j$-quadrant I, and since the point ${\widehat{D}}_m$ is unique for each ray, it follows that ${\displaystyle m\angle T_{m_1}{P}_j{T}_{m_2}}$ is well defined. We define angle measures in a similar way for $P_j$-quadrant $II$ and $P_j$-quadrant $IV$.

Given two points $P_u$ and $P_v$ such that the three points $P_u$, $P_v$, and $P_j$ are distinct and non-collinear, we define ${\displaystyle m\angle P_u{P}_j{P}_v}$ using the above method together with angle addition if $P_u$ and $P_v$ lie in different $P_j$-quadrants. With these definitions, we see that that vertical angles are congruent, and angle addition holds in $\mathbb{A}$. However, both $SAS$ and $SAA$ do not hold in $\mathbb{A}$, and $\mathbb{A}$ is not a continuous model.

Some angles in $\mathbb{A}$ have angle bisectors. For example, $\angle E_j{P}_j{K}_2$ has angle bisector $\overrightarrow{P_j{K}_1}$. However, many angles do not have angle bisectors. For example, $\angle E_j{P}_j{K}_1$ does not have an angle bisector. Suppose, to the contrary, that $\angle E_j{P}_j{K}_1$ does have an angle bisector $\overrightarrow{P_{j}D}$, where D is a point in $\left(D_m\right)$. Thus, ${\displaystyle m\angle E_j{P}_{j}D=\frac{1}{2}m\angle E_j{P}_j{K}_1=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\left(m\angle E_j{P}_j{N}_j\right)=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\left(\frac{\pi }{2}\right)}$. Since D is a point in $\left(D_m\right)$, it follows that ${\displaystyle m\angle E_j{P}_{j}D}$ = ${\displaystyle \left(\sum _{i=1}^h\frac{q_i}{3^{r_i}}\right)\left(\frac{\pi }{2}\right)}$, where, for each $i=1,\dots h$, we have that $q_i$ and $r_i$ are positive integers. Thus, ${\displaystyle \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\left(\frac{\pi }{2}\right)}$ = ${\displaystyle \left(\sum _{i=1}^h\frac{q_i}{3^{r_i}}\right)\left(\frac{\pi }{2}\right)}$. Let $3^t$ denote the least common denominator of ${\displaystyle \frac{q_1}{3^{r_1}},\dots ,\frac{q_h}{3^{r_h}}}$, so that ${\displaystyle \left(\sum _{i=1}^h\frac{q_i}{3^{r_i}}\right)}$ can be written in the form ${\displaystyle \left(\sum _{i=1}^h\frac{s_i}{3^t}\right)}$, where for each $i=1,\dots h$, $s_i$ is equal to $q_i$ times a multiple of 3. Therefore, it follows that ${\displaystyle \left(\sum _{i=1}^h\frac{s_i}{3^t}\right)}$ = ${\displaystyle \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}$, which implies that ${\displaystyle 3^{t-1}=2\sum _{i=1}^h{s}_i}$, where each $s_i$ is a positive integer. However, since 2 is not a factor in the prime factorization of $3^{t-1}$, we have a contradiction.

• The Model $\mathbb{B}$:

For the model $\mathbb{B}$, we define points and lines exactly as in the model $\mathbb{A}$ above. Given a point $P_j$ = $(a,b)$, where $a,b\in \mathbb{Q}$, we again let $E_j$= $(a+1,b)$, $W_j$= $(a-1,b)$, $N_j$ = $(a,b+1)$, and $S_j$ = $(a,b-1)$. We also define the four $P_j$-quadrants as above.

We again define the distance in the model $\mathbb{B}$ (in a non-continuous way) using the taxicab metric t. We also define the angle measure in the model $\mathbb{B}$ similar to the method used in $\mathbb{A}$, namely by dividing each $P_j$ quadrant into nested sequences of sub-angles and using angle addition when necessary. However, in $\mathbb{B}$, the measure of each angle in a nested sequence is defined to be exactly ${\displaystyle \frac{1}{2}}$ of the angle measure of the preceding angle in the sequence. It follows immediately that each angle in $\mathbb{B}$ has an angle bisector. Also, we see that vertical angles are congruent, and angle addition holds in $\mathbb{B}$. However, both $SAS$ and $SAA$ do not hold in $\mathbb{B}$, and $\mathbb{B}$ is not a continuous model.