A Finite, Feasible, Quantifier-free Foundation for Constructive Geometry
AA Finite, Feasible, Quantifier-Free Foundationfor Constructive Geometry
John BurkeSeptember 18, 2020
Abstract
In this paper we will develop an axiomatic foundation for the geo-metric study of straight edge, protractor, and compass constructions,which while being related to previous foundations such as [4], [8], [18],[1], and [13], will be the first to have all axioms written and all proofsconducted in quantifier-free first order logic. All constructions withinthe system will be justified to be feasible by basic human faculties. Nostatement in the system will refer to infinitely many objects and onecan posit an interpretation of the system which is in accordance toour free, creative process of geometric constructions. We are also ableto capture analogous results to Euclid’s work on non-planar geometryin Book XI of
The Elements .This paper primarily builds on the work of Suppes in [14] and [15]and draws from Beeson’s work in [1]. By further developing Sup-pes’ work on parallel line segments, we are able to develop analogsto most theorems about parallel lines without assuming an equivalentto Euclid’s Fifth Postulate which we deem as introducing non-feasibleconstructions. In [1] Besson defines the characteristics such a geomet-ric foundation should have to called constructive. This work satisfiesthese characteristics. Additionally this work would be considered con-structive as Suppes defined it (see [12] and [13]). a r X i v : . [ m a t h . M G ] S e p ontents Pasch’s Axiom, Plane Separation, and Angle Orientation 21
11 Conclusion 90
In [14] Suppes introduced a set of quantifier-free axioms for constructive affineplanar geometry. The main features of this axiomatic system are a relationfor when points are colinear and two constructions which had interpretationsrelating to the doubling of a line segment and finding the midpoint of a linesegment. From these primitive concepts Suppes was able to define when twoline segments are parallel. The great advantage of this axiomatic system isthat from a finite set of points, to start with, one can prove theorems aboutthe possible finite configurations one can create using the basic geometrictools of halving and doubling a line segment. Furthermore, this axiomaticsystem allows one to prove theorems about the parallel relation between linesegments created from these constructions. It is compelling that from suchbasic tools one would be able to mathematically discuss concepts such asparallel line segments. Additionally, In [15] Suppes was able to give a finiteconstructive model for the axioms in which points are given rational numbercoordinates. It is natural to ask, as Van Bendegem did in his article
Finistismin Geometry [3], if this affine theory can be expanded all the way into a full-fledged geometrical theory. The objective of this work is to expand Suppes’4heory to create a finite, feasible, constructive axiomatic theory which is ananalog to Hilbert’s axiomatic system presented in his work
Grundlagen derGeometrie (tr.
The Foundations of Geometry ) [8].
Geometry is one of the oldest form of mathematics. During the reign ofPtolemy I (323283 BCE), an Alexandrian Greek name Euclid wrote a textcalled
The Elements which was one of the first well presented (and enduring)texts collecting the vast amount of geometric knowledge up to that time inhis part of the world. The most important feature of the work was that allgeometric statements were proved from a small list of assumptions. Whilealmost all of the assumptions were ‘indisputable’ truths about points, linesand angles in space and magnitudes of these objects, one of the assumptionscalled Euclid’s fifth postulate, or the parallel postulate, was not as obviouslytrue. The postulate states that if two lines are drawn which are intersected bya third line, called a transversal, in such a way that the sum of the measuresof the two inner angles on the same side of the the transversal is less than tworight angles, then the two lines inevitably must intersect each other on thatside if extended far enough (see Figure 1). Once one has made relativelyintuitive interpretations for the meaning of the geometric objects such aspoints, line and angles, one realizes that this postulate asserts a claim thatis not realistically testable or observable in all instances. One can constructexamples where the length of the line segment from a to b , see Figure 1, wouldbe larger than any magnitude one could construct or traverse. This is doneby simply making the interior angles deviate from being supplemental byextraordinarily small amounts (see Figure 2). In fact this issue was pointedout by Proclus in his Commentary on Euclid’s Elements around 700 hundredyears after Euclid lived. Proclus writes“This [fifth postulate] ought even to be struck out of the Pos-tulates altogether; for it is a theorem involving many difficultieswhich Ptolemy, in a certain book, set himself to solve, and itrequires for the demonstration of it a number of definitions aswell as theorems. And the converse of it is actually proved by5uclid himself as a theorem. It may be that some would be de-ceived and would think it proper to place even the assumption inquestion among the postulates ... . So in this case the fact that,when the right angles are lessened, the straight lines converge istrue and necessary; but the statement that, since they convergemore and more as they are produced, they will sometime meetis plausible but not necessary, in the absence of some argumentshowing that this is true in the case of straight lines. For thefact that some lines exist which approach indefinitely, but yet re-main non-secant, although it seems improbable and paradoxical,is nevertheless true and fully ascertained with regard to otherspecies of lines [for example curves like the hyperbola that hasasymptotes]. May not then the same thing be possible in thecase of straight lines that happens in the case of the lines referredto? Indeed, until the statement in the Postulate is clinched byproof, the facts shown in the case of other lines may direct ourimagination the opposite way. And, though the controversial ar-guments against the meeting of the straight lines should containmuch that is surprising, is there not all the more reason why weshould expel from our body of doctrine this merely plausible andunreasoned (hypothesis)?”It is evident that Euclid also had some amount of misgivings about thispostulate given the fact that he choose to hold off using this assumption,when it would have been useful, early on in the text. a b
Figure 1:6igure 2: Two non-parallel lines whose point of intersection will be very farto the right of the transversal.In 1890’s Hilbert developed a set of axioms for Euclidean geometry. UnlikeEuclid, Hilbert started with undefined terms such as points, lines and anglesand defined relations that these terms can satisfy between each other. Hethen listed axioms which these terms and relations must satisfy. Hilbert alsoused a more modern and less constructive axiom about parallel lines. Hisaxiom simply stated that given a line, (cid:96) , and a point, p , not on that line, thereis one and only one line, (cid:96) (cid:48) , through p that does not intersect the originalline. Using classical logic one can prove that this statement is equivalent toEuclid’s fifth postulate. It is important to note that Hilbert also added anaxiom of completeness to his set of assumptions. This axiom dealt with theinability to make extensions point-wise of any line. From this Hilbert wasable to show that the only set-theoretic model of his axiomatic system wasthe Real Cartesian plain.Later in the 1920’s Tarski developed a set of axioms for what he called elemen-tary geometry. Important characteristics of his axiomatic set was that theonly undefined term was ‘point’. All linear structure was dealt with via thecolinear relation among points. Angle congruence was supplanted with onlyspeaking about a triple of points as a triangle and using a clever 5-segmentaxiom instead of the more familiar Side-Angle-Side triangle congruence.In [1], Beeson developed constructive versions of Tarski’s axiomatic systemsome of these versions had quantifier-free axioms while others did not. Thetwo main works that this current work is influenced by are [1] by Beesonand [14] by Suppes. They both developed axioms for constructive geometry.It is important to point out here that they are using the term constructive7eometry in two very different ways. Suppes would define constructive ax-ioms for geometry as a set of quantifier-free axioms that have operationswhich have intuitive interpretations as geometric constructions such as find-ing the intersection of two non parallel lines or finding the midpoint of a linesegment. In fact, Suppes along with Moler in [12] were the first to develop-ment such a set of geometric axioms. Many others have done related workand Pambuccian summed up these works in [13]. Meanwhile Beeson woulddefine a constructive theory (which potentially could be axiomatized usingconstructions/functions instead of quantifiers) as a theory where the methodof proof using this axioms involves using only intuitionistic logic. When considering the viable models to interpret these axiomatic systems onecan create set-theoretic models in which the points are interpreted as tuplesof numbers called coordinates. In [15] Suppes defines a constructive modelin which one starts with a finite number of coordinates and then is able togive coordinates to the points which are produced by applying constructionsto elements of the original collection. This model is finite because one is onlyallowed to produce one new point at a time. Additionally, each formula in thequantifier-free language only discusses finitely many terms at a time. Some ofBeeson’s axiomatic systems are quantifier-free while others are not. Note thata geometric system containing an axiom with an existential quantifier whichasserts the existence of a point between any two points would only have settheoretic models which contain infinitely many points. Beeson in fact onlyconsiders infinite set-theoretic models for all of his axiomatic systems. It isimportant to point out that although some of Beeson’s axiomatic system arequantifier free he never stipulates that the language of the theory is quantifierfree. Thus he freely uses definitions that involve existential quantifiers.In his text
New Foundation for Physical Geometry: The Theory of LinearStructures [11], Maudlin make a distinction between metaphorical and geo-metrical spaces. In the category of metaphorical spaces Maudlin would placetopological groups such as the integers, the real number line, R and R . Inthe category of geometrical spaces Maudlin would place 2 and 3 dimensionalEuclidean space and Minkowski space as well as Riemannian geometries.This may at first seem like a nonsensical distinction to some mathematicians8nd physicists given that many would identify 2-dimensional Euclidean spacewith R , but in fact R is a set-theoretic model of the theory of Euclideanspace (using Hilbert’s axioms for instance). R has more structure than Eu-clidean space. The point (0 ,
0) called the origin has an elevated importanceamong all of the points in the space R . Given a point/element of R youcan discuss properties of it such as if one of its coordinates is negative orpositive or rational or irrational. One can perform arithmetical operationsto two given points/elements of R . The points of 2-dimensional Euclideanspace have none of this structure. All points are intrinsically identical. Thespace is homogeneous and isotropic. Most physicists would agree that welive in a three dimensional euclidean space (note the three spacial dimen-sions of spacetime are not curved, to the best of our knowledge, hence spaceis euclidean), but most would not agree that we live in the set of all orderedtriples of real numbers called R . In geometrical spaces the structure betweenpoints is intrinsic, but for metaphorical spaces the structure between pointsis emergent from other arithmetic structure between elements. Maudlin saysthe the following:“Through most of the history of mathematics, mathemati-cians would never have thought of even comparing a fundamen-tally arithmetic (and set-theoretic) object such as R with a fun-damental geometical object such as Euclidean space. So confusionbetween the two, which is now so prevalent, is relatively recentin historical vintage.”He goes on to say:“Even if it is possible to use a metaphorical space such a thespace of ordered triples of real numbers to represent a geometri-cal space, this imposes a screen of mathematical representationbetween us and the object in which we are interested.”Interestingly Hilbert also felt the need to circumvent the use of real numberscoordinates in his Grundlagen der Geometrie . He wrote that “The presentgeometrical investigation seeks to uncover which axioms, hypotheses or aidsare necessary for the proof of a fact in elementary geometry ...” Hilbertreferred to this as “purity in the methods of proof”.9he ancient Greeks made this distinction between the rigorous study of theactual physical space we live in, Geometry, and the rigorous study of num-bers, Artihmetic. They referred to geometrical magnitudes as megethos andnumbers as arithmos . There is a fascinating history dating from the ancientGreeks up to Dedekind in the nineteenth century where the supremacy ofgeometry was supplanted by arithmetic culminating in the modern conceptof a set and the foundational theory of sets. The works of
Greek Mathemati-cal Thought and the Origins of Algebra [9] by Jacob Klien and
MathematicalThoughts from Ancient to Modern Times [10] by Morris Kline are wonderfulresources to better understanding the ancient Greek philosophy of mathe-matics and how from the seventeenth through the nineteenth century thearithmetizing of space and all of mathematics took hold.As for this work, we will focus on the study of geometric space (not metaphor-ical space). We will claim that there is a significant and important distinctionbetween geometry and arithmetic. We do not simply want to have a finite,feasible foundation for arithmetic and then attempt to arithmetize geometrywith it. In fact the need to have arithmetic accomplish the feat of being afoundation to geometry might be seen as a reason for introducing extraordi-narily nonfinitistic means to arithmetic (for instance the need for irrationalnumbers as distances between coordinatized points). Therefore we will notpresent any arithmetical models of our theory. The only model to be consid-ered will be the real-world physical phenomenon of using tools to constructpoints and intuitive relations between these points.
We hold that all construction must be feasible. By feasible we simply meanthat the intuitive interpretation of the construction is executable (by simpletools) for all circumstances where it is applicable. We introduce seven un-defined constructions in our system. We introduce a midpoint construction(like Suppes did) which produces a point that is understood to be equidistantand colinear with two given points. This construction is executable in thereal-world by folding a string or straight edge whose ends are first held downat the original two points. We also introduce a segment extension construc-tion which extends a given line segment in one of two directions by a length ofsome other given segment. This is executable using a marked straight edge.10nce we have define ‘sides’ of a line there is an angle transport (on sameside of segment) construction. Given triples abc and def this constructionproduces a point x such that angle dex is congruent to abc , segment ex iscongruent to bc , and x and f are on the same side of de . (See Figure 19)This is executable using a marked protractor and marked straight edge. InSection 9, we introduce a circle-circle intersection construction which pro-duces a point of intersection between two circles if one circle contains a pointinside and a point outside of the other circle. This construction is executablewith a compass. We also introduce a construction called the crossbow con-struction. Given a line segment bd and two other points a and c which areon opposite sides of bd The construction produces a point between a and c which is also colinear with b and d . (See Figure 9) This construction is alsoexecutable with a straight edge. Since the length of segment from b to theconstructed point is not known prior to the construction, one might worrythat the length of the straight edge needed to construct the point might bemany magnitudes greater than any constructed up to that point especially as a and c get farther and farther away from b . This concern is a serious one fora feasible foundation. Fortunately within our system it is possible to provethat the line segment from b to this constructed point is ‘less than’ ba or bc where ‘less than’ is defined as it would be in Hilbert’s system. The proof ofthis fact is sketched out in Section 8 Thus there is no worry of not being ableto produce a straight edge with a suitable length to create the desired pointof intersection with segment ac . Lastly, we introduce a construction we callthe orthogonal construction which given three points a , b and c , where a (cid:54) = b and c (cid:54) = b , produces a point o such that oba and obc are right angles [seeFigure 28]. To demonstrate the feasibility of such a construction we designeda tool called the orthogonator which is sketched and discussed in section 10. As in most axiomatic system where constructions take the place of existen-tial quantifiers, there are circumstances where one could theoretically applya construction to a collection of points in which the intuitive real-world con-struction would not apply. Two examples in our system would be to applythe crossbow construction where the points a and c are on the same sideof bd or to apply the angle transport construction where f is colinear with11 and e or where a , b , and c are colinear (and thus do not form an anglein our system). Formally, in our theory, there are no statements that canbe proved about constructions applied to these non-intuitive settings. Thusone could easily ignore these applications as being outside the scope of thetheory and allow the constructions to be undefined in these instances. Onecould also choose to define the construction in some trivial unimportant waysuch as letting the crossbow construction in the application described aboveproduce the point b . This later choice need only be made if one finds itphilosophically important that all constructions are formally defined ‘every-where’. Mathematically either choice will not affect the work contended infuture sections. Having said that, our stance philosophically is to leave theconstruction undefined. We claim that our system is constructive in two ways. As has been mentionedalready our axiomatic system and the theory developed from these axiomsis completely in the formal language of quantifier-free first-order logic. Inthe place of existential quantifiers we will introduce functions which willbe referred to as constructions since they have interpretations of physicalconstructions an individual can carry out in the physical-world. This impliesthat our system is constructive as stipulated by Suppes.In [1] Beeson defined the properties a constructive theory of geometry shouldsatisfy. He claimed that by including certain stability axioms the only sig-nificant difference between a classical theory of geometry and a constructivetheory was that in a constructive theorem one is not able to make case dis-tinctions when constructing the object in an existential claim. Thus oneshould always have uniform (case-free) constructions. We in fact only needuniform constructions to achieve our goals. Because of this, we claim, with-out stipulating the stability axioms, that our theory is constructive as Beesondefines it. 12 .6 Sides of a Line Segment and Angle Orientation
Hilbert’s system satisfy what is called the plane separation property. Onedefines two points not on a given line as being on the same side of that line ifthe segment between those two points does not intersect the given line. Onethen defines these points as being on opposite sides of that line if the segmentbetween them does intersect the line. In a modern set-theoretic frameworkthe property claims that all of the points of the plane not on a line can beseparated into two equivalence classes called sides. One will note that ourcrossbow construction from earlier is related to these concepts. One will alsonotice that it would not be possible to smoothly transition the totality ofthese concepts over to a quantifier-free language. Having said that Tarskiand others were able to define a relation of four points for when a and b areon the same or opposite side(s) of a segment cd [1]. Both of these relationswere defined using only the between relation and an existential quantifier.It is impressive that Tarski and others were able to avoid using universalquantifiers when defining sides of a line segment. Having said that we willnot follow Tarski and others on this issue. Although the truth values ofsame/opposite side(s) relations can be determined by simply verifying theorder relations between points, we have two issues with incorporating theserelations. First we want our entire theory to be quantifier-free. Secondlyincorporating these relations will take us too far away form a Hilbert-styletheory. Hilbert was able to prove many theorems from only his order, inci-dence, and congruence axioms without relying on a parallel postulate. Thispart of the theory that excludes the parallel postulate is referred to as aHilbert Plane. Tarski theory, on the other hand uses an axiom equivalentto the Euclid’s fifth postulate and a line-circle continuity axiom (related tocompass constructions) in order to obtain similar results. Given that we havemajor issues with traditional axioms about parallel lines and wish to hold offon involving compass construction, we have chosen to go a different rout.One way that we could mimic a Hilbert-style axiomatic system would be tosimply introduce an undefined relation for two points being on the same sideof a line segment as was done by Greenberg in [5]. One could then includeseveral axioms to insure that this relation had the necessary properties. Wehave philosophical issue with this method. Having an undefined ‘same side’relation might lead one to think we are relying on an intuition about how linespartition the plane into two disjoint regions. This intuition is not acceptable13or our goals. We will therefore aim to develop the concept of sides of a linefrom a foundation more in line with our goals. We can all agree that anindividual is able to determine if two points a and b are on the ’same side’of line segment cd . So how does one determine this. We conclude that themethod actually boils down to some intuitive grasp of orientation and not theinfinite plane being partitioned by the line through c and d . Anyone who hasstudied elementary trigonometry has been exposed to concept that anglesin the plane can be positively or negatively oriented. We thus introduce anundefined same orientation relation. Given two non-colinear ordered triples abc and def the relation has a truth value of true when both ordered anglesare either positively oriented or both negatively oriented. By introducingaxioms pertaining to this relation we are able to fully develop a relation of‘same side’ from which we are able to prove many analogous results to Hilbert.Additionally all of these assumptions only pertain to the orientation of anglesin triangles or pertain to points involved in feasible constructions. We findthis a more philosophically appealing foundation to build from. This is alsoof interest in a purely mathematical sense. We have not found any previousworks where angle orientation is treated in synthetic geometry on its own oras it relates to sides of a line. Suppes’ approach formally codifies the everyday reality that humans onlymanipulate finite objects in finite ways. We claim that it may be possibleto interpret Beeson’s axiomatic systems which use skolem functions as beingconstructively finite in a similar way. Given this claim, one might supposethat a path to a finite, feasible, constructive foundation for synthetic geome-try would be best forged by conducing a deeper study of Beeson’s work. Wehold that although some of Beeson’s systems have the potential to be con-structive and finite, in the sense that they only speak about finitely manyobjects and finitely many construction there are features of his systems thatare not feasible. His use of axioms which are equivalent to Euclid’s fifthpostulate being a prime example.The current work is distinct from Beeson’s work is three important ways.First we claim that constructions producing a point of intersection for twonon-parallel line (segments) are not feasible. Thus we will build on the14ork of Suppes to codify the concept of parallel line segments without sucha construction. Secondly we will be developing our system as an analogto Hilbert’s axiomatic system. Thirdly, while some of Beeson’s axiomaticsystems have quantifier-free axioms, Beeson still uses definition which involveexistential quantifiers. This means that the over all theory is not quantifier-free. Our system and the theory developed from it is completely in quantifier-free first-order logic as was Suppes’.
As the long history of failed attempts to prove Euclid’s fifth postulate andthe discovery of non-euclidean geometry at the end of the nineteenth cen-tury would convince you of, one must assume a somewhat non-trivial featureof space that is not derived from simpler concepts of order, incidence, con-gruence of segments and angles, and basic constructions using straightedge,compass, or protractor. We have chosen to follow the work of Suppes andassume that the midpoint construction satisfies the property of bisymmetry.Formally this axiom states that for any four points a , b , c , and d we have mid ( mid ( a, b ) , mid ( c, d )) = mid ( mid ( a, c ) , mid ( bd )) where mid ( − , − ) is themidpoint construction. We would like to explain why we believe this axiomdoes not violate our desire for a finite, feasible, constructive foundation.It is easy to see why this axiom is finite and constructive given that it onlydiscusses four points and a construction. So we must ask ourselves if it is fea-sible. As was stated early our only requirement for the quantifier-free theoryto be feasible is for the constructions to have standard interpretation whichcan be performed by human faculties. Obviously the midpoint constructionis feasible. Additionally we strive for the concepts and objects of our sys-tem be accessible in terms of constructions that can be performed. UnlikeEuclid’s fifth postulate which makes claims that can not be verified by hu-man experience, for any four points one can actually verify if the midpointconstruction satisfies the bisymmetry condition in that instance. Thus ourtheory makes a assumption about space which could actually be experimen-tally tested leading to probabilistic confidence in the theory while also beingpotentially falsifiable. 15 .9 Number of Axioms In total our system has 36 axioms. It is natural for quantifier-free axiomaticsystem to have more axioms than their first-order logic counterparts [See [2]for an example]. Traditionally Hilbert’s axiom for the Hilbert plane arelisted thirteen in total, but this is misleading since many of them would bepresented as separate axioms in a more modern treatment. For exampleHilbert’s first congruence axiom discusses not only the existence of pointson either sides of a particular point on given line, but also states that theline segment congruence relation is reflexive. In a modern system this wouldbe two separate axioms. By our counting Hilbert’s axiomatic system for theHilbert plane totals 19 axioms. Furthermore if one chooses to have a planeseparation axiom instead of a Pasch’s axiom, as Greenberg has in [5], the totalwould go up to 20. Our system would take 29 axioms to codify an appropriateanalog to the Hilbert plane. This is surely more, but there are instances wherehaving a system with the only objects being points is actually more efficient.For instance we have no need for an analog to Hilbert’s first incidence axiomwhich states that any two points determine a line. Fortunately, our systemonly needs one additional axiom to extend our analog of the Hilbert plane toa analog of Hilbert’s treatment of the Euclidean plane sans continuity axioms.(Note that Hilbert’s axioms make no reference to compass constructions orcircle continuity.) In the end we are pleased that our system is still able to bepresented with such economy given that it is presented in such a minimalistand restrictive language.
Section 2 covers Hilbert’s order axioms except for Pasch’s axiom, Section3 discusses Hilbert’s incidence axioms, and Section 5 discusses his congru-ence axioms. In Section 5 we introduce the extension and angle transportconstructions. In Section 4 we focus on Pasch’s axiom and plane separationproperties. In this section we introduce our crossbow construction and angleorientation relation. We define and prove properties of sides of a line andalso prove an analog to Hilbert’s line separation property. In Section 6 wepoint out results from Hilbert’s theory which directly translate over to oursystem. In Section 7 we point out a few results which do not translate over as16ffectively. In particular we introduce a midpoint construction and uniformconstructions for erecting a perpendicular segment and dropping a perpen-dicular segment. By the end of Section 7 we have fully develop an analogto Hilbert’s theory of a Hilbert plane. In section 8 we introduce conceptof parallel line segment developed by Suppes. We then expand on Suppes’affine plane to develop a theory for a flat geometric plane. It is in this sectionthat we introduce our parallel axiom to take the place of traditional parallelpostulates. Given that we do not have the full might of a traditional parallelpostulate, there are results of Hilbert’s pertaining to parallel lines for whichwe do not have analogs. Having said that, there are many results pertainingto parallel lines that we do have analogs of. For example we have analogsof the Alternate Interior Angle Theorem and its converse. We are also ableto prove that parallel line segments are equidistant and that extensions ofnon-parallel lines either come closer together or move farther apart. Usingthese properties of parallel lines we are able to define and prove propertiesabout convex quadrilaterals and also define a uniform construction for tri-secting a segment. Lastly in this section we sketch out a proof of our earlierclaim that our crossbow construction is feasible. In Section 9 we introducea circle-circle intersection construction, we assume a version of circle-circlecontinuity, and prove uniform constructions for versions of line-circle conti-nuity and segment-circle continuity. In Section 10 we define a relation forwhen four points are coplanar. We then develop analogs to many solid ge-ometry results of Hilbert’s and Euclid’s. We also discuss the modificationsneeded to fully incorporate the planar results of the earlier sections into thisnew context. Lastly, we prove many results pertaining to sides of a plane.
Hilbert introduced an undefined between relation for three points, said ‘ b isbetween a and c ’, which has the interpretation of stating that b is on theinterior of the line segment with a and c as endpoints. We will introduce theundefined relation B ( a, b, c, ). 17 .1 Hilbert’s Order Axiom 1 Hilbert’s first order axiom states that for three distinct points, a , b , and c ,on a line, if b is between a and c , then b is between c and a .We thus introduce the following two axioms:Axiom 1: B ( a, b, c ) → a (cid:54) = b and a (cid:54) = c Axiom 2: B ( a, b, c ) → B ( c, b, a )Note that from these two axioms one can also prove that if b is between a and c , then b (cid:54) = c . Hilbert’s second order axiom states that given two points a and b there existsa point c such that b is between a and c Our axiomatic system will include a construction for segment extension.This construction, denoted ext ( ab, cd ), will be understood to be a point con-structed so that B ( a, b, ext ( ab, cd )) and the line segment from b to ext ( ab, cd )is congruent to the line segment from c to d . (Line segment congruence willbe defined in the next section.) There we will introduce an axiom whichstates these properties of this construction.For the time being, we can satisfy Hilbert’s axiom by noting that ext ( ab, ab )will function as the needed point c . Hilbert’s third order axiom states that given three distinct points on a line,one and only one lies between the other two. We thus introduce the followingaxiom.Axiom 3: ¬ ( B ( a, b, c ) ∧ B ( a, c, b ))Note if B ( a, b, c ), then B ( c, b, a ). Therefore ¬ B ( a, c, b ), ¬ B ( b, c, a ), ¬ B ( c, a, b ),and ¬ B ( b, a, c ). 18 .4 Hilbert’s Order Axiom 4 Hilbert’s fourth order axiom is known as Pasch’s axiom. We will be devotingall of section 4 to this topic as well as the closely related topic of plane sep-aration. For now we will turn our attention to Hilbert’s axioms of incidence.
Hilbert’s axiomatic system for planar geometry refers to two classes of objects(points and lines). His system would maybe best be formalized today by usingthe language of set theory and second order logic. To define the relationsamong these two types of objects Hilbert had two incidence axioms describinghow points and lines can be ’on’ and ’contain’ each other.Our axiomatic system in the tradition of Tarski will only have one type ofobject (points) all aspects of linearity will have to be dealt with by a relationof colinearity among three points.We now give a definition for the colinearity relation, defined L ( − , − , − ), ofthree points:Definition 1: L ( a, b, c ) ≡ B ( a, b, c ) ∨ B ( a, c, b ) ∨ B ( b, a, c )Note by Axiom 2 , a , b , and c are also colinear when B ( c, b, a ), B ( b, c, a ), and B ( c, a, b ). By Axiom 1, if L ( a, b, c ), then a , b , and c are all distinct points.Lastly one can prove that if L ( a, b, c ), then any permutation of a , b , and c isstill colinear. Hilbert’s first incidence axiom states for any two points there is a line that‘contains’ them. Again since we do not have the goal of formalizing linesas objects, we will not be introducing an analogous axiom. Our system willonly have an intuitive interpretation of line segments with two endpoints.19 .2 Hilbert’s Incidence Axiom 2
Hilbert’s second incidence axiom states that for every two points there existsno more than one line that ‘contains’ them. We will need a modified versionof this axiom. In particular we will not want that points a , b , and c are co-linear and points a , b , and d are co-linear while a , c , and d are not co-linear.Thus we have the following Axiom:Axiom 4: L ( a, b, c ) ∧ L ( a, b, d ) → L ( a, c, d )From this axiom one can prove that any distinct triple of a , b , c , or d iscolinear. Thus if two linear triples share two points then any (distinct) tripleformed from those four points is colinear. Hilbert’s third incidence axiom states that every line contains at least twopoints and that there are at least three points not all on the same line.Note that co-linearity is defined among three distinct points. In section 5 wewill define congruence of lines segments (defined by two endpoints). Becauseof this, our system will not allow for the discussion of any linear structuresfor less than two points.In order to codify the second point we will be follow Beeson and Suppes byhaving an axiom that states that there are three distinct constants which arenot colinear. First, though, we will defined a new relation which states threepoints are distinct and non-colinear.Definition 2: T ( a, b, c ) ≡ ¬ L ( a, b, c ) ∧ a (cid:54) = b ∧ b (cid:54) = c ∧ a (cid:54) = c This definition simply states that a , b , and c form a triangle.Next we introduce the following axiom.Axiom 5: T ( α, β, γ ) 20 Pasch’s Axiom, Plane Separation, and An-gle Orientation
Hilbert’s fourth order axiom is known as Pasch’s axiom. It states that if a , b , and c are three distinct non-colinear points and (cid:96) is a line that does notpass through any of the three points, but does intersect the line segmentconnecting a and b , then (cid:96) contains a point between b and c or a and c . Inother words, if a line intersects the interior of one side of a triangle (and doesnot intersect any of the vertices), then it must interest the interior of exactlyone of the two remaining sides. Figure 3 illustrates this concepts. Pasch’saxiom assures that the axiomatic system is one of planar geometry and nota higher dimensional geometry. (cid:96) (cid:96)aa b b cc Figure 3: Pasch’s axiomAs the reader will note, that it is not a simple task to capture this axiom inthe a formal language that does not allow for the mentioning of lines. Tarski’sdid this very thing by defining two subversions of Pasch’s axiom called innerPasch and outer Pasch. (The original version of Tarski’s axioms assumedboth inner and outer Pasch. It was later proved by Gupta in [6] that outerPasch could be proved from inner Pasch.)
There is a property of the Euclidean plane that states that every line dividedthe points on the plane not on the line into two sets called sides having the21ollowing properties: If a and b are on different sides of the line, then the linesegment between a and b contains a point on the line and if a and b are onthe same side of the line then the line segment between a and b contains nopoints of the line.In some presentation of Hilbert’s system (in [5] for example) a plane sepa-ration axiom was assumed and Pasch’s ‘axiom’ was shown to be a result ofthis assumption. As has been stated, Hilbert assumed Pasch’s axiom. InHartshorne’s text [7] one can find a proof of the plane separation propertybased off of Hilbert’s original system. The proof relies heavily on Pasch’saxiom. The proof shows that the relation ‘same side’ is an equivalence rela-tion for the points not on a given line. He then shows that if points a and b are not on the same side of some line and b and c are also not on the sameside then a and c are on the same side. This shows that there are only twoequivalence classes.Since Tarski’s system does not allow for the discussion of sets or lines for thatmatter, some work must be done to translate this ideas over in an analogousways. Since we will not be following Tarski et al in their methods, we willtry to avoid going too deep into this topic. We will simply point out a fewimportant details about their work. They define ‘opposite side’ as a 4-aryrelation. They then define ‘same side’ by having a witness point on the otherside of the line. One is then able to prove a plane separation theorem whichstates that if a and b are on the same side and a and c are on opposite sidesthen b and c are on opposite sides. See [1] for details.We will be taking a very different track from Hilbert and Tarski et al. First,we will not be assuming any version of Pasch’s axiom. We will insteadintroduce a relation whose interpretation is understood as two angles havingthe same orientation as well as one new construction. From axioms about thisrelation and construction we will be be able to prove analogous statementsto Hilbert’s plane separation axiom and Pasch’s axiom. We desire to formally capture humans’ intuitive understanding of orientationof the plane without resorting to concepts such as an infinite line dividing the22hole plane into two disjoint regions. We thus choose to have an undefinedrelation for when two non-colinear triples of points (two angles) have thesame orientation. Since we are not able to discuss angles as Hilbert would,we will have an undefined 6-ary relation for two triples of points.By
SameOrientation ( a, b, c, d, e, f ) we will interpret that the rotation fromsegment ab to segment bc has the same orientation as the rotation fromsegment de to segment ef . See Figure 4. For ease of reading we will use thenotation SO ( abc, def ) instead of SameOrientation ( a, b, c, d, e, f ). b a e dc f Figure 4: Diagrammatic representation of angles abc and def having sameorientationWe have quite a few axioms pertaining to this relation. nearly all have inter-pretations that are elementary about the relationship between line segmentsand angle orientation.Axiom 6: SO ( abc, def ) → T ( a, b, c ) ∧ T ( d, e, f )This axiom simply states that if two angles have the same orientation thenthey are both distinct non-colinear triples. This allows us to not have todeal with non-distinct or colinear triples. Much like Hilbert, we will only bediscussing angles less than a straight angle.Axiom 7: SO ( abc, abc )Axiom 8: SO ( abc, def ) ∧ SO ( abc, ghi ) → SO ( abc, ghi )These two axioms assure that the relation is reflexive, symmetric, and tran-sitive. 23e now define a relation, SameDirection ( a, b, c ), shortened SD ( a, b, c ), forwhen two points of a colinear triple are on the same side of the line in relationto the third point. This is analogous to Hilbert’s concept of a ray.Definition 3: SD ( a, b, c ) ≡ B ( a, b, c ) ∨ B ( a, c, b ) ∨ ( a (cid:54) = b ∧ c = b ) ∨ ( a = b ∧ c (cid:54) = b )The following axioms states that extending or shortening a side of an angledoes not change it’s orientation. This takes the place of Hilbert’s use of rayswhen defining angles.Axiom 9: T ( a, b, c ) ∧ SD ( b, a, e ) → SO ( abc, ebc )We now define two angles having opposite orientation to mean that they donot have the same orientation and are both distinct non-colinear triples.Definition 4: OO ( abc, def ) ≡ ¬ SO ( abc, def ) ∧ T ( a, b, c ) ∧ T ( d, e, f ) [See Fig-ure 5.] b a e dc f Figure 5: Diagrammatic representation of angles abc and def having oppositeorientationGiven the transitivity and symmetry properties of the
SameOrientation re-lation we can prove the following theorem.Theorem 1: SO ( abc, def ) ∧ OO ( def, ghi ) → OO ( abc, ghi )The following axiom states that switching the initial and terminal side of anangle changes its orientation. 24xiom 10: OO ( abc, cba )The next axiom states that there is parity among angles orientation. Thisaxiom is a critical property to build toward plane separation.Axiom 11: OO ( abc, def ) ∧ OO ( def, ghi ) → SO ( abc, ghi )One can prove the following lemma which states given two angles with same(opposite) orientations if one changes the initial and terminal sides of bothangles the two resulting angles will have the same (opposite) orientation.Lemma 1: SO ( abc, def ) → SO ( cba, f ed ) and OO ( abc, def ) → OO ( cba, f ed )We now introduce some axioms dealing with the interplay between angleorientation and order.Axiom 12: B ( a, b, c ) ∧ T ( a, c, d ) → OO ( dba, dbc ) [See Figure 6.] a cb d Figure 6:Given a line segment db , OO ( dba, dbc ) is our proto-version of stating a and c are on ‘opposite sides’ of db . Thus the previous axiom states somethinganalogous to the idea that if a line segment contains one endpoint between a and c then a and c are on opposite sides of the line segment. (A fullyflushed out definition of opposite sides of a line segment will be discussed insubsection 4.3.)The next theorem shows that there is some relationship between colinearalityand angle orientation. 25 cbde Figure 7:Theorem 2: L ( a, b, c ) ∧ OO ( abd, abe ) → OO ( cbd, cbe ) [See Figure 7] Proof.
Since L ( a, b, c ), either B ( a, b, c ) or SD ( b, a, c ). If SD ( b, a, c ), then byAxiom 9 and Theorem 1 we have OO ( cbd, cbe ). Since L ( a, b, c ), a (cid:54) = c . Since OO ( abd, abe ), T ( a, b, d ) and thus a (cid:54) = d and d (cid:54) = c . Since a , c , and d are alldistinct, T ( a, c, d ) otherwise L ( a, c, d ). This would imply L ( a, b, d ) by Axiom4, but T ( a, b, d ). Similarly T ( a, c, e ). Thus if B ( a, b, c ) then OO ( abd, cbd )and OO ( abe, cbe ) by Axiom 12. Since OO ( abd, cbd ) and OO ( abd, abe ), we caninfer SO ( cbd, abe ) by Axiom 11. And lastly, since OO ( abe, cbe ), OO ( cbd, cbe )by Theorem 1.We now define the relation that states when a point is in the interior of anangle.Definition 5: Int ( d, abc ) ≡ SO ( cbd, cba ) ∧ SO ( abd, abc ) [See Figure 8.] dc ab Figure 8:From the definition one can prove the following theorem.26heorem 3:
Int ( d, abc ) → OO ( dba, dbc ) Proof.
By Axiom 10 we have OO ( abc, cba ). Since we have SO ( abd, abc ), byTheorem 1 we obtain OO ( abd, cba ). Since we also have SO ( cba, cbd ), by thesame theorem we now have OO ( abd, cbd ). By applying Lemma 1 above wecan infer OO ( dba, dbc ).We now wish to introduce our first construction. First we will define the non-strict colinear relation which will make the first axiom about this constructionsignificantly easier to state.Definition 6: (cid:101) L ( a, b, c ) ≡ L ( a, b, c ) ∨ a = b ∨ a = b ∨ a = c Do note that (cid:101) L ( a, b, b ) and (cid:101) L ( a, a, a ) are always true.Our first construction is called the crossbar construction. It is denoted crossbow ( d, a, b, c ) which will be write as cb ( d, abc ) for ease of reading. Thereis only one axiom pertaining to this construction. This axiom states that if a and c are on ‘opposite sides’ of bd , then the constructed point cb ( d, abc )is non-strict colinear with b and d and is between a and c . Note that theassumption of the axiom guaranties that b (cid:54) = d . See Figure 9.Axiom 13: OO ( dba, dbc ) → (cid:101) L ( b, cb ( d, abc ) , d ) ∧ B ( a, cb ( d, abc ) , c ) cb ( d, abc ) a a cb ( d, abc ) b bd dcc Figure 9:This axiom gives us an analog of the concept that if a and c are on oppositesides of bd then the line bd intersects the interior of ac .27he next theorem states that given a triangle abc with a point d between a and c we have that d is in the interior of angle abc .Theorem 4: T ( a, b, c ) ∧ B ( a, d, c ) → Int ( d, abc ) Proof.
Note that given the assumptions all four point a , b , c , and d aredistinct. In particular if b = d then ¬ T ( a, b, c ). Suppose ¬ T ( a, b, d ), then L ( a, b, d ). But since L ( a, d, c ) also, we obtain L ( a, b, c ). This is a contradic-tion. Thus T ( a, b, d ) and by similar arguments T ( c, b, d ).Suppose ¬ Int ( d, abc ), then ¬ SO ( cbd, cba ) or ¬ SO ( abd, abc ). Given the finalsentence of the previous paragraph, if ¬ SO ( cbd, cba ) then OO ( cbd, cba ) Thus, B ( a, cb ( c, abd ) , d ). Since L ( a, d, c ), we have L ( a, cb ( c, abd ) , c ). Because ofthis we know that cb ( c, abd ) (cid:54) = c . Since we also have L ( b, cb ( c, abd ) , c ) or cb ( c, abd ) = b , we can conclude L ( a, b, c ). This is a contradiction. Similarly,supposing ¬ SO ( abd, abc ) leads to a contradiction. Thus Int ( d, abc ).We are now able to prove our analog to Hilbert’s Crossbar Theorem. Thefollowing theorem states that if d is in the interior of angle abc , then cb ( d, abc )is the same direction from d as b .Theorem 5: Int ( d, abc ) → SD ( b, cb ( d, abc ) , d ) [See Figure 10] Proof.
Let cb ( d, abc ) = x . If x = d we are done. Suppose x (cid:54) = d . Since Int ( d, abc ) we have d (cid:54) = b . By Axiom 13 we know (cid:101) L ( b, x, d ). We can infer L ( b, x, d ). This implies that either SD ( b, x, d ) or B ( x, b, d ) (See Theorem14.) Suppose B ( b, x, d ). One can show T ( x, d, a ). Thus by Axiom 12 weobtain OO ( abx, abd ). Since Int ( d, abc ), we have SO ( abd, abc ). By Theorem4, we know Int ( x, abc ) and thus SO ( abx, abc ). Using the symmetry andtransitivity of the same orientation relation we can infer SO ( abx, abd ). Thisis a contradiction. Thus SD ( b, x, d ).The next theorem states that if angle cbd and angle cba have the same ori-entation and a and d are not on the same ’ray’, then either d is interior toangle abc or a is interior to angle dbc . This theorem will be used when angleinequalities are defined.Theorem 6: SO ( cbd, cba ) ∧ T ( a, d, b ) → Int ( d, abc ) ∨ Int ( a, dbc )28 b ( d, abc ) = xb d a c Figure 10:
Proof. If SO ( abd, abc ), then Int ( d, abc ) and we are done. Suppose ¬ SO ( abd, abc ).Since SO ( cbd, cba ), we have T ( c, b, d ) and T ( c, b, a ). Since T ( c, b, a ), wehave T ( a, b, c ). We have also assumed T ( a, b, d ), thus OO ( abd, abc ). Thus SO ( abd, cba ). Since we assume SO ( cbd, cba ), we now have SO ( abd, cbd ). So SO ( dba, dbc ). Therefore Int ( a, dbc ).We now introduce our last axiom pertaining to angle orientation.Axiom 14: T ( a, b, c ) → OO ( abc, bac ) [See Figure 11.] a bc Figure 11:This axiom is the final assumption needed to obtain our analogs to planeseparation. The following theorem shows that if two points are on the ‘sameside’ of a line segment, then no point can be co-linear with the line segmentand be between the two points.Theorem 7: SO ( cbd, cba ) ∧ B ( a, x, d ) → ¬ (cid:101) L ( b, x, c ) [See Figure 12 ]29 b a x d cb a x d Figure 12:
Proof.
Start by noting that T ( c, b, d ) and T ( c, b, a ). We will first show that x (cid:54) = b and x (cid:54) = c . Suppose x = b . Then B ( a, b, d ). If L ( a, d, c ), then L ( b, c, d )which contradicts T ( c, b, d ). Thus we can infer T ( a, d, c ). By Axiom 12 wehave OO ( cbd, cba ). This is a contradiction.Now suppose x = c . Then B ( a, c, d ). If L ( a, d, b ), then L ( b, c, d ) whichcontradicts T ( c, b, d ). Thus we can infer T ( a, d, b ). By Axiom 12 we have OO ( bcd, bca ) and by an application of Axiom 14 we have OO ( cbd, cba ) This isa contradiction. Thus x (cid:54) = b and x (cid:54) = c . If we suppose (cid:101) L ( b, x, c ) we can theninfer L ( b, x, c ). We will show that this assumption leads to a contradiction.There are two cases to consider.Case 1: Suppose T ( a, b, d ). By Theorem 4, we have Int ( x, abd ) and thus OO ( xba, xbd ). Since we are supposing L ( b, x, c ), By Theorem 2, we have OO ( cba, cbd ). This is a contradiction.Case 2: Suppose ¬ T ( a, b, d ). Since T ( c, b, d ), T ( c, b, a ) and a (cid:54) = d , we havethat a , b , and d are all distinct. Thus L ( a, b, d ). Since L ( a, x, d ) also, we have L ( b, a, x ). And since L ( b, x, c ), we have L ( a, b, c ). This is a contradiction.We now turn to proving our analog to Pasch’s axiom. But first we prove alemma.Lemma 2: T ( a, b, c ) ∧ B ( a, d, b ) ∧ T ( a, b, e ) ∧ T ( e, d, c ) → OO ( edb, edc ) ∨ OO ( edc, eda ) [See Figure 13.] 30 b cd ea b cde a b cd e Figure 13:
Proof.
We first want to show T ( e, d, b ). One can show that e , d , and b areall distinct. Thus if ¬ T ( e, d, b ), then L ( e, d, b ). But since L ( a, d, b ), we wouldhave L ( a, e, b ). This is a contradiction. Thus T ( e, d, b ).Now if OO ( edb, edc ) we are done. So suppose ¬ OO ( edb, edc ). Since T ( e, d, b )and T ( e, d, c ), and ¬ OO ( edb, edc ), we have SO ( edb, edc ). By Axiom 12, wehave OO ( ebd, eda ). Therefore by Theorem 1, we obtain OO ( edc, eda ).This next lemma states something analogous to the idea that a line can’tintersect the interior of three sides of a triangle. It is used in the proof of atheorem in Section 10.Lemma 3: T ( a, b, c ) ∧ B ( a, d, b ) ∧ B ( b, e, c ) ∧ B ( a, f, c ) → T ( d, e, f ) Proof.
First we will show that d , e , and f are distinct. If d = e then B ( a, e, b ).Since L ( b, e, c ) and L ( b, e, a ), we have L ( a, b, c ). This is a contradiction. Thus e (cid:54) = d . In similar fashion we can show the d (cid:54) = f and e (cid:54) = f . Furthermorewe can show that all points a , b , c , d , e , and f are all distinct. We can show T ( a, f, b ). If ¬ T ( a, f, b ), then L ( a, f, b ). This would imply L ( a, b, c ) which isa contradiction. Likewise we can show that T ( c, f, b ), T ( c, d, e ) and T ( a, d, c ).Suppose ¬ T ( d, e, f ). Since d , e , and f are distinct, we have L ( d, e, f ). Since L ( d, e, f ) without loss of generality we can say B ( d, f, e ). Since T ( a, f, b )31nd B ( a, d, b ) by Theorem 4 we have Int ( d, af b ) and thus SO ( af d, af b ).Likewise we have Int ( e, cf b ) and thus SO ( cf e, cf b ). By axiom 12, we have OO ( af b, cf b ). Since SO ( af d, af b ) and OO ( af b, cf b ) by Theorem 1 we have OO ( af d, cf b ). Since SO ( cf e, cf b ) and OO ( af b, cf b ) by Theorem 1 we have OO ( cf e, af b ). Since B ( d, f, e ) we have OO ( cf e, cf d ) via an application ofAxiom 12. Thus by Axiom 11 we have SO ( af d, cf d ). But by Axiom 12 weknow OO ( af d, cf d ). This is a contradiction. Thus T ( d, e, f ).We now state our version of Pasch’s axiom as a theorem:Theorem 8: Let A ≡ T ( a, b, c ) ∧ B ( a, d, b ) ∧ T ( a, b, e ) ∧ T ( e, d, c ), P ≡ (cid:101) L ( d, cb ( e, adc ) , e ) ∧ B ( a, cb ( e, adc ) , c ), Q ≡ (cid:101) L ( d, cb ( e, bdc ) , e ) ∧ B ( b, cb ( e, bdc ) , c ), R ≡ B ( b, x, c ) → ¬ (cid:101) L ( e, d, x ), and S ≡ B ( a, x, c ) → ¬ (cid:101) L ( e, d, x ), then A → ( P ∧ R ) ∨ ( Q ∧ S ), Proof.
Assuming A , Lemma 2 tells us that OO ( edb, edc ) or OO ( edc, eda ).If OO ( edb, edc ), then by Axiom 13 we have P . As was pointed out in theproof of the previous lemma, by Axiom 12 we know that OO ( eda, edb ). Thusby Axiom 1 and the previous lemma we have SO ( eda, edc ). Therefore byTheorem 7 we have R . If OO ( edc, eda ) by similar reasoning we have Q and S Up to this point we have using the relation SO ( abc, abd ) to be our proto-definition for c and d being on the ‘same side’ of segment ab . It is quiteobvious that more work still needs to be done. Although we were able toprove quite a few results about the crossbow construction and the interior ofangles, we have not explicitly shown that if c and d are on the ‘same side’ of32 b , then c and d are on the ‘same side’ of ba . Furthermore we would desirethat if points a , b , c , and d were colinear then e and f being on the ‘sameside’ of ab would imply that e and f were on the ‘same side’ of cd . We canaccomplish this.First we will define a same side relation, SameSide ( c, d, a, b ), shortened to SS ( c, d, ab ).Definition 7: SS ( c, d, ab ) ≡ SO ( abc, abd )The first issue pointed in the previous paragraph is addressed by part three ofthe following theorem. From the reflexivity of the angle orientation relationwe can prove part 1 of the theorem below. By the symmetry of angle orien-tation we can prove part 2 and by using the Axiom 14 we can prove part 3.Part 4 can be proved from the transitivity property of the same orientationrelationTheorem 9:1. T ( a, b, c ) → SS ( c, c, ab )2. SS ( c, d, ab ) → SS ( d, c, ab )3. SS ( c, d, ab ) → SS ( c, d, ba )4. SS ( c, d, ad ) ∧ SS ( d, e, ab ) → SS ( c, e, ab )To address the second concern pointed out in the first paragraph of thissection we have the following theorem.Theorem 10: Given a (cid:54) = b , c (cid:54) = d , (cid:101) L ( a, b, c ), and (cid:101) L ( b, c, d ), if SS ( e, f, ab ) then SS ( e, f, cd ). Proof.
Case 1) Let a = c and b = d . Then we are done.Case 2) Let a = d and b = c . This can be proved by an application of part3 of the previous theorem.Case 3) Let a = c and b (cid:54) = d . If B ( b, a, d ) (see left side of Figure 14),then SS ( e, f, ab ) implies SO ( abe, abf ) which in turn implies SO ( dbe, dbf )33y Axiom 9. Using Axiom 14 we can infer SO ( bde, bdf ) and this implies SO ( cde, cdf ). Thus by definition we have SS ( e, f, cd ). If ¬ B ( b, a, d ), we have SD ( a, b, d ) (see theorem 14). (See right side of figure 14.) If SS ( e, f, ab ), thenwe have SO ( abe, abf ). This implies SO ( bae, baf ) by axiom 14. From this wecan obtain SO ( bce, bcf ) and then SO ( dce, dcf ) by Axiom 9. Thus we have SS ( e, f, cd ). a = cb de f a = c b de f Figure 14:The next three cases follow from similar methods.Case 4) Let a = d and b (cid:54) = d .Case 5) Let b = c and a (cid:54) = d .Case 6) Let b = d and a (cid:54) = c .Case 7) Let a , b , c , and d all be distinct. Suppose SD ( b, a, c ). Since SS ( e, f, ab ) we then have SO ( abe, abf ) which will imply SO ( cbe, cbf ) byAxiom 9. By Axiom 14 we obtain SO ( bce, bcf ).Now suppose ¬ SD ( b, a, c ). We can infer that B ( a, b, c ) (see Theorem 14).Since SS ( e, f, ab ), we have SO ( abe, abf ). By Theorem 2 we SO ( cbe, cbf )and by Axiom 14 we have SO ( bce, bcf ) also.Since SO ( bce, bcf ), if SD ( c, b, d ) also then we have SO ( dce, dcf ). If ¬ SD ( c, b, d ),then B ( d, c, b ). Since we have SO ( bce, bcf ), by theorem 2 we obtain SO ( dce, dcf )also. We can then can infer SS ( e, f, dc ) and by applying Theorem 9 we have SS ( e, f, cd ).We can define an opposite sides relation in a similar fashion. This relation34an be denoted OppositeSides ( c, d, a, b ), but as with other relations will beshortened to OS ( c, d, ab ).Definition 8: OS ( c, d, ab ) ≡ OO ( abc, abd )One can then prove the following theorem.Theorem 11: T ( a, b, c ) ∧ T ( a, b, d ) ∧ ¬ SS ( c, d, ab ) → OS ( c, d, ab )There are analogous versions of the two main theorems of this subsectionwhere the same side relation is replaced with the opposite sides relation. Weomit the proofs.Theorem 12:1. OS ( c, d, ab ) → OS ( d, c, ab )2. OS ( c, d, ab ) → OS ( c, d, ba )3. OS ( c, d, ad ) ∧ OS ( d, e, ab ) → SS ( c, e, ab )Theorem 13: Given a (cid:54) = b , c (cid:54) = d , (cid:101) L ( a, b, c ), and (cid:101) L ( b, c, d ), if OS ( e, f, ab ) then OS ( e, f, cd ). It is possible that the reader may be asking themselves why we did not justintroduce a ’same side’ relation to start with and avoid a discussion of angleorientation. The reader may also wonder if the number of axioms assumedwould be minimized and the road to proving Pasch’s axiom would simpler ifwe could just assume a small set of assumptions about the same side relationsto obtain the plain separation property and then prove Pasch’s axiom as atheorem in the style of Greenberg in [5]. They may think that this would leadto not having need for a construction such as the crossbar construction. Onemay wonder if such a strategy would be more economical. We would replythat in fact one is not able to greatly reduce the number of axioms assumedand constructions used by introducing a ‘same side’ relation to start with.35f one were to try to obtain the plane separation properties and Pasch’s axiomsimply from an undefined ‘same same’ relation they would in fact have toassume quite a few axioms. They would have to assume that the relationhad the three properties stated in Theorem 9. They would have to definean ‘opposite side’ relation. They would have to assume that if a and b wereon the same side and b and c were on opposite sides then a and c wouldbe on opposite sides. They would have to assume an axiom analogous toTheorem 7 which would interpret the property that lines do not intersect aline segment between two points on the same side of it. They would have toassume an axiom like Axiom 12 which states that if a line intersects a segmentthen the endpoints of the segment are on opposite sides. They would haveto introduce a crossbow-like construction and assume an axiom like Axiom13 to produce the point for where a line intersects the interior of a segmentwith points on opposite sides of it. They would have to assume an axiom likeTheorem 10 in order to preserve the sides of two line segments when they arecolinear. They would then need to prove Pasch’s axiom and then also provethe Crossbar Theorem. Because of this we see no formal reason why havingan undefined ’same side’ relation would be a great improvement. An important result in any presentation of Hilbert’s system is proving a lineseparation property. In [7] Hartshorne defines line separation as the propertythat the set of points on a line (cid:96) not equal to a particular point a on the linecan be divided into two sets called sides such that b and c are on the sameside of a if and only if a is not in the segment bc and b and c are on differentsides of a if and only if a is in the segment bc .We already have relations which codify the idea of b and c being on the sameor opposite sides of a . The relation B ( b, a, c ) will take the place of b and c being on ‘opposite sides’ of a and SD ( a, b, c ) will take the place of b and c being on the ‘same side’ of a .It can be shown that if L ( a, b, c ) then ¬ SD ( a, b, c ) implies B ( b, a, c ) and ¬ B ( b, a, c ) implies SD ( a, b, c ).Theorem 14: 36. L ( a, b, c )) ∧ ¬ SD ( a, b, c ) → B ( b, a, c )2. L ( a, b, c ) ∧ ¬ B ( b, a, c ) → SD ( a, b, c ) Proof.
To prove part 1 note that ¬ SD ( a, b, c ) implies that ¬ B ( a, b, c ) and ¬ B ( a, c, b ). Since we have L ( a, b, c ) and ¬ B ( a, c, b ), ¬ B ( a, b, c ) we can con-clude B ( b, a, c ).Part 2 is proved via a straight forward proof based on definitions.Hilbert’s line separation property is captured by the following theorem.Theorem 15:1. SD ( a, b, b )2. SD ( a, b, c ) → SD ( a, c, b )3. SD ( a, b, c ) ∧ SD ( a, c, d ) → SD ( a, b, d )4. SD ( a, b, c ) ∧ B ( d, a, c ) → B ( d, a, b )5. B ( b, a, c ) ∧ B ( d, a, c ) → SD ( a, b, d ) cba dγ d b a cγ Figure 15:Parts 1 and 2 follow directly from the definition of the same direction relation.The first three parts show that the two points being of the ‘same side’ of a is a reflexive, symmetric, and transitive relation. One can use the symmetricand transitive properties along with the previous theorem to prove part 4.Thus we will only prove parts 3 and part 5.37 roof. By Axiom 5 we know that at least one of α , β , or γ are not colinearwith a and b . Without loss of generality let T ( a, b, γ ). The assumptions ofboth part 3 and part 5 imply L ( a, b, c ) and L ( a, c, d ). From this it can beshown that γ is not colinear with any pair of points among a , b , c , or d .To prove part 3 (see right side of Figure 15)note that SD ( a, b, c ) implies ¬ B ( b, a, c ) by the previous theorem. We now wish to show that ¬ B ( b, a, c )implies SO ( aγb, aγc ). Suppose ¬ SO ( aγb, aγc ), then by the last sentencein the previous paragraph we can infer OO ( aγb, aγc ). We will show thatthis leads to a contradiction. Given OO ( aγb, aγc ) we can infer certain prop-erties about cb ( a, bγc ). By axiom 13 we know B ( b, cb ( a, bγc ) , c ). Now if cb ( a, bγc ) = γ , then L ( b, γ, c ). This is a contradiction. Thus cb ( a, bγc ) (cid:54) = γ .If cb ( a, bγc ) (cid:54) = a , then L ( γ, cb ( a, bγc ) , a ). We already know L ( b, cb ( a, bγc ) , c )and L ( a, b, c ). By applying axiom 4 repeatedly we can show L ( b, a, cb ( a, bγc )),followed by L ( γ, b, cb ( a, bγc )), then followed by L ( γ, b, c ). This is a con-tradiction. Thus we can infer that SO ( aγb, aγc ). By similar methods wecan show that SD ( a, c, d ) implies SO ( aγc, aγd ). Thus by axiom 8 we have SO ( aγb, aγd ). We want to show that this implies ¬ B ( b, a, d ). Suppose B ( b, a, d ), then by axiom 12 we could infer OO ( aγb, aγd ). This is a contra-diction. Thus ¬ B ( b, a, d ). By the previous theorem this implies SD ( a, b, d ).To prove part 5 (see left side of Figure 15) note that B ( b, a, c ) implies OO ( aγb, aγc ) and B ( d, a, c ) implies OO ( aγd, aγc ). By Axiom 11 we can infer SO ( aγb, aγd ). This in turn implies ¬ B ( b, a, d ) which implies SD ( a, b, d ) bythe previous theorem. Hilbert introduced two congruence relations. One relation was the congru-ence of line segments. The interpretation was that two line segments werecongruent if they had the same length. The other relation was for two anglesbeing congruent. Angles for Hilbert were formed by two rays that shared anendpoint. All angles were less than a straight angle. For us an angle is an or-dered triple of three non-colinear points. Tarski was able to avoid discussingangles all together by a use of “five-segment” axiom that took the place ofthe Side-Angle-Side triangle congruence axiom used by Hilbert. In [1] it was38hown that a Hilbert style angle congruence can be derived from Tarski’saxioms.
In his first congruence axiom Hilbert states two important properties. Firstis that every line segment is congruent to itself.We introduce a relation for line segment congruence C ( a, b, c, d ) which maybe shortened as C ( ab, cd ). Its interpretation will be that the line segment ab is the same length as the line segment cd . The following axiom capturesHilbert’s reflexive properties as well a stating that segment congruence isunaffected by reversing endpoints.Axiom 15: C ( ab, ab ) and C ( ab, ba )It is important to point out that we will be allowing for the congruence ofnull segments such as aa ∼ = bb . In Hilbert’s system segment congruence isonly referred to when discussing distinct pairs of points. In [7] Hartshornefollows this convention. Because of this, our definition of segment inequalitywill differ slightly for that of Hilbert and in turn Hartshorne. There are prosand cons to either convention, but we have chosen our because it will aidin defining and proving results about compass constructions later on. (Itappears that Hartshornes section of circle continuity is in fact flawed in thatit’s defining of circle-circle continuity and this definition’s application in theproof of line-circle continuity because the two different version of segmentinequality are conflated.)The second property of Hilbert’s first congruence axiom is that given a raywith endpoint a and a line segment cd , there is a unique point x on the raysuch that ax is congruent to cd .Back in subsection 2.2 we stated that we would be introducing an extensionconstruction ext ( a, b, c, d ) (which may be shortened to ext ( ab, cd )). The con-structed point ext ( ab, cd ) will be interpreted as being a point that extendsthe line segment ab by the length of the line segment cd in the direction from a to b . These properties are formally codified in the following axiom [seeFigure 16]. 39xiom 16: a (cid:54) = b → C ( b, ext ( ab, cd ) , c, d )Axiom 17: a (cid:54) = b ∧ c (cid:54) = d → B ( a, b, ext ( ab, cd )) c da b ext ( ab, cd )Figure 16:One will note that the condition a (cid:54) = b is necessary since we conclude that B ( a, b, ext ( ab, cd )). Also one would not be able to make sense of a directionfrom a to b when a = b . We will be allowing extensions of non-null segmentsby null segments.We will now define a new construction from the extension. This constructionwill have a central importance in later sections about parallel line segments.Definition 9: If a (cid:54) = b , then doub ( a, b ) ≡ ext ( ab, ab ). Also doub ( a, a ) = a .We will call this construction the doubling construction. Its interpretation isa point that extends the line segment ab , in the direction from a to b , by thelength of ab if a (cid:54) = b . In the case a = b the construction is simply the point a .We now define a construction which will help satisfy the second property ofHilbert’s first congruence axiom.Definition 10: lf ( ab, cd ) ≡ ext ( doub ( b, a ) , a, c, d ) [See Figure 17]This construction will be called the layoff construction. One can easily verify C ( a, lf ( ab, cd ) , c, d ). One can also verify SD ( a, lf ( ab, cd ) , b ) by using resultsabout line separation in subsection 4.4.We want the points constructed by the extension construction to be uniquein terms of order and congruence properties so we introduce the followingaxiom. 40 ddoub ( b, a ) a ext ( doub ( ba ) , a, c, d ) b Figure 17:Axiom 18: ( B ( a, b, x ) ∨ b = x ) ∧ C ( bx, cd ) → x = ext ( ab, cd )This last definition and last axiom give us a layoff construction with analo-gous properties to Hilbert’s first congruence axiom.We can now define an inequality relation between two line segments.Definition 11: ab < cd ≡ B ( c, lf ( cd, ab ) , d ) ∨ ( a = b ∧ c (cid:54) = d )It should be pointed out that by the uniqueness of the extension construction,if B ( a, b, c ), then ab < ac .In the future we will have need of the following theorem. We omit the proof.Theorem 16: ab < cd ∧ cd < ef → ab < ef The following axiom is identical to Hilbert’s second congruence axiom.Axiom 19: C ( ab, cd ) ∧ C ( ab, ef ) → C ( cd, ef )From Axiom 15 and Axiom 19 we see that line segment congruence is reflex-ive, symmetric, and transitive. Hilbert’s third congruence axiom states that if ab and bc are two segmentson the same line that have no points in common other than b and de and41 f are two segments one the same line that have no points in common otherthan e , then if ab is congruent to de and bc is congruent to ef , we have ac is congruent to df . This is know as the addition of segments axiom. Wecapture this axiom as follows.Axiom 20: B ( a, b, c ) ∧ B ( d, e, f ) ∧ C ( ab, de ) ∧ C ( bc, ef ) → C ( ac, df ) Hilbert’s fourth congruence axiom deals with the construction of congruentangles. First we will introduce an angle congruence relation. The relationwill be notated AC ( a, b, c, d, e, f ) (shortened to AC ( abc, def ) for readability).In order to avoid having to speak about the congruence of colinear or nondistinct triples will have the following axiom.Axiom 21: AC ( abc, def ) → T ( a, b, c ) ∧ T ( d, e, f )Hilbert’s fourth axiom states that given an angle abc and a line de , thenon a given side of the line de there is a unique ray dx such that angle edx is congruent to angle abc . [Hilbert’s original axiom also included proper-ties about the interior of angles. We will be following the convention ofHartshorne and not included this. We have done the necessary work to cod-ify the interior of angles in Section 4.2] We capture this axiom by introducinga construction. Our angle transport (same side) construction will be denoted ats ( a, b, c, d, e, f ) (shortened to ats ( abc, def )). We interpret this constructedpoint to have two important properties. We want that the angle formedfrom d to e to ats ( abc, def ) to be congruent with the angle abc . We alsowant that the constructed point is on the same side of de as f . The followingaxiom captures these properties as well as requiring that the segment bc iscongruent to the segment from e to ats ( abc, def ).Axiom 22: T ( a, b, c ) ∧ T ( d, e, f ) → AC ( a, b, c, d, e, ats ( abc, def )) ∧ SS ( ats ( abc, def ) , f, de ) ∧ C ( b, c, e, ats ( abc, def )).In order to fully capture Hilbert’s axiom we will need to define an angletransport opposite side construction .42 e ats ( abc, def ) f ab c Figure 18: de ats ( abc, def ) doub ( f, e ) f ab c Figure 19:Definition 12: ato ( abc, def ) ≡ ats ( a, b, c, d, e, doub ( f, e )) [See Figure 19]By axiom 12, if T ( d, e, f ), then OS ( doub ( f, e ) , f, de )Lastly we need to guaranty that angle construction are unique up to segmentcongruence, angle congruence, and angle orientation. The following axiomdoes this.Axiom 23: AC ( abc, dex ) ∧ SS ( f, x, de ) ∧ C ( bc, ex ) → x = ats ( abc, def )From these axioms one can prove the following theorem. This theorem canbe interpreted as saying that two congruent angles that have a shared initialside and both of their terminal sides on the same side of the shared initialside have their terminal sides lying on the same ray.Theorem 17: AC ( abc, abd ) ∧ SS ( c, d, ab ) → SD ( b, c, d )We can fully capture Hilbert’s properties of angle transportation and anglecongruence by additionally assuming the following axiom. This axiom states43hat an angle constructed by shortening or extending the terminal side of anangle is congruent to the original angle.Axiom 24: SD ( b, c, d ) ∧ T ( a, b, c ) → AC ( abc, abd )We will now take a quick aside in order to define inequality relations forangles.Definition 13: abc < def ≡ Int ( ats ( abc, def ) , def )Note if Int ( d, abc ), then abd < abc .For the proof of the feasibility of the crossbow construction we have need ofthe following Theorem.Theorem 18: If abc < def and def < ghi , then abc < ghi .We omit the proof of this lemma given that one only has to check that thethe correct line segment is interior to the correct angle. This simply boilsdown to using methods of proof about angle orientation that were coveredextensively in Section 4.2.In Hilbert’s fourth congruence axiom Hilbert also states that every angleis congruent to itself. We in addition will also need to state that anglecongruence is unaffected by reversing the order of triples.Axiom 25: AC ( abc, abc ) ∧ AC ( abc, cba ) Hilbert’s fifth congruence axiom is nearly identical to the following axiom.Axiom 26: AC ( abc, def ) ∧ AC ( abc, ghi ) → AC ( def, ghi )The last two axioms show that the angle congruence relation is reflexive,symmetric, and transitive. 44 .6 Hilbert’s Congruence Axiom 6 Hilbert’s sixth and final congruence axiom is a weak form of the side-angle-side congruence for triangles. The following axiom is identical to Hilbert’ssixth congruence axiom.Axiom 27: T ( a, b, c ) ∧ T ( d, e, f ) ∧ C ( ab, de ) ∧ C ( ac, df ) ∧ AC ( bac, edf ) → AC ( abc, def ) In this section we will be referring heavily to Hartshorne’s text [7], in partic-ular sections 8, 9 and 10. At this time we have captured (without the notionof a line) what is referred to as a Hilbert Plane which is a Euclidean planewithout the notion of parallel lines and circles. In Section 8 of [7] Hartshornedevelops the concept of the difference of two segments. Note the sum of twosegments was discussed above [See [7] Propositions 8.2 and 8.3.] Fortunatelyhe does not present this topic by defining operations on congruence classesof segments. His methods carry over to our system Thus his proofs that thedifferences of congruent segments are congruent can be incorporated into oursystem. In section 9 Hartshorne develops the sum and difference of angles.[See [7] Proposition 9.4 and Exercise 9.1.] In a similar fashion his conceptsand results that the sum and difference of congruent angle are congruentcarry over to our system with simple modifications. In particular one im-portant modification needed is using the SameDirection relation instead ofthe concept of rays. In this work we have not needed to invoke use of theseresults thus we not be explicitly developing them here.In the same section Hartshorne also defines and then develops the proper-ties of supplemental angles, vertical angles, right angles, and in Section 10he proves the various triangle congruence theorems such as Side-Angle-Side,Angle-Side-Angle and Side-Side-Side. We will define appropriate versions ofsupplemental angles, vertical angles, and right angles. From these defini-tions one can directly carry over Hartshorne’s theorems about these topics45s well as the triangle congruence theorems into corresponding proofs for oursystem. Thus we will not be developing the basic properties of supplemen-tal, vertical and right angles or the basic triangle congruence theorems. Wewill however provide a uniform construction of a right angle, the uniformconstruction of erecting a perpendicular, and the uniform construction ofdropping a perpendicular in the next section.Definition 14: If B ( a, b, c ) and T ( a, c, d ) we say angle dba and angle dbc aresupplementary. d ca b Figure 20:Definition 15: If B ( a, c, b ), B ( d, c, e ) and T ( a, c, d ) we say angle acd is verticalto angle bce . c ba d e Figure 21:Definition 16: Angle abc is a right angle if it is congruent to one of its sup-plements.In [7] it is proved that supplements of congruent angles are congruent, verticalangles are congruent, and all right angles are congruent.46
Midpoints, Perpendiculars, and Angle Bi-sectors
From Hilbert’s axioms there is a proof stating that given a line segment ab one can construct a point c such that triangle acb is isosceles with ac congruent to bc (Theorem 10.2 in [7]). This construction is a vital ingredientfor proving the existence of midpoints and angle bisectors. The constructiondoes have a case distinction depending on the angle congruence relationshipbetween angles dab and dba for some point d which is non-colinear with a and b . The three case distinctions are dab > dba , dab < dba , or dab ∼ = dba .This means that there is a different procedure for constructing the point c depending on this relationship. Because of this, stating such a theorem wouldbe complicated in our system. Additionally, this will violate our goal of haveonly case free constructions. TFirst we state a theorem pertaining to uniformly constructing a point whichis our analog to dropping a perpendicular from a point off a segment to thatgiven segment.Theorem 19: Given T ( a, b, c ) there is a uniform construction of a point p suchthat (cid:101) L ( a, p, b ) and apc or bpc is right. p = ac (cid:48) c b a pcc (cid:48) b Figure 22:
Proof. (Reference Figure 22.) Let T ( a, b, c ). Let c (cid:48) = ato ( abc, abc ). (Thus c (cid:48) is the point resulting from reflecting c over ab .) Let p = ( b, cac (cid:48) ). (Note thatthere is no requirement for cac (cid:48) to be an angle.) If p = a , then bpc ∼ = bpc (cid:48) and L ( c, p, c (cid:48) ). Thus we can conclude bpc is right. If p (cid:54) = a , then we can47onclude T ( c, a, p ) since (cid:101) L ( a, p, b ). Similarly we have T ( c (cid:48) , a, p ). Given how c (cid:48) was constructed, we can conclude that triangle cap is congruent to triangle c (cid:48) ap by Side-Angle-Side. Since B ( c, p, c (cid:48) ) we have apc is right.In order to provide uniform constructions of erecting a perpendicular linesegment, constructing a isosceles triangle with a given base, bisecting an an-gle, and others we will need to add only one more undefined construction.We introduce the midpoint construction denoted mid ( a, b ). Its interpreta-tion is simply a point c that is colinear with a and b where ac is congruentto bc when a and b are distinct. The following axiom captures these proper-ties. Note that we could have replace the conclusion B ( a, mid ( a, b ) , b ) with L ( a, mid ( a, b ) , b ) in the first axiom.Axiom 28: a (cid:54) = b → B ( a, mid ( a, b ) , b ) ∧ C ( ac, bc )Axiom 29: mid ( a, a ) = a We can now prove a theorem that states that one can create a right angle.Theorem 20: Angle αdβ where d = mid ( α, lf ( βγ, αβ )) is a right angle. d c γβα Figure 23:
Proof.
Recall that α , β , and γ are three constants that by Axiom 5 forma triangle. To construct our right angle we will first construct the point lf ( βγ, αβ ) = c . We then construct the point m ( α, lf ( βγ, αβ )) = d . SeeFigure 23. Note that angle αdβ and angle cdβ are supplemental. Alsotriangles αdβ and angle cdβ are congruent and by the side-side-side trianglecongruence theorem we have that the two supplemental angles which arecongruent. Thus angle αdβ is a right angle.48iven this theorem we can erect a perpendicular segment at a given point ona line segment on the same side as some point off of the line segment usingour angle transport construction. We can also construct a perpendicularbisecting segment on either side of a given segment by using the midpointconstruction and the angle transport construction. Furthermore we can provethat given a line segment ab there is a point c such that abc is a isoscelestriangle with base ab . This is accomplished by constructing a perpendicularbisector segment of ab and letting c be the endpoint of the bisector thatis not the midpoint of ab . We can also construct a segment that bisects agiven angle by first using the lay off construction to create congruent sidesof the angle and then constructing the midpoint of the segment between thetwo non-shared endpoint of these sides. The segment from the vertex of theoriginal angle to this midpoint can be proved to bisect the original angle [seefigure 24]. m ( d, c ) a lf ( ba, bc ) bc Figure 24:Lastly it should be noted that in the system devised by Tarski the construc-tion of angle bisetors, midpoints and perpendiculars relied on the construc-tions involving circles (compass constructions) and a circle circumscriptionconstruction which is logically equivalent to Eulcid’s parallel postulate. Eu-clid also invokes compass constructing for some of this constructions.
This section, although modified and extended, is based off of Suppes’ workin [14] which was heavily influenced by the work of Szmielew in [16]. In [14]Suppes defined what he called a constructive affine plane. The system wasquantifier-free and the only objects were points. The primitive constructions49ere that of finding the midpoint between two points and the (directed) dou-bling of a segment. From these constructions Suppes was able to define whentwo line segments were parallel. We will follow Suppes in his definition ofparallel lines segments, but it should be noted that that we will be extendingmay of his results since our system contains the concepts of angle orientationand angle congruence.For his constructive affine plane, Suppes assumed axioms pertaining to therelation of three points being colinear and properties of the midpoint anddoubling constructions. Suppes used an undefined non-strict linear relation.We defined this relation from a strict linear relation in definition 6.All but one of Suppes’ axioms about the midpoint and doubling constructionsare provable in our system. We list these results in the following theorem.Theorem 21:1. mid ( a, a ) = a mid ( a, b ) = mid ( b, a )3. mid ( a, b ) = mid ( a, b (cid:48) ) → b = b (cid:48) doub ( a, b ) = doub ( b, a ) → a = b doub ( a, b ) = doub ( a, b (cid:48) ) → b = b (cid:48) doub ( a, b ) = doub ( a (cid:48) , b ) → a = a (cid:48) mid ( a, doub ( a, b )) = b (cid:101) L ( a, b, mid ( a, b ))As we said above, there is one crucial axiom of Suppes that we can not provein our system at this point. At this time we have captured many of the resultspertaining to a Hilbert Plane and have thus made no reference to parallellines. If we wish to fully capture the nature of the plane and in particular theproperties of parallel line segments we will need another assumption. Giventhe objective of this paper, we cannot use an assumption like Euclid’s fifth orPlayfair’s axiom. Based off of the work of Suppes and Szmielew, we will show50hat by assuming the midpoint construction has the property of bisymmetry(which was called bicommutativity by Suppes and Szmielew) we can captureall the necessary features of parallel line segments.We assume the following axiom.Axiom 30: mid ( mid ( a, b ) , mid ( c, d )) = mid ( mid ( a, c ) , mid ( b, d ))We now define a parallelogram relation for four ordered points.Definition 17: Points a , b , c , and d form a parallelogram, denoted P ( a, b, c, d ),if and only if mid ( a, c ) = mid ( b, d ). [See Figure 25] ab c d m ( a, c ) m ( b, d ) m ( a, c ) = m ( b, d ) d ba c Figure 25:Note that we are allowing for flat parallelograms where a , b , c , and d areall pairwise colinear with the midpoint mid ( a, c ) = mid ( b, d ). (Which wasnot allowed by Suppes, but was allowed by Szmielew.) From the above def-inition one can prove the following results. We reference Suppes’ analogoustheorems, but do note that he proofs, while containing the necessary ingre-dients for proving these theorem, would need slight modifications since ourdefinition of the parallelogram relation is not identical to Suppes’.Theorem 22: P ( a, b, c, d ) → P ( c, b, a, d )Theorem 23: ( [14] Theorem 5) b (cid:54) = c ∧ P ( a, b, c, d ) → ¬ P ( a, c, b, d )Theorem 24: ( [14] Theorem 9 and 10) P ( a, b, c, d ) → P ( b, c, d, a ) ∧ P ( c, d, a, b ) ∧ P ( d, a, b, c ) ∧ P ( c, b, a, d ) ∧ P ( b, a, d, c ) ∧ P ( a, d, c, b ) ∧ P ( d, c, b, a )51he following two theorems show that given three distinct non colinear pointsone can construct a fourth point to form a parallelogram and that this pointis unique.Theorem 25: ( [14] Theorem 7) P ( a, b, c, doub ( b, m ( a, c ))) [See Figure 26.] ab c d ( b, m ( a, c )) m ( a, c ) Figure 26:Theorem 26: ( [14] Theorem 6) P ( a, b, c, d ) ∧ P ( a, b, c, d (cid:48) ) → d = d (cid:48) The next theorem will be used to show that the parallel relation between linesegments is transitive, but for now it will be stated in terms of parallelograms.Theorem 27: ( [14] Theorem 12) P ( a, b, c, d ) ∧ P ( c, d, e, f ) → P ( a, b, f, e ) [SeeFigure 27] bfae d c Figure 27:We now define a relation for when two line segments are parallel.Definition 18: ( [14] Definition 3) ab (cid:107) cd ≡ T ( a, b, c ) ∧ c (cid:54) = d ∧ (cid:101) L ( c, doub ( b, mid ( a, c )) , d )Note the definition does not allow for ab (cid:107) cd if (cid:101) L ( a, b, c ).52 b c d ( b, m ( a, c )) m ( a, c ) d Figure 28:In Theorem 16 of [14], Suppes lists a set of properties of parallel line segmentswithout providing proofs. We have unfortunately not been able to produceproofs for all of these properties from his system. Fortunately are able toprove these properties in our system with the aid of angle congruence.Theorem 28:1. T ( a, b, c ) ∧ P ( a, b, c, d ) → ab (cid:107) cd ∧ bc (cid:107) ad
2. If ab (cid:107) cd , then a , b , c , and d are all distinct.3. ab (cid:107) cd → cd (cid:107) ab ab (cid:107) cd → ba (cid:107) cd ab (cid:107) pq ∧ ab (cid:107) rs ∧ T ( p, q, r ) → pq (cid:107) rs Change of Notation:
Before proving this theorem we will make a changeof notation for the ease of the reader. We will use ab ∼ = cd in place of C ( ab, cd )and we will use abc ∼ = def in place of AC ( abc, def ). Proof.
To prove part 1, one simply has to verify the non-linearity of varioustriples of points. To prove part 2, one uses the non-colinearity of points toprove that they are distinct.To prove part 3 (see Figure 29 ), let doub ( b, mid ( c, a )) = x . Since ab (cid:107) cd , wehave (cid:101) L ( c, x, d ) and c (cid:54) = d . Now let doub ( d, mid ( c, a )) = y . We need to showthat (cid:101) L ( a, b, y ) and T ( c, d, a ) (we already know that a (cid:54) = b ).Note that ac , bx , and yd have a shared midpoint. Let this midpoint be called z . Since T ( a, b, c ), using Theorem 4 and Axiom 9 we can show that z and x abc . Thus T ( a, b, z ). By Theorem 4 we can also inferthat z is in the interior of cxa . Thus T ( c, x, a ) and T ( c, x, z ). From T ( a, b, z )we can infer that T ( a, y, z ). From T ( c, x, z ) we can infer that T ( c, d, z ) andfrom T ( c, x, a ) we can infer that T ( c, d, a ). Further more, if d (cid:54) = x and b (cid:54) = y ,then we can show that T ( d, x, z ) and T ( y, b, z ).Since we know T ( c, d, a ), if y = b , then (cid:101) L ( a, b, y ) and we can done. Thus as-sume y (cid:54) = b . If x = d by properties of midpoints and extension constructionswe can show that b = y which would be a contradiction. Thus we also have x (cid:54) = d .So we have L ( c, x, d ). Therefore one of these three points must be betweenthe other two. Without loss of generality let B ( c, x, d ). Note that angles zxc and zxd are supplemental. Using vertical angles, segment congruence, andside-angle-side, one can show that zby ∼ = zxd and zba ∼ = zxc . Note that since c and d are on opposite sides of bx , by axiom 3 and results about the same siderelation (theorems 9 and 10), we have that a and y are on opposite sides of bx . Thus OO ( zby, zba ). Note that zba and zb ( doub ( a, b )) are supplemental.Since zba ∼ = zxc we have zxd ∼ = zb ( doud ( a, b )). Thus zby ∼ = zb ( doub ( a, b ))and y and doub ( a, b ) are on the same side of bz . Therefore, by the uniquenessof angle constructions, we have SD ( b, y, doub ( a, b )). Thus L ( a, b, y ). ay bc x m ( a, c ) = z d Figure 29:To prove part 4 (See Figure 30), let doub ( b, mid ( a, c )) = x and let doub ( a, mid ( b, c )) = y . By the assumption (cid:101) L ( c, x, d ), c (cid:54) = d , and T ( a, b, c ). Note T ( b, a, c ). Wewish to show that (cid:101) L ( y, c, d ). By theorem 26 P ( c, x, a, b ) and P ( a, b, y, c ) andby theorem 27 we have P ( c, x, c, y ). Thus mid ( x, y ) = mid ( c, c ) = c . Thuswe can infer that L ( x, c, y ) and thus (cid:101) L ( y, c, d ).54 y b c x m ( a, c ) m ( b, c ) d Figure 30:To prove part 5 (see Figure 31), let doub ( b, mid ( a, p )) = x and let doub ( b, mid ( a, r )) = y . We know that (cid:101) L ( p, x, q ) and (cid:101) L ( r, y, s ). We are also assuming T ( p, q, r ).From this last assumption we can infer that T ( x, p, r ). Using part 2 of thistheorem we can also infer that r (cid:54) = s and p (cid:54) = q .Since P ( p, x, a, b ) and P ( a, b, r, y ) by Theorem 26 we have P ( p, x, y, r ) bytheorem 27. Since P ( p, x, y, r ), T ( x, p, r ), and r (cid:54) = s we have px (cid:107) rs . Bypart 3 we have rs (cid:107) px . Thus T ( s, r, p ). From this and the fact that (cid:101) L ( p, x, q )and p (cid:54) = q we have rs (cid:107) pq and by part 3 again we have pq (cid:107) rs . ys ab x qpr m ( a, p ) m ( a, r ) Figure 31:The following theorem shows that given two parallel line segments one can55reate new pairs of parallel line segments by simply extending or truncatingone or both of the original lines.Theorem 29:1. ab (cid:107) cd ∧ (cid:101) L ( c, d, x ) → ab (cid:107) cx ab (cid:107) cd ∧ (cid:101) L ( a, b, y ) → ay (cid:107) cd Proof.
Part 1 is easily inferred from the definition of parallel segments andnon-strict linearity. Part 2 can be shown by invoking part 3 of the perivoustheorem and part 1 of this theorem.Given a line segment ab and another point c , the following theorem statesthat all line segments parallel to ab with endpoint c must be colinear. Theproof is simply a reinterpretation of the definition of parallel line segments.Theorem 30: T ( c, d, doub ( b, mid ( a, c, ))) → ab ∦ cd From part 5 of Theorem 28 we can prove the following lemma.Lemma 4: If ab (cid:107) cd and ab (cid:107) rs and pq ∦ rs , then (cid:101) L ( p, q, r ).Up to this point we have been allowing for flat parallelograms. In order torefer to traditional results about parallelograms we will now define a non-flatparallelogram.Definition 19: (cid:98) P ( a, b, c, d ) ≡ P ( a, b, c, d ) ∧ T ( a, b, c )Using common well-known methods, such as triangle congruence theorems,one can prove the following.Theorem 31: The opposite sides of a non-flat parallelogram are congruentand the opposite angles of a non-flat parallelogram are congruent. (Whereopposite sides and opposite angles are defined traditionally.)56 .1 Alternate Interior Angle Theorem and its Con-verse Theorem 32: Converse of the Alternate Interior Angle Theorem ab (cid:107) cd ∧ OS ( b, d, ac ) → cab ∼ = acd xb a c dy Figure 32:
Proof.
See Figure 32. Let doub ( c, mid ( a, d )) = x and let doub ( a, mid ( b, c )) = y . By the definition of parallel segments we know that L ( b, a, x ). Similarlywe know that L ( y, c, d ).Note that Int ( mid ( a, y ) , acy ) by theorem 4. Since P ( a, b, y, c ), mid ( a, y ) = mid ( b, c ) and we can infer that Int ( mid ( b, c ) , acy ). Thus Int ( b, acy ) and SS ( b, y, ac ). Therefore OS ( y, d, ac ). We can then infer that B ( y, c, d ) for ifnot then we would contradict OS ( y, d, ac ). By a similar argument we canconclude that B ( b, a, x ).We can show (cid:98) P ( b, y, c, a ) and (cid:98) P ( c, a, x, d ). From Theorem 27 we can show (cid:98) P ( b, y, x, d ). Thus by Theorem 31 and Axiom 24 acd ∼ = axd ∼ = bxd ∼ = bx (cid:48) d ∼ = bx (cid:48) c ∼ = bac ∼ = cab .Theorem 33: Alternate Interior Angle Theorem cab ∼ = acd ∧ OS ( b, d, ac ) → ab (cid:107) cd a c dxy Figure 33:
Proof.
See Figure 33. Let doub ( a, mid ( bc )) = y . By arguments similar to oneused in the previous proof we can show SS ( b, y, ac ) and thus OS ( y, d, ac ). Let doub ( y, c ) = x . We can show SS ( x, d, ac ). Since L ( y, c, x ) and (cid:98) P ( a, b, y, c )we can infer that ab (cid:107) cx .We now what to show that ab (cid:107) cd . We will do this by showing (cid:101) L ( c, x, d )and invoking Theorem 29. We have enough to show OS ( b, x, ac ). Thereforesince ab (cid:107) cx we have cab ∼ = acx by the previous theorem. Additionally since cab ∼ = acd , we have acx ∼ = acd . Thus SS ( x, d, ac ) and acx ∼ = acd and byTheorem 17 we have SD ( c, x, d ). We can then conclude (cid:101) L ( c, x, d ) and we aredone.In standard elementary geometry the Alternate Interior Angle Theorem isprovable without invoking the parallel postulate and the converse of thestatement is proved using the original statement and the parallel postulate.Interestingly we needed to invoke our version of the parallel postulate toprove the Alternate Interior Angle Theorem by using its converse to prove it.This shows that there is indeed a striking difference between our definitionof parallel line segments and the standard definition of parallel lines.58 .2 Convex Quadrilaterals The property of convexity is an important one in planar geometry. Whatfollows is a definition for convex quadrialterials and a few results pertainingto this definition.Definition 20: We say abcd is a convex quadrilateral if
Int ( d, abc ) and Int ( a, bcd ).Theorem 34: If abcd is a convex quadrilateral then Int ( b, cda ), and Int ( c, dab ). Proof.
See Figure 34. Since
Int ( d, abc ) we have B ( a, cb ( d, abc ) , c ) and SD ( b, cb ( d, abc ) , d )by Theorem 10. Additionally we know T ( d, b, a ). Let cb ( d, abc ) = x . Since Int ( a, bcd ) we have B ( b, cb ( a, bcd ) , d ) and SD ( c, cb ( a, bcd ) , a ) by Theorem 10.We also have T ( c, d, a ). Let cb ( a, bcd ) = y . Suppose x (cid:54) = y . Then by Axiom4 we can show L ( a, b, d ). This is a contradiction. Thus x = y . Therefore B ( b, x, d ) and B ( a, x, c ). Since T ( c, d, a ) and T ( d, a, b ), by Theorem 4 wehave Int ( x, cda ) and Int ( x, dab ). Using Axiom 9 we can infer Int ( b, cda ),and Int ( c, dab ) b c adx = y Figure 34:We now state two observations that will be helpful in the following section.Observation 1: If abcd is a convex quadrilateral then SS ( c, d, ab ).Observation 2: If (cid:98) P ( a, b, c, d ), then abcd is a convex quadrilateral.Given B ( b, x, d ) and B ( a, x, c ) by using triangle congruence theorems, the-orems about vertical angles, and the Alternate Interior Angle Theorem andits converse one can prove the following statements.59 e xa b d Figure 35:Theorem 35:1. If abcd is a convex quadrilateral whose opposite sides are congruent,then ab (cid:107) cd and bc (cid:107) ad . Thus (cid:98) P ( a, b, c, d ).2. If abcd is a convex quadrilateral whose opposite angles are congruent,then ab (cid:107) cd and bc (cid:107) ad . Thus (cid:98) P ( a, b, c, d ). In the introduction we claimed that given a and c on opposite sides of bd theline segment from b to cb ( d, abc ) is less than ab or cb . We will now sketch theoutline of the proof of this fact.We have need for the following lemma.Lemma 5: If T ( a, b, c ) and abc < acb , then ac < ab .In Euclid’s The Elements this is Proposition 19. It states that in a trianglethe greater side is opposite the greater angle. The proofs of this result and theresults needed to prove this result carry over to our system without difficulty.The following theorem is the result we need to justify our claim about thefeasibility of the crossbow construction.Theorem 36: If T ( a, b, c ) and B ( a, x, c ), then bx < ba ∨ bx < bc .60 roof. See Figure 35. Let d = doub ( a, mid ( b, c )) and let e = doub ( d, b ). Firstwe want to show that cbd < xbd . Note cabd is a parallelogram and thus aconvex quadrilateral. There by Theorem 35 we have Int ( c, abd ). We alsohave Int ( x, abc ) by Axiom 4. By observing the angle orientations of variousangles, one can show Int ( c, xbd ). Thus cbd < xbd . One can then modify thesemethods to show that abe < xbe . By the converse of the Alternate InteriorAngle Theorem we know cbd ∼ = bcx , abe ∼ = bax , axb ∼ = dbx , and cxb ∼ = ebx .Thus we have xab < cxb and xcb < axb . There are three cases to consider.If xab ∼ = xcb , then xcb < cxb and by the previous lemma above bx < bc . If xab < xcb , then by the Theorem 18 above we have xab < axb . Thus bx < ba .If xcb < xab , then we have xcb < cxb . This implies bc < bc .Observation 3: In the previous proof it was shown that cab ∼ = eba and acb ∼ = dba . If one so chooses they could defined the angles eba , abc , and dbc as a socalled linear triple and observe that the angles of the triangle are congruentto these angles in a one to one correspondence. This would be our analog tothe result that the angles of a triangle sum to 180 degrees. In Section 9 we will need to apply a version of the Triangle Inequality Theo-rem. Before we state and proof such this theorem, we will need to state andprove our version of the Exterior Angle Theorem.Theorem 37: The Exterior Angle Theorem: T ( a, b, c ) → ac ( doub ( b, c )) > abc ∧ ac ( doub ( b, c )) > acb Proof.
The proof is identical to Euclid’s proof of Proposition 16 in Book Iof
The Elements . The one critical issue is to verify that certain points areon the same side of a segment when claiming that a particular angle is lessthan another. These claims are verifiable in our system and the details areomitted.Below we have stated our version of the Triangle Inequality Theorem. Thekey distinction between our statement is that we have avoided defining thesum operation on the congruence classes of line segments. Because of this, we61ave written out the details of the proof even though the proof is relativelystandard.Theorem 38: The Triangle Inequality Theorem: T ( a, b, c ) → a ( ext ( ab, bc )) > ac Proof.
Let d be the point constructed by dropping a perpendicular from b to ac . There are three cases to consider.Case 1: If B ( a, c, d ), then since adb is right by Theorem 37 abd > adb . Thus ab > ad by Lemma 5. Therefore a ( ext ( ab, bc )) > ac .Case 2: If B ( d, a, c ) then a similar argument to the previous case will show a ( ext ( ab, bc )) > ac .Case 3: If B ( a, d, c ) then by Lemma 5 ab > ad and bc > dc . We canuse the layoff construction to construct points s and t such that (cid:101) L ( s, a, c ), ds ∼ = ab > da , (cid:101) L ( t, a, c ), dt ∼ = bc > dc . We can then infer B ( s, a, d ) and B ( a, d, t ). We then have st ∼ = a ( ext ( ab, bc )). Therefore a ( ext ( ab, bc )) > ac . In this section we will derive an analogous result to the idea that two lineare parallel if and only if they are equidistant. We will also show that if twosegments are not equidistant then when one extends one of the line segmentsin one direction the segments grow more distant while growing closer orcrossing when that segment is extended the other direction.Traditionally in axiomatic geometry it is said that parallel lines are equidis-tant and that lines that are equidistant are parallel. The concept of two lines (cid:96) and (cid:96) being equidistant is formalized as follows: line (cid:96) is equidistant toline (cid:96) if for any points a and b on (cid:96) when one drops perpendiculars from a and b down to (cid:96) intersecting (cid:96) at x and y respectively ax ∼ = by .62ome modification must be made to translate this concept into our system.Importantly we can not make a statement such as ’for all points’. We cancapture this idea in our system as follows.Theorem 39: Consider points a , b , c , and d such that T ( b, c, d ) and T ( a, c, d ).Let x be the point constructed when dropping the perpendicular from a to cd and let y be the point constructed when dropping the perpendicular from b to cd . Then ab (cid:107) cd if and only if SS ( a, b, cd ) and ax ∼ = by . Proof. (= ⇒ ) Suppose that ab (cid:107) cd . By Theorem 29 we have ab (cid:107) xy . Firstwe will show that SS ( a, b, cd ). Let doub ( a, mid ( b, y )) = t . (See Figure 36.)Then abty is a non-flat parallelogram and thus ab (cid:107) xy and ab (cid:107) ty . FromTheorem 4 we can infer (cid:101) L ( y, x, t ). Since abty is a non-flat parallelogram wecan infer SS ( a, b, yt ) from Observations 1 and 2. Therefore SS ( a, b, yx ) and SS ( a, b, cd ) by Theorem 10.Now let doub ( x, mid ( a, y )) = z . Note that axyz is a non-flat parallelogram.Thus az (cid:107) xy and therefore az (cid:107) cd . Since ab (cid:107) cd and az (cid:107) cd by usingTheorem 4 we can infer (cid:101) L ( a, z, b ).Suppose b (cid:54) = z . Since az (cid:107) cd and ab (cid:107) cd we know a (cid:54) = z and a (cid:54) = b and thuswe have L ( a, z, b )Using the definition of right angles and the converse of the Alternate InteriorAngle Theorem we can show that since angle yxa is a right angle xyz isalso a right angle. From above we have SS ( a, b, xy ). Since axzy is a non-flat parallelogram we have SS ( a, z, xy ). Therefore by Theorem 9 we have SS ( z, b, xy ). Since we additionally know that xyb and xyz are both rightangles and therefore congruent, we have SD ( y, z, b ) by Theorem 17. Sincewe are supposing that b (cid:54) = z , we have L ( y, z, b ). Additionally since L ( a, z, b )we can infer L ( a, z, y ) which is a contradiction. Thus b = z . Therefore axyb is a strict parallelogram and ax ∼ = by by Theorem 35.( ⇐ =) Assume that SS ( a, b, cd ) and ax ∼ = by . We want to show that ab (cid:107) cd . Let doub ( x, mid ( a, y )) = s . (See Figure 37.) Since axys is a strictparallelogram we know SS ( a, s, xy ) and thus SS ( a, s, cd ). Also we knowthat ax ∼ = sy and thus sy ∼ = by . Furthermore, since angle yxa is rightangle, we have that xys is a right angle. Thus by the uniqueness of angle63 cd y tb = za Figure 36:construction (Axiom 23 ) we have b = s . Thus axyb is a strict parallelogramand in particular ab (cid:107) xy . Therefore ab (cid:107) cd . x cd yb = sa Figure 37:Let a , b , c , d , x , and y be defined as they were in the previous theorem with a and b on the same side of cd . If ab ∦ cd , then by that theorem we can inferthat ax (cid:54)∼ = by . Without loss of generality we can say that ax < by . We wishto show that if we extend the segment ab in the direction of a to some point e and z is the point constructed by dropping a perpendicular from e to cd ,then either a and e are on opposite sides of cd , e is co-linear with c and d ,or ez < ax . Also we wish to show that if ab is extended in the direction of b to some point g and w is the point constructed by dropping a perpendicularfrom g to cd , then gw > by .The following theorem proves these results.64heorem 40: Let a , b , c , d , x , and y be defined as they were in Theorem39 where SS ( a, b, cd ) and ax < bx . Let B ( e, a, b ) and B ( a, b, g ). Theneither OS ( a, e, cd ), (cid:101) L ( c, d, e ), or SS ( a, e, cd ). If SS ( a, e, cd ) and z is thepoint constructed from dropping a perpendicular from e to cd , then ez < ax .We can also conclude that SS ( b, g, cd ) with gw > by where w is the pointconstructed from dropping a perpendicular from w to cd . c z x y w dat sb ge Figure 38:
Proof.
To prove the first conclusion note that the following three relations aremutually exclusive: OS ( a, e, cd ), (cid:101) L ( c, d, e ), and SS ( a, e, cd ). We will supposethat a and e are not on opposite sides of cd and c , d , and e are not (non-strict)colinear. We can then infer that SS ( a, e, cd ). We now desire to show that ez < ax . Let doub ( x, mid ( a, z )) = t . (See Figure 38.) By methods used inthe proof of Theorem 39 we can infer that SD ( z, e, t ). If e = t then ae (cid:107) xz and thus ab (cid:107) cd . This is a contradiction. Thus e (cid:54) = t . So either B ( z, e, t )or B ( z, t, e ). Suppose B ( z, t, e ). By axiom 12 we have OS ( z, e, at ). Let doub ( x, mid ( a, y )) = s . We can show that SD ( y, s, b ) and since ax < by and ax ∼ = sy we know that B ( y, s, b ). Thus OS ( y, b, sa ) and therefore OS ( y, b, at ).By theorem 27 we know tzys is a non-flat parallelogram and thus SS ( y, z, at ).We then can infer SS ( e, b, at ). This contradicts axiom 12. Thus we have B ( z, e, t ) and since tz ∼ = ax we can infer that ez < ax . Using similar methodsone can show that gw > by .The following theorem shows us that as the sides of an angle are extendedthe distance between the sides increases.Theorem 41: Given T ( b, a, d ), B ( a, b, c ), bd (cid:107) ce , and L ( a, d, e ), bd < ce .65 ea b cx Figure 39:
Proof.
Given the assumptions, one can show T ( b, d, e ). Thus we can con-struct a non-flat parallelogram edbx where doub ( d, mid ( b, e )) = x . (See Fig-ure 39.) Note that (cid:101) L ( e, x, c ) since db (cid:107) ec . If x = c then bc (cid:107) de and thus ac (cid:107) de . This is a contradiction so x (cid:54) = c . Since edbx is a non-flat parallel-ogram x (cid:54) = e . Thus L ( e, x, c ). One can show T ( a, x, c ). Thus by Axiom 12we have OS ( a, c, bx ). Since bx (cid:107) de we have bx (cid:107) ae . Therefore by Theorem39 we have SS ( a, e, bx ). So OS ( e, c, bx ). From this we can conclude that B ( e, x, c ) and thus ex < ec . We know that db ∼ = ex . Thus we can concludethat db < ec . In his work
The Methods , Descartes describes a method of defining addition,subtraction, multiplication, and division of magnitudes of line segments viageometry procedures. Although we are not considering interpretations of ourconstructions as total functions or total operations, the procedures for addi-tion and subtraction can be carried out in our system by use of the extensionand laying off constructions. The necessary property of Euclidean Geometrythat proves that Descartes’ procedure for multiplication is well defined reliesof the Parallel Postulate. Our system is does not have the necessary con-structions needed to produce the required points (In Euclidean Geometrythese points are understood as the intersections of non-parallel lines.) In-terestingly Decartes’ procedure for division is accomplishable in our system.This leads to the potential for interesting philosophical discussion as to thegeometrization of arithmetic and to potential argument that multiplication66oes not satisfy feasibility properties that the other three operations do.Here we will use Descartes’ procedure for division to define a uniform con-struction for trisecting a line segment. This trisection construction alludes toa method for constructing points on a given line segment that divide the seg-ment up into n congruent parts. Our language does not encoded the naturalnumbers so a separate proof would be needed for each n . This constructionis closely related to Suppes’s trapezoid construction discussed in [14].Theorem 42: Given a line segment ab there is a uniform construction forconstructing points c and d such that ac ∼ = cd ∼ = db . y xa c d bγ γ (cid:48) γ (cid:48)(cid:48) Figure 40:
Proof.
Consider two distinct points a and b . By Axiom 5 we know thatone of the points α , β , or γ must be distinct and non-colinear to a and b . Without loss of generality let T ( a, b, γ ). Let doub ( a, γ ) = γ (cid:48) and let doub ( γ, γ (cid:48) ) = γ (cid:48)(cid:48) . (See Figure 40.) Also let doub ( γ (cid:48)(cid:48) , mid ( γ (cid:48) , b )) = x . Since bγ (cid:48)(cid:48) γ (cid:48) x is a non-flat parallelogram we have OS ( γ (cid:48)(cid:48) , x, γ (cid:48) b ). Additionally wecan show T ( a, γ (cid:48)(cid:48) , x ) and T ( γ (cid:48) , x, b ). By lemma 2 either Int ( x, bγ (cid:48) γ (cid:48)(cid:48) ) or Int ( x, aγ (cid:48) b ). If Int ( x, bγ (cid:48) γ (cid:48)(cid:48) ), then we have SS ( γ (cid:48)(cid:48) , x, γ (cid:48) b ). This is a con-tradiction. Thus Int ( x, aγ (cid:48) b ). Therefore B ( a, cb ( x, aγ (cid:48) b ) , b ) and SD ( γ (cid:48) , d, x )(Theorem 10). Let cb ( x, aγ (cid:48) , b ) = d . We now wish to show that B ( γ (cid:48) , d, x ).67e will accomplish this by showing that γ (cid:48) d < γ (cid:48) x . Note that γ (cid:48) x ∼ = γ (cid:48)(cid:48) b .Thus we want that γ (cid:48) d < γ (cid:48)(cid:48) b . This can be shown by an application of ofTheorem 41.Next we will construct the point doub ( γ (cid:48) , mid ( γ, d )) = y . Using similar meth-ods as in the previous paragraph we can construct a point c such B ( γ, c, y )and B ( a, c, d ). By the converse of the Alternate Interior Angle Theorem andTheorem 35 one can conclude aγc ∼ = dyc ∼ = γγ (cid:48) d ∼ = bxd . Give how γ (cid:48) and γ were constructed by Theorem 35 we can show aγ ∼ = γ (cid:48) γ (cid:48)(cid:48) ∼ = dy ∼ = γ (cid:48) γ (cid:48)(cid:48) ∼ = bx .Using the observation that vertical angles are congruent and the converse ofthe Alternate Interior Angle Theorem again we can show acγ ∼ = ycd ∼ = γ (cid:48) dc ∼ = xdb . Lastly, using the Angle-Angle-Side triangle congruence on triangles acγ , dcy , and bdx we can conclude that ac ∼ = cd ∼ = db . Circles and constructions involving them were central to Euclid’s
Methods .One of the earliest theorems proved in Euclid’s text was the construction ofan equilateral triangle give a base. This construction was heavily reliant onfacts about the continuity of circles which Euclid did not correctly assume.In section 11 of [7], Hartshorne discusses circle-circle continuity and line-circle continuity. These continuity properties determine when, how often,and under what conditions circles intersect circles and lines intersect circles.One version of stating circle-circle continuity is as follows: If Λ and ∆ aretwo circles and there are points a and b such that a and b are on ∆, a is insideof Λ, and b is outside of Λ then there exist two points which are on both Λand ∆. Line-circle continuity can be stated as follows: If ∆ is a circle, (cid:96) isa line, and a is a point on (cid:96) which is also inside of ∆, then there exit twopoints which are on (cid:96) and ∆.Assuming the axioms for a Hilbert plane, Hartshorne proves line-circle con-tinuity from circle-circle continuity. He also states that the equivalence ofboth over an arbitrary Hilbert plane follows from a classification theorem ofPejas of Cartesian planes over fields. Hartshorne states that he is unaware68f any direct proof. We will assume an analog to circle-circle continuity andmodify Hartshore’s proof to derive an analog to line-circle continuity.It is worth pointing out for the reader that given Hartshornes definition ofsegment inequality and it use in defining when a point is inside of a circle,Hartshorne’s statement of circle-circle continuity does not allow for a to bethe center of Λ. Furthermore this issue causes problems in his proof of line-circle continuity. Our definition of segment inequality does away with theissues of stating circle-circle continuity and our proof of line-circle continuityis both modified for our system and corrects the error of Hartshorne’s proof.In Tarski’s system a different continuity property was assumed. Possiblybecause lines are not objects in Tarski’s system a segment-circle continuityaxiom was assumed. Segment-circle continuity states that if a is a pointinside of a circle and b is a point outside of the circle, then there exists a pointbetween a and b which is on the circle. In [1] Beeson states all three continuityaxioms in the language of Tarski’s system using existential quantifiers. Hepoints out that line-circle and segment-circle continuity are equivalent inTarski’s system without invoking a parallel postulate. The proof of this factis highly non-trivial. He then shows how circle-circle continuity is derivablefrom line-circle continuity. Again the proof of this observation is complex.Beeson does not point out that line-circle continuity is derivable from circle-circle continuity. The reason for this might lie in the fact that Beeson usesline-circle continuity to define the construction of perpendiculars and usesperpendiculars to define a construction for midpoints. The construction ofmidpoints and perpendiculars are central to Hartshorne’s proof of derivingline-circle continuity from circle-circle continuity.In this section we will introduce an undefined construction for producing apoint which can be interpreted as an intersection of two circles given the con-ditions similar to those stated for circle-circle continuity above but modifiedfor the language of our system. We will then provide a uniform construc-tion of a second point of intersection. We will then derive an analog ofline-circle continuity by defining a uniform construction for producing twodistinct points which can be interpreted as the intersection of the line andthe circle. Lastly we will derive an analog of segment-circle continuity bydefining a uniform construction. The details of these results will be laid outsince the use of undefined constructions, the absence of existential quanti-fiers, and the focus on uniform constructions differs from [1] and [7] to a high69nough degree.We introduce an undefined construction circlecircleintersect ( c , a, b, c , d )which will be shortened to cci ( c , a, b, c , d ). We want cci ( c , a, b, c , d ) tohave the interpretation of a point of intersection of two circles where the firstcircle has center c and radius c a ∼ = c b , the second circle has center c withradius c d and where c a < c d ( a is inside the second circle) and c d < c b ( b is outside the second circle). [See Figure 41.] As one will note in thephysical world, there are two points that this undefined construction couldbe referring to. In the language of our system there is no distinct betweenthe two points. We will show that given one such point we can constructthe other in a uniform fashion. This strategy is needed if we wish to avoidthe use of existential quantifiers and have all of our constructions be uniformwith no case distinctions. c c bd a cci ( c , a, b, c , d )Figure 41:What follows is the only axiom pertaining to the circlecircleintersect con-struction.Axiom 31: c a ∼ = c b ∧ c a < c d ∧ c d < c b → c cci ( c , a, b, c , d ) ∼ = c a ∧ c cci ( c , a, b, c , d ) ∼ = c d Theorem 43: Given c a ∼ = c b , c a < c d , and c d < c b , there is a uni-form construction of a point x such that c x ∼ = c a , c x ∼ = c d , and x (cid:54) = cci ( c , a, b, c , d ). 70 apc c db x Figure 42:
Proof.
First note that the assumptions imply c (cid:54) = c . Suppose c = c .Since c a < c d and c a = c b , we have c b < c d . This is a contradiction.Let y = cci ( c , a, b, c , d ). It can also be shown T ( c , y, c ). Suppose ¬ T ( c , y, c ) ≡ (cid:101) L ( c , y, c ), then L ( c , y, c ). There are 3 cases to consider.Case 1: Let B ( c , y, c ). Note a (cid:54) = y and b (cid:54) = y by the uniqueness of extensionconstruction. Let d be the point uniformly constructed such that c yd is rightand SS ( a, d, c y ). We claim SS ( c , a, yd ). If (cid:101) L ( a, y, d ), then by Lemma 5 wehave c a > c y . Note c ay < c ya by the Exterior Angle Theorem (Theorem37). This is a contradiction. If OS ( c , a, yd ), then we can use the crossbowconstruction to construct a point z such that B ( c , z, a ) and (cid:101) L ( z, y, d ). Bysimilar reasoning to the previous step we have c z > c y . Since c a > c z wehave c a > c y . This is a contradiction. Therefore SS ( c , a, yd ). Similarly wecan prove SS ( c , a, yd ) since c < c d = c y . Since B ( c , y, c ) and T ( c , y, d )we have OS ( c , c , yd ). This is a contradiction. y c (cid:48) c c b d (cid:48) Figure 43:Case 2: Let B ( y, c , c ). (Reference Figure 43.) Note c y ∼ = c b and c y ∼ = c d .Recall c d < c b . Thus we can construct a point d (cid:48) such that SD ( c , d (cid:48) , b )71nd c d (cid:48) ∼ = c d . Note B ( c , d (cid:48) , b ). Similarly we can construct a point c (cid:48) such that B ( c , c (cid:48) , d ) and c c ∼ = c c (cid:48) . Notice c y ∼ = c ext ( c c , c b ) ∼ = c d (cid:48) where B ( c , d (cid:48) , b ). Thus c ext ( c , c , c b ) < c b . This contradicts the TriangleInequality Theorem (Theorem 38).Therefore we can conclude T ( c , y, c ).We may now construct the point x . Let p be the point constructed bydropping a perpendicular from y to c c and let x = doub ( y, p ). [See Figure42] We know that ypc is a right angle. Therefore ypc ∼ = xpc . We also knowthat yp ∼ = xp . Therefore by Side-Angle-Side triangle congruence theorem wehave c y ∼ = c x . Since c y ∼ = c a , we have c x ∼ = c a . By similar methods wecan show c x ∼ = c d . Lastly since x = doub ( y, p ) we have B ( y, p, x ). Thus x (cid:54) = y .We now state a prove our version of line-circle continuity.Theorem 44: Given ca < cd , c (cid:54) = d , and a (cid:54) = b , there is a uniform constructionof two distinct points x and y such that (cid:101) L ( a, b, x ), (cid:101) L ( a, b, y ), and cd ∼ = cy ∼ = cx . xaa (cid:48) pgc c (cid:48) ed by Figure 44:
Proof.
We begin by defining points satisfying the conditions of Axiom 31.Let p be the produced by dropping a perpendicular from c to ab . Let c (cid:48) = doub ( c, p ). Let e = ext ( cc (cid:48) , cd ) and g = ext ( ec (cid:48) , cd ). See Figure 44.72e will now show that cg < cd (cid:48) and ce < cd . Thus g will be ‘inside’ and e will be ‘outside of the ‘circle’ with center c and radius cd . Given properties ofthe extension construction we know B ( c, c (cid:48) , e ). Thus c and e are on oppositesides of c (cid:48) . Given how g was constructed we know B ( g, c (cid:48) , e ). Thus e and g are on opposite sides of c (cid:48) . Therefore by Theorem 15 we have SD ( c, g, c (cid:48) ).Let a (cid:48) = doub ( a, p ). Note T ( a, a (cid:48) , c ) and B ( a, p, a (cid:48) ). Thus by the Theorem36 we have cp < ca or cp < ca (cid:48) . By Side-Angle-Side triangle congruence wehave ca ∼ = ca (cid:48) . Thus cp < ca . Since ca < cd ∼ = c we have cp < cd . Since c (cid:48) p ∼ = cp we have c (cid:48) p < cd ∼ = c (cid:48) g (cid:48) . Thus B ( c (cid:48) , p, g (cid:48) ). Since B ( c (cid:48) , p, c ) we have SD ( p, g (cid:48) , c ). If g (cid:48) = c , then cg (cid:48) < cd since c (cid:54) = d . Thus either B ( c, g (cid:48) , p )or B ( g (cid:48) , c, p ). If B ( c, g (cid:48) , p ), then cg (cid:48) < cp < cd . If B ( g (cid:48) , c, p ) we can infer B ( g (cid:48) , c, c (cid:48) ) by Theorem 15 since we have SD ( c, p, c (cid:48) ). So cg (cid:48) < c (cid:48) g (cid:48) ∼ = cd .Since B ( c, c (cid:48) , e ) we have ce > c (cid:48) e ∼ = cd .Let cci ( c (cid:48) , e, g, c, d ) = x . By Axiom 31 we know cx ∼ = cd ∼ = c (cid:48) x . We wishto show T ( c, c (cid:48) , x ). If ¬ T ( c, c (cid:48) , x ), then L ( c, c (cid:48) , x ) since all three points aredistinct. Since c (cid:48) x ∼ = cd , by the uniqueness of the extension constructioneither x = e or x = g . In either case it can be shown that cx (cid:54) = cd . Thisis a contradiction. Thus T ( c, c (cid:48) , x ). By Side-Angle-Side triangle congruencewe have xpc ∼ = xpc (cid:48) . Thus xpc is a right angle. Therefore one can show L ( a, p, x ) (since a (cid:54) = p ) whether a and x are on the same or opposite side(s)of cc (cid:48) . Thus (cid:101) L ( a, b, x ).Let y = doub ( x, p ). By methods used in the proof of the previous theoremwe can show x (cid:54) = y , (cid:101) L ( a, b, y ), and cy ∼ = cd .We now prove our version of segment-circle continuity. (It should be pointedout that Tarski’s system uses a non-strict between relation to define segment-circle continuity. Thus our version differs slightly.)Theorem 45: If ca < cd and cd < cb , then there is a uniform construction ofa point z such that B ( a, z, b ) and cz ∼ = cd . Proof.
By the proof of the previous theorem we know that there is a uniformconstruction of a point x such that cx ∼ = cd . Let z = lf ( pa, px ) where p isdefined as it was in the previous proof. (It may be that z = x , but in order73o avoid case distinctions we will construct z as stated.) Since ca < cz wehave a (cid:54) = z . Thus L ( p, a, z ). If B ( p, z, a ), we would have T ( p, a, c ), cz < cp ,and cz < ca which would contradict Theorem 36. Thus B ( p, a, z ). We canthen use Theorem 15 to show SD ( a, z, b ). If B ( a, b, z ), we will have cb < ca and cb < cz which is a contradiction. Thus B ( a, z, b ).As we have seen, it is possible for two distinct points, x and y , to both beequidistant to two distinct points, c and c . The only case that has beenconsidered up to this point is when both x and y are not colinear with c and c . The following theorem states that if x is equidistant and colinear with c and c and y is some point equidistant to c and c , then x and y are thesame point. Note that x = mid ( c , c ) would satisfy the conditions for x inthe following theorem.Theorem 46: L ( c , c , x ) ∧ c x ∼ = c y ∧ c x ∼ = c y → x = y Proof.
The proof is given by Hartshorne in the first half of the proof ofProposition 11.4 in [7].There is a well known result in Euclidean Geometry that any three distinctnon colinear points, a , b , and c , lie on a unique circle with a unique centerat the intersection of the perpendicular bisectors of ab and bd . In order toconstruct this center one must invoke a parallel postulate to find the intersec-tion of the two perpendicular bisectors. Given that we classify these type ofconstructions as non feasible, we are not able to define such a construction.What we are able to prove is that three distinct points cannot be equidis-tant to two different points. First we state our analog to the result that if c is equidistant to distinct points a and b , then c lies on the perpendicularbisector of ab .Lemma 6: Let x = mid ( a, b ) and a (cid:54) = b . If ca ∼ = cb , then c = x ∨ axc is rightwhere Proof.
There are two cases to consider.Case 1) If (cid:101) L ( a, c, b ), then L ( a, c, b ) given that a (cid:54) = b . Thus c = x by theproperties of the midpoint construction and the uniqueness of the extensionconstruction. 74ase 2) If T ( a, c, b ), then let y be the point constructed by dropping a per-pendicular from c to ab . First we will show B ( a, y, b ). Suppose ¬ B ( a, y, b ).Then either B ( y, a, b ) or B ( a, b, y ). Without loss of generality let B ( y, a, b ).By Lemma 5 ca > cy and by Theorem 36 cy > ca or cb > ca . Therefore, cb > ca . This is a contradiction. Thus B ( a, y, b ). Since ca ∼ = cb we know cab ∼ = cba . We can then use Angle-Angle-Side triangle congruence (prooffound in Euclid’s Proposition 26 in Book I) to show ya ∼ = yb . We can theninfer x = y .Theorem 47: Let a (cid:54) = b , a (cid:54) = d , and b (cid:54) = d . Then c a ∼ = c b ∼ = c d ∧ c a ∼ = c b ∼ = c d → c = c . Proof.
Suppose c (cid:54) = c . Let x = mid ( a, b ) and let y = mid ( b, d ). By theprevious lemma c xa , c yd , c xa , and c yd are all right. Also note that x (cid:54) = y .There are 3 cases to consider.Case 1) Let c and c both not equal x or y . By the uniqueness of angleconstructions we have L ( x, c , c ) and L ( y, c , c ). Thus L ( x, y, c ) Therefore, yxa and xyd are both right. Thus by the Alternate Interior Angle Theorem, ab (cid:107) bd . This is a contradiction.Case 2) Let c equal x or y . Without loss of generality let c = x . Note L ( a, c , b ) . Therefore yc a and c yd are both right. Thus by the AlternateInterior Angle Theorem, ab (cid:107) bd . This is a contradiction.Case 3) Let c equal x or y . This case is analogous to the previous case.Therefore c = c .
10 Coplanar Relation
In this section we aim to develop an analog to part of Book XI of Euclid’sElements in which solid geometry is discussed. Similar to how we avoidintroducing lines as objects by introducing a colinear relation, we will avoiddiscussing planes directly by introducing a coplanar relation. We introduce75he relation
P L ( a, b, c, d ), with a condensed written form of P L abc ( d ), to havethe interpretation that d is coplanar with a , b , and c where a , b , and c arenot colinear. Additionally we have the need to introduce a new constructioncalled the orthogonal construction which is denoted o ( a, b, c ). The point o ( a, b, c, ) (where a (cid:54) = b and c (cid:54) = b ) is understood to be a point such that theangles abo ( a, b, c ) and cbo ( a, b, c ) are right angles. To display the feasibility ofthis construction we have designed a tool we call the orthogonator which willproduce such a point given three points a , b and c (where a (cid:54) = b and c (cid:54) = b ).In figure 45, we have displayed a diagram of such a tool. The instrument hasthree segments all with a common endpoint. The segments 1 and 3 form a(fixed) right angle. Segment 2 also forms a right angle with segment 3 whilebeing able to swing a full rotation is such a way that segments 1 and 2 actmuch like the hands of a clock (where segment 1 is fixed). The length of thethree segments is not stipulated. Label the endpoints of the first segment y and x . Label the endpoints of the second segment y and z . Label theendpoints of the third segment y and o . The tool is put into use by placing y at b , placing x at lf ( ba, yx ), and placing z at lf ( bc, yz ). The resultingposition of endpoint o is the desired point. y = bo a x = lf ( ba, yx ) c z = lf ( bc, yz )Figure 45:We have five axioms pertaining to coplanar relation and the orthogonal con-struction.Axiom 32: P L abc ( d ) → T ( a, b, c )Axiom 33: T ( a, b, c ) → P L abc ( b )Axiom 34: P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ) ∧ P L abc ( x ) → P L def ( x )76xiom 35: a (cid:54) = b ∧ c (cid:54) = b → abo ( a, b, c ) is right ∧ cbo ( a, b, c ) is right.Axiom 36: T ( a, b, c ) ∧ dbo ( a, b, c ) is right → P L abc ( d )Axiom 32 states that for d to be coplanar with a , b and c , a , b and c mustnot be colinear. Axiom 33 simply states that if a , b and c are non colinear,then b is coplanar with a , b and c . Axiom 34 states that if three non colinearpoints d , e , and f are all coplanar with a , b and c and x is a point whichis coplanar with a , b , and c then x is coplanar with d , e and f . Axiom 35states that if a (cid:54) = b and c (cid:54) = b then the segment b ( o ( a, b, c )) is perpendicularto both ab and cb . Lastly, Axiom 36 states that if a , b and c are non colinearand segment bd is perpendicular to segment b ( o ( a, b, c )), then d is coplanarwith a , b , and c .Surprisingly there are minimal modifications that need to be made to pre-vious axioms and theorems to fully incorporate the results from previoussections about planar geometry into the setting of solid geometry. Before weintroduce those modifications, we will work out a set of results which followwithout these modifications.Hilbert’s fourth through eighth incidence axioms deal with planes. His fourthand fifth axioms state that three non colinear points determine a plane andthat only one plane can contain these three points. Our use of a coplanarrelation and Axiom 32 will account for analogous features. Hilbert’s sixthincidence axiom states that if two points of a line are contained in a planethen the entire line is contained in the plane. We will prove an analogousresult in Theorem 51. Hilbert’s seventh incidence axiom states that if two(distinct) planes have a point in common, then they have another point incommon. This axiom of Hilbert’s is nonconstructive. In Theorem 53 we willprovide a uniform construction for creating such a point. Lastly, Hilbert’seighth incidence axiom states that there exist at least four points not lyingin the same plane. In Theorem 52 we will prove that for three non colinearpoints a , b and c (in particular α , β and γ ), o ( a, b, c ) is not coplanar with a , b and c .Euclid did not clearly state all the assumptions he made about planes. Mean-while Hilbert did not develop many results pertaining to solid geometry fromhis axioms. In this section we will develop numerous results analogous to77ropositions of Euclid’s while also proving results analogous to Hilbert’s ax-ioms.Theorem 48: T ( a, b, c ) → P L abc ( a ) ∧ P L abc ( c ) Proof.
An application of Axiom 35 followed by an application of Axiom 36.Theorem 49:
P L abc ( x ) → P L bac ( x ) ∧ P L cba ( x ) Proof.
An application of Axiom 32 to obtain T ( a, b, c ). Then apply Axiom33 and Theorem 48 followed by Axiom 34.The previous theorem implies that we can also obtain P L acb ( x ), P L bca ( x )and P L cab ( x ) given P L abc ( x ). Thus we obtain all permutations of a , b and c for the coplanar relation notation.Theorem 50: P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ) → [ P L abc ( x ) ↔ P L def ( x )] Proof.
Assume
P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ).(= ⇒ ) Given P L abc ( x ), we obtain P L def ( x ) by an application of Axiom 34.( ⇐ =) Given the assumptions above we obtain P L abc ( a ), P L abc ( b ) and P L abc ( c )by Axiom 33 and Theorem 48. From Axiom 34 we have P L def ( a ), P L def ( b )and P L def ( c ) By Axiom 32 we also have T ( a, b, c ). Thus another applicationof Axiom 34 give us P L def ( x ) → P L abc ( x ).The next Theorem is our analog to Hilbert’s axiom that if two points of aline are contained in a plane then the entire line is contained in the plane.Theorem 51: P L abc ( d ) ∧ P L abc ( e ) ∧ L ( d, e, f ) → P L abc ( f ) Proof.
By Axiom 32 we have T ( a, b, c ). Because of this we have T ( a, d, e ), T ( b, d, e ) or T ( c, d, e ). Without loss of generality let T ( a, d, e ). By Axiom 35,angle edo ( a, d, e ) is right. Since L ( d, e, f ), note f (cid:54) = d , f do ( a, d, e ) is right.Thus by Axiom 36 we obtain P L ade ( f ). Lastly, since P L abc ( a ), P L abc ( d ), P L abc ( e ), T ( a, d, e ) and P L ade ( f ) we have P L abc ( f ) by Axiom 34.78 We will now discuss the modifications that are needed to lift our system fromone about planar geometry to one about solid geometry. Surprisingly, thereare only two modifications that need to be made before compass constructionsare introduced. We modify Axiom 6 by stimulating that angles abc and def not only be non colinear triples (as is stated in the original version), but alsothat d , e , and f are also coplanar with a , b , and c . In a similar vein, westipulate that d , e , and f are also coplanar with a , b , and c in the Definition4.Modified Axiom 6: SO ( abc, def ) → T ( a, b, c ) ∧ T ( d, e, f ) ∧ P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f )Modified Definition 4: OO ( abc, def ) ≡ ¬ SO ( abc, def ) ∧ T ( a, b, c ) ∧ T ( d, e, f ) ∧ P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f )From these two modification many statements in Sections 2 through 8 inher-ently imply that the points spoken about are coplanar with the same triple.For example if two points are said to be on the same side of a segment, thenthe definition of same side will imply that they are coplanar. If a point issaid to be in the interior of an angle, then the definition of interior will implythat the point is coplanar with the angle. Coplanar relations can be shownin the statments of Alternate Interior Angle Theorem and its converse. Onecan also verify that the constructions discussed are producing points whichare coplanar with the starting points. For example the ats construction’sdefinition implies this. Theorem 51 is critical in justifying certain coplanarrealtions. We can use Theorem 51 to make the following observation.Observation about Definition 18: ab (cid:107) cd → P L abc ( d )In Section 9 we have need to modify both the premise and conclusion ofAxiom 31 to state that all points introduced and constructed are coplanarwith the same triple. This fact will be used in the proof of Theroem 53.Modified Axiom 31: c a ∼ = c b ∧ c a < c d ∧ c d < c b ∧ P L ac b ( d ) ∧ P L ac b ( c ) → c cci ( c , a, b, c , d ) ∼ = c a ∧ c cci ( c , a, b, c , d ) ∼ = c d ∧ P L ac b ( cci ( c , a, b, c , d ))We can then add P L ac b ( x ) to the conclusion of Theorem 41.79odified Theorem 41: Given c a ∼ = c b , c a < c d , and c d < c b , there isa uniform construction of a point x such that c x ∼ = c a , c x ∼ = c d , and x (cid:54) = cci ( c , a, b, c , d ). We will now be assuming the modifications stated above in our proofs. Firstwe will state and prove a theorem that implies that there are four points forwhich one is not coplanar with the other three.Theorem 52: T ( a, b, c ) → ¬ P L a,b,c ( o ( a, b, c )) [In particular ¬ P L α,β,γ ( o ( α, β, γ )).] Proof.
Let o = o ( a, b, c ). Suppose ¬ P L abc ( o ). Since T ( a, b, c ) we have a (cid:54) = b and c (cid:54) = b . Thus by Axiom 35 we have abo is right and cbo is right. Since abo is right we have T ( o, b, a ). Additionally, since P L abc ( o ) we can infer P L abc ( o ), P L abc ( b ), and P L abc ( a ). Likewise we can obtain T ( o, b, c ), P L abc ( o ), P L abc ( b ),and P L abc ( c ). From Modified Axiom 6 and Modified Definition 4 we canconclude SS ( oba, obc ) or OS ( oba, obc ). Therefore SS ( a, c, ob ) or OS ( a, c, ob ).Since both abo and cbo are right, by the uniqueness of angle constructionswe have L ( a, b, c ). This is a contradiction. Thus ¬ P L abc ( o ).The following theorem states that given a point which is coplanar with a , b and c and d , e , and f there is a uniform construction for producing anotherpoint which is also coplanar with a , b and c and d , e , and f . This is ouranalog for Hilbert’s seventh incidence axiom.Theorem 53: Given P L abc ( x ) and P L def ( x ) there is a uniform constructionof a point p ( abc,def,x ) such that P L abc ( p ), P L def ( p ) and p (cid:54) = x where p = p ( abc,def,x ) . Proof. (As was stated above the constructions used will result in certaincoplanar relations. We will clearly state this relations.)We will being by uniformly constructing two points u and v such that T ( x, u, v ), P L abc ( u ) and P L abc ( v ). Let y be a point uniformly constructed such that P L abc ( y ), aby is right and by = ab . Let u = ext ( ba, xy ). Note P L abc ( u ) byTheorem 51. We will show that x (cid:54) = u .80 hb x = ua Figure 46:Suppose x = u = ext ( ba, xy ).(Reference Figure 46.) Note ax ∼ = yx and B ( b, a, x ). Since triangle aby is isosceles, bay ∼ = bya . Similarly xay ∼ = xya .Note yab is supplemental to yax . Let h = doub ( b, t ). Note P L abc ( h ) by The-orem 51. By Axiom 6 we have OS ( b, h, da ). We also have ayb supplementalto ayh with ayb ∼ = yab . Thus ayx ∼ = ayh . By the uniqueness of angle con-structions we then have L ( t, h, x ) and thus L ( b, y, x ). From this we can inferthat L ( b, a, y ). This contradicts how y was constructed. Therefore x (cid:54) = u and with P L abc ( u ).Let v = cci ( x, xu, u, ux ). We can show P L abc ( x ), P L abc ( u ), P L abc ( v ) and T ( x, u, v ). In a similar fashion we can uniformly construct points s and t such that P L def ( x ), P L def ( s ), P L def ( t ) and T ( x, s, t ).Let p ( abc,def,x ) = o ( o ( uxv ) , x, o ( sxt )). By Axiom 35, pxo ( uxv ) is right and pxo ( sxt ) is right. Therefore by Axiom 36 we have P L uxv ( p ) and P L sxt ( p ).Via an application of Axiom 34 we can obtain P l abc ( p ) and P L def ( p ). Lastly,since pxo ( uxv ) is right we know p (cid:54) = x .The following theorem is our analog to the statement that distinct planesthat intersect have a line as their intersection (Proposition 3 of Book XI).Theorem 54: P L abc ( x ) ∧ P L def ( x ) ∧ P L abc ( y ) ∧ P L def ( y ) ∧ ¬ P L abc ( d ) → (cid:101) L ( x, y, p ( abc,def,x ) ) Proof.
Let p = p ( abc,def,x ) . Suppose ¬ (cid:101) L ( x, y, p ). Then T ( x, y, p ) (note P L abc ( x ), P L abc ( y ) and P L abc ( p )). Since P L def ( x ), P L def ( y ), P L def ( p ) and P L def ( d )81e have P L xyp ( d ) by Axiom 34. Since P L abc ( x ), P L abc ( y ), P L abc ( p ) and P L xyp ( d ), we have P L abc ( d ) by Theorem 50. This is a contradiction. Thus (cid:101) L ( x, y, p ).We will need the following two lemmas in order to prove the next theorem.Lemma 7: If abc and a (cid:48) b (cid:48) c (cid:48) are vertical angles [ B ( a, b, a (cid:48) ), B ( c, b, c (cid:48) ), and T ( a, b, c )], T ( a, b, x ) and T ( c, b, x ), then Int ( x, abc ), Int ( x, abc (cid:48) ), Int ( x, a (cid:48) bc ),or Int ( x, a (cid:48) bc (cid:48) ). Proof.
Note T ( x, b, a ) and T ( c, b, a ). Thus either SS ( x, c, ba ) or OS ( x, c, ba ).Likewise, since T ( x, b, c ) and T ( a, b, c ), either SS ( x, a, bc ) or OS ( x, a, bc ).There are then four possibilities:Case 1: Suppose SS ( x, c, ba ) and SS ( x, a, bc ). By definition we Int ( x, a, bc ).Case 2: Suppose OS ( x, c, ba ) and SS ( x, a, bc ). Note SS ( x, a, bc (cid:48) ) by Theorem10. Also note that OS ( c (cid:48) , c, ba ) by Axiom 12. Thus, by Theorem 12 we have SS ( x, c (cid:48) , ba ). By definition we then have Int ( x, abc (cid:48) ).Case 3: Suppose SS ( x, c, ba ) and OS ( x, a, bc ). Using similar methods to theprevious case we can show SS ( x, c, ba (cid:48) ) and SS ( x, a (cid:48) , bc ) and this infer that Int ( x, a (cid:48) bc ).Case 4: Suppose OS ( x, c, ba ) and OS ( x, a, bc ). Then we have SS ( x, c (cid:48) , ba )and SS ( x, a (cid:48) , bc ) which implies SS ( x, c (cid:48) , ba (cid:48) ) and SS ( x, a (cid:48) , bc (cid:48) ). Thus we ob-tain Int ( x, a (cid:48) bc (cid:48) ).Lemma 8: If abc and a (cid:48) b (cid:48) c (cid:48) are vertical angles [ B ( a, b, a (cid:48) ), B ( c, b, c (cid:48) ), and T ( a, b, c )], Int ( x, abc ) and B ( x, b, x (cid:48) ), then Int ( x (cid:48) , a (cid:48) bc (cid:48) ). Proof.
Since
Int ( x, abc ) we have SS ( x, a, bc ) and SS ( x, c, ba ). By methodsuse in the previous proof we know SS ( x, a, bc ) implies OS ( x (cid:48) , a, bc ). Whichin turn implies SS ( x (cid:48) , a (cid:48) , bc ). Which finally implies SS ( x (cid:48) , a (cid:48) , bc (cid:48) ). Similarly SS ( x, c, ba ) implies SS ( x (cid:48) , c (cid:48) , ba (cid:48) ). Thus Int ( x (cid:48) , a (cid:48) bc (cid:48) ).82e now turn our attention to the interplay between the orthogonal construc-tion and our version of parallel segments. In particular we will be focusing onresults related to Propositions 4 through 8 and Proposition 13 for Book XIof The Elements . The following theorem has a corollary which is a converseof sorts to Axiom 35 and is closely related to Proposition 4 from Book XI ofEuclid’s Elements.Theorem 55: abd is right ∧ cbd is right ∧ P L abc ( x ) ∧ x (cid:54) = b → xbd is right Proof. If (cid:101) L ( b, a, x ) or (cid:101) L ( b, c, x ) then we are done. Suppose ¬ (cid:101) L ( b, a, x ) and ¬ (cid:101) L ( b, c, x ). Let e = lf ( ba, bc ). Since P L abc ( x ) we have T ( a, b, c ). Thus, T ( e, b, c ). Let e (cid:48) = doub ( eb ) and c (cid:48) = doub ( cb ). Then by Lemma 7 wehave Int ( x, ebc ) ∨ Int ( x, e (cid:48) bc ) ∨ Int ( x, cbc (cid:48) ) ∨ Int ( x, e (cid:48) bc (cid:48) ). Without loss ofgenerality let Int ( x, ebc ). Also let x (cid:48) = doub ( xb ). by Lemma 8 Int ( x (cid:48) , e (cid:48) bc (cid:48) ).Let y = cb ( x, ebc ) and y (cid:48) = cb ( x (cid:48) , e (cid:48) bc (cid:48) ).To summarize we have P L abc ( c ), P L abc ( c (cid:48) ), P L abc ( e ), P L abc ( e (cid:48) ), P L abc ( y ), P L abc ( y (cid:48) ), P L abc ( b ), ebc is vertical to e (cid:48) bc (cid:48) , B ( y, b, y (cid:48) ), B ( c, y, e ), B ( c (cid:48) , y (cid:48) , e (cid:48) ),and ebd , cbd , e (cid:48) bd , c (cid:48) bd are all right. The proof can then be finished off byslightly modifying Euclid’s proof of Proposition 4 from Book XI.From the previous theorem we obtain the following corollary.Corollary 1: P L abc ( x ) ∧ x (cid:54) = b → xbo ( abc ) is rightEuclid’s Proposition 5 from Book XI is captured by our Axiom 36. The nextresult is analogous to Proposition 13 in Book XI. (It can be noted that Euclidshould have proved Proposition 13 before Proposition 6.)Theorem 56: T ( a, b, c ) ∧ abd is right ∧ cbd is right → (cid:101) L ( b, d, o ( abc )) Proof.
Let o = o ( abc ) and suppose ¬ (cid:101) L ( b, d, o ) ≡ T ( b, d, o ). Let p = p ( abc,obd,b ) .Since P L abc ( p ), Theorem 55 implies pbd is right. Likewise by Corollary 1 wehave pbo is right. Since P L obd ( b ), P L obd ( d ), P Lobd ( o ), and P L obd ( p ) with pbd is right and pbo is right and ¬ (cid:101) L ( b, d, o ) we have a contradiction. Thus (cid:101) L ( b, d, o ( abc )). 83he following Theorem is our analog to Proposition 6 from Book XI. Thisproof is an analogous version of Euclid’s proof of Proposition 6, but the detailsof the proof are provided given there is enough of a distinction between ourproof and Euclid’s.Theorem 57: P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ) → bo ( abc ) (cid:107) eo ( def ) bo g o (cid:48) e Figure 47:
Proof. (Reference Figure 47.) Let g be the point uniformly constructed sothat beg is right, eg ∼ = ab , and P L abc ( g ). Let o ( abc ) be o and let o ( def ) be o (cid:48) .By Corollary 1 gbo and ebo are right. By Axiom 34 P L def ( b ) and P L def ( g ).Thus beo (cid:48) and geo (cid:48) are right. Since bo ∼ = eg , eb ∼ = be , and obe ∼ = geb , anapplication of Side-Angle-Side implies triangle ebo is congruent to triangle beg . We can then infer eo ∼ = bg . Since bg ∼ = eo , ob ∼ = ge , and go ∼ = og , andapplication of Side-Side-Side implies triangle obg is congruent to triangle geo . We can then infer obg ∼ = geo . Therefore geo is right. By the previoustheorem, we have (cid:101) L ( e, g, o ( beo (cid:48) )). Since oeg is right, we have oeo ( beo (cid:48) ) is right.Therefore P L beo (cid:48) ( o ). By the Alternate Interior Angle Theorem [Theorem 33],we have bo (cid:107) bo (cid:48) .In Book XI Euclid uses Proposition 7 to prove Proposition 8. The analog ofProposition 7 which is needed for us to prove out analog of Proposition 8 isthe observation we made about Definition 18 when discussing modificationsearlier in this section. This observation was the following implication: ab (cid:107) cd → P L abc ( d ). The next Theorem is our analog to Proposition 8.Theorem 58: P L abc ( d ) ∧ bo ( abc ) (cid:107) de → [ P L abc ( x ) → xde is right]84 o f ed Figure 48:
Proof. (Reference Figure 48.) Let o = o ( abc ). Since P L abc ( d ), by Corollary 1we have dbo is right. By the observation made about Definition 18 (discussedbefore the statement of the theorem) and the Converse of the AlternateInterior Angle Theorem [Theorem 32] we can infer bde is right also.Similar to the construction used in the proof of the previous theorem, con-struct a point f such that bdf is right, df ∼ = bo , and P L abc ( f ). Note f bo isright. Using similar reasoning as in the proof of the previous theorem, wecan use Side-Angle-Side to conclude triangle f db is congruent to triangle obf ,thus obtaining do ∼ = bf . We can then use Side-Side-Side to conclude triangle obf is congruent to triangle f do , thus obtaining f bo ∼ = f do . Therefore f do is right. By the observation made about Definition 18 we know P L dbo ( e ).Therefore by Theorem 55 we have f de is right.Since P L abc ( b ), P Labc ( f ), P L abc ( d ) and T ( b, f, d ) by Axiom 34 we have P L bdf ( x ). Since bde and f de are right, by Theorem 55 we have xde is right.Lastly we have a theorem which states something analogous to the idea thatif two planes have parallel normals and share a point in common then theyare the same plane.Theorem 59: bo ( abc ) (cid:107) eo ( def ) ∧ P L abc ( x ) ∧ P L def ( x ) → [ P L abc ( y ) → P L def ( y )] Proof.
Since T ( a, b, c ), then T ( a, b, x ) or T ( a, x, c ) or T ( x, b, c ). Withoutloss of generality let T ( a, x, c ). Likewise since T ( d, e, f ), then T ( d, e, x ) or85 ( d, x, f ) or T ( x, e, f ). Without loss of generality let T ( d, x, f ). By Theo-rem 57 bo ( abc ) (cid:107) xo ( axc ) and eo ( def ) (cid:107) xo ( dxf ). Note xo ( axc ) ∦ xo ( dxf )since x is an endpoint of both segments. Therefore by Lemma 4 we have (cid:101) L ( x, o ( axc ) , o ( dxf )). Since P L abc ( y ) by Corollary 1 we have ybo ( abc ) is right.By Theorem 58 we have yxo ( axc ) is right. Since (cid:101) L ( x, o ( axc ) , o ( dxf )), wehave yxo ( dxf ) is right. By invoking Theorem 58 we have yxo ( def ) is right.Therefore P L def ( y ).From this theorem one can prove the following Corollary.Corollary 2: bo ( abc ) (cid:107) eo ( def ) ∧ P L abc ( x ) ∧ (cid:54) P L def ( x ) → ¬ [ P L abc ( y ) ∧ P L def ( y )] In Theorem 7 of [8], Hilbert states a separation property for space in whicha plane partitions all points not on the plane into two sets called sides. Twopoints are said to be on the same side of the plane if the segment betweenthem does not intersect the plane and are said to be on opposite sides if thesegment between them does intersect the plane. We will develop our analogresults by first defining what it will mean for two points to be on oppositesides of three non coplanar points a , b , and c . This approach differs from howangle orientation and sides of a line were defined in Section 4 where sameorientation/side was defined before opposite orientation/side.Definition 21: OS ( x, y, abc ) ≡ T ( a, b, c ) ∧ T ( x, b, y ) ∧¬ P L abc ( x ) ∧¬ P L abc ( y ) ∧ OS ( x, y, bp ) (where p = p ( abc,xby,b ) )We then define what is meant by two points being on the same side of threenon coplanar points in terms of the previous definition.Definition 22: SS ( x, y, abc ) ≡ T ( a, b, c ) ∧¬ P L abc ( x ) ∧¬ P L abc ( y ) ∧¬ OS ( x, y, abc )Note that is may be the case (cid:101) L ( x, b, y ).The following theorem shows that the coplanar relation in some sense pre-serves the planar opposite sides relation.86heorem 60: P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ) ∧ OS ( x, y, abc ) → OS ( x, y, def ) Proof.
We need to show ¬ P L def ( x ) and ¬ P L def ( y ). If these were false, thenwe could use Theorem 34 to show ¬ P L abc ( d ) and ¬ P L abc ( e ) which would bea contradiction. Thus all we have left to show is T ( x, e, y ) and OS ( x, y, ep (cid:48) )where p (cid:48) = p ( def, xey, e ).Since OS ( x, y, abc ), we have OS ( x, y, bp ). Let z = cb ( p, xby ). Then B ( x, z, y )and (cid:101) L ( b, z, p ). Since (cid:101) L ( b, z, p ), by Theorem 51 we have P l abc ( z ). By Axiom34 we obtain P L def ( z ). We claim ¬ (cid:101) L ( x, z, e ).If (cid:101) L ( x, z, e ), then P L def ( x ) by Theorem 51. This is a contradiction. Thuswe have ¬ (cid:101) L ( x, z, e ). This implies T ( x, e, y ) and by Axiom 12 we also have OS ( x, y, ez ).Note P L xey ( z ) since L ( x, z, y ). Also recall P L def ( z ). Thus by Theorem 51we have (cid:101) L ( e, p (cid:48) , z ). Then by Theorem 13 we obtain OS ( x, y, ep (cid:48) ). Finally byDefinition 21 we have OS ( x, y, def ).We now justify an analogous theorem for the planar same side relation.Theorem 61: P L abc ( d ) ∧ P L abc ( e ) ∧ P L abc ( f ) ∧ T ( d, e, f ) ∧ SS ( x, y, abc ) → SS ( x, y, def ) Proof.
By Theorem 50 we have
P L def ( a ), P L def ( b ) and P L def ( c ). By Axiom32 we have T ( a, b, c ). Suppose OS ( x, y, def ). Then by the previous theoremwe would have OS ( x, y, abc ). This is a contradiction. Thus ¬ OS ( x, y, def ).Therefore by Definition 21 we have SS ( x, y, def ).The two previous theorem are planar analogs for Theorem 10 and Theorem13. The following Theorem is an analog to Theorem 12.Theorem 62:1. OS ( x, y, abc ) → OS ( y, x, abc )2. OS ( x, y, abc ) → OS ( x, y, bac ) ∧ OS ( x, y, cba )87. OS ( x, y, abc ) ∧ OS ( y, z, abc ) → SS ( x, z, abc ) Proof.
Part 1) is justified by part 1) of Theorem 12. Part 2) is justified withAxiom 32, Axiom 33, Theorem 48, Theorem 49 and Thereom 60.The proof of Part 3) is a bit more work. we have T ( a, b, c ) from OS ( x, y, abc ).Also since OS ( x, y, abc ) we have ¬ P L abc ( x ). Similarly since OS ( y, z, abc )we have ¬ P L abc ( z ). Thus all we need to prove is ¬ OS ( x, z, abc ). Suppose OS ( x, z, abc ). Since OS ( x, z, abc ), we have T ( x, b, z ) and OS ( x, z, bp ( abc,xbz,b ) ).Likewise we have T ( x, b, y ), OS ( x, y, bp ( abc,xby,b ) ), T ( y, b, z ), and OS ( y, z, bp ( abc,ybz,b ) ).Let u = cb ( p ( abc,xby,b ) , xby ), v = cb ( p ( abc,ybz,b ) , ybz ), and w = cb ( p ( abc,xbz,b ) , xbz ).There are two cases to consider.Case 1) Let T ( x, y, z ). Since B ( x, u, y ), B ( y, v, z ), and B ( x, w, z ), we have P L xyz ( u ), P L xyz ( v ), P L xyz ( w ) by Theorem 51. One can also show P L abc ( u ), P L abc ( v ), P L abc ( w ). Thus by Theorem 54 we can infer (cid:101) L ( u, v, w ). If u = v , u = w or v = w , then ¬ T ( x, y, z ). This is contradiction. If u (cid:54) = v , u (cid:54) = w ,and v (cid:54) = w , then by Lemma 3 we have T ( u, v, w ). This is a contradiction.Thus ¬ OS ( x, z, abc ).Case 2) Let (cid:101) L ( x, y, z ). By Theorem 51 we can infer P L xby ( z ). One can theninfer that b , u , v , and w are all non-strict colinear by Theorem 54. By usingAxiom 4 we can conclude u = v = w . Using Theorem 12 and Theorem13 we can make the following implications: OS ( x, y, bw ), OS ( y, z, bw ), and SS ( x, z, bw ). But this would imply SS ( x, z, bp ( abc,xbz,b ) ) which is a contradic-tion.The next Theorem is an analog to Theorem 9.Theorem 63:1. T ( a, b, c ) ∧ ¬ P L abc ( x ) → SS ( x, x, abc )2. SS ( x, y, abc ) → SS ( y, x, abc )3. SS ( x, y, abc ) → SS ( x, y, bac ) ∧ SS ( x, y, cba )4. SS ( x, y, abc ) ∧ SS ( y, z, abc ) → SS ( x, z, abc )88 roof. To justify part 1) we only need to show ¬ OS ( x, x, abc ). Note that ¬ T ( x, x, b ). Thus ¬ OS ( x, x, abc ). The proofs of part 2) and 3) are omitted.Thus we turn to the proof of part 4). Since SS ( x, y, abc ) we have T ( a, b, c ).Also since SS ( x, y, abc ) we have ¬ P L abc ( x ) and since SS ( y, z, abc ) we have ¬ P L abc ( z ). Thus we only need to show ¬ OS ( x, z, abc ). We consider fourcases.Case 1) Let (cid:101) L ( x, y, b ) and (cid:101) L ( y, z, b ). One can show (cid:101) L ( x, b, z ). Thus ¬ T ( x, b, z )and therefore ¬ OS ( x, z, abc ).Case 2) Let (cid:101) L ( x, y, b ) and ¬ (cid:101) L ( y, z, b ). Since T ( y, b, z ), we can construct p ( abc,ybz,b ) . Given SS ( y, z, abc ) we can infer SS ( y, z, bp ( abc,ybz,b ) ). Since (cid:101) L ( x, y, b )we can conclude SS ( x, z, bp ( abc,ybz,b ) ) and P L yzb ( x ). Given P L yzb ( x ) we canconclude P L yzb ( p ( abc,xzb.b ) ) and by using Theorem 54 infer (cid:101) L ( b, p ( abc,ybz,b ) , p ( abc,xzb.b ) ).Therefore we have SS ( x, z, bp ( abc,xzb.b ) ) which implies ¬ OS ( x, z, abc ).Case 3) Let ¬ (cid:101) L ( x, y, b ) and (cid:101) L ( y, z, b ). The justification is similar to Case 2.Case 4) Let ¬ (cid:101) L ( x, y, b ) and ¬ (cid:101) L ( y, z, b ). If L ( x, z, b ) then ¬ OS ( x, z, abc ) andwe are done. Suppose ¬ L ( x, z, b ) ≡ T ( x, b, z ). Note we can construct the fol-lowing points: p xy = p ( abc, xby, b ), p yz = p ( abc,ybx,b ) , and p xz = p ( abc,xbz,b ) . Alsoobserve that we have SS ( x, y, bp xy ) and SS ( y, z, bp yz ). Let u = cb ( p xz , xbz ).Note that P L abc ( u ).Suppose OS ( y, z, bu ). Then cb ( u, ybz ) is (non-strict) colinear with b and p yz .This would imply OS ( y, z, bp yz ) which is a contradiction. Thus ¬ OS ( y, z, bu )and since T ( y, z, b ) and T ( y, z, u ) (because P L abc ( u )) we can conclude SS ( y, z, bu ).Similarly we can conclude SS ( x, y, bu ). We can then infer SS ( x, z, bu ). Fromthis we can then show SS ( x, z, bp xz ). Finally we can conclude ¬ OS ( x, z, abc ).The following theorem is provable from parts 2) and 4) of the previous the-orem.Theorem 64: SS ( x, y, abc ) ∧ OS ( y, z, abc ) → SS ( x, z, abc )89 In [3], Van Bendegem ask if Suppes’ quantifier-free axioms for constructiveaffine plane geometry could be expanded all the way into a full-fledged geo-metric theory. We claim this work answers that question in the affirmative.Furthermore we claim that this work is the first work to build a full geometrictheory which contains only feasible construction and who’s concepts are inline with the bounded experience of real-world constructions.