Journal of Mathematical Sciences, Vol. 119, No. 1, 2004

METRIC AXIOMATICS OF EUCLIDEAN SPACE

A. A. Ivanov UDC 515.124, 514.12.01

A version of axioms of Euclidean space based on a single initial notion, namely on the notion of distance, isconsidered. The notions of straight line and plane are introduced in terms of distance. Thus, Euclidean space isregarded as a metric space with metric satisfying the corresponding axioms. Bibliography: 3 titles.

§1. Introduction

We describe a system of axioms for the usual Euclidean 3-space based upon the notion of metric as the onlyinitial notion. The axioms are simple geometric statements. The truth of these statements is proved here. Theopposite procedure of deducing Hilbert’s axioms (see [1]) from metric axioms is also rather simple. However, thepassage from the usual axioms of Euclidean space to metric axioms allows us to look at geometry from anotherpoint of view, as well as to consider different generalizations of Euclidean space by weakening the correspondingmetric axioms. One more advantage of metric axiomatics suggested here is its finiteness, which allows us toconsider finite objects only: segments, triangles, tetrahedra, etc.

As was already said, the notion of metric is our initial notion. Assume that we are given a set X and adistance d(x, y), which is a nonnegative function of points x and y of X.

Definition. A function d : X ×X −→ R+, where R+ is the set of nonnegative real numbers, is a metric on X if1. d(x, y) = 0 ⇐⇒ x = y (axiom of identity);2. d(x, y) = d(y, x) (axiom of symmetry);3. d(x, z) ≤ d(x, y) + d(y, z) (axiom of triangle).The pair (X, d) is called a metric space.

Definition. A bijective mapping f : (X, d) → (X′, d′) is an isometry if d(x, y) = d′(f(x), f(y)). In this case, thespaces are said to be isometric.

Remark. Our metric system of axioms of Euclidean space does not include the axioms of metric.

§2. Linear sets of metrical spaces

First, we consider a general notion of linear set in an arbitrary metric space (X, d).

Definition. Let A, B, and C be three points in a metric space (X, d). We say that {A, B, C} is a linear set in(X, d) if

±d(A, C) ± d(C, B) = d(A, B).

If both summands on the left-hand side are not zero and have sign plus, then we say that C lies between A andB.

Remark. If three different points form a linear set, then, obviously, one of them lies between the others. Alsonotice that in [2] such points were said to be arranged rectilinearly.

Definition. A set L in (X, d) is linear if any three different points in L form a linear set.

Definition. The segment [AB] with ends A and B is the set consisting of A and B and all points C lyingbetween A and B. The distance d(A, B) between A and B is the length |AB| of [AB].

Remark. [AB] and [BA] denote the same segment because A and B occur in the definition of segment sym-metrically.

Definition. The union AB of all segments containing a segment [AB] is the (straight) line through A and B.

Remark. If indication of A and B is unessential, then we say simply about a line.

Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 279, 2001, pp. 89–110. Original article submitted January 9,2001.

1072-3374/04/1191-0045 $25.00 c©2004 Plenum Publishing Corporation 45

Definition. If there is no line containing three points A, B, and C, then we say that they are noncollinear.

Definition. Let A �= B. The ray [AB of AB outgoing from A towards B is the set consisting of [AB] and allpoints C with |AC|− |BC| = |AB|. The ray A]B of AB emanating from A in the direction opposite to B is theset of all points C such that |CB| − |CA| = |AB|.

The theory of linear sets in a metric space without additional assumptions is rather poor. For example, a lineis not nesessarily a linear set, as the following example shows.

Example. Let X = {A, B, C, D}, d(A, B) = 1, d(A, C) = 3, d(A, D) = 5, d(B, C) = 3, d(B, D) = 4, andd(C, D) = 2. We easily see that X satisfies the axioms of metric and [AD] = X. On the other hand, the set{B, C, D} is not linear, and therefore [AD] is not linear.

Remark. This example also demonstrates that in the general case three points can lie on a line which is nota linear set, while a line is not nesessarily the union of two rays emanating from a point in opposite directions.Indeed, in this example, we have AB = [AB], [CD = {C, D}, and C]D = {A, C}, whence AB �= [CD ∪ C]D.

On the other hand, if a finite set X is linear, then there exists a segment containing X, and its ends arepoints of X. Indeed, among all pairs belonging to X, there exists a pair A, B with maximal distance. Let C beanother point of X. Then {A, B, C} is a linear set, and so one of these points lies between the others. Since|AB| is the maximal distance among |AB|, |AC|, and |BC|, we have |AC| + |CB| = |AB|, i.e., C ∈ [AB].

Unfortunately, the linearity of a metric space (X, d) does not imply the uniqueness of points M satisfying therelation ±|AM | ± |MB| = |AB| (the signs are assumed to be fixed). Indeed, let X = {A, B, C, D}, d(A, C) =d(B, D) = 2, and d(A, B) = d(B, C) = d(C, D) = d(D, A) = 1. Any three points of X form a linear set since oneof the distances between them is 2 and the other distances are 1. On the other hand, B and D lie between A andC, and we have d(A, B) = d(A, D) and d(B, C) = d(D, C). Therefore, it is natural to introduce the followingcondition.

Axiom E1 (uniqueness axiom). For any two points A and B, a point M satisfying the condition ±|AM | ±|MB| = |AB| (|AM |, |MB|, and the signs are fixed) is uniquely determined.

Euclidean space obviously satisfies this axiom.Another condition is related to the notion of linearly coupled sets.

Definition. Two sets are linearly coupled if their intersection contains at least two points.

Definition. A finite system of sets in (X, d) is linearly coupled if they can be numbered, possibly with repetitions,so that every two neighboring sets are linearly coupled.

Axiom E2 (linearity axiom). The union of sets of any linearly coupled system of linear sets is a linear set.

Theorem. Euclidean space satisfies Axiom E2.

Proof. First of all, notice that any linear set L in Euclidean space is contained in a certain line. Indeed, if Lis one-point, the statement is trivial. Let L contain at least two points A and B, and let C ∈ L. Then, bylinearity, we have |AB| = ±|AC| ± |CB|, whence C ∈ AB. It follows that L ⊂ AB. Now, if two linear sets Xand Y are linearly coupled, then there are two points A and B such that X, Y ⊂ AB and X ∪ Y is a linearset. When considering a linearly coupled system of linear sets, we pass to the corresponding finite sequence oflinearly coupled linear sets. Replacing adjacent sets of the sequence by their union, we obtain a similar sequence.Repeating this procedure, we obtain a linear set, which is the union of the sets of the considered system. �

If a metric space satisfies Axiom E2, then its segments have some properties of the usual segments.

Theorem. If a metric space (X, d) satisfies the linearity axiom, then the following statements hold true:1. each segment is a linear set;2. if C, D ∈ [AB], then [CD] ⊂ [AB];3. the union of two linearly coupled segments is contained in a segment the ends of which are some ends of

the initial segments.4. each line is the union of an increasing chain of an segments.

Proof. 1. Let [AB] be a segment in X. We can assume that A �= B since all one-point sets are linear. LetC, D, E ∈ [A, B]. By the definition of a segment, {A, C, B}, {A, D, B}, and {A, E, B} are linearly coupled linearsets, and so, by Axiom E2, their union {A, B, C, D, E} is linear set as well as its subset {C, D, E}.

46

2. Let C, D ∈ [AB] and let M ∈ [CD]. To prove this part, it is sufficient to consider the case whereA = C or D = B. For the sake of definiteness, let D = B. In this case, we have |AC| + |CB| = |AB|and |CM | + |MB| = |CB| if M ∈ [CB]. These equalities imply |AC| + |CM | + |MB| = |AB|, whence|AM |+ |MB| = |AB|, and so M ∈ [AB].

3. Consider two linearly coupled segments [AB] and [CD]. Their union is a linear set by Axiom E2. Inparticular, {A, B, C, D} is a linear set. However, then it is contained in a segment with end points in {A, B, C, D}.Without loss of generality, we assume that these ends are either A, B or A, D. In the first case, [CD] ⊂ [AB],while in the second case we have [AB] ⊂ [AD] and [CD] ⊂ [AD].

4. Consider any line AB. By definition, it consists of segments containing [AB]. By assertion 3, any twoof these segments lie in a segment of AB. Among systems of such segments included one into another there isa maximal system, which we fix later on. Let [DE] be a segment in this system, and let M ∈ AB. For fixedM , either there exists [DE] with |DE| ≥ |DM | and |DE| ≥ |EM |, or for each [DE] we have |DE| < |DM |or |DE| < |EM |. In the first case, we have M ∈ [DE] ⊂ AB, while in the second case each [DE] is strictlycontained either in [DM ] or in [EM ], which is strictly contained, by assertion 3, in a certain segment containing[AB], which contradicts the maximality of the system considered. �

The following condition is also important for the progress of the corresponding theory.

Axiom E3 (axiom of filling a line). For any two points A and B and any relation

±d1 ± d2 = d(A, B), d1 ≥ 0, d2 ≥ 0,

there exists a point C such that

d(A, C) = d1 and d(C, B) = d2.

Theorem. Euclidean space satisfies axiom E3.

Proof. Consider the three possible cases:d1 + d2 = d(A, B),

d1 − d2 = d(A, B),

−d1 + d2 = d(A, B).

In the first case, [AB] contains a unique point C with |AC| = d1 and |CB| = d2. In the second case, [ACcontains a unique point C with |AC| = d1 and |BC| = d2. Finally, in the third case, AB] contains a uniquepoint C with |CA| = d1 and |CB| = d2. �

Remark. Axiom E3 alone does not imply the linearity of line. Indeed, the points A, B, C, and D of the aboveexample can be added to a metric space satisfying this axiom.

We can prove an even stronger statement.

Theorem. Any metric space is contained in a metric space satisfying Axiom E3.

Proof. Let (X, d) be a metric space. Each pair a, b of distinct points in X is assigned a set La,b isometric to thereal line such that the distances between a and b in (X, d) and in La,b coincide and La,b ∩ Lc,d = {a, b} ∩ {c, d}for any a, b, c, and d. Put X′ =

⋃a,b

La,b, where a, b ∈ X and a �= b.

Define a metric d′ on X′. Let x ∈ La,b and y ∈ Lc,d. Consider the following four sums:

|xa|+ |ac|+ |cy|, |xa|+ |ad|+ |dy|, |xb|+ |bc|+ |cy|, and |xb|+ |bd|+ |dy|.

The least of them is equal to d′(x, y). We easily see that d′ is a metric.Axiom E3 may be violated in the space (X′, d′), but it is true for points in X. Now, repeating this extension

procedure for (X′, d′), we obtain a metric space (X′′, d′′), and so on. Repeating this procedure n times, weobtain a metric space (X(n), d(n)). The union

⋃n∈N

X(n) is an extension of (X, d) satisfying axiom E3. �

Axioms E1–E3 already imply that a linear geometry is Euclidean.

47

Theorem. If a metric space (X, d) satisfies Axioms E1–E3, then any line of this space is isometric to the realline with the usual metric.

Proof. Consider any line AB ⊂ X and a mapping

φ : AB −→ R, φ(M) =

d(A, M) if M ∈ [A, B],d(A, B) + d(B, M) if M ∈ A[B,

−d(M, A) if M ∈ A]B.

(We obviously have φ(A) = 0 and φ(B) = d(A, B).) This mapping is well defined since Axiom E2 implies thatAB = A]B ∪ [AB] ∪A[B, Axiom E1 implies that φ is one-to-one, and Axiom E3 implies that φ is onto.

The mapping φ is an isometry. Indeed, let M, N ∈ X. It suffices to consider the following six cases:

M ∈ A]B, N ∈ A]B, or N ∈ [AB], or N ∈ A[B;M ∈ [AB], N ∈ [AB] or N ∈ A[B;M ∈ A[B, N ∈ A[B.

All these cases are considered in the same way. For example,

φ(M) = −d(M, A) and φ(N) = −d(N, A).

Without loss of generality, we can assume that d(M, A) > d(N, A), whence

d(M, N) = d(M, A) − d(N, A) = φ(N) − φ(M). �

§3. Planar sets in metric spaces

Now, we consider the notion of planar set in an arbitrary metric space (X, d).

Definition. Let A, B, and C be some points of X. We say that they define a triangle [ABC]. The segments[AB], [BC], and [CA] are the sides of the triangle, and A, B, and C are the vertices. The number |ABC| =√

p(p − a)(p − b)(p − c), where a, b, and c are the lengths of the sides and p = 12 (a + b + c) is the half-perimeter

of the triangle, is the area of [ABC]. Triangles with zero area are degenerate.

Remark. We easily see that if a triangle [ABC] is degenerate, then {A, B, C} is a linear set. Indeed, in thiscase, one of the factors in p(p − a)(p − b)(p − c) vanishes. For example, let p − a = 0, i.e., a = b + c. Withoutloss of generality, we can assume that

a = |BC|, b = |CA|, c = |AB|, and |CB| = |CA|+ |AB|,

i.e., A lies between C and B.

Definition. We say that a point M ∈ (X, d) belongs to a triangle [ABC] if |ABC| = |MAB|+ |MBC|+ |MCA|.In what follows, any triangle is regarded as the set of the points belonging to it.

Remark. By definition, each vertex of the triangle [ABC] belongs to it. For example, A ∈ [ABC] since

|AAB| = |ACA| = 0, whence |AAB| + |ACA|+ |ABC| = |ABC|.

The situation with sides of triangles is more intricate.

Axiom E4. For an arbitrary fixed line l and any point C, we have |ABC| = h|AB|, where h does not dependon the choice of the segment [AB] ⊂ l.

Theorem. Euclidean space satisfies Axiom E4.

Proof. In the relation |ABC| = h|AB|, the number h is the altitude of the triangle [ABC] with base [AB], whichis equal to the distance from C to AB. �

Remark. If a metric space (X, d) satisfies Axiom E4, then any triangle contains its sides. Indeed, let M ∈ [AB].Then |AM |+ |MB| = |AB|, and by E4 we have |AMC|+ |MBC| = |ABC|. Then |AMB|+ |AMC|+ |MBC|=|ABC| since |AMB| = 0, whence, by definition, M ∈ [ABC]. �

48

Definition. The union ABC of all triangles containing a nondegenerate triangle [ABC] is the plane throughA, B, and C.

Remark. If the indication of A, B, and C is not necessary, then we speak simply about a plane.

Definition. Let A, B, C, D ∈ (X, d). We say that {A, B, C, D} is a planar set if

±|DAB| ± |DBC| ± |DCA| = |ABC|.If A, B, and C are noncollinear and all summands on the left-hand side are not zero and have sign plus, then wesay that D lies between A, B, and C.

Definition. A set in (X, d) is planar if each quadruple of its points is a planar set.

Definition. Two sets are planarly coupled if their intersection contains three noncollinear points.

Definition. A finite system of sets in (X, d) is called planarly coupled if the sets can be numbered, possiblywith repetitions, so that every two neighbor sets are planarly coupled.

The following condition is of importance for linear sets.

Axiom E5 (axiom of plane). The union of any planarly coupled system of planar sets is a planar set.

Theorem. Euclidean space satisfies Axiom E5.

Proof. It is similar to the proof of Axiom E3. �

Remark. Axiom E5 implies that any triangle is a planar set. Indeed, by definition, each point M ∈ [ABC]satisfies |ABC| = |MAB| + |MBC| + |MCA|, and so {M, A, B, C} is a planar set. For arbitrary pointsM1, M2, M3, M4 ∈ [ABC], we consider the planarly coupled planar sets {A, B, C, Mi}. By Axiom E4, theirunion {A, B, C, M1, M2, M3, M4} is a planar set. Therefore, {M1, M2, M3, M4} is a planar set.

Axiom E6 (axiom of filling a plane). For any three noncollinear points A, B, and C and any nonnegativenumbers S1, S2, and S3 such that

±S1 ± S2 ± S3 = |ABC|,there exists a unique point M such that

S1 = |MAB|, S2 = |MBC|, and S3 = |MCA|.

Remark. As in the linear case, Axiom E6 can be replaced by two axioms separately considering the conditionof uniqueness.

Theorem. Euclidean space satisfies Axiom E6.

Proof. This is trivial. �

Remark. Actually, this means that barycentric coordinates determine a one-to-one correspondence betweenpoints of plane and triplets of numbers with unit sum.Remark. If Axiom E6 is fulfilled, then every line AB divides each plane α containing it into two parts. Theseparts are distinguished by fixing some point C ∈ α \AB. Consider a triangle [ABC] and a point M ∈ ABC. ByAxiom E6, we have

±|MAB| ± |MBC| ± |MCB| = |ABC|.If |MAB| occurs in this sum with the sign +, then M belongs to the half-plane containing C. Otherwise, Mbelongs to the second half-plane.

The following condition also relates to axioms of plane.

Axiom E7 (axiom of similarity). Let

d(A, A′) + d(A′, O) = d(A, O) and d(B, B′) + d(B′, O) = d(B, O),

where all distances differ from zero, and let

d(A′, O)d(A, O)

=d(B′, O)d(B, O)

.

Thend(A′, O)d(A, O)

=d(B′, O)d(B, O)

=d(A′, B′)d(A, B)

.

49

Theorem. Euclidean space satisfies Axiom E7.

Proof. This follows from the corresponding criterion of similarity of triangles and proportionality of theirsides. �

This axiom plays an important role in the theory of parallelism of lines and planes.

§4. Axioms of volume

The volume group of metric axioms of the usual (3-dimensional) Euclidean space differs by the structurefrom the two previous ones. First of all, notice that the corresponding theory is interesting only if the followingcondition is fulfilled.

Axiom E8. There are four points A, B, C, and D not forming a planar set.

Euclidean space obviously satisfies this axiom.Fundamental objects here are tetrahedra [ABCD], which are determined by four points. Each tetrahedron

has volume |ABCD|, which is defined in Euclidean geometry as a function of the lengths |AB|, |AC|, |AD|,|BC|, |BD|, |CD| of the edges. Hence, the volume of a tetrahedron is a metric notion.

Axiom E9 (axiom of filling space). For any four noncoplanar points A, B, C, and D and any nonnegativenumbers V1, V2, V3, and V4 such that

±V1 ± V2 ± V3 ± V4 = |ABCD|,

there exists a unique point M with

|MBCD| = V1, |AMCD| = V2, |ABMD| = V3, |ABCM | = V4,

and such points fill all space.

The proof of this axiom for Euclidean space is similar to that of axiom E6.

§5. Axioms of congruence

Definition. Let (X, d) be a metric space. Two sets U, V ⊂ (X, d) are congruent or equal if there exists anisometry f : (X, d) → (X, d) such that f(U) = V .

Axiom E10. Let |AB| = |A′B′|. Then there exists an isometry f : (X, d) → (X, d) such that f(A) = A′ andf(B) = B′.

Remark. Let f : X → X be an isometry such that f(A) = A′ and f(B) = B′. Then f [AB] = [A′B′]. Indeed,if f(A) = A′, f(B) = B′, and C ∈ [AB], i.e., |AC|+ |CB| = |AB|, then |f(A)f(C)|+ |f(C)f(B)| = |f(A)f(B)|,whence f(C) ∈ [f(A)f(B)].

Axiom E11. Let two planes α and β intersect along a line l, and let A ∈ α. Then there exists an isometry(X, d) → (X, d) identical on l and taking A to a certain point B ∈ β. The point B can lie in any half-plane of βwith respect to l.

Euclidean space obviously satisfies axioms E10 and E11.Remark. By analogy with the system of metric axioms of 3-dimensional Euclidean space, we can formulate asystem of axioms for the n-dimensional Euclidean space.

§6. The first group of Hilbert’s axioms

Now, we prove that the metric axioms for Euclidean space imply all Hilbert’s axioms. We start with the firstgroup, i.e., with axioms of connection. This group contains eight axioms.

Axiom I1. Any two points A and B lie on a line.

Proof. This follows from the existence of a metric line containing the segment [AB] and, therefore, passingthrough A and B. �

50

Axiom I2. For any two points A and B, there exists at most one line through A and B.

Proof. By definition, any line through A and B, for example, AB, contains [AB]. If a line CD also contains[AB], then there exists a segment of CD containing both [AB] and [CD]. Therefore, it is sufficient to considerthe case where [CD] ⊂ [AB]. In this case, we have AB ⊂ CD. Prove the inreverse inclusion. Let M ∈ CD.Then {C, D, M} is a linear set. On the other hand, {A, B, C, D} is a linear set linearly coupled with {C, D, M},and so, by Axiom E2, the set {A, B, C, D, M} is also linear. Then the set {A, B, M} is linear, i.e., one of thepoints lies between the others. If M lies between A and B, then M ∈ [AB] ⊂ AB. If A lies between M and B,then [AB] ⊂ [BM ], whence M ∈ AB. The case where B lies between M and A is considered similarly. �Axiom I3. Any line contains at least two points. There are at least three noncollinear points.

Proof. This follows from the definition of a line and Axiom E8. �Axiom I4. Any three noncollinear points lie in a plane. Any plane contains at least one point.

Proof. This follows from the definition of a plane. �Axiom I5. Any three noncollinear points A, B, and C lie in at most one plane.

Proof. Consider the plane ABC and any plane DEF containing A, B, and C. As in the linear situation, wemay assume that [DEF ] ⊂ [ABC], whence ABC ⊂ DEF . Prove the inverse inclusion. Let M ∈ DEF .Then {M, D, E, F} is also a planar set. Hence, {A, B, C, D, E, F} ⊂ ABC. The intersection of these pla-nar sets is, by the condition of the theorem, the planar set {D, E, F}. Therefore, by Axiom E5, their union{A, B, C, D, E, F, M} is a planar set. It follows that M ∈ ABC, whence DEF ⊂ ABC.

Axiom I6. If two points on a line a lie in a plane α, then a ⊂ α.

Proof. Let A, B, C ∈ α, and let C /∈ AB. We consider an arbitrary point M ∈ AB. Then {A, B, M} is a linearset, i.e., ±|AM | ± |MB| = |AB|. By Axiom E4, we have ±|AMC| ± |MBC| = |ABC|. Since |ABM | = 0, weobtain

±|ABM | ± |AMC| ± |MBC| = |ABC|,whence, by axiom E6, M ∈ α. �Axiom I7. If two planes α and β have a common point M , then they have another common point N .

Proof. Let M belong to α and β. Let [ABC] be a triangle in α containing the interior point M and a pointD ∈ β \ α. Then the rays starting at D that are sufficiently near to [DM intersect [ABC] at points M ′. Someof these rays lie in β, and therefore there is another point N belonging to both planes. �Axiom I8. There are at least four points not lying in a plane.

Proof. See Axiom E8. �

§7. Second group of Hilbert’s axioms

Axioms of this group define the notion “between.”

Axiom II1. If a point B lies between points A and C, then A, B, and C are different points of a line and Balso lies between C and A.

Proof. This immediately follows from the definition of the metric notion “between.”

Axiom II2. For any two distinct points A and C, the line AC contains at least one point B lying between Aand C.

Proof. Consider any number d ∈ (0, |AC|). Then, by Axiom E3, there is a point B such that |AB| = d and|AB| + |BC| = |AC|, i.e., B lies between A and C.

Axiom II3. For any three points on a line, at most one of them lies between the two others.

Proof. Let A, B, and C lie on a line. Without loss of generality, we can assume that B lies between A and C,i.e., |AB|+ |BC| = |AC|. Since both terms are not zero, we have |AB| < |AC| and |BC| < |AC|. It follows thatthe equalities |AC|+ |CB| = |AB| and |BA| + |AC| = |BC| are impossible, i.e., neither C lies between A andB, nor A lies between B and C.

To prove the last axiom of this group (the Pasch axiom), we need the following additional statements.

51

Theorem. Let a be a line in a plane α, and let B ∈ α \ a. Then α contains a line b through B disjoint with a.

Proof. Let a be a line, let A ∈ a, let B /∈ a, let O ∈ [AB \ [AB], and let k = |OB|/|OA|. For any M ∈ a,there is a unique point D = D(M) ∈ MO such that |OD|/|OM | = k. We easily see that the set b = {D(M) |M ∈ a} is a line through B. Indeed, let Di = D(Mi), i = 1, 2, 3. Without loss of generality, we assume that|M1M2| + |M2M3| = |M1M3|. Then, by the axiom of similarity,

|D1D2|+ |D2D3| = k|M1M2| + k|M2M3| = k|M1M3| = |D1D3|,

i.e., {D1, D2, D3} is a linear set. Since the points are arbitrary, b is a line, and if M2 = A, then D2 = B, i.e.,B ∈ b. The line b does not intersect a. Indeed, let N be a point of intersection of a and b. Then D(N) ∈ ON ,whence ON = b. However, this is impossible because in this case O ∈ b, which contradicts the choice of O.

Definition. Two disjoint lines a and b are parallel to each other.

Remark. It was proved above that any point B /∈ a lies on a line parallel to a. We cannot state till nowthat such a line is unique. Therefore, speaking on parallel lines, we mean the lines that are determined by theconstruction given in this proof. The point O used there will be called the center of similarity.

Theorem. Let a be a line in a plane α, let B ∈ α \ a, let b be the line through B parallel to a. Then the regionS between a and b in α is filled by lines parallel to a.

Proof. Let M ∈ S \ (a∪ b). The segment [MO], where O is the center of similarity, intersects b at M ′, and henceMO intersects a, by construction of b. Using a similarity with center O, we obtain a line through M parallel toa. �

Theorem. Let two lines a and b in a plane α intersect at a point A. Then for each of these lines the oppositerays with initial point A are contained in different half-planes into which the other line divides α.

Proof. Assume that the opposite rays of b lie in the same half-plane. Then the half-plane contains a line cparallel to a and intersecting both rays at points C and C ′. However, then b = c, which is impossible since c isparallel to a, while b and a intersect. �

Axiom II4 (Pasch axiom). Let A, B, and C be three noncollinear points, and let l be a line in ABC passingthrough none of the points A, B, and C. If l passes through a point of the segment [AB], then l passes througha point of [AC] or [BC].

Proof. Let a denote the line AB. Let c be a line through C parallel to a. Then the region between c and a isfilled by lines parallel to a. Now, consider a line l passing through none of the vertices of the triangle [ABC],but passing through a point M of [AB]. Then the rays into which M divides l lie on different sides of a in ABC.Therefore, one of these rays lies on the same side of a as the triangle [ABC]. Each point of this ray lies on a lineparallel to a. Among these points, we choose those belonging to segments of parallel lines with ends on [AC]and [BC]. We easily see that these points form a segment with an end belonging either to [AC] or to [BC], butnot to both simultaneously. The point N is the point of intersection of l with one of the sides [AC] or [BC] of[ABC]. �

§8. Third group of Hilbert’s axioms

Axioms of this group define the notion of congruence.

Axiom III1. If a segment [AB] lies on a line a and A′ is a point on the same or other line a′, then there is apoint B′ on a prescribed side of A′ in a′ such that the segment [A′B′] is congruent to [AB] ([A′B′] = [AB]).

Proof. There is a point B′ ∈ a′ lying on a prescribed side of A′ and such that |AB| = |A′B′|. The rest followsfrom Axiom E10. �

Axiom III2. If two segments are congruent to a third one, then they are congruent to each other.

Proof. If isometries f and g transfer [AB] and [CD] onto [EF ], then the isometry g−1f transfers [AB] onto[CD]. �

52

Axiom III3. If segments [AB] and [BC] lie in a line a on different sides of a point B and segments [A′B′] and[B′C ′] lie on a line a′ on different sides of a point B′, then [AB] = [A′B′] and [BC] = [B′C ′] imply [AC] = [A′C ′].

Proof. We easily see that |AB| = |A′B′| and |BC| = |B′C ′|. It follows that |AC| = |A′C ′|, whence [AC] =[A′C ′]. �

Axiom III4. Let ∠lm be an angle in a plane α, let a′ be a line in a plane α′, and let us fix a certain side ofα′ with respect to a′. Let l′ be a ray in a′ starting at O′. Then α′ contains a unique ray m′ with the followingproperty: ∠l′m′ is congruent to ∠lm and all interior points of ∠l′m′ lie in α′ on the given side of a′.

Proof. Axiom E10 allows us to transfer the ray l into a ray l′ emanating from O′. It is possible that the ray mis transferred not into α′, but into another plane β. Using Axiom E11, we transfer it into m′. �

Axiom III5. If [ABC] and [A′B′C ′] are two triangles such that AB = A′B′, AC = A′C ′, and ∠BAC =∠B′A′C ′, then [ABC] = [A′B′C ′].

Proof. Using Axiom E10, we transfer [AB] into [A′B′]. Then the triangle [ABC] is transferred into [A′B′C ′′].After that, using Axiom E10, we transfer C ′′ into C ′, while [A′B′] is fixed. Since in both cases we use isometries,[ABC] is transferred into [A′B′C ′]. �

§9. Fourth group of Hilbert’s axioms

This group contains only one axiom, which Euclid (in other formulation) called the Fifth postulate. Thefamous fifth postulate of Euclid more than 2000 years attracted attention of many mathematicians and simplydilettantes.

Axiom IV . Let a be a line and let O /∈ a. Then the plane through a and O contains at most one line throughO not intersecting a.

Proof. Let A ∈ a. Using O as a center of similarity, we construct lines parallel to a. These parallels lB wereconstructed earlier for all points B ∈ [AO] with the exception of O. However, it is possible to draw a line lOthrough O not intersecting a as a limit line of the lines lB. The region between a and lO is filled by lines lBdefined for all B ∈ [AO]. Consider any line c through O. If c �= lO, then c passes through a point C ∈ lB forsome B ∈ [AO]. But every point of lB lies in a certain segment [OA′], where A′ ∈ a, and so c intersects a. Itfollows that lO is a unique line through O not intersecting l. �

§10. Fifth group of Hilbert’s axioms

This group contains two axioms.

Axiom V1 (Archimedean principle). Let [AB] and [CD] be two segments in a line a. Then a contains afinite number of points A1, A2, . . . , An such that the segments [A, A1], [A1A2], . . . , and [An−1An] are isometricto CD and B lies between A and An.

Proof. This immediately follows from the fact that two segments are isometric if and only if their lengths areequal. �

Axiom V2 (axiom of linear completeness). The points of any line form a system that cannot be extendedwith preservation of linear order, the first axiom of congruence, and the Achimedean principle.

Proof. This immediately follows from the fact that any Euclidean line is isometric to the real line with the usualmetric. �

§11. Conclusion

Here, we described a system of metric axioms for Euclidean space. There may exist simpler systems of axiomsand simpler passages from metric axioms to the classical ones, but the system of axioms described here forEuclidean space seems to be simpler than Hilbert’s axiomatic. Presentation of Euclidean geometry based uponmetric axioms is not customary, but it is worth trying. It is also interesting to look what we obtain if we dropor weaken some of the metric axioms.

Translated by A. A. Ivanov

53

REFERENCES

1. D. Hilbert, Grundlagen der Geometrie (Wissenschaft Hypothese, VII), 7th edn., Teubner, Leipzig–Berlin(1930).

2. V. F. Kagan, Sketches on Geometry [in Russian], Moscow State University , Moscow (1963).3. A. D. Aleksandrov and N. Yu. Netsvetaev, Geometry [in Russian], Nauka, Moscow (1990).

54