CONGRUENCE AXIOMS FCR ABSOLUTE GEOMETRY

J. F. Rigby

(communicated by H. G. Forder, 1 March, 1974)

Introducti on

The modem axiomatic foundations of Euclidean geometry were laid by Pasch (1882), Peano (18S9), Hilbert (1899), Veblen (1904), Moore (1908), etc, There are four groups of axioms: axioms of order, axioms of congruence, the axiom of continuity, and the Euclidean parallel axiom [5], (Hilbert’s extra axioms of incidence can be deduced from properly formulated axioms of order.) If we omit the parallel axiom, the remaining axioms give either Euclidean or hyperbolic geometry. Many important theorems can be proved using only the axioms of order and congruence, and the name absolute geometry is given to geometry in which we assume only these axioms.

Over the years, some of the original modem axioms of congruence have been shown to be unnecessary, and other axioms have been weakened (see §3). Between 1967 and 1969, I wrote two papers in which the process of weakening was continued [12, 13] and two others giving counterexamples necessary for the discussion of the independence of the axioms [14, 15].

More recently I have realized that the slightly unsatisfactory concept of isometric points [12,§1] can be replaced by a more natural concept of related points, and as a result I have succeeded in replacing axiom C3* [12] by a special case of the axiom; I have discarded part of axiom C5a [12] which was purely negative in the sense that it applied only to a geometrical configuration which was later shown not to exist; also I have found an elementary way of proving thatevery segment has a mid-point, using only the weak axioms (§8); this new proof replaces my proof in 12 [ ,§6], which used rather more sophisticated concepts and theorems from Bachmann's account of metric planes [l]. (I was previously unable to find a simple proof of this theorem even assuming

Math. Chronicle 4(1975), 13-44 axiom Cl instead of Cl* [12].)

I have incorporated these new developments in the present paper, which consists of a self-contained account of the proofs of the basic theorems of absolute geometry starting from the weak axioms of congruence Cl1, C2', C3', C4', C4a , C5‘, C5", which are stated below. We shall also consider the construction of congruent segments, perpendiculars and mid-points, using weak methods of construction. Some of the proofs from [12, 13] are repeated here for the sake of completeness with occasional modifications and improvements, and a few other parts of the text of [12, 13] are used here verbatim; other proofs, and the order of presentation of certain theorems, have had to be changed.

We shall assume without comment various results of ordered geometry. For the axioms and theorems of ordered geometry see [5]. The symbol 1 [.ABC]' means ' 3 and C lie in the geometrical order ABC ' .

1. Initial axioms of congruence

An absolute space is an ordered space [5, Chapter 2] together with a relation of congruence satisfying certain axioms of congruence. Congruence is a relation between ordered pairs of distinct points. (The term 'ordered pair' is used in its set-theoretical sense, and has no connection with geometrical order.) Intuitively, two segments are congruent if one can be picked up and placed exactly on the other, and the axioms of congruence are based on this idea; it is more con­ venient to use ordered pairs rather than segments in the axioms, but we must then bear in mind that 04>B) is not the same as (BjA). Euclid's use of 'equal' where we now use 'congruent' confuses the significance of his postulates and common notions', in m o d e m terminology two ordered pairs 04,3) and (C>D) are equal if and only if A coincides with C and B with D, so two ordered pairs can be congruent without being equal. If 04,3) is congruent to (CSD) we shall write AB CD.

As we introduce the axioms to be used in the present paper, we shall compare them with my previous axioms [12], Forder's axioms [5] and Euclid’s postulates and common notions [7].

14 Euclid's Postulate 3 states that 'a circle can be drawn with any centre and distance1 [7, pp. 154, 199], In other words, we can draw a circle with centre A and any radius AB . He does not assume that we can draw a circle with centre and radius congruent to a given segment OC ; this is a consequence of his Proposition 2: To place at a given point (as an extremity) a straight line equal to a given straight line [7, p. 244]. To prove Proposition 2 he assumes that certain circles intersect. Modern geometers have preferred not to mention circles or their properties in the axioms, so they have taken Proposition 2 as an axiom. Thus we have Forder’s axiom Cl: If A , B are distinct points and 0 is any point, then on any ray from 0 (i.e. any ray with origin ) 0 there exists just one point C such that AB -*■ OC . We shall assume a weaker axiom Cl 1 [see 12, Cl*], similar to Postulate 3.

Cl'. If A ,B are distinct points, then on any ray from A there exists just one point C such that AB -> AC ,

From an intuitive point of view we can regard postulate 3 as providing a means of physically constructing the circle with centre A and radius AB , The physical implement with which we do this has been described as ’collapsing compasses': we cannot place the point of these compasses on 0 and the pencil on C , then transfer the point to A , as they will then collapse. In Cl' we use these collapsing compasses; but not to draw a circle, only to mark off C on the given ray such that A3 -> AC . The subsequent axioms, defini­ tions and proofs in this section and the next are based on an intuitive investigation of what we can construct by using the collapsing compasses in this restricted way and also using a straight edge to join any two given points. We shall discuss later what constructibility means in abstract geometry C§4).

Our axiom C2' is the same as Forder's C2 and corresponds to Euclid’s Common notion 1: Things which are equal to the same thing are also equal to one another [7, p. 155].

15 C 2 '. If AB + CD and CD + EF then AB + EF

Forder's axiom C3 is: If [ABC] and [A'B'C'] } and if AB -*■ A'B' , BC -»•B'C' , then AC -*■ A'C' . This corresponds to Euclid's Common notion 2: If equals be added to equals, the wholes are equal [7, p. 155]. In [12] I used C3*, which, with slight variation, stated: If B ., C lie on a ray from A , if B ' , C' lie on a ray from A' , and if [ABC] , AB •+ A'B' , AC A'C' , then [A'B'C'] and BC -* B'C' . This corresponds to Euclid's Common notion 3: If equals be subtracted from equals, the remainders are equal. The extra conclusion 1[A'B'C'] ' does not appear to be explicitly stated by Euclid, but if it had been it might have appeared in Heath's translation as: That which is equal to the greater is greater than that which is equal to the less. Here we shall use only a special case of C3*, in which A = A' .

C3* (Figure 1A). If [ABC] , if B' , C' lie on a ray from A t and if AB + AB' , AC + AC' , then [AB'C'] and BC -* B'C' .

These three axioms will suffice for the present.

The usual proof of our first result below requires Forder's axiom C4: AB -> BA . Our C 4 1 (§3) would be of no use here.

1.1 [12, 1.2]. If A * B , then AB -* AB .

Proof. On ray AB , there exist B' and B~ such that AB -> AB' and AB' -*■ AB" (Cl'). Hence AB AB~ (C2‘). Since AB + AB' also, we have B' = B~ (Cl1). Suppose B f B' and [ABB'] . Since AB + AB' and AB' + AB' , we have [AB'B'] (C31), a contradiction. Similarly if B t B' and [AB'B] . Hence B = B' , so AB -*■ AB .

1.2. If B , B' lie on the same ray from A and if AB + AB' then B = B' . (Use 1.1 and the uniqueness in Cl'.)

1.3. If AB + AC then AC AB .

ifi. Proof. There exists B' € ray AB such that AC -»■AB' (Cl1); then AB -> AB' (C21). Hence B = B' (1.2), so AC + AB .

2. Related points

We now define related -points inductively. The term 'related points’ was suggested to me by Professor Forder as a preferable alternative to my 'isometric points' in 12 [ ]; I have used it here in a similar but somewhat different sense.

Definition. If there exists a point P such that PA PB , then A is 1-related to B ;

if A is n-related to C and C is 1-related to B , then A is (n+1)-related to B ;

A is 0-related to A ;

if A is n-related to B for some integer n > 0 , then A is related to B and we write A ~ B .

Note that it is possible for A to be both m-related and n-related to B for m f n \ for instance A is 2-related to A .

2.1. (i) If A is m-related to C and C is n-related to B , then A is (m+n)-related to C .

(ii) The relation ~ is an equivalence relation between points. (Use (i) and 1.3.)

In many of the subsequent figures we shall depict related points by the same type of dot.

2.2. If A ~ B , if P is distinct from A , and if b is any ray from B , then there exists Q € b such that AP BQ .

Proof. We use induction. If A is 0-related to B , the result is just Cl1. Assume the result true for ?c-related points, and suppose A is (k+1)-related to B . Then there exists C such that A is Zc-related to C and C is 1-related to B (Figure 2A); this means

17 that there exists 0 such that OC OB . On ray C/O (the ray opposite to CO ) there exists R such that AP -* CR , by the induction hypothesis. On ray OB there exists S such that OR -*■ OS (Cl1); then CR -> BS (C3 1). On ray b there exists Q such that BS BQ (Cl'). Then AP + CR -> BS -> BQ , so AP -»■ BQ (C2'). The result now follows by induction.

c

0

2.3. If A ~ B and AP ■+ BQ then BQ ^ AP .

Proof. By 2.1 and 2.2, B ~A so there exists P' € rayAP such that BQ AP' . Then AP -*■ AP' (C2'), so P = P' (1.2) and BQ ^ AP .

2.4. In the statement of 2.2, Q is unique.

Proof. If AP -> BQ and AP + BQ' , we have BQ -* AP(2.3) + BQ' , so Q = Q' (C2' and 1.2).

2.5. If A ~ B and! >iP BQ P ~ $ .

Proof. We shall show by induction that if A is n-related to B and AP BQ , then P is (2n+l)-related to Q .

When n = 0 the result is trivial. Assume the result true for n = k , and suppose A is (k+1)-related to B . Then there exists C such that A is ^-related to C and C is 1-related to B . We can use Figure 2A again; P is (2fc+l)-related to R by the induction hypothesis, R is 1-related to S and S is 1-related to Q . Hence P is (2&+3)-related to Q . The result now follows by induc­ tion.

18 2.6. If [ABC] , if A ~A' , if B' , C' lie on a ray from A' > and if AB ■+ A'B' , AC + A'C' , then [.A'B'C'] and BC + B'C' .

Proof (Figure 2B). We use induction. If A is 0-related to A' the result is just C3* . Assume the result true for k-rclaizi t:c-v and suppose A is (fc+1)-related to A' . Then there exists A~ such that A is ^-related to A" and A~ is 1-related to A ' ; there exists 0 such that OA" -> OA' .

On ray A~/0 there exist B " , C" such that AB A"B~ , AC A~C~ . By the induction hypothesis we have [A"B~C~] and BC •+ B~C~ ; hence [OA~B~C~] .

On ray OA* there exist D , E such that OB" -> OB , 0C~ -> OE . By C3' we have A"B" -*■ A'D , B~C~ -> DE and A"C~ + A'E ; also [OA'D] , [ODE] and [OA'E] . Hence [A'DE] .

We now have A'B' -> AB (2.3) -* A~B~ A'D , A'C' AC + A~C~ -> A'E and BC B~C~ DE . Hence A'D -> A'B' and A'E -*■ A'C' (2.3) . Hence [A'B'C'] and DE + B'C' (C3') . Hence BC + DE -* B'C' . The result now follows by induction.

2.7. If [ABC] , [A'B'C'] , if A ~A' , and if AB -> A'B' , BC + B'C' , then AC -*» A'C' .

Proof. There exists D' € ray^'5' such that AC + A'D' (2.2). Then [A'B'D'] and BC + B'D' (2.6). Hence B'C' BC (2.5 and 2.3) -> B'D' , so C' = D' (1.2). Hence AC + A'C' .

We neither postulate nor prove the existence of congruent point- pairs AB and CD except when A ~ C . An example of a one-dimensional geometry satisfying axioms Cl', C2', C31 is given in [12, p. 162]; we need only replace ’isometric' by 'related'. In this example there are two distinct classes of related points and we do not have AB ■+ CD if A and C are not related. Various two-dimensional examples can be similarly adapted from examples in [14].

19 Let us now write }AB = C D X to mean 'A ~ C and AB CD' . We can summarise the main results of §1 and §2 in 2.8 below.

2.8. (i) The relation = is an equivalence relation between ordered pairs of distinct points. (ii) If A ~ B , if P f A , and if b is any ray from B , then there exists a unique point Q Z b such that AP = BQ , and also P ~ Q . (iii) If [ABC] , if B' , C' lie on a ray from A' , and if AB = A'B' , AC = A'C' , then [A'B'C'] and BC e B'C' . (iv) If [ABC] , [A'B'C'] , and if AB = A'B' , BC = B'C' , then AC = A'C' .

It would be possible to define the transitive relation ~ and the symbol MB = CD ' before introducing axioms C2 1 and C31. Axiom C2' could then be given in the form: If AB = CD and CD = EF then AB = EF . This refinement is unnecessary; it does not give us a weaker axiom in any significant sense.

3. Further axioms of congruence

Definition. A point M on the line AB , where A t B , such that MA = 14B , is a mid-point of the segment AB , or of the point-pair AB , and M bisects AB .

Consider a one-dimensional geometry whose points are ordered pairs

(p,x) of real numbers. We define a geometrical order on this line by saying that B[q,y) lies to the right of A(p,x) if either q > p , or q = p and y > x . The distance from A to B is

'(0, \y -x\) if q = p ,

dAB = - {q - p,y) if q > p ,

S p if <7 < p •

on We define AB CD if dAB = dCD . Axioms Cl', C2 1 , C3' are satisfied. The segment AB has a mid-point if p = q , or if p £ q and y = -x , so any two points are at most 2-related. In general AB f BA . If A is (0,0) and B is (2,0) , then (1,2) is a mid-point of AB for all values of z .

We see then that we cannot prove the uniqueness of mid-points using only Cl', C2‘, C31. Hence we shall introduce a further axiom C4*. Using this extra axiom we prove 3.1 and 3.3.

C4'. If A ~ B then AB = BA .

3.1. If A , B lie on the scare ray from 0, and if AC = BO , then A = B .

Proof (Figure 3A). There exists C on ray •0/B such that {e.g., take C such that OB = OC). We then have AO e BO , OC = OC , hence AC = BC (2.8iv). Hence CA = AC e BC = CB (C4'). Hence A = B (1.2).

3.2. If M is a mid-point of seg AB , then [AMB] .

A 3 o C 3A *------O-- 1-- 0------* A N M N' B 3B

3.3. A segment cannot have more than one mid-point.

Proof [12, 2.4] (Figure 3B). We follow the lines of the usual proof, but we need 3.1. Suppose there exist distinct mid-points M ,N of AB , where [ABM] without loss of generality (using 3.2). There exists N' € ray 143 such that MN = MN' . Also MA = MB and [MM] . Hence [Mll'B] and NA = N'B . But HA e NB since N is a mid-point of AB . Hence N'B = NB . Hence N' = N (3.1), a contradiction since [ANM] and [MN'B] .

The next result is not essential to the development of the theory, so we refer the reader to 12 [ , 2.2] for the proof.

21 3.4. If A , B , C , D are oollinear, and if AB = CD , then BA - DC .

Example (c) on p. 188 of [14] shows that we cannot prove that if AB = CD then BA = DC for general non-collinear points A , B , C , D using Cl' -C4', so we must introduce this as an axiom.

C4". If AB = CD then BA e DC .

3.5. If [ABC} > if A' , B' , C' are collinear, and if BC = B'C' CA = C'A' , AB = A'B' , then [A'B'C'] .

Definition. If ABC , A'B'C' are triangles, and if BC e B'C' , CA = C'A' , AB = A'B' , then we write M.BC = AA'B'C' and say that the triangles are congruent.

Without axiom C4", any definition of congruent triangles would be somewhat artificial. Our definition implies that corresponding vertices of congruent triangles are related. Some authors define congruent triangles in a different way [e.g., 5, p. 93, p. 97; 9, p. 116] they then prove that two triangles are congruent if, and only if, corresponding sides are congruent.

The remaining traditional axioms of congruence are given below. These axioms presuppose axioms Cl - C4, so that 'AB = C D ’ means M B + C D ' .

C5 (Figure 3C). If AABC = LA'B'C' , and if [ABX] and [A'B'X'] , and if BX = B'X' , then CX = C'X' .

C6. If the triangles ABC , ABC' lie in the same plane, if C , C' lie on the same side of line AB in this plane3 and if kABC = LABC' , then C = C' .

C7. If ABC is a triangle, if AB e A'B' , and if tr' is any plane containing A'B' , then there exist just two points C' , C' in it ' , one on either side of A'B' ,■ such that LABC = LA'B'C' = LA'B'C' . 1 2

M. Every segment has a mid-point. R. All Tight arjgles are congruent [see 5, p. 97 et seq.~\.

Axiom R is Euclid's Postulate 4 [7, p. 154]. Forder's book The foundations of Euclidean geor.etry [5], based partly on the work of such authors as Veblen [18, 19] and Moore [10], appeared in 1927. Forder used Cl - C 6 and a weaker form of R as his axioms of congruence [5, Chapter IV]. Axioms C7 and R are equivalent [5, p. Ill et seq.], and C7 clearly implies C6. Some authors, such as Kerekjarto [9, §14] use C7 without mentioning C6, but Forder used the weaker axiom Co to prove the existence of right angles and various properties of them. Hilbert [8, §6] used congruence of angles as an undefined relation, but his axioms are easily seen to be equivalent to Cl - C5 and C7.

It can be shown that Cl - C5 and M imply Co and R in a plane [5, p. 132] and Dorroh showed in 1930 that the same is true in spaces of higher dimension [4]. In 1927 Dorroh showed that Cl - C5 imply M [3]. This result was also proved independently by Piesyk in 1965 [11], using axioms due to Tarski [17] that are equivalent to Cl - C5. Piesyk starts from axioms of order and congruence, and proves that every point- pair has a mid-point without defining any of the terms line, plane and right angle.

In 1947 Forder showed that C5 could be replaced by foux’ special cases, in which the congruent triangles ABC , A'B'C' have a common side [6; 12, p. 165]. In [12] I used only one of Forder's special cases (C5a) and two weaker forms of another of his special cases (C5b, C5c); in [13] I discarded C5c. Axiom C5a states (Figure 3D) that if AABC = AABC' , and if [ABX'\ , then XC = XC' . We can split this axiom into two cases. Either (i) the triangles ABC , ABC' lie in distinct planes, or in the same plane with C , C' on opposite sides of AB , or (ii) the triangles ABC , ABC' lie in the same plane with C , C' on the same side of AB . In case (ii), either C - C* in which case the axiom is a tauto- logy, or C i C' . We can use the axioms to prove that it is impossible to have C f C' in case (ii). Now, in any argument byreductio ad absurdum we apply an axiom

23 or a theorem to a situation that we want to show to be impossible, and we obtain a contradiction. However, axiom C5a(ii) is an axiom that applies only to a situation that is later proved to be impossible. It seems preferable to avoid using such an axiom: we shall therefore use C5a(i) only, and shall rename it C5‘.

C5' (Figure 3D) . I f hABC e hABC' , where the two triangles lie either in distinct iplanes or in the same plane with C , C' on opposite sides o f AB , a n d i f , then XC e XC' .

We shall give the remaining axiom C5b a new name, C5".

C5n (Figure 3E). I f OA = OB , a n d i f Y , Y' lie on rays OA, OB , or on rays 0/A , 0/B , and if OY = OY' , then BY e AY* .

Axiom C5" is the special case of C5 in which the congruent triangles are OAB and OBA .

We shall now use the term absolute space for any ordered space satisfying axioms Cl', C21, C3‘, C4', C4", C51, C5".

3.6 [an extension of C5 ‘; 5, p. 97]. If the congruent triangles ABC , ABC' satisfy the conditions of axiom C51, and i f [.AYB] , then

YC E YC' .

Proof (Figure 3F). There exists X such that [ABX] . Then

XC e XC' (C5‘). Hence AX B C = AXBC' . Also [XBY] , so Y C = YC' (C51).

For convenience we shall refer to both C51 and 3.6 as C 5 ‘. Definition. An isometry of an absolute space (of any number of dimensions) is a one-one mapping of the space onto itself that maps each point-pair onto a congruent point-pair.

c

Because we have discarded axiom C5a(ii) we cannot prove at this stage that three collinear points cannot be congruent to a triangle. Thus we cannot yet prove that an isometry maps collinear points onto collinear points.

Definition. Let 0 , P be distinct points of an absolute space. If P' is the unique point such that 0 bisects PF' , then P' is the reflection or image of P in 0 . The reflection of 0 in 0 is defined to be 0 itself. The mapping P -+ P' (for fixed 0") is called the reflection in 0 .

3.7. The reflection in 0 is an isometry.

Proof [6, §4] (Figure 3G). Let P,Q be any two distinct points. If 0 = P or if 0 = Q or if 0 , P , Q are collinear, then c l e a r l y PQ = P'Q' (2.3, 3.4 or C411). If not, then there exists X on ray OP such that OQ = OQ' = OX , and there exists Y on ray CQ such that OP = OP' = OY . Then YZ = PQ and = P'Q' (C5"). Hence PQ = P'Q' .

4. Perpendiculars; constructions

Definition (Figure 4A). Let the lines £ ,m intersect in 0 . If, for all points Q € m , and for all pairs of points .3 , B' € & such that 0 bisects SB' , we have QB = QB' , then m is perpendicular to H , and we write m ! I .

25 4.1. If 11 , m intersect in 0 , and if there exist P (. m and A fA' € I such that P j- 0 , A f A' , 0 bisects AA' and PA = PA' , then % and rn are distinct and m J_ i .

Proof. The fact that m f I follows from the uniqueness of mid-points. Let Q be any point on m , and let B , B' be any pair of distinct points on I such that 0 bisects BB' (Figure 4A). Let P' be the reflection of P in 0 . Then PA e PA' = P'A (reflecting in 0), so AOAP = AOAP' . Hence BP = BP' (C5‘) = B'P (reflecting in 0). Hence AOPB = LOPB' . Hence QB = QB' (C5‘). Hence m J_ I by defi­ nition.

Corollary. Perpendicular lines are distinct.

a

4.2. If m J_ SL then I J_ m .

Proof [12, 4.2. of. 6, 14; 16, Theorem 1](Figure 4A). There exist A,A' € I and P,P' € m such that 0 bisects both AA' and PP' . Then AP' = A'P (reflecting in 0) = AP (since &?_[_£.). Hence I J_ m (4.1).

My proofs of the next two results are simpler than those usually given [12, 4.3, 4.4. of. 6, 13; 16, Theorems 4, 6].

4.3. In any plane containing a line I , there exists at most one line perpendicular to % through a given point 0 on I .

Proof (Figure 4B). Suppose that there exist two such lines, m and « , m f n . Take P € m , Q Z n , on opposite sides of I . Then PQ meets I at R , say, R f 0 . There exists A € SL such ;hat

26 [ORA] . There exists A' € SL such that 0 bisects AA' . Since m J_ i and n J_ I , we have PA = PA/ and QA = QA' . Hence kPQA = APQA' , so iM = RA' (C51). Hence R bisects AA' , contra­ dicting 3.3.

4.4. Through any point P not on the line I , there exists at most one line perpendicular to £ .

Proof (Figure 4C). Suppose that PM , PN are both perpendicular to H , M f N . There exists P' such that M bisects PP' . There exists M' £ I such that N bisects AM' . Then P'M' = PM' (since i 1 PP') = PM (since PN _L I) = P'M . Hence P'N 1 iI (4.1). This contradicts 4.3, since P'N t PN are distinct lines.

Definition. Let & be a line and P a point. If P { H , and if there exists a perpendicular PM from P to % , unique by 4.4, where Mil, let P' be the point such that M bisects PP' . If P € I , let P' = P . Then P* is the reflection or image of P in Si . If, for fixed H , every point P (or every point P in some plane containing Z ) has a reflection in H , then the mapping P ■+ P' is called the reflection in I .

The reflection in a line, if it exists, is clearly one-one and onto; also {P')' = P , so the reflection is a mapping of period 2.

It is of interest to consider the constructibility of certain points and lines, as distinct from their mere existence. We assume that we can construct (i) the ray AB , where A , B are given distinct points, (ii) the point of intersection of two lines that are known to intersect, (iii) the point C , on any given ray through A , such that AB e AC , where A , B are given distinct points.

It follows from (i) that we can construct the line AB . We also assume that we can decide, in the course of a construction,

27 whether or not a point coincides with another point or lies on a given line, ray, or segment; and that, in the course of a construction, we can make use of an arbitrary point, either on a given line, ray, or segment, or not on it.

Constructions are usually met at an early stage in geometry as practical operations performed with straight edge, compasses, collapsing compasses (§1) etc. The meaning of constructibility in abstract geometry is roughly this: certain basic 'construction1 symbols are introduced, such as '£flm' for the point of intersection of two lines £ and m , when I and m are known to intersect, MB' for the line joining A and B , '{AB’,m}' (not a standard notation) for the point C on the ray m from A such that AC = AB , etc.', if a finite set S of points, lines etc. is given, then any point, line etc. that can be expressed as a finite sequence of the symbols of the elements of S and other arbitrary points, together with construction symbols, is said to be constructible from S , provided that the final result is independent of the arbitrary points.

Many of the subsequent proofs will be constructive proofs: we shall shew for instance not only that perpendicular lines and mid-points exist but that they can be constructed. In such cases, a statement such as 'We can construct the unique line through P perpendicular to i ' means 'There exists a unique line through P perpendicular to I , and we can construct it'.

In the course of a constructive proof, if we are given (or if we construct) points A , B , P such that PA = PB , we can say that A and B are constructibly \-related\ we can similarly define constructibly n-related points. The following results, corresponding to 2.2 and 2.5, are easily verified.

If A is constructibly related to B , if P is distinct from A , and if b is any ray from B , then we can construct Q € b such that AP = BQ . If A is constructibly n-related to B , and if AP = BQ then P is constructibly [2n+1)-related, to Q ,

4.5. If we can construct one line m perpendicular to a given line I in a given plane tt , then (i) through any point P € it , P / I , we can construct the unique line perpendicular to I ; (ii) the reflection in a , confined to the points of it , exists, and the reflection of any given point is constructible.

V P'

Proof [cf. 12, 4.5; 6, 10; 16, Theorem 5] (Figure 4D). (i) Let iCirr: = M . If P £ m there is nothing more to construct. If not, let R be any point of m on the same side of 2, as P . Construct R' € m such that M bisects RR' . PR' meets I at 0 , say; we can construct 0 . We have OR = OR' (4.2). Construct P 1 € ray0/R such that OP^ s OP ; P?1 meets II at I , say. There exists R* 6 I such that OR = OR' e OR* (we choose one of the two possible positions for R* ). There exists P* € rayO/R* such that OP e OP = OP* . Tnen PR* = P*R' (C5") e P*R (since I 1 m) e P ^ * (C5"). Hence A0R*P = AOR*P , so

XP e XP (C5'). Hence £ J_ P? x (4.1). Hence P P 1 H (4.2). The uniqueness of the perpendicular follows from 4.4. (ii) The existence and constructibility of the reflection follow immediately from (i).

4.6. If l is a line in a plane it , and if the reflection in I , confined to the points of ir , exists, then this reflection is an isometry.

Proof [12, 4.6; 6, 10; 16, Theorem 10]. Let P ,Q be distinct points

29 of ir . If either P or Q lies on £ , then PQ = P'Q' by the definition of perpendicularity. Suppose that neither P nor Q lies on Jt . Let E be any point on the same side of % as P . We shall show that PE = P'E' and PE' = P'E . If P , Q lie on the same side of . £ , the result then follows if we take E = Q . If P ,Q lie on opposite sides of I , the result follows if we take E = Q' .

If PE J_ £ , the result is immediate. If not, let PE' (1 H = 0 . (See Figure 4D, in which the line m is no longer a special line perpendicular to I as in 4.5.) Define M , P 1 and X as in the proof of 4.5. Then PP J_ £ and X bisects PP^. , as in the proof of 4.5.

Hence Pj = P' . Hence PE' e P ^ = P'E . Finally PE = P ^ ' (C5") = P'E' .

5. The existence of perpendiculars

In this section, all the points and lines referred to lie in a given plane tt .

Because we are not assuming axiom C5a(i), we cannot yet prove that if we can construct one line m perpendicular to a given line I , then through any point L of i we can construct a line perpen­ dicular to i (6.3). Hence we shall slightly alter the meaning of the terminology of [13]: if, through every pointnot on a line JI , we can construct a line perpendicular to I , we shall say that we can construct all perpendiculars to .1 . We can now restate 4.5(i).

4.5(i). • If we can construct one line m perpendicular to a given line I (in a given plane it ) then we can construct all perpendiculars to H .

5.1 [cf. 13, 1.2]. If we can construct one line perpendicular to a given line I , and if m is a line that meets I , then we can construct all perpendiculars to m .

Proof (Figure 5A). Let Z d m = 0 . If m j_ i , then i J_ m ; hence we can construct all lines perpendicular to m (4.5i). If not, let P € m , P t 0 ; construct the perpendicular PQ from P to i (4.5i), where Q £ Z . Then Q i 0 . Construct R € Z such that Q bisects OR . Construct X € rayOQ such that OP = OX and construct Y , Z € rayOP such that OQ = OY , OR = OZ (so that Y bisects OZ ). Then XZ = PR (C5") = PO (by the definition of perpendicularity) = XO . Hence XY J_ m (4.1). Hence we can construct all perpendiculars to m (4.5i).

(We do not need to construct R and Z ; we require only their existence in the proof.)

5.2 [of. 13, 1.3]. If we can construct one pair of perpendicular lines, then we can construct all perpendiculars to all lines.

Proof. Let I be one of the pair of perpendicular lines. Let n be any other line; construct a line m joining a point of I to a point of n . Then m meets Z , so that we can construct all perpendicu­ lars to m (5.1), and n meets m , so that we can construct all perpendiculars to n (5.1).

In 4.5, 5.1 and 5.2, if we assume only the existence of perpen­ diculars and not their constructibility, we obtain three existence theorems which we shall denote by 4.5*, 5.1* and 5.2*. For instance, 4.5*(i) states that if there exists a line perpendicular to a line Z , then all perpendiculars to Z exist.

Corollary of 5.2*. Either all perpendiculars to all lines exist, or there exists no pair of perpendicular lines.

5.3 [13, 1.4]. If there exists no pair of perpendicular lines, then

31 there exists no pair of congruent triangles ABC and ABC' with C and C' on opposite sides of AB .

Proof (Figure 5B). Suppose that such a pair of triangles exists. Let C C ' H A B = X and suppose, without loss of generality, that X t A . Then XC = XC' (C51). Also AC = AC' ; thus AX ]_ CC' (4.1), a contradiction.

It will be helpful to consider the basic idea behind the proof of 5.4 below. The simplest way to construct a pair of perpendicular lines in ’elementary' geometry is this: we take a pair of distinct, non­ opposite, rays OTA , OQB , with OP e OQ , OA e OB (Figure 5C); then AQ meets PS at S , say. We have SA = SB and hence OS J_ AB as in the proof of 5.3. Hence, if there exists no pair of perpendicular lines, we must have SA £ SB . If we investigate what happens when SA £ SB we can obtain a contradiction as in the proof below. We need to make use of the mid-point M of AQ , so we start the proof by taking three collinear points A , M ,Q such that M bisects AQ .

5.4 [13, 1.5]. All perpendiculars to all lines exist.

Proof (Figure 5C). Suppose the contrary; then there exists no pair of perpendicular lines (5.2*, cor.). Let A , M be distinct points. There exists Q such that M bisects AQ . Let 0 be any point not on line AM . Then OA t OQ , for otherwise OM J_ AQ , a contradiction. There exists P € rayOA such that OQ = OP , and there exists B € rayOQ such that OA = OB . Then P / A , Q f- B , and either [OPA] and [OQB] (as shown) or [OAP] and [05$] . In either case, seg/4$ meets seg PB at 5 , say. The points A , Q , B , P are all related; hence AP = BQ = QB , Also AQ = BP (C5").

There exists X € raySA such that SB = SX . Now SA f SB , for otherwise LOSA = AOSB , contradicting 5.3; hence X t A . Suppose [5Xi4] ; the proof is easily adapted if [SAX] . There exists Y € rayQ/A such that AX = QY . Since X and Y are related to A , B , P and Q , we easily see that XY = AQ . Hence XY = BP . Also SX = SB ; thus SY = SP . Hence PX = YB (C5") = BY . Hence LAXP = AQYB .

There exists Y' € AQ such that A bisects XY' . Then M bisects YY' . Let B' be the reflection of 5 in M , and let B" be the reflection of B' in A . The reflections in M and A are isometries (3.7); thus A QYB = hAY'B' = kAXB~ . Hence MXP = tAXB~ . However, P and BM lie on opposite sides of AX . This contradicts 5.3. Hence our supposition is incorrect. Hence all perpendiculars to all lines exist.

(Although we have not yet proved that every isometry maps triangles onto triangles (and not onto three collinear points) it is easily seen that the four triangles in this proof really are triangles.)

Note. It is tempting to try to turn this proof into a direct existence proof, in the following way. We have MA = MQ . If OA = OQ then OM \_AQ . If not, we define P,S,5 as before; if SA = SB then OS _[_ AB . If SA f SB then we proceed as before and finally obtain PB~ J_ AX . Hence in any case there exists a pair of perpendicular lines. However, we must now continue the argument: we deduce from 5.2* that all perpendiculars to all lines exist, and then (as in the proof of 6.1i) we deduce that in the situation of 5.4 we always have SA = SB ; so eventually we obtain a contradiction. By giving an indirect proof of 5.4 we confine the contradiction to this single proof, instead of letting it spread from 5.4 to 6.1.

33 5.5. The reflection in every line exists and is an isometry. (Use 5.4, 4.5*, 4.6.)

5.6. It is not ■possible to have distinct points C , C* on the same side of a line AB such that LABC = LABC* .

Proof (Figure 5D). Suppose that two such congruent triangles do exist.

Let C' be the reflection of C in AB . Then hABC' e AABC e LABC* . Hence, as in the proof of 5.3, C* is the reflection of C' in AB , contradicting the uniqueness of reflections. Thus congruent triangles such as ABC , ABC* do not exist.

5.7 [12, 3.2]. Three collinear points cannot be congruent to a triangle i.e. if ABC is a tria-ngle and if [A'B'C'] , then we cannot have

AB e A'B' , BC e B'C' and CA e C'A' .

•------1------o A ' 8 / C ' 5E

Proof (Figure 5E). Suppose such a situation occurs. There exists D on segment AC such that A'B' e AD and B'C' = DC . We then have

AB E A'B' e AD and CB = C'B' e CD .

There exists E such that [ABE] and AE e AC ; hence EB = CD .

There exists F such that [CBF] and CF e CA ; hence FB = AD . Hence

ED e CB and FD E AB (C5"). Hence we have EB = CD = CB = ED and

FB e AD e AB e FD . Thus AEFB = hEFD , a contradiction (5.6) since it is easily proved that B and D lie on the same side of EF .

5.8. Every isometry maps collinear points onto collinear points and preserves order', hence it maps lines onto lines. (Use 5.7 and 3.5.) 5.9 [12, 4.8], Every isometry maps perpendicular tines onto perpendicular lines.

Proof (Figure 4A). Let Z ,m be perpendicular lines, and let £, flm = 0 . Let A , A.* be distinct points on Z such that 0 bisects AA* , and let P € m , P ^ 0 . Suppose that an isometry maps Z , m , 0 , A , A.* , P onto , m-^ , 0 l, A ^ , A' 1, Pj . Then OA 5 etc.; but OA = OA* and PA = PA* ; hence 0^^ = O^A* ^ and • Hence J_£1 (4.1).

6. The construction of perpendiculars

6.1. If OA and OB are distinct congruent segments, then (i) the mid-point M of AB exists, (ii) M is constructible, and (iii) OM J_ AB if 0 , A , B are not collinear.

Proof. (i) [cf. 13, 1.7] (Figure 6A ) . If 0,A,B are collinear, then 0 is the mid-point of AB . If not, then there exists a line through 0 perpendicular to AB (5.4) meeting AB at M , say. Let 0' be the reflection of 0 in AB . Then O'A = OA = OB = O'B . If M - A , then the collinear points 0 , A , 0' are congruent to the triangle 0B0' . This contradicts 5.7; thus M t A . Similarly M ± B . Hence A00*A = AOO'B . Hence A , B lie on opposite sides of 00* (5.6) and M bisects AB as in the proof of 5.3. (ii) [13, 1.7] (Figure 6B). If 0,A,B are collinear, there is nothing more to construct. If not, let 7 be a point between 0 and A , and construct Z on ray OB such that OY = OZ . Let M be the mid-point of AB , and let OMClAZ = S . Since OM \_AB by the proof of (i), the reflection in OM interchanges A and B . Hence this reflection maps Z onto Y (5.5, 5.8). However, A , S , Z are collinear, therefore their reflections B , S , J are collinear (5.5, 5.8). Hence S - AZ (1 BY ; thus S is constructible. There­ fore M =0S[\AB is constructible. (iii) This has already been proved in (i).

6.2. (i) Through any point not on a line Jl , we can construct the

35 unique line perpendicular to £ . (ii) The reflection in every line is constructible.

Proof. This result is a simple consequence of 6.1 and 5.2. For an easy method of constructing a particular line perpendicular to £ , see [12, 5.3].

0

e

A M/Vi B P' 6C Q' 6B

6.3. Through any point on a line £ , we can construct, in any plane containing £ , the unique line perpendicular to £ .

Proof [cf. 12, 4.7](Figure 6C ) . Let L be the given point on £ . Construct a line m perpendicular to £ in the given plane; let SLf)m = M . If L - M there is nothing more to construct. If not, let P be any point of m , P £ M , and construct the reflection P' of P in £ ; P' £ m . Construct Q , Q' such that L bisects P'Q and PQ' . Construct the point N where QQ' meets £ . Then, as in the proof of 4.5(i), Q' is the reflection of Q in £ . Let M' be the reflection of M in L . Then Q' , M' , Q (the reflections of P , M , P ' in I) are collinear (5.8), so M' - N . Hence L bisects MN .

Lines MP , NQ are both perpendicular to £ , so they cannot meet (4.4); P ,Q lie on the same side of £ ; hence segments PN , QM meet, at S , say; we can construct S . The product of the reflec­ tion in £ and the reflection in L maps P , Q , M , N onto Q , P , N , M , This product is an isometry, so S = PN fl QM is mapped onto itself (5.8), and hence SM = SN . Clearly S t L , so LS J_ £ (4.1). 7. Acute and obtuse angles

In the proof that every segment has a mid-point (8,5) we shall use the ideas of acute and obtuse angles, and of inequalities between segments. An angle consists simply of two distinct non-opposite rays with a common origin. We shall not need here to define congruent angles [5, p. 99],

Definition. The angle /_ bOc is obtuse if one of the rays through 0 perpendicular to b lies inside [_ bOo and also one of the rays through 0 perpendicular to o lies inside l_ bOo .

7.1. If a. ray through 0 perpendicular to b lies inside L bOc then L bOc is obtuse.

Proof (Figure 7A). Let B € b ; there exists C € c such that OB = OC . Let the ray OD perpendicular to b meet segment BC at D . The mid-point M of BC exists, and OM j_ BC (6.1). The reflection in OM interchanges B and C , and maps D onto E , say, where [CEB] since [BBC] (5.8). Since OD ]_ OB , we have OE 1 OC (5.9) and OE lies inside Z bOc . Hence L b O c is obtuse.

Definition. If ray Ob is perpendicular to ray Oc , then /_ bOc is a right angle. An angle that is neither obtuse nor right is acute.

From the existence of right angles, and from results in ordered geometry, we easily deduce the next result.

7.2. If Ob , Ob' are opposite rays, and if ray Oc is distinct from Ob and Ob' then either L bOc and L b'Oc ax>e both right angles or one of them is acute and the other obtuse.

37 7.3. If the triangle ABC has a right angle at B f then L BAC and L BCA are both acute.

Proof (Figure 7B). If /_ BAC is obtuse, then a ray through A perpen­ dicular to AB lies inside L BAC and hence meets segment BC at P , say. We then have two perpendiculars through P to AB , a contradiction (4.4), Similarly [_ BAC cannot be a right angle. Hence L BAC is acute. Similarly /_ BCA is acute.

8. Inequalities between segments

Definition (Figure 8A ) . Let AB , PQ be directed segments with A ~P . Then there exists R on ray PQ such that AB = PR (2.2). If [PRQ'\ , then AB is less than PQ and PQ is greater than AB , and we write AB < PQ or PQ > AB .

The following results are easily proved.

8.1. (i) If AB , PQ are directed segments with A ~ P , then either AB < PQ or AB = PQ or AB > PQ ; these three -possibilities are mutually exclusive. (ii) If AB < PQ and PQ < XY , or if AB < PQ and PQ = XY , or if AB = PQ and PQ < XY , then AB < XY .

Since we have not yet shown that all points in a plane are related, this relation of inequality is essentially a relation between directed segments. If AB < PQ it may be meaningless to try to compare BA and PQ , or BA and QP . I have constructed a one­ dimensional geometry [15, §4] in which AB = PQ with [APQB] as shown in Figure 8B. In this geometry we have AB = PQ < PB but BA > BP . 8.2. If the triangle ABC has a right angle at B , and if [BDC]

or if D = B , then AD < AC.

Proof (Figure 8C). Suppose [BDC] and AD > AC . Then there exists E on segment AD such that AC = AE . Line CE meets segment AB at F , where [CBF] . The mid-point M of CE exists and AM J_ CE (6.1). Then Z BFC is acute from the right-angled triangle BFC , and Z AFC is acute from the right-angled triangle AFM . This contradicts 7.2.

We obtain a similar contradiction if AD = AC , and the proof is similar if D = B .

The proof of the next result is easy.

8.3. If the triangle ABC has a right angle at B , and if D is the foot of the perpendicular from B to AC , then D lies "between • * A and C .

8.4. If the triangle ABC has a right angle at B , if BC < BA and if P lies between A and C , then BP < BA.

Proof (Figure 8D). Define D as in 8.3; then [A.DC] . If [APD] or

if P = D, then BP < BAby 8.2. If [DPC] as in the figure, then

BP < BC < BAby 8.2.

8D SE

8.5. In an absolute geometry of dimension at least tw o, every segment has a mid-point, which is construotible, so that all points are 1-related.

39 Proof (Figure 8E). Let A ,B be any two points, and tt a plane con­ taining them. On the perpendicular through A to AB in tt , there exists P such that AB = AP . On the perpendicular through B to BA in ir , there exists Q on the opposite side of AB from P , such that BA 5 BQ .

Since the lines AP , BQ do not meet (4.4), we can use theorems of ordered geometry to show that segment PQ meets segment AB at R , say. Suppose R is not the mid-point of AB ; then without loss of generality RA < RB . Hence there exists A' such that RA = RA' and [RA'B] . Also there exists P' on ray RQ such that RP = RP' . The reflection in R shows that A'P' J_ A'R (5.9); hence we must have [RP'Q] for otherwise the perpendiculars to RB at A' and B will meet each other.

Now BP' < BQ (8.4, since BR < BA = BQ ) and P'A' < P'B (8.2). Hence there exists P" such that [BP"Q'\ and BP' = BP" , and there exists A" such that [P"A"B~\ and P'A' = P~A~ (since P'A' < P'B = P"B) . We now have P~A~ = P'A' = PA (from the reflection in R) = BA = BQ .

Now B , P~ are related, so that P~B = BP" ; hence there exists X such that [BXP"] and P"A" = BX . Hence BX = BQ , a contradiction.

Hence R is the mid-point of AB . Clearly R is construc- tible since P and Q are constructible; since RA = RB , A and B are 1-related.

9. Conclusion

Since all points are related, we can now easily prove axioms Cl - C4, and using reflections in points and lines it is not difficult to prove axioms C5 and C7 in a plane. We need some further theory to prove C5 and C7 in spaces of higher dimension; for this the reader is referred to [12, §8]. Absolute geometry can now be built up in the usual way.

A r\ 10. Appendix

I have not found an answer to the following interesting question in any of the standard texts. Let A be a point not on the line a in an absolute plane, and let L be the foot of the perpendicular from A to & . Let P be a point such that AP > AL . Does there

exist a point Q € 1 such that AP = AQ ? In other words, does the circle with centre A and radius AP intersect I ?

If we assume the axiom of continuity, the answer is "yes' [5, p. 308], but let us analyse the general situation algebraically for planes of Euclidean type.

Let F be a totally ordered field. In the usual way we can define an ordered plane tt whose points are ordered pairs of elements of F and whose lines are determined by linear equations. We use distance to define congruence: the distance between (x^z/j) and (x 2 ,y2) is

[Gcx -^ 2)2 + G/j - y 2)2Y 2 , and we shall impose a condition on F to ensure that this expression is an element of F . We define AB = CD if the distance from A to B equals the distance from C to D .

If the plane is to satisfy axiom Cl, then the distance from the origin to (1,20 must be measurable along the x-axis, so that the field F must satisfy condition (A):

(A) If b € F then Cl + b2) 2 ( F, i.e., 1 + b 2 has a square root in F . Conversely, if (A) is satisfied, then distance is always defined as an element of F and axiom Cl is satisfied. For

with an obvious modification if x = x 2 . Also, let I be the line through the origin 0 with gradient m ; then there exist points on

41 I at any given distance d from 0 , namely

+ f d md y (1 + m 2) /(I + m2)_

Now let o Z F , a > 1 . The circle with centre 0 and radius o intersects the line x = 1 if and only if /(c2 - 1) € F .

I am indebted to Dr M. N. Huxley for the following example of a totally ordered field F satisfying (A) but containing an element a > 1 such that o2 - 1 has no square root in F .

Let Q(:c) be the field of rational functions of the indeterminate x with coefficients in the rational field Q . Define a sequence of fields inductively as follows: KQ = Q(:c) ; K^ is the smallest field containing all elements of the form /(l + b2) where b € K ° ^ n (the elements of each are formal expressions involving quotients and roots). Let Q = y K . Then Q satisfies condition (A). x n=0 n x

Let t be any real transcendental number, and let Q be the field of all elements /(£) where fix) € . It is easily seen by induction that fit) is always real, so is a subfield of the real field R . Moreover fit) = 0 in R implies fix) = 0 in Q , otherwise t would be algebraic over Q . Hence the mapping fix) -> fit) is a one-one mapping of onto and into R .

Consider the field F = Q , where it now has its usual 7T transcendental meaning and no longer stands for a plane; F is totally ordered, being a subfield of R . Also ir € F and tt > 1 .

10.1. tt2 - 1 has no square root in F .

Proof. Suppose it2 - 1 = [/(tt)]2 > where fix) £ . Then

x 1 - 1 = [/Or)]2 in because of the one-one mapping of Q onto Q . Hence, 'putting = i t £ - 1 • [ / ©2 I in «1/t • a contradiction since the left hand side is negative.

Thus the answer to the question at the beginning of this section is in general ’no ’.

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