A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY WILLIAM RICHTER Abstract. This note is intended to be useful to good high school students wanting a rigorous treatment of Hilbert s axioms for plane geometry. 1. Introduction Euclid s Elements [6] is very creative geometry, which e.g. proves the triangle inequality (Prop. I.20) without the Pythagorean theorem or the parallel postulate. However Euclid s work contains errors involving angle addition, which were found and fixed by Hilbert [14], by adding betweenness axioms. It seems that Euclid s errors are made in every widely used high school Geometry text in the US today. A good high school student should be able to read Moise s or Venema s college geometry texts [17, 19], which have both rigorous and elegant proofs. A more interesting textbook choice might be Hartshorne s book [12], which gives an excellent exposition of Euclid s Elements [6] together with the Hilbert fixes needed [12, Thm. 10.4]. However Hartshorne s book contains gaps that should trouble a high school student. In 7 we fill in needed details to his proofs of Euclid s Propositions I.7, I.9, I.21 and I.27, and work his overly difficult parallelogram exercise [12, Ex. 10.10]. Venema lists the minimal version of Hilbert s axioms [19, App. B1], but he uses stronger axioms, which obviate much of 3 and 4. The geometry texts for graduate students [7, 12, 15] also use stronger Hilbert s axioms. We prove that the minimal Hilbert s axioms suffice. This result was proved by Hilbert, Moore, Wylie, and Greenberg [14, 11, 20, 7], but our proofs seem elegant, and are collected in one source. Our hope is that a good high school student who reads our Hilbert proofs and our fixes of Hartshorne s proofs can then read Hartshorne s book. Learning rigorous axiomatic geometry is especially a good idea now because of new developments in formal theorem-provers explained by Hales [10], who is attempting to check his proof of the Kepler sphere-packing conjecture [9] on a computer. Students can code up their axiomatic geometry proofs and check them in one of the proof-checking compilers like HOL Light, Mizar, Isabelle or Coq. See http://www.math.northwestern.edu/~richter/tarskiaxiomgeometry.ml for 1000 lines of HOL Light code partially proving Tarksi s theorem [18] that his much weaker geometry axioms imply Hilbert s axioms. The code is largely a much more human readable port of Narboux s proofs in the form of Coq psuedo-code http://dpt-info.u-strasbg.fr/~narboux/tarski.html. Venema discusses a widely used high school Geometry text [2], but not its lack of rigor. A high school student reading Venema s text might ask why they should not read [2] instead. In 10 we discuss Venema s proof of the triangle sum theorem (Lemma 10.1) and contrast it with the inadequate proof of [2]. Venema also uses 1

2 WILLIAM RICHTER Birkhoff s axioms [4] that the real line R measures segments and angles, but he does not explain Birkhoff s work. In 9 we discuss Birkhoff s work, explained by MacLane [16], which gives an alternative to Hilbert s work by using directed angles and taking the crossbar theorem and its converse as an axiom. Hilbert s axiom [14] are actually for 3-dimensional space, and in 8 we discuss a minimal version of Hilbert s entire set of axioms, and prove various 3-dimensional results of Euclid [6], profiting from the rigorous solid geometry proofs in [1]. Greenberg s text [7] seems too difficult for high school students, but it seems to be the fundamental text on Hilbert s work, and it was a big influence on our paper (and [12, 15]) and and we cite his results freely. We reprove some of his results. Greenberg s survey [8] gives reasons to use Hilbert s axioms instead of, following Birkhoff, assuming R, and contains interesting mathematical logic. In 2 we explain Hilbert s simplest axioms, e.g. that two points determine a line, and explain how we, using set theory, do not give strictly axiomatic proofs. In 3 we give Hilbert s axiom for betweenness of points and show that a line l defines an equivalence relation l. Our proof of the crossbar theorem (Thm. 3.8), a tool that shows two lines intersect, is simpler than the proofs of Greenberg, Moise and Venema [7, 17, 19], as we do not use their plane separation axiom (Prop. 4.13). In 4 we show a line separates the plane into two half-planes ([7, 12, 17, 19] take this as an axiom). First we give in Prop. 4.1 Wylie s proof [20, 4] that Greenberg s strengthened version of Hilbert s axiom II.5 follows from Hilbert s (our axiom B4). Next we reprove (Prop. 4.10) Moore s result [11] that Hilbert s axiom II.4 is redundant, with essentially Moore s proof. (Wylie [20, 6] gave an interesting and different proof of Moore s result.) These two results imply Prop. 4.13. In 5, we begin proving the surprisingly difficult results (Prop. 7.9 and 10.2) that a quadrilateral with opposite sides (or angles) congruent is a parallelogram. In 6 we give Hilbert s axioms for congruence of both segments and angles. We deduce Greenberg s angle addition result from an angle subtraction result and also an angle ordering result which Greenberg deduces from the angle addition result. The SAS axiom in [7, 12, 19] is stronger than Hilbert s, and we give Hilbert s proof (Theorem 6.2) that the weaker SAS axiom C6 suffices. In 7 we explain how Hartshorne s rigorization of Euclid s Prop. I.10 shows that Hilbert s axiom II.2 follows from our weaker axioms. In 7 we adopt the Euclidean parallel postulate for the remainder of the paper. Thanks to Bjørn Jahren, who recommended [7, 12], found [20] and discovered the proof of Prop. 4.1 independently. Thanks to Benjamin Kordesh, the high school Geometry student test pilot for the paper, Miguel Lerma, who found [11], Takuo Matsuoka, and Stephen Wilson, who recommended [19], for helpful conversations. 2. Hilbert s simplest axioms Following Venema, we shall give a set-theoretic version of Hilbert s axioms. We are given a set which is called a plane. Elements of this set are called points. We are given certain subsets of a plane called lines. There are axioms that a plane and its subset lines must satisfy, explained here and also 3 and 6. In 8 we will consider many different planes, but for now we discuss one plane, referred to as the plane. A line l is then the set of all point P in the plane so that P l. Two lines l and m are then equal if P l iff P m, for all points P. As is usual we will write iff for if and only if. We have as usual the intersection l m of two lines l and m,

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 3 the set of points P in the plane such that P l and P m. We will call points collinear if there is a line in the plane containing them. Hilbert s simplest axioms (I for incidence) are I1. Given distinct points A and B in the plane, there is a unique line l in the plane such that A l and B l. I2. Given a line l in the plane, there exists two distinct points A and B in the plane such that A l and B l. I3. There exist three distinct non-collinear points in the plane. We will call AB the unique line in the plane containing A and B. We will need Lemma 2.1. Given two distinct lines l and m, if X l m, then l m = {X}. Proof. Suppose that A X and A l m. By axiom I1, l = AX = m. This contradicts our assumption, so there can not exist any such A. We are not working in a purely axiomatic framework, where proofs are given using only the axioms and logic. The plane is what logicians call a model of Hilbert s plane geometry axioms, and we use Hilbert s axioms together with set theory to give proofs in the model. The advantage of a model is that we can use our set theory skills, and this simplifies proofs, and even the statement of axioms. In a purely axiomatic framework, Lemma 2.1 would be written in this hard to read fashion. For all lines l and m with l m, and for every point X, if X l and X m, then for every point Y with Y X, either Y l or Y m. So it s very convenient to use sets freely and not write purely axiomatic proofs. Later we will define an angle as a set of two elements, which we could not define in a purely axiomatic framework, and more circumlocution would be required. So we will use set-theoretic constructions like l m, {X}, and complements. So for any line l, the complement of l is the set of points X in the plane not on l. Given two lines l and m, the complement l m is the set of points P with P l and P m. Set theory allows us to make arguments like the following. Given two sets L and M with L M and M L, it follows that L = M. By using set theory, we make our proofs easier and practice an important mathematical skill. There is a trade-off in our free use of set theory, though. Axiomatic geometry is important is for artificial intelligence (AI), because a robot supplied with the Hilbert axioms can prove everything about points, lines and planes that we can prove using Hilbert s axioms. This follows from what is called Goedel s Completeness Theorem in First Order Logic (FOL). But to achieve this this AI advantage, we would have to phrase our proofs entirely in FOL, and not freely use set theory. See http://www.math.northwestern.edu/ richter/richtertarskimizar.tar for essentially FOL Mizar code which partially proving a theorem of Tarski [18]. The formal theorem provers like HOL Light and Mizar make it quite attractive to learn and write up axiomatic proofs which can then be checked on a compiler. 3. The betweenness relation and the crossbar theorem We have an undefined betweenness relation. For all points A, B and C we have the statement A B C, which is supposed to capture the idea that B lies between A and C on some line. The betweenness axioms are B1. If A B C, then C B A, and A, B and C are distinct points on a line. B2. Given two distinct points A and B, there exists a point C such that A B C.

4 WILLIAM RICHTER B3. If A, B and C are distinct points on a line, then exactly one of the statements A B C, A C B and B A C is true. B4. Let A, B and C be non-collinear points not on line l. If there exists D l so that A D C, there exists an X l such that A X B or B X C. Using, we define the open interval (A, B) as the subset of the plane (A, B) = {C A C B} AB. Then A C B iff C (A, B). Note that (A, A) =. All we will ever use of B1 is B1. For any points A and C, there is a subset (A, C) of the plane. If A C, then (A, C) AC {A, C} and (A, C) = (C, A). Furthermore (A, A) =. We use axiom B1 quite often without mentioning it. We often refer to the equality of sets (A, C) = (C, A) as symmetry. Using open intervals, we have equivalent versions of axioms B2 and B3. B2. Given distinct points A and B, there exists a point C such that B (A, C). B3. Given distinct collinear points A, B and C, exactly one of the statements A (B, C), B (A, C) and C (A, B) is true. Axiom B4 is Hilbert s axiom II.5. Hilbert draws a picture of a triangle with vertices A, B and C with a line l intersecting the open intervals (A, C) and (A, B). Take a line l. We define the relation l on the complement of l. Take two points A, B l. We write A l B if (A, B) l =. We write A l B if (A, B) l. We have two equivalent versions of B4. B4. Take a line l and three non-collinear points A, B, C l. If A l C, then A l B or B l C. B4. Take a line l and three non-collinear points A, B and C not on l. If A l B and B l C, then A l C. B4 is a restatement of B4 using open intervals. B4 is equivalent to B4 because B4 follows from B4 by a short proof by contradiction, and vice versa. In Proposition 4.13 we prove a stronger version of axiom B4 that Greenberg [7, p. 64] essentially takes as an axiom. A simple result using the relation l is Lemma 3.1. Take a point on a line X m and a line l with m l = {X}. Take points A, B m {X}, and suppose that X (A, B). Then A, B l and A l B. Proof. A l, since A X and m l = {X}. Similarly B l. Assume (for a contradiction) that A l B. Then there exists a point G (A, B) l. By axiom B1, (A, B) m. So G m l = {X}. Thus G = X. But this contradicts X (A, B). Hence A l B is false, so A l B. A relation is an equivalence relation if it is reflexive, symmetric and transitive. Lemma 3.2. Take a line l. Then the following statements are true. (1) The relation l is an equivalence relation on the complement of l. (2) Take points A, B and C not on l. If A l B and B l C, then A l C. Proof. l is obviously reflexive and symmetric, as (A, A) = and (A, B) = (B, A). Transitivity is the second assertion (2), which we now prove. So take points A, B and C not on l with A l B and B l C. We must prove A l C. By axiom B4, A l C if A, B and C are not collinear. Since l is reflexive, A l C if A, B and C are not distinct. So we can assume that A, B and C are distinct collinear points contained in a line m. By axiom I1 we can write m = AC.

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 5 If the lines l and m do not intersect, then clearly A l C. So assume that l and m intersect. The lines l and m are distinct, because our three points belong to m but not l. Therefore, m l = {X}, for some point X, by Lemma 2.1. By axiom I2, there exists a point E X on l. Thus E m. There exists a point B with B (E, B ), by axiom B2, since B E, since E l and B l. Then B B, by axiom B1. Let n = EB, so B n, by axiom B1. By Lemma 2.1, n l = {E}, because B l. Therefore B l. By axiom B3, E (B, B ). So B l B, by Lemma 3.1. The lines m and n are distinct, because E n. Thus by Lemma 2.1, m n = {B}. Thus A n, and so A, B and B are not collinear. Since A l B and B l B, axiom B4 implies that A l B. By an argument similar to the previous paragraph, C, B and B are not collinear, and C l B. Since l is symmetric, B l C. But A, B and C are also not collinear, since m n = {B}. Thus by axiom B4, A l C. This proves (2). Since this shows l is transitive, this proves (1). We use Moise s definition of a ray [17, p. 55], which is simpler than Greenberg s. Definition 3.3. For distinct points O and P, the ray OP is the subset of the plane (1) OP = {X OP O (X, P )}. Let O, A and B be non-collinear points, and let a = OA and b = OB. Define the angle AOB as the (unordered) set consisting of the two rays AOB = { OA, OB}. We define the interior of the angle AOB as the subset of the plane int AOB = {P P a, P b, P a B and P b A}. We prove a result of Greenberg s [7, Prop. 3.7] that is in a sense a converse to the Crossbar Theorem 3.8 we prove below. Lemma 3.4. Take non-collinear points O, A and B in the plane and a point G (A, B). Then G int AOB. Proof. Let a = OA, b = OB, and l = AB. Then a l = {A}, by Lemma 2.1, since B a. Hence G a, as G l and G A, by axiom B1. A (G, B), by axiom B3, since G (A, B). Thus G a B, by Lemma 3.1, since a l = {A} and A (G, B). Similarly G b and G b A, as B (G, A). Thus G int AOB. We prove Greenberg s result [7, Prop. 3.8(a)]. Lemma 3.5. If X int AOB, then OX {O} int AOB. Proof. Take P OX {O}. Let a = OA, b = OX and x = OX. x a = {O}, by Lemma 2.1, since X a. Similarly x b = {O}. Since X int AOB, X a B and X b A. Then P a and P a X, by Lemma 3.1, since O (P, X). Similarly P b and P b X. By Lemma 3.2, P a B and P b A. Thus P int AOB. We prove an easy trichotomy law for interiors of angles, inspired by [7, Prop. 3.21]. Lemma 3.6. Take distinct points O and A. Let a = OA. Take points P, Q a with P a Q and P, Q and O non-collinear. Then P int QOA or Q int P OA.

6 WILLIAM RICHTER Proof. Let p = OP and q = OQ. Then q a = {O}, by Lemma 2.1, since Q a. Thus A q, since A O. Then p q = {O}, by Lemma 2.1, since P q. Assume P int QOA. Then P q A, since P q, P a and P a Q. Take a point G (P, A) q. Then G int P OA by Lemma 3.4, so G a P, G a, and G O. Since P a Q, Lemma 3.2 implies Q a G. Thus O (Q, G), and then Q OG {O}. Hence Q int P OA, by Lemma 3.5. We give a simpler proof of Greenberg s result [7, Prop. 3.8(c)]. Lemma 3.7. Take non-collinear points O, A and B, a point D int AOB and a point A with O (A, A ). Then B int DOA. Proof. Let a = OA and b = OB. By axiom I1, a = OA. Then b a = {O}, by Lemma 2.1, since B a. Then A b and A b A, since A a {O}. Since D int AOB, D a, D b, D a B, and D b A. Thus D b A, by Lemma 3.2 and a short proof by contradiction. Hence D int BOA. Thus B int DOA, by Lemma 3.6, since B, D a, D a B, and D b. We give a simpler proof of Greenberg s crossbar theorem [7, p. 69]. Theorem 3.8. Given non-collinear points O, A and B, and a point D int AOB, the ray OD intersects the interval (A, B). Proof. By axiom I1 there exists lines a = OA and b = OB, since O, A and B are distinct, since they are non-collinear. B a by non-collinearity. D int AOB implies D a B, D a, D b, and D O. Let l = OD. Then a l = {O} = b l, by Lemma 2.1. Thus A, B l, since A O and B O. By axiom B2 there is a point A a such that O (A, A ). Then A l, since A O, and A l A. B int DOA, by Lemma 3.7, so B l A. Then B l A, since A l A, by Lemma 3.2 and a proof by contradiction. Thus there is a point G (A, B) l. Then G int AOB by Lemma 3.4, so G a and G a B. Thus G a D by Lemma 3.2, since D a B. Then O (G, D), so G OD. Thus G (A, B) OD. Our proof of the Crossbar Theorem 3.8 is simpler than the proofs of Greenberg [7, p. 69], Venema [19, Thm. 5.7.15] and Moise [17, p. 69], which use Proposition 4.13, but only really need Lemma 4.11, based on Lemmas 4.4 4.8 which order four points. 4. plane separation We prove below (Prop. 4.13) what Moise and Venema call the plane separation postulate, which is also essentially Greenberg s axiom B4. Wylie s argument [20, 4] proves what Greenberg calls Pasch s theorem [7, p. 67]. Proposition 4.1. For non-collinear points A, B, C l, either A l B, A l C or B l C. Proof. Suppose (for a contradiction) that A l B, A l C and B l C. There exist points X, Y, Z l so that X (A, B), Y (A, C) and Z (B, C). By axiom B3, one of X, Y, Z l lies between the other two. We may assume that Y (X, Z). Let p = BA, q = BC, and m = AC. Then p q = {B}, by Lemma 2.1, since A q. Hence X q. But q = BZ by axiom I1. Thus X, B and Z are non-collinear. p m = {A}, by Lemma 2.1, since B m. Thus X m. By axiom B3, A (B, X). Thus X m B, by Lemma 3.1. Similarly Z m and B m Z. Thus X m Z by axiom B4. But Y (X, Z), so X m Z. This is a contradiction.

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 7 Now we prove, in a series of exercises, the collinear case of Proposition 4.1. We specialize Lemma 3.2 to a fixed line. Lemma 4.2. Take a point on a line O m and points P, Q, R m {O}. If O (P, Q) and O (Q, R), then O (P, R). Proof. By axiom I3, there exists E m. Let l = OE. Then m l = {O}, by Lemma 2.1, because E m. Then P l Q and Q l R, by Lemma 3.1 and the hypothesis. Thus P l R, by Lemma 3.2. Therefore O (P, R). We now show that the ray OP is in a sense independent of P. Lemma 4.3. Take points O, P and Q O, with P OQ {O}. Then OP = OQ. If R (O, Q), then OR = OQ. Proof. Let m = OQ, so P, Q m {O}. By axiom I1, m = OP, so OP m. Take X OP {O}. Then O (X, Q), by Lemma 4.2, since P, Q, X m {O}, O (X, P ), and O (P, Q). Hence X OQ. Thus OP OQ. But O (Q, P ), so Q OP {O}. The same argument shows that OQ OP. Thus OP = OQ. Now assume R (O, Q). Then O, R and Q are distinct collinear points, and O (R, Q), by axiom B3. Thus R OQ {O}. By the first part, OR = OQ. We will prove Hilbert s axiom II.4 is redundant using only the following result. Lemma 4.4. Given points A, O and B with O (A, B), we have OA OB = {O}. If X OB {O}, then X OA. Proof. O OA OB. So the second statement implies the first statement. So take X OB {O}. A, O and B are distinct by axiom B1. Let m = AB. Then m = OB by axiom I1. Then A, X, B m {O}. Since O (X, B) and O (A, B), we have O (A, X), by Lemma 4.2 and a simple proof by contradiction. Thus X OA. Given distinct collinear points A, B, C and D, we will say that the 4-tuple A, B, C, D is an ordered sequence if B (A, C), C (B, D), B (A, D) and C (A, D). Note that by symmetry, if A, B, C, D is an ordered sequence, then D, A, B, C is also an ordered sequence. An an easy corollary of Lemma 4.4 is Lemma 4.5. Take points A, B, C and D with B (A, C) and C (B, D). Then B (A, D), and A, B, C, D is an ordered sequence. Proof. B, C and D are distinct by axiom B1, since C (B, D). Let l = BC. Then A, D l by axiom B1, since B (A, C). B (D, C), by axiom B3. So D BC {B}. Thus D BA, by Lemma 4.4. Thus B (D, A), since l = BA, since A B by axiom B1. By symmetry, B (A, D). C (D, B) and B (C, A), by symmetry. By the first statement, C (D, A). Thus C (A, D). This proves the second statement. With our plane separation Proposition 4.13 in mind, we now prove Lemma 4.6. Take points B (A, C). Then (A, B) (A, C). Proof. Take X (A, B). Then X BA {B}, since X B and B (X, A), by axiom B3. So X BC, by Lemma 4.4, since B (C, A). Thus B (X, C). Hence X (A, C), by Lemma 4.5, since X (A, B) and B (X, C).

8 WILLIAM RICHTER The proof of the above result proves also that Lemma 4.7. Let A, X, B and C be distinct points on a line, and suppose that X (A, B) and B (A, C). Then A, X, B, C is an ordered sequence. This implies Lemma 4.8. Let A, B, C and X be distinct collinear points. Then at least one of the statements X (A, B), X (B, C) and X (A, C) is true. Proof. By axiom B3, one of the distinct points A, B and C is between the other two. We may assume that B (A, C), as our result is symmetric in A, B and C. If X (A, B), we are done, so assume X (A, B). Then A, X, B, C is an ordered sequence by Lemma 4.7, so B (X, C). Thus X (B, C), by axiom B3. Now we have the 4-tuple ordered sequence result we have been aiming for. Lemma 4.9. Let A, B, C and X be distinct collinear points, with B (A, C). (1) Either A (X, B) or X (A, B) or X (B, C) or C (B, X). (2) One of the 4-tuples X, A, B, C, A, X, B, C, A, B, X, C or A, B, C, X is an ordered sequence. Proof. A (X, B) or X (A, B) or B (A, X), by axiom B3 applied to A, B and X. Assume B (A, X). Since B (A, C), we have B (C, X) by Lemma 4.8. Then by axiom B3, X (B, C) or C (B, X). This proves (1). To prove (2), we consider four cases. If A (X, B), then X, A, B, C is an ordered sequence by Lemma 4.5, since B (A, C). If X (A, B), then A, X, B, C is an ordered sequence by Lemma 4.7. If X (B, C), then X (C, B) and B (C, A), by symmetry. Thus C, X, B, A is an ordered sequence by Lemma 4.7, and so by symmetry, A, B, X, C is also an ordered sequence. If C (B, X), then A, B, C, X is an ordered sequence by Lemma 4.5. Hilbert s axiom II.4 is now an easy corollary of Lemma 4.9. Proposition 4.10 (Moore). Any four distinct collinear points can be renamed P 1, P 2, P 3 and P 4 so that P 1, P 2, P 3, P 4 is an ordered sequence. Proof. Call three of the points A, B and C and the fourth point X. By axiom B3, one of the points A, B and C must be between the other two. Since we are allowed to rename the points, we can assume that B (A, C). By Lemma 4.9, we have one of four possible ordered 4-tuples. In all four cases we can rename the points P 1, P 2, P 3 and P 4 to obtain our ordered sequence P 1, P 2, P 3, P 4. Lemma 4.8 also implies a result suggested by Lemma 4.4. Lemma 4.11. Take points O (A, B) on a line l. Then l = OA OB. Proof. We prove each set is a subset of the other. OA l, since OA = l, by axioms B1 and I1. Similarly OB l. Thus OA OB l. For the reverse inclusion, take X l with X OB. We must show that X OA. If X = A we are done, so assume that X A. Then O, X, A and B are distinct points in l, since O (X, B) and O (A, B). Thus O (X, A), by Lemma 4.8, so X OA. Hence l OA OB. Therefore l = OA OB. OA is called the opposite ray of OB in l. Proposition 4.1 and Lemma 4.8 imply

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 9 Corollary 4.12. For points A, B, C l, either A l B, A l C or B l C. Proof. If A, B and C are non-collinear, we are done by Proposition 4.1. If two of A, B and C are equal, we are done since l is reflexive. So assume A, B and C are distinct points in a line m. If l m =, then we are done, as all three statements are true. So assume l m. Then l m = {X}, for some point X, by Lemma 2.1, since A l. Then either X (A, B), X (A, C) or X (B, C), by Lemma 4.8. Thus either A l B, A l C or B l C, by Lemma 3.1. We call a set U convex if A, B U implies (A, B) U. We prove what Moise [17, PS-1, p. 62] and Venema [19, Axiom 5.5.2] call the plane separation postulate. Proposition 4.13. The complement in the plane of a line l is a disjoint union of two nonempty convex sets H 1 and H 2. If P H 1 and Q H 2, then (P, Q) l. Proof. By axiom I3, there is a point A l. Take a point E l. Then AE l = {E} by Lemma 2.1, since A l. By axiom B2, there exists a point B such that E (A, B). Then B l, since B E. Thus A l B. Define H 1 = {X l X l A} and H 2 = {X l X l B}. H 1 and H 2 are disjoint nonempty sets, since l is an equivalence relation (Lemma 3.2). If C l, then by Corollary 4.12, either A l C or B l C, as A l B. Thus the complement of l is the disjoint union H 1 H 2. If P H 1 and Q H 2, then P l Q, by Lemma 3.2. Thus (P, Q) l. Take P, Q H 1, so P l A and Q l A. By Lemma 3.2, P l Q. Let X (P, Q). Then (P, X) (P, Q) by Lemma 4.6. So (P, X) l (P, Q) l =. Thus X l P, and so X l A,, by Lemma 3.2. Thus X H 1. Hence (P, Q) H 1. Thus H 1 is convex. Similarly H 2 is convex. The disjoint convex sets H 1 = {X l X l A} and H 2 = {X l X l B} above are called the half-planes bounded by l. Since l is an equivalence relation, H 1 and H 2 do not depend on the choice of points A H 1 and B H 2. Take non-collinear points O, A and B and a point D. The ray OD is between OA and OB if D int AOB. By Lemma 4.3, the betweenness relation of rays does not depend on the points A, B and D, but only the rays OA, OB and OD. A triangle ABC is an ordered triple A, B, C of three non-collinear points. Lemma 4.14. Take a triangle AOB, and points G int AOB and F AOG. Then F int AOB. Proof. By the Crossbar Theorem 3.8, there exists a point G 0 (A, B) OG. By Lemma 4.3, OG 0 = OG, and AOG0 = AOG. So we may replace G by G 0. Then G (A, B). By the Crossbar Theorem 3.8, there exists a point F 0 (A, G) OF. By Lemma 4.3, OF 0 = OF. Then F0 (A, B), since Lemma 4.6 implies that (A, G) (A, B). Thus F 0 int AOB, by Lemma 3.4. Hence F int AOB, by Lemma 3.5, since F OF 0 {O}. 5. quadrilaterals Moise has a nice treatment on quadrilaterals [17, 4.4] which we expand on. With Proposition 7.5 in mind, we first generalize the notion of a quadrilateral. We define a tetralateral to be an ordered 4-tuple ABCD of distinct points in the plane, no three of which are collinear. Take a tetralateral ABCD. We define (for this section) the lines a = AB, b = BC, c = CD, d = DA, l = AC, and

10 WILLIAM RICHTER m = BD. Since no three points of the tetralateral ABCD are collinear, we have C, D a, A, D b, A, B c, B, C d, B, D l, and A, C m. There are six intersections of the four open intervals (A, B), (B, C), (C, D) and (D, A). Then Lemma 5.1. For a tetralateral ABCD, we have four empty intersections: (A, B) (B, C) = (B, C) (C, D) = (C, D) (D, A) = (D, A) (A, B) =. Proof. (A, B) (B, C) a b = {B}, by Lemma 2.1, since A b. Since B (A, B), by axiom B1, (A, B) (B, C) =. The other intersections are similarly empty. We define a quadrilateral ABCD [17, p. 69] to be a tetralateral ABCD where (A, B) (C, D) = and (B, C) (A, D) =. Given a quadrilateral ABCD, all six intersections of the four open intervals (A, B), (B, C), (C, D) and (D, A) are empty, by Lemma 5.1. A quadrilateral ABCD is convex [17, p. 69] if A int BCD, B int CDA, C int DAB, and D int ABC. The term convex quadrilateral indicates that the inside of the quadrilateral is a convex set, but we will not pursue this. Moise explains the term convex quadrilateral is inconsistent: the union of four non-collinear line segments (or four points) cannot be a convex set. We begin by proving an exercise of Moise [17, Ex. 4, 4.4]. Lemma 5.2. Take a quadrilateral ABCD. Then either A c B or C a D. Proof. Assume that C a D. Then there exists a point G a (C, D). Then by Lemma 2.1, a c = {G}, since A c. But G (A, B), since (A, B) (C, D) =. Furthermore A, B a {G}, since A, B c. Thus A c B, by Lemma 3.1. We generalize Moise s [17, Thm. 1, 4.4] slightly, with the same proof. Lemma 5.3. Take a tetralateral ABCD with B l D and A m C. Then there is a point G (A, C) (B, D), and we have a convex quadrilateral ABCD. Proof. Since A m C, there exists a point G (A, C) m. So G l m. Then l m = {G}, by Lemma 2.1, since A m. Thus B, D m {G}, since B, D l. Hence G (B, D), since B l D, by Lemma 3.1, and a short proof by contradiction. Thus G (A, C) (B, D). Then B (D, G), by axiom B3, so D BG {B}. Now G int ABC, by Lemma 3.4. Thus D int ABC, by Lemma 3.5. A similar argument shows that A int BCD, B int CDA, and C int DAB. We prove Moise s result on the diagonals of a convex quadrilateral. Lemma 5.4. Take a convex quadrilateral ABCD. Then B l D and A m C. There is a point G (A, C) (B, D), and ABDC is not a quadrilateral. Proof. By the Crossbar Theorem 3.8, B l D, since A int BCD. A m C. Thus there exists a point G (A, C) (B, D), by Lemma 5.3. ABDC is not a quadrilateral, because (C, A) (B, D). Similarly We need this tetralateral result for Proposition 7.5. Lemma 5.5. Take a tetralateral ABCD with C a D. Then either ABCD is a convex quadrilateral, ABDC is a convex quadrilateral, D int ABC, or C int DAB.

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 11 Proof. Either C int DAB or D int CAB, by Lemma 3.6, since C, D a, C a D, and C d. We study these two cases. Case 1: Assume that C int DAB. Then C d B. Either D int CBA or C int DBA, by Lemma 3.6, since D b. We study these two cases. Case 1.1: Assume D int CBA. Then D int ABC and A b D, by Lemma 3.2. By the Crossbar Theorem 3.8, C m A, so C int BDA. Thus B int CDA, by Lemma 3.6, since B, C d, B d C, and C m. Thus A c B, by Lemma 3.2. Hence A int BCD, since A b, and A c. Thus ABCD is a convex quadrilateral. Case 1.2: Assume C int DBA. Then C int DAB, since C int DAB. Thus in case 1, either ABCD is a convex quadrilateral or C int DAB. Case 2: Assume that D int CAB. By the proof of case 1, either ABDC is a convex quadrilateral or D int ABC. We can obtain the proof by switching C and D is the above proof, because our lemma is symmetric in C and D. We specialize Lemma 5.5 to quadrilaterals. Lemma 5.6. Take a quadrilateral ABCD. Then either ABCD is a convex quadrilateral, A int BCD, B int CDA, C int DAB or D int ABC. Furthermore either B l D or A m C. Proof. Either A c B or C a D, by Lemma 5.2. We will apply to Lemma 5.5 to the quadrilaterals ABCD and CDAB. CDAB is a quadrilateral. ABDC is not a convex quadrilateral, by Lemma 5.4 and a short proof by contradiction, since ABCD is a quadrilateral. Similarly, CDBA is a not a convex quadrilateral, since CDAB is a quadrilateral. If C a D, then either ABCD is a convex quadrilateral, D int ABC, or C int DAB, by Lemma 5.5. Similarly, if A c B, then either CDAB is a convex quadrilateral, B int CDA, or A int BCD. But CDAB is convex iff ABCD is convex. This proves our first assertion. Thus either ABCD is a convex quadrilateral, A int BCD, B int ADC, C int BAD, or D int ABC. For each of these five possibilities, either B l D or A m C, or both, by Lemma 5.4 and the Crossbar Theorem 3.8. 6. The congruence relation = For points A and B, the segment AB is defined as the subset of the plane AB = {A, B} (A, B). Note that AB = BA. We have the undefined binary relation = on segments. If AB = CD, we say that the segment AB is congruent to the segment CD. Hilbert s axioms for congruence of segments are C1. Take distinct points A and B and a ray OZ defined by distinct points O and Z. Then there is a unique point P OZ {O} with OP = AB. C2. = is an equivalence relation on the set of segments. C3. If B (A, C), B (A, C ), AB = A B and BC = B C, then AC = A C. We call C1 the segment construction axiom and C3 the segment addition axiom. We write AB < CD if there exists a point E (C, D) such that AB = CE. Greenberg [7, Prop. 3.13] proves the segment order relation < satisfies a trichotomy law, a transitive property and is well defined up to congruence of segments. We have the undefined binary relation = on angles, and Hilbert s axioms are

12 WILLIAM RICHTER C4. Take an angle AOB and a ray O A. Let l = O A, and take a point Y l. Then there is a unique ray O B so that B l Y and A O B = AOB. C5. = is an equivalence relation on the set of angles. C6. Take triangles ABC and A B C. If AB = A B, AC = A C and BAC = B A C, then ABC = A B C. We prove a result about axiom C4 analogous to Lemma 2.1. Lemma 6.1. Take distinct points O and A and let l = OA. Take points B, P l with P l B. If AOP = AOB, then OB = OP. Proof. B l B by Lemma 3.2. AOB = AOB by axiom C5. Then OB = OP by the uniqueness of axiom C4 applied to AOB, ray OA, and the point B l. We write ABC = A B C, and say the two triangles are congruent, if the corresponding segments and angles are congruent. Now we prove that the SAS criterion implies congruent triangles, following Hilbert s proof. Theorem 6.2 (SAS). Take triangles ABC and A B C. If AB = A B, AC = A C and BAC = B A C, then ABC = A B C. Proof. Let l = AB, m = BC and n = AC. Since A, B and C are non-collinear points, l m = {B} and m n = {C}, by Lemma 2.1. Then ABC = A B C, by axiom C6. Axiom C6 applied to the triangles ACB and A C B implies ACB = A C B, because BAC = { AB, AC} = CAB, and thus CAB = C A B. Note that the axiom C5 reflexive property of = is not needed here. By axiom C1, there is a unique point D BC {B} with BD = B C. Then BD = BC, by Lemma 4.3. Thus ABD = ABC, so ABD = A B C. Thus BAD = B A C, by axiom C6. Hence BAD = BAC, by hypothesis and axiom C5. But B (D, C), since D BC. Therefore D l C, by Lemma 3.1, since D, C m {B}. Thus AD = AC, by Lemma 6.1, since BAD = BAC. Then D m n. Thus C = D, so BC = B C. Hence ABC = A B C. Greenberg takes Theorem 6.2 as an axiom instead of C6. His proof [7, Prop. 3.17] of the ASA theorem is similar to Hilbert s proof of Theorem 6.2. His proof [7, Prop. 3.19] of the angle addition result seems difficult, and does not follow Hilbert s treatment. Trying to follow Hilbert, we instead prove an angle subtraction result. Lemma 6.3. Take triangles AOB and A O B, and points G int AOB and G int A O B. Suppose we have the angle congruences AOB = A O B and AOG = A O G. Then BOG = B O G. Proof. For any point B 0 OB {O}, Lemma 4.3 implies AOB0 = AOB and B 0 OG = BOG. We may replace B by B 0, because this does not change the statement of our lemma. Similarly we may replace A, G, A, B, and G. By axiom C1, and the Crossbar Theorem 3.8, we may assume that OA = O A, OB = O B, G (A, B) and G (A, B ). By the SAS Theorem 6.2, AOB = A O B, so AB = A B, OAB = O A B and OBA = O B A. By ASA [7, Prop. 3.17], AOG = A O G, so AG = A G. Then BG = B G by the segment subtraction result [7, Prop. 3.11]. By SAS, BOG = B O G. Thus BOG = B O G. We prove the angle analogue of a segment result of Greenberg [7, Prop. 3.12].

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 13 Lemma 6.4. Take triangles AOB and A O B with AOB = A O B and a point G int AOB. There exists G int A O B such that AOG = A O G. Proof. We only sketch the proof, as it is similar to the above proof. We may assume OA = O A, OB = O B, AB = A B, OAB = O A B and G (A, B). By Greenberg s result [7, Prop. 3.12], there is a point G (A, B ) with AG = A G. Then G int A O B by Lemma 3.4. By SAS, OAG = O A G. We deduce the angle addition result from Lemma 6.3 and Lemma 6.4. Lemma 6.5. Take triangles AOB and A O B, and points G int AOB and G int A O B. Suppose we have the angle congruences AOG = A O G and BOG = B O G. Then AOB = A O B. Proof. Let l = O A. Then G l B by Definition 3.3. By axiom C4, there exists a unique ray O X so that X l G and AOB = A O X. By Lemma 6.4, there is a point Y int A O X so that AOG = A O Y. Then Y l X. So Y l G by Lemma 3.2. By axioms C4 and C5, O G = O Y, so G O Y {O }. Thus G int A O X by Lemma 3.5. Thus by Lemma 6.3, BOG = XO G. By hypothesis and axiom C5, XO G = B O G, so G O X = G O B. Let m = O G. By the Crossbar Theorem 3.8, the intervals (A, X) and (A, B ) both intersect the ray O G. So A m X and A m B. Thus X m B by Lemma 4.12. Hence O X = O B, by Lemma 6.1. Therefore AOB = A O B. Given an angle AOB and a point A with O (A, A ), the angle BOA is called the supplement of AOB, and we say AOB and BOA are supplementary angles. We say an angle is a right angle if it is congruent to its supplement. Then Lemma 6.6. Supplements of congruent angles are congruent. An angle congruent to a right angle is a right angle. All right angles are congruent. Take a line h = OH and two right angles AOH and HOB, with A h B. Then O (A, B). Proof. Greenberg proves the first statement [7, Prop. 3.14]. This implies the second statement, by axiom C5. The third statement, due to Hilbert, is [7, Prop. 3.23]. Let v = OA. Then h v = {O}, by Lemma 2.1, since A h. By axiom B2, there exists a point X v with O (A, X). Then X h, since X O, and A h X. Thus X h B, by Corollary 4.12, since B h. HOX = AOH, since AOH is a right angle with supplement HOX. By axiom C5, HOX = HOB, since all right angles are congruent. Thus OX = OB, by Lemma 6.1, since X h B. Thus B v. Hence O (A, B), by Lemma 3.1 and a short proof by contradiction, since A, B v {O}. Following Greenberg, we define [7, p. 69 & 77] triangle interiors and angle order. Definition 6.7. Given non-collinear points P, Q and R, let p = QR, q = P R, and r = P Q. We define the interior of the triangle as the set of points in the plane int P QR := {X X p, X q, X r, X p P, X q Q, and X r R}. We say ABC < P QR if there exists G int P QR so that ABC = P QG. We reprove Greenberg s trichotomy and congruence angle ordering result [7, Prop. 3.21], as his exercise proof seems to require Lemma 6.4, which he did not state. We first prove the parts of [7, Prop. 3.21] we actually use.

14 WILLIAM RICHTER Lemma 6.8. Take angles α and β with α < β. Then α = β. Proof. Assume (for a contradiction) that α = β. Write β = AOB. Since α < β, there is a point G int AOB so that α = AOG. Let a = OA and b = OB. Then G a, G b, and G a B. By axiom C5, AOG = AOB, since α = β. Thus OB = OG, by Lemma 6.1, so G b. This is a contradiction. Therefore α = β. Next we have an obvious corollary of Lemma 6.4. Lemma 6.9. Take angles α, β and γ with α < β and β = γ. Then α < γ. Proof. Let β = AOB and γ = A O B. Since α < β, there exists a point G int AOB so that α = AOG. By Lemma 6.4, there exists G int A O B such that AOG = A O G. By axiom C5, α = A O G. Thus α < γ. Now we prove transitivity for the angle order relation <. Lemma 6.10. Take angles α, β and γ with α < β and β < γ. Then α < γ. Proof. Let γ = AOB. Since β < γ, there is a point G int AOB with β = AOG. Then α < AOG, by Lemma 6.9, since α < β. There is a point F int AOG with α = AOF. Then F AOB, by Lemma 4.14. Hence α < γ. We finish proving [7, Prop. 3.21], although we only use the simpler results above. Lemma 6.11. Given angles α and β, exactly one of the following statements is true: α = β, α < β, and β < α. Proof. If α < β or β < α, then α = β, by Lemma 6.8 and axiom C5. Assume (for a contradiction) that α < β and β < α. Then α < α and α = α, by Lemma 6.10 and axiom C5. By Lemma 6.8, this is a contradiction. Thus α β or β α. Hence at most one of the statements α = β, α < β, and β < α is true. Let α = P OA, and let a = OA. By axiom C4, there exists a point Q such that Q a, Q a P and β = QOA. We have two cases. First assume that P, Q and O are collinear. O (Q, P ), since Q a P. Thus Q OP {O}, so OP = OQ, by Lemma 4.3. Hence QOA = P OA. Thus α = β, by axiom C5. Assume P, Q and O are non-collinear. Then P int QOA or Q int P OA by Lemma 3.6. Then P OA < QOA or QOA < P OA, by axiom C5. By axiom C5 and Lemma 6.9, α < β or β < α. The two cases show that α = β, α < β, or β < α. 7. Euclid, Hilbert and the parallel postulate Hartshorne [12] does an excellent job explaining how Hilbert s work rigorizes book I of Euclid s Elements [6]. Here we fill some gaps in Hartshorne s proofs. Hartshorne gives the mild Hilbert rigorization [12, p. 97] needed for Euclid s Prop. I.5 (an isosceles triangle has congruent base angles). A proof requiring no Hilbert rigorization [17, 6.2, Thm. 1] is due to Pappus [13, p. 254] in 300 C.E. Lemma 7.1. Take a triangle with BAC with AB = AC. Then CBA = ACB. Proof. BAC = CAB, by SAS, since BAC = BAC, by axiom C5. Greenberg [7, Prop. 3.22] and Moise [17, 6.2, Thm. 3] give the same Hilbert rigorization of Euclid s Prop. I.8, the SSS Theorem about congruent triangles. Theorem 7.2. Take triangles ABC and A B C with AB = A B, AC = A C, and BC = B C. Then ABC = A B C.

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 15 Hartshorne s proof [12, Prop. 10.1] of the SSS Theorem 7.2 is perhaps too sketchy. Hartshorne s Hilbert rigorization of Euclid s Prop. I.9 [12, p. 100], the existence of angle bisectors, has a gap which we fill in the second paragraph of our proof Lemma 7.3. Given an angle BAC, there exists a point F int BAC so that BAF = F AC. Proof. By axiom B2, there exists a point D with B (A, D), so D AB {A}. By axiom C1, there exists a point E AC {A} with AE = AD. Then AD = AB and AE = AC, by Lemma 4.3. Thus BAC = DAE. Then EDA = AED, by Lemma 7.1. Let d = AD, e = AE, and h = DE. Then d e = {A}, by Lemma 2.1, since C e = AC, and C d = AB. Thus E d, since E A. Hence d h = {D}, by Lemma 2.1, so A h, since A D. By axioms C4 and C1, there exists a point F so that F DE = EDA, F h, F h A, and DF = DA. By the SAS Theorem 6.2, DF E = DAE, since DE = DE by axiom C2, so F E = AD and DEF = AED. Let v = AF. We must show D, E v and D v E. Assume (for a contradiction) that D v. Then v h = {D}, by Lemma 2.1, since A D. Then D (A, F ), by Lemma 3.1 and a short proof by contradiction, since A, F v {D}. Thus EDA is a right angle, as it is congruent to its supplement F DE. By Lemma 6.6, AED and DEF are right angles and E (A, F ). Thus D = E, by axiom C1, since AE = AD and D, E AF, since A (D, F ) and A (E, F ), by axiom B3. This is a contradiction, since D E. Thus D v. By a similar proof by contradiction, E v. Thus D, E v. By the SSS Theorem 7.2, F AD = F AE. Thus DAF = F AE. Since D, E v, either D v E or D v E. Assume (for a contradiction) that D v E. Then D = E by axioms C4 and C1, since DAF = EAF. This is a contradiction, so D v E. Thus there exists a point G (D, E) v. Then G int DAE, by Lemma 3.4, and h v = {G}, by Lemma 2.1, since A h. Thus G (A, F ), by Lemma 3.1 and a short proof by contradiction, since A, F v {G}, and F h A. By axiom B3, F (A, G), so F AG {A}. Thus F int DAE, by Lemma 3.5. Thus, F int BAC and BAF = F AC. Take distinct points B and C. Hartshorne constructs [12, Prop. 10.2] an isosceles triangle BAC, and then proves the existence of midpoints by using the Crossbar Theorem 3.8, Lemma 7.3 and SAS to find the midpoint of (A, B). This, together with axiom B2, proves Hilbert s axiom II.2, which is stronger than our axiom B2: II.2. Given two distinct points A and C, there exists points B and D such that B (A, C) and C (A, D). Greenberg, who takes II.2 as an axiom himself, gives a much simpler but less appealing proof [7, p. 86, Ex. 6] of II.2 using only axiom B2. Hartshorne gives a nice Hilbert rigorization of Euclid s Prop. I.16 [12, Prop. 10.3] Theorem 7.4. Take a triangle ABC with a point D such that C (B, D). Then CAB < ACD. This result (see also [19, Thm. 6.3.2]) is called the Exterior Angle Theorem, because ACD, the supplement of BCA, is called an exterior angle of ABC. Hartshorne s Hilbert rigorization of Euclid s Prop. I.7 [12, p. 35 & Ex. 9.4] is insufficient, so we reprove Prop. I.7, adding details to Hartshorne s proof. Afterward we will give a much simpler proof of the result.

16 WILLIAM RICHTER Proposition 7.5. Take distinct points A and B in the plane, and let a = AB. Take distinct points C and D in the plane with C, D a and C a D, and suppose that AC = AD. Then BC = BD. Proof. CAD is isosceles, so CDA = ACD, by Lemma 7.1. Define the lines b = BC, c = CD, d = AD, l = AC and m = BD. By Lemma 2.1, l a = {A}, since C a. Hence B l. Similarly B d, A m, and A b. Assume (for a contradiction) that C d. Then A (D, C), since C a D, so C AD. Then C = D, by axiom C1, since AC = AD. But C D, so we have a contradiction. Hence C d. We similarly prove that D l. Since C d, Lemma 2.1 implies c d = {D}. Hence A c, since A D, since D a. If D b or C m, then a similar argument shows that BC = BD, and we are done. Hence we may assume that D b and C m. Then b c = {C}, by Lemma 2.1, since D b. Thus B c, since B C, since C a. Thus A, B, C and D are distinct points, and C, D a, A, D b, A, B c, B, C d, B, D l, and A, C m. Thus ABCD is a tetralateral with C a D. By Lemma 5.5, either ABCD is a convex quadrilateral, C int DAB, ABDC is a convex quadrilateral, or D int ABC. We study the four cases. Case 1: Assume ABCD is a convex quadrilateral. Then B int CDA and A int BCD. By Definition 6.7, CDB < CDA and ACD < BCD, and CDA = ACD. Thus CDB < BCD, by Lemmas 6.9 and 6.10. Thus CDB = BCD, by Lemma 6.8. Lemma 7.1 implies that BD = BC. Case 2: Assume C int DAB. By axiom B2 there exists a points E d so that D (A, E). By axiom I1, d = DE. Since C d and D b, the angles CDE and BCD exist. B int CDE, by Lemma 3.7, since C int ADB. By the Crossbar Theorem 3.8, there exists a point F (B, D) AC, since C int BAD. Thus A (F, C), and F int BCD, by Lemma 3.4. A, C, F l are distinct points, because ABDC is a tetralateral. F (A, C), since C m A, since C int ADB. Thus C (A, F ) by axiom B3. Supplements of congruent angles are congruent [7, Prop. 3.14], so CDE = F CD, since CDA = ACD. Thus CDB < CDE = F CD < BCD, since B int CDE and F int BCD. The argument of case 1 shows that BC = BD. We have cases 3 and 4, where ABDC is a convex quadrilateral or D int ABC. In both cases we prove that BC = BD by the arguments of cases 1 and 2. This works because our proposition is symmetric in C and D. Remark 7.6. Hartshorne explains Euclid s proof of Prop. I.7 has a gap which must be filled by showing that A int DCB and B int CDA in case 1, but he ignores the other 3 cases. Fitzpatrick s literal translation of the Elements (http://farside.ph.utexas.edu/euclid.html) gives only Hartshorne s one case. The version of the Elements we use [6] gives our cases 1 and 2. In what is considered to be the definitive version of the Elements, Heath explains that [13, p. 259 260] a literal translation of Euclid s statement of Prop. I.7 is apparently obscure and vague 1, that this second case of [6] is due to the 5th Century mathematician Proclus, and that Euclid had the general practice of giving only one case, and 1 Fitzpatrick translation is On the same straight-line, two other straight-lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a different point on the same side (of the straight-line), but having the same ends as the given straight-lines.

A MINIMAL VERSION OF HILBERT S AXIOMS FOR PLANE GEOMETRY 17 that the most difficult, and leaving the others to be worked out by the reader. For Prop. I.7 this seems unfortunate, especially because of extra Hilbert theory needed. Now we give a simple proof of Euclid s Prop. I.7 which we think is well known. Proof of Proposition 7.5. Assume (for a contradiction) that BC = BD. We have triangles ACB and ADB because C, D a. Then ACB = ADB, by the SSS Theorem 7.2, since AB = AB, by axiom C2. Hence BAC = BAD. By axiom C4, AC = AD, since C b D. Then C = D, by axiom C1, since AC = AD. Since C and D are distinct, we have a contradiction. Hence BC = BD. We discuss Hartshorne s treatment of circles [12, 11]. Given a point O and a segment AB, let Γ be the circle with center O and radius AB defined as Γ = {C OC = AB}. A point X is inside Γ if OX < OA and outside Γ if OA < OX. As Hartshorne explains, Euclid s Prop. I.7 says that two circles can not have two intersection points in the same half-plane of the line between their centers. We work Hartshorne s circle convexity exercise [12, Ex. 11.1(a)], which Greenberg proves in an exercise [7, Ex. 4.29]. Our proof is similar but slightly simpler. Lemma 7.7. Let Γ be a circle points with center O and radius OR. Assume the points A and B are inside Γ. Then the open interval (A, B) lies inside Γ. Proof. We have OA < OR and OB < OR. Take X (A, B). We must show that OX < OR. As in [12, Prop. 11.2], we use Theorem 7.4 and Euclid s Proposition I.19 (which relates sides of a triangle to their opposite angles), and we also need I.18 [6, p. 16 22]. We have two cases depending on whether OA = OB or not. Assume OA = OB. We may assume that OA < OB. Applying I.18 to OAB yields OBA < OAB. Applying I.16 to OAX yields OAB < OXB. Then OBA < OXB, by Lemma 6.10. Applying I.19 to OXB yields OX < OB. Then OX < OR by the transitive property of segment ordering [7, Prop. 3.13]. Assume OA = OB. Then AOB is isosceles, so OBA = OAB, by Lemma 7.1. The proof is now similar to the case above, but we do not need I.18. We say lines l and m in a plane are parallel, and write l m, if l m =. The Exterior Angle Theorem 7.4 allows us to construct parallel lines. Take lines l and m, distinct points A, C l m and distinct points B, E m l. The line x = AB is called a transversal which cuts l and m. Suppose also that E x C. Then CAB and EBA are called alternate interior angles [19, p. 107]. Venema [19, Thm. 6.8.1] gives a nice proof of the Alternate Interior Angles Theorem Lemma 7.8. If CAB = EBA, then l m. Proof. Assume (for a contradiction) that l m, so there exists a point G l m. Then G A and G B. By Lemma 2.1, x l = {A} and x m = {B}, since B l and A m. Thus C x, G x, and E x. Assume (for a contradiction) that G x C. Then A (G, C). Thus AG = AC, by Lemma 4.3, since G AC {A}. Hence GAB = CAB, so GAB = EBA. By Lemma 3.2 and a short proof by contradiction, G x E, since C x G and C x E. Thus B (E, G), by Lemma 3.1 and a short proof by contradiction, since x m = {B} and E, G m {B}. Hence EBA is the supplement of ABG. Thus

18 WILLIAM RICHTER GAB < EBA, by the Exterior Angle Theorem 7.4. Hence GAB = EBA, by Lemma 6.8. This is a contradiction. Hence G x C. Then G x E, by Corollary 4.12, since G x C and C x E. By axiom C5, EBA = CAB. We obtain a contradiction by the argument above, switching the pairs l and m, A and B, and E and C. This switch works because of the symmetry in our lemma. Hence l m. As Hartshorne points out [12, Thm. 10.24], the result is Euclid s Prop. I.27. Hartshorne however fails to point out that the proof requires Hilbert betweenness fixing, and that Euclid s Prop. I.27 is vaguely stated, as Euclid lacks the betweenness axioms needed to define alternate interior angles. As Hartshorne points out [12, Ex. 10.10], Lemma 7.8 implies, without using the parallel postulate, Proposition 7.9. A quadrilateral ABCD with AB = CD and BC = AD is a parallelogram. Proof. By Lemma 5.6, either B x D or A y C. We may assume that B x D. So when we cut the lines l and r by the transversal x, CAB and ACD are alternate interior angles. Also cut the lines t and b by the transversal x. Then B x D again implies that CAD and ACB are alternate interior angles. By SSS, we have congruent triangles CAB = ACD so CAB = ACD and CAD = ACB. Thus AB CD and AD BC by Lemma 7.8. A typical reader of Hartshorne s could not be expected to work this exercise [12, Ex. 10.10], as Hartshorne does not mention the betweenness issues we dealt with, and merely says Join the midpoints of AB and CD, then use (I.27). We prove part of Euclid s Prop. I.21, as Hartshorne [12, Thm. 10.4] fails to explain that Euclid s proof requires Hilbert betweenness fixing. Lemma 7.10. Take a triangle ABC with a point D int ABC. Then BAC < BDC. Proof. Let m = AC. Then D int ABC and D m B, since D int ABC. There exists a point E (A, C) BD, by the Crossbar Theorem 3.8. Apply the Exterior Angle Theorem 7.4 to BAE. Then BAE < BEC, since E (A, C). By Lemma 4.3, AC = AE, so BAC = BAE. Thus BAC < BEC. B, D and E are distinct, and B (E, D), since E BD. But E (B, D), since B m D and (A, C) m. Thus D (E, B), by axiom B3. Apply the Exterior Angle Theorem 7.4 to DEC. Then DEC < BDC. EB = ED, by Lemma 4.3, since D (E, B). Thus BEC = DEC. Hence BAC < BDC, by Lemma 6.10. For the rest of the paper we adopt Venema s version of Euclid s parallel postulate. P. Given a line l in the plane and a point P l, there exists a unique line m in the plane so that P m and m l. We prove a version of the triangle sum theorem (Lemma 10.1). Lemma 7.11. Take a triangle ABC, with lines l = AC, x = AB and y = BC. There exists a line m B and points E, F m {B} with m l, B (E, F ), E x C, F y A, C int ABF, EBA = CAB and CBF = BCA.