Selected Topics from GeometryA project under construction

Franz RotheDepartment of Mathematics

University of North Carolina at Charlotte

Charlotte, NC 28223

frothe@uncc.edu

January 9, 2011

Contents

0 A Synopsis of Euclid 20.1 Book I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Book II. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . 60.3 Book III. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . 60.4 Book IV. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . 70.5 Book V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80.6 Book VI. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . 80.7 Book X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90.8 Book XI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90.9 Book XII. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . 90.10 Book XIII. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . 10

1 On the History of Translations and Editions of Euclid’s Elements 11

I Neutral Geometry 13

1 Incidence and Order 141.1 Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2 David Hilbert’s Axiomatization of Euclidean Geometry . . . . . . . . 16

1.2.1 Introduction from Hilbert’s Foundations of Geometry . . . . . . . 161.2.2 Hilbert’s Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3 Incidence Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1

1.4 The Axioms of Order and their Consequences . . . . . . . . . . . . . . 23

3 Congruence of Segments, Angles and Triangles 393.1 Congruence of Segments . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2 Some Elementary Triangle Congruences . . . . . . . . . . . . . . . . . 443.3 Congruence of Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.4 SSS Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.5 The Equivalence Relation of Angle Congruence . . . . . . . . . . . . . 603.6 Constructions with Hilbert Tools . . . . . . . . . . . . . . . . . . . . . 673.7 The Exterior Angle Theorem and its Consequences . . . . . . . . . . . 763.8 SSA Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.9 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4 Measurement and Continuity 1034.1 The Archimedean Axiom . . . . . . . . . . . . . . . . . . . . . . . . . 1034.2 Axioms related to Completeness . . . . . . . . . . . . . . . . . . . . . 112

4.2.1 Cantor’s axiom . . . . . . . . . . . . . . . . . . . . . . . . . . 1124.2.2 Dedekind’s axiom . . . . . . . . . . . . . . . . . . . . . . . . 1134.2.3 Hilbert’s axiom of completeness . . . . . . . . . . . . . . . . 117

5 Legendre’s Theorems 1195.1 The First Legendre Theorem . . . . . . . . . . . . . . . . . . . . . . . 1195.2 The Second Legendre Theorem . . . . . . . . . . . . . . . . . . . . . . 1215.3 The Alternative of Two Geometries . . . . . . . . . . . . . . . . . . . . 1275.4 What is the Natural Geometry? . . . . . . . . . . . . . . . . . . . . . . 132

8 Towards a Natural Axiomatization of Geometry 1368.1 The Uniformity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 1368.2 A Hierarchy of planes . . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.3 Wallis’ Axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1438.4 Proclus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.5 More about Aristole’s Axiom . . . . . . . . . . . . . . . . . . . . . . . 148

II Euclidean Geometry 154

1 Some Euclidean Geometry of Circles 1551.1 Thales’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1551.2 Rectangles and the Converse Thales Theorem . . . . . . . . . . . . . . 1621.3 Angles in a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1671.4 Secands in a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

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2 Pappus’, Desargues’ and Pascal’s Theorems 1822.1 Pappus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1822.2 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1872.3 Pascal’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

3 Euclidean Geometry and Ordered Fields 1963.1 Ordered Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1963.2 A Hierarchy of Cartesian planes . . . . . . . . . . . . . . . . . . . . . . 199

8 Inversion by a Circle 2008.1 Definition and Construction of the Inverse Point . . . . . . . . . . . . . 2008.2 The Gear of Peaucollier . . . . . . . . . . . . . . . . . . . . . . . . . . 2018.3 Invariance Properties of Inversion . . . . . . . . . . . . . . . . . . . . . 203

Area in Euclidean Geometry 200

11 Euclidean Constructions with Restricted Means 21111.1 Constructions by Straightedge and Unit Measure . . . . . . . . . . . . 21111.2 Tools equivalent to Straightedge and Compass . . . . . . . . . . . . . . 21611.3 Hilbert Tools and Euclidean Tools Differ in Strength . . . . . . . . . . 221

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0 A Synopsis of Euclid

0.1 Book I

Definitions

1. A point is that which has no part.

2. A line is length without breadth.

3. ?

4. A straight line lies evenly with its points.

5. ?

6. ?

7. ?

8. A plane angle is the inclination of two lines.

9. ?

10. When two adjacent angles are equal, they are right angles.

11. ?

12. ?

13. ?

14. ?

15. A circle is a line all of whose points are equidistant from a point.

16. ?

17. ?

18. ?

19. ?

20. A triangle with two equal sides is isosceles.

21. ?

22. ?

23. Parallel straight lines are lines in the same plane that do not meet, no matter howfar extended in either direction.

4

Postulates

1. To draw a line through two points.

2. To extend a given line.

3. To draw a circle with given center through a given point.

4. All right angles are equal.

5. If a line crossing two other lines makes the sum of the interior angles on the sameside less than two right angles, then these two lines will meet on that side whenextended far enough.

Common Notions

1. Things equal to the same thing are equal to each other.

2. Equals added to equals are equal.

3. Equals subtracted from equals are equal.

4. Things which coincide are equal.

5. The whole is greater than the part.

Propositions

1. To construct an equilateral triangle on a given segment.

2. To draw a segment equal to a given segment at a given point.

3. To cut off a smaller segment from a larger segment.

4. Side-angle-side (SAS) congruence for triangles.

5. The base angles of an isosceles triangle are equal.

6. If the base angles are equal, the triangle is isosceles.

7. It is not possible to put two triangles with equal sides on the same side of a segment.

8. Side-side-side (SSS) congruence for triangles.

9. To bisect an angle.

10. To bisect a segment.

11. To construct a perpendicular to a line at a given point on the line.

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12. To drop a perpendicular from a point to a line not containing the point.

13. A line standing on another line makes angles with sum equal to two right angles.

14. ?

15. Vertical angles are equal.

16. The exterior angle of a triangle is greater than either opposite interior angle.

17. Any two angles of a triangle have a sum less than two right angles.

18. If one side of a triangle is greater than another, then the angle opposite to thegreater side is greater than the angle opposite to the smaller side.

19. If one angle of a triangle is greater than another, then the side opposite to thegreater angle is greater than the side opposite to the smaller angle.

20. The sum of any two sides of a triangle is greater than the third side.

21. ?

22. To construct a triangle from all three sides given— provided the sum of any twosides is greater than the remaining side.

23. To produce a given angle at a given point and side.

24. For two triangles with two pairs of equal sides, the triangle with the greater anglebetween them has the greater opposite side.

25. For two triangles with two pairs of equal sides, the triangle with the greater thirdside has the greater angle between the two matched sides.

26. Angle-side-angle (ASA) and angle-angle-side (AAS) congruence for triangles.

27. If the two alternate interior angles between a transversal and two lines are equal,then these two lines are parallel.

28. If the exterior angle between a transversal and the first line and the opposite interiorangle between the transversal and the second line have sum two right angles, thenthe two lines are parallel.

28a. If the two interior angles between a transversal and two lines have sum two rightangles, then the two lines are parallel.

29. A transversal of two parallel lines makes equal alternate interior angles with thetwo parallels.

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30. Lines parallel to the same line are parallel.

31. To draw a line parallel to a given line through a given point.

32. The sum of angles of a triangle is two right angles.

32a. An exterior angle equals the sum of the two opposite interior angles of a triangle.

33. The lines joining the endpoints of two parallel segments are parallel and the seg-ments cut out by the first two parallels are congruent.

34. The opposite sides and angles of a parallelogram are equal.

35. Parallelograms on the same base and in the same parallels have equal areas.

36. Parallelograms on equal bases and in the same parallels have equal areas.

37. Triangles on the same base and in the same parallels have equal areas.

38. Triangles on equal bases and in the same parallels have equal areas.

39. Triangles with equal areas on the same base lie in the same parallels.

40. Triangles with equal areas on equal bases lie in the same parallels.

41. The area of a parallelogram is twice the area of a triangle on the same base and inthe same parallels.

42. To construct a parallelogram with a given angle and area equal to the area of agiven triangle.

?43. The triangles on opposite sides of a diagonal of a parallelogram are equal.

44. To construct a parallelogram with given side and angle and area equal to the areaof a given triangle.

45. To construct a parallelogram with a given angle and area equal to the area of agiven figure.

46. To construct a square on a given segment.

47. (Theorem of Pythagoras) The square on the hypothenuse is equal in area to thesum of the squares on the legs of the triangle.

48. If the sum of the areas of squares on two sides of a triangle is equal to the area ofthe square on the third side, then the triangle is right.

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0.2 Book II. Propositions

?1. The rectangle contained by two lines is the sum of the rectangles contained by oneand the segments of the other.

4. The area of the square on the whole segment is equal to the sum of the areas on thetwo parts of the segment plus twice the area of the rectangle on these two parts.

5. The area of the square on the first longer part of a segment is equal to the area ofthe rectangle with the difference of the parts and the entire segment as sides, plusthe area of the square on the second part. 1

6. The area of a rectangle with one side a longer segment and an added piece, the secondside the added piece plus the area of the square on half of the longer segment addup to the area of the square on the half plus the added piece. 2

11. To cut a longer segment so that the rectangle on the whole and the part is equal tothe square on the other part. 3

14. To construct a square with area equal to that of a given figure.

0.3 Book III. Propositions

1. To find the center of a circle.

2. The segment joining two points of a circle lies inside the circle.

5. If two circles intersect, they do not have the same center.

6. If two circles are tangent, they do not have the same center.

10. Two circles can intersect in at most two points.

11. 12. If two circles are tangent, their centers lie in a line with the point of tangency.

16. The line perpendicular to a diameter at its endpoint is tangent to a circle, andthe ”angle between the tangent line and the circle” 4 is less than any ”rectilinealangle”.

17. To draw a tangent to a circle from a point outside the circle.

18. A tangent line to a circle is perpendicular to the radius at the point of tangency.

1p2 = [p− (c− p)]c + (c− p)22(c + p)p +

(c2

)2 =(

c2 + p

)2

3Solve cx = (c− x)2 for the part x.4The notion in quotes is not well defined by today’s standard.

8

19. The perpendicular to a tangent line at the point of tangency will pass through thecenter of a circle.

20. The angle at the center is twice the angle at a point of the circumference subtendinga given arc of a circle.

21. Two angles from points of a circle subtending the same arc are equal.

22. The opposite angles of a quadrilateral in a circle have sum equal to two right angles.

31. The angle in a semicircle is a right angle.

32. The angle between a tangent line and a chord of a circle is equal to the angle onthe arc cut off by the chord.

35. If two chords cut each other inside the circle, the rectangle on the segments on onechord is equal in area to the rectangle on the segments on the other chord.

36. From a point outside a circle, let a tangent and a secant be drawn. The square onthe segment on the tangent is equal in area to the rectangle formed by the twosegments from the point outside of the circle to the endpoints of the secant.

37. From a point outside a circle are drawn two lines cutting or touching the circle.If the square on the segment on one line is equal in area to the rectangle formedby the two segments from the point outside to the endpoints of the secant on theother line, then the first line is tangent to the circle.

0.4 Book IV. Propositions

1. To inscribe a given segment in a circle.

2. To inscribe a triangle, equiangular to a given triangle, in a circle.

3. To circumscribe a triangle, equiangular to a given triangle, around a circle.

4. To inscribe a circle in a triangle.

5. To circumscribe a circle around a triangle.

10. To construct an isosceles triangle whose base angles are twice the vertex angle.

11. To inscribe a regular pentagon in a circle.

12. To circumscribe a regular pentagon around a circle.

15. To inscribe a regular hexagon in a circle.

16. To inscribe a regular 15-sided polygon in a circle.

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0.5 Book V

Definitions

4. Magnitude are said to have a ratio if either one, being multiplied, an exceed theother.

5. Four magnitudes a, b, c, d are in the same ratio a : b equal to c : d, if and only if forany whole numbers m and n either one of these three cases occurs:

both ma > nb and mc > nd,

both ma = nb and mc = nd,

both ma < nb and mc < nd.

0.6 Book VI. Propositions

1. The areas of triangles of the same height are in the same ratio as their bases.

2. A line is parallel to the base of a triangle if and only if it cuts the two other sides inthe same ratio—proportionately.

3. A line from the vertex of a triangle to the opposite side bisects the angle if and onlyif it cuts the opposite side in proportion to the remaining sides of the triangle.

4. The sides of equiangular triangles are proportional.

5. If the sides of two triangles are proportional, their corresponding angles are pairwiseequal.

6. If two triangles have one pair of equal angles, and the sides containing these pair areproportional, then the triangles are similar.

8. The altitude from the right angle of a right triangle divides it into two trianglessimilar to each other and to the whole triangle.

12. To find a fourth proportional for three given segments.

13. To find a mean proportional between two given segments.

16. Four segments are proportional if and only if the rectangle on the extremes is equalin area to the rectangle on the means.

?30. To cut a segment in extreme and mean ratio. 5

31. Any figure on the hypothenuse of a right triangle has area equal to the sum of theareas of similar figures on the legs of the triangle.

5Solve c : x = x : p for x.

10

0.7 Book X

1. Given two unequal quantities, if one subtracts from the greater one the smaller one,and repeats this process enough times, there will remain a quantity lesser thanthe smaller of the two original quantities.

117. 6 The diagonal of a square is incommensurable with its side.

0.8 Book XI

Definitions

25. A cube is a polyhedron made of six equal squares.

26. An octahedron is a polyhedron made of eight equal equilateral triangles.

27. An icosahedron is a polyhedron made of twenty equal equilateral triangles.

28. An dodecahedron is a polyhedron made of twelve equal regular pentagons.

Propositions

21. The plane angles in a solid angle make a sum less than four right angles.

28. A parallelepiped is bisected by its diagonal plane.

29. 30. Parallelepipeds on the same base and of the same height are equal in volumes.

31. Parallelepipeds on equal bases and of the same height are equal in volumes.

0.9 Book XII. Propositions

2. The areas of circles are in the same ratio as the squares on their diameters.

3. A pyramid is divided into two pyramids and two prisms.

5. Pyramids of the same height on triangular bases have volumes in the ratio of theareas of the bases.

6. A prism with triangular bases is divided into three triangular pyramids with equalvolumes.

6not in Heath, but in Commandino

11

0.10 Book XIII. Propositions

7. If at least three angles of an equilateral pentagon are equal, the pentagon is regular.

10. In a circle, the square on the side of the inscribed pentagon is equal in area tothe square on the side of the inscribed hexagon plus the square on the inscribeddecagon. 7

13. To inscribe a tetrahedron in a sphere.

14. To inscribe a octahedron in a sphere.

15. To inscribe a cube in a sphere.

16. To inscribe an icosahedron in a sphere.

17. To inscribe a dodecahedron in a sphere.

18. (Postscript) Besides these five figures there is no other contained by equal regularpolygons.

7a25 = R2 + a2

10

12

1 On the History of Translations and Editions of

Euclid’s Elements

The highly acclaimed Elements is simply a huge collection—divided into 13 books—of 465 propositions from plane and solid geometry. Today, it is generally agreed thatrelatively few of these theorems were Euclid’s own invention. Rather, by compiling,editing, and perfecting the known body of Greek mathematics, he created a superblyorganized treatise. The Elements were so successful and revered that they thoroughlyobliterated all preceding works of its kind. They have become a lasting contribution tomankind.

It is a misfortune that no copy of Euclid’s Elements has been found that actuallydates from the author’s own time. Hence his writings have had to be reconstructedfrom numerous recensions, commentaries, and remarks by other writers. Proclus makesit clear that the Elements were highly valued in Greece and refers to numerous com-mentaries on it. Among the most important must have been those by Heron (100 B.C.- A.D.100), Porphyry and Pappus (third century A.D.).

Modern editions of the work are based upon a revision that was prepared by theGreek commentator Theon of Alexandria (end of the fourth century A.D). He livedalmost 700 years after the time of Euclid. Until 1808, Theon’s revision was the oldestedition of the Elements known. In 1808, Napoleon ordered valuable manuscripts tobe taken from Italian libraries and send to Paris. At that occasion, Francois Peyrardfound, in the Vatican library, a tenth-century copy of an edition of Euclid’s Elementsthat predidates Theon’s recension. The historians J.L.Heiberg and Thomas L. Heathhave used principally this manuscript. A study of this older edition and a careful siftingof citations and remarks made by early commentators indicate that the introductorymaterial of Euclid’s original treatise undoubtedly underwent some editing in the subse-quent revisions. But the propositions and their proofs have remained as Euclid wrotethem, except for minor additions and deletions.

There are also Arabic translations of Greek works and Arabic commentaries, pre-sumably based on Greek manuscripts no longer available. The first really satisfyingArabic translation of the Elements was done by Tabit ibn Qorra (826-901). These, too,have been used to decide what was in Euclid’s original. But the Arabic translations andrevisions are on the whole inferior to the Greek manuscripts.

The first complete Latin translation of the Elements were not made from the Greekbut from Arabic. In the eighth century, a number of Byzantine manuscripts of Greekworks were translated by the Arabians. In 1120, the English scholar Adelard of Barth(ca. 1075-1160) made a Latin translation of the Elements from one of the older Ara-bian translations. Other Latin translations were made from the Arabic by Gherardo ofCremona (1114-1187) and, 150 years after Adelard, by Johannes Campanus.

Euclid’s Elements were widely read, translated and changed by people all over theworld. From 1601 to 1607, Mattes Ricci (1552-1610) from Italy and Hsu Kuang-ching(1562-1634) translated the first six books into Chinese. This played a significant role in

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the subsequent development of mathematics in China.The first printed edition of the Elements was made at Venice in 1482 and contained

Campanus’ translation. This very rare book was beautifully executed. It was thefirst mathematical book of any consequence to be printed. In 1574, Clavius (1537-1612) published an edition that is valuble for its extensive scholia. An important Latintranslation from the Greek was made by Commandino in 1572. This translation servedas a basis for many subsequent translations, including the very influential work byRobert Simson. From the latter, many English editions were derived. The first completeEnglish translation of the Elements was the monumental Billingsley translation issuedin 1570.

Nicolaus Mercator (ca. 1620-1687) published an edition, along with his works intrigonometry, astronomy and other topics. Adrien-Marie Legendre (1752-1833) is knownfor his very popular Elements de geometrie, in which he attempted a pedagogical im-provement of Euclid’s Elements by considerably rearranging and simplifying many of thepropositions. He used some algebra not in the original. The high school versions mostwidely used during our century are patterned on Legendre’s modification of Euclid’swork.

Euclid’s Elements is to geometry what the Bible is to Christianity. No work, exceptthe Bible, has been more widely used, edited, or studied than Euclid’s Elements. Overone thousand editions of Euclid’s Elements have appeared since the first one printed in1482. For more than two millennia, this work has dominated all teaching of geometry.Abraham Lincoln is credited with learning logic by studying Euclid’s Elements, as hisbiographer Carl Sandburg tells. In his autobiography, Bertrand Russell (1872-1970)penned this remarkable recollection: ”At the age of eleven, I began Euclid, with mybrother as tutor. This was one of the great events of my life, as dazzling as first love.”Probably no work has exercised a greater influence on scientific thinking.

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Part I

Neutral Geometry

15

1 Incidence and Order

1.1 Logic

10 Problem 1.1. Find the pair of equivalent statements. The find all pairs ofstatement and its negation and mark them with matching color. How many pairs (colors)are there? Find the statements, of which the negation is not listed and mark them by acircle around them.

1. "If the sum shines, the streets are dry."

2. "All men like to drive."

3. "If the sun shines, the streets are wet."

4. "Some men do not like to drive."

5. "It rains and the streets are dry."

6. "No woman likes to drive."

7. "Some women like to drive."

8. "There exists a triangle with sum of its angles equal to two right angles."

9. "Every triangle has sum of angles either less or more than two right

angles."

10. "No triangle is equilateral."

11. "All triangles are equilateral."

12. "No point has three or more lines passing through it."

13. "There exists a point through which at most two lines pass."

14. "There exists a point through which three or more lines pass."

15. "Every point has at most two lines passing through it."

Answer. Four pairs of statement and negation are [(2)(4)] [(6)(7)] [(8)(9)] [(12)(14)], orequivalently [(14)(15)]. Statements (1)(3)(5)(10)(11)(13) are not negated. Statements(12) and (15) are equivalent.

Remark. It helps to write the last four statements in symbols. Let Pn denote thepredicate

"Through point P pass exactly n lines."

Here are the last four statements in symbolic logic:

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12. "No point has three or more lines passing through it."

¬∃P∃n (n ≥ 3 ∧ Pn)

13. "There exists a point through which at most two lines pass."

∃P∃n (n ≤ 2 ∧ Pn)

14. "There exists a point through which three or more lines pass."

∃P∃n (n ≥ 3 ∧ Pn)

15. "Every point has at most two lines passing through it."

∀P∃n (n ≤ 2 ∧ Pn)

10 Problem 1.2. Give the converse of the following sentences:

"If Peter drives fast, he does not read the traffic signs."

Answer. ”If Peter does not read the traffic signs, then he drives fast.”

"If Tom reads the traffic signs, then he drives fast."

Answer. ”If Tom drives fast, then he reads the traffic signs.”

10 Problem 1.3. Give the negation of the following sentences in a clear andsimple form. You are not supposed to simply say that the statement is false!

"Bill drives slowly and he reads the traffic signs."

Answer. ”Either Bill drives fast or he does not read the traffic signs.”

A second possible answer. ”If Bill drives slowly, he does not read the traffic signs.”

"If Peter drives fast, he does not read the traffic signs."

Answer. ”Peter drives fast and he reads the traffic signs.”

"If Tom reads the traffic signs, then he drives fast."

Answer. ”Tom reads the traffic signs, but he does not drive fast.”

"If Paul drives fast, then he does not read the traffic signs."

Answer. ”Paul drives fast and he reads the traffic signs.”

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1.2 David Hilbert’s Axiomatization of Euclidean Geometry

1.2.1 Introduction from Hilbert’s Foundations of Geometry

Geometry needs—similar to arithmetic—only a few simple basic principlesfor its consequential development. These basic principles are called theaxioms of geometry. Starting with Euclid, the setup of the axioms ofgeometry and the investigation of their mutual connections has beenthe subject of many excellent treatises in the mathematical literature.The problem in question is basically a logical analysis of our spatialimagination.

The present—meaning Hilbert’s—investigation is a new attempt to set up acomplete and as simple as possible system of axioms. And, furthermore,to deduct from them the most important geometric theorems, such thatthe meaning and importance of the different axioms and their conse-quences become clear.

1.2.2 Hilbert’s Axioms

As suggested in this paragraph, Hilbert introduces the modern habit to develop the con-sequences of the different groups of axioms immediately after introducing them. But—because I guess it is more convenient for the reader—we shall now state the completeaxiomatic system at once, at the beginning.

0. Undefined Elements and Relations

Elements:

• A class of undefined objects called points, denoted by A, B, C, . . . .

• A class of undefined objects called lines, denoted by a, b, c, . . . .

• A class of undefined objects called planes, denoted by α, β, γ, . . . .

Relations:

• Incidence (being incident, lying on, containing)

• Order (lying between) (for points on a line)

• Congruence (for segments and angles)

Remark. Planes are only needed to include three dimensional geometry.

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I. Axioms of Incidence

I.1 For every two points A and B there exists a line that contains each of the pointsA, B.

I.2 For every two points A and B there exists no more than one line that contains eachof the points A, B.

I.3a There exist at least two points on a line.

I.3b There exist at least three points that do not lie on a line.

Remark. I did separate axiom I.3 into I.3a and I.3b, in order to stress there is no directlogical connection between the two sentences intended.

Remark. Additional axioms I.4 through I.8 are only needed to include three dimensionalgeometry.

I.4 For any three points A, B, C that do not lie on the same line, there exists a planeα that contains each of the points A, B, C. For every plane there exists a pointwhich it contains.

I.5 For any three points A, B, C that do not lie on the same line, there exists no morethan one plane that contains each of the points A, B, C.

I.6 If two points A and B of a line a lie on a plane α, then every point of a lies in thesame plane α.

I.7 If two planes α, β have a point A in common, then they have at least one morepoint B in common.

I.8 There exist at least four points which do not lie in a plane.

II. Axioms of Order

II.1 If a point B lies between a point A and a point C, then the points A, B, C arethree distinct points of a line, and B also lies between C and A.

II.2 For two points A and C, there also exists at least one point B on the line AC suchthat C lies between A and B.

II.3 Of any three points on a line there exists no more than one that lies between theother two.

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DEFINITIONS: Segment, point of segment, inside of segment, outside of segment, sideof line in a plane, ray, polygon, side of polygon, vertex of polygon, triangle, quadrilateral,n-gon.

Remark. A segment AB is assumed to have two different endpoints A and B.

II.4 (Pasch’ Axiom) Let A, B, C be three points that do not lie on a line and let abe a line in the plane ABC which does not meet any of the points A, B, C. If theline a passes through a point of the segment AB, it also passes through a point ofthe segment AC, or through a point of the segment BC.

III. Axioms of Congruence

III.1 If A, B are two points on a line a, and A′ is a point on the same or anotherline a′, then it is always possible to find a point B′ on a given side of the linea′ through A′ such that the segment AB is congruent to the segment A′B′. Insymbols AB ∼= A′B′.

III.2 If a segment A′B′ and a segment A′′B′′ are congruent to the same segment AB,then segment A′B′ is also congruent to segment A′′B′′.

III.3 On a line, let AB and BC be two segments which except for B have no point incommon. Furthermore, on the same or another line a′, let A′B′ and B′C ′ be twosegments which except for B′ also have no point in common. In that case,

if AB ∼= A′B′ and BC ∼= B′C ′ , then AC ∼= A′C ′

DEFINITION: Angle.

III.4 Let ∠(h, k) be an angle in a plane α and a′ a ray in a plane α′ that emanatesfrom the point O′. Then there exists in the plane α′ one and only one ray k′ suchthat the angle ∠(h, k) is congruent to the angle ∠(h′, k′) and at the same time allinterior points of the angle ∠(h′, k′) lie on the given side of a′. This means that

∠(h, k) ∼= ∠(h′, k′)

Every angle is congruent to itself, thus it always holds that

∠(h, k) ∼= ∠(h, k)

III.5 If for two triangles ABC and A′B′C ′ the congruences

AB ∼= A′B′ , AC ∼= A′C ′ , ∠BAC ∼= ∠B′A′C ′

hold, then the congruence ∠ABC ∼= ∠A′B′C ′ is also satisfied.

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IV. Axiom of Parallels

IV.1 Let a be any line and A a point not on a. Then there exists at most one line inthe plane determined by a and A that passes through A and does not intersect a.

V. Axioms of Continuity

V.1 (Axiom of Archimedes) If AB and CD are any segments, then there exists anumber n such that n segments congruent to CD constructed contiguously fromA, along a ray from A through B, will pass beyond B.

V.2 (Axiom of completeness) An extension of a set of points on a line, with itsorder and congruence relations existing among the original elements as well as thefundamental properties of line order and congruence that follow from Axioms I-IIIand from V.1, is impossible.

Remark. In Hilbert foundations of geometry, the questions of consistency, categorialnature, and independence of his axioms are addressed. 8

Independence of the SAS-axiom, the parallel axiom, and the Archimedean axiom isproved. Relative consistency is proved, once consistency of the real number system istaken for granted—which turned out to be the really deep unsolvable problem!

Hilbert proves that his axiom system is categorial, once his axiom (V.2) of complete-ness is assumed, but states clearly that the system without this axiom is not categorial.

As one realizes, there are infinitely many geometries which satisfy the axiomgroups I through IV and (V.1). On the other hand, there is only one—namely the Cartesian geometry—which satisfies the completeness axiom(V.2), too.

1.3 Incidence Geometry

Among the consequences of the axioms of incidence, Hilbert spells out two Propositions.As far as two dimensional geometry is concerned, they reduce to one simple statement,distinguishing the possibilities of intersecting or parallel lines.

Proposition 1.1 (From Hilbert’s Proposition 1). Any two different lines haveeither one or no point in common.

Definition 1.1. Two different lines which do not intersect are called parallel.

Definition 1.2. We say that an incidence geometry has the Euclidean parallel propertyiff for every line l and point P not on line l, there exists a unique parallel to line lthrough point P .

8As of today, the word completeness has several meanings in precise foundations of mathematics.

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Definition 1.3. We say that an incidence geometry has the elliptic parallel property iffany two lines do intersect—necessarily at a unique point.

Definition 1.4. We say that an incidence geometry has the hyperbolic parallel propertyiff for every line l and point P not on line l, there exist at least two parallels to line lthrough point P .

Question. Give at least two further useful formulations of Proposition 1.1.

Answer. Here are two possible answers:

• Any two different lines which are not parallel, have a unique point of intersection.

• If two lines have two or more points in common, they are equal.

10 Problem 1.4 (The four-point incidence geometries). Find all non iso-morphic incidence geometries with four points. Which parallel property (elliptic, Eu-clidean, hyperbolic, or neither) does hold? Which one is the smallest affine plane?

Figure 1.1: There are two four-point incidence geometries.

Answer. There exist two non-isomorphic four-point geometries.

(a) Six lines with each one two points. It has the Euclidean parallel property, and isthe smallest affine plane.

(b) There are four lines, one of which has three points. It has the elliptic parallelproperty.

10 Problem 1.5 (The five-point incidence geometries). Find all non iso-morphic incidence geometries with five points. Describe the properties of their points andlines. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold?

Answer. There exist four different five-point geometries.

(a) Ten lines have each two points. This model has the hyperbolic parallel property.

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Figure 1.2: There are four five-point incidence geometries.

(b) Exactly one line with three points. Altogether, there are eight lines. Neither parallelproperty holds.

(c) Two intersecting lines with three points. Altogether, there are six lines. Neitherparallel property holds.

(d) There are five lines, one of which has four points. This model has the ellipticparallel property.

10 Problem 1.6. Which ones of the four or five point incidence geometriessatisfy the statement:

On every line there exist at least two points, and furthermore, there existthree points not on the this line.

Why is this statement different from Hilbert’s axioms?

Answer. The statement of the problem implies existence of five different points. Henceit does not hold for any four point incidence geometry. The statement postulates threeextra points lying not on any given line, additionally to at least two points on this line.Hence it can only hold for a five point geometry, where every line has exactly two points.Indeed it holds only for model (a) above.

10 Problem 1.7. Why is the statement from the last problem different fromHilbert’s axioms?

Answer. Hilbert’s axiom of incidence for plane geometry are (I.1)(I.2) and (I.3). Axiom(I.3) consists of two sentences:

I.3a There exist at least two points on a line.

I.3b There exist at least three points that do not lie on a line.

Here part (I.3a) and (I.3b) do not refer to each other. Part (I.3a) is a universal statementabout any line. But part (I.3b) is a purely existential statement about three points.

On the other hand, the statement given in the problem is a different universalstatement about a line. The first part repeats Hilbert’s (I.3a). It goes on requiringadditionally existence three points which are not lying on the same line.

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10 Problem 1.8 (A nine point incidence geometry). Find a model of inci-dence geometry with nine points, which satisfies

(I3+) "Every line contains exactly three points."

and for which the Euclidean parallel axiom does hold:

"For every line l and every point P not on l, there exists exactly

one parallel to the line l through P."

It is enough to provide a drawing to explain the model. Use colors for the lines. Howmany lines does the model have?

Remark. The model can be constructed using analytic geometry and arithmetic modulo3. Take as ”points” the ordered pairs (a, b) with a, b ∈ Z3. As in analytic geometry,lines are given by linear equations ax + by + c = 0 with a, b, c ∈ Z3 and a, b are notboth zero. One defines the slope of a line in the usual way. Lines are parallel if andonly if they have the same slope. To find the parallel to a given line through a givenpoint, one uses the point slope equation of a straight line. This procedure shows thatthe Euclidean parallel property holds.

Answer. There is exactly one such model. It has 12 lines.Name any point A and let two lines through it consist of the points l = A, B, C and

m = A, D, G (lines l and m are drawn horizontally and vertically). Let D, E, Fbe the parallel to line A, B, C through D. Two different parallels to a given linecannot intersect, because of the uniqueness of parallels. Hence the parallel to l throughpoint G contains the remaining points and necessarily is G, H, I. The parallel toline A, D, G through B contains one of the points E, F and one of the points G, H.Possibly by exchanging names of those four points, we can assume that B, E, H is theparallel to line A, D, G through B. Finally, C, F, I is the parallel to line A, D, Gthrough C.

Up to now, we have mentioned and drawn six lines. Because of incidence axiom (I.1),there is a unique line through any two points. Hence there need to exist further linesin the model. As an example, we find the line through points A and E. It cannot passthrough any of the points B, C, D, F,H, I, because otherwise we get a line with four ormore points on it. Hence A, E, I is a line. Similarly, one finds lines H, E, C, andfinally B, D, I,A, F,H,B, F, G,C, D, I. (It does not matter that the last fourlines cannot be drawn as straight Euclidean lines. Neither does it matter that they havemore intersection points—those are not included as points of the model.) The modelcontains 12 lines. It is the unique model satisfying all requirements.

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Figure 1.3: A nine-point incidence geometry

1.4 The Axioms of Order and their Consequences

Definition 1.5 (Segment). Let A and B be two distinct points. The segment AB isthe set consisting of the points A and B and all points lying between A and B. Thepoints A and B are called the endpoints of the segment, the points between A and Bare called the interior points, and the remaining points on the line AB are called theexterior points of the segment.

Definition 1.6 (Triangle). We define a triangle to be union of the three segmentsAB, BC and CA. The three points A, B and C are assumed not to lie on a line. Thesethree points are the vertices, and the segments BC, AC, and AB are the sides of thetriangle.

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For a segment, it is assumed that the two endpoints A and B are different. For atriangle 4ABC, it is assumed that the three vertices do not lie on a line.

Proposition 1.2 (Hilbert’s Proposition 3). For any two points A and C, thereexists at least one point B on the line AC lying between A and C.

Figure 1.4: How to get a point inside the given segment AC.

Proof. By the axiom of incidence I.3, there exists a point E not lying on the line AC,and by axiom of order II.2, there exists a point F such that E is a point inside thesegment AF . By the same axiom, there exists a point G such that C is a point insidesegment FG. Now we use Pasch’s axiom for the triangle 4ACF and line EG.

Question. Why is 4ACF a triangle?

Answer. Points A and C are two different points by assumption. The third vertex Fdoes not lie on line AC— pother point E would lie on that line,too, contrary to theconstruction.

Question. Why are E and G two different points?

Answer. Assume towards a contradiction E = G. In that case, lines AF and CF wouldintersect in that point, hence F = E = G, contradicting the definition of point F .

For the triangle 4ACF and line EG, Pasch’s axiom yields that the line intersectsa second side of the triangle besides side AF . But line EG does not intersect segmentFC, because the intersection point of the lines EG and FC is G, which lies outsidethe segment FC. Hence line EG intersects the third side of triangle 4ACF , which issegment AC. The intersection point B is a point between A and C, existence of whichwas to shown.

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Definition 1.7 (Ray). Given two distinct points A and B, the ray−→AB is the set

consisting of the points A and B, the points inside the segment AB, and all points Pon the line AB such that the given point B lies between A and P . The point A is calledthe vertex of the ray.

The axiom of order II.2 tells that the ray−→AB contains points not lying in the segment

AB.

(Hilbert’s Proposition 4, also called ”Three-point Theorem”). Among any threepoints A, B and C lying on a line, there exists and only one lying between the two otherpoints.

Figure 1.5: Assume that neither A not C lies between the two other of the three pointsA,B and C. The construction shows that B does lie between A and C.

Proof. The axiom or order II.3 states that at most one of the three points lies betweenthe two others. We need to prove that actually one of the three points does lie betweenthe two others. Assume that neither A not C lies between the two other points. Theconstruction shows that B does lie between A and C. One chooses a point D not lying

on line AC. Next we choose a point G on the ray−−→BD such that D lies between B and

G. Being used repeatedly, we to mention the follow consequence of axioms II.3 and II.4:

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A consequence of Pasch’s axiom. If a line cuts one side of a triangle, and the extensionof a second side, then the line cuts the third side.

Now we use Pasch’s axiom for the triangle 4BCG and the line AD. Because thisline intersects side BG, but not side BC, the line intersects the third side CG, say inpoint E.

A similar application of Pasch’s axiom, now for triangle 4ABG and line CD yieldsexistence of a point F between A and G.

Now use Pasch’s axiom a third time—for triangle 4AEG and the line CF . Becausethis line intersects side AG, but not side GE, the line intersects the third side, which isthe segment AE. But this intersection point has to be the intersection of line CF withline AE, which his point D. Hence point D lies between A and E.

Finally use Pasch’s axiom a fourth time—for triangle 4AEC and the line GB.Because this line intersects side AE, but not side CE, the line intersects the third sideAC. But this intersection point has to be the intersection of lines GB and AC, whichhis point B. Hence point B lies between A and C.

Proposition 1.3 (Hilbert’s Proposition 5, also called ”Four-point Theorem”).Any four points on a line can be notated in a way that all four order relations that keepthe alphabetic order hold.

We begin the proof with several Lemmas. Let four points A, B, C,D on a line g begiven.

Lemma 1.

A∗B∗C and

A∗ C∗D imply

B∗C∗D

Reason for Lemma 1. The construction done in the figure on page 27 is used to provethe claim that C does lie between B and D. One chooses a point G not lying on line g.

Next we choose a point F on the ray−−→BG such that G lies between B and F .

The line FC intersects neither segment AB nor segment BG. Hence Pasch’s axiomimplies that line FC does not intersect segment AG.

The existence of intersection point H of segment GD and line FC is shown byapplying Pasch’s axiom to this line and triangle 4ADG.

The ”coup de grace” is to apply Pasch’s axiom to triangle 4BGD and line FC.This line does not intersect segment BG, but does intersect segment GD. Hence theline FC intersects the segment BD.

It is already know that line FC and line BD intersect in point C. The argumentabove now confirms that this intersection point lies on the segment BD. Hence C liesbetween B and D.

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Figure 1.6: The construction used in Lemma 1.

Lemma 2.

A∗B∗C and

B∗C∗D imply

A∗ C∗D

Reason for Lemma 2. One uses the same figure on page 27 as in Lemma 1.The existence of intersection point H of segment GD and line FC is shown by

applying Pasch’s axiom to this line and triangle 4BDG.The ”coup de grace” is to apply Pasch’s axiom to triangle 4AGD and line FC. and

confirms that intersection point C lies on the segment AD. Hence C lies between A andD.

Lemma 3.

A∗B∗C and

B∗C∗D imply

A∗B∗ D

Reason for Lemma 3. This follows from Lemma 2 by exchanging A and D as well as Band C.

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Lemma 4.

A∗B∗C and

A∗ C∗D imply

A∗B∗ D

Reason for Lemma 4. Lemma 1 yields B ∗ C ∗D. Hence Lemma 3 yields the assertionA ∗B ∗D.

Lemma 5. The endpoints of a segment cannot lie between two interior points.

Reason for Lemma 5. Assume towards a contradiction that the assertion does not holdfor a segment AC and its interior points B and D.

In that case, A ∗B ∗C and A ∗D ∗C and B ∗A ∗D. Now, by Lemma 3, B ∗A ∗Dand A ∗D ∗C imply B ∗A ∗C. This is a contradiction to A ∗B ∗C, because, by axiomII.3, at most one of three points on a line can lie between the other two.

Definition 1.8. Given any n points lying on a line. We say that the list [A1, A2, . . . , An]represents the order of the points A1, A2, . . . , An iff the order relations Ai ∗Aj ∗Ak holdfor all 1 ≤ i < j < k ≤ n.

Proof of Proposition 1.3. Let P, Q,R, S be any four points on a line. By the three-pointTheorem, exactly one among the points P, Q and R is lying between the other two. Wechoose the notation such that

P ∗Q ∗R

Secondly, distinguish three cases based on the order of points P, R and S.

(1) P ∗R ∗ S

(2) R ∗ P ∗ S

(3) P ∗ S ∗R

In the first two cases, a list representing the order of the points can be deducted usingLemma 1 and 4 and turns out to be

(1) [P, Q,R, S]

(2) [R,Q, P, S]

In the third case (3), one needs still to take the order of points P, Q and S into account.Since both points Q and S lie inside the segment PR, Lemma 5 implies that P cannotlie between Q and S. We are left with only two subcases (3a) and (3b):

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(3a) P ∗ S ∗R and P ∗Q ∗ S

(3b) P ∗ S ∗R and P ∗ S ∗Q

in which a list representing the order of the points turns out to be

(3a) [P, Q, S, R]

(3b) [P, S, Q, R]

as can be seen via Lemma 1 and 4, once more.

Corollary. Any four points on a line can be put in exactly two ways into a list thatrepresents their order. These two lists are just reversed to each other.

Question. There are three pairs of order relations among four points A, B, C,D ona line that imply all four alphabetic order relations. (The order representing list is[A, B, C,D].) Which are these three pairs?

Answer. If either

A ∗B ∗ C and B ∗ C ∗D or(1)

A ∗B ∗ C and A ∗ C ∗D or(2)

A ∗B ∗D and B ∗ C ∗D(3)

then all four alphabetic order relations follow.

Let A, B and C be three points on the given line with B lying between A and C.

Definition 1.9. The two rays−→BA and

−−→BC. are called the opposite rays if their common

vertex B lies between A and C. We say that two points in the same ray lie on the sameside of the vertex. Two points in the opposite rays lie on different sides of the vertex.

Proposition 1.4 (”Line separation Theorem”). Given is a line and a vertex lyingon it. Each point of the line except the vertex is contained in exactly one of the oppositerays originating from the vertex.

Proof. The complement of the ray−−→BC consists of all points P such that P ∗ B ∗ C, as

follows directly from the definition of a ray and the three-point Theorem.We want to check that an arbitrary point P 6= B is contained in exactly one of

the rays−→BA and

−−→BC. If P = A or P = C, the assertion follows from the three-point

Theorem.Otherwise, P 6= A, P 6= B and P 6= C, and the assertion follows from the four-point

Theorem, and its Corollary.The four points can be put into an order representing list, with A preceding B. In

that list, B preceeds C, since A ∗B ∗ C.If the natural order of these four points turns out to be [P, A, B, C] or [A, P, B, C],

the point P is contained in the ray−→BA, but not in the opposite ray

−−→BC.

If the natural order of these four points is [A, B, P, C] or [A, B, C, P ], the point P is

contained only in the ray−−→BC.

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Lemma 6. Given are any n points lying on a line. If two lists both represent the orderof n points A1, A2, . . . , An, the two lists are either equal or reversed of each other.

Proposition 1.5 (Hilbert’s Proposition 6, the ”n-point theorem”). Given afinite number of points on a line, they can be labelled as A, B, C,D,E, . . . , K in a waysuch that B lies between A and C, D,E, . . . ,K, C lies between A, B and D, E, . . . , K,and so on. Except this notation, only the reversed one K, . . . , E, D, C,B,A yields theseorder relations.

Induction start. For three points, the Proposition follows from the three-point Theorem.

Induction step ”n 7→ n + 1”: Assume that Proposition 1.5 holds up to a given numbern ≥ 3 of points. Let A1, A2, . . . , An, An+1 be any n + 1 points on a line. We distinguishthree cases based on the order of points A1, An and An+1.

(1) A1 ∗ An ∗ An+1

(2) An+1 ∗ A1 ∗ An

(3) A1 ∗ An+1 ∗ An

In the first two cases, a list representing the order of the n + 1 points can be deductedusing Lemma 1 and 4 and turns out to be

(1) [A1, A2, . . . , An, An+1]

(2) [An+1, A1, A2, . . . , An]

Consider case (1). For any 1 < j < n, Lemma 4 confirms that

A1∗Aj∗An and

A1∗ An∗An+1 imply

A1∗Aj∗ An+1

For any i such that 1 < i < j, Lemma 1 confirms that

A1∗Ai∗Aj and

A1∗ Aj∗An+1 imply

Ai∗Aj∗An+1

Hence all order relations represented by list (1) hold. Case (2) can be handled using thereversed lists.

In case (3), one puts at first the n points A2, . . . , An, An+1 into an order representinglist. By Lemma 6, the two possible choices are lists reversed of each other. We choose thelist which keeps the already established order of the points A2, . . . , An without reversal.Hence the order representing list of n points A2, . . . , An, An+1 is obtained by insertingpoint An+1 into the list [A2, . . . , An]— between Aj and Aj+1 for some 1 ≤ j < n. Thelist representing the order of all n + 1 points turns out to be

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(3) [A1, A2, . . . , Aj, An+1, Aj+1, . . . , An]

We now check that all order relations represented by list (3) hold. The open casesconcern the order of A1 and An+1, and any point Ai with 1 < i ≤ j or Ak withj < k < n. For any i such that 1 < i < j, Lemma 3 confirms that

A1∗Ai∗Aj and

Ai∗Aj∗An+1 imply

A1∗Ai∗ An+1

Too, the ordering A1 ∗ Aj ∗ An+1 follows by Lemma 2.For any k such that j < k < n, Lemma 1—reversed—confirms that

A1∗An+1∗ An and

An+1∗Ak∗An imply

A1∗An+1∗Ak

Hence all order relations represented by the new list hold.

Proposition 1.6 (Hilbert’s Proposition 7). Between any two points of a line lieinfinitely many points.

Proof. We construct a sequence of points A1, A2, . . . , all of which are different and liebetween the two given points A and B. The existence of the point A1 between A andB was proved in Proposition 1.2.

Assume that n points A1, A2, . . . , An between A and B have been constructed, andthere order is represented by the list

[A, A1, A2, . . . , An, B]

We choose any point An+1 between An and B. As shown in Proposition 1.5, the orderof the n + 1 points is now represented by the list

[A, A1, A2, . . . , An, An+1, B]

Hence one has obtained n + 1 different points between A and B.

Question. Consider the following model: The "points" are the integers modulo 5. Thereis one "line" through all five points. The order relation is defined by requiring

a ∗ b ∗ c if and only if 2b ≡ a + c mod 5

Give an order representing cyclic list of the points. Which axioms of incidence and orderhold in this model, which do not hold?

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Answer. A cyclic list is0, 2, 4, 1, 3, 0, 2, 4, 1, 3, . . .

The axioms of incidence I.1 and I.2 hold, because there is only one line. But axiom I.3is violated for the same reason. The axioms of incidence II.1, II.2 and II.3 hold, butPasch’s axiom does not.

One can say that Pasch’s axiom tells: a line entering a triangle has to leave it, too.It can be proved that a line cannot intersect all three sides of a triangle. We havepostponed this proposition up to now in order to allow for a streamlined proof.

Proposition 1.7 (Bernays’ Lemma). A line cannot cut all three sides of a triangle.

Figure 1.7: Bernay’s Lemma would not hold— and line A,E′, E, C would be equal to lineE′, C, D, B—all points collapse onto one line.

Proof. 9 We assume that line l intersects all three sides of triangle 4ABC and derivea contradiction. If the sides BC, CA and AB would intersect line l in the points D, Eand F , these three points would be all different.

By Proposition 4, one of the three points lies between the two others. Without lossof generality, we can assume that point D lies between E and F . We now apply Pasch’saxiom to the triangle 4AEF and the line BC. This line cut the side EF in point D.Hence it intersects a second side, say

side AE in point E ′ or

9Hilbert credits this proof to his collaborator Paul Bernays.

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side AF in a point F ′.

In both cases we get a contradiction.Here are the details for the first case: E ′ 6= C, because E ′ lies inside the segment

AE, but C lies on the ray−→AE outside that segment. The four points A, E ′, E, C lie

on one line. Too, the four points E ′, C,D, B lie on one line. But these two lines havethe two points E ′ and C in common. Hence they are identical. Thus all three pointsA, B, C lie on one line, contradicting the assumption that 4ABC is a triangle.

In the second case, one gets a similar contradiction: F ′ 6= B, because F ′ lies inside

the segment AF , but B lies on the ray−→AF outside that segment. The four points

A, F ′, F, B lie on one line. Too, the four points F ′, B, D,C lie on one line. But thesetwo lines have the two points F ′ and B in common. Hence they are identical. Thusall three points A, B, C lie on one line, contradicting the assumption that 4ABC is atriangle. Hence a line cannot intersect all three sides of a triangle.

Proposition 1.8 (Hilbert’s Proposition 8, the ”Plane separation Theorem”).Each line a lying in a plane A separates the points of this plane not lying on the lineinto two regions R and S with the following properties:

• for any points A in region R and B in region S, the segment AB intersects linea;

• for any two points A and A′ in the same region, the segment AA′ does not intersectthe line a.

Question. Why is a plane mentioned in the plane separation theorem?

Answer from a student. It must be stated that the line lies in a specific plane to be ableto separate the plane.

Answer. These are just the points of any given plane that are separated into the pointson the line and in the two half planes. Too, to keep all rigor, one needs the plane justto tell which points are separated by the theorem.

The theorem is not true for a line in the three dimensional space.

Definition 1.10. We say that the points A and A′ lie on the same side of line a—or inthe same half plane— in case that segment AA′ does not intersect the line a. The pointsA and B lie on different sides of line a in case that the segment AB does intersect theline a.

The two open regions one gets in this way are called the half planes bounded by theline a in the plane A.

Proof. We check the following facts:

(a) If two points A and B lie on the same side of line a, and points B and C lie on thesame side of line a, the points A and C lie on the same side.

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(b) If two points A and A′ lie on the same side of line a, and points A′ and C lie ondifferent sides of line a, the points A and C lie on different sides.

(c) If two points A and B lie on different sides of line a, and points B and C lie ondifferent sides of line a, the points A and C lie on the same side.

In each argument, we need to distinguish the case that the three points in question lieon a line, or they form a triangle.

If the three points do not lie on a line, assertion (a) and (b) follow directly fromPasch’s axiom. Assertion (c) follows from Bernay’s Lemma.

If the three points do lie on a line g, we use the line separation theorem. Considerassertion (a). If line g does not intersect line a, the assertion is obviously true. Now letP be the intersection point of lines a and g. Because P does not lie between A and B

the rays−→PA =

−−→PB are equal. Similarly, because P does not lie between B and C the

rays−−→PB =

−→PC. Hence

−→PA =

−→PC, which implies that P does not lie between A and C.

Consider assertion (b). Now let P be the intersection point of lines a and g. Because

P does not lie between A and A′ the rays−→PA =

−−→PA′ are equal. Similarly, because P

does lie between A′ and B the rays−−→PA′ and

−−→PB are opposite. Hence the rays

−→PA and−−→

PB are opposite. and hence P does not lie between A and B.Assertion (c) follows similarly, because there exist only two opposite rays. from a

given vertex.

Definition 1.11 (Angle). An angle ∠BAC is the union of two rays−→AB and

−→AC with

common vertex A not lying on one line. The point A is called the vertex of the angle.

The rays−→AB and

−→AC are called the sides of the angle.

Figure 1.8: Interior and exterior of an angle are intersection and union of two half planes.

Definition 1.12 (Interior and exterior of an angle). The interior of an angle lyingin a plane A is the intersection of two corresponding half planes—bordered by the sidesof the angle, and containing points on the other side of the angle, respectively. Theexterior of an angle is the union of two opposite half planes—-bordered by the sides of

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the angle, and not containing the points neither in the interior nor on the legs of theangle. Half planes, interior and exterior of an angle all do not include the lines or rayson their boundary.

Thus the interior of ∠BAD is the intersection of the half plane of AB in which D lies,and the half plane of AD in which B lies. The exterior of ∠BAD is the union of thehalf plane of AB opposite to D, and the half plane of AD opposite to B.

Remark. Note the ”exterior of an angle” is defined differently from ”exterior angle of atriangle”. The latter is any supplementary angle to an interior angle of the triangle.

Figure 1.9: A ray interior of an angle intersects a segment from side to side.

Proposition 1.9 (The Crossbar Theorem). A segment with endpoints on the twosides of an angle and a ray emanating from its vertex into the interior of the angleintersect.

Proof. Let ray r =−→AC lie in the interior of the given angle ∠BAD, and let B and D

be arbitrary points on the two sides of this angle. We have to show that the ray−→AC

intersects the segment BD.

Let F be any point on the ray opposite to−→AB. We apply Pasch’s axiom to triangle

4FBD and line l = AC. The line intersects the side FB of the triangle in point A,and does not pass through neither one of the vertices F, B, D. We check that side FDdoes not intersect line AC.

Indeed, the points inside segment FD and the points inside ray−→AC lie on different

sides of line AD. But the points inside segment FD and inside the opposite ray lie ondifferent sides of line AB. (Points A, D, F are exceptions, but neither can segment FDand line AC intersect in any of these points.)

Hence—by Pasch’s axiom—the third side BD intersects line AC, say at point Q.Segment BD, and hence the intersection point Q are in the interior of ∠BAD. Since

only the ray−→AC, but not its opposite ray, lies in the interior of ∠BAD, the intersection

point Q lies on the ray−→AC. Here is a drawing, to show how Pasch’s axiom is applied.

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Figure 1.10: The Crossbar Theorem is proved using Pasch’s axiom.

Warning. A point in the interior of an angle may not lie on any segment going fromside to side of the angle.

To spell it out precisely: Given an angle ∠BAD and a point Q in its interior, there

does not always exist any two points G and H on the rays−→AB and

−−→AD such that Q lies

on the segment GH.Indeed, in hyperbolic geometry, the union of all segments from side to side of an

angle span only the interior of an asymptotic triangle—and this is a proper subset ofthe interior of the angle.

Definition 1.13. A list of segments AB, BC, CD, . . . , KL is called a polygonal curve.The points inside the segments and their endpoints are called the points of polygonalcurve. If the endpoint L is equal to the beginning point A = L, the curve is calledclosed or a polygon. The points A, B, C, . . . , K are called the vertices, and the segmentsAB, BC, CD, . . . , KL are called the sides of the polygon.

Definition 1.14. A polygon or polygonal curve is called simple, if the vertices—withpossible exception of the first and last—are all different, and any two sides do notintersect except at their common endpoint when they follow each other in the listAB, BC, CD, . . . , KL.

Proposition 1.10 (Restricted Jordan Curve Theorem). A simple closed polygonalcurve C partitions the points of the plane not lying on the polygon into two regions, calledthe interior and exterior of the polygon, which have the following properties:

• If A is a point of the interior, and B a point of the exterior region, then eachpolygonal curve connecting A to B has at least one point in common with thecurve C.

• If A and A′ are two points of the interior region, then there exists a polygonalcurve connecting A to A′ which has no point in common with the curve C.

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Figure 1.11: The restricted Jordan Curve Theorem.

• Similarly, if B and B′ are two points of the exterior region, there exists a polygonalcurve connecting B to B′ which has no point in common with the curve C.

• There exists a line which does not cut the interior region.

• Every line cuts the exterior region.

Sketch of a proof for the Euclidean plane. 10 The set R2 \ C consists of finitely manycomponents. As one follows the polygon C, the points near to the right, are all in thesame component as some reference point R. Similarly, the points near to the left are allin the same component as some reference point L. Furthermore, one can choose pointsR and L near to the same side of the polygon C and to each other. Any point X inR2 \ C can be connected by a polygonal curve inside this set R2 \ C to one of these thesetwo points R and L. Hence R2 \ C consists of at most two connected components.

To prove that the points on the left and the right lie in different components, considerrays. We call a ray exceptional, if it contains a vertex of the curve C. For a given vertexP , there exist at most finitely many directions of exceptional rays.

Each ray that is not exceptional, has only finitely many intersection points with thecurve C. As one turns the ray around a fixed vertex P , the number of intersection pointschanges only at the exceptional directions. Nearby, before and after passing through an

10This proof follows H. Tverberg [?].

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exceptional direction, the parity of the number of intersection points is the same. Hencewe can assign a even parity or odd parity to the vertex P , depending of the number ofintersection points being even or odd.

Next we show that the parity is constant inside each component of the set R2 \ C.Let X and Y be two points in the same component. There exists a polygonal curve Pconnecting X and Y inside this component. We can assume that the rays along the sidesof curve P are not exceptional. This can be achieved by an arbitrarily small adjustmentof the curve P .

After this adjustment, the two endpoints of each segment of P have the same parity,since the intersection set of the ray with C changes continuously as one moves the vertexof the ray along one side of P , keeping the ray parallel. Hence any two points in thesame component of R2 \ C have the same parity.

The endpoints of a small segment crossing the polygon C have opposite parities.Points far away from the polygon have even parity, because there exists a ray notintersecting the curve C. Hence the unbounded outside component of R2 \ C has evenparity, and the inside bounded component has odd parity.

Remark. A complete proof in the present context is given by G. Feigl [?]. The reallyinteresting point is that none of the further axioms of congruence, continuity or parallelsare needed. Especially, the Jordan Curve Theorem holds for the hyperbolic plane, too.

On the other hand, the Restricted Jordan Curve Theorem 1.10 is not valid for theprojective plane, and hence neither in double elliptic geometry. The reason is that theimproper points and line of the projective plane change the topological structure.

Remark. The more general Jordan curve theorem says that any closed simple continuouscurve separates the Euclidean plane into an interior and exterior domain. It is creditedto Camille Jordan (1836-1922), but Jordan’s original argument was in fact inadequate.The first correct proof is credited to Oswald Veblen (1880-1960).

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3 Congruence of Segments, Angles and Triangles

This section deals with David Hilbert’s axiomatization of neutral geometry.

3.1 Congruence of Segments

Proposition 3.1 (Congruence is an equivalence relation). Congruence is an equiv-alence relation on the class of line segments.

Question. Which three properties do we need to check for a congruence relation?

Answer. For an arbitrary relation to be a congruence relation, we need to check

(a) reflexivity : Each segment is congruent to itself, written AB ∼= AB.

(b) symmetry : If AB ∼= A′B′, then A′B′ ∼= AB.

(c) transitivity : If AB ∼= CD, and CD ∼= EF , then AB ∼= EF .

Proof of reflexivity, for a bottle of wine from Hilbert personally. 11 We need to show thateach segment is congruent to itself. Take any given segment AB. We transfer the seg-ment to a ray starting from any point C. By Hilbert’s axiom III.1, there exists a pointD on that ray such that AB ∼= CD. Now use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

Hence, for a case with other notation, AB ∼= CD and AB ∼= CD imply AB ∼= AB.

Proof of symmetry. Assume that AB ∼= CD. Because of reflexivity CD ∼= CD. Oncemore, we use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

Hence CD ∼= CD and AB ∼= CD imply CD ∼= AB.

Proof of transitivity. Assume AB ∼= CD and CD ∼= EF . Because of symmetry EF ∼=CD. Now we use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

For a case with other notation, this means that AB ∼= CD and EF ∼= CD implyAB ∼= EF , as to be shown.

11which one can take gracefully as a thank-you—for translating from page 15 of the millenium editionof ”Grundlagen der Geometrie”–etc.

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Proposition 3.2 (Existence and uniqueness of segment transfer). Given a seg-ment AB and given a ray r originating at point A′, there exists a unique point B′ onthe ray r such that AB ∼= A′B′.

Question. Which part of this statement is among Hilbert’s axioms? Which axiom isused?

Answer. The existence of the segment A′B′ is postulated in Hilbert’s axiom of congru-ence III.1.

Question. How does the uniqueness of segment transfer follow? Which axioms areneeded for that part?

Answer. The uniqueness of segment transfer follows from the uniqueness of angle trans-fer, stated in III.4, and the SAS Axiom III.5.

Figure 3.1: Uniqueness of segment transfer

Proof of uniqueness. Assume the segment AB can be transferred to the ray r from A′

in two ways, such that both AB ∼= A′B′ and AB ∼= A′B′′. We choose a point C ′ not onthe line A′B′. Using problem 0.1, one obtains the congruences

A′B′ ∼= A′B′′ , A′C ′ ∼= A′C ′ , ∠B′A′C ′ ∼= ∠B′′A′C ′

By the axiom axiom III.5 for SAS, this implies ∠A′C ′B′ ∼= ∠A′C ′B′′. By theuniqueness of angle transfer, as stated in axiom III.4, this implies that the rays−−→C ′B′ =

−−−→C ′B′′ are equal. Hence B′ = B′′ is the unique intersection point of the two

different rays r =−−→A′B′ and

−−→C ′B′. We have shown that AB ∼= A′B′ and AB ∼= A′B′′

imply B′ = B′′. Thus segment transfer is unique.

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Proposition 3.3 (Subtraction of segments). Given are three points on a line suchthat A ∗B ∗ C, and two points B′ and C ′ on a ray emanating from A′. Suppose that

AB ∼= A′B′ , AC ∼= A′C ′

then BC ∼= B′C ′ and A′ ∗B′ ∗ C ′ follow.

Figure 3.2: Segment subtraction

Proof. On the ray originating from B′, opposite to−−→B′A′, we transfer segment BC. We

get the seventh point S such that BC ∼= B′S, and A′ ∗B′ ∗ S. By Hilbert’s axiom III.3on segment addition, AB ∼= A′B′ and BC ∼= B′S imply now AC ∼= A′S.

On the other hand, it is assumed that AC ∼= A′C ′. At first, it seems that the locationof C ′ could be like C ′? or C ′′? in the drawing. But notice, as explained in problem 0.2:the transfer of segment AC is unique. Hence we get C ′ = S. Finally BC ∼= B′S andA′ ∗B′ ∗ S imply BC ∼= B′C ′ and A′ ∗B′ ∗ C ′, as to be shown.

Definition 3.1 (Segment comparison). For any two given segments AB and CDwe say that AB is less than CD, iff there exists a point E between C and D such thatAB ∼= CE and C ∗E ∗D. In this case, also say that CD is greater than AB. We writeCD > AB or AB < CD. equivalently.

Proposition 3.4 (Segment comparison holds for congruence classes). AssumingAB ∼= A′B′ and CD ∼= C ′D′, we get: AB < CD if and only if A′B′ < C ′D′.

Proof. Transfer segment AB onto ray−−→CD, and get AB ∼= CE. The assumption AB <

CD implies that C ∗ E ∗ D. Too, we transfer segment A′B′ onto ray−−→C ′D′, and get

A′B′ ∼= C ′E ′.We now use segment subtraction for points C, E, D and C ′, E ′, D′. Since CD ∼= C ′D′

by assumption, and CE ∼= AB ∼= A′B′ ∼= C ′E ′—by assumption and construction andtransitivity—, and C ∗E ∗D; segment subtraction yields ED ∼= E ′D′ and C ′ ∗E ′ ∗D′.Thus E ′ lies between C ′ and D′, from which we conclude A′B′ < E ′D′.

Proposition 3.5 (Transitivity of segment comparison). If AB < CD and CD < EF ,then AB < EF .

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Figure 3.3: Segment comparison for congruence classes

Proof. I can assume that all three segments lie on the same ray−→AB, and A = C = E.

(Some transferring of segments will produce segments congruent with the given oneswhich satisfy these requirements.) Now, having done this, we get AB < AD andAD < AF . By definition this means A ∗B ∗D and A ∗D ∗ F .

Question. Express these order relations in words.

Answer. Point B lies between A and D, and point D lies between A and F .

By Theorem 5 from Hilbert’s foundations, any four points on a line can be notatedin a way that all four alphabetic order relations hold. (I shall that statement the ”Fourpoint Theorem”). Now the four points A, B, D, F satisfy the order relations A ∗ B ∗Dand A ∗D ∗ F , which implies they are already notated in alphabetic order. Hence thetwo other order relations follow: A ∗ B ∗ F and B ∗D ∗ F . But A ∗ B ∗ F means bydefinition that AB < AF , as to be shown.

Figure 3.4: Transitivity of segment comparison

Corollary. If AB < AD and AD < AF , then BD < BF .

Proposition 3.6 (Any two segments are comparable). For any two segments ABand CD, one and only one of the three cases (i)(ii)(iii) occurs:

Either (i) AB < CD or (ii) AB ∼= CD or (iii) CD < AB.

Proof. Transfer segment AB onto the ray−−→CD. We get AB ∼= CB′. Now the two

segments CB′ and CD start at the same vertex and lie on the same ray.

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By Theorem 4 from Hilbert’s foundations, for any given three points on a line, exactlyone lies between the other two. Hence the three points C, B′, D can satisfy

either (i) C ∗ B′ ∗ D. or (ii) B′ = D. or (iii) C ∗ D ∗ B′. These cases

correspond to the three cases as claimed. Indeed, in case (i), C ∗ B′ ∗ D implies bydefinition AB < CD. Indeed, in case (ii), B′ = D implies AB ∼= CB′ = CD byconstruction. An explanation may be needed in case (iii): One transfers segment CD

Figure 3.5: All segments are comparable

back onto the ray−→AB and gets CD ∼= AD0. Since CB′ ∼= AB, we can now use segment

subtraction, and get DB′ ∼= D0B and A ∗ D0 ∗ B. Because of the last order relation,

transferring segment CD back to ray−→AB confirms CD < AB.

Proposition 3.7 (About sums of segments). The order, and the sum of segmentsis defined on equivalence classes of congruent segments. These equivalence classes havethe following properties:

Commutativity a + b = b + a.

Associativity (a + b) + c = a + (b + c)

Comparison Any two segments a and c satisfy either a < c, or a = c or a > c.

Difference a < c if and only if there exists b such that a + b = c.

Comparison of Sums If a < b and c < d, then a + c < b + d.

Proof. By Hilbert’s axiom of congruence III.3, the sum is defined on equivalence classesof congruent segments. Now we check the items stated:

Commutativity: Let segment AB represent the equivalence class a. By axiom III.1,we can choose point C such that A ∗ B ∗ C, and segment BC represents theequivalence class b. Since BC = CB, and BA = AB, these two segments representthe classes b and a. By the definition of segment addition CA = CB + BA, andthe segment represents b + a. Hence AC = CA implies that a + b = b + a.

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Associativity: We construct (a + b) + c. To this end, choose a segment AB of con-gruence class a, in short AB ∈ a. Again, choose point C such that A ∗B ∗C, andsegment BC ∈ b. Furthermore, choose point D such that A ∗C ∗D, and segmentCD ∈ c. We now know that

AD ∈ (a + b) + c

On the other hand, we begin by constructing at first b + c. By the four-pointtheorem, A ∗ B ∗ C and A ∗ C ∗ D imply B ∗ C ∗ D. Hence BD ∈ b + c. Theconstruction of a + (b + c) is finished by finding point H such that A ∗B ∗H, andsegment BH ∈ b + c. We now know that

AH ∈ a + (b + c)

The uniqueness of segment transfer implies H = D. Finally, we see that AD =AH, and hence (a + b) + c = a + (b + c).

Comparison: Let any segments a and c be given. These equivalence classes can be

represented by segments AB ∈ a and AC ∈ c on the same ray−→AB =

−→AC. The

three point theorem implies that either B = C, or A ∗B ∗C, or A ∗C ∗B. Henceeither a = c, or a < c, or c < a. In the last case a > c as shown above.

Difference: Assume a < c. These equivalence classes can be represented by segmentsAB ∈ a and AC ∈ c such that A ∗ B ∗ C. Now BC represents the class b suchthat a + b = c. The converse is as obvious.

Comparison of Sums The proof is left to the reader.

Proposition 3.8 (Comparison of supplements). Let a segment PQ and two pointsA, B in it be given. Assume that PA < PB. Then BQ < AQ.

Proof. The reader has to try himself.

3.2 Some Elementary Triangle Congruences

Definition 3.2 (Triangle, Euler’s notation). For a triangle 4ABC, it is assumedthat the three vertices do not lie on a line. I follow Euler’s conventional notation forvertices, sides and angles: in triangle 4ABC, let A, B and C be the vertices, let thesegments a = BC, b = AC, and c = AB be the sides and the angles α := ∠BAC,β := ∠ABC, and γ := ∠ACB be the angles.

For a segment AB, it is assumed that the two endpoints A and B are different. Fora triangle 4ABC, it is assumed that the three vertices do not lie on a line.

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Proposition 3.9 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert’s theo-rem 11] An isosceles triangle has congruent base angles.

Question. Formulate the theorem with specific quantities from a triangle 4ABC. Pro-vide a drawing.

Answer. If a ∼= b, then α ∼= β.

Figure 3.6: An isosceles triangle

Proof. This is an easy application of SAS-congruence. Assume that the sides AC ∼= BCare congruent in 4ABC. We need to show that the base angles α = ∠BAC andβ = ∠ABC are congruent. Define a second triangle 4A′B′C ′ by setting

A′ := B , B′ := A , C ′ := C

(It does not matter that the second triangle is just ”on top” of the first one.) To applySAS congruence, we match corresponding pieces:

(1) ∠ACB = ∠BCA = ∠A′C ′B′ because the order of the sides of an angle is arbitrary.By axiom III.4, last part, an angle is congruent to itself. Hence ∠ACB ∼= ∠A′C ′B′.

(2) AC ∼= A′C ′.

Question. Explain why this holds.

Answer. AC ∼= BC because we have assumed the triangle to be isosceles, andBC = A′C ′ by construction. Hence AC ∼= A′C ′.

(3) Similarly, we show that (3): BC ∼= B′C ′:

Proof. BC ∼= AC because we have assumed the triangle to be isosceles, and con-gruence is symmetric, and AC = B′C ′ by construction. Hence BC ∼= B′C ′.

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Finally, we use axiom III.5. Items (1)(2)(3) imply ∠BAC ∼= ∠B′A′C ′ = ∠ABC. Butthis is just the claimed congruence of base angles.

The next Proposition is Theorem 12 of Hilbert’s Foundations of Geometry: Hereonly the weak form of the SAS-axiom (Hilbert’s Axiom III.5) is assumed.

Question. Why is Hilbert’s Axiom III.5 weaker than the SAS congruence theorem?

Answer. In Hilbert’s Axiom, only congruence of a further pair of angles is postulated.In the SAS congruence theorem, all pieces of the two triangles are stated to be pairwisecongruent.

Figure 3.7: SAS congruence

Proposition 3.10 (Hilbert’s SAS-axiom implies the full SAS Congruence The-orem). [Theorem 12 in Hilbert] Given are two triangles. We assume that two sides andthe angle between these sides are congruent to the corresponding pieces of the second tri-angle. Then the two triangles are congruent, which means that all corresponding piecesare pairwise congruent.

Proof. Let 4ABC and 4A′B′C ′ be the two triangles. We assume that the angles at Aand A′ as well as two pairs of adjacent sides are matched:

c = AB ∼= A′B′ = c′ , b = AC ∼= A′C ′ = b′ , α = ∠CAB ∼= ∠C ′A′B′ = α′

We need to show that

β = ∠ABC∼=∠A′B′C ′ = β′(a)

γ = ∠BCA∼=∠B′C ′A′ = γ′(b)

a = BC∼=B′C ′ = a′(c)

Note that by Hilbert’s weaker SAS-axiom, we can concluded only part (a). Part (b)follows immediately by applying Hilbert’s axiom to triangles 4ACB and 4A′C ′B′. We

48

need still to show part (c). Transferring segment BC onto ray−−→B′C ′, we get a seventh

point D′ on that ray such that

(1) BC ∼= B′D′

Applying Hilbert’s (weak) SAS axiom to the two triangles 4ABC and 4A′B′D′ yields∠BAC ∼= ∠B′A′D′. On the other hand, by assumption ∠BAC ∼= ∠B′A′C ′. By unique-ness of angle transfer (see Hilbert’s axiom III.4), there exists one and only one ray, that

forms with the ray−−→A′B′ the given ∠BAC, and lies on the same side of the line A′B′ as

point C ′. Hence rays−−→A′D′ =

−−→A′C ′ are equal, in other words D′ lies on the ray

−−→A′C ′.

By construction of 4A′B′D′, point D′ lies on the ray−−→B′C ′, too. Since the intersection

point of the two lines A′C ′ and B′C ′ is unique, one concludes

(2) C ′ = D′

From (1) and (2) we get BC ∼= B′C ′, as to be shown.

Proposition 3.11 (Extended ASA-Congruence Theorem). Given is a triangleand a segment congruent to one of its sides. The two angles at the vertices of this sideare transferred to the endpoints of the segment, and reproduced in the same half plane.Then the newly constructed rays intersect, and one gets a second congruent triangle.

Figure 3.8: Extended ASA congruence

Proof. Let the triangle be 4ABC and the segment A′B′ ∼= AB. The angle β = ∠ABC

is transferred at B′ along ray−−→B′A′, and the angle α = ∠BAC is transferred at A′ along

ray−−→A′B′, both are reproduced on the same side of line A′B′. One gets two new rays r′A

and r′B such that

α = ∠(−→AB,−→AC) ∼= ∠(

−−→A′B′, r′A) , β = ∠(

−→BA,−−→BC) ∼= ∠(

−−→B′A′, r′B)

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It is claimed that the two new rays r′A and r′B do intersect at some point C ′′. Further-more, it is claimed that the two triangles 4ABC and 4A′B′C ′′ are congruent.

On the newly produced ray r′A, we transfer segment AC and get a point C ′′ suchthat AC ∼= A′C ′′. Now SAS congruence (Theorem 12 of Hilbert, see Proposition 3.10above) implies

(1) 4ABC ∼= 4A′B′C ′′

The three pairs of matching pieces used to prove the congruence are stressed in matching

colors. Congruence (1) implies β = ∠ABC ∼= ∠A′B′C ′′ = ∠(−−→B′A′,

−−−→B′C ′′). On the other

Figure 3.9: How to get extended ASA congruence

hand, ∠ABC ∼= ∠(−−→B′A′, r′B) was assumed, too. Hence by uniqueness of angle transfer,

we get two equal rays:−−−→B′C ′′ = r′B. Thus the point C ′′ lies on the newly produced ray

r′B, too. Thus the two rays r′A and r′B intersect in point C ′′, and (1) holds, as to beshown.

Proposition 3.12 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles witha pair of congruent sides, and pairwise congruent adjacent angles are congruent.

Proof of ASA congruence. Given the triangles 4ABC and 4A′B′C ′, with congruentsegments AB ∼= A′B′ and two pairs of congruent adjacent angles at A, A′ and B, B′.

The segment AC is transferred onto the ray−−→A′C ′. On this ray, one gets a point X

such that AC ∼= A′X. Now SAS congruence is applied to the triangles 4ABC and4A′B′X. The three pairs of matching pieces used to prove the congruence are stressedin matching colors. One concludes

4ABC ∼= 4A′B′X(3.1)

∠ABC ∼= ∠A′B′X

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Figure 3.10: How to get ASA congruence

On the other hand, ∠ABC ∼= ∠A′B′C ′ was assumed, too. Hence by uniqueness of angletransfer, we get two equal rays: −−→

B′X =−−→B′C ′

Thus the point X lies on both this ray, and the ray−−→A′X =

−−→A′C ′, too. These two rays

have a unique intersection point, since they do not lie on the same line. Hence X = C ′,and (3.1) is just the required triangle congruence.

Short proof of ASA congruence. Given the triangles 4ABC and 4A′B′C ′, we applyextended ASA congruence to the first 4ABC and segment A′B′. The uniqueness of

angle transfer implies r′A =−−→A′C ′ and r′B =

−−→B′C ′. The uniqueness of the intersection

point of these two rays implies point C ′′ = C ′, and (1) yields the congruence to beshown.

Question. In which point does the extended ASA congruence theorem extend the usualASA congruence theorem?

Answer. The extended ASA theorem differs from the usual ASA-congruence theorem,because existence of a second triangle is not assumed—only a segment is given. It isproved that the two newly produced rays do intersect.

Proposition 3.13 (Preliminary Converse Isosceles Triangle Proposition). Ifthe two base angles of a triangle are congruent to each other, the triangle is isosceles.

Question. Formulate the theorem with specific quantities from a triangle 4ABC. Pro-vide a drawing.

Answer. If α ∼= β and β ∼= α, then a ∼= b.

Remark. We can avoid that awkward assumption, ”If α ∼= β and β ∼= α”, later inProposition 3.29. Hilbert proves the converse isosceles triangle theorem as his Theorem24, after having the exterior angle theorem at his disposal.

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Figure 3.11: An isosceles triangle, two ways to look at it

Proof. We use ASA-congruence to prove the Proposition. Assume that the angles α =∠CAB and β = ∠ABC are congruent in 4ABC. We need to show that the two sidesa = BC and b = AC are congruent. Define a second triangle 4A′B′C ′ by setting

A′ := B , B′ := A , C ′ := C

(It does not matter that the second triangle is just ”on top” of the first one.) To applyASA congruence, we match corresponding pieces:

(1) AB = B′A′ = A′B′. Hence AB ∼= A′B′, because the order of the endpoints of asegment is arbitrary, and a segment is congruent to itself.

(2) α ∼= α′.

Question. Explain why this holds.

Answer. α ∼= β by assumption, and β = ∠ABC = ∠B′A′C ′ = α′ by construction.Hence α ∼= α′.

(3) Similarly, one shows that β ∼= β′: β ∼= α by assumption, and α = ∠BAC =∠A′B′C ′ = β′ by construction. Hence β ∼= β′.

Via ASA congruence, items (1)(2)(3) imply that 4ABC ∼= 4A′B′C ′, and hence espe-cially AC ∼= A′C ′ = BC as to be shown.

3.3 Congruence of Angles

Definition 3.3 (Supplementary Angles). Two angles are called supplementary an-gles, iff they have a common vertex, both have one side on a common ray, and the twoother sides are the opposite rays on a line.

Definition 3.4 (Vertical Angles). Two angles are called vertical angles, iff they havea common vertex, and their sides are two pairs of opposite rays on two lines.

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Proposition 3.14 (Congruence of Supplementary Angles). [Theorem 14 of Hilbert]If an angle ∠ABC is congruent to another angle ∠A′B′C ′, then its supplementary angle∠CBD is congruent to the supplementary angle ∠C ′B′D′ of the second angle.

Proof. The three steps of the proof each identify a new pair of congruent triangles.Step 1: One can choose the points A′, C ′ and D′ on the given rays from B such that

AB ∼= A′B′ , CB ∼= C ′B′ , DB ∼= D′B′

Because of the assumption ∠ABC ∼= ∠A′B′C ′, SAS congruence (Theorem 12 of Hilbert,see Proposition 3.10 above) now implies that 4ABC ∼= 4A′B′C ′. In the drawing, thethree pairs of matching pieces used to prove the congruence are stressed in matchingcolors. Congruence of the two triangles implies

Figure 3.12: Congruence of supplementary angles, the first pair of congruent triangles

(1) AC ∼= A′C ′ and ∠BAC ∼= ∠B′A′C ′

Step 2: By axiom III.3, adding congruent segments yields congruent segments. Hencethe segments AD and A′D′ are congruent. Now SAS congruence (Theorem 12 of Hilbert,see Proposition 3.10 above) implies that the (greater) triangles, too, are congruent inthe two figures below. In the drawing, the three pairs of matching pieces used to provethe congruence are stressed in matching colors. The congruence 4CAD ∼= 4C ′A′D′

Figure 3.13: Congruence of supplementary angles, the second pair of congruent triangles

implies

(2) CD ∼= C ′D′ and ∠ADC ∼= ∠A′D′C ′

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Figure 3.14: Congruence of supplementary angles, the third pair of congruent triangles

Step 3: At last we consider the two triangles 4BCD and 4B′C ′D′ on the right side.In the drawing, the three pairs of matching pieces used to prove the congruence arestressed in matching colors. Again by using SAS congruence (Theorem 12 of Hilbert,see Proposition 3.10 above), we see the two triangles are congruent. Finally ∠CBD ∼=∠C ′B′D′, as to be shown.

Corollary. Adjacent angles congruent to supplementary angles are supplementary, too.

Figure 3.15: Supplementary angles yield points on a line

Proof. Given are supplementary angles ∠ABC and ∠DBC, a further congruent angle∠ABC ∼= ∠A′B′C ′, and a point D′ with A′ and D′ lying on different sides of line B′C ′.

We shown that the angles ∠A′B′C ′ and ∠D′B′C ′ are supplementary if and only if∠CBD ∼= ∠C ′B′D′. Above, we have already shown one direction: If the angles ∠A′B′C ′

and ∠D′B′C ′ are supplementary, then ∠CBD ∼= ∠C ′B′D′.Now we show the converse. Assume that ∠CBD ∼= ∠C ′B′D′. We have to check

whether point B′ lies between A′ and D′. Choose any point D′′ on the ray opposite to−−→B′A′. Congruence of supplementary angles (Hilbert’s Theorem 14, see Proposition 3.14above) implies ∠CBD ∼= ∠C ′B′D′′. On the other hand, ∠CBD ∼= ∠C ′B′D′ is assumed.

Angle ∠CBD is transferred uniquely along ray−−→B′C into the half plane opposite to A′.

Indeed, by axiom III.4, angle transfer produces a unique new ray. Hence−−−→B′D′′ =

−−→B′D′.

and the four points A′, B′, D′, D′′ lie on one line, with point B′ between A′ and D′.Hence angles ∠A′B′C ′ and ∠D′B′C ′ are supplementary, too.

Proposition 3.15 (Congruence of Vertical Angles). [Euclid I.15] Vertical anglesare congruent.

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Proof. This is an easy consequence of Hilbert’s Theorem 14 about supplementary angles(see Proposition 3.14 above). Take any two vertical angles ∠ABC and ∠A′BC ′. Weassume that vertex B lies between points A and A′ on one line, as well as betweenthe two points C and C ′ on a second line. As shown in the figures, angle ∠ABC ′ hasangle (i) ∠ABC as supplementary angle. Secondly, angle ∠ABC ′ has (ii) ∠A′BC ′ as

Figure 3.16: Two pairs of supplementary angles yield vertical angles

supplementary angle, too.An angle is congruent to itself, as stated in Axiom III.4. Hence especially ∠ABC ′ ∼=

∠ABC ′. By Theorem 14 of Hilbert (see Proposition 3.14 above), angles supplementaryto congruent angles are congruent, too. Hence we conclude congruence of the two verticalangles: ∠ABC ∼= ∠A′BC ′, as to be shown.

Definition 3.5 (The sum of two angles). Let an angle ∠ABC and a ray−−→BG in its

interior. Angle ∠ABC is called the sum of the angles ∠ABG and ∠GBC. One writes∠ABC = ∠ABG + ∠GBC.

Figure 3.17: Putting together two angles, three cases are possible:(i) They are supplementary (ii) They can be added (iii) They cannot be added

Lemma 3.1. Given are two angles ∠ABG and ∠GBC with same vertex B lying on

different sides of a common ray−−→BG. Exactly one of three possibilities occur about the

angles ∠ABG and ∠GBC:

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(i) They are supplementary. The three points A, B and C lie on a line.

(ii) They can be added. Their sum is ∠ABG+∠GBC = ∠ABC. Points A and Glie on the same side of line BC. Points C and G lie on the same side of line AB.

(iii) They cannot be added. Points A and G lie on different sides of line BC.Points C and G lie on different sides of line AB. (The sum would be an over-obtuse angle.)

Figure 3.18: If A and G lie on different sides of BC, then C and G lie on different sides ofAB.

Proof. Suppose that neither case (i) nor (ii) occurs. Under that assumption, either

(a) points A and G lie on different sides of line BC—or

(b) points C and G lie on different sides of line AB.

Suppose case (a) occurs. Segment AG intersects line BC, say in point H. We usePasch’s axiom for triangle 4CHG and line AB. This line does not intersect side GH,but intersects side CH. Indeed, point B lies between C and H, since C and A, andhence C and H are on different sides of line BG.

Now Pasch’s axiom—for triangle 4CHG and line AB—implies that this line inter-sects side CG, say at point Pasch. Thus points C and G lie on different sides of lineAB. We have shown case (iii) to occur.

Suppose case (b) occurs. The same argument—with A and C exchanged— showsthat case (iii) occurs once more.

Proposition 3.16 (Proposition for Angle-Addition). [Theorem 15 in Hilbert]

Given is an angle ∠ABC and a ray−−→BG in its interior, as well as a second angle ∠A′B′C ′

and a ray−−→B′G′ in its interior. Furthermore, assume that ∠CBG ∼= ∠C ′B′G′ and

∠ABG ∼= ∠A′B′G′. Then the two angle sums are congruent, too: ∠ABC ∼= ∠A′B′C ′.

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Figure 3.19: Angle addition

Proof. By the Crossbar Theorem, a segment going from one side of an angle to theother, and a ray in the interior of that angle always intersect. Hence there exists a point

H such that−−→BG =

−−→BH and A ? H ? C. For simplicity, we choose G = H from the

beginning. Too, we may assume that A′, C ′ and G′ are chosen such that BA ∼= B′A′,BC ∼= B′C ′ and BG ∼= B′G′. The proof uses three pairs of congruent triangles, in step(1)(2)(3), respectively.

Step (1): The SAS-congruence axiom implies 4BGC ∼= 4B′G′C ′ because of∠GBC ∼= ∠G′B′C ′ , GB ∼= G′B′ and BC ∼= B′C ′ which hold by assumption and theremarks above. In the drawing, the three pairs of matching pieces used to prove thecongruence are stressed in matching colors. From the congruence of these two triangles,

Figure 3.20: the first pair of congruent triangles

we conclude that

∠BCG ∼= ∠B′C ′G′(1)

∠BGC ∼= ∠B′G′C ′(1b)

Step (2): As a second step, the SAS-congruence axiom implies 4BGA ∼= 4B′G′A′

because of ∠GBA ∼= ∠G′B′A′ , GB ∼= G′B′ and BA ∼= B′A′ which hold by assumption

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and the remarks above. Again, in the drawing, the three pairs of matching pieces usedto prove the congruence are stressed in matching colors. By construction, the two angles

Figure 3.21: the second pair of congruent triangles

∠BGA and ∠BGC are supplementary angles. The congruence ∠BGA ∼= ∠B′G′A′ fol-lows from step (2), and ∠BGC ∼= ∠B′G′C ′ as stated by (1b). As derived in our Corollaryto Hilbert’s Theorem 14 (see Proposition 3.14 above), adjacent angles congruent to sup-plementary angles are supplementary, too. Hence the two angles ∠B′G′A′ and ∠B′G′C ′

are supplementary. Hence the three points A′, G′ and C ′ lie on a straight line. Bysegment addition (Hilbert’s axiom III.3), AG ∼= A′G′ and GC ∼= G′C ′ imply

(2) AC ∼= A′C ′

Step (3): To set up a third pair of congruent triangles, we use (1) (2) and the assumption

(3) BC ∼= B′C ′

The SAS axiom shows that 4ABC ∼= 4A′B′C ′ because of (1)(2)(3). I have stressedthese three pairs of matching pieces in matching colors. Finally, ∠ABC ∼= ∠A′B′C ′

follows from 4ABC ∼= 4A′B′C ′, as to be shown.

Remark. The other case that ray−−→BG does not lie in the interior of ∠ABC, but ray−−→

BC lies in the interior of angle ∠ABG, leads to a corresponding theorem about anglesubstraction.

Remark. Because of these theorems, angle addition and subtraction are defined forcongruence classes of angles.

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Figure 3.22: the third pair of congruent triangles

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3.4 SSS Congruence

This is Theorem 17 in Hilbert. It is needed as a preparation to get SSS congruence.

Proposition 3.17 (The Symmetric Kite). [Theorem 17 of Hilbert] Let Z1 and Z2 betwo points on different sides of line XY , and assume that XZ1

∼= XZ2 and Y Z1∼= Y Z2.

Then the two angles ∠XY Z1∼= ∠XY Z2 are congruent.

Proof. The congruence of the base angles of isosceles4XZ1Z2 yields ∠XZ1Z2∼= ∠XZ2Z1.

Similarly, one gets ∠Y Z1Z2∼= ∠Y Z2Z1. Now angle addition (or substraction) implies

(*) ∠XZ1Y ∼= ∠XZ2Y

Angle addition is needed in case ray−−−→Z1Z2 lies inside ∠XZ1Y , angle subtraction in case

ray−−−→Z1Z2 lies outside ∠XZ1Y .In the special case that either point X or Y lies on the line Z1Z2, one gets the same

conclusion even easier. Here are drawings for the three cases. One now applies SAS

Figure 3.23: The symmetric kite

congruence to 4XZ1Y and 4XZ2Y . Indeed the angles at Z1 and Z2 and the adjacentsides are pairwise congruent. Hence the assertion ∠XY Z1

∼= ∠XY Z2 follows.

Before getting the general SSS-congruence, I consider one further special case.

Lemma 3.2 (Lemma for SSS Congruence). Assume that the two triangles 4ABCand 4AB′C have a common side AC, and all three corresponding sides are congruent,and the two vertices B and B′ lie on the same side of line AC. Then the two trianglesare identical.

Proof. We transfer the angle ∠BAC onto the ray−→AC, on the side of line AC opposite

to B and B′. On the newly produced ray, we transfer segment AB, starting at vertexA. Thus we get point B′′, and segment AB′′ ∼= AB. From SAS-congruence (Theorem12 of Hilbert, see Proposition 3.10 above), one concludes that BC ∼= B′′C. As stressedin the first drawing, one has constructed a symmetric kite with the four points

X := A, Y := C, Z1 := B, Z2 := B′′

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Figure 3.24: Which kite is symmetric?

Hence by Theorem 17 (see Proposition 3.17 above), ∠B′′AC ∼= ∠BACBut wait! Another choice of four points to get a kite is

X := A, Y := C, Z1 := B′, Z2 := B′′

—replacing B by B′. Because of the assumptions, and the construction of the first kite,AB′ ∼= AB ∼= AB′′ and B′C ∼= BC ∼= B′′C. We see that the second kite, stressed in thesecond drawing, satisfies the assumptions of Hilbert’s Theorem 17 (see Proposition 3.17above), too. Now we conclude from Hilbert’s Theorem 17 (see Proposition 3.17 above)

that ∠B′′AC ∼= ∠B′AC. Finally, the uniqueness of angle transfer implies that−→AB =

−−→AB′. Since AB ∼= AB′, and segment transfer was shown to produce a unique point, wehave confirmed that B = B′. Thus the two triangles 4ABC and 4AB′C are identical,as to be shown.

Remark. We do not need to assume that angle congruence is an equivalence relation.But therefore we have to consider two kites, by choosing two of the three triangles, asshown in two drawings. Then we can use that transferring an angle gives a unique ray.

From Hilbert’s Theorem 17 (see Proposition 3.17 above), and the Lemma 3.2 above,we see that SSS-congruence holds for any two triangles with a common side. Now wecan easily get the general case of

Proposition 3.18 (SSS Congruence). [Theorem 18 in Hilbert’s Foundations] Twotriangles with three pairs of congruent sides are congruent.

Proof. Assume the two triangles 4ABC and 4A′B′C ′ have corresponding sides which

are congruent. We transfer the angle ∠BAC onto the ray−−→A′C ′, at vertex A′, to the

same side of A′C ′ as B′. On the newly produced ray, we transfer segment AB, startingat vertex A′. Thus we get point B0, such that A′B0

∼= AB. Because of SAS-congruence(Theorem 12 of Hilbert, see Proposition 3.10 above), we get

(*) 4ABC ∼= 4A′B0C′

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Figure 3.25: Which two triangles are congruent?

Hence especially, AB ∼= A′B0∼= A′B′ and BC ∼= B0C

′ ∼= B′C ′. We can now applythe Lemma‘3.2 to the two triangles 4A′B′C ′ and 4A′B0C

′. Hence B′ = B0, and theassertion follows from (*).

3.5 The Equivalence Relation of Angle Congruence

Proposition 3.19 (Theorem 19 in Hilbert’s Foundations). If two angles ∠(h′, k′)and ∠(h′′, k′′) are congruent to a third angle ∠(h, k), then the two angles ∠(h′, k′) and∠(h′′, k′′) are congruent, too.

Proof for a bottle of wine from Hilbert to A. Rosenthal. 12 Let the vertices of the anglesbe O,O′, O′′. Choose points A, A′, A′′ on one side of the three angles, respectively,such that O′A′ ∼= OA and O′′A′′ ∼= OA. Similarly choose points B, B′, B′′ on thethree remaining sides, respectively, such that O′B′ ∼= OB and O′′B′′ ∼= OB. Now theassumption of SAS congruence (Theorem 12 of Hilbert, see Proposition 3.10 above) aremet for both 4A′O′B′ and 4AOB, as well as 4A′′O′′B′′ and 4AOB. Hence

A′B′ ∼= AB , A′′B′′ ∼= AB

We use axiom III.2:

"two segments congruent to a third one are congruent to each other".

Hence 4A′B′O′ and 4A′′B′′O′′ have three pairs of congruent sides. By the SSS-congruence Theorem 18, we conclude that ∠(h′, k′) ∼= ∠(h′′, k′′), as to be shown.

Proposition 3.20 (Congruence is an equivalence relation). Congruence is anequivalence relation on the class of angles.

Question. Which three properties do we need to check for a congruence relation?

Answer. For an arbitrary relation to be a congruence relation, we need to check

(a) reflexivity: Each angle is congruent to itself, written α ∼= α.

12Hilbert gives credit for this proof to A. Rosenthal (Math. Ann. Band 71)

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(b) symmetry: If α ∼= β, then β ∼= α.

(c) transitivity: If α ∼= β, and β ∼= γ, then α ∼= γ.

Question. How can we claim reflexivity?

Answer. Reflexivity is given by the last part of axiom III.4

Proof of symmetry, for a bottle of wine from A. Rosenthal to Hilbert. 13 By Theorem 19(see Proposition 3.19 above), ∠(h′, k′) ∼= ∠(h, k) and ∠(h′′, k′′) ∼= ∠(h, k) imply ∠(h′, k′) ∼=∠(h′′, k′′). Hence, with just other notation, we see that β ∼= β and α ∼= β implyβ ∼= α.

Proof of transitivity. Assume α ∼= β and β ∼= γ. Because of symmetry γ ∼= β. Now weuse Theorem 19 (see Proposition 3.19 above). Hence, just with other notation, α ∼= βand γ ∼= β imply α ∼= γ.

Definition 3.6 (Angle comparison). Given are two angles ∠BAC and ∠B′A′C ′.

We say that ∠BAC is less than ∠B′A′C ′, iff there exists a ray−−→A′G in the interior of

∠B′A′C ′ such that ∠BAC ∼= ∠B′A′G. In this case, we also say that ∠B′A′C ′ is greaterthan ∠BAC. We write ∠B′A′C ′ > ∠BAC and ∠BAC < ∠B′A′C ′, equivalently.

Proposition 3.21 (Angle comparison holds for congruence classes). Assumethat α ∼= α′ and β ∼= β′. α < β if and only if α′ < β′.

Proof. The reader should do it on her own.

Proposition 3.22 (Transitivity of angle comparison). If α < β and β < γ, thenα < γ.

Proof. After having done some transfer of angles, I can assume that all three angles

α, β, γ have the common side−→AB, and lie on the same side of AB. Choose any point E

on the second side of γ = ∠BAE.Because β < γ, the second side of β is in the interior of the largest angle γ. Hence,

by the Crossbar Theorem the segment BC intersects that second side of β, say at pointD, and β = ∠BAD as well as B ∗D ∗ E.

Because α < β, the second side of α is in the interior of β . Hence, by theCrossbar Theorem the segment BD intersects that second side of α, say at point E, andα = ∠BAE as well as B ∗ C ∗ D. Any four points on a line can be ordered in a waythat all four alphabetic order relations hold. (see Theorem 5 in Hilbert, which I callthe ”four-point Theorem”). Now the four points B, C, D,E satisfy the order relationsB ∗ C ∗ D and B ∗ D ∗ E. Therefore they are already put in alphabetic order. HenceB ∗ C ∗ E . This shows by definition that α = ∠BAC < ∠BAE = γ.

13Hilbert gave credit and got back a good bottle

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Figure 3.26: Transitivity of comparison of angles

Proposition 3.23 (All angles are comparable). For any two angles α and β, oneand only one of the three cases (i)(ii)(iii) occurs:

Either (i) α < β or (ii) α ∼= β or (iii) β < α.

Proof. Not more than one of the cases (i)(ii)(iii) can occur at once. Case (ii) excludeseither (i) or (iii) by definition of angle comparison. But case (i) and (iii) cannot holdboth at the same time, neither: By transitivity, α < β and β < α together would implyα < α. This is impossible, because an angle is congruent to itself: α ∼= α by axiom III.4.

Now we show that actually one of the cases (i)(ii)(iii) does occur. Transfer angle αonto one side of β. Thus we get the two angles α = ∠BAC ∼= α and β = ∠BAD which

have the common side−→AB, and lie on the same side of AB. The second side

−→AC of

angle α can

either (i) lie in the interior of angle β.

or (ii) be identical to the second side of β.

or (iii) lie in the exterior of angle β.

These cases correspond to the three cases as claimed. An explanation may be needed in

Figure 3.27: All angles are comparable

case (iii): The ray−→AC lies in the exterior of angle β = ∠BAD. What does that mean?

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Answer. Either points B and C lie on different sides of line AD, or points C and D lieon different sides of line AB.

The points C and D lie on the same side of AB by the arrangement of the angles. HenceB and C lie on different sides of AD. By the Crossbar theorem, segment BC intersects

ray−−→AD. In term to comparing angles, we get β = ∠BAD < ∠BAC = α ∼= α.

Proposition 3.24 (Comparison of supplements). If α < β, then their supplementsS(α) and S(β) satisfy S(α) > S(β).

Figure 3.28: Comparison of supplements

Proof. Transfer angle α onto one side of β, in the same half plane. Thus we get the two

angles α ∼= α and β = ∠BAD which have the common side−→AB, and lie on the same

side of AB.Because of α < β, the second side of angle α lies in the interior of angle β. By the

Crossbar theorem it intersects the segment BD, say at point Q. We have arranged that

α ∼= α = ∠BAQ, and β = ∠BAD. Let F be any point on the ray opposite to−→AB.

Apply Pasch’s axiom (Hilbert’s axiom of order II.4) to 4FBQ and line AD. That linedoes not intersect side BQ because of B ∗Q ∗D, but does intersect side FB because ofF ∗A∗B. Hence line AD intersects the third side FQ, say at point S. Indeed F ∗S ∗Q,

and S lies on the ray−−→AD because only that ray, and not its opposite ray lies in the

interior of ∠FAQ.The supplementary angle of β is S(β) = ∠FAD = ∠FAS. The supplementary angle

of α is S(α) = ∠FAQ. From F ∗ S ∗Q we get

S(β) = ∠FAS = ∠FAD < ∠FAQ = S(α)

The final step uses Hilbert’s Theorem 14 (see Proposition 3.14 above): the supplementsof congruent angles are congruent, too. Hence α ∼= α implies S(α) ∼= S(α). FromS(β) < S(α) and S(α) ∼= S(α), we conclude S(β) < S(α) as to be shown.

Definition 3.7 (acute, right and obtuse angles). A right angle is an angle congruentto its supplementary angle. An acute angle is an angle less than a right angle. An obtuseangle is an angle greater than a right angle.

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Remark (Remark about supplements of acute and obtuse angles). If an angle α is acute,its supplement S(α) is obtuse. Furthermore, the angle is less than its supplement.

If an angle β is obtuse, its supplement S(β) is acute. Furthermore, the angle is less thanits supplement.

Reason. Assume α < R, where R denotes a right angle. From the comparison ofsupplements done in Proposition 3.24, we conclude S(α) > S(R). But by the definitionof a right angle R ∼= S(R). Hence S(α) > S(R) ∼= R, and because comparison isfor congruence classes, we get S(α) > R. Hence α < R < S(α), and by transitivity(Problem 10.2), we conclude α < S(α).

Similarly, we explain that β > R implies S(β) < R and S(β) < β.

Proposition 3.25 (All right angles are congruent).

Question. How is a right angle defined?

Answer. A right angle is an angle congruent to its supplementary angle.

Proof in the conventional style. Let α be a right angle, and β = S(α) ∼= α be itscongruent supplement. Similarly, we consider a second pair of right angles α′ andβ′ = S(α′) ∼= α′ being its congruent supplement.

The question is whether both right angles α and α′ are congruent to each other. ByProposition 3.23 all angles are comparable. Hence the two angles α and α′ satisfy justone of the following relations.

Either (i) α < α′ or (ii) α ∼= α′ or (iii) α′ < α.

We need to rule out cases (i) and (iii) by deriving a contradiction. I only need to explaincase (i), because (iii) is similar. Now assume α < α′ towards a contradiction. By thecomparison of supplements from Proposition 3.24, we get S(α) > S(α′). Because a rightangle is congruent to its supplement we get α ∼= S(α) > S(α′) ∼= α′. As explained inProposition 3.21, angle comparison holds for congruence classes. Hence we get α > α′.

Thus, from the assumption α < α′, we have derive α > α′. Transitivity would implyα < α which is impossible. (Axiom III.4 states α ∼= α.)

Thus case (i) leads to a contradiction. Similarly, case (iii) leads to a contradiction.The only remaining possibility is case (ii): Any two given right angles α and α′ arecongruent.

Proposition 3.26 (About sums of angles). The sum of angles is defined on equiv-alence classes of congruent angles. It satisfies the following properties:

Commutativity If α + β exists, then β + α exists and β + α ∼= α + β.

Associativity (α + β) + γ ∼= α + (β + γ) and existence of one side implies existenceof the other one.

Difference α < γ if and only if there exists β such that α + β ∼= γ.

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Comparison of Sums 1 If α ∼= β and γ < δ and β + δ exists, then α + γ exists andα + γ < β + γ.

Comparison of Sums 2 If α < β and γ < δ and β + δ exists, then α + γ exists andα + γ < β + γ.

Proof. By Hilbert’s Theorem 15 and Proposition 3.20, the sum of angles is defined onequivalence classes of congruent angles. Now we check the items stated:

Commutativity: Let α ∼= ∠ABG and β ∼= ∠GBC where points G and C lie both onthe same side of line AB. With that setup, α + β ∼= ∠ABC, which is assumed toexist. Since the order of the two rays of an angle is defined to be chosen arbitrarily,

∠ABC = ∠CBA ∼= ∠CBG + ∠GBA ∼= β + α

hence the latter angle exists and β + α ∼= α + β.

Associativity: The proof is left to the reader.

Difference: Let α ∼= ∠ABG and γ ∼= ∠ABC where points G and C lie both on the

same side of line AB. With that setup, α < γ iff the ray−−→BG lies inside the angle

∠ABC iff α + β ∼= γ with β ∼= ∠GBC.

Comparison of Sums 1: Assuming that α ∼= β and γ < δ and β + δ exists, we knowthere exists ε such that γ + ε ∼= δ. Hence

(α + γ) + ε ∼= α + (γ + ε) ∼= β + δ

and α + γ < β + γ where the former is shown to exist.

Comparison of Sums 2: Assuming α < β and γ < δ, we know there exist angles ηand ε such that α + η ∼= β and γ + ε ∼= δ. Hence

(α + γ) + η + ε ∼= (α + η) + (γ + ε) ∼= β + δ

and α + γ < β + γ where the former is shown to exist.

Proposition 3.27 (The Hypothenuse Leg Theorem). Two right triangles for whichthe two hypothenuse, and one pair of legs are congruent, are congruent.

Question. Of which one of SAS, SSA, SAA, ASA, SSS congruence contains the hypothenuse-leg theorem as a special case? Is there a corresponding unrestricted congruence theorem?

Answer. The hypothenuse-leg theorem is a special case of SSA congruence. There is nounrestricted SSA congruence theorem.

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First proof of the hypothenuse-leg theorem 3.27 . Take two right triangles 4ABC and4A′B′C ′. As usual, we put the right angles at vertices C and C ′. We assume congruenceof the hypothenuses AB ∼= A′B′ and of one pair of legs AC ∼= A′C ′. We can build a kiteout of two copies of triangle 4ABC and two copies of triangle 4A′B′C ′.

To this end, one transfers angle ∠ABC onto the ray−−→BC into the opposite half plane

and gets ∠ABC ∼= ∠DBC. Point D can be chosen on the newly produced ray suchthat BA ∼= BD. By SAS-congruence it is easy to get 4ABC ∼= 4DBC. Too, the rightangle at C implies that points A, C and D lie on a line. (Why?)

Next we transfer from the second given triangle the angle ∠C ′A′B′ onto the ray−→AC into the half plane opposite to B, and get ∠C ′A′B′ ∼= ∠CAB′′. Point B′′ can bechosen on the newly produced ray such that A′B′ ∼= AB′′. The triangle congruence4A′B′C ′ ∼= 4AB′′C is easily checked by SAS-congruence. Congruence of all rightangles implies that points B, C and B′′ lie on a line.

Finally we transfer the same angle ∠C ′A′B′ onto the ray−−→DC, and get ∠C ′A′B′ ∼=

∠CAB′′′. Point B′′′ can be chosen on the newly produced ray such that A′B′ ∼= DB′′′.By SAS-congruence the triangle congruence 4A′B′C ′ ∼= 4DB′′′C is easily checked.Hence C ′B′ is congruent to both CB′′ and CB′′′. Thus these two segments are congruent(Why?), and B′′ = B′′′.

As shown in the drawing, one has constructed a symmetric kite with the four points

X := D, Y := A, Z1 := B, Z2 := B′′

Because of the congruence of the segments on the upper and the lower half of the figure

Figure 3.29: Four right triangles yield a kite. When is it even a rhombus?

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AB ∼= AB′′ and DB ∼= DB′′, we can apply Hilbert’s kite-theorem 3.17 and conclude

∠DAB = ∠XY Z1∼= ∠XY Z2 = ∠DAB′′

Now SAS-congruence easily yields4CAB ∼= 4CAB′′. Since by construction4C ′A′B′ ∼=4CAB′′, we conclude 4CAB ∼= 4C ′A′B′, as to be shown.

3.6 Constructions with Hilbert Tools

Now we show that a right angle actually exists, and do some further basic constructionswith the Hilbert tools. The only means of construction are those granted by Hilbert’saxioms.

Definition 3.8 (Hilbert tools). By incidence axiom I.1 and I.2, we can draw a uniqueline between any two given points. By axiom III.1 and Proposition 3.2 from the verybeginning, we transfer a given segment uniquely to a given ray. Finally by axiomIII.4, we transfer a given angle uniquely on a given ray into the specified half plane.Constructions done using only these means are called constructions by Hilbert tools.

10 Problem 3.1 (Drop a Perpendicular). Given is a line OA and a point Bnot on this line. We have to drop the perpendicular from point B onto line OA.

Construction 3.1. Draw ray−−→OB. Transfer angle ∠AOB, into the half plane opposite

to B, with ray−→OA as one side. On the newly produced ray, transfer segment OB to

produce a new segment OC ∼= OB. The line BC is the perpendicular, dropped frompoint B onto line OA.

Figure 3.30: Drop the perpendicular, the two cases

Proof of validity. The line OA and the segment BC intersect, because B and C lie ondifferent sides of OA. I call the intersection point M . It can happen that O = M . Inthat special case A 6= M and

(**) ∠AMB ∼= ∠AMC

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But these is a pair of congruent supplementary angles, because M lies between B andC. By definition, an angle congruent to its supplementary angle is a right angle. Hence∠AMB is a right angle.

In the generic situation O 6= M , we distinguish two cases

(i) On the given line OA, points M and A lie on the same side of O.

(ii) Point O lies between M and A.

In both cases we show by the SAS-congruence theorem that the two triangles 4OMBand 4OMC are congruent. Indeed, the angles at vertex O are congruent, both in case(i) and (ii):

In case (i) , the rays−→OA =

−−→OM are equal, and ∠MOB = ∠AOB ∼= ∠AOC = ∠MOC

follows from the construction.

In case (ii) , the rays−→OA and

−−→OM are opposite. Thus ∠MOB and ∠AOB, as well as

∠MOC and ∠AOC are supplementary angles. Again ∠AOB ∼= ∠AOC becauseof the angle transfer done in the construction. Now Theorem 14 of Hilbert (seeProposition 3.14 above) tells that supplements of congruent angles are congruent.Hence we get ∠MOB ∼= ∠MOC once again.

The adjacent sides OB ∼= OC are congruent by construction, too. Finally the commonsides OM is congruent to itself. The drawing stresses the pieces matched to prove thecongruence. Because the two triangles are congruent, the corresponding angles at vertex

Figure 3.31: Congruences needed in the two cases

M are congruent, too. Hence

(*) ∠OMB ∼= ∠OMC

which is a pair of congruent supplementary angles. A right angle is, by definition, anangle congruent to its supplementary angle . Hence either ∠OMB is a right anglebecause of formula (*), or, in the special case O = M , angle ∠AMB is a right anglebecause of (**).

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Proposition 3.28 (Supplements of acute and obtuse angles). Let R denote aright angle. For any angle γ and its supplement S(γ) exactly one of the following threecases occurs: Either (1) or (2) or (3).

(1) γ < R, S(γ) > R and γ < S(γ)

(2) γ ∼= R, S(γ) ∼= R and γ ∼= S(γ)

(3) γ > R, S(γ) < R and γ > S(γ).

Proof. All angles are comparable (see Proposition 3.23). For angle γ and the right angleR, exactly one of the three cases holds:

Either (i) γ < R or (ii) γ ∼= R or (iii) γ > R.

From SAS congruence (Theorem 12 of Hilbert, see Proposition 3.10 above), we know thatα ∼= β implies S(α) ∼= S(β). From Proposition 3.24 about comparison of supplements,we know that α < β implies S(α) > S(β). Hence with the help of Propositions 3.20and 3.21 we get:

in case (i), one concludes S(γ) > S(R) ∼= R > γ, and hence (a) γ < S(γ),

in case (ii), one concludes S(γ) ∼= S(R) ∼= R ∼= γ, and hence (b) γ ∼= S(γ),

in case (iii), one concludes S(γ) < S(R) ∼= R < γ, and hence (c) γ > S(γ).

By Proposition 3.23 above, all angles are comparable. Hence the two angles γ and itssupplement S(γ) satisfy either (a) or (b) or (c). This observation allows one to get theconverse of the conclusions shown above. For example, (a) excludes both (ii) and (iii),and hence implies (i). Thus we get

(a) implies (i) , (b) implies (ii) , (c) implies (iii); and finally(a) if and only if (i) , (b) if and only if (ii) , (c) if and only if (iii).

This leads to the mutually exclusive cases (1) (2) (3), as originally stated.

Corollary (All right angles are congruent).

Proof. The proposition 3.28 about supplements of acute and obtuse angles yields aneasy proof that all right angles are congruent:

Let R be the right angle as has been constructed in Construction 3.1. Now supposeγ = R′ is another right angle. By definition of a right angle, this means that γ ∼= S(γ).Hence case (b) above occurs, which implies (ii): γ ∼= R. You see that existence of aright angle make proving its uniqueness a bit easier.

Proposition 3.29 (Converse Isosceles Triangle Proposition). [Euclid I.6, Theo-rem 24 of Hilbert] A triangle with two congruent angles is isosceles.

Question. Formulate the theorem with specific quantities from a triangle 4ABC.

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Answer. If α ∼= β, then a ∼= b.

Proof. Given is a triangle 4ABC with α ∼= β. By Hilbert’s Theorem 19 (see Proposi-tion 3.19 above) and Proposition 3.20, congruence of angles is an equivalence relation.Hence α ∼= β implies β ∼= α. In the preliminary version given as Proposition 3.13, wehave shown that α ∼= β and β ∼= α together imply a ∼= b. Hence the 4ABC is isosceles,as to be shown.

Proposition 3.30 (Existence of an Isosceles Triangle). For any given segmentAB, on a given side of line AB, there exists an isosceles triangle.

Of course, this triangle is not unique.

Construction 3.2 (Construction of an isosceles triangle). Given is a segmentAB. Choose any point P not on line AB, in the half plane specified. Compare the twoangles ∠BAP and ∠ABP . In the drawing, ∠BAP is the smaller angle. Transfer the

smaller angle, with ray−→BA as one side, and the newly produced ray rB in the same half

plane as P . Ray rB and segment AP do intersect. The intersection point C lies in thehalf plane as required, and 4ABC is isosceles.

Figure 3.32: Construction of an isosceles triangle

Proof of validity. By Proposition 3.23, all angles are comparable. Comparison of thetwo angles ∠BAP and ∠ABP leads to one of the three possibilities:

Either (i) The two angles are congruent

or (ii) α = ∠BAP < ∠ABP = β

or (iii) α = ∠BAP > ∠ABP = β.

It is enough to consider cases (i) and (ii). In case (i) let C = P . In case (ii) we have

transferred the smaller angle ∠BAP , onto ray−→BA, and the newly produced ray rB in

the same half plane as P . By the Crossbar Theorem, a segment going from one side

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of an angle to the other, and a ray in the interior of an angle always intersect. By theCrossbar Theorem, ray rB and segment AP intersect. The intersection point C liesbetween A and P , hence both points C and P lie in the same half plane of line AB.Hence point C lies in the half plane as required.

By construction, the base angles of triangle4ABC are congruent. Hence the triangleis isosceles by Euclid I.6 (Converse Isosceles Triangles).

Remark. One may suggest to make it part of the construction how to compare the twoangles ∠BAP and ∠ABP . One possibility is to transfer both angles to the other vertex,B or A, respectively. Only the newly produced ray from the smaller angle does intersectthe opposite side of 4ABP .

10 Problem 3.2 (Erect a Perpendicular). Given is a line l and a point R onthis line. We have to erect the perpendicular at point R onto line l.

Construction 3.3. Choose any point A 6= R on line l. Transfer segment AR onto

the ray opposite to−→RA to get a segment RB ∼= AR. As explained in Construction 3.2,

construct any isosceles triangle 4ABC. The line CR is the perpendicular to the givenline l through point R.

Figure 3.33: Erect a perpendicular

Reason for validity. The two triangles 4RAC and 4RBC are congruent by SAS con-gruence. Indeed, ∠RAC ∼= ∠RBC by the construction of the isosceles triangle. Fur-thermore, we have a pair of congruent adjacent sides: AR ∼= BR by the constructionabove, and AC ∼= BC because 4ABC is isosceles.

Now the triangle congruence 4RAC ∼= 4RBC implies that ∠ARC ∼= ∠BRC. Butthese two angles are supplementary angles. Because congruent supplementary anglesare right angles, we have confirmed that ∠ARC is a right angle, as to be shown.

Definition 3.9 (The perpendicular bisector). The line through the midpoint of asegment and perpendicular to the segment is called the perpendicular bisector.

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10 Problem 3.3. Given is any segment AB. Construct the perpendicular bisec-tor.

Construction 3.4. Construct two different isosceles triangles over segment AB. Theline connecting the third vertices of the two isosceles triangles is the perpendicular bi-sector.

Figure 3.34: The perpendicular bisector via the kite

Proof of validity. Let 4ABC be the first isosceles triangle, and 4ABD be the secondone. We want to proceed as in Hilbert’s theorem 17 (see Proposition 3.17 above).

Question. Why do the points A and B lie on different sides of CD? We need just touse the Lemma 3.2 to SSS-congruence!

Answer. If points A and B would lie on the same side of line CD, the congruence4ACD ∼= 4BCD would imply A = B, contradicting to the endpoints of a segmentbeing different.

Because A and B lie on different sides of line CD, the segment AB intersects theline CD, say at point M . Now we have the symmetric kite ACBD with two congruenttriangles (left and right in the figure): 4ACD ∼= 4BCD.

Question. For convenience, please repeat how this follows just as in Theorem 17 (seeProposition 3.17 above).

Answer. Addition or subtraction of the base angles of the two isosceles triangles 4ABCand 4ABD yields the congruent angles ∠CAD ∼= ∠CBD—either using sums or dif-ferences of congruent base angles. The triangles 4ACD and 4BCD have congruentangles at A and B, and two pairs of congruent adjacent sides AC ∼= BC and AD ∼= BD.Now SAS congruence implies 4ACD ∼= 4BCD.

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Of the three points M, C, D exactly one lies between the two others. We can assumethat C does not lie between D and M , this assumption can be achieved by possiblyinterchanging the names C and D. Next we show that 4ACM ∼= 4BCM . Indeed,

Figure 3.35: Get the perpendicular bisector via: (a) a convex kite, (b) a nonconvex kite

the triangles 4ACM and 4BCM have congruent angles at the common vertex C, andtwo pairs of congruent adjacent sides AC ∼= BC and MC ∼= MC. Now we get by SAScongruence that 4ACM ∼= 4BCM . Matching pieces are stressed in the drawing.

From the last triangle congruence we get ∠AMC ∼= ∠BMC. Because M lies betweenA and B, these are two congruent supplementary angles. Hence they are right angles.Too, the triangle congruence implies AM ∼= BM . Hence M is the midpoint of segmentAB. We have shown that the segment AB and the line CD intersect perpendicularlyat the midpoint of segment AB. Hence CD is the perpendicular bisector.

Remark. There are the possibilities of an (a) convex kite, or (b) a non convex kite.

(a): If the two isosceles 4ABC and 4ABD lie on different sides of AB, points C andD are on different sides of AB, and it is clear that the segment CD intersects theline AB.

(b): Too, it is possible to use two different isosceles triangles on the same side of AB.To get the second isosceles triangle 4ABD, one transfers as base angles anytwo congruent angles which are less than the base angles of the first triangle4ABC. The figure ACBD is still a symmetric—but non convex—kite. As provedin Hilbert’s Theorem 17 (see Proposition 3.17 above), the two triangles left andright in the figure are still congruent: 4ACD ∼= 4BCD. The line CD stillintersects the segment AB, because points A and B are on different sides of CD,but the two segments AB and CD do not intersect each other.

Construction 3.5 (Solution— the way one really wants it). Construct any isoscelestriangle over the given segment. Drop the perpendicular from its third vertex.

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Figure 3.36: Construction of the perpendicular bisector—the natural way

Independent proof of validity. Let 4ABC be the isosceles triangle constructed at thefirst step. In the next construction step, the base angle ∠BAC is reproduced along the

ray−→AB to the other side of line AB, opposite to point C. Finally, one transfers segment

AC onto the newly produced ray, and gets the congruent segment AD ∼= AC.

Figure 3.37: (up-down) 4ABC congruent 4ABD (left-right) 4ACD congruent 4BCD(upleft-upright) 4ACM congruent 4MCB

Question. Show the congruence

(up-down) 4ABC ∼= 4ABD

Answer. This follows by SAS congruence. Indeed, the two triangles have the commonside AB, the two sides AC ∼= AD are congruent, and the angles ∠BAC and ∠BAD arecongruent, both by construction.

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Hence the construction has produced two isosceles triangles with base AB, whichare congruent to each other. Next we show that the triangles

(left-right) 4ACD ∼= 4BCD

are congruent.

Question. Explain how this congruence is shown.

Answer. This follows by SAS congruence. All four sides AD ∼= AC ∼= BC ∼= BD arecongruent. Furthermore, the two triangles have congruent angles ∠CAD ∼= ∠CBD, asfollows from the congruences ∠CAB ∼= ∠DAB, and ∠ABC ∼= ∠ABD of base angles ofthe two isosceles triangles shown in (up-down)—and angle addition.

The line AB and the segment CD intersect, because C and D lie on different sidesof AB. I call the intersection point M . Finally we show that the triangles

(upleft-upright) 4ACM ∼= 4BCM

are congruent.

Question. Explain how this congruence is shown.

Answer. This follows by SAS congruence. Indeed, the two triangles have the commonside CM , the two sides AC ∼= AB are congruent, and the angles ∠ACM and ∠BCMare congruent because of congruence (left-right).

Hence the segments AM ∼= BM are congruent— line CD bisects the segment AB.

Question. Why does M lies between A and B.

Answer. We know that the segments AM ∼= BM are congruent—this is impossible fora point on the line AB outside the segment AB.

Too, the angles ∠AMC ∼= ∠BMC are congruent supplementary angles, and henceright angles. Thus we have confirmed that CD is the perpendicular bisector of segmentAB, as to be shown.

Another attempt to prove validity, it does not yet work! As already shown in the con-struction for dropping a perpendicular, the line CD is perpendicular to the given lineAB. Let M be the intersection point of these two lines.

We need still to show that M is the midpoint of segment AB. To this end, we provethat the triangles 4AMC ∼= 4BMC are congruent. This can be shown only usingSAA-congruence from Proposition 3.38 below!

By the converse isosceles triangle Proposition, we know that AC ∼= BC. Becauseof congruence of vertical angles, we know that ∠BMC ∼= ∠AMD, the latter angle wasalready shown to be a right angle. Hence, all put together, we get ∠BMC ∼= ∠AMD ∼=R ∼= ∠AMC as expected. Finally, we have the congruence base angles ∠BAC ∼= ∠ABCby construction.

Hence via SAA-congruence from Proposition 3.38 below, we look forward to es-tablishing the triangle congruence 4AMC ∼= 4BMC, and hence confirming AM ∼=MB.

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Question. Why does this proof not work, at this point of the development?

Answer.

3.7 The Exterior Angle Theorem and its Consequences

Proposition 3.31 (The Exterior Angle Theorem). [Euclid I.16. Theorem 22 inHilbert] The exterior angle of a triangle is greater than both nonadjacent interior angles.

Proof. For the given 4ABC, we can choose point D on the ray opposite to−→AB, such

that AD ∼= CB. We compare the two nonadjacent interior angles γ = ∠ACB andβ = ∠ABC to the exterior angle δ = ∠CAD. As a first step, we show

Lemma 3.3. The exterior angle δ = ∠CAD is not congruent to the interior angle∠ACB = γ.

Proof of Lemma‘3.3. Assume towards a contradiction that δ ∼= γ.The supplementary angle of δ is α = ∠CAB. The supplementary angle of γ is

∠ACE, with a point E on the ray opposite to−−→CB. Supplementary angles of congruent

angles are congruent (Theorem 14 in Hilbert, see Proposition 3.14 above). Hence

?1 ∠CAB ∼= ∠ACE

On the other hand, we use SAS-congruence for the two triangles 4ABC and 4A′B′C ′

Figure 3.38: The impossible situation of a congruent exterior angle

withA′ := C , B′ := D , C ′ := A

Those two triangles would be congruent by SAS congruence. Indeed the angles atC and C ′ are congruent by our assumption γ = ∠ACB ∼= ∠CAD = δ. Too, theadjacent sides are pairwise congruent because of CB ∼= AD = C ′B′ by construction,

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and CA ∼= AC = C ′A′. From Axiom III.5, we conclude that the corresponding anglesat A and A′ = C are congruent:

?2 ∠CAB ∼= ∠C ′A′B′ = ∠ACD

We have shown that angle ∠CAB is congruent to both (?1) ∠ACE and (?2) ∠ACD.Now we use the uniqueness of angle transfer. Transferring angle ∠CAB with one

side−→CA, into the half plane not containing B, yields as second side of the angle once

the ray−−→CE and the second time

−−→CD. Hence these two rays are the same.

Hence points E and D both lie on the line la := BC. In other words, the four pointsB, C, D,E all lie on one line. Because B and D lie on opposite rays with vertex A,these two points are different: B 6= D. Note that just this simple remark is not true inelliptic geometry!

Because a line is uniquely specified by two of its points, this implies la = BC = BD.We have just seen that point C lies on this line. By construction, point A lies on theline BD, too. Hence all the points A, B, C,D,E lie on the same line. This contradictsthe definition of a triangle. By definition of a triangle, its three vertices do not lie onone line. This contradiction confirms the original claim δ γ.

Remark. To get a reminiscence to point symmetry about the midpoint M of AC, we canchoose AB ∼= CE. With the choice AB ∼= CE, we get even E = D, but still D 6= B.

Remark. In spherical geometry, the figure constructed does exist. We do not get acontradiction, the conclusion is just that the two points B and D are antipodes, andthe vertices A and C lie on two different lines BAD and BCD through these twoantipodes. Too, the sum of the segments is congruent going both ways from B to D.

Figure 3.39: Two ways from B to D

The triangles 4ABC ∼= 4A′B′C ′ = 4CDA are congruent, by SAS congruence (givenin Proposition 3.10 above). Hence BC ∼= DA and CD ∼= AB, and hence the sumsBC + CD ∼= DA + AB are congruent.

The next Lemma rules out the possibility δ < γ.

Lemma 3.4. The exterior angle δ = ∠CAD is not smaller that the interior angle∠ACB = γ.

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Figure 3.40: The impossible situation of a smaller exterior angle

Proof of Lemma 3.4. Suppose towards a contradiction that δ < γ. We transfer the

exterior angle δ = ∠CAD with one side−→CA, into the half plane containing B. The

second side of the angle we get is a ray−−→CB′ inside the angle ∠ACB. By the Crossbar

Theorem, it meets the segment AB in a point B′.We can now apply the first Lemma 3.3 to 4AB′C. This smaller triangle would have

exterior angle δ congruent to the interior angle γ′ = ∠ACB′. This is impossible byLemma 3.3.

Thus we have both ruled out the possibility that δ = γ, or that δ < γ. By Propo-sition 3.23, all angles are comparable. Hence there remains only the possibility thatδ > γ.

Lemma 3.5. The exterior angle δ = ∠CAD is greater than the interior angleβ = ∠ABC.

Proof of Lemma 3.5. We compare angle β = ∠ABC to the exterior angle δ = ∠CAD.

Choose any point F on the ray opposite to−→AC. The vertical angles δ = ∠CAD ∼=

Figure 3.41: Comparing the other nonadjacent angle

∠FAB = δ′ are congruent by Euclid I.15. Now we are back to the case already covered,

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because the interior angle β = ∠ABC and exterior angle ∠FAB lie on opposite sidesof triangle side AB, which is also part of one side of each of these angles. By Lemma3.3and 3.4, we conclude ∠FAB > ∠ABC and hence δ > β.

Remark. Here is an other way to prove Lemma 3.5: Define 4A3B3C3 by setting A3 :=A, B3 := C, C3 := B. Now we can apply Lemma3.3 and 3.4 to this new triangle andget δ′ = δ3 > γ3 = β. By congruence of the vertical angles δ = ∠CAD ∼= ∠FAB = δ′

we get δ > β.

Thus the proof of the exterior angle theorem is finished.

Proposition 3.32 (Immediate consequences of the exterior angle theorem).

(i) Every triangle can have at most one right or obtuse angle.

(ii) The base angles of an isosceles triangle are acute.

(iii) The foot point of a perpendicular is unique.

(iv) Given a line l and a point O not on l. At most two points of l have the samedistance from O. Hence a circle an a line can intersect in at most two points.

Proof. (i): Suppose angle α of 4ABC is right or obtuse: α ≥ R. By Proposition 3.28,its supplement is right or acute: S(α) ≤ R. But the supplement is an exteriorangle: δ = S(α). By the exterior angle theorem, the two other nonadjacentinterior angles of the triangle are less than that exterior angle. Hence they areacute: β, γ < δ = S(α) ≤ R, and hence β, γ < R.

(ii): The two base angles being congruent, by (i), they cannot be both right or obtuse.

(iii): Given a line l and a point O not on l. Suppose there are two perpendiculars, withfoot points F and G. The 4OFG would have two right angles, contradicting (i).

(iv): Given a line l and a point O not on l. Suppose there are three points A, B, Con the line l such that OA ∼= OB ∼= OC. Of any three points on a line, one liesbetween the two others (Theorem 4 in Hilbert’s foundations). We can supposethat A ∗ B ∗ C. Let the isosceles 4OAB and 4OBC have base angles α and γ,respectively. Because these are the base angles of the isosceles 4OAC, too, theyare congruent. Congruent base angles α ∼= γ occur as supplements at the middlevertex B. This would imply the base angles are right angles, ruled out in (ii).

Definition 3.10 (Alternate interior angles or z-angles). Let a transversal t inter-sect two lines a and b at the points A and B. A pair of alternate interior angles or simplyz-angles are two angles with vertices A and B, lying on different sides of the transversal.They have as one pair of sides lies on the transversal and contains the segment AB, theremaining two sides are rays lying on a and b.

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Figure 3.42: No three points have the same distance from a line

Proposition 3.33 (Congruent z-angles imply parallels). [Euclid I.27] If two linesform congruent z-angles with a transversal, they are parallel.

Proof. This is an immediate consequence of the exterior angle theorem. We argue bycontradiction. Suppose the two lines would intersect in point C. One of the z-anglesis an exterior angle of triangle 4ABC, the other one is a nonadjacent interior angle.Lemma 3.3 does imply that the two z-angles are not congruent— contradicting theassumption from above. Hence the two lines a and b cannot intersect.

Warning. The converse statement is not true is neutral geometry: Indeed, in hyperbolicgeometry, parallels can form non-congruent z-angles with a transversal.

10 Problem 3.4 (Addendum the the extended ASA-Congruence Theo-rem). In the situation of the extended ASA-congruence theorem, what happens in casethat A′B′ < AB or A′B′ > AB?

Question. Using the exterior angle theorem and Pasch’s axiom, get some results for thefirst case.

Answer. In the case A′B′ < AB, one can conclude that the two rays r′A and r′B still dointersect.

Detailed proof. I show that the two rays do intersect. One transfers segment AB to the

ray−−→A′B′ and gets a segment A′B2

∼= AB with A′ ∗B′ ∗B2.We apply the extended ASA theorem to4ABC and segment A′B2, and get a triangle

4A′B2C2∼= 4ABC.

Now apply Pasch’s axiom to triangle 4AB2C2 and line l on the ray newly producedr′B. This line intersects the triangle side A′B2 in B′, by construction. Hence Pasch’saxiom tells that line l intersects a second side of 4AB2C2, too, or goes through pointC2.

Line l does not go through point C2. Otherwise, one would get a contradiction tothe exterior angle theorem.

Question. For which triangle do you get a this contradiction?

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Answer. 4B′B2C would have both the interior angle β at vertex B2, and the exteriorangle β at vertex B′.

Line l does not intersect line B2C2. Again, one would get a contradiction to theexterior angle theorem. Alternatively, one can say that the two lines l and B′C ′ areparallel, because they form congruent z-angles with line A′B′.

Hence Pasch’s axiom implies that line l intersects the third side of 4AB2C2, whichis segment A′C2. We call the intersection point C ′. Thus we have produced a 4A′B′C ′,the angles of which at vertices A′ and B′ are congruent to the corresponding angles of4ABC.

Figure 3.43: The extended ASA congruence once more

Remark. In Euclidean geometry, the two rays r′A and r′B always intersect, no matterwhether A′B′ < AB, A′B′ ∼= AB, or A′B′ > AB. In all three cases, the 4A′B′C ′ issimilar to 4ABC.

Remark. Here is what happens in hyperbolic geometry: In hyperbolic geometry, similartriangles are always congruent, this implies that 4A′B′C ′ and 4ABC are not similarif either A′B′ < AB or A′B′ > AB!

In the case just considered, A′B′ < AB and γ′ > γ. In the opposite case thatA′B′ > AB, the second triangle 4A′B′C ′ has either a different angle γ′ < γ, or doesnot exist at all. The last possibilities occurs because the two rays r′A and r′B do notintersect at all. This happens always, once the segment A′B′ is long enough.

Proposition 3.34 (Comparison of Sides implies comparison of Angles). [EuclidI.18, Theorem 23 of Hilbert] In any triangle, across the longer side lies the greater angle.

Proof. In 4ABC, we assume for sides AB and BC that c = AB > BC = a. The issueis to compare the angles α = ∠CAB and γ = ∠ACB across these two sides.

We transfer the shorter side BC at the common vertex B onto the longer side. Thusone gets a segment BD ∼= BC, with point D between B and A. Because the 4BCD isisosceles, it has two congruent base angles

δ = ∠CDB ∼= ∠DCB

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Because B ∗D ∗ A, we get by angle comparison at vertex C

δ = ∠DCB < γ = ∠ACB

Now we use the exterior angle theorem for 4ACD. Hence

α = ∠CAB < δ = ∠CDB

By transitivity, these three equations together imply that α < γ. Hence the angle αacross the smaller side CB is smaller than the angle α lying across the greater side AB.In short, we have shown that c > a⇒ γ > α.

Figure 3.44: Across the longer side lies the greater angle

Proposition 3.35 (Comparison of Angles implies Comparison of Sides). [EuclidI.19] In any triangle, across the greater angle lies the longer side.

Proof. In 4ABC, we assume for two angles that α = ∠CAB < γ = ∠ACB. The issueis to compare the two sides BC and AB lying across the angles and show a = BC <AB = c. As shown in Proposition 3.6, any two segments are comparable.

Question. Please repeat the reasoning for convenience.

Answer. We transfer segment BC along the ray−→BA, and get a segment BD ∼= BC. By

Theorem 4 in Hilbert, of the three points A, B, D on a line AB, exactly one lies betweenthe two others. This leads to the following three cases: Either

(i) A ∗D ∗B and a < c. or (ii) A = D and a = c. or (i) D ∗ A ∗B and a > c.

We can now rule out cases (ii) and (iii).

In case (ii), the 4ABC is isosceles, and Euclid I.5 would imply α = γ, contrary tothe hypothesis about these two angles.

In case (iii), Euclid I.18 or Theorem 23 of Hilbert, would imply α > γ, contrary tothe hypothesis.

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Hence only case (i) is left, and a < c, as to be shown. In short, we have shown thatγ > α⇒ c > a.

Question. Explain how Euclid I.18 and Euclid I.19 are logically related.

Answer. Euclid I.18 in shorthand: c > a⇒ γ > α.

Euclid I.19 in shorthand: γ > α⇒ c > a.

Euclid I.19 is the converse of Euclid I.18.

Question. Does Euclid I.19 follow from Euclid I.18 by pure logic? Why not?

Answer. No, the converse does not follow purely by logic.

Question. Which fact does the proof above work nevertheless?

Answer. Because any two segments are comparable, we get the converse, nevertheless.

Corollary. A triangle with two congruent angles is isosceles.

Proof. Assume α ∼= γ for the given triangle. Since c > a⇒ γ > α and c < a⇒ γ < α,and any two segments are comparable, only a ∼= c is possible.

Proposition 3.36 (The foot point has the shortest distance). Given is a lineand a point O not on the line. The foot point F of a perpendicular from O to the lineis the unique among all points of the line, which has the shortest distance.

Figure 3.45: The foot point has the shortest distance

Proof. Call the given line l and let O be the given point not on l. Take any point A 6= Fon the line l. The 4AFO has a right angle at vertex F . Hence, by Proposition 3.32(i),its other two angles are acute. As shown in Proposition 3.35 (Euclid I.19), comparisonof angles of a triangle implies comparison of its sides. Hence ∠FAO < ∠AFO impliesOF < AO.

Because all other point on l except the foot point have strictly greater distance fromO, we can once more conclude that the foot point of the perpendicular is unique.

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Figure 3.46: The triangle inequality

Proposition 3.37 (The Triangle Inequality). [Euclid I.20] In any triangle, the sumof two sides is greater than the third side.

Proof. In 4ABC, we compare the sum CA + AB to the third side BC. To get the(congruence equivalence class) for the length of the sum CA+AB, we transfer segment

CA to the ray opposite to−→AB. We get segment DA ∼= AC and an isosceles 4DAC. Its

two congruent base angles are

δ = ∠CDA ∼= ∠DCA

Because point A lies between D and B, angle comparison at vertex C yields

δ ∼= ∠DCA < ∠DCB = η

Now we use Euclid I.19 for the larger 4DCB. Hence the side BC, across the smallerangle δ is smaller than the side DB lying across the greater angle η: DB > BC. Forthe original 4ABC, this shows that indeed CA + AB > CB.

Proposition 3.38 (SAA-Congruence Theorem). [Theorem 25 in Hilbert] Assumetwo triangles have a pair of congruent sides, one pair of congruent angles across thesesides, and a second pair of congruent angles adjacent to these sides. Then the twotriangles are congruent.

Proof. Let the triangles be4ABC and4A′B′C ′, and assume that AB ∼= A′B′, ∠BAC ∼=∠B′A′C ′ and ∠ACB ∼= ∠A′C ′B′. We choose a point C ′′ on the ray

−−→A′C ′ such that

AC ∼= A′C ′′. By axiom III.5

γ = ∠ACB ∼= ∠A′C ′′B′

Indeed, the SAS congruence even implies

(*) 4ABC ∼= 4A′B′C ′′

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Figure 3.47: SAA congruence

On the other hand, by assumption

γ = ∠ACB ∼= ∠A′C ′B′

If C ′ 6= C ′′, an exterior angle of 4C ′C ′′B′ would be congruent to a nonadjacent interiorangle, which is impossible by the exterior angle theorem. Hence we can conclude thatC ′ = C ′′. Because of (*), this implies 4ABC ∼= 4A′B′C ′, as to be shown.

Question. How does this theorem differ from the ASA congruence (Hilbert’s Theorem13, and Proposition 3.12 above)?

Answer. Of the two pairs of angles that are given (or compared), one pair lies acrossthe pair of given sides.

Figure 3.48: The hypothenuse leg theorem

Second proof of the hypothenuse-leg theorem 3.27. Take two right4ABC and4A′B′C ′

with b = b′, c = c′ and γ = γ′ = R. One transfers segment CB and segment C ′B′, both

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on the ray opposite to−−→C ′B′, and get new segments C ′D0

∼= CB as well as C ′E ′ ∼= C ′B′.From the construction and SAS congruence, we conclude

(1) 4ABC ∼= 4A′D0C , 4A′B′C ′ ∼= 4A′E ′C ′

Hence especially

(2) A′B′ ∼= A′D0∼= A′E ′

By Proposition 3.32(iv), at most two points can have the same distance from a line.Hence not all three points B′, D0, E

′ can be different. The only possibility left is D0 = B′,because they both lie on the opposite side of A′C ′ than B′.

Question. For completeness, explain once more. Assume that D0 6= E ′ towards a con-tradiction. Take the case that B′ ∗D0 ∗E ′ shown in the drawing. (The other cases canbe dealt with similarly.)

Independent answer. There are two isosceles 4B′A′D0 and 4DA′E ′.

Question. What would happen with their base angles at vertex D0?

Answer. The two base angles would be supplementary.

Question. Why is this impossible?

Answer. This is impossible, because the exterior angle theorem would imply that eachone of them is larger than the other one.

This contradiction leave only the possibility that D0 = E ′.

Now the required congruence 4ABC ∼= 4A′B′C ′ follows from D0 = E ′ and (1).

Definition 3.11 (The angular bisector). The ray in the interior of an angle whichbisects the angle into two congruent angles is called the angular bisector.

Proposition 3.39 (Uniqueness). The angular bisector is unique.

Proof. Suppose both rays−−→BG and

−−→BG′ bisect angle ∠BAC. The angles α := ∠ABG and

α′ := ∠ABG′ are comparable, because all angles are comparable by Proposition 3.23.Suppose towards a contradiction that α < α′. As specified in Proposition 3.26, one

can add inequalities of angles. Hence we conclude that

(?) ∠ABC = ∠ABG + ∠GBC ∼= α + α < α′ + α′ ∼= ∠ABG′ + ∠G′BC ∼= ∠ABC

which is impossible. Similarly, the case α > α′ can be ruled out. Hence α ∼= α′, and

hence by uniqueness of angle transfer−−→BG =

−−→BG′, as to be shown.

10 Problem 3.5. Given is any angle ∠BAC. Construct the angular bisector.

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Figure 3.49: The angular bisector

Construction 3.6. We choose the segments on its sides to be congruent, thus assuming

AB ∼= AC. Draw the line BC, and transfer the base angle ∠ABC to the ray−−→BC, on

the side of line BC opposite to vertex A. On the new ray, we transfer segment AB to

get the new segment BD ∼= BA. The ray−−→AD is the bisector of the given ∠BAC.

Question. Reformulate the description of this construction precisely, and as short aspossible.

Answer. One transfers two congruent segments AB and AC onto the two sides of theangle, both starting from the vertex A of the angle. The perpendicular, dropped fromthe vertex A onto the segment BC, is the angular bisector.

Proof of validity. At first note that point D does not lie on the ray−→AB, as can be ruled

out by means of the exterior angle theorem. Similarly, one confirms that point D does

not lie on the ray−→AC. Because, by construction, points A and D lie on different sides

of line BC, segment AD intersects line BC, say at point M .Now, in step (1)(2), we get three congruent triangles.

Step (1): We confirm that 4AMB ∼= 4DMB. The matching pieces used for the proofare stressed.

Figure 3.50: The first pair of congruent triangles

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Answer. Indeed, by construction, ∠ABC ∼= ∠DBC. Hence ∠ABM ∼= ∠DBM . (It

does not matter whether M lies on the ray−−→BC or the opposite ray.) Too, we have a

pair of congruent adjacent sides: Indeed BD ∼= BA by construction, and BM ∼= BM .Now SAS congruence implies 4AMB ∼= 4DMB.

Question. Explain really carefully why ∠AMB is a right angle.

Answer. Because of 4AMB ∼= 4DMB, we get ∠AMB ∼= ∠DMB. Because point Mlies between A and D, these are two supplementary angles. Hence they are right angles.

Step (2): Finally, we confirm that 4AMB ∼= 4AMC. Again the pieces needed tomatch for this theorem are stressed in the drawing.

Figure 3.51: The second pair of congruent triangles

Answer. Indeed, I use the hypothenuse-leg theorem. ∠AMB ∼= ∠AMC ∼= R, becausea right angle is congruent to its supplement. (Again, it does not matter whether M lies

on the ray−−→BC or the opposite ray.) Too, we have a pair of congruent sides: Indeed

AB ∼= AC by construction, and AM ∼= AM .

From the triangle congruence 4AMB ∼= 4AMC, we get ∠MAB ∼= ∠MAC, andMB ∼= MC. Since point M lies on the line BC, the last congruence shows that M

lies between B and C, too. Hence ray−−→AM =

−−→AD lies inside the given ∠BAC. Too,

∠DAB = ∠MAB ∼= ∠MAB = ∠DAC confirms that the given angle is bisected.

Proposition 3.40 (Existence of the angular bisector). For any angle ∠BAC,

there exists a ray−−→AD inside the given angle such that ∠DAB ∼= ∠DAC.

Proposition 3.41 (The Hinge Theorem). [Euclid I.24] Increasing the angle betweentwo constant sides increases the opposite side of a triangle.

Corollary 3.7 states the equivalence we get by taking from [Euclid I.24] and [EuclidI.25] together:

Corollary. Given are two triangles with two pairs of congruent sides. In the first tri-angle, the angle between them is smaller, congruent, or greater than in the second oneif and only if the opposite side is smaller, congruent, or greater in the first triangle.

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Figure 3.52: The Hinge Theorem

Proof. Given are 4ABC and 4A′B′C ′ with a = a′ and c = c′. Assuming β < β′, wehave to check whether b < b′. One can assume that A = A′, B = B′, and put the twopoints C and C ′ lie on the same side of line AB. By the hypothesis β < β′, the twopoints A and C ′ lie on different sides of line BC. Hence this line intersects the segmentAC ′.

Indeed, as stated by the Crossbar theorem, the ray−−→BC intersects the segment AC ′.

Let D be the intersection point. We now distinguish three cases:

(a) B ∗D ∗ C. This case occurs if the triangle side BC to be rotated is long enough.

Figure 3.53: The Hinge Theorem: turning a long side

(b) C = D.

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Figure 3.54: The Hinge Theorem: the borderline case

(c) B ∗ C ∗D. In this case the triangle side BC to be rotated is rather short.

Figure 3.55: The Hinge Theorem: turning a short side

In the border line case (b), we see directly that β < β′ implies A ∗ C ∗ C ′ and henceAC < AC ′, as to be shown. In the two other cases, we apply Euclid I.19 to triangle4ACC ′. Thus it is enough to show that this triangle has a larger angle ε = ∠ACC ′ atvertex C than the angle ε′ = ∠AC ′C at vertex C ′. The 4BCC ′ is isosceles. Hence, byEuclid I.5, it has two congruent base angles, which we denote by ϕ.

In case (a), we proceed as follows: Because B and C lie on opposite sides of AC ′,comparison of angles at vertex C ′ yields

(1) ε′ < ϕ

And because A and C ′ lie on opposite sides of line BC, comparison of angles at vertexC yields

(2) ϕ < ε

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Because of transitivity, (1)(2) together imply ε′ < ε and hence AC ′ > AC.

In case (c), we show that ε is an obtuse, and ε′ is an acute angle. Because A and C ′ lieon opposite sides of line BCD, we get

(3) ∠C ′CD < ∠C ′CA = ε

but ∠C ′CD is an exterior angle of the isosceles 4BCC ′. Because the base angles ofan isosceles triangle are acute, its supplement ∠C ′CD is obtuse. Hence by (3), ε isobtuse, too. Since the triangle 4ACC ′ can have at most one right or obtuse angle, theangle ε′ is acute. Now ε′ < R < ε implies again ε′ < ε, and hence Euclid I.19 yieldsAC ′ > AC.

Proposition 3.42 (The Midpoint of a Segment). [Theorem 26 in Hilbert] Anysegment has a midpoint.

Construction 3.7. Given is a segment AB. Transfer congruent angles with the end-points of the given segment AB as vertices, on different sides of AB. Next we transfercongruent segments AC ∼= BD onto the newly produced legs of these two angles. Finally,

lines←→AB and

←→CD intersect at the midpoint M .

Figure 3.56: Hilbert’s construction of the midpoint

(a) Here is a drawing.

(b) Explain why lines AB and CD intersect.

Answer. By construction, points C and D lie on different sides of line AB. Hence,by the plane separation theorem, line AB and segment CD intersect.

Let M be the intersection point. From the plane separation theorem, too, it followsthat the intersection point M lies between C and D. But it turns out to be harderto see why M lies between A and B!

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Figure 3.57: M = A is impossible

(c) Show that M = A is impossible.

Answer. In that case line l = AC would go through point D. The ray−→AC would

be an extension of side AD of the triangle 4ABD. This triangle would have theinterior angle ∠ABD congruent to the exterior angle ∠CAB, contradicting theexterior angle theorem. (Remember: an exterior angle is always greater than anonadjacent interior angle.)

(d) Show that M ∗ A ∗B is impossible.

Figure 3.58: Point A lying between M and B is impossible

Answer 1. We use Pasch’s axiom for 4MBD and line l = CA. Which conclusiondo you get?

Answer. The line CA enters 4MBD on the side MB. By Pasch’s axiom, thisline either (i) intersects side DB, or (ii) goes through point D, or (iii) intersectsside MD.

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In all three cases, we derive a contradiction:

Case (i): suppose line l intersects segment DB, say at point G. The 4ABGwould have the interior ∠ABG congruent to the exterior ∠CAB. This con-tradicts the exterior angle theorem, which tells an exterior angle is alwaysgreater than a nonadjacent interior angle.

Case (ii): suppose line l goes through point D. The 4ABD would have theinterior ∠ABD congruent to the exterior ∠CAB. This contradicts the ex-terior angle theorem, which tells an exterior angle is always greater than anonadjacent interior angle.

Case (iii): suppose line l intersects segment MD, say at point F . Points C 6= Fare different, because they lie on different sides of AB. The lines AC andMD intersect both in point C and in point F ,. Because this are two differentpoints, they determine a line uniquely. Hence all five points C, M, D,A, F lieon one line. Hence we are back to the case M = A, ruled out earlier.

Figure 3.59: Point A lying between M and B is impossible

Answer following Hilbert. We apply the exterior angle theorem twice. Here is asketchy drawing for that impossibility. 14 In triangle 4ABD, the interior angleat vertex B is β = ∠DBM , which is smaller than the exterior angle at vertexε = ∠BMC. In triangle 4AMC, the angle ε from above is interior angle atvertex M , and hence smaller than the exterior angle α = ∠BAC at vertex A.

Now transitivity yields β < ε < α. On the other hand, the angles ∠DBA = β andα = ∠CAB are congruent by construction. This contradiction rules out the caseM ∗ A ∗B.

14The drawing, too, occurs in the millenium edition of ”Grundlagen der Geometrie”, page 26.

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(e) Now we know that M lies between A and B, we finally can prove that M is themidpoint.

Question. Which congruence theorem is used for which triangles?

Answer. One uses SAA congruence for4AMC and4BMD. Indeed, the angles at

Figure 3.60: Apply the SAA congruence

A and B are congruent by construction, and the angles at vertex M are congruentvertical angles. (Both statements are only true because M lies between A and B!)The sides AC and BD opposite to those angles are congruent by construction.Hence the two triangles are congruent, and especially AM ∼= MB

Figure 3.61: The generic situation for Proposition 3.43, for which we prove: Two segmentsCX and DY on different sides of XY are congruent if and only if midpoint M of segment CDlies on line l. The third figure shows the case with both conditions true.

Proposition 3.43. The midpoint of a segment lies on a given line l if and only if thetwo endpoints of a segment have the same distance to the line, and lie on the oppositesides of it.

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Proof. Assume that the two endpoints C and D of the given segment lie on oppositesides of l. Furthermore, assume CX ∼= DY are the congruent segments to the footpoints X and Y .

Let Q be the intersection point of line l and segment CD, which exists by planeseparation. In the special case X = Y , we are ready immediately. Otherwise, we needto see why Q lies between X and Y ! This is done is the same way as in Hilbert’sconstruction 3.7 of the midpoint, which is now applied to the segment AB = XY .

Finally, one obtains the triangle congruence

(oneflier) 4CQX ∼= 4DQY

via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,and the congruent segments CX ∼= DY . Hence CQ ∼= DQ, confirming that Q = M isthe midpoint of segment CD.

Conversely, assume that the midpoint M of segment CD lies on the line l. Inthe special case X = Y we are ready immediately. Otherwise, we need to confirm, oncemore, why M lies between X and Y ! Again one needs to use the exterior angle theorem.Finally, one obtains the triangle congruence

(oneflier) 4CMX ∼= 4DMY

via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,and the congruent segments CX ∼= DY , as to be shown.

3.8 SSA Congruence

Next we study the possibilities and difficulties with SSA congruence. Thus the matchingpieces of the two given triangles are two pairs of sides, and one pair of angles oppositeto one of these sides. I follow Euler’s conventional notation: in triangle 4ABC, leta = BC, b = AC, and c = AB be the sides and α := ∠BAC, β := ∠ABC, andγ := ∠ACB be the angles.

10 Problem 3.6. Investigate an example Use straightedge and compass toconstruct in Euclidian geometry all triangles (up to congruence) with γ = ∠ACB = 30,side AC = 10 units and side AB given as below. How many non-congruent solutions(none, one, two?) do you get in each case? How many of them are acute, right orobtuse triangles? Measure and report the angle β = ∠ABC for all your solutions. Makeclear by the drawings what happens in all cases, especially how many acute, right, andobtuse triangles you get as solutions.

Hint: It is convenient to construct all triangles with one common side AC. Checkcarefully to find the obtuse angles! (a) AB = 4 units.

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Figure 3.62: A Euclidean example for SSA triangle construction

Answer (a). One has to begin the construction by drawing a segment AC of length 10,

and a ray r with vertex C, forming an angle of 30 with←−CA. The point B has to lie on

the ray r, as well as on a circle of radius AB around A. In case (a), the circle does notintersect this ray, hence there is no solution.

(b) AB = 5 units.

Answer. In case (b), the circle just touches the ray r, hence there is one solution, witha right angle at B.

(c) AB = 5.5 units.

Answer. In case (c), the circle intersect the ray r in two points, hence there are two noncongruent solutions. At vertices B and B′, I measure the angles about β = 67 andβ′ = 113. One solution is an acute, the other an obtuse triangle.

(d) AB = 6 units.

Answer. Again in case (d), the circle intersects the ray r in two points, and there aretwo non congruent solutions. At vertices B and B′, I measure the angles about β = 57

and β′ = 123. Both solutions are obtuse triangles. Indeed, the first solution has theobtuse angle α = 180 − β − γ = 93.

Proposition 3.44 (SSA Matching Proposition). Given are two triangles. Assumethat two sides of the first triangle are pairwise congruent to two sides of the secondtriangle, and that the angles across to one of these pairs are congruent.

Then either the two triangles are congruent, or the angles across the other pair ofcongruent sides add up to two right angles.

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Proof. It is assumed that two sides and the angle across one of these sides of 4ABCcan be matched to congruent pieces of 4A′B′C ′:

(SSA) c = AB ∼= A′B′ = c′, b = AC ∼= A′C ′ = b′, γ = ∠ACB ∼= ∠A′C ′B′ = γ′

We have to show that either (a) or (b) holds:

(a) 4ABC ∼= 4A′B′C ′.

(b) The two angles across the second matched side β = ∠ABC and β′ = ∠A′B′C ′ are(congruent to) supplementary angles. One of them is acute and the other one isobtuse.

We begin by reducing the problem to the special case that A = A′, C = C ′ and thatB and B′ = D lie on the same side of AC. We use the SAS congruence to construct atriangle 4ADC, such that 4A′B′C ′ ∼= 4ADC, with points B and D on the same sideof AC. The drawing below shows that procedure.

Figure 3.63: Matching two triangles with SSA

Question. Explain how you get the triangle 4ADC congruent to 4A′B′C ′.

Answer. I just transfer segment C ′B′ onto the the ray−−→CB and get C ′B′ ∼= CD. The

congruence 4A′B′C ′ ∼= 4ADC, follows from SAS, given in Theorem 12 of Hilbert (seeProposition 3.10 above).

Assume that congruence (a) does not hold. We have to prove that case (b) occurs.We know that α 6= α′, since otherwise the SAS-congruence theorem implies 4ABC ∼=4A′B′C ′, which again would be case (a) just ruled out. Without loss of generality, wecan assume α′ < α. Since ∠DAC = α′ < α = ∠BAC, the point D lies between B andC, as shown in the drawing.

Since AD ∼= A′B′ ∼= AB by assumption, triangle 4ABD is isosceles, with baselineBD. By Euclid I.5, the two base angles of an isosceles triangle are congruent. One ofthem is the angle β = ∠ABD. By Proposition 3.32(ii), these base angles are alwaysacute.

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The angle β′ = ∠A′B′C ′ ∼= ∠ADC is the supplement to the second base angle∠ADB ∼= ∠ABD = β. Hence β and β′ are congruent to supplementary angles. Thefirst one is acute, the second one is obtuse, as claimed in (b).

Corollary (SSAA Congruence). Two triangles which have two pairs of congruentsides, and two pairs of congruent angles across to the latter, are congruent.

Proposition 3.45 (Restricted SSA Congruence, case of a unique solution). Asin the matching Proposition above, we assume that two sides and the angle across oneof these sides of 4ABC can be matched to congruent pieces of 4A′B′C ′:

(SSA) AB ∼= A′B′ , AC ∼= A′C ′ and ∠ACB ∼= ∠A′C ′B′

Under the additional assumption that either (i) or (ii) hold,

(i) The given angle γ lies across a side longer or equal to the other given side: c =AB ≥ AC = b.

(ii) The given angle γ = ∠ACB is right or obtuse.

congruence holds among all triangles matched by (SSA). In all these cases there donot exist any non congruent triangles with congruent sides b, c and angle γ. Actuallyassumption (ii) implies (i).

Proof of Proposition 3.45. Assume that (ii) holds. Indeed, by Proposition 3.32(i), atriangle can have at most one angle which is right or obtuse. Hence assumption (ii)implies γ ≥ β. By Euclid I.19, across the greater angle of any triangle lies the longerside. Hence γ ≥ β implies c ≥ b, which is assumption (i).

We now assume that c = AB ≥ AC = b holds, as stated by (i). By Euclid I.18,across the longer side of a triangle lies the greater angle. Hence c ≥ b implies γ ≥ β.A triangle cannot have two angles which are right or obtuse. Hence β and β′ are bothacute or right. Now congruence follows because case (b) in the matching Proposition isruled out.

Corollary (SSA Congruence for isosceles and right triangles). (i’) Two isosce-les triangles with congruent legs and congruent base angles are congruent.

(ii’) Any two right triangles with congruent hypothenuses and one pair of congruent legsare congruent.

Proof. Assumption (i’) is a special case of (i), and (ii’) a special case of (ii).

Proposition 3.46 (Restricted SSA Congruence Theorem, case with non unique-ness). Again we assume that two sides and the angle across one of these sides of 4ABCcan be matched to congruent pieces of 4A′B′C ′:

(SSA) AB ∼= A′B′ , AC ∼= A′C ′ and ∠ACB ∼= ∠A′C ′B′

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Complementary to Proposition 3.45, we assume that neither (i) nor (ii) nor β ∼= R holds.In other words, we assume that the given angle γ = ∠ACB is acute and lies across theshorter given side: c = AB < AC = b, and angle β is not a right angle. Under theseadditional assumptions (a) There exist two non congruent triangles matched by (SSA).

(b) Nevertheless, equivalent are

(1) The two given triangles 4ABC and 4A′B′C ′ are congruent.

(2) (SSAA) The two triangles can be matched in two sides and the two angles acrossboth given sides.

(3) For both triangles, the angles β and β′′ across the longer given side b are both acuteor both right or both obtuse.

(4) Both triangles are acute, or both are right, or both triangles are obtuse with the twoobtuse angles at corresponding vertices.

In this case just the two given triangles 4ABC and 4A′B′C ′—not all other triangleswith those given angle and sides γ, c, b—are congruent.

Figure 3.64: Establish two solutions for SSA

Proof for part (a). We drop the perpendicular from vertex A onto the opposite sideCB. The foot point F 6= B, because otherwise β would be a right angle. Transfer

segment FB on the ray opposite to−−→FB to produce segment FB′′ ∼= FB. The two

right triangles 4ABF and 4AB′′F are congruent by the Hypothenuse-Leg Theorem15. Hence AB ∼= AB′′. Thus we got two non congruent triangles 4ABC and 4AB′′Cwhich nevertheless satisfy

(SSA”) AB ∼= AB′′ , AC ∼= AC and ∠ACB ∼= ∠ACB′′

For these non congruent triangles, the two angles β and β′′ are supplementary. One ofthem is acute, the other one is obtuse.

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Proof for part (b). Clearly (1) implies (2) implies (3) implies (4). If (4) holds, thematching Proposition excludes non congruent solutions.

Indeed, if both triangles are acute, both angles β and β′ are acute. Congruencefollows by the SSA matching Proposition. If both triangles are right, both angles βand β′ are acute or right. In both cases, congruence follows by the SSA matchingProposition.

If both triangles are obtuse, and obtuse angle occurs at corresponding vertices, theneither both angles β and β′ are obtuse, or both angles α and α′ are obtuse. In the secondcase both angles both angles β and β′ are acute. In both cases, congruence follows bythe SSA matching Proposition.

Remark. Earlier on, we have carefully studied triangles with

c = AB = 5 , 5.5 . 6 b = AC = 10 and γ = ∠ACB = 30

in Euclidean geometry. In the first case c = 5, one gets a unique right triangle assolution. In the second case c = 5.5 and c = 6, there are two non-congruent solutions,one of which is an acute triangle, the other one is an obtuse triangle with β′ > R.

In the third case c = 6, there are two non-congruent solutions again. But both areobtuse triangles! One gets one triangle with α > R and β < R, as well as a second onewith α′ < R and β′ > R.

3.9 Reflection

Definition 3.12 (Reflection across a Line). Given is a line l. The reflection acrossline l maps an arbitrary point P to a point P ′ meeting the following requirements:

If P lies on the symmetry axis l, we put P ′ := P .If P does not lie on the symmetry axis, the point P ′ is specified by requiring

(i) P and P ′ lie on different sides of l.

(ii) the lines l and PP ′ are perpendicular, intersecting at F .

(iii) PF ∼= P ′F .

Proposition 3.47. Given are two points P, Q on the same side of l and their imagesP ′, Q′. We show that

(a) PQ ∼= P ′Q′

(b) If the lines m :=←→PQ and l intersect at M , then P ′, Q′ and M lie on a second line

m′.

Question. For this exercise, you can if you like, do drawing and proof, in neutral geom-etry, on your own.

102

Figure 3.65: Reflection by a line l

Proof of (a). Let F and G be the foot points of the perpendiculars dropped from P andQ onto the symmetry axis l. We use SAS-congruence for the two triangles 4QGF and4Q′GF .

Question. Indeed both triangles have a right angle at G—. Explain why.

Answer. ∠QGF is right by definition of a reflection. Since a right angle is congruent toits supplementary angle, the supplementary angle ∠Q′GF is a right angle, too.

—a common side GF and congruent sides GQ ∼= GQ′, by definition of a reflection.Hence SAS congruence implies

(1) 4QGF ∼= 4Q′GF

Because of congruence (1), we get

α = ∠QFG ∼= ∠Q′FG ,(2)

QF ∼= Q′F(3)

Since ∠GFP ∼= ∠GFP ′ is a right angle, angle subtraction yields

(4) β = ∠QFP ∼= ∠Q′FP ′

By construction

(5) PF ∼= P ′F

Using (3)(4) and (5), SAS-congruence implies that

(6) 4QFP ∼= 4Q′FP ′

Hence especially (a) holds.

103

Proof of (b). We use SAS-congruence for the two triangles 4QGM and 4Q′GM . In-deed both triangles have a right angle at G: ∠QGM is right by definition of a reflection.Since a right angle is congruent to its supplementary angle, the supplementary angle∠Q′GM is a right angle, too. Furthermore, they have a common side GM and congruentsides GQ ∼= GQ′, by definition of a reflection. Hence SAS congruence implies

4QGM ∼= 4Q′GM

For the angle x between the line l of reflection and the given line m, one gets

(8) x = ∠QMG ∼= ∠Q′MG

Now we argue similarly for the two triangles 4PFM and 4P ′FM . Thus we get

(9) y = ∠PMF ∼= ∠P ′MF = ∠P ′MG

Now by assumption, the three points P, Q and M lie on a line m.

(10) x = ∠QMG = ∠PMF = y

From (8)(9),(10) and transitivity of angle congruence, we get

(11) ∠Q′MG = ∠P ′MG

Now by Hilbert’s axiom, the angle transfer produces a unique ray. Hence−−→MQ′ =

−−→MP ′.

Thus the uniqueness of angle transfer implies that the three points M, P ′ and Q′ lie onone line.

An alternative for the proof of (b) . (6) above implies

(7) γ = ∠QPF ∼= ∠Q′P ′F

We apply the extended ASA congruence to triangle 4PFM and segment FP ′. Theangles to be transferred to the endpoints of that segment are a right angle at F and γat P ′. Transferring produces as second sides of these angles part of the line l and the

ray−−→P ′Q′. Hence those rays intersect, say at point M ′. By ASA-congruence

4MFP ∼= 4M ′FP ′

Hence FM ∼= FM ′. The points Q and Q′ and hence M and M ′ lie on the same side ofPP ′. Hence uniqueness of segment transfer implies M = M ′ as to be shown.

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4 Measurement and Continuity

Hilbert postulates two axioms of continuity: the axiom of Archimedes, and the axiomof completeness.

4.1 The Archimedean Axiom

Given are any segments AB and CD. We now use the second one as a measure unit.The Archimedean axiom states existence of natural number n and as finite sequence of

points A = A0, A1, A2, . . . , An on the ray−→AB such that

(4.1) CD ∼= Ak−1Ak for k = 1, 2, . . . n and A ∗ Ak−1 ∗ Ak for k = 2, 3 . . . n

and the point B either lies between An−1 and An, or B = An−1. In Hilbert’s words,

"n segments congruent to CD constructed contiguously from A, along

a ray from A through B, will pass beyond B."

As a shorthand notation, we shall write

(4.2) n · CD > AB

More colloquially, one can say that no segment is too long as to be measured in termsof a given unit segment. As a first consequence of this fact, we note that no segment is

Figure 4.1: The Archimedean axiom. For the case drawn in the figure, it turns out thatn = 7.

so short that it cannot be measured in terms of a given unit segment.

(Repeated bisection produces arbitrarily short segments). Given any segmentsPQ and EF , there exists a natural number s such that s successive bisections produce

a segment PsQs =PQ

2sshorter that segment EF , written as a formula

PQ

2s< EF

Especially, not both points E and F can lie inside segment PsQs or on its endpoints.

105

Proof. We apply the Archimedean axiom to the segments AB := EF and CD := PQ.Hence there exists a natural number n such that

(4.3) n · EF > PQ

Let s be an integer such that 2s ≥ n. We get

2s · EF > PQ

Bisection of all segments involved leads inductively to the inequalities

2s−1 · EF >PQ

2, 2s−2 · EF >

PQ

4, 2s−3 · EF >

PQ

8, . . . , EF >

PQ

2s

as to be shown.

The Archimedean axiom allows the measurement of segments and angles using realnumbers. These real numbers occur during the measurement process in the form ofbinary fractions. Since Hilbert, this axiom is also known as the axiom of measurement.To start the measuring process, one assigns the length one to some arbitrary—butconvenient—segment. We shall call it the unit segment.

Main Theorem (Measurement of segments). Given is a unit segment OI. Thereexists a unique way of assigning a length to any segment. We denote the length ofsegment AB by |AB|. 15 Too, this length is called the distance of points A and B. Thelength has the following properties:

(1) Positivity If A 6= B, then |AB| is a positive real number, and |OI| = 1.

(2) Congruence |AB| = |CD| if and only if AB ∼= CD.

(3) Ordering |AB| < |CD| if and only if AB < CD.

(4) Additivity The distances are additive for any three points on a line. With pointB lying between A and C, we get |AC| = |AB|+ |BC|.

Proof. We begin by assuming existence of the measurement function, and really con-struct it. In a second step, we confirm that the construction has produced the resultas claimed. Let M be the midpoint of the unit segment OI. Since OM ∼= MI, theadditivity of segment lengths implies |OM | + |MI| = 1, and hence |OM | = |MI| = 1

2.

Successive bisections now produce segments of lengths 14, 1

8, . . . . A length is assigned

to any given segment AB by the process indicated in the figure on page 105. Asin the Archimedean axiom, one begins by constructing a finite sequence of points

A0 = A, A1, A2, . . . , An on the ray−→AB such that

(4.4) OI ∼= Ak−1Ak and−−→AAk =

−→AB for k = 1, 2, . . . n

15In projective geometry, one uses a signed length, which I shall denote by AB.

106

Figure 4.2: This measurement yields |AB| = 1.1101 . . . as a binary fraction.

This part of the process stops, and the integer part of the length |AB| is determined, assoon as point B either lies between An−1 and An, or B = An−1. In the latter case, onegets the length |AB| = n − 1 exactly, and the measure process is finished. In general,the former case occurs. One concludes that n − 1 < |AB| < n, and needs to constructthe digits ds of an finite or infinite binary fraction

n− 1 . d1d2d3 . . .

in order to achieve a measurement with more and more precision. 16 To begin theprocess, we set the lower bound L0 := An−1, and the upper bound U0 := An. Themeasurement uses sequences for the lower bound, digit, and upper bound, which areconstructed inductively by repeated bisections.

Inductive measurement step to determine ds. For s = 1, 2, 3, . . . , let Hs−1 be midpointof segment Ls−1Us−1.

If point B lies between Ls−1 and Hs−1, one puts

new digit ds := 0

lower bound Ls := Ls−1

upper bound Us := Hs−1

If point B lies between Hs−1 and Us−1, or B = Hs, one puts

new digit ds := 1

lower bound Ls := Hs−1

upper bound Us := Us−1

16Of course, any real life measure has to stop such an infinite process, and hence cannot confirm anexact real number as result, but only a finite number digits, limited by the technical possibilities.

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Hence the digit ds depends on whether point B lies left or right of Hs−1. In both cases,the approximation to |AB| obtained by the s-th step is

|ALs| = n− 1 +d1

2+

d2

4+

d3

8+ · · ·+ ds

2s

It can happen that B = Hs−1, in which case this approximation is exact, and themeasure process is finished. In general, the measurement process does not stop, butproduces an infinite fraction. By Cantor’s principal of boxed intervals, the infinitebinary fraction is indeed a real number.

Question. Explain once more, how the first digit d1 is determined.

Answer. One puts the lower bound L0 := An−1 and the upper bound bound U0 := An,with An−1 and An obtained via the Archimedean axiom. Let H0 be the midpoint ofsegment L0U0.

If point B lies between L0 and H0, one puts

new digit d1 := 0

current approximation |AL1| = n− 1

lower bound L1 := L0

upper bound U1 := H0

If point B lies between H0 and U0, or B = H0, one puts

new digit d1 := 1

current approximation |AL1| = n− 1 + 12

lower bound L1 := H0

upper bound U1 := U0

Finally, one wants to be convinced that the congruence, ordering, and additivityproperties stated as (2),(3) and (4) are really satisfied. We start by confirming

(2a) If AB ∼= CD, then |AB| = |CD|.

(3a) If AB < CD then |AB| < |CD|. If AB > CD then |AB| > |CD|.

108

It is rather straightforward to deduct (2a) from Hilbert’s axiom III.3. Let any twosegments AB < AC be given. Repeated bisection produces arbitrarily short segments.Hence there exists a number s such that

(4.5) LsUs∼=

OI

2s< BC

Because we go through the measurement process explained above for segment AB, weknow that B = Ls or Ls ∗B ∗Us. By the segment comparison above, not both points Band C can lie in the interval LsUs, but indeed Ls ∗ Us ∗ C. Hence the binary fractionsfor points B and C differ in the sth place, and hence have less than s common places,furthermore |AB| 6= |AC|.

Assume that t is the first place where the measurement fractions of points B and Care different:

(4.6) b1 = c1, b2 = c2, . . . , bt−1 = ct−1, but bt 6= ct

It can happen by accident that t from equation (4.6) is much smaller than any numbers for which equation (4.5) holds. The four-point theorem yields the natural order Lt−1 ∗B ∗ C ∗ Ut−1, and its generalization to five or more points leads to the natural orderLt−1 ∗ B ∗Ht−1 ∗ C ∗ Ut−1, or C = Ht−1. Hence |AB| < |AC| follows by using the t-thapproximations. The second part of (3a) is quite similar to check.

Since all segments are comparable, items (2a) and (3a) implies the converse state-ments, and hence (2) and (3) are confirmed.

The additivity (4) can at first be checked by induction to hold for all segments ofinteger lengths. Next, one can prove additivity for segments of which lengths are finitebinary fractions, using induction on the number of digits occurring. I only explain thefurther case that the length of segment AB is given by a finite binary fraction, butsegment BC by an infinite fraction. For any arbitrary lower and upper bound finitefractions BL < BC < BU , we get

BL < BC < BU

|BL| < |BC| < |BU ||AB|+ |BL| < |AB|+ |BC| < |AB|+ |BU |

|AL| < |AB|+ |BC| < |AU |

and on the other hand

AB + BL < AB + BC < AB + BU

AL < AB + BC < AU

|AL| < |AB + BC| < |AU |

Both statements|AL| < |AB|+ |BC| < |AU ||AL| < |AB + BC| < |AU |

109

hold for all lower and upper bounds L and U . Hence the Archimedean axiom for thereal numbers implies |AB|+ |BC| = |AB + BC|.

Corollary. Given any three points A, B and C, the equality

|AB|+ |BC| = |AC|

holds if and only if the three points lie on a line and either two of them are equal or Blies between A and C.

The measurement of angles is done quite similar to the measurement of segments.Contrary to the situation for segments, there exists a well defined and convenient unitof measurement, which is the right angle. The traditional measurement in degrees isobtained by assigning the value 90 to the right angle.

Main Theorem (Measurement of angles). There exists a unique way of assigninga degree measurement to any angle. We denote the value of angle ∠ABC by (∠ABC).17

For additivity, we consider a ray−−→BG in the interior angle ∠ABC. Following def-

inition 3.5, the angle ∠ABC is the sum of the angles ∠ABG and ∠GBC, written as∠ABC = ∠ABG + ∠GBC.

The measurement has the following properties:

(1a) Positivity For any angle ∠ABC, the degree measurement is a positive numberand 0 < (∠ABC) < 180.

(1b) Right angle The measurement of a right angle is 90.

(2) Congruence (∠ABC) = (∠A′B′C ′) if and only if ∠ABC ∼= ∠A′B′C ′.

(3) Ordering (∠ABC) < (∠A′B′C ′) if and only if ∠ABC < ∠A′B′C ′.

(4) Additivity If ∠ABC = ∠ABG + ∠GBC, then (∠ABC) = (∠ABG) +(∠GBC).

Question. Why does bisecting any angle produce an acute angle?

Answer. It is shown in the figure on page 109, how to bisect an angle. Let A be thevertex of the angle. One transfers two congruent segments AB and AC onto the twosides of the angle, both starting from the vertex A. The perpendicular, dropped fromthe vertex A onto the segment BC, is the angular bisector.

Let F be the foot point of the perpendicular. As a consequence of the exterior angletheorem, we have shown that the two further angles in a right triangle are acute. Hencethe angle ∠BAF is acute, but this is just the bisected angle.

17In calculus and differential geometry, one uses often measurement using the arc length on the unitcircle. In that case, the right angle is assigned the value π

2 .

110

Figure 4.3: The angular bisector

(The Archimedean property for angles, Hilbert’s Proposition 34). Assume thatthe Archimedean Axiom holds. For any two angles α and ε there exists a natural numberr such that

(4.7)α

2r< ε

Proof of Proposition 34. The angle α2

is acute. If α2≤ ε, assertion (3.2) holds with

r = 1, or r = 2 in case of equality. We need a construction for the remaining case α2

> ε.Choose any point C on one side of the angle α

2and drop the perpendicular on the other

side of that angle. Let B be the foot point of that perpendicular, and let A be thevertex of α

2. Next we transfer angle ε at vertex A, with one side AB, and the other side

inside the angle α2

= ∠CAB. By the Crossbar Theorem, there exists a point D wherethis other side intersects segment BC.

By the Archimedean axiom, there exists a natural number n such that

(4.8) n ·BD > BC

Now, one transfers the angle ε, repeating n times, always with vertex A, using commonsides, and turning away from segment AB. Let C0 = B, C1 = D. Let the new sides of

transferred angle ε intersect−−→BC at points C1 = D, C2, C3, . . . . We distinguish two cases

(4.9) or (4.10):

(4.9) It can happen that one of the new sides do no longer intersect ray−−→BC. Let m be

the smallest number of angle transfers for which this happens. Since the second

leg−→AC of angle α

2does intersect ray

−−→BC, we conclude that

(4.9) m · ε >α

2

Otherwise the Crossbar Theorem would lead to a contradiction.

111

Figure 4.4: Many consecutive small angles surpass any angle.

(4.10) The other possible case is that the new sides produced by transferring angle ε

all n times intersect ray−−→BC, say at points C1 = D, C2, C3, . . . , Cn.

Let E = Cn be the point where the new side of nth angle ε intersects−−→BC. The segments

C0C1, C1C2, C2C3, . . . , Cn−1Cn cut on line BC by these angles ε = ∠CkACk−1 satisfy

Ck−1Ck < CkCk+1

for k = 1, 2, 3, . . . , n − 1, as follows from Proposition 33 given below. Hence a simpleinduction argument yields Ck−1Ck ≥ BD and BCk > k ·DB for k = 1, 2, 3, . . . , n − 1.Hence (4.8) implies BCn > n ·BD > BC. Because ∠BACn = n · ε, we get

(4.10) n · ε >α

2

Let r be an integer such that 2r−1 ≥ m or 2r−1 ≥ n, in case (a) or (b), respectively.Now (4.9) or (4.10) imply

2r−1 · ε >α

2

Bisection of those two angles leads inductively to the inequalities

2r−2 · ε >α

4, 2r−3 · ε >

α

8, . . . , ε >

α

2r

This finishes the proof of Proposition 34.

Here is still the missing proposition, already used above.

(Hilbert’s Proposition 33). Let triangle 4OPZ have a right angle at vertex P . LetX and Y be two points on segment PZ such that ∠XOY = ∠Y OZ. Then XY < Y Z.

112

Figure 4.5: Congruent angles cut longer and longer segments from a line.

Proof of Proposition 33. One transfers segment OX onto the ray−→OZ at vertex O. One

gets the segment OX ′ ∼= OX. Because segment OZ is the side opposite to the obtuseangle in 4OXZ, and the obtuse angle is the largest angle of any triangle, we getOZ > OX from Euclid I.19. Hence the point X ′ lies between O and Z.

From the exterior angle theorem (Euclid I.16) for4OY Z one gets ∠OZY < ∠OY X.From the exterior angle theorem for 4OX ′Y one gets ∠OY X ′ < ∠Y X ′Z. Hence

(4.11) ∠X ′ZY = ∠OZY < ∠OY X ∼= ∠OY X ′ < ∠Y X ′Z

In figure 4.1, we see that α < β < γ. By Euclid I.19, the side opposite to a larger angleis larger. We use this theorem for 4X ′Y Z. Hence (4.11) implies X ′Y < Y Z and henceXY ∼= X ′Y < Y Z as to be shown.

Sketch of the proof for the measurement of angles. Let the given angle be ∠BAC. We

erect the perpendicular to the ray−→AB at vertex A. Depending on whether the per-

pendicular is inside the supplementary right angle, or inside the given angle, or the

perpendicular coincide with the second side−→AC, the given angle is acute, obtuse or

right. The measurement process is now most easily explained by taking the right angleas unit. Starting with that one of the two complementary right angles, inside of which

the ray−→AC lies, one successively bisects angles—always keeping the ray

−→AC inside of

them. The left bounds of the bisected angles correspond to an infinite fraction

0 . d1d2d3 . . .

for an acute, or1 . d1d2d3 . . .

for an obtuse angle. To obtain a measurement in the traditional degrees, these binaryfractions have to be multiplied by 90 18 The further details are so similar to the case ofsegment measurement that we do not need to repeat.

18In binary notation 90 = 64 + 16 + 8 + 2 = 1011010

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4.2 Axioms related to Completeness

The clear-cut understanding of this material was achieved only in the late nineteenthcentury. There are several axioms for completeness, with very similar implications,having slight but deep differences. It is hard to say what is the most natural one ofthese axioms. Even Hilbert has suggested different axioms of continuity in differenteditions of his foundations of geometry.

4.2.1 Cantor’s axiom

Following Baldus and Lobell [7], p.43, I state my favorite version of Cantor’s axiom.

Cantor’s Axiom. There exists at least one segment A1B1 with the following property:Given any sequence AiBi of boxed subsegments for i = 2, 3, . . . such that

(4.12) Ai−1 ∗ Ai ∗Bi ∗Bi−1 for all i = 2, 3, . . .

there exists a point X∗ such that

(4.13) Ai ∗X∗ ∗Bi for all i = 1, 2, 3, . . .

For any every Hilbert plane, the following similar, but a bid more general statement isan immediate consequence:

Cantor’s principal of boxed intervals. For every sequence of boxed segments AiBi

such that either(4.14)Ai−1∗Ai∗Bi∗Bi−1 or Ai−1 = Ai , Ai−1∗Bi∗Bi−1 or Ai−1∗Ai∗Bi−1 , Bi−1 = Bi,

for all i = 2, 3, . . . , there exists a point X∗ such that

(4.15) either Ai ∗X∗ ∗Bi or X∗ = Ai or X∗ = Bi for all i = 1, 2, 3, . . .

10 Problem 4.1. Convince yourself, and explain that Cantor’s axiom impliesCantor’s principal of boxed intervals, just assuming the axioms of incidence, order andcongruence.

Main Theorem (Completeness of measured segments and angles). Assume theaxioms of incidence, order and congruence, as well as the Archimedean and Cantor’saxiom.

Given is a unit segment OI of length |OI| = 1. For every real number r > 0, thereexists a segment segment AB of length |AB| = r.

For every real number 0 < d < 180, there exists an angle, which has the measurement∠ABC = d, in traditional degrees.

114

Indication of reason. The real number r can be given as an finite or infinite binaryfraction

r = n− 1 . d1d2d3 . . .

The boxed intervals are constructed as explained in the measuring process. Let L0U0 bea segment of unit length such that |AL0| = n− 1 and |AU0| = n. We define a sequenceof boxed intervals LsUs by setting

Ls := Ls−1 , Us := Hs−1 if ds = 0

Ls := Hs−1 , Us := Us−1 if ds = 1

for s = 1, 2, 3, . . . . Here Hs−1 is midpoint of segment Ls−1Us−1. In both cases, the leftlower approximation to |AB| obtained by the s-th step is

|ALs| = n− 1 +d1

2+

d2

4+

d3

8+ · · ·+ ds

2s

In the special case of a finite fraction, one gets X∗ = Hs−1, in which case this approxi-mation is exact, and the measure process is finished.

For an infinite binary fraction, we need Cantor’s principal of boxed intervals: Thereexists a point X∗ lying is in the infinite intersection of all intervals LsUs. The intervalL0X

∗ has length |L0X∗| = r equal to the originally given number r.

4.2.2 Dedekind’s axiom

This is now most popular axiom for continuity. Let us start with a definition.

Definition 4.1 (Dedekind Cut). A Dedekind cut of the line l is a pair of sets Σ1, Σ2such that the set of points on a line l is the disjoint union of the two nonempty sets Σ1

and Σ2, which have the following property:(P) If P1 and P3 are any two points in Σi, and the third point P2 lies between P1

and P3, then P2 is in the same set Σi, for i = 1, 2.

Dedekind’s Axiom. Assume that Σ1, Σ2 is a Dedekind cut of the line l. Thenthere exists a cut point K∗ on the line l such that Σ1 ∪ K∗ and Σ2 ∪ K∗ arethe two opposite rays on the line l with vertex K∗.

In his recent book [1], Greenberg writes on p.260 about this axiom:

We bring in our deus ex machina, as classical Greek theater called it—a godcomes down from heaven to save the day.

This axiom was proposed by J.W.R. Dedekind in 1871. Here is what Dedekind saysabout continuity in ”Stetigkeit und irrationale Zahlen” (1872):

115

I find the essence of continuity in the following principle: ”If all the pointsof a line fall into two classes in such a way that each point of the firstclass lies to the left of each point of the second class, then there existsone and only one point that gives rise to this division of all the pointsinto two classes, cutting of the line into two pieces.”

As mentioned before, I believe I am not wrong if I assume that everyone willimmediately admit truth of this assertion; most of my readers will be verydisappointed to realize that by this triviality the mystery of continuitywill be revealed. I am very glad if everyone finds the above principle soclear and so much in agreement with our own conception of a line; for Iam not in a position to give any kind of proof of its correctness; nor isanyone else.

The assumption of this property of a line is nothing else than an axiom bywhich we first recognize continuity of the line, through which we thinkcontinuity into the line.

If space has any real existence at all, it does not necessary need to be con-tinuous; countless properties would remain the same if it were discon-tinuous. And if we knew for certain that space was discontinuous, stillnothing could hinder us, if we so desired, from making it continuous inour thought by filling up its gaps; this filling up would consist in thecreation of new point-individuals, and would have to be carried out inaccord with the above principle.

Indeed, Dedekind’s axiom is a very strong axiom. It essentially introduces the realnumbers into our geometry, which is not in the spirit of Euclid, but useful and evennecessary from the engineering point of view.

Dedekind’s axiom is by no means necessary to do interesting mathematics— onthe contrary—it spoils many fine points of algebra and set theory. Many more subtledistinctions and questions about constructibility are obscured by Dedekind’s axiom.

Too, the strength of Dedekind’s axiom becomes apparent because of its many con-sequences, some of which are now explained.

Theorem 4.1. Dedekind’s axiom implies the Archimedean axiom.

Theorem 4.2. Dedekind’s axiom implies Cantor’s axiom.

Main Theorem (Dedekind’s axiom implies completeness). The elements of geometry—the points, lines and planes— which satisfy the axioms of incidence, order, congruence,and Dedekind’s axiom have no extension to any larger system, for which all these axiomsstill hold.

To facilitate the proofs of theorem 4.1 as well as theorem 4.2, we introduce stillanother suggested axiom:

116

Weierstrass’ Axiom. Let Ai for i = 2, 3, . . . be sequence of a points , and B a point,lying all on one line, such that

(4.16) Ai−1 ∗ Ai ∗B for i = 2, 3, . . .

Then there exists a point K∗ such that

(i) either K∗ = B or

(4.17) Ai ∗K∗ ∗B for i = 1, 2, 3, . . .

(ii) Furthermore, every other point X such that

(4.18) Ai ∗X ∗B for i = 1, 2, 3, . . .

satisfies

(4.19) Ai ∗K∗ ∗X ∗B for i = 1, 2, 3, . . .

Proposition 4.1. Dedekind’s axiom implies Weierstrass’ axiom.

Proof that Dedekind’s axiom implies Weierstrass’s axiom. We define a Dedekind cut asfollows: Let Σ1 consist of all points P such that

either P ∗ A1 ∗B or P = A1 or there exists i ≥ 1 such that A1 ∗ P ∗ Ai

As follows logically from Hilbert’s four-point theorem (1.3), the complement Σ2 on theline A1B consists of all points such that

A1 ∗ Ai ∗ P for all i = 2, 3 . . .

The cut point K∗ of the Dedekind cut Σ1, Σ2 needs to lie in Σ2. It is either K∗ = B orsatisfies Ai ∗K∗ ∗B for i = 1, 2, 3, . . . . This confirms item (i) of the Weierstrass axiom.Furthermore, if one assumes that any point X satisfies

(4.20) Ai ∗X ∗B for i = 1, 2, 3, . . .

then X ∈ Σ2. Hence K∗ ∗X ∗B because K∗ is the vertex of the ray producing Σ2. Nowrelation (4.20) and the four-point theorem imply

(4.19) Ai ∗K∗ ∗X ∗B for i = 1, 2, 3, . . .

as to be shown, in order to confirm item (ii) of Weierstrass’ axiom.

Proposition 4.2. Weierstrass’ axiom implies Cantor’s axiom.

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Proof that Weierstrass’s axiom implies Cantor’s axiom. Within the assumptions to setupCantor’s axiom, Hilbert’s n-point theorem (1.5) implies inductively

A1 ∗ Ai ∗Bj ∗B1 for all i, j ≥ 2

The first item (i) of Weierstrass’ axiom implies

Ai ∗K∗ ∗B1 for i = 1, 2, 3, . . .

Fix some index j. Since Ai ∗ Bj ∗ B1 for all i = 1.2. . . . , the second item (ii) fromWeierstrass’ axiom with X := Bj implies

Ai ∗K∗ ∗Bj ∗B1 for i, j = 1, 2, 3, . . .

as to be shown.

Proposition 4.3. Weierstrass’ axiom implies the Archimedean axiom.

Proof. As in the Archimedean axiom, two segments CD and AB are given. We use CDas unit of measurement and make the following

Definition 4.2. We say that a point P on the ray−→AB can be reached with the unit of

measurement CD if and only if—as in the Archimedean axiom—, there exists a naturalnumber n for which construction of the finite sequence of points A0 = A, A1, A2, . . . , An

such that

(4.4) CD ∼= Ak−1Ak and−−→AAk =

−→AB for k = 1, 2, . . . n

leads up to a point An such that point P either lies between An−1 and An, or P = An−1.

Let A1, A2, A2, . . . be the sequence of points constructed by the Archimedean mea-surement (4.4). If the Archimedean axiom would not hold, then there would exist apoint F which cannot be reached with the unit measurement CD. We can now applyWeierstrass’ Axiom for the sequence Ai and point F . Hence there exists a point K∗

such that

(4.21) Ai ∗K∗ ∗ F for i = 1, 2, 3, . . .

In other words, the point K∗ cannot be reached by measurement. Let K− and K+ bethe points on both sides of K∗ such that K∗K− ∼= K∗K+

∼= CD, and K− ∗K∗ ∗ F . Itis clear that these points K+ and K− cannot be reached by measurement, neither.

Because of item (ii) of Weierstrass’ axiom, K∗ is the point most to the left thatcannot be reached by measurement: every other point X such that

(4.20) Ai ∗X ∗B for i = 1, 2, 3, . . .

satisfies

(4.19) Ai ∗K∗ ∗X ∗B for i = 1, 2, 3, . . .

Such a reasoning would now imply both for X := K+ and X := K−, which is impossible.This contradiction implies that the Archimedean axiom does hold.

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In order to defend my preference of Cantor’s axiom—and trying to take some of thedeus ex machina image away from Dedekind—I prove now:

Main Theorem. Assuming both the Archimedean axiom, as well as Cantor’s axiom,Dedekind’s axiom follows.

Proof. Given is a line AB with a Dedekind cut Σ1, Σ2 of it. Let OI be any mea-surement unit. We may assume A ∈ Σ1 and B ∈ Σ2. As explained in the theorem ofmeasurement, one measures the distance from A ∈ Σ1 to the cut point. Thus one canconstruct a sequence of approximations to the cut point: Ls ∈ Σ1 and Us ∈ Σ2 suchthat

|ALs| = n− 1 +d1

2+

d2

4+

d3

8+ · · ·+ ds

2s

and |LsUs| = 2−s for s = 0, 1, 2, . . . . Cantor’s principle of boxed intervals yields a pointX such that Ls ∗X ∗ Us, which is the cut point.

4.2.3 Hilbert’s axiom of completeness

The following axiom of completeness is suggested in the millenium edition of Hilbert’sfoundations:

V.2 (Axiom of linear completeness) An extension of a set of points on a line, withits order and congruence relations existing among the original elements as well asthe fundamental properties of line order and congruence that follow from AxiomsI-III and from V.1, is impossible.

Remark. The insight that it is enough to require the extension of a set of points on aline goes back to Paul Bernays.

(Theorem of completeness, Hilbert’s Proposition 32). The elements of geometry—the points, lines and planes— have no extension to any larger system, under the assump-tions that the axioms of incidence, order, congruence, and the Archimedean axiom stillhold for the extension.

The parallel axiom may be assumed or not, it does not interact with completeness atall. The axiom of completeness is not a consequence of the Archimedean axiom. But,on the contrary, the Archimedean axiom needs to be assumed for completeness to bemeaningful to hold.

Main Theorem. Together, the Archimedean axiom (V.1), and the axiom of linearcompleteness (V.2) imply Cantor’s axiom and Dedekind’s axiom.

There exist infinitely many models for the axioms I through IV, and V.1. Butonly one model satisfies the axiom of completeness, too—this the Cartesian geometry.

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Historic context. In the very earliest edition, Hilbert proposed the conclusion of theTheorem of completeness (4.2.3) as an axiom. Only the German edition of the foun-dation of 1903 contains the axiom of linear completeness. Even earlier, the axiom oflinear completeness had already appeared in May 1900 in the French translation of thefoundation. It appeared for the very first time on October 12th 1899, in Hilbert’s lecture”Uber den Zahlbegriff”.

Hilbert’s axiom of completeness has given rise to many positive as well as negativecomments by important mathematicians.

”An axiom about axioms with a complicated logical structure” (Schmidt)

”An unhappy axiom” (Freudenthal)

”The axioms of continuity are introduced by Hilbert, to show that they arereally unnecessary.” (Freudenthal)

Hilbert’s completeness axiom is obviously not a geometric statement, andnot a statement formalizable in the language used previously—so whatdoes it accomplish? (M.J. Greenberg, 2010)

”The foundations of geometry contain more than insight in the nature ofaxiomatic.” (Freudenthal)

”The most original creation in Hilbert’s axiomatic” (Baldus)

”Hilbert has made the philosophy of mathematics take a long step in ad-vance.” (H. Poincare)

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5 Legendre’s Theorems

Recall that a Hilbert plane is a geometry, where the axioms of incidence, order, andcongruence are assumed, as stated in Hilbert’s Foundations of Geometry. Neither theaxioms of continuity (Archimedean axiom and the axiom of completeness), nor theparallel axiom needs to hold for an arbitrary Hilbert plane. The proposition numberingis taken from Hilbert’s Foundations of Geometry. For simplicity, I am using the letterR to denote a right angle.

5.1 The First Legendre Theorem

(The First Legendre Theorem, Hilbert’s Proposition 35). Given is a Hilbertplane, for which the Archimedean Axiom is assumed to hold. Then every triangle hasangle sum less or equal two right angles.

Proof. The proof relies on three ideas:

(a) The exterior angle theorem Euclid I.16, from which one concludes Euclid I.17: thesum of any two angles of a triangle is less than two right angles.

(b) A construction, given by Lemma 1. From a given triangle, this construction yieldsa second triangle, with the same angle sum as the first one; and, additionally, oneof its angles is less or equal half of an angle of the original triangle.

(c) The Archimedean property for angles given in Hilbert’s Proposition 34.

Lemma 5.1. For any given 4ABC, there exists a 4A′B′C ′ such that

(5.1) α′ + β′ + γ′ = α + β + γ and α′ ≤ α

2

Proof of the Lemma. Let D be the midpoint of side BC. Extend the ray−−→AD and

transfer segment AD to get point E such that AD ∼= DE. We need to consider twocases:

(i) If ∠EAB ≤ ∠CAE, the new 4A′B′C ′ has vertices A′ = A, B′ = B and C ′ = E.

(ii) If ∠EAB > ∠CAE, the new 4A′B′C ′ has vertices A′ = A, B′ = C and C ′ = E.

Other equivalent conditions all leading to case (i) are

α′ ≤ γ′ ⇐⇒ EB ≤ AB ⇐⇒ AC ≤ AB ⇐⇒ β ≤ γ

With these choices of the new triangle, the inequality α′ ≤ α2

holds in both cases.

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Figure 5.1: Two triangles with same angle sum—and area.

We explain the details for case (i). By SAS congruence, 4ADC ∼= 4EDB, becausethe vertical angles at vertex D are congruent, and the adjacent sides are pairwise congru-ent by construction. The congruence of the two triangles yields two pairs of congruentangles

γ = ∠ACD ∼= ∠DBE and γ′ = ∠DEB ∼= ∠DAC

From angle addition at vertices A and B, respectively, and a final addition of formulas,one concludes

α = α′ + γ′

β + γ = β′

α + β + γ = α′ + β′ + γ′

A similar result is concluded in the second case (ii) via the congruence 4ADB ∼=4EDC.

Figure 5.2: Two triangles with same angle sum, cases (i) and (ii).

Corollary. The two triangles 4ABC and 4A′B′C ′ have the same area.

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Reason. To produce the new triangle 4A′B′C ′ from the old triangle 4ABC, one needsto take away triangle 4ADC and add the congruent triangle 4EDB.

End of the proof of the First Legendre Theorem. Let 4ABC be any triangle. We usethe first Lemma repeatedly to get a sequence of triangles

4A′B′C ′ = 4A1B1C1 , 4A2B2C2 , 4A3B3C3 , . . .4AnBnCn , . . .

such that α + β + γ = αn + βn + γn and αn ≤ α2n for all natural numbers n ≥ 0. The

exterior angle theorem implies βn + γn < 2R and hence

(5.2) α + β + γ = αn + βn + γn <α

2n+ 2R

for all n ≥ 0. Now a limiting process with n→∞ implies the result.An accurate version of this part of the argument uses the Archimedean property for

angles. We argue by contradiction, assuming that the angle sum would be α+β+γ > 2R.Let

(5.3) ε := α + β + γ − 2R

which, because of the assumption α + β + γ > 2R, would be an angle ε > 0. By theArchimedean property for angles (Proposition 33), there would exist a natural numberr such that

(5.4)α

2r< ε

Now (5.2), (5.4) and (5.3) together would imply

α + β + γ = αr + βr + γr <α

2r+ 2R ≤ ε + 2R = α + β + γ

which is impossible. Because any two angles are comparable, α+β +γ ≤ 2R is the onlypossibility left, as was to be shown.

5.2 The Second Legendre Theorem

(The Second Legendre Theorem, Hilbert’s Proposition 39). Given is any Hilbertplane. If one triangle has angle sum 2R, then every triangle has angle sum 2R.

It turns out to be easier, to prove at first a Proposition about quadrilaterals. Wedefine some special quadrilaterals.

Definition 5.1. A Saccheri quadrilateral ABCD has right angles at two adjacent ver-tices A and B, and the opposite sides AD ∼= BC are congruent. A Lambert quadrilateralhas three right angles. A rectangle has four right angles.

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Figure 5.3: A Saccheri quadrilateral is bisected into two Lambert quadrilaterals.

(Saccheri and Lambert quadrilaterals, Hilbert’s Proposition 36). Assume thatABCD is a Saccheri quadrilateral with right angles at vertices A and B—and congru-ent opposite sides AD ∼= BC. Let M be the midpoint of segment AB. The perpendicularbisector m of segment AB intersects the opposite side CD at right angles, say at pointN . One gets two congruent Lambert quadrilaterals AMND and BMNC.

Proof. To show that m intersects the opposite side CD, one can use plane separationwith line m. Indeed, points A and B lie on different sides of m by construction. ByEuclid I.27, the three lines AD and m and BC are parallel. Hence points A and D lieon the same side of m. By the same reasoning, B and C lie on the same side of linem. Hence C and D lie on different sides of m. Thus segment DC intersects m. By

Figure 5.4: The steps to get symmetry of a Saccheri quadrilateral.

SAS congruence, 4MAD ∼= 4MBC, since the right angles at vertices A and B andthe adjacent sides match pairwise. Next we see that the triangles 4MDN ∼= 4MCN ,again by SAS congruence, since the angles at the common vertex M and the adjacentsides match. Hence we conclude that ∠ADM ∼= ∠BCM and ∠MDN ∼= ∠MCN . Nowangle addition yields

∠ADC ∼= ∠BCD

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Hence the quadrilateral ABCD has two congruent angles at vertices C and D. At vertexN , the angles ∠MND ∼= ∠MNC are congruent supplementary angles. Hence they areboth right angles. This finishes the proof that AMND and BMNC are congruentLambert quadrilaterals.

Definition 5.2. The reflection by line l is defined as follows: From a given point Pthe perpendicular is dropped onto l, and extended by a congruent segment FP ′ ∼= PFbeyond the foot point F . Then P ′ is called the reflected point of P .

Corollary. By the perpendicular bisector of its base line, a Saccheri quadrilateral isbisected into two Lambert quadrilaterals, which are reflection symmetric to each other.

(Hilbert’s Proposition 37). Assume ABCD is a rectangle. Drop the perpendicular

from a point E of line←→CD onto the opposite side AB. Let F be the foot point. Then

the quadrilaterals ADEF and BCEF are both rectangles.

Figure 5.5: Getting rectangles of arbitrary width— Is ζ a right angle?

Proof. One reflects the segment EF , both by line←→AD and

←→BC. Let EiFi for i = 1, 2

be the mirror images. By Proposition 36, both segments are congruent to EF . BecauseABCD was assumed to be a rectangle, both points Ei lie on line CD and both Fi

lie on line AB. The assumptions of Proposition 36 hold for all three quadrilateralsEFF1E1, EFF2E2 and E1F1F2E2.

Hence all three are Saccheri quadrilaterals. We get four congruent angles with ver-tices E1, E2 and E—denoted in figure 5.2 by ε, θ, ϕ and κ.

One of these three points E1, E2 and E lies between the other two. At that vertex,one gets a pair of supplementary angles, which are congruent. Hence they are rightangles, and the other angles just mentioned are right angles, too.

(Hilbert’s Proposition 38). If there exists a rectangle ABCD, then every Lambertquadrilateral is a rectangle.

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Figure 5.6: ε is congruent to both θ and ϕ, hence there are two supplementary right anglesθ and ϕ!

Figure 5.7: If one rectangle exists, why are all Lambert quadrilaterals rectangles?

Proof. Assume that the Lambert quadrilateral A1B1C1D1 has its three right angles atvertices A1, B1 and D1. One may assume that A = A1, and the three points A, B, B1 aswell as A, D, D1 lie on a line. We have drawn the case that A ∗B ∗B1 and A ∗D1 ∗D.The easy modification of the proof to the other possible orders of points A, B, B1 andA, D1, D is left to the reader. As in Proposition 36, we show that segment D1C1 andline BC do intersect, say at point F . Now Proposition 37 implies that these two linesare perpendicular to each other. Hence ABFD1 is a rectangle. Applying Proposition37 a second time, we conclude that the quadrilateral AB1C1D1 is a rectangle, too.

Corollary. If there exists a rectangle ABCD, then every Saccheri quadrilateral is a

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rectangle. There exists rectangles of arbitrarily prescribed height and width.

Reason. Because of Proposition 36, a Saccheri quadrilateral is subdivided into two Lam-bert quadrilaterals by its midline. By Proposition 38, those are rectangles. Hence theSaccheri quadrilateral is a rectangle, too. The height and width of a Saccheri quadri-lateral can be prescribed arbitrarily, and they are all rectangles. Hence there existsrectangles of arbitrarily prescribed height and width.

Note of caution. In hyperbolic geometry, the height and width of a Lambert quadrilat-eral cannot be prescribed arbitrarily. Indeed if one chooses the width for a Lambertquadrilateral, the height has an upper bound.

(Proposition 39). To every 4ABC with angle sum 2ω, there corresponds a Saccheriquadrilateral with two top angles ω.

Figure 5.8: To every triangle corresponds a Saccheri quadrilateral.

Proof. The Saccheri quadrilateral is ABGF , with right angles at F and G and con-gruent opposite sides AF ∼= BG, and congruent angles ω at A and B.

The drawing indicates, how the Saccheri quadrilateral is obtained. One connects themidpoint D of segment AC and midpoint E of segment BC, by line l. Then one dropsthe perpendiculars from all three vertices A, B, C onto line l. Let F, G and H be thefoot points of the perpendiculars, respectively.

By the SAA congruence theorem, 4AFD ∼= 4CHD, because of the right anglesat vertices F and H, the congruent vertical angles at vertex D, and because segmentsAD ∼= DC are congruent by construction. By similar reasoning we get 4BGE ∼=4CHE. From these triangle congruences we conclude that HC ∼= FA, and HC ∼= GB.Hence FA ∼= GB, which implies that ABGF is a Saccheri quadrilateral.

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Figure 5.9: How to get two pairs of congruent triangles, and how to get the top angles.

Its top angles at vertices A and B were shown to be congruent in Proposition 36. Wedenote them by ω. From the triangle congruences, too, we get γ1 = ∠DCH ∼= ∠DAFand γ2 = ∠ECH ∼= ∠EBG . The sum of the angles of 4ABC is

α + β + γ = α + γ1 + β + γ2 = ∠FAB + ∠GBA = 2ω

as to be shown.

Corollary. A triangle has angle sum 2R if and only if the corresponding Saccheri quadri-lateral is a rectangle.

Proof. This is an immediate special case of Proposition 39.

End of the proof of the Second Legendre Theorem. Assume triangle 4ABC has anglesum 2R.

Question. What can you say about the corresponding Saccheri quadrilateral?

Answer. The Saccheri quadrilateral corresponding to4ABC is a rectangle by the Corol-lary.

Hence, by Proposition 38, every Lambert quadrilateral is a rectangle. Now let4XY Z be any triangle.

Question. What can you say about the corresponding Saccheri quadrilateral?

Answer. As explained in Proposition 36, the Saccheri quadrilateral corresponding to4XY Z is bisected into two congruent Lambert quadrilaterals. By Proposition 38, theseare both rectangles, since a rectangle already exists. Hence the Saccheri quadrilateralcorresponding to 4XY Z is a rectangle.

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We have just shown that the Saccheri quadrilateral corresponding to the arbitrarilychosen 4XY Z is a rectangle, too. Hence by Proposition 39b, the angle sum of 4XY Zis 2R.

5.3 The Alternative of Two Geometries

Finally, one wants to link the angle sum of triangles to uniqueness of parallels.

Definition 5.3 (Euclidean Parallel Postulate). For every line l and for every pointP lying not on l, there exists a unique parallel m to l through point P .

Note that Euclid I.2 through I.28 are theorems that hold for every Hilbert plane.Indeed Euclid I.29 is the first theorem in Euclid’s elements that uses the Euclideanparallel postulate. Indeed, the existence of a parallel can be proved in neutral geometry.One parallel to l through point P is conveniently constructed as ”double perpendicular”.

Proposition 5.1 (Existence of a parallel). For every line l and for every point Plying not on l, there exists at least one parallel m to l through point P .

Proof. Given is line l and a point P not on l. One drops the perpendicular from point Ponto line l and denotes the foot point by F . Next, one erects at point P the perpendicularto line PF . Thus, one gets a line, which we call m. Because the two lines l and m form

congruent z-angles with the transversal←→PF , Euclid I.27 or I.28 imply that l and m are

parallel. This leaves open the question whether or not m is the unique parallel to line lthrough point P .

Definition 5.4 (Hilbert’s Parallel Postulate). For every line l and for every pointP lying not on l, there exists at most one parallel m to l through point P .

10 Problem 5.1. Explain why the following is an easy consequence of Hilbert’sparallel postulate:

If one of two parallel lines is intersected by a third line, the other one is intersected,too.

Solution. Suppose towards a contradiction that the transversal t intersects one of theparallel lines l and m, but not the other one.

Say that P is the intersection point of lines t and m. If lines t and l would notintersect, then t and m would be two different parallels of line l through point P . Thiscontradicts the uniqueness of parallels.

Because of Proposition 5.1, Euclid’s and Hilbert’s parallel postulate turn out to beequivalent for any Hilbert plane.

Definition 5.5 (Pythagorean plane). A Pythagorean plane is a Hilbert plane forwhich the Euclidean parallel postulate holds.

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Proposition 5.2. If a triangle 4ABC has angle sum α + β + γ < 2R, then there existtwo parallels to line AB through point C.

Conversely, if there is a unique parallel to line AB through point C, then the triangle4ABC has angle sum 2R.

Corollary. In a Pythagorean plane, each triangle has angle sum 2R. In other words,each Pythagorean plane is semi-Euclidean.

Proof. We form two congruent z-angles α = ∠BAC by transferring that angle with

one side ray−→CA as transversal. The second side m1 of the new angle α has to lie the

opposite side of−→CA as point B. Similarly one transfers angle β = ∠ABC with one side

ray−−→CB, and gets as second side the ray m2.The angle formed by the rays m1 and m2 is α + β + γ < 2R. Hence m1 and m2 do

not lie on the same line. On the other hand, the rays m1 and m2 are both (parts of)parallels to line l through point P . This follows from Euclid’s I.27 (Alternate interiorangles imply parallels). We have thus constructed two different parallels to line l = ABthrough point C.

(The Third Legendre Theorem). Given is a Hilbert plane, for which the ArchimedeanAxiom is assumed to hold. If the angle sum of every triangle is 2R, then the EuclideanParallel Postulate holds.

Idea of the proof. Given is line l and a point P not on l. We need to check the uniquenessof the parallel to line l through point P .

Inductively, there is constructed a sequence of isosceles triangles 4PFn−1Fn. Tostart, let F0 = F , and let F1 be any of the two points on line l such that F0F1

∼= PF .

Next let F2 be the point on ray−−→FF1 such that F1F2

∼= PF1. Inductively, assume that

F1, F2, . . . , Fn−1 have been constructed, and let Fn is the point on ray−−→FF1 such that

Fn−1Fn∼= PFn−1 and F ∗ Fn−1 ∗ Fn.

The angles φn = ∠Fn−1PFn can all be calculated by means of the following

Proposition 5.3. Assume that the angle sum for every triangle is 2R. Then the baseangle of an isosceles triangle is half of the exterior angle at the top.

Reason. It was assumed that the angle sum of any triangle is 2R. Let δ be the exteriorangle at the top vertex A of any 4ABC. Thus δ is the supplement of the interior angleα at that vertex. Hence

δ = 2R− α = α + β + γ − α = β + γ

which is the sum of the two nonadjacent interior angles. For an isosceles triangle withtop F , the two base angles are congruent by Euclid I.5, I call them β. Hence the exteriorangle at the top is ϕ = 2β double the base angle. Hence the base angle β = ϕ

2is half of

the exterior angle ϕ.

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Figure 5.10: The angle sum 2R implies uniqueness of the parallel.

Figure 5.11: Halfing an angle with an isosceles triangle.

End of the proof of the third Legendre Theorem. 4F0PF1 has a right angle on top. (Itis a right isosceles triangle, for which all three angles can be calculated in Euclideangeometry.) Indeed the Proposition yields

(1.1) φ0 = ∠F0PF1 =R

2

The other base angle of 4F0PF1 is the exterior angle on the top of 4F1PF2. Since thistriangle is isosceles, too, the Lemma implies that its base angle is

(1.2) φ1 = ∠F1PF2 =φ0

2=

R

4

Inductively, we get that

(1.n) φn = ∠Fn−1PFn =R

2n

131

for all n = 0, 1, 2, . . . . Here is the induction step:Assume that (1.n-1) has been shown. The exterior angle on the top of 4Fn−1PFn

is also a base angle of 4Fn−2PFn−1, which is φn−1 = R2n−1 by the induction assumption.

Since 4Fn−1PFn is isosceles, too, the Proposition implies that its base angle is half ofthat angle. Hence

φn =φn−1

2=

R

2n−1 · 2=

R

2n

which confirms (1.n). By angle addition, formulas (1.n) imply

(2) ∠FPFn = φ1 + φ2 + · · ·+ φn = R

(1

2+

1

4+ · · ·+ 1

2n

)= R

(1− 1

2n

)One parallel m to l through point P is conveniently constructed as ”double perpen-dicular”, as explained above. We now give an argument to show that m is the uniqueparallel to l through P . We assume towards a contradiction that m′ 6= m is a different

parallel to l through P . Let−→m′ be one of the two rays on m′ starting at point P that

forms an acute angle with the perpendicular−→PF . Let −→m be the ray on m starting at

point P that lies on the same side of PF as−→m′.

Legendre’s triangle construction needs to be done on that side of PF where the two

rays −→m and−→m′ lie. Let ε > 0 be the angle between −→m and

−→m′. By the Archimedean

axiom for angles, there exists a natural number r such that

R

2r< ε

Hence the (complementary) angle between the rays−→PF and

−→m′ satisfies

R− ε < R− R

2r= ∠FPFr

This shows that the ray−→m′ lies in the interior of ∠FPFr. Hence, by the Crossbar

Theorem, the ray−→m′ intersects the segment FFr, say at point Q. Thus the line m′ is

not a parallel to line l, which is a contradiction. Hence there exists only one parallel toline l through point P .

For completeness, we restate

Proposition 5.4 (The Crossbar Theorem). If ray−−→AD lies in the interior of ∠BAC,

and B and C are arbitrary points on the two different sides of that angle, then the ray−−→AD intersects the segment BC.

Taking the three Legendre Theorems together one gets the

Main Theorem (Alternative of Two Geometries). In a Hilbert plane, for whichthe Archimedean Axiom is assumed, there occurs one of the two possibilities:

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(a) All triangles have angle sum two right angles. All Lambert and all Saccheri quadri-laterals are rectangles. The Euclidean parallel postulate holds. One has arrived atthe Euclidean geometry.

(b) All triangles have angle sum less than two right angles. All Lambert quadrilateralshave an acute angle. All Saccheri quadrilaterals have two congruent acute topangles. Rectangles do not exist. For every line l and point P not on l, there existtwo or more parallels to line l through point P . In this case, one gets the hyperbolicgeometry.

Figure 5.12: For angle sum less 2R, there exist two different parallels.

Reason. Pure logic tells that either (1) or (2) holds:

(1) All triangles have angle sum two right angles.

(2) There exists a 4ABC with angle sum not equal than two right angles.

By the third Legendre Theorem, alternative (1) implies the Euclidean parallel postu-late. By Proposition 39, we conclude that all Lambert and Saccheri quadrilaterals arerectangles. Thus case (a), the usual Euclidean geometry occurs.

Now we assume alternative (2), and derive all the conclusions stated in (b). Bythe first Legendre Theorem, the 4ABC has angle sum less than two right angles. Bythe contrapositive of the second Legendre Theorem, we conclude that no triangle canhave angle sum two right. Hence, again by the first Legendre Theorem, all triangleshave angle sum less than two right angles. The statement about quadrilaterals followsfrom Proposition 39. The Proposition below gives the well known construction, leadingtwo different parallels. Thus all assertions of case (b) hold—we have arrived at thehyperbolic geometry.

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5.4 What is the Natural Geometry?

A legitimate question remains open at this point:

Is there a clear cut, suggestive or self-evident postulate that would betterreplace the Euclidean postulate?

Legendre’s contribution to this discussion is his investigation of the following postulate:

Definition 5.6 (Legendre’s Postulate). The exists an angle such that every pointin its interior lies on a segment going from one side of the angle to the other one.

(The Fourth Legendre Theorem). Given is a Hilbert plane, for which the Archi-medean Axiom is assumed to hold. If Legendre’s postulate holds, there exists a trianglewith angle sum two right.

By combining the result with Legendre’s second Theorem, we conclude that every trianglehas angle sum two right.

Corollary. A Hilbert plane were the Archimedean Axiom and Legendre’s postulate holdis semi-Euclidean.

Proof. The proof relies on three ideas:

(a) The additivity of the defect of triangles.

(b) A construction doubling the defect of a triangle.

(c) The Archimedean property for angles given in Hilbert’s Proposition 34.

Definition 5.7 (Defect of a triangle). The defect of a triangle is the deviation of itsangle sum from two right angles. We writeδ(ABC) = 2R− α− β − γ.

Figure 5.13: The defect and the area of triangles are both additive.

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Lemma 5.2 (Additivity of the defect). Let 4ABC be any triangle and D be a pointon segment AB. Then

δ(ABC) = δ(ADC) + δ(DBC)

Suppose that a triangle is partitioned into four smaller triangles by three points lying onits sides. Then the defect of larger triangle is the sum of the defects of the four smallertriangles.

Proof. It is left to the reader to check these simple facts.

Lemma 5.3 (Doubling the defect). Given is a triangle 4BAC where Legendre’spostulate holds for the angle ∠BAC. Then there exists a triangle 4B′AC ′, with thesame vertex A and angle at A such that

(5.5) δ(B′AC ′) ≥ 2 δ(BAC)

Figure 5.14: Doubling the defect of a triangle.

Proof. One starts similarly to the procedure for the first Legendre Theorem. Let D be

the midpoint of side BC, and extend the ray−−→AD and transfer segment AD to get point

E such that AD ∼= DE. The congruences

(5.6) 4ADC ∼= 4EDB and 4ADB ∼= 4EDC

are shown by SAS congruence. Next, the congruence

(5.7) 4ABC ∼= 4ECB

is shown by ASA congruence. By Legendre’s postulate, there exists a point B′ on ray−→AB and a point C ′ on ray

−→AC such that point E lies on the segment B′C ′.

We next claim point B lies between A and B′. This follows from the exterior angletheorem: Indeed, the triangle 4AEC ′ has the interior angle ∠EAC ′, which is less thana non adjacent exterior angle, thus

∠EAC ′ < ∠AEB′

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On the other hand, the triangle congruence (5.6) implies

∠EAC ′ = ∠DAC ∼= ∠BED = ∠AEB

Hence ∠AEB < ∠AEB′. Since B and B′ lie on the same side of line AE, this impliesthat B lies between A and B′. Similarly, one shows that C lies between A and C ′.

The larger triangle4B′AC ′ is partitioned into four smaller triangles by the segmentsbetween the three points B, E and C lying on its sides. The additivity and positivityof the defect, and finally the congruence (5.11) yield

(5.8)δ(AB′C ′) = δ(ABC) + δ(ECB) + δ(EBB′) + δ(ECC ′)

≥ δ(ABC) + δ(ECB) = 2 δ(ABC)

as to be shown.

End of the proof of the Fourth Legendre Theorem. Let 4ABC be such that Legendre’spostulate holds for the angle ∠BAC. We use Lemma 5.3 repeatedly to inductively geta sequence of triangles

4AB′C ′ = 4AB1C1 , 4AB2C2 , 4AB3C3 , . . .4ABnCn , . . .

all with the same vertex A and angle ∠BAC. Since the defect has obviously the upperbound 2R, one concludes

2R ≥ δ(ABnCn) ≥ 2n δ(ABC)

and after dividing one gets

δ(ABC) ≤ 2R

2n

for all natural numbers n. Now the Archimedean property for angles implies δ(ABC) =0 and hence α + β + γ = 0, as to be shown.

Proof of the Corollary. If Legendre’s postulate holds for one angle ε, one easily checksthat it holds for any smaller angle, For the doubled angle 2ε, the postulate holds, too.

Given any other angle α, we use the Archimedean property. There exists a naturalnumber r such that

(5.9)α

2r< ε

Now we successively conclude that the Legendre postulate holds for the angles

α

2r, 2

α

2r, . . . ,

α

2, α

. Hence Legendre’s postulate holds for every angle. By the proof above, we see thatevery triangle has angle sum two right.

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Figure 5.15: Hyperbolic geometry produces an angle at vertex E.

10 Problem 5.2. Start a similar construction in hyperbolic geometry. Insteadof relying on Legendre’s postulate, it is natural to choose the points B′ and C ′ on thetwo sides of angle ∠BAC such that BB′ ∼= CE and CC ′ ∼= BE. Which congruenceshold among the four triangles appearing in formula (5.8)? Show that the quadrilateralAB′EC ′ is convex and that its total defect is 2(2R− ∠B′EC ′).

Solution. As above,

(5.10) 4ABC ∼= 4ECB

is shown by ASA congruence and

(5.11) 4BB′E ∼= 4CEC ′

is shown by SAS congruence. The total defect is

δ(AB′EC ′) = δ(ABC) + δ(EBC) + δ(EBB′) + δ(ECC ′) = 2δ(ABC) + 2δ(EBB′)

Angle addition at vertex B yields 2R = α′ − β − γ and hence

δ2 :=δ(AB′EC ′)

2= (2R− α− β − γ) + (2R− α′ − β′ − γ′)

= (2R− α′ − β − γ) + (2R− α− β′ − γ)

= 2R− α− β′ − γ′ = 2R− ∠B′EC ′

Half of the defect of quadrilateral AB′EC ′ equals its exterior angle at vertex E.

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8 Towards a Natural Axiomatization of Geometry

8.1 The Uniformity Theorem

Main Theorem (Uniformity Theorem). Any Hilbert plane is of either one of thefollowing three types:

semi-Euclidean The angle sum of every triangle is two right angle, and every Lambertor Saccheri quadrilateral is a rectangle.

semi-hyperbolic The angle sum of every triangle is less than two right angle, andevery Lambert or Saccheri quadrilateral has one respectively two acute angles.

semi-elliptic The angle sum of every triangle is larger than two right angle, and everyLambert or Saccheri quadrilateral has one respectively two obtuse angles.

Figure 8.1: Across the longer side BC is the larger angle δ.

Lemma 8.1. Given is any quadrilateral ABCD with right angles at vertices A andB, vertices C and D lying on the same side of line AB. The angles γ at vertex C andδ at vertex D satisfy

γ < δ if and only if BC > AD

γ ∼= δ if and only if BC ∼= AD

γ > δ if and only if BC < AD

Proof. As shown in the figure, we assume BC > AD. We have to check whether γ < δ.By transferring segment AD, we produce the congruent segment BE ∼= AD such thatABED is a Saccheri quadrilateral. Let its top angles be congruent to ε.

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In the triangle 4DEC, the exterior angle theorem yields γ < ε. Since points A andC lie on different sides of line DE, angle comparison at vertex D implies ε < δ. Henceγ < ε < δ, and transitivity of angle comparison yields γ < δ, as to be shown.

By a similar argument, we prove that BC < AD implies γ > δ. Finally BC ∼= ADimplies γ ∼= δ, since the top angles of a Saccheri quadrilateral are congruent.

Figure 8.2: For an acute top angle, the perpendicular dropped from a point P inside thetop segment CD is shorter than the opposite sides.

Lemma 8.2. Given is a Saccheri quadrilateral ABCD, with right angles at verticesA and B. From any point P inside the top segment CD, the perpendicular is droppedonto the segment AB, with foot point Q. The following three cases can occur:

γ < R and PQ < AD

γ = R and PQ ∼= AD

γ > R and PQ > AD

Proof. At vertex P there occur the supplementary angles α = ∠DPQ and β = ∠CPQ.We begin by assuming PQ < AD, and look for a result about the top angle γ. Using

the previous Lemma 8.1 for the quadrilateral AQPD, we conclude δ < α. Using theLemma 8.1 once more for the quadrilateral QBCP , we conclude γ < β. Hence angleaddition yields

2γ = γ + δ < β + α = 2R

and hence γ < R.By a similar argument, the assumption PQ > AD implies γ > R, and finally, indeed,

the assumption PQ ∼= AD implies γ = R.

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Figure 8.3: For an acute top angle γ, the perpendicular PQ dropped from a point outsidethe top segment is longer than the opposite sides BC.

Lemma 8.3. Given is a Saccheri quadrilateral ABCD, with right angles at vertices

A and B. From any point P on the ray−−→DC outside of the top segment CD, the

perpendicular is dropped onto the segment AB, with foot point Q. The following threecases can occur for the top angle:

γ < R and PQ > AD

γ = R and PQ ∼= AD

γ > R and PQ < AD

Proof. By transferring segment AD, we produce the congruent segment QE ∼= AD suchthat AQED is a Saccheri quadrilateral. Let its top angles be congruent to ε. We getthe third Saccheri quadrilaterals BQEC, denote its top angles by β. The line BCintersects segment DE (Why?). Let F be the intersection point.

At first, we assume QP > AD, and check that γ < R. The three points A, B and Qon the base line lie on the same side of top line DE, since baseline and top line of theSaccheri quadrilateral AQED have the middle line as their common perpendicular,and hence are parallel. Since QP > AD ∼= QE, points Q and P lie on different sidesof line DE. Hence points P and C lie on the (upper) side of DE, whereas A, B and Qlie on the lower side. The intersection point F of line DE with line BC lies between Band C.

Angle addition at vertex C yields γ + β + χ = 2R. In the triangle 4DEC, theexterior angle χ > δ − ε = γ − ε. Finally, comparison of angles at vertex E shows that

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β > ε. Put together, we get

2R = γ + β + χ > γ + β + γ − ε > 2γ

and hence γ < R, as to be shown.

Figure 8.4: For an obtuse top angle γ, the perpendicular PQ dropped from a point outsidethe top segment is shorter than the opposite sides BC and AD.

Under the assumption PQ < AD, several modifications occur, as can be seen in thefigure on page 139. Indeed, because of QP < AD ∼= QE, points Q and P lie on thesame side of line CE. Hence all five points A, B, Q, E and D lie on the same (lower) sideof top line DE. The intersection point F of line DE with line BC lies outside segmentBC.

Angle addition at vertex C yields γ + β − χ = 2R. In the triangle 4DEC, theexterior angle χ > −δ + ε = −γ + ε. Finally, comparison of angles at vertex E showsthat β < ε. Put together, we get

2R = γ + β − χ < γ + β + γ − ε < 2γ

and hence γ > R, as to be shown.

Lemma 8.4. Given are two Saccheri quadrilaterals ABCD and A′B′C ′D′ with acommon middle segment MN . Then there top angles γ and γ′ are either both acute,both right, or both obtuse.

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Figure 8.5: Two Saccheri quadrilaterals with a common middle line MN have either bothacute, both obtuse, or both right top angles.

Proof. Without loss of generality we may assume A ∗ A′ ∗M ∗ B′ ∗ B as order of thevertices on the base line. Using the previous Lemma 8.2 for the Saccheri quadrilateralABCD, we get the three equivalences

γ < R if and only if A′D′ < BC

γ ∼= R if and only if A′D′ ∼= BC

γ > R if and only if A′D′ < BC

Using Lemma 8.3 for the Saccheri quadrilateral A′B′C ′D′, we get the three equiva-lences

γ′ < R if and only if A′D′ < BC

γ′ ∼= R if and only if A′D′ ∼= BC

γ′ > R if and only if A′D′ < BC

Put together, we see that angles γ and γ′ are either both acute, both right, or bothobtuse, as to be shown.

End of the proof of the Uniformity Theorem. Given a Saccheri quadrilateral ABCDwith middle line MN , and any second Saccheri quadrilateral. We transfer the middle line

of the second Saccheri quadrilateral onto the ray−−→MB. By means of congruent triangles,

it is straightforward to verify that we can produce a Saccheri quadrilateral QPRSwhich is congruent to the second given one, and has the bottom line SQ = MN andthe middle line MK = AB.

The two lines CD and PR do intersect (Why?). With the intersection point L, weget a Lambert quadrilateral MKLN . Let λ = ∠KLN be its top angle.

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Figure 8.6: Any two Saccheri quadrilaterals can be put into a position such that the middleline of the first one is the base line of the second one, and vice versa.

By Lemma 8.4, we conclude that the top angles γ and λ of the quadrilateralsABCD and MKLN are either both acute, right, or obtuse. Similarly, we con-clude that the top angles ϕ and λ of the quadrilaterals QPRS and MKLN areeither both acute, right, or obtuse. Hence the top angles γ and ϕ of the two givenSaccheri quadrilaterals are either both acute, right, or obtuse.

8.2 A Hierarchy of planes

Recall that a Hilbert plane is a geometry, where Hilbert’s axioms of incidence, order, andcongruence, as stated in Hilbert’s Foundations of Geometry, are assumed. Neither theaxioms of continuity—Archimedean axiom and the axiom of completeness—nor the par-allel axiom need to hold for an arbitrary Hilbert plane. The Uniformity Theorem bringsconfidence that these few axioms are a correct start towards a natural axiomatizationof full-blown Euclidean geometry.

Which postulates do we want to be added to the Hilbert plane axioms? This sectiongives an account of some recent achievements, mainly of M.J. Greenberg in the axiom-atization of geometry. We begin with the axioms of a Hilbert plane, and successivelypostulate few additional axioms with the effort to arrive at a system as close as possibleto Euclid’s system. By further axioms—mainly of continuity—the system is narrowed

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to a system that has the real Cartesian plane as its unique model.The Uniformity Theorem motivates the following definition:

Definition 8.1 (Three basic types of Hilbert planes). According to the three casesoccurring in the Uniformity Theorem—

A semi-Euclidean plane is a Hilbert plane for which the angle sum of every triangleis two right angles.

A semi-hyperbolic plane is a Hilbert plane for which the angle sum of every triangleis less than two right angles.

A semi-elliptic plane is a Hilbert plane for which the angle sum of every triangle islarger than two right angles.

In the next step, we postulate the uniqueness of parallels.

Definition 8.2 (Pythagorean plane). A Pythagorean plane is a Hilbert plane forwhich the Euclidean parallel postulate holds.

In a third step, we need to add some axiom justifying the classical use of ruler andcompass.

Definition 8.3 (Euclidean plane). A Euclidean plane is a Hilbert plane for whichboth the Euclidean parallel postulate and the circle-circle intersection property hold.

In a fourth step, we use Archimedes’ axiom in order to justify measurements. Finally,we introduce a suitable axiom of completeness.

Definition 8.4 (Archimedean plane). A Archimedean plane is a Hilbert plane forwhich the Archimedean Axiom hold.

Definition 8.5 (Real Euclidean plane). A real Euclidean plane is a Hilbert planefor which both the Euclidean parallel postulate and the Dedekind axiom hold.

Main Theorem. We thus get a chain of theories of strictly increasing strengths:

The real Euclidean plane is a Euclidean plane.

Any Euclidean plane is a Pythagorean plane.

Any Pythagorean plane is a semi-Euclidean plane.

Any semi-Euclidean plane is a Hilbert plane.

The converse does not hold in any of these steps.

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8.3 Wallis’ Axiom

Already in the section on Legendre’s theorem, we have addressed the question to finda clear cut, suggestive or self-evident postulate that would better replace the Euclideanpostulate.

John Wallis (1616-1703), in a treatise on Euclid published in 1693, was astute enoughto propose a new postulate that he believed to be more plausible than Euclid’s parallelpostulate. He phrased it as follows:

Finally (supposing the nature of ratio and of the science of similar figuresalready known), I take the following as a common notion: to every figurethere exists a similar figure of arbitrary magnitude.

We want to make Wallis’ idea precise in the context of axiomatic geometry, beginningwith Hilbert’s axiom. Euclid’s theorems about similar triangles depends on the Archi-medean axiom—via his definition of equality of ratios. Hence it is better to restrictattention to equiangular triangles instead of similar figures. Recall that equiangulartriangles have three pairs of congruent angles. Both Greenberg and Hartshorne suggesta modification of Wallis’ axiom along these lines.

(Wallis-Greenberg Postulate). Given any triangle 4ABC and segment DE thereexists a triangle with DE as one side such that the triangles 4ABC and 4DEF areequiangular.

(Wallis-Hartshorne Postulate). Given any triangle 4ABC and segment DE thereexists a triangle 4A′B′C ′ have side A′B′ ≥ DE such that the triangles 4ABC and4A′B′C ′ are equiangular.

In this paragraph, we show the following theorem:

Theorem 8.1. The following three postulates are equivalent in any Hilbert plane:

(a) the Euclidean parallel postulate

(b) the Wallis-Greenberg postulate

(c) the Wallis-Hartshorne postulate

Proposition 8.1. The Euclidean parallel postulate implies the Wallis-Greenberg postu-late.

Proof. Given is triangle 4ABC and segment DE. We transfer segment AB onto the

ray−−→DE and get the congruent segments AB ∼= DE ′.By the extended ASA-theorem, we construct the congruent triangles 4ABC ∼=

4DE ′F ′. Indeed, as already explained in the proof of the extended ASA-theorem, the

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Figure 8.7: Uniqueness of parallels implies Wallis’ postulate.

ray rD forming the angle ∠BAC with ray−−→DE and the ray r′E forming the angle ∠ABC

with ray−−→E ′D, both constructed in the same half-plane of line DE, intersect in point F ′.

Additionally, we construct the ray rE forming the same angle ∠ABC with ray−−→ED,

again in the same half-plane of line DE. The rays rE and r′E are parallel by [Euclid I.27],see 3.33. The ray rD intersects one of these parallel rays. Hence, as already explainedin Problem 5.1, it intersects the line of the second ray rE, too. Indeed, otherwise, theline of rE would have the two different parallels rD and r′E both through point F ′.

Let now F be the intersection point of rD and rE. The triangles 4ABC and 4DEFhave two pairs of congruent angles at vertices A and D, as well as vertices B and E, byconstruction.

We have seen in Proposition 5.2 above that uniqueness of parallels implies that everytriangle has angle sum 2R—every Pythagorean plane is semi-Euclidean.

Hence the triangles 4ABC and 4DEF have a third pair of congruent angles atvertices C and F and are equiangular, as to be shown.

Proposition 8.2. The Wallis-Hartshorne postulate implies the Euclidean parallel pos-tulate.

Proof. Given is line l and a point P not on l. One parallel can be obtained as the ”doubleperpendicular” as already explained in Proposition 5.1. One drops the perpendicular pfrom point P onto line l and denotes the foot point by F . Next, one erects at point Pthe perpendicular to line PF . Thus, one gets a line m parallel to the given line l.

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Figure 8.8: Wallis’ postulate implies uniqueness of parallels.

We now suppose towards a contradiction that there exists a second line t throughpoint P which is parallel to line l, too. Let

−→t be the ray on line t with vertex P and

lying between the two parallels l and m. We choose any point Q 6= P on this ray, anddrop the perpendicular onto the line PF . The foot point R lies in the segment PF ,since lines t and l do not intersect.

We now apply the Wallis-Hartshorn axiom to the triangle 4PQR and the segmentPF . Hence there exists an equiangular triangle 4P ′Q′R′ with side P ′R′ ≥ PF .

We transfer this segment and then the triangle onto the ray−→PF and get a triangle

4PQ′′R′′, which is congruent to 4PQ′R′ and hence equiangular to triangle 4PQR.We can put points Q and Q′′ on the same side of line PF . Because of congruence of theangles α = ∠R′′PQ′′ ∼= ∠RPQ, the uniqueness of angle transfer imply that these rays

are equal:−→PQ =

−−→PQ′′. Hence point Q′′ lies on the line t = PQ.

Since PR′′ ≥ PF , we conclude that either R′′ = F or R′′ and P lie on differentsides of F . We consider the second case. The foot points R and R′′ lies on differentsides of F and hence the line l. The lines RQ, l and R′′Q′′ are parallel, being all threeperpendicular to PF . Hence points R and Q lie on one side of line l and points R′′ andQ′′ lie on the other side. Hence the segment QQ′′ intersects the line l, say in point T .In the first case, we conclude that point Q′′ =: T lies on line l.

Hence, in both cases, the lines l and t do intersect. We have confirmed no secondparallel t to line l through point P can exist.

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8.4 Proclus’ Theorem

What is the missing link leading from the semi-Euclidean plane to a plane with unique-ness of parallels? Based on Proclus commentaries to Euclid, Greenberg has suggestedthe following angle unboundedness axiom. This axiom goes back to Aristole’s book I ofthe treatise De Caelo (”On the heavens”).

Definition 8.6 (Aristole’s Angle Unboundedness Axiom). For any acute angleθ and any segment PQ, there exists a point X on one side of the angle such that theperpendicular XY dropped onto the other side of the angle is longer than the givensegment: XY > PQ.

Main Theorem (Proclus’ Theorem). A Hilbert plane is Pythagorean if and only ifit is semi-Euclidean and Aristole’s axiom holds.

We have seen in Proposition 5.2 above that uniqueness of parallels implies that everytriangle has angle sum 2R. Hence every Pythagorean plane is semi-Euclidean. Indeed,uniqueness of parallels implies Aristole’s axiom, too.

Proposition 8.3. Hilbert’s parallel postulate implies Aristole’s axiom.

Figure 8.9: Uniqueness of parallels implies unbounded opening of an angle.

Proof. Let the acute angle θ = ∠(m,n) and a segment be given. We transfer the segmentonto the perpendicular erected at the vertex of the angle onto the side m and get thesegment PQ. Next we erect the perpendicular l onto line PQ at Q. The lines l and mare parallel since they are both perpendicular to line PQ.

Because uniqueness of parallels has been assumed, the lines n and l are not parallel.Let S be their intersection point. Take any point X on the line n such that S is between

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P and X. We drop the perpendicular from X onto line m and let Y be the foot point.Since X and Y are on different sides of line l, the segment XY intersects the line l. CallT the intersection point.

The Lambert quadrilateral QTY P is indeed a rectangle, as shown by the Unifor-mity Theorem. Its opposite sides are congruent by Lemma 8.1. Hence we get

PQ ∼= Y T < Y X

as claimed by Aristole’s angle unboundedness axiom.

By Proposition 8.3 and Proposition 5.2, we know that Aristole’s Axiom and theangle sum 2R for triangles are both necessary for uniqueness of parallels to occur. Tocomplete the proof of Proclus’ Theorem, we now show that these two assumptions aresufficient.

Proposition 8.4. A semi-Euclidean Hilbert plane, for which the Aristole’s Axiom holdsis Pythagorean.

Figure 8.10: The second line n 6= m, different to the double perpendicular m, drawn throughpoint P is along the side of a triangle 4PXR, and hence by Pasch’s axiom intersects line l.

Proof. Given is a line l and a point P not on this line. As explained in Proposition 5.1,we use the standard ”double perpendicular” m to get a parallel to line l through pointP . Let Q be the foot point of the perpendicular dropped from point P onto line l, andlet m be the perpendicular erected at point P onto the first perpendicular PQ.

We need to show that m is the unique parallel through point P to line l. Take anyother line n through point P . Let θ = ∠(rm, rn) be the acute angle between two rayson the lines n and m from vertex P . We can choose these rays such that the angle θ is

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firstly acute, and secondly, the interior of the angle lies in the same half plane of line mas line l.

We now explicitly prove existence of an intersection point of ray rn and line l. LetXY > PQ be the segment between the sides of angle θ = ∠XAY , as postulated byAristole’s Axiom. We can choose point X to lie on the ray rn, and drop the perpendicularfrom point X onto the line PQ, obtaining the foot point R. We have obtained a Lambertquadrilateral RXY P . In a semi-Euclidean plane, this is a rectangle. Its opposite sidesare congruent by Lemma 8.1. Hence we conclude

PR ∼= XY > PQ

and points P and R lie on different sides of point Q.We now use Pasch’s axiom for line l and triangle 4PXR. Line l intersects the

side PR in point Q and is parallel to side RX, since the lines l and RX are bothperpendicular to PQ. Hence by Pasch’s axiom, the line l intersects the third side PXof the triangle, say in point S.

Thus we have explicitly checked existence of an intersection point lines n = PX withline l, as to be shown.

8.5 More about Aristole’s Axiom

Proposition 8.5. In a semi-elliptic Hilbert plane, Aristole’s Axiom does not hold. Es-pecially, it does not hold for the angle between any two parallels to the same line, norfor any angle given by the excess of the angle sum of a triangle over two right angles.

Figure 8.11: For the angle between two parallels m and n to line l, intersecting at P , thedistance from a point on one side to the other side of angle ∠(m,n) is always smaller than theperpendicular PQ.

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Proof. Given is a line l and a point P not on this line. Again, let Q be the foot pointof the perpendicular dropped from point P onto line l, and let m be the perpendicularerected at point P onto the first perpendicular PQ. As explained in Proposition 5.1,”double perpendicular” m is a parallel to line l through point P .

Let n be a second parallel to line l through point P . Take any point X on the linen such that X and Q lie on the same side of line m. We drop the perpendicular frompoint X onto the line PQ and let Z be its foot point. All three line m, ZX and l areperpendicular to PQ, and hence parallel. Hence points X and Z both lie between theparallels l and m. Hence point Z lies between P and Q, and hence.

PQ > PZ

The perpendicular from point X onto the line m has foot point Y . Thus we haveobtained the Lambert quadrilateral ZXY P . In a semi-elliptic plane, its fourth angleχ = ∠ZXY is obtuse. From Lemma 8.1 we conclude

PZ > XY

Thus we have obtained that PQ > PZ > XY holds for any point X on one of therays from P on line n. Hence the distance from any point of line n to line m is strictlysmaller than PQ, and Aristole’s angle unboundedness axiom does not hold.

Proposition 8.6. In a semi-Euclidean as well as in a semi-hyperbolic Hilbert plane,Archimedes’ Axiom implies Aristole’s Axiom.

Figure 8.12: Doubling the distance from a point on one side to the other side of angle θ.

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Proof. Let the acute angle θ = ∠(m, n) be given. We choose any point C on side

m =−→AC, and drop the perpendicular onto n to get the foot point D. In order to double

the segment CD, we construct point E such that C is the midpoint of segment DE andpoint F on ray m such that C is the midpoint of segment AF . The triangles

4ACD ∼= 4FCE

are congruent by SAS-congruence, because of the vertical angles at point C and twopairs of congruent sides AC ∼= CF and DC ∼= CE.

Because of the triangle congruence, ∠CDA ∼= ∠CEF = ∠DEF are both rightangles. We drop the perpendicular from point F onto the other side n and get footpoint G.

Now DEFG is a Lambert quadrilateral, with an acute or right angle ∠EFG. ByLemma 8.1 either side adjacent to the acute angle is longer than the respectively oppositeside. Hence we conclude

FG ≥ ED = 2 · CD

We need to assure that doubling the original segment CD is sufficient to get another seg-ment XY longer than any arbitrarily segment PQ. To this end, we use the ArchimedeanAxiom. Let FG := C1D1. By induction, we can construct segments

CnDn ≥ 2n · CD > n · CD

for all natural numbers n.Given any segment PQ, the Archimedean axiom tells there exists a natural number

n such thatn · CD > PQ

Define XY := CnDn for such a number n. Both inequalities together imply

XY > n · CD > PQ

as required for Aristole’s angle unboundedness axiom to hold.

From Proposition 8.5, the first Legendre Theorem and the Uniformity Theorem wecan recapitulate:

Corollary. In a semi-elliptic Hilbert plane, neither Aristole’s Axiom nor ArchimedesAxiom does hold.

Together with Proposition 8.6, we get the remarkable result—for which it would benice to have a direct proof:

Corollary. In every Hilbert plane, the Archimedean Axiom implies Aristole’s Axiom.

Too, from Proposition 8.6 and Proposition 8.4, we get once more the third LegendreTheorem—the direct proof of which we have given earlier.

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Figure 8.13: Existence of asymptotically parallel rays implies Aristole’s Axiom.

Corollary. A semi-Euclidean plane for which the Archimedean axiom holds is Pythagorean.

Proposition 8.7. In a semi-hyperbolic Hilbert plane, the existences of asymptoticallyparallel rays implies Aristole’s Axiom.

Proof. Let the acute angle θ = ∠(m, n) be given. Reflection across the side n yieldsthe double angle 2θ = ∠(m,m′). As shown in the proof of Hilbert’s foundation ofhyperbolic geometry, the two rays m and m′ have an inclosing line l. One checks that land n intersect perpendicularly, say at point P .

Let p be the ray on l with vertex P which is asymptotically parallel to ray m. Let µdenote their common end. We transfer any given segment onto this ray to produce the(arbitrarily long) segment PQ. We drop the perpendicular from Q onto ray m. Sincewe are staying on one side of n, we obtain the foot point F . The angle ∠FQµ is anacute angle of parallelism.

Hence the angle ∠PQF is obtuse. We erect onto line l the perpendicular ray g atQ inside this angle. This ray is inside the angle ∠PQF , and by the crossbar theorem,intersects segment PF , say at point G.

We apply Pasch’s axiom to triangle4APF and line g, where A is the vertex of angleθ. Since g and n are both perpendicular to PQ, they are parallel. But ray g and sidePF intersect in point G. Hence by Pasch’s axiom, line g intersects the third side AF .Call X the intersection point.

We drop the perpendicular from point H onto line n and obtain the foot point Y .Finally, we have the Lambert quadrilateral QXY P , with an acute angle χ = ∠QXY

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at vertex X. By Lemma 8.1 either side adjacent to the acute angle is longer than therespectively opposite side. Hence we conclude

XY > PQ

as required for Aristole’s angle unboundedness axiom to hold.

Proposition 8.8. Given is a semi-hyperbolic Hilbert plane, for which Aristole’s Axiomholds.

It two lines have a common perpendicular, then the distance between two lines isarbitrarily large. In other words, for any given segment, there exists a point on one ofthe two lines from which the distance to the other line is longer than the given segment.

Corollary. In a hyperbolic plane, the distance between two lines with a common per-pendicular is arbitrarily large.

Figure 8.14: In a hyperbolic plane, the distance between two lines with a common perpen-dicular is arbitrarily large.

Proof. Let PQ the segment on the common perpendicular from line m onto line l. Fromany second point A of line m, we drop the perpendicular onto line l and get the footpoint B. Now QBAP is a Lambert quadrilateral, with an acute angle ∠PAB.

We drop the perpendicular from point P onto line AB and get the foot point C.We have produced a second Lambert quadrilateral QBCP , which has an acute angle

∠QPC. The ray−→PC lies inside the right angle ∠QPA, and hence point C lies between

A and B.We use Aristole’s axiom for the angle θ = ∠APC. Hence there exists a point X on

the side−→PA for which the segment XY dropped onto the other side is longer than the

given segment ST :XY > ST

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We drop from point X the perpendicular onto line l and get the foot point D. Thepoints X and D lie on different sides of line PC, since points A and X lie on one side,and the points Q, B and D lie on the other side. Hence the segment XD intersects linePC in a point E, and

XD > XE

The segment XE is the hypothenuse of the right triangle 4XY E, and hence its longestside

XE > XY

Hence for any given segment ST , there exists a point X on line m such that the distance

XD > XE > XY > ST

from point X the foot point D on line l is longer than the given segment ST .

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Part II

Euclidean Geometry

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1 Some Euclidean Geometry of Circles

Some interesting properties of circles and rectangles in Euclidean geometry are investi-gated in this section. The material is from book III of Euclid, and includes furthermorethe converse of Thales’ theorem, and the basic properties of rectangles. Recall that inEuclidean geometry, the parallel axiom and its consequences are now assumed to hold.

1.1 Thales’ Theorem

Figure 1.1: Thales Theorem

Theorem 1.1 (Thales’ Theorem). ”The angle in a semicircle is a right angle.” Moreprecisely stated: If an angle has its vertex C on a circle, and its sides cut the circle atthe two endpoints A and B of a diameter, then angle γ = ∠ACB is a right angle.

Thales’s lived ca. 624-546 B.C., in Miletus, a Greek island along the coast of AsiaMinor. These dates are known rather precisely, because, as reported by Herodotus,he predicted a solar eclipse, which has been determined by modern methods to haveoccurred on May 28th of 585 B.C.

Tradition names Thales of Miletus as the first Greek philosopher, mathematician andscientist. He is known for his theoretical as well as practical understanding of geometry.Most important, he is credited with introducing the concept of logical proof for abstractpropositions.

Traditions surrounds him with legends. Herodotos mentioned him as having pre-dicted a solar eclipse, which put an end to fighting between the Lydians and the Medes.

Aristotle tells this story about him:

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Once Thales somehow deduced that there would be a great harvest of olivesin the coming year; so, having a little money, he gave deposits for theuse of all the olive presses in Chios and Miletus, which he hired at a lowprice because no one bid against him. When the harvest-time came, andmany were wanted all at once and of a sudden, he let them out at anyrate he pleased, and made a quantity of money.

Plutarch tells the following story:

Solon who visited Thales asked him the reason which kept him single. Thalesanswered that he did not like the idea of having to worry about children.Nevertheless, several years later Thales anxious for family adopted hisnephew Cybisthus.

Thales went to Egypt and studied with the priests. While he was in Egypt, he wasable to determine the height of a pyramid by measuring the length of its shadow at themoment when the length of his own shadow was equal to his height. Thales is said tohave proved some simple theorems of geometry, as well as the not so obvious theoremabout the right angle in a semicircle. As stressed by David Park [?], the story raises animportant point, whether or not Thales really invented the proof:

Babylonians and Egyptians had a number of mathematical tricks. For ex-ample, Babylonians knew this proposition, as well as the Pythagoreantheorem a thousand years before Thales and Pythagoras found them. Ifthey were known, they must have been proved, but there is no sign thatanyone thought the proofs were important enough to preserve. Whoeverset the process of proof at the center of the stage is the founder of all themathematics since then, and if it is not Thales it was someone who livednot long afterwards.

Proof of Thales’ Theorem in modern manners. Draw 4ABC and, as an extra for theproof, the line from the center O to the vertex C. The base angles of an isosceles triangleare equal by Euclid I.5 . In modern parlance, we say:

The base angles of an isosceles triangle are congruent.

We use that fact at first for 4AOC. Hence

α = ∠OAC ∼= ∠OCA

Secondly, we use Euclid I.5 for 4COB. Hence

β = ∠OBC ∼= ∠OCB

By angle addition at vertex C

(1.1) γ = ∠ACB = α + β

Next we use Euclid I.32, which tells us:

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The sum of the angles in a triangle is two right angles.

Because α, β, γ are just the angles in 4ABC, we conclude that

(1.2) α + β + γ = 2R

I have used the letter R to denote a right angle. Pulling formulas (1.1) and (3.2) togetheryields γ + γ = α + β + γ = 2R, and hence γ = R , as to be shown.

If a Mathematician learns such a nice theorem, as Thales’ theorem is for sure, what doeshe want to do with it? 19 How does mathematics and other clever people benefit fromit, after putting all issues of priority aside? Here are some general considerations:

Generalize the theorem. This can mean either getting along with less assumptions,or just putting the given assumptions into a more general context.

Simplify the statement of the theorem.

Sharpen the conclusions. Expressing the conclusions in another way and simplifyingthe statement can both help to sharpen them.

Ask whether a converse holds. There can be more than one version for the con-verse, depending one how the theorem is formalized. Too, in case the conversefails to be true, one can ask for similar statements of which the converse doeshold.

Consider special cases. They can look surprisingly different—for example, some as-sumption may turn out to be too obvious to be stated explicitly.

Find applications. Are there constructions or algorithms which follow from the the-orem.

Simplify the proof of the theorem.

Build a theory. Find the natural place for the theorem in a larger context.

10 Problem 1.1 (Tangents to a circle). Given is a circle C, with center O anda point P outside of C. Construct the tangents from point P to the circle C. Actuallydo and describe the construction!

Construction 1.1 (Tangents to a circle). One begins by constructing a second circleT with diameter OP . (I call this circle the Thales circle over the segment OP ). TheThales circle intersects the given circle in two points T and S . The lines PT and PSare the two tangents from P to circle C.

19Well, in case he has discovered the theorem himself, one should publish it—with the possibility inmind that somebody else has already discovered something similar before.

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Figure 1.2: Tangents to a circle

Validity of the Construction. By Thales theorem, the angle ∠PTO is a right angle,because it is an angle in the semicircle over diameter OP . Since point T lies on thecircle C, too, the segment OT is a radius of that circle. By Euclid III. 16,

The line perpendicular to a diameter is tangent to a circle.

Since TP is perpendicular to the radius OT , and hence to a diameter, it is a tangent ofcircle C.

Question. Which one of construction (1.1) and construction (??) —explained in thesection on neutral geometry of circles and continuity—remains valid in hyperbolic ge-ometry? Explain why.

Answer. Construction (1.1) is no longer valid in hyperbolic geometry. On the otherhand, construction (??) remains valid in hyperbolic geometry.

Construction (1.1) is not valid in hyperbolic geometry, because the angle sum of atriangle is less than two right, and hence Thales’ theorem does not hold in hyperbolicgeometry.

On the other hand, construction (??) depends only on SAS-congruence, and thefact that tangent and radius being perpendicular. These statements hold in hyperbolicgeometry, too.

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10 Problem 1.2. Do and describe a Euclidean construction to find the perpen-dicular p to a given line l through a given point P lying on l which depends on Thales’theorem.

Figure 1.3: Erect the perpendicular via Thales’ Theorem. The numbers indicate the orderof the steps.

Answer. One chooses an arbitrary point O not on line l, and draws a circle c around itthrough point P . This circle intersects the given line l at point P , and a second point,

which is called A. Next one draws the line←→OA. It intersects the circle c at A and a

second point, called B. Finally, the line←→BP is the perpendicular to be erected on line

l at point P .

Theorem 1.2 (A strengthening of Thales’ Theorem). Given is a triangle 4ABC,and a circle with its side AB as diameter.

(i) If the third vertex C of the triangle lies inside the circle, the angle at vertex C isobtuse.

(ii) If the third vertex C of the triangle lies outside the circle, the angle at vertex C isacute.

Corollary (The Converse of Thales’ Theorem). Given is a triangle 4ABC anda circle C with its side AB as diameter. If the triangle is right, the vertex C with theright angle lies on the circle with the hypothenuse AB as diameter.

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Proof. Following Euler’s convention, let α = ∠BAC, β = ∠ABC and γ = ∠ACBdenote the angles at the vertices A, B and C, respectively. Let O be the midpoint ofthe triangle side AB. In the figures on page 161 and page 161, there are given drawingsof the 4ABC and the circle for the two cases:

(i) OC < OA or

(ii) OC > OA.

We use Euclid I.18:

If one side of a triangle is greater than another, then the angle

opposite to the greater side is greater than the angle opposite

to the smaller side.

First consider the case that (i) OC < OA holds. Using Euclid I.18 for 4AOC, weconclude that

(1.3) α < α′ = ∠OCA

Now OA ∼= OB, since O is the midpoint of the hypothenuse AB. Hence, because of (i),OC < OB holds, too. Using Euclid I.18 once more, this time for triangle 4BOC , weconclude that

(1.4) β < β′ = ∠OCB

Adding (1.3) and (1.4) yields α + β < α′ + β′. Angle addition at vertex C yieldsα′ + β′ = γ. Now we use that the sum of the angles in a triangle is two right angles, asstated in Euclid I.32. Hence

(1.5) 2R = α + β + γ < α′ + β′ + γ = 2γ

and hence γ > R. Hence the triangle is obtuse, as to be shown.By a similar reasoning, one shows that in case (ii), the assumption OC > OA implies

that the angle γ is acute.

10 Problem 1.3. Provide a drawing for case (i). Provide two drawings for case(ii), one where the triangle 4ABC is acute, and, as a catch, a second one where the4ABC is obtuse, nevertheless. (The triangle is obtuse because of a different obtuseangle.)

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Figure 1.4: A strengthening of Thales’ Theorem, with vertex C inside the circle

Figure 1.5: A strengthening of Thales’ Theorem, with vertex C outside the circle

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Figure 1.6: A strengthening of Thales’ Theorem, still another case with vertex C outsidethe circle

1.2 Rectangles and the Converse Thales Theorem

As already stated in the section on Legendre’s Theorems, a rectangle is defined to be aquadrilateral with four right angles. The segments connecting the opposite vertices arecalled the diagonals of the rectangle. Obviously, a diagonal bisects a rectangle into tworight triangles. They turn out to be congruent. Conversely, one can built a rectanglefrom two congruent right triangles. At first, we use this idea for an independent proof ofthe converse of Thales’ Theorem 1.1. Secondly, the same idea helps to prove the ratherobvious, but important properties of a rectangle.

Independent proof of the converse of Thales’ Theorem 1.1. Let4ABC be the given righttriangle, with the right angle at vertex C. The idea of this independent proof is con-structing a rectangle from the right triangle.

Step 1: We transfer the angle α = ∠BAC to vertex B in order to produce the congruentangle ∠ABD = α on the side of hypothenuse AB opposite to vertex C. Furthermore,we transfer the segment AC to obtain the congruent segment BD ∼= AC on the newlyproduced ray.

Question. Show the congruence

(up-down) 4ABC ∼= 4BAD

Which congruence theorem do you use?

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Figure 1.7: (up-down) 4ABC congruent 4BAD (left-right) 4CAD congruent 4DBC(scissors) 4AMC congruent 4BMD

Answer. This follows by SAS congruence. Indeed, the two triangles have the commonside AB, the two sides AC ∼= BD are congruent, and the angles ∠BAC ∼= ∠ABD arecongruent, both by construction.

Step 2: The quadrilateral ADBC obtained from the two triangles has remarkablesymmetry. It is now bisected along its other diagonal CD to obtain a second pair ofcongruent triangles

(left-right) 4CAD ∼= 4DBC

Question. Explain how this congruence is shown.

Answer. This follows by SAS congruence. Because of the first congruence (up-down),we have two pairs of congruent sides: AC ∼= BD by construction, and BC ∼= AD asa consequence of step 1. Furthermore, there are two pairs of congruent z-angles: theangles ∠BAC ∼= ∠ABD = α are congruent by construction. The angles ∠ABC ∼=∠BAD = β are congruent as a consequence of (up-down). Hence angle addition yields∠CAD ∼= ∠DBC. Finally, we get the claim (left-right) via SAS congruence.

The congruence (left-right) now yields two further pairs of congruent z-angles:

∠ACD ∼= ∠BDC = α′ and ∠CDA ∼= ∠DCB = β′

Step 3: Since points C and D lie on different sides of line AB, the segment CDintersects this line at midpoint M .

Question. Use the congruence

(scissors) 4AMC ∼= 4BMD

in order to show that the diagonals AB and CD bisect each other at their commonmidpoint M . By which theorem is this congruence proved?

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Answer. The congruence (scissors) is obtained with the SAA congruence theorem. Touse this theorem, we need the congruent sides AC ∼= BD, the adjacent pair of congruentz-angles α′, and the vertical angles at vertex M . Hence one gets the congruent triangles(scissors). Finally, we get the segment congruences

AM ∼= BM = m′ and CM ∼= DM = m

Clearly point M lies between C and D, because these two points are on differentsides of line AB by construction. But it is quite hard to confirm that M lies betweenA and B. Indeed, the congruence AM ∼= BM implies that point M lies between A andB. 20

Figure 1.8: The congruence of 4BCA to 4CBD implies that the diagonals of a rectangleare congruent

Step 4: We now get to the part of the reasoning which is valid only in Euclideangeometry. Because the angle sum in triangle 4ABC is two right, and the angle atvertex C is assumed to be a right angle, we conclude that α + β = R. Similarly, oneconfirms that the quadrilateral ADBC has a right angle at vertex B. Hence it is arectangle.

The right angle at vertex A yields still another pair of congruent triangles

(overlap) 4ABC ∼= 4DCB

20See also figure on page 93 which is taken from the millenium edition of ”Grundlagen der Geometrie”,page 26.

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Question. Which theorem do you use to confirm this claim?

Answer. This follows by SAS congruence. Indeed, the two triangles have the commonside BC = CB, the two sides AC ∼= DB are congruent by construction, and the angles∠BCA and ∠DBC are both right. Indeed, ∠BCA was assumed to be a right angle,whereas

∠DBC = ∠DBA + ∠ABC ∼= α + β = R

because of congruent z-angles α by construction, and the angle sum.

As a consequence of the congruence (overlap), we get AB ∼= CD. Thus we haveshown that the diagonals of the quadrilateral ADBC are congruent.

Step 5: This is the final step to get the converse Thales Theorem. Because the diagonals

AB ∼= CD

are congruent, and point M bisects them both:

AM ∼= MB and CM ∼= MD

Hence all four segments from point M to the vertices A, B, C and D are congruent.Hence vertex C lies on a semicircle with diameter AB, as to be shown.

Remark. As a variant for the final step 5, we compare the distance AM ∼= BM = m′

with CM ∼= DM = m.Assume that m′ > m. Across the longer side of a triangle lies the larger angle.

Using this fact for triangle 4AMC, we conclude that α′ > α. Similarly, using triangle4BMC, we get β′ > β. Hence by addition, we get R = α′ + β′ > α + β. Hence theangle sum in the right triangle 4ABC would be less than two right angles, which isfalse in Euclidean geometry.

Similarly, the assumption m′ < m leads to a contradiction, too. The only remainingpossibility is m = m′. Hence vertex C lies on a semicircle with diameter AB and centerM , as to be shown.

10 Problem 1.4. Which steps are still valid in hyperbolic geometry. Comparethese quantities: m with m′, α with α′, and β with β′.

Answer. Steps 1,2 and 3 are still valid. We still get α′ +β′ = 90, because the construc-tion was started with a right triangle 4ABC with right angle at vertex C. But, becausethe angle sum of a triangle is less than two right angles, one gets α + β < 90. Sincem < m′ occurs if and only if α < α′ and β < β′, these inequalities are all three true.The figure on page 166 indicates the differing angles and segments using different colors.Remarkably enough, one can take the proof a step further without appeal to Euclideangeometry: The perpendiculars dropped from M onto the four sides of the quadrilateralADBC lie on two lines, but these two lines are not perpendicular to each other.

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Figure 1.9: What steps 1,2 and 3 yield in hyperbolic geometry

We shall now recapitulate the same ideas, in order to derive the basic properties ofa rectangle.

Corollary (The basic properties of the rectangle). Given is a rectangle. This isa quadrilateral, about which it is only assumed that it has four right angles.

(i) A diagonal partitions the rectangle into two congruent triangles.

(ii) The opposite sides of a rectangle are congruent.

(iii) The opposite sides of a rectangle are parallel.

(iv) The two diagonals of a rectangle bisect each other.

(v) The diagonals of a rectangle are congruent.

(vi) The diagonals of a rectangle intersect at the center of its circum circle.

Proof. By definition, a rectangle is just a quadrilateral with four right angles—that isall what is assumed. The diagonal AB bisects the rectangle ADBC into two righttriangles 4ABC and 4ABD. Again, let α = ∠CAB and β = ∠ABC be the angles ofthe first of these triangles at vertices A and B, respectively.

We now use the fact that the angle sum in a triangle is two right angles. 21 Henceα+β +R = 2R and α+β = R. The given rectangle has a right angle at vertex A, hence

21Actually, while proving the second Legendre Theorem, we have shown that the existence of rectanglealready implies this fact about the angle sum.

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angle subtraction at vertex A now confirms that the second lower triangle 4ADB hasthe ∠BAD = R− α = β at vertex A.

Thus we have obtained a pair of congruent z-angles β = ∠ABC ∼= ∠BAD. Sim-ilarly, we get a second pair of congruent z-angles α = ∠BAC ∼= ∠ABD. With theASA congruence theorem, we can now confirm the two triangles 4ABC ∼= 4BAD arecongruent. Finally, the construction done above can be used once more to rebuild therectangle from these two congruent right triangles.

We see that a diagonal partitions the rectangle into two congruent right triangles,and the opposite sides of the rectangle are congruent. Too, they are parallel because ofthe congruent z-angles.

The two diagonals of a rectangle are congruent because of

(overlap) 4BCA ∼= 4DAC

Let M be the intersection point of the diagonals. The congruence (overlap) implies thatthe triangle 4AMC has congruent base angles α. Hence the converse isosceles theoremimplies MA ∼= MC. Similarly, one can show that the triangle 4BMC has congruentbase angles β and the converse isosceles theorem implies MC ∼= MB. Hence the twodiagonals of a rectangle bisect each other. and their intersection point is the center ofits circum circle.

10 Problem 1.5. In Euclidean geometry, one defines a parallelogram to be aquadrilateral, the opposite sides of which are parallel. Which parts of the Corollary 1.2are true for all parallelograms, in Euclidean geometry.

Answer. Items (i) true (iv) are true for any parallelogram, but items (v) and (vi) arenot true in general.

1.3 Angles in a Circle

Theorem 1.3 (Angles in a circle). The angle subtending any circular arc—withvertex at the center of the circle—is twice the angle subtending the same arc with vertexon the circle (Euclid III.20). Hence, if two angles inscribed in a circle subtend the samearc, they are congruent (Euclid III.21).

We call the angle with vertex at the center central angle, and the angle with vertexon the circle circumference angle of the given arc. In short, Euclid III.20 says:

The central angle is twice the circumference angle.

Remark. Usually, one considers the case that the arc AB is less or equal half of thecircle. In the special case that the arc AB is equal to a half circle, the chord AB getsa diameter. In that case, we get Thales’ theorem. In the case the arc AB is more thana half circle, the central angle is greater than 180. Allowing such angles, the theoremstill remains valid. This can be shown using Euclid III.22.

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Figure 1.10: Central and circumference angle of a circular arc

10 Problem 1.6. Provide a drawing with appropriate notation. Mark your arcAB in some color, then choose your third point C outside that arc. Mark the circum-ference angle γ = ∠ACB in the same color. Too, mark the central angle ω = ∠AOB.Avoid that the center of the circle lies on any of the chords involved. Prove the claim forthe situation occurring in your drawing. The simpler version of the proof uses base an-gles of isosceles triangles (Euclid I.5), and the exterior angle of a triangle (Euclid I.32).(There are several possibilities, with angle addition or subtraction, but it is enough todo the proof for the situation in your drawing. The other cases are all quite similar.)

Reason for Euclid III.20. It is enough to use just the two isosceles triangles 4BOCand 4COA, each of which has a pair of congruent base angles, called α and β. Inthe case drawn, the two triangles do not overlap, since the center O lies inside triangle

4ABC. One extends ray−→CO to the other side of center O. Let C ′ be any point on

this extension. Now we see the exterior angles χ = ∠AOC ′ and ν = ∠BOC ′ of the twotriangles 4BOC and 4COA. By Euclid I.32:

The exterior angle of a triangle is the sum of the two nonadjacent

interior angles.

We apply this theorem to the isosceles 4AOC. Because the two nonadjacent angles arethe two congruent base angles α, we conclude that

(1.6) χ = 2α

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Figure 1.11: Angles in a circle—still another case

Similarly, using triangle 4BOC, we conclude that

(1.7) ν = 2β

By angle addition at vertex C, one gets γ = α + β. By angle addition at vertex O,one gets ω = χ + ν. (In the case that center O lies outside of 4ABC, one gets anglesubtraction in both cases.) In the end, one gets

(1.8) ω = χ + ν = 2α + 2β = 2γ

as to be shown.

The next two problems deal with Euclid III.22. At first, I do a special case, and thenext problem completes the proof of Euclid III.22.

10 Problem 1.7. Let ABCD be a quadrilateral with a circum circle, and as-sume that two vertices give a diameter BD, and the other two vertices A and C lie ondifferent sides of it. Use Thales’ theorem to show that the sum of the angles at B andD is two right angles. Provide a drawing with appropriate notation. The quadrilateralneed not be a rectangle!

Solution. The diameter BD partitions the quadrilateral into two triangles 4BAD and4BCD. The sum of the angles of the quadrilateral ACBD equals the sum of the

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Figure 1.12: A special quadrilateral

angle sums of the two triangles. Since the angle sum in a triangle is two right angles,the angle sum of a quadrilateral is 4R.

By Thales’ theorem, these are both right triangles, with right angles at vertices Aand C. After subtraction of the two right angles at vertices A and C, the sum of thetwo remaining angles of the quadrilateral at vertices B and D is 2R.

Theorem 1.4 (Euclid III.22). The opposite angles of a quadrilateral ABCD witha circum circle add up to two right angles.

Proof. In the circum circle, either one of the arcs ABC or ADC is less or equal theother one, and hence less or equal a semicircle. We can assume that arc ABC is less orequal a semicircle. Let B2 be the second endpoint of diameter BB2. Now B2 lies on thearc ADC. As shown in the last problem, the angles at opposite vertices B and B2 addup to two right angles in the quadrilateral ABCB2

(1.9) ∠ABC + ∠AB2C = 2R

By Euclid III.21, the circumference angle of arc ABC is

(1.10) ∠ADC ∼= ∠AB2C

Together, we conclude

(1.11) ∠ABC + ∠ADC ∼= ∠ABC + ∠AB2C = 2R

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Figure 1.13: Opposite angles in a quadrilateral with a circum circle

Remark. Here is an argument to get Euclid III.22 at once. We draw the four radialsegments OA, OB, OC and OD and partition the quadrilateral ABCD into fourisosceles triangles. Let p, q, r and s be their respective base angles.

Because of angle addition at each vertex, the sum of the two angles of the quadri-lateral ABCD at the opposite vertices A and C is

(1.12) α + γ = (p + q) + (r + s) = (p + s) + (q + r) = β + δ

Since the sum of angles at all four vertices is α + β + γ + δ = 4R, we conclude

(1.13) β + δ = α + γ = 2R

Theorem 1.5 (Euclid III.32). The angle between a tangent line and a chord is con-gruent to the circumference angle of the arc corresponding to the chord.

10 Problem 1.8. Provide a drawing with appropriate notation.

Proof. Let AC be the chord and T be a point on the tangent at point A, such thatτ = ∠TAC is acute or right. Let AB be a diameter of the circle. By Thales’ theorem,triangle 4ABC is a right triangle. Because the angle sum of a triangle is 2R, its twoangles at vertices A and B, add up to a right angle:

α + β = R

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Figure 1.14: Use isosceles triangles

Figure 1.15: The angle between tangent and chord is congruent to the circumference angle

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On the other hand, the tangent is orthogonal to a diameter by Euclid III.19. Hence, byangle addition at vertex A,

α + τ = R

From these two equation we conclude that τ = β. Thus the angle between the tangentand the chord τ = ∠TAC equals the circumference angle β = ∠ABC of that chord. ByEuclid III.21:

Two angles from points of the circle subtending the same arc are

congruent.

Hence it does not matter that we have chosen a special position for point B.

Remark. Using the notion of limits, there is an easy argument to show Euclid III.32: Inthe limit B → A, one side of the angle ABC becomes the chord AC, and the other sideBA becomes the tangent t to the circle at point A. The circumference angle of chordAC constantly stays β, and becomes the angle between the chord and the tangent t atits endpoint.

The following lemma is remarkable by itself, and shall be used again in the proof ofPappus’ and Pascal’s Theorems.

Figure 1.16: The sides of an angle cutting two circles at the endpoints of their commonchord cut them in two further parallel chords.

Lemma 1.1 (Two Circle Lemma). If the endpoints of the common chord of twocircles lie on two lines, these lines cut the two circles in two further parallel chords.

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Proof. Let C and D be two circles with the common chord CD. Let AB be the secondchord, where the two given lines cut circle C, and EF be the additional third chord,where these two lines cut circle D. We get a circular quadrilateral ABCD inside circleC, and a second circular quadrilateral CDEF inside the second circle D. By EuclidIII.22, the opposite angles of a circular quadrilateral sum up to two right angles.

Question. Formulate this statement as a congruence.

Answer. The interior angle of a convex circular quadrilateral is congruent to the exteriorangle at the opposite vertex.

Question. How can we check whether the segments AB and EF are parallel.

Answer. We check whether they form congruent angles with one of the given lines.

To get a definite picture, we consider the situation drawn in the figure on page 173.In that case, the two given lines intersect in point O, and the points mentioned aboveoccur on the two lines in the order O ∗B ∗C ∗F , and O ∗A∗D ∗E. We need to comparethe two angles ∠OBA and ∠OFE.

The angle ∠OBA is exterior angle in the circular quadrilateral ABCD. By theremark above, it is congruent to the opposite interior angle:

∠OBA ∼= ∠ODC

Now the second angle is exterior angle in the second circular quadrilateral CDEF .Hence it is congruent to the opposite interior angle:

∠ODC ∼= ∠OFE

From these two congruences, we conclude ∠OBA ∼= ∠OFE. Hence, by Euclid I.27, thelines AB and FE are parallel.

Question. Explain how the proof has to be modified in the case drawn in the figure onpage 175.

Answer. In this case, the circular quadrilateral has intersecting opposite sides AB andCD. In this case, the interior angles at opposite vertices are congruent. Hence, theexterior angles at opposite vertices are congruent, too.

1.4 Secands in a Circle

Theorem 1.6 (Theorem of chords (Euclid III.35)). If two chords cut each otherinside a circle, the product of the segments on one chord equals the product of the seg-ments on the other chord.

Assume two chords intersect each other outside a circle. The product of the segments,measured from the point of intersection to the two intersection points with the circle areequal for both chords.

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Figure 1.17: Another example for the two-circle lemma.

10 Problem 1.9. Provide a drawing, with appropriate notation.

Proof using proportions. Let AB and CD be two chords of circle γ intersecting at pointP inside the circle. We have to check whether |PA| · |PB| = |PC| · |PD|. FollowingLegendre, I am using proportions. Equiangular triangles are, by definition, triangles withcongruent angles. Congruent circumference angles, as shown in Euclid III.21, occur atvertices A and D, as well as at vertices C and D. Furthermore, we get congruent verticalangles at vertex P . The vertices of the equiangular triangles have to be listed in suchan order that these congruent angles occur at corresponding vertices. Hence we get theequiangular triangles

4PAC ∼ 4PDB

By Euclid VI.4

The sides of equiangular triangles are proportional.

Hence the ratios of corresponding sides are the same for two equiangular triangles.

(1.14)|PA||PC|

=|PD||PB|

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Figure 1.18: Theorem of chords, case of chords intersecting inside the circle

Now multiplying with the denominators yields

|PA| · |PB| = |PC| · |PD|

which is just the claim of Euclid III.35.The proof for the case of the segments intersecting outside the circle is almost iden-

tical. But note that the two segments QA and QB now overlap.

For the case where the extension of the chords intersect outside the circle, one obtainsa further result by considering the tangent as a limit of a small secant. In the figures onpage 177, we have illustrated the limit C → T,D → T , where T is the touching pointof a tangent drawn from point Q to the circle. Thus one is lead to claim Euclid’s nextresult.

Theorem 1.7 (Theorem of chord and tangent (Euclid III.36)). From a pointoutside a circle, a tangent and a second are drawn. The square of the tangent segmentequals the product of the segments on the chord, measured from the point outside to thetwo intersection points with the circle.

10 Problem 1.10. Provide a drawing, with appropriate notation.

Independent proof, again using proportions. Let Q be a point outside circle γ, let T bethe touching point of the tangent to the circle through Q, and let AB be a chord of thecircle the extension of which runs through Q. We assume that B lies between Q and A.

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Figure 1.19: Theorem of chords, case of chords intersecting outside the circle

Figure 1.20: Getting the limit C → T,D → T

I have to check whether |QT | 2 = |QA|·|QB|. To this end, one compares the triangles4QAT and 4QTB. We use Euclid III.32:

The angle between a tangent line and a chord is congruent to the

circumference angle of the arc corresponding to the chord.

Hence the angles at vertices A and T , for the two triangles, respectively, are congruent.

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Figure 1.21: The common notation for a right triangle

With vertices listed in an order that these congruent angles occur at correspondingvertices, we get equiangular triangles 4QAT ∼ 4QTB. By Euclid VI.4:

The sides of equiangular triangles are proportional.

Hence the ratios of corresponding sides are the same for two equiangular triangles.

(1.15)|QA||QT |

=|QT ||QB|

and multiplying with the denominators yields the result.

10 Problem 1.11. Explain for which special situation Euclid III.35 implies thetheorem about the altitude of a right triangle, usually stated as h2 = pq.

Answer. Take for segment AB a diameter of circle γ and chose CD to be any segmentperpendicular to that diameter. One gets a right 4ABC, with P as foot point ofits altitude. Because of PC ∼= PD and |PC| = |PD|, Euclid III.35 implies pq =|PA| · |PB| = |PC| · |PD| = h2.

10 Problem 1.12. Explain how Euclid III.36 implies the leg theorem a2 = pc.

Answer. In Euclid III.36, as given in theorem 1.7, we choose points A and T on adiameter of circle γ, and point Q on the tangent to the circle at point T . We seethat 4ATQ is a right triangle, because tangent and radius are perpendicular to eachother. The segment AQ intersects the circle in a second point B. The triangle 4ATQ

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Figure 1.22: The altitude theorem for a right triangle

Figure 1.23: The leg theorem

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Figure 1.24: A pair of triangles |4A′QB| .= |4C ′QD| of equal area directly confirms thetheorem of chords

has altitude TB, because Thales’ theorem shows that the angle ∠ABT is right. NowEuclid III.36 tells that the square of one leg QT equals the product of the hypothenusetime the projection of that leg onto the hypothenuse. After renaming T → C, B →P, Q→ B, A→ A, one gets the statement in its usual form a2 = pc.

Remark. In the spirit of Euclid, one has to understand the product of segments occurringin the theorems above as areas of rectangles or squares. The equality of their areas canreally be shown by obtaining one from the other in a finite sequence of cuts and pastes.In the figure below, I want to visualize this interpretation. Take the case of chordsintersecting at point Q outside the circle. We extend the two chords to the other side oftheir intersection point and transfer the segments QA and QC to these opposite rays toobtain a quadrilateral A′ACC ′ with axial symmetry. It is not hard to check that thisquadrilateral has a circum circle and the triangles 4QAC ∼= 4QA′C ′ are congruent.

The two circle lemma implies that the chords A′C ′ ‖ BD are parallel. Hence wehave obtained a new pair of similar triangles

4QA′C ′ ∼ 4PDB

which have the center of similarity Q.The figure also contains two triangles of equal area, which confirm Euclid’s theorem

III.35 directly as an equality among areas. Since A′C ′ ‖ BD, the triangles

|4A′C ′B| .= |4A′C ′D|

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have equal base A′C ′ and equal height, and hence equal areas. Their intersection isthe triangle 4QA′C ′. We subtract this triangle from both sides, and obtain the pair oftriangles

|4A′QB| .= |4C ′QD|

again of equal areas. Now the claim of Euclid III.35—

|QA′| · |QB| = |QC ′| · |QD|

can easily be obtained as an equality among areas.

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2 Pappus’, Desargues’ and Pascal’s Theorems

2.1 Pappus Theorem

Figure 2.1: For Pappus’ configuration: If BC ′ ‖ B′C and AC ′ ‖ A′C, then AB′ ‖ A′B.

Theorem 2.1 (Pappus Theorem). Let A, B, C and A′, B′, C ′ be both three points ontwo intersecting lines, all different from the intersection point. If the lines BC ′ and B′Care parallel, and the lines AC ′ and A′C are parallel, then the lines AB′ and A′B andparallel, too.

Remark. In Hilbert’s foundations [?], this theorem is named after Pascal. Pascal’s nameis now generally associated with the theorem about the hexagon in a circle or conicsection. Following Stillwell’s book [?], I prefer to use the name Pappus for the theoremwhich does not involve any circle or conic section.

Following the third proof of Pappus’ Theorem from Hilbert’s foundations. Define three cir-cles: circle CA through the three points A′, B and C; circle CB through the three pointsA, B′ and C; finally circle CA through the three points A, B and C ′. We keep the nota-tion as in the two-circle Lemma. Let D′ be the intersection point of line OA′ with thecircle CA. Let C∗C be the circle through the three points D′ and A, B. Finally let C∗ bethe intersection point of line OA′ with the circle C∗C . Recall the two-circle Lemma 1.1:

If the endpoints of the common chord of two circles lie on two lines, theselines cut the two circles in two further parallel chords.

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Figure 2.2: Producing parallel chords

Figure 2.3: Three circles help to produce the third pair of parallel segments.

We shall use the two-circle Lemma 1.1 three times.As a first step, the two-circle Lemma 1.1, yields that the chords A′C and AC∗ are

parallel. Hence we conclude that C∗ = C ′, since by assumption, the chords A′C andAC ′ are parallel, too. Hence C∗C = CC , and the salient point D′ lies on both circles CAand CC .

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Figure 2.4: Circles CA and CC intersect the angle in the parallel chords A′C and AC ′

A similar reasoning is now done replacing B 7→ A, B′ 7→ A′ and CA 7→ CB. Thus onegets that the salient point D′ lies on both circles CB and CC .

Figure 2.5: Circles CB and CC intersect the angle in the parallel chords B′C and BC ′

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Question. This fact is indeed just another instance of two-circle Lemma 1.1, and thereasoning above. Use the figure on page 184 and go over all details, once more.

Answer. Let C∗B be the circle through the three points D′ and A, C. Finally, let B∗ bethe intersection point of line OA′ with this circle. By the lemma, the chords BC ′ andB∗C are parallel. Hence B∗ = B′, and the four points D′, B′, C and A lie on a circleC∗B = CB.

Figure 2.6: Finally, circles CA and CB help to confirm that AB′ and A′B are parallel.

Question. Use the Lemma 1.1 a third time, now for the circles CA and CB, and confirmthat A′B ‖ AB′.

Answer. Since point D′ lies on all three circles CA, CB and CC , the circles CA and CBhave the common chord CD′. The remaining intersection points of the two lines ABCand A′B′C ′ with these two circles yields the parallel chords A′B ‖ AB′.

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Theorem 2.2 (Simplified Pappus’ Theorem). Let A, B, C and A′, B′, C ′ be boththree points on two intersecting lines, different from the intersection point. If the linesBC ′ and B′C are parallel, and 4OAC ′ and 4OA′C are similar isosceles the triangles,then the lines AB′ and A′B and parallel, too.

Figure 2.7: Pappus’ theorem in the special case with two isosceles similar triangles 4OAC ′

and 4OA′C

Following the second proof from Hilbert’s foundations. Let D′ be the point of line OA′

such that OB ∼= OD′. Let CB through the three points A, C and D′. Finally let B∗ bethe intersection point of line OA′ with the circle CB.

We confirm that the chords BC ′ and B∗C are parallel. Hence B∗ = B′, and the fourpoints D′, B′, C and A lie on a circle CB.

The congruences ∠OCB∗ ∼= ∠AD′O and ∠OD′A ∼= ∠OBC ′ follow from Euclid’scongruence of circumference angles, applied to the circles CB and SAS-congruence of4OAD′ ∼= 4OBC ′, respectively.

Hence ∠OCB∗ ∼= ∠OBC ′, and Euclid I.27 implies that the lines B′C and B∗C areparallel. This confirms that B∗ = B′, as claimed.

The congruences ∠OBA′ ∼= ∠CD′O and ∠CD′O ∼= ∠OAB′ follow from SAS-congruence and Euclid’s congruence of circumference angles in circle CB, respectively.

Hence ∠OBA′ ∼= ∠OAB′, and Euclid I.27 implies AB′ ‖ AB′.

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2.2 Desargues’ Theorem

Figure 2.8: Desargues’ configuration

Definition 2.1 (Triangles in perspective). Two triangles are in perspective froma point O means that each of the three lines through a pair of corresponding verticespasses through point O.

Theorem 2.3 (Desargues’ Theorem). If two triangles are in perspective, and, fur-thermore, two pairs of corresponding sides are parallel, then the third pair of sides areparallel, too.

Theorem 2.4 (Converse Desargues Theorem). If the sides of two triangles arepairwise parallel, then the two triangles are either in perspective from a point, or thethree lines through pairs of corresponding vertices are parallel.

Question. Convince yourself that Desargues’ Theorem implies the Converse DesarguesTheorem.

Answer.

Question. Convince yourself that the Converse Desargues’ Theorem implies the Desar-gues Theorem.

Answer.

There are remarkably different routes to a proof of this theorem!

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Figure 2.9: Pappus’s Theorem implies Desargues’ Theorem

Theorem 2.5 (Theorem of Hessenberg). For an affine plane, validity of Pappus’sTheorem implies Desargues’ Theorem.

Proof. We shall proof that the second part of Desargues’ Theorem holds under the givenassumptions. Furthermore, we shall assume that the two triangles4ABC and4A′B′C ′

are in perspective from point O, and that AB ‖ A′B′ and AC ‖ A′C ′. We give theproof under the following simplifying assumption:

OB′‖ A′C ′(2.1)

Draw the parallel to line OB through point A. Let L be the intersection point of thisparallel with line A′C ′. Let M be the intersection point of the parallel with line OC.Let N be the intersection point of lines LB′ and AB.

Question. Why do points L, M and N exist?

Answer.

We now use Pappus’ Theorem for three different configurations:Step 1: Use Pappus’ Theorem in configuration ONALA′B′. Because AB = NA ‖A′B′ and AL ‖ B′O, we conclude ON ‖ LA′.

Now ON ‖ LA′ = A′C ′ and A′C ′ ‖ AC imply ON ‖ AC.Step 2: Use Pappus’ Theorem in configuration ONMACB. Because ON ‖ AC andMA ‖ BO, we conclude NM ‖ CB.Step 3: Use Pappus’ Theorem in configuration ONMLC ′B′. Because ON ‖ LC ′ =LA′ and ML ‖ B′O, we conclude NM ‖ C ′B′.

Finally, NM ‖ CB and NM ‖ C ′B′ imply CB ‖ C ′B′, as to be shown.

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Figure 2.10: The case with OABC a parallelogram can be handled, too

Figure 2.11: The Little Pappus Theorem asserts for a hexagon AC ′BA′CB′ that has itsvertices alternating on two parallel lines: if BC ′ ‖ B′C and AC ′ ‖ A′C, then AB′ ‖ A′B.

Theorem 2.6 (Little Pappus Theorem). Let A, B, C and A′, B′, C ′ be both threepoints on two parallel lines. If the lines BC ′ and B′C are parallel, and the lines AC ′

and A′C are parallel, then the lines AB′ and A′B and parallel, too.

Theorem 2.7 (Little Desargues Theorem). If corresponding vertices of two trian-gles lie on three parallel lines, and, furthermore two pairs of corresponding sides areparallel, then the third pair of sides are parallel, too.

Theorem 2.8 (”Little Hessenberg Theorem”). For an affine plane, validity of thelittle Desargues Theorem implies the little Pappus Theorem.

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Figure 2.12: The Little Desargues Theorem asserts for two triangles with vertices on threeparallel lines: if AB ‖ A′B′ and BC ‖ B′C ′, then AC ‖ A′C ′.

Figure 2.13: The Little Desargues Theorem implies the Little Pappus Theorem.

Proof. Given are points A, B, C and A′, B′, C ′ on two parallel lines such that BC ′ ‖ B′Cand AC ′ ‖ A′C.

We draw the parallel to line AC ′ through point B′, and the parallel to line BC ′

through point A′. These two lines intersect in a point D. The little Desargues Theoremis now applied to the two triangles 4ACB′ and 4C ′A′D. Corresponding vertices areindeed joint by three parallel lines. Too, there are two pairs of parallel sides: AC ‖ C ′A′

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and CB′ ‖ A′D. Hence the little Desargues Theorem assures that

AB′ ‖ C ′D

Secondly, one applies the little Desargues Theorem to triangles 4CBA′ and 4B′C ′D.Again, corresponding vertices are indeed joint by three parallel lines. Two pairs ofparallel sides are CB ‖ B′C ′ and CA′ ‖ B′D. Hence the little Desargues Theoremassures that

BA′ ‖ C ′D

Both instances of the little Desargues theorem together imply AB′ ‖ BA′, as to beshown.

Remark. For simplicity all four configurations— Pappus, Desargues and Little Pappus,Little Desargues— were given above in the affine version. The affine version deals withparallel lines. Hilbert’s foundations [?] use the affine version.

As explained in definition ??, improper elements can be adjoined to produce a pro-jective plane from a given affine plane. The bundles of parallel lines are denoted asimproper points for the different directions of these bundles. The line through all im-proper points is called the improper line.

All four configurations and theorems— Pappus, Desargues and Little Pappus, LittleDesargues— have a corresponding version in the projective plane. In the projectiveversion, the statement that any lines are parallel has to be replaced by the statementthat they intersect on the improper line. Stillwell’s book [?] states only the projectiveversions.

2.3 Pascal’s Theorem

Theorem 2.9 (Pascal’s Hexagon Theorem). The three pairs of opposite sides of anarbitrary circular hexagon intersect in three points lying on one line.

Corollary (Pascal’s Hexagrammum Mysticum). The three pairs of opposite sidesof an arbitrary hexagon inscribed into any conic section intersect in three points lying onone line. With points and lines of the projective plane, there are no exceptional cases.

Proof. This theorem follows from Pascal’s circular hexagon theorem by applying a conve-nient projective transformation to the configuration. Indeed, for any given conic section,there exists a projective transformation mapping the given conic section into a circle.This transformation preserves collinearity and incidence, in the projective sense. Hencethe given hexagon inscribed into any conic section is mapped into a circular hexagon.By the inverse transformation, the entire configuration of Pascal’s circular hexagon ismapped back into Pascal’s Hexagrammum Mysticum.

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Figure 2.14: Pascal’s circular hexagon

Figure 2.15: Pascal’s circular hexagon tangled

Remark. In the setting of projective geoemtry, Pascal’s Hexagrammum Mysticum isreally the natural common generalization of several theorems from this section. Notethat both the circle, and pairs of parallel or intersecting lines are special cases for

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Figure 2.16: Hexagrammum Mysticum

conic sections. The corresponding special cases for the Theorem of the HexagrammumMysticum are Pascal’s circular hexagon theorem 2.9, and the projective versions of thePappus and Little Pappus theorems, respectively. Furthermore, in the second case,choosing the Pascal line as improper line, we can further specialize to the affine versionsof the Pappus theorem 2.1 and Little Pappus theorem 2.6 as stated above.

Proof of Pascal’s circular hexagon. Occurring in counterclockwise order, let the circularhexagon have the vertices A, B′, C,A′, B, C ′. Let Pc := AB′ ∩ A′B, Pa := B′C ∩ BC ′

and Pa := CA′ ∩ C ′A be the intersection points of opposite sides, extended.For the proof we need a second circle Q through the points B, B′ and Pa. Lemma 1.1

is now used with the originally given circle, and this second circle Q. The common chordof these two circles is BB′. Through both points B and B′, the given configuration hastwo lines drawn. Thus one is led to four instances of Lemma 1.1. One of these does notlead to a result, because lines BC ′ and B′C intersect the circle Q in the same point Pa.The other three instances yield three pairs of parallel chords:

(2.2)

AC ′ ‖ Q′Pa

AA′ ‖ Q′Q

A′C ‖ QPa

where Q is the second intersection point of circle Q with line BA′, and point Q′ is thesecond intersection of circle Q with line B′A. The segments AC ′ and A′C are extended,and intersect in point Pb. Thus one gets two triangle 4AA′Pb and 4Q′QPa for which

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Figure 2.17: Proving Pascal’s circular hexagon configuration

three pairs of corresponding sides are pairwise parallel. By the converse DesarguesTheorem, these two triangles are in perspective. Hence the three lines AQ′ = AB′,A′Q = A′B and PbPa intersect in one point, which is Pc = AB′ ∩A′B. Hence the threepoint Pa, Pb and Pc lie on one line, as to be shown.

The exceptional cases because of parallel lines can be eliminated by using the pro-jective plane.

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Figure 2.18: Hexagrammum Mysticum works on a hyperbola!

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3 Euclidean Geometry and Ordered Fields

3.1 Ordered Fields

Definition 3.1 (Field). A field is a system of undefined elements (symbols, numbers),two binary relation symbols + and · and two constants 0 and 1 with the followingproperties:

(1) (i) For any two numbers a and b exists the sum a + b.

(ii) (a + b) + c = a + (b + c) for any a, b, c.

(iii) a + b = b + a for any a, b.

(iv) a + 0 = a for any a.

(v) for each number a there exists a number −a such that a + (−a) = 0.

(2) (i) For any two numbers a and b exists the product a · b.(ii) (a · b) · c = a · (b · c) for any a, b, c.

(iv) a · 1 = 1 · a = a for any a.

(v) for each number a 6= 0 there exists a number a−1 such that a·a−1 = a−1 ·a = 1.

(3) (i) 0 6= 1.

(ii) (a + b) · c = a · c + b · c for any a, b, c.

(iii) c · (a + b) = c · a + c · b for any a, b, c.

(4) a · b = b · a for any a, b.

A system of elements satisfying only (1)(2)(3) is called a skew field.

Any model for the field axioms is simply called a field, too. In this context, wedenote by F the set of elements. Clearly 0, 1 ∈ F.

Definition 3.2 (Ordered field). An ordered field is a field with an order relation <having the following properties:

(i) For any two numbers a and b, one of the three relations holds: either a < b or b < aor a = b.

(ii) For any two numbers a and b, no more than one of the three relations a < b orb < a or a = b hold.

(iii) a < b and b < c imply a < c, for any a, b, c.

(iv) a < b implies a + c < b + c, for any a, b, c.

(iv) a < b and 0 < c imply a · c < b · c and c · a < c · b, for any a, b, c.

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Every ordered field contains the rational numbers as a subfield: Q ⊆ F. Hence everyordered field has the characteristic zero, and especially it is infinite.

Definition 3.3 (Archimedean field). A field or skew field is called Archimedean iffor any a, c > 0 there exists a natural number n such that a < n · c.

Recall that each nonempty set of natural numbers has a smallest one. Hence itis easy to see that for any a, c > 0, there exists a natural number n ≥ 0 such thatn · c < a ≤ (n + 1) · c.

Proposition 3.1 (Theorem 59 of Hilbert). Every Archimedean skew field is com-mutative.

Proof. We get by an easy induction that

n · a = a · n

for any natural number n and any a of the skew field. Take two arbitrary elements a andb, for which we need to check the commutative law. It is easy to see that it is enoughto consider—towards a contradiction—the case

(3.1) a > 0 , b > 0 , ab− ba > 0

From the existence of the inverse, we conclude that there exists an element c > 0 suchthat

(3.2) (a + b + 1) · c = ab− ba

There exists a number d such that

0 < d < 1 , d < c

Let n,m ≥ 0 be the natural numbers such that

m · d < a ≤ (m + 1) · d and n · d < b ≤ (n + 1) · d

Multiplication and subtraction yield

mn · d2 < ba < ab ≤ mn · d2 + (m + n + 1) · d2

ab− ba < (m + n + 1) · d2

(m + n + 1) · d2 < (a + b + 1) · d < (a + b + 1)c

ab− ba < (a + b + 1)c

contradicting relation (3.2). Hence the assumption (3.1) cannot be true, and the com-mutative law is confirmed.

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Definition 3.4 (Pythagorean field). A Pythagorean field is an ordered field with theproperty

(3) If a, b ∈ F, then√

a2 + b2 ∈ F.

Definition 3.5 (Hilbert field). The Hilbert field Ω is the smallest real field with theproperties

(1) 1 ∈ Ω.

(2) If a, b ∈ Ω, then a + b, a− b, ab ∈ Ω.

(2a) If a, b ∈ Ω and b 6= 0, then ab∈ Ω.

(3) If a, b ∈ Ω, then√

a2 + b2 ∈ Ω.

Definition 3.6 (constructible field). The constructible field K is the smallest realfield with properties

(1) 1 ∈ K.

(2) If a, b ∈ K, then a + b, a− b, ab ∈ K.

(2a) If a, b ∈ K and b 6= 0, then ab∈ K.

(3) If a ∈ K and a ≥ 0, then√

a ∈ K.

It is also called the surd field.

Definition 3.7 (Euclidean field). A Euclidean field is an ordered field with the fol-lowing property:

(3) If a ∈ F and a ≥ 0, then√

a ∈ F.

Remark. The Hilbert field Ω is smallest Pythagorean field. The Hilbert field countable.It is a uniquely defined object.

Quite similarly, the constructible field K , also called surd field, is smallest Euclideanfield. The constructible field is countable. It is a uniquely defined object, too.

As shown earlier,√√

2− 1 ∈ K \ Ω, hence the two fields are different.There are many very different Pythagorean fields, and many different Euclidean

fields, among them the real number field R. There exist non-Archimedean Pythagoreanfields, as well as non-Archimedean Euclidean fields, too.

Definition 3.8 (real closed field). A field is called real closed if every polynomial ofodd order has at least one zero.

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3.2 A Hierarchy of Cartesian planes

The introduction of Cartesian coordinates turns out to be a perfect way to producemany different models of Pythagorean planes.

Main Theorem. The Cartesian plane F2 over a Pythagorean field F is a Pythagoreanplane.

Conversely, any Pythagorean plane is isomorphic to the Cartesian plane F2 oversome Pythagorean field F.

Main Theorem. The Cartesian plane F2 is Euclidean if and only if the field F isEuclidean.

The Cartesian plane F2 is Archimedean if and only if the field F is Archimedean.

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8 Inversion by a Circle

8.1 Definition and Construction of the Inverse Point

Let D be a open circular disk of radius R and center O, and denote its boundary circleby ∂D.

Definition 8.1 (Inversion by a circle). The inversion by the circle ∂D is definedto be the mapping from the plane plus one point ∞ at infinity to itself, which mapsan arbitrary point P 6= O, to its inverse point P ′—defined to be the point on the ray−→OP such that |OP | · |OP ′| = R2. Hence, especially, all the points of ∂D are mapped tothemselves. The inversion maps the origin O to∞, and∞ to O. We denote the imagesby inversion with primes.

10 Problem 8.1. Do an example for the construction of the inverse point. Usethe theorems related to the Pythagorean theorem.

Figure 8.1: Construction of the inverse point

Construction 8.1 (Inversion of a given point). Let P be the given point. One

erects the perpendicular onto ray−→OP at point P . Let C be an intersection point of the

perpendicular with ∂D. Next one erects the perpendicular on radius OC at point C, andgets a tangent to circle ∂D. The inverse point P ′ is the intersection of that tangentwith the ray OP .

Reason. Indeed, by the leg theorem, |OP | · |OP ′| = |OC|2 = R2.

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Remark. Here is an alternative justification of the construction. Put in Thales’ circlewith diameter CP ′. By the converse Thales’ theorem, P lies on that circle. Then usethe chord-tangent theorem Euclid III.36. One concludes |OP | · |OP ′| = |OC|2 = R2,once more.

8.2 The Gear of Peaucollier

Figure 8.2: The gear of Peaucollier.

The gear of Peaucellier allows a construction of the inverse point.

10 Problem 8.2 (The gear of Peaucellier). Six stiff rods are linked and canbe turned flexibly by each other within a plane. Two rods of length a are linked to eachother at point O, the remaining endpoints are A and C. The four remaining rods have adifferent length b—they are linked to a rhombus, and the points A and C on one diagonalare linked to the first two rods.

(i) Assume that a > b. Prove that the endpoints of the other diagonal of the rhombusare inverse points by a suitable circle ∂D with center O. Find the radius of thiscircle. You can use the Theorem of chords Euclid III.35 for a second circle L withcenter A.

(ii) Check that the assumption a > b implies that point O lies outside circle L. Let OTbe a tangent to the same circle. Use Euclid III.36 to get the length of OT .

(iii) Check the Theorem of Pythagoras for the right triangle 4OTA.

(iv) Prove that the circles ∂D and L intersect each other perpendicularly.

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(v) What happens in the case a < b?

Figure 8.3: The Theorem of chords explains the gear of Peaucollier.

Answer. We draw the circle L around A through points P and P ′. This circle intersectsline OA in two endpoints E and F of a diameter. Too, the three points O,P and P ′ lieon a line.

(i) The assumption a > b implies that P and P ′ lie on the same side of O. We use theTheorem of chords Euclid III.35 (see 1.7) to conclude

|OP | · |OP ′| = |OE| · |OF | = (a + b)(a− b)

Hence P and P ′ are inverse points by a circle of radius R, and

R2 = (a + b)(a− b) = a2 − b2

(ii) By the Theorem of chord and tangent Euclid III.36 (see 1.7), the length of thetangent from point O to the circle L satisfies

|OT |2 = |OP | · |OP ′|

Together, we get |OT |2 = a2 − b2 = R2, and hence |OT | = R.

(iii) The right triangle4OTA has legs |OT | = R, |TA| = b, and hypothenuse |OA| = a.Hence the Theorem of Pythagoras tells that a2 = R2 + b2, as we have already seenabove.

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(iv) The segment AT is a radius of circle L. The perpendicular segment OT is a radiusof circle ∂D, and at the same time a tangent of circle L. Hence AT is a tangentof circle ∂D, and the two circles intersect perpendicularly.

(v) In the case b > a, the point O lies between P and P ′. Now P and P ′ are antipodalpoints by a circle ∂D of radius Ra, but R2

a = b2− a2. Point O lies now inside thiscircle. The drawing on page 203 provides an example.

Figure 8.4: In case b > a, the gear of Peaucollier constructs the antipodal point.

8.3 Invariance Properties of Inversion

After two further definitions, we can state the main result of this section.

Definition 8.2. A generalized circle is defined to be either a circle or a straight line.

Definition 8.3. The cross ratio of four point A, B, C,D is defined as

(AC, BD) =|AB| · |CD||CB| · |AD|

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Remark. Remember:A→ C B → D

C ← A B → D

Main Theorem. The inversion by a circle maps generalized circles to generalized cir-cles, conserves angles, and conserves the cross ratio.

Definition 8.4. The power of a point O with respect to a circle C is defined by

p = |OA| · |OB|

Here A and B are the two intersection points of any line l through O with the circle C.The power is negative for points inside the circle—and positive for points outside thecircle.

Soundness of definition. Let k be any other line intersecting the circle C, now in thepoints P and Q. We need to confirm that we get the same value for the power, usingline k. Indeed |OA| · |OB| = |OP | · |OQ| by Euclid III.35 and III.36. Hence

p = |OP | · |OQ|

which shows that p does not depend on the choice of the line. Therefore the power iswell defined.

Proposition 8.1. The inversion by a circle maps generalized circles to generalized cir-cles.

Circles not through O are mapped to circles. Given is a circle C which does not go throughcenter O. We prove that its inverse image is a circle, too.

Take any two lines l and k through O. Let A, B and P, Q be their intersection pointswith C. By definition of inversion, |OA| · |OA′| = |OB| · |OB′| = R2. And hence

|OA′| · |OB′| = R4

|OA| · |OB|=

R4

p

where p is the power of point O relative to circle C. For the points P and Q on thesecond line k, one calculates again

|OP ′| · |OQ′| = R4

|OP | · |OQ|=

R4

p

Since this is the same value

|OP ′| · |OQ′| = |OA′| · |OB′|

Euclid III.35 and III.36 imply that the four points A′, B′, P ′ and Q′ lie on a circle C ′.Indeed, since the line k is arbitrary, we have shown that the images of all points of C lieon that circle C ′.

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Figure 8.5: A circle, not going through O is mapped to a circle.

Question. What is the power of point O relative to the inverted circle C ′.Answer. The calculation above show that R4

pis the power of point O relative to the

inverted circle C ′.In the case that point O lies outside of circle C, one can conclude even more. In

the limiting case that P moves to Q, line k becomes a tangent from point O to circle

C. In the same process, P ′ moves to Q′. Hence the ray−→OP =

−−→OP ′ becomes a common

tangent of the two circles C and C ′. Hence both common tangents of circles C and C ′intersect at point O.

This construction suggests that circles C and C ′ are preimage and image for a centraldilation with center O. We now prove this claim in both cases that O lies outside orinside of C.

Lemma 8.1. A central dilation z with center O and ratio

k =R2

p

maps the circle C to the circle C ′. The intersection points P and Q of circle C with anycentral ray k are mapped as

P 7→ Q′ and Q 7→ P ′

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Figure 8.6: The common tangent of a circle and its inverse image intersect at O.

—in a different way as inversion by circle ∂D maps them.The central dilation z maps the center Z of circle C to the center Z∗ of circle C ′, and

the touching point C of a common tangent to touching point C2. Hence

|OC2||OC|

=|OZ∗||OZ|

= k

Proof. We use proportions. Indeed

|OQ′||OP |

=|OP ′||OQ|

=|OP ′| · |OP ||OQ| · |OP |

=R2

p= k

Hence the dilation z maps P 7→ Q′ and Q 7→ P ′. Too, the center Z of circle C is mappedto he center Z∗ of circle C ′, and the touching point C of a common tangent to touchingpoint C2.

A circle through O is mapped to a line. We now consider the exceptional case that thepoint O lies on C, and prove that its image by inversion is a line. Let OA be a diameterof circle C and P be an arbitrary point on that circle. We erect the perpendicular c∗ on

ray−→OA at the inverse point A′. Let P∗ be the intersection of c∗ with the ray

−→OP . The

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Figure 8.7: A circle through O is mapped to a line.

right triangles 4OAP and 4OP∗A are similar. Hence by Euclid VI.6, correspondingsides have the same ratio:

|OP ||OA|

=|OA′||OP∗|

and hence |OP | · |OP∗| = |OA| · |OA′| = R2. Thus P∗ = P ′ is the inverse image of pointP , and the line c∗ = C ′ is the inverse image of C.

Proposition 8.2. The inversion by a circle conserves angles.

The angle between a radial ray and a circle not through O is conserved. To show anglesare conserved, we need to map further objects by the dilation z. Let t be the tangent tocircle C at point P . The dilation z maps the tangent t to the tangent t∗ to the circle C ′at point Q′. By simple facts about central dilations, the tangents t and t∗ are parallel.

Hence by Euclid I.29, the lines t and t∗ intersect the ray−→OP in congruent angles α = α′′.

Now let t2 be the tangent to circle C ′ at the point P ′. (Why is t2 not the image of tunder the inversion? This question distracts a bid, but see: the image of tangent t is acircle though O, and has a common tangent with C ′ at point P ′.)

In general, the tangents t∗ and t2 intersect, say at point S. We get an isosceles4SP ′Q′ with two congruent base angles

α = ∠SQ′P ′ ∼= ∠SP ′Q′

Thus the tangent t to circle C at P , and the tangent t2 to circle C ′ at P ′ both intersect

the projection ray−→OP at angle α.

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Figure 8.8: The angle between a circle and the radial ray is conserved.

In the exceptional case that the tangents t∗ and t2 are parallel, they both intersect

the ray−→OP at right angles. Since the tangents t∗ and t are always parallel, both the

tangent t to circle C at P , and the tangent t2 to circle C ′ at P ′ intersect the projection

ray−→OP at right angle.

10 Problem 8.3. Show that the angle between a central ray and a circle throughO is conserved by inversion. To this end, prove that the three angles α, α′ and α′′ in thefigure on page 207 are congruent.

Answer.

The angle between generalized circles is conserved. It is now easy to see that angles be-tween circles are conserved. One maps two circles C1, C2 intersecting at point P into twogeneralized circles C ′1, C ′2. They intersect at point P ′. One puts in the common projec-

tion ray−→OP and uses angle addition. The angle between the two generalized circles C1

and C2 is congruent to the angle between C ′1 and C ′2. It is left to the reader to check theremaining cases—involving generalized circles.

Proposition 8.3 (Conservation of the Cross Ratio). The inversion by a circleconserves the cross ratio of any four points A, B, C,D. (The four points need not lie ona circle.)

210

Figure 8.9: Relation of the ratio of three points, and ratio of their inverted images.

Reason. In figure on page 209, the circle through points A, B, C is mapped by inversionto the circle through the inverted points A′, B′, C ′. On the inverted circle C ′, both theinverted points A′, B′, C ′, and the dilated points A2, B2, C2 are marked.

The first goal is to show that

((*))|A′B′||C ′B′|

=|AB||CB|

· |OC2||OA2|

The easy part is to use the central dilation z. Because central dilations conserve ratios,we get

(1)|AB||CB|

=|A2B2||C2B2|

We need now to relate the distances |A2B2| and |C2B2| to the distances |A′B′| and|C ′B′|.To this end, we use angles in circle C ′ to find similar triangles

4OA2B2 ∼ 4OB′A′

Clearly 4OA2B2 and 4OB′A′ have a common angle at vertex O. By Euclid III.35,we get supplementary ∠B′B2A2 and ∠B′A′A2, because these angles subtend the twodisjoint arcs from B′ to A2 on circle C ′. Because the angles ∠OB2A2 and ∠B′B2A2 aresupplementary, we get congruent angles ∠OB2A2

∼= ∠B′A′A2∼= ∠B′A′O. Hence the

two triangles have two pairs of congruent angles. Because the angle sum in any triangle is

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two right angles (Euclid I.32), all three angles of the two triangles are pairwise congruent.Hence, by Euclid VI.4, these triangles are similar. Hence

|A′B′||OB′|

=|A2B2||OA2|

This ratio is actually sin α′

sin ω. Similarly one gets that

|C ′B′||OB′|

=|C2B2||OC2|

and dividing the two ratios yields

(2)|A′B′||C ′B′|

=|A2B2||C2B2|

· |OC2||OA2|

Now (1) and (2) imply the claim (*). Now one can use a similar relation for the threepoints A, D and C, just replacing point B by point D:

((**))|A′D′||C ′D′|

=|AD||CD|

· |OC2||OA2|

(It is not required that all four points A′, B′, C ′, D′ lie on one circle, so that secondrelationship uses possibly a different pair of circles.) Division of the two relations (*)

and (**) cancels out the last fraction |OC2||OA2| on the left, and leads to an equation between

of cross ratios:(A′C ′, B′D′) = (AC, BD)

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11 Euclidean Constructions with Restricted Means

11.1 Constructions by Straightedge and Unit Measure

Given is a segment UV , which we call the unit segment. The axiom (III.1) about transferof segments is restricted to the transfer of the unit segment. Transfer of angles is notpostulated at all. We thus consider the weakened axioms:

III.1 unit If A′ is a point on the line a′, then it is always possible to find a point B′ ona given side of the line a′ through A′ such that the unit segment UV is congruentto the segment A′B′. In symbols UV ∼= A′B′.

III.4 unit Let the angle ∠(h, k) and the ray a′ be given. Then there exists at mostone ray k′ such that the angle ∠(h, k) is congruent to the angle ∠(h′, k′) and atthe same time all interior points of the angle ∠(h′, k′) lie on the given side of a′.Every angle is congruent to itself, thus it always holds that

∠(h, k) ∼= ∠(h, k)

Definition 11.1 (Hilbert plane with restricted transfer). A geometry with theaxioms of incidence (I.1),(I.2),(I.3), the axioms of order (II.1) through (II.5), but theaxioms of congruence (III.1) and (III.4) replaced by the weaker axioms (III.1 unit) and(III.4 unit) is called a Hilbert plane with restricted transfer.

In the case of Euclidean geometry, these restricted axioms turn out to be equivalent tothe Hilbert’s original axioms.

Main Theorem (Straightedge and unit measure are equivalent to Hilberttools (Theorem of Hilbert and Kurschak)). 22 For a Hilbert plane with parallelaxiom (IV.1), all constructions that can be done with Hilbert tools, can be done usingonly straightedge and unit measure, too.

In a Hilbert plane with restricted transfer, the Hilbert tools originally postulated byaxioms (III.1) and (III.4) have been forbidden. In order to emulate these Hilbert tools,we are going to solve the following construction problems with straightedge and unitmeasure: 23

Construction 11.1. Transfer a given segment onto a given ray, starting the new con-gruent segment at the vertex of the ray.

Construction 11.2. Transfer a given angle onto a given ray, with the vertex of thenew congruent angle at the vertex of the ray, one side of the new angle on the given ray,and the other side in a prescribed half plane.

22This Theorem is Hilbert’s Proposition 63—who gives credit to J. Kurschak [?].23We need the parallel axiom (IV.1) to achieve that.

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Construction 11.3. Construct the parallel to a given line through a given point.

Construction 11.4. Construct any perpendicular to a given line.

Construction 11.5. Construct the perpendicular to a given line through a given point.

Figure 11.1: Another construction of the Euclidean parallel— the lines are numbered inthe order they are constructed.

Construction (11.3). By transfer of the unit segment UV to any point M of the givenline l, one gets a point A such that UV ∼= AM . A second transfer yields a point B suchthat UV ∼= MB, and M becomes the midpoint of segment AB.

Let P be the given point through which we have to get the parallel. Draw line APand choose on it any point C such that P lies between A and C. Draw segments CM

and PB. They intersect in some point S. Draw the segment BC and the ray−→AS. They

intersect in some point Q. Line PQ is the required parallel to AB through point P .

Reason using the harmonic quadrilateral. The construction uses a special case of thecomplete quadrilateral from projective geometry.

Figure 11.2: The harmonic quadrilateral.

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Theorem 11.1 (Theorem of the harmonic quadrilateral). Take any quadrilateralABQP . Let a pair of opposite sides AB and PQ intersect in point Y , and the otherpair of opposite sides in point C. Let the diagonals intersect in point S. Let the line CSintersect side AB in point X. Then the four points A, B, X, Y are harmonic points. Inother words, the cross ratio

(11.1) (AB, XY ) =AX ·BY

BX · AY= −1

Via the Theorem 11.1 of the harmonic quadrilateral, we show the lines AB and PQcannot intersect for special case occurring in the construction. Since X = M is themidpoint of segment AB, with lines AB and PQ intersecting in any point Y , the crossratio would be

(11.2) (AB, MY ) =AM ·BY

BM · AY= −BY

AY6= −1

Hence the harmonic property (11.1) could not be satisfied. The only remaining possi-bility is that the lines AB and PQ are parallel.

Figure 11.3: Transfer of a given segment AB onto line l at point P .

Construction (11.1). Given is a segment AB to be transferred onto the ray l at pointP . Draw the parallel to AB through point P , and the parallel to AP through point B.The two parallels intersect in point Q. Use the unit measure UV to get the segment

PC on the parallel ray−→PQ and the segment PD on the given ray l such that UV ∼= PC

and UV ∼= PD. Draw the parallel to line CD through point Q. It intersects the givenray l in point E such that AB ∼= PE.

Question. There are some special cases:

(1) The given ray l is parallel to AB, but does not lie on AB.

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(2) The given point P lies on line AB, but the ray l does not lie on AB.

(3) The given point P and the given ray l both lie on line AB.

Modify the construction to cover these special cases.

Figure 11.4: Constructing a perpendicular line, using only straightedge and unit measure.

Construction (11.4). Choose an arbitrary point M on the given line l, and transfer theunit segment UV onto both opposite rays of l with vertex M . One gets the congruentsegments UV ∼= BM ∼= MC.

Now choose two further rays with vertex M in the same half plane of line l, andtransfer the unit segment UV onto them to produce the congruent segments UV ∼= MDand UV ∼= ME.

The lines BD and CE intersect in a point F , and the lines BE and CD intersect ina point H. Now the line FH is perpendicular to the given line l.

Reason for validity. The four points B, D, E, C lie on a circle with diameter BC. Henceby Thales’ theorem, ∠BDC and ∠BEC are right angles. The two altitudes BE andCD of triangle 4BFC intersect in H. Because all three altitudes of a triangle intersectin one point, FH is the third altitude and hence perpendicular to BC.

The construction problem 11.5 can now be solved using the just established con-struction 11.4 of a perpendicular, and construction 11.3 of the parallel through a givenpoint.

Before proceeding to construction 11.2, we solve a special case of that problem—justrotating a given angle.

216

Construction 11.6. For a given β = ∠D′AB′ and line l = AC ′ 24 through its vertex,on both sides of line l, construct the congruent angles ∠C ′AE1

∼= ∠C ′AE2∼= β.

Figure 11.5: Rotating a given angle about its vertex.

Construction (11.6). Let D be the foot point of the perpendicular dropped from pointB onto AD′. Similarly, let C be the foot point of the perpendicular dropped from pointB onto line AC ′.

Next we drop the perpendicular from vertex A onto line DC, and let E be its foot

point. Finally, ∠CAE ∼= β is the angle to be constructed. Its legs are the given ray−−→AC ′

and the perpendicular dropped from A onto line DC.

Remark. Both points C and D lie on a circle with diameter AB. Hence the constructionof the perpendiculars can easily be done with rusty compass, by choosing AB twice theradius of the rusty compass and drawing a circle with diameter AB.

Reason for validity. The four points A, B, D, E lie on a circle C with diameter AB.Hence ∠DAB ∼= ∠DCB because they are angles in circle C over the same arc DB.Finally, ∠DCB ∼= ∠EAC because their sides are pairwise perpendicular. (Let F bethe intersection lines BC and AE. One gets three similar right triangles 4AFC ∼4ACE ∼ 4CFE.)

Remark. To transfer the angle to the other side of line l, one interchanges points B′ andD′ on the sides of the given angle β. Then one proceeds as above.

We are now in the position to solve the construction problem 11.2.

24Primes have been used where new points will be needed below, producing the same rays and angles.

217

Figure 11.6: Validity of construction to rotate an angle.

Figure 11.7: Transferring a given angle to both sides of the given ray.

Construction (11.2). Given is an angle β = ∠DAB, and a ray−−→A′B′. At first we con-

struct the parallel to the line A′B′ through vertex A. Next we rotate the given anglearound its vertex A to produce a congruent angle with one side on this parallel. Finally,we need to construct the parallel to the other side of the rotated angle through pointA′. Now we have the required angle at vertex A′.

11.2 Tools equivalent to Straightedge and Compass

Proposition 11.1 (Collapsible and noncollapsible compass are equivalent). Fora Hilbert plane, we assume Euclid’s postulate to draw a circle with given center througha given point, and the circle-circle intersection property.

Under these assumptions, it is possible to construct a circle around a given center

218

with radius congruent to a given segment, using only the collapsible compass.

Remark. The parallel axiom is not needed in proposition 11.1. Neither do we need anytransfer of segments.

Figure 11.8: Emulation of the non-collapsible compass.

Emulation of the non-collapsible circle. Let the segment AB and center C 6= A, B begiven. We draw a circle with center A through point C, and a second circle with centerC through point A. By the circle-circle intersection property, these two circles intersectin two points X and Y . Now we draw two further circles around points X and Y throughthe second point B.

The two circles intersect in a second point D such that the segments AB ∼= CD arecongruent. In the special case that these two circles intersect only in point B, the threepoints X, Y and B lie on a line, and D = B. Still the congruence AB ∼= CD = CBholds.

By incidence axiom I.1 and I.2, we can draw a unique line between any two givenpoints. By proposition 1.1, any two different lines have either one or no point in common.By Euclid’s postulate, we can draw a circle with given center through a given point.By the circle-line intersection property and the circle-circle intersection property, thereexist intersection points of of a line or circle if one goes from the inside to the outsideof a second circle.

Definition 11.2 (Traditional Euclidean tools). Constructions done using straight-edge and compass only for the performance of the steps mentioned above, are calledconstructions by traditional Euclidean tools.

219

Main Theorem (Constructions by compass only (Theorem of C. Mohr andLorenzo Mascheroni)). For the Euclidean geometry with parallel axiom (IV.1) andcircle-circle intersection property, all constructions that can be done with straightedgeand compass, can be done using only a (collapsible) compass.

Figure 11.9: Construction of the midpoint with compass only.

Example 11.1 (Construction of the midpoint with compass only). Given is thesegment AC to be bisected. We draw a circle A with center A through point C, and asecond circle C with center C through point A. By the circle-circle intersection property,these two circles intersect in two points X and Y .

Now we find the second endpoint of diameter AA′ of circle C. One uses simply theregular hexagon inscribed into this circle. This step is a Euclidean construction.

We need the the intersection points Z and Z ′ of circle A with the circle around A′

through point A. Finally, the the two circles around Z and Z ′ through point A intersectin the midpoint M . This step, too, is a Euclidean construction.

Reason for the construction. The isosceles triangles 4A′AZ and 4ZAM are equiangu-lar. By Euclid VI.4, their sides are proportional. Hence

|A′A||AZ|

=|ZA||AM |

= 2

which confirms that |AM | = 12|ZA| = 1

2|AC|.

Main Theorem (Constructions with one circle (Theorem of Poncelet andSteiner)). For the Euclidean geometry with parallel axiom (IV.1) and circle-circle in-tersection property, all constructions that can be done with straightedge and compass,can be done using only the straightedge, and only one circle with its center given.

220

The Arabian mathematician Abul-Wefa suggested to use straightedge and a rustycompass This is a compass with fixed radius. From the Theorem of Poncelet and Steiner,we can deduct that rusty compass and straightedge are equivalent to the traditionalEuclidean tools. But, since the actual constructions with fixed circle are rather awkward,I prefer to pose some problems using the rusty compass.

10 Problem 11.1. Find the intersection of a given line l with a circle, of whichthe center O and one point B are given, using straightedge and rusty compass.

Here is the idea how to proceed. We build two images, which are mapped to eachother by a central dilation with center O. One image contains the line l and the circleC, the other one contains a parallel line l′ and a circle C ′ around O of radius one unitwhich can be drawn using the given compass. You see that this is a problem of Euclideangeometry. It makes ample use of parallels!

Figure 11.10: Intersection of line and circle with the rusty compass.

Answer (Description of the construction). Begin by drawing a circle C ′ of the given unit

radius around point O. Let B′ be the intersection point of C ′ with the radial ray−−→OB.

Next choose an arbitrary point A on the line l. Then construct two similar triangles:

4OAB and 4OA′B′. This is easy, because A′ is just the intersection of−→OA with the

parallel to line AB through point B′.Next we get line l′. It is the parallel to line l through point A′. Let H ′

1 and H ′2 be

the intersections of circle C ′ with line l′. (If they do not exist, the circle C and line l donot intersect, neither.) The intersections H1 and H2 of the circle C and line l are now

easy to get, because they lie on the rays−−→OH ′

1 and−−→OH ′

2.

221

Reason for validity. Here is the reason why Hi, for i = 1 or 2, lie on the circle C:Because of the similar triangles 4ABO ∼ 4A′B′O one gets the proportion

|OA||OA′|

=|OB||OB′|

Because of the similar triangles 4AHiO ∼ 4A′H ′iO one gets the proportion

|OHi||OH ′

i|=|OA||OA′|

By construction OH ′i = OB′ = 1 is the radius of the rusty compass. Hence the propor-

tions imply OHi = OB, again for i = 1, 2.

Figure 11.11: What the teacher wanted.

10 Problem 11.2 (David at school 1). David Hilbert sees that the geometryteacher has draw the figure on page 220 on the board. The teacher tells that the squareABCD have sides 1 unit, the two curved lines are circular arcs, and that G is thecenter of the semicircle. Then the teacher poses the following problems:

(a) Calculate the length of segment AH.

(b) Use compass and straightedge to reproduce the drawing for your notebook.

Answer. AH2

= AF ·AB can be seen using the right 4FBH. Because of AB = 1 and

AF =√

2− 1, be get AH2

=√

2− 1 and hence AH =√√

2− 1.

222

10 Problem 11.3 (David at school 2). At that point, David realizes that he haslost his compass. Luckily enough, he still has a straightedge and finds a rusty compass,with opening just one unit in his back pocket. David manages to get points A throughG with a score of circles of his rusty compass, and the straightedge. Really do similarconstructions, even if you need more (or less!) circles. Report what you have done, bestby leaving some obvious details aside. But count how many circles you did need.

Answer. The unit square ABCD can be drawn with a rusty compass. using 7 circles,to get a stack of equilateral triangles.

To get point F , one can bisect the 45 angle ∠ABD, and then drop the perpendicularfrom point D onto the bisector. One needs 5 extra circles for that purpose.

To get point G, note that it is the midpoint of segment AF . Hence one drops theperpendicular from the midpoint M of the square ABCD onto the same bisector. Oneneeds only 3 circles.

10 Problem 11.4 (David at school 3). Use an extra drawing to explain howto get point H with the same rusty compass. Count how many circles you need. Onecan use a central dilation with center B, mapping G 7→ A.

Figure 11.12: What Hilbert—at school 3—could have done.

Answer. Images for the same central dilation are A 7→ F and H 7→ H ′, since

|BG||BA|

=|BA||BF |

Erect the perpendicular at point F onto line AB. Get H ′ is the intersection of thisperpendicular with the circle around A through point B. Point H is the intersection oflines BH ′ and AD.

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11.3 Hilbert Tools and Euclidean Tools Differ in Strength

Proposition 11.2 (The traditional Euclidean tools can do at least all construc-tions possible with Hilbert tools). We assume only Hilbert’s axioms (I.1)–(I.3),(II.1)–(II.4), (III.2), (III.3) and (III.5). Furthermore, we assume Euclid’s postulate todraw a circle with given center through a given point. We assume that a circle intersectsa diameter in exactly two points, and finally the circle-circle intersection property.

Under these assumptions, it is also possible to transfer any segment as well as anyangle—, using only the traditional straightedge and collapsible compass. In other words,traditional straightedge and compass are stronger than the Hilbert tools postulated byHilbert’s axiom (III.1) and (III.4).

Remark. The parallel axiom is not needed in proposition 11.2.

Figure 11.13: Transfer of a segment with Euclidean tools.

Construction to transfer a segment. Construct an equilateral triangle 4AA′C, gettingpoint C as an intersection of two circles with center A through point A′, and center

A′ through point A. A segment AB1∼= AB on the ray

−→CA can be obtained from the

intersection point B1 of this ray with the circle around A through point B. A second

circle around C through point B1 enable one to get segment CB2∼= CB1 on the ray

−−→CA′,

from the intersection point B2 of ray and circle. Finally, two segments A′B′ ∼= A′B2 andA′B′′ ∼= A′B2 on the given line a′ are obtained from the two intersection points B′ andB′′ of line a′ with the circle around A′ through point B2. One of these two segments lieson the side of A′ as required.

224

Reason for validity of the construction. Using segment addition or subtraction, it iseasy to check that AB1

∼= A′B2, since by construction both CB1∼= CB2 and CA ∼= CA′.

Hence transitivity of congruence implies AB ∼= AB1∼= AB2

∼= AB′ ∼= AB′′.

Figure 11.14: Alternative construction to transfer of a segment with Euclidean tools.

An alternative construction using more circles, but less lines is shown in the figureon page 223.

Construction to transfer an angle. Given angle ∠BAD, we draw a circle around Athrough point B on one vertex, and choose point D on the other vertex such thatAB ∼= AD. After that initial step, we go on and transfer segment AB onto the line a′,at the side of point A′ required, by means of the construction above. We get a segmentA′B′ ∼= AB.

Finally we transfer segment BD, and get a circle around B′ of radius B′D2∼= BD.

This step needs a second equilateral triangle 4BB′E. We construct the intersectionpoints D′ and D′′ of the circles around A′ of radius AB, and around B′ of radius BD.One gets the required angle as either ∠B′A′D′ ∼= ∠BAD, or ∠B′A′B′′ ∼= ∠BAD,depending on half plane to which the angle has to be transferred.

Finally, we explain, why Hilbert tools are different, and indeed weaker than thetraditional Euclidean tools. In the following discussion, it is agreed that some unitsegment AB of length |AB| = 1 is given.

225

Figure 11.15: Transfer of an angle segment with Euclidean tools.

Definition 11.3 (Hilbert field). The Hilbert field Ω is the smallest real field withproperties

(1) 1 ∈ Ω.

(2) If a, b ∈ Ω, then a + b, a− b, ab ∈ Ω.

(2a) If a, b ∈ Ω and b 6= 0, then ab∈ Ω.

(3) If a, b ∈ Ω, then√

a2 + b2 ∈ Ω.

Proposition 11.3 (The Hilbert field are the segment lengths constructiblewith Hilbert tools ). The set of all lengths constructible with straightedge and unitmeasure are exactly those in the Hilbert field.

Indication of reason. Let Ω be the Hilbert field, which is just the smallest field withproperties (1)(2)(3). Let Ωconstr be the set of lengths constructible with Hilbert tools.All constructions with Hilbert tools can only produce lengths which are obtained viathe algebraic operations mentioned above. Hence Ωconstr ⊆ Ω. Too, it is easy to check

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that Ωconstr is a field. Since the domain Ω is the smallest domain with the properties(1)(2)(3), we get Ωconstr = Ω.

Proposition 11.4. All numbers in the Hilbert field are totally real algebraic numbers.That means that they are roots of a irreducible polynomial of (possibly high) degree Nwith integer coefficients, all N roots of which are real.

By Proposition 11.3 and Proposition 11.4, all lengths constructible by Hilbert toolsare totally real.

The domain T of all totally real Euclidean numbers has the properties (1),(2),(2a)and (3), too. Hence the minimality of Ω, shown in Proposition 11.3 implies Ω ⊆ T ⊆ K,where K is the Euclidean field. 25

Proposition 11.5 (Counterexample). The number z =√√

2− 1 is not totally real.A segment of that length cannot be constructed by Hilbert tools, just starting from agiven unit segment.

In the figure on page 220, points A through G are constructible with Hilbert tools,but not point H is not constructible.

Reason for the counterexample. The number z =√√

2− 1 is a root of P (z) = (z2 +1)2 − 2 = 0. This polynomial is irreducible over the rational numbers. Hence z is notroot of any integer polynomial of degree two or three. But the polynomial P (z) has twocomplex roots. Hence z /∈ T and z /∈ Ω.

Let the segment AB in the figure on page 220 be the unit segment. All segments con-structible starting from AB —of length |AB| = 1—with straightedge and unit measure

are totally real algebraic numbers. But the number z =√√

2− 1 is not totally real.Hence a segment of that length cannot be constructed by Hilbert tools, just startingfrom a given unit segment.

Finally, here is the algebraic field of numbers constructible by the traditional Eu-clidean tools.

Definition 11.4 (constructible field). The constructible field K is the smallest realfield with properties

(1) 1 ∈ K.

(2) If a, b ∈ K, then a + b, a− b, ab ∈ K.

(2a) If a, b ∈ K and b 6= 0, then ab∈ K.

(3) If a ∈ K and a ≥ 0, then√

a ∈ K.

25Emil Artin has proved that even Ω = T , but we do not use that much stronger result here.

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Main Theorem. The lengths constructible with straightedge and compass are exactlythe numbers in the constructible field K. The lengths constructible with Hilbert tools areexactly the numbers in the Hilbert field Ω ⊂ K.

The Hilbert field Ω ⊂ K is a proper subset of the constructible field K. Especially,√√2− 1 ∈ K \ Ω.

Proof. The algebraic operations in the definition of the constructible field can be emu-lated by the traditional Euclidean tools— straightedge and compass. No other algebraicoperations can be emulated by these construction tools.

Similarly, the algebraic operations in the definition of the Hilbert field are exactlythose which can be emulated with Hilbert tools.

Straightedge and rusty compass are stronger than straightedge and unit measure. Whatis possible, what is impossible, concerning the construction of point H in the figure onpage 220 from given points A and B?

Point H in this figure —and a segment of length z =√√

2− 1— can be constructedwith straightedge and rusty compass. The reason is that these tools are equivalent tothe traditional Euclidean tools, as follows from the Theorem of Poncelet and Steiner.

Clearly ,the length z =√√

2− 1 can be constructed with traditional Euclidean tools.It would even be possible to restrict the use of the compass to drawing only a singlecircle.

On the other hand, nor point H in the figure on page 220 —and neither the segment of

length z =√√

2− 1— can be constructed with straightedge and unit measure. Indeed,by the Theorem of Hilbert and Kurschak, straightedge and unit measure are equivalent

to Hilbert tools. But a segment of length z =√√

2− 1 cannot be constructed with

Hilbert tools, since√√

2− 1 is not totally real.

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