The Elements, Book I MATH 392 – Geometry Through History Feb. 1 and 3, 2016

1. The Elements, Book IMATH 392 – Geometry Through HistoryFeb. 1 and 3, 2016

2. Want to follow Euclid's logical structure, without necessarily sticking that closely to the specific ways he said things. • So in particular, we'll freely use modern symbolic forms of geometric statements. • Example: Congruence of figures (line segments, angles, triangles, etc.) We'll write something like ΔAEB ≅ ΔBDA to say triangles are congruent. Euclid doesn't do this – just says line segments, angles, triangles are “equal.” • So, this might be a bit different from the online translations of Euclid on the course homepage.

3. The 5 Common Notions Things that are equal to the same thing are equal to one another. If equals be added to equals, the wholes are equal. If equals be subtracted from equals, the remainders are equal. Things that coincide with one another are equal to one another. The whole is greater than the part.

4. The First Four Postulates • [It is possible] to draw a straight line from any point to any point. • [It is possible] to produce any finite straight line continuously in a straight line. • [It is possible] to describe a circle with any center and given distance as radius. • All right angles are equal (i.e. congruent) to one another.

5. The Fifth Postulate 5. If a straight line falling on two straight lines makes the angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

6. The situation in Postulate 5

7. After Producing the lines sufficiently far, …

8. Comments Postulates 1, 2, and 3 describe the constructions possible with an (unmarked) straightedge and a “collapsing” compass – that is the compass can be used to draw circles but not to measure or transfer distances Postulate 4 is a statement about the homogeneous nature of the plane – every right angle at one point is congruent to a right angle at any other point Postulate 5 is both more complicated than, and less “obvious” than the others(!)

9. Content of Euclid's Text • Following the Common Notions (Axioms) and Postulates comes a sequence of Propositions and their proofs – 48 in all in Book I. • Two traditional types of Propositions: • The “problems” show how to construct something (and show the construction works) • The “theorems” claim something (and prove those claims) • Manuscript copies of the Elements typically also contained extensive scholia, or commentaries, added by later scholars. They were effectively part of the Elements for later readers.

10. Proposition 1 • To construct, on a given line segment AB, an equilateral triangle. Proof: AC = AB (this was stipulated in the definition of a circle). Similarly BC = AB. Therefore, AC = BC (Common Notion 1) Hence ∆ABC is equilateral (again, a definition). QEF

11. Strictly speaking, we don't know the circles must intersect somewhere. That does not follow from the Common Notions and Postulates. Euclid is appealing to our intuition about physical circles here and either does not appreciate that there is something missing, and/or does not want to address that point at this stage (he doesn't come back to it later either(!)) This is a flaw, but perhaps justified because of the pedagogical orientation of the text. Quite a few similar issues elsewhere. A reworking of Euclid's foundations done by D. Hilbert (early 20th century – see Chapter 2 in McCleary) – adding quite a few additional postulates – was designed to overcome all of these, and it was entirely successful. A Possible Criticism

12. Propositions 2 and 3 • These are somewhat technical constructions aimed at showing that the straightedge and “collapsing” compass are sufficient for routine tasks such as measuring off a given length on a given line. • Proposition 2. Given a line segment AB and a point P, construct a point X such that PX = AB. • The construction uses Proposition 1 and Postulate 3 • Proposition 3. Given two unequal line segments, lay off on the greater a line segment equal to the smaller. • This construction uses Proposition 2 and Postulate 3 again, but without using the compass to “transfer” the length

13. A natural Question • It’s natural to ask: Why did Euclid go to the trouble of making these somewhat involved constructions for relatively simple tasks that would be easy if we had an implement like the modern compass that can be used to measure and transfer lengths in a construction? • The answer seems to be that his goal was to show that a very small set of simple starting assumptions was sufficient to develop the basic facts of geometry. • So some technical stuff would be acceptable at the start to establish “routines” for those tasks under the more restrictive working conditions or hypotheses.

14. The “SAS” Congruence Criterion • Proposition 4. Two triangles are congruent if two sides of one triangle are congruent with two sides of the other triangle, and the included angles are also congruent. • The proof given amounts to saying: move the first triangle until the sides bounding the equal angles coincide, then the third sides must coincide too (since there's just one line joining two distinct point – another unstated assumption). • This idea gives a valid proof, of course, but it again raises a question: What in the Postulates says we can move any figure? Common Notion 4 seems to be hinting at this, but that is pretty ambiguous. Again, there is physical intuition and/or an unstated assumption being used here(!)

15. The “Pons Asinorum” • Proposition 5. In any isosceles triangle, the base angles are equal; also the angles formed by the extensions of the sides and the base are equal.

16. Outline of proof of Proposition 5 Diagram is constructed by extending AC to D (Postulate 2), extending AB to E with BE = CD (Proposition 3), then joining AE and BD (Postulate 1) First, ΔDCB ≅ ΔECA (Proposition 4) Therefore, ΔAEB ≅ ΔBDA (Proposition 4 again – note the angle at D is the same as the angle at E by Common Notion 2.) This shows ᐸ DAB = ᐸ EBA Then by the two triangle congruences, ᐸ BAC = ᐸ EAC - ᐸ EAB = ᐸ DBC - ᐸ DBA = ᐸ ABC (Common Notion 3). QED

17. Propositions 6 and 7 • First, a converse of Proposition 5: • Proposition 6. If two angles of a triangle are equal, then the sides opposite those angles are equal. • Euclid gives a proof by contradiction, using Propositions 3 and 4. Next: • Proposition 7. If in the triangles ΔABC and ΔABD, with C and D on the same side of AB, we have AC = AD and BC = BD, then C = D. Another proof by contradiction, using Proposition 5; Euclid gives only one case out of several.

18. Proposition 8 • Proposition 8. If the three sides of one triangle are equal to the three sides of another triangle, then the triangles are congruent. • This is the “SSS” congruence criterion • Proof is based on Proposition 7; in fact can almost see that Euclid wanted to present the reasoning “broken down” into easier steps by doing it this way. • Modern mathematicians call a result used primarily to prove something else a “lemma.”

19. Next, a sequence of “bread-and-butter” constructions • Proposition 9. To bisect a given angle.

20. Line Segment Bisection • Proposition 10. To bisect a given line segment. Construction: Let AB be the given line segment Using Proposition 1, construct the equilateral triangle ΔABC Using Proposition 9, bisect the angle at C Let D be the intersection of the angle bisector and AB. Then D bisects AB.

21. “Erecting” a perpendicular • Proposition 11. To construct a line at right angles to a given line from a point on the line. • Construction is closely related to Proposition 10: • Given point A on the line, use Postulate 3 to construct two other points on the line B, C with AB = AC. • Construct an equilateral triangle ΔBCD (Proposition 1) • Then DA is perpendicular to the line at A.

22. “Dropping” a perpendicular • Proposition 12. To drop a perpendicular to a given line from a point not on the line. • Construction: Given point A not on the line and P on the other side of the line, use Postulate 3 to construct a circle with radius AP and center A – it intersects the line in points B, C with AB = AC. • Let D be the midpoint of BC (Proposition 10) • Then DA is perpendicular to the line at D. • Proof: ΔADB and ΔADC are congruent by Proposition 8 (“SSS”). Hence <ADB = <ADC are right angles. QEF

23. Propositions about angles • Proposition 13. If from a point on a line a ray is drawn, then this ray forms with the line two angles whose sum is the same as two right angles. • Proposition 14. If two angles have a side in common, and if the noncommon sides are on different sides of the common side, and if the angles are together equal to two right angles, then the noncommon sides lie along the same straight line. This is a converse of Proposition 13. The reasoning is similar in that it is based just on the Common Notions. Note: Euclid did not consider 180˚ (“straight”) angles as angles.

24. Propositions about angles, continued Proposition 15. Vertical angles are equal. Proof: <BPC + <CPD is the same as two right angles by Proposition 13. Similarly for <APB + <BPC. Hence <CPD + <BPC = <APB + < BPC by Postulate 4. Therefore, <CPD = <APB by Common Notion 3. QED

25. A group of propositions about angles, continued • Proposition 16. In a triangle, an exterior angle is greater than either of the nonadjacent interior angles. • The statement is that <DCB is greater than <CAB, <CBA:

26. Proof of Proposition 16 • Euclid's proof is clever! To show <DCA is greater than <BAC: • Construct the midpoint E of AC (Proposition 10) and extend BE to BF with BE = EF (Postulate 2 and Proposition 3). Construct CF (Postulate 1). Note that <AEB = <FEC by Proposition 15.

27. Proof of Proposition 16, concluded • Therefore ΔAEB and ΔCEF are congruent (Proposition 4 – “SAS”). Hence <BAE = <ECF. • But <ECF is a part of the exterior angle <DCA. So the exterior angle is larger (Common Notion 5). QED • A similar argument shows <DCA is larger than < ABC.

28. Proposition 17 • Proposition 17. In any triangle, the sum of any two angles is less than two right angles. • This follows pretty immediately from Proposition 16 and Proposition 13 (which says that the interior and exterior angles sum to two right angles). • Note that Euclid has not yet proved that the sum of all the angles in a triangle is equal to two right angles (or 180˚). So he cannot use any facts related to that yet. • The angle sum theorem is coming later, in Proposition 32, after facts about parallel lines have been established. • We will skip lightly over the next group of propositions – important for geometry, but off our main track here(!)

29. Sides in triangles • Proposition 18. In a triangle, if one side is greater than another side, then the angle opposite the larger side is larger than the angle opposite the smaller side. • Proposition 19. In a triangle, if one angle is greater than another angle, the side opposite the greater angle is larger than the side opposite the smaller angle. • Proposition 20. In a triangle, the sum of any two sides is greater than the third side. • We will omit Propositions 21, 24 entirely.

30. Constructing triangles and angles • Proposition 22. To construct a triangle if the three sides are given. • The idea should be clear – given one side, find the third corner by intersecting two circles (Postulate 3). This only works if the statement of Proposition 20 holds. • Proposition 23. To construct with a given ray as a side an angle that is congruent to a given angle • This is based on finding a triangle with the given angle (connecting suitable points using Postulate 2), then applying Proposition 22.

31. Additional triangle congruences Proposition 26. Two triangles are congruent if • One side and the two adjacent angles of one triangle are equal to one side and the two adjacent angles of the other triangle • One side, one adjacent angle, and the opposite angle of one triangle are equal to one side, one adjacent angle, and the opposite angle of the other triangle. • Both statements here are cases of the “AAS” congruence criterion as usually taught today in high school geometry. Euclid's proof here does not use motion in the same way that his proof of the SAS criterion (Proposition 4) did.

32. Theory of parallels • Proposition 27. If two lines are intersected by a third line so that the alternate interior angles are congruent, then the two lines are parallel. • As for us, parallel lines for Euclid are lines that, even if produced indefinitely, never intersect • Say the two lines are AB and CD and the third line is EF as in the following diagram

33. Proposition 27, continued • The claim is that if <AEF = <DFE, then the lines AB and CD, even if extended indefinitely, never intersect. • Proof: Suppose they did intersect at some point G

34. Proposition 27, concluded • Then the exterior angle <AEF is equal to the opposite interior angle <EFG in the triangle ᐃEFG. • But that contradicts Proposition 16. Therefore there can be no such point G. QED

35. Parallel criteria Proposition 28. If two lines AB and CD are cut by third line EF, then AB and CD are parallel if either • Two corresponding angles are congruent, or • Two of the interior angles on the same side of the transversal sum to two right angles.

36. Parallel criteria Proof: (a) Suppose for instance that <GEB = <GFD. By Proposition 15, <GEB = <AEH.So <GFD = <AEH (Common Notion 1). Hence AB and CD are parallel by Proposition 27.

37. Parallel criteria Proof: (b) Now suppose for instance that <HEB + <GFD = 2 right angles. We also have <HEB + < HEA = 2 right angles by Proposition 13. Hence <GFD = <HEA (Common Notion 3). Therefore AB and CD are parallel by Proposition 27. QED

38. Familiar facts about parallels Proposition 29. If two parallel lines are cut by a third line, then (a) the alternate interior angles are congruent, (b) corresponding angles are congruent, (c) the sum of two interior angles on the same side is equal to 2 right angles.

39. Familiar facts about parallels Proof of (c): The claim is that, for instance, if AB and CD are parallel, then <BEH+<DFG = 2 right angles. Suppose not. If the sum is less, then Postulate 5 implies the lines meet on that side of GH. But this is impossible since AB and CD are parallel. If the sum is greater, then since <BEH = <GEA and < DFG = <HFC (Proposition 15), while <GEA + <AEH = 2 right angles = <HFC + CFG (Proposition 13), then <AEH + <CFG is less than 2 right angles, and Postulate 5 implies the lines meet on the other side. QED

40. Comments about Proposition 29 • This is the first use of Postulate 5 in Book I of the Elements • It is almost as if Euclid wanted to delay using it as long as possible (maybe to see if it was necessary??) • Recall how much less intuitive and “obvious” the statement is – there was a long tradition of commentary that ideally Postulate 5 should be a Proposition with a proof derived from the other 4 Postulates and the Common Notions – for instance in quotation from Proclus on page 27 of McCleary's book. • The other parts of Proposition 29 are proved similarly.

41. Further properties of parallels • Proposition 30. If two lines are parallel to the same line then they are parallel to one another. • That is, in modern language, parallelism is a transitive relation on lines(!) • The proof Euclid gives depends on intuitive properties of parallels that are “obvious” from a diagram, but that do not follow directly from the other Postulates and previously proved theorems. • In this case, though, the gap can be filled with additional reasoning – will appear on a future problem set(!).

42. Construction of parallels • Proposition 31. To construct a line parallel to a given line and passing through a given point not on that line. • Construction: Say AB is the line and F is the given point. • Pick any point E on AB and construct EF (Postulate 1) • Construct FG so that <GFE = <BEF (Proposition 23) • Proof: Then FG and AB, produced indefinitely, are parallel lines (Proposition 27). QED • Note: in some modern geometry textbooks, the statement that there is exactly one such parallel line is used as a substitute for Euclid's Postulate 5. Not hard to see they are equivalent statements – we'll return to this(!)

43. The angle sum theorem • Proposition 32. In any triangle, (a) each exterior angle is equal to the sum of the two opposite interior angles and (b) the sum of the interior angles equals 2 right angles. • Construction: Given ᐃABC, extend AB to D and construct BE parallel to AC (Proposition 31).

44. The angle sum theorem, proof • Proof: (a) The exterior angle <DBC is equal to <DBE + <EBC. But <DBE = <BAC and <EBC = < ACB by Proposition 29 parts (a) and (b). • (b) <ABC + <DBC = 2 right angles by Proposition 13. Therefore, using part (a), the sum of the three angles in the triangle equals 2 right angles. QED

45. Parallelograms • Proposition 33. If two opposite sides of a quadrilateral are equal and parallel, then the other pair of opposite sides are also equal and parallel. • Let AB and CD be the given parallel sides and construct CB (Postulate 1)

46. Parallelograms • Proof: We have <BCA = <DCB by Proposition 29 (a). Hence ᐃABC and ᐃDCB are congruent (Proposition 4 – SAS). Therefore BD = AC and <BCA = <DBC. But then BD and AC are also parallel by Proposition 27. QED

47. More on parallelograms • Proposition 34. In a parallelogram, the opposite sides are congruent and the opposite angles are congruent. Moreover a diagonal divides the parallelogram into two congruent triangles. • This follows from Propositions 33 and 29.

48. Comparing areas • Proposition 35. Two parallelograms with the same base and lying between the same parallel lines are equal in area.

49. Comment • This depends on knowing that the entire segment EF lies on one side of CD. • There are other possible arrangements too! Euclid does not address this, but it is not too difficult to adjust the argument to handle the case where the upper sides of the parallelograms overlap too (to appear on a future problem set!)

50. A corollary • Proposition 36. Two parallelograms with congruent bases and lying between the same two parallel lines are equal in area. • This follows from Proposition 35 and Common Notion 1 – say AB = GH. Then areas ABDC = ABFE = GHFE.