The%20Elements,%20Book%20I%20

1
The Elements, Book I Highlights and Comments
2
The 5 Common Notions

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

3
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 distance.
• All right angles are congruent to one another.

4
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.

5
The situation in Postulate 5
6
After Producing the lines sufficiently far,
7
Comments

1. 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
2. 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
3. Postulate 5 is both more complicated than, and
   less obvious than the others(!)

8
Proposition 1

• To construct on a given line segment AB an
  equilateral triangle.
• Construction
• Using Postulate 3, construct the circle with
  center A and radius AB, then the circle with
  center B and radius AB.
• Let C be one of the points of intersection of the
  two circles
• Using Postulate 1, connect AC and BC by line
  segments
• Then ?ABC is equilateral.

9
The Accompanying Figure
10
The Proof

• AC AB, since radii of a circle are all equal
  (this was stipulated in the definition of a
  circle).
• Similarly BC AB.
• Therefore, AC BC (Common Notion 1)
• Hence ?ABC is equilateral (again, this was given
  as a definition previously). QEF

11
A Question

• How do we know that such a point C exists?
• We don't (!) It does not follow from any of
  the Postulates that Euclid set out at the start
  we would need additional postulates to assure
  this.
• Euclid is clearly appealing to our intuition
  about physical circles here and either does not
  realize that there is something missing, or does
  not want to address that point at this stage of
  the development. He doesn't come back to it
  later either(!)

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

• Its 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 two
  equal angles coincide, then the third sides must
  coincide too.
• This idea gives a valid proof, of course, but it
  again raises a question What in the Postulates
  says we can move any figure? Again, there are
  unstated assumptions 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.
• Construction
• Let the triangle be ?ABC with CA CB
• Extend CA to D (Postulate 2)
• On the extension of CB, lay off a line segment CE
  with CE CD (Proposition 3)
• Draw BD and AE (Postulate 1)

16
Figure for Proposition 5

17
Start of proof of Proposition 5

• By the construction CD CE, and CA CB by
  hypothesis. Moreover ? ACE ? BCD.
• Hence ?DCB ? ?ECA (Proposition 4)
• It follows that BD AE, ? CBD ?CAE, and
  ? CDB ? CEA (corresponding parts of
  congruent triangles)
• Then since CE CD and CA CB, we also have AD
  BE (Common Notion 3)
• Therefore, ?AEB ? ?BDA (Proposition 4 note the
  angle at D is the same as the angle at E from the
  third line above.)

18
Conclusion of proof of Proposition 5

• This shows that ? DAB ? EBA, which is the
  second part of the statement of the Proposition.
• To finish the proof we must show that the base
  angles in ?CAB are equal.
• ? EAB ? DBA from the congruence of the small
  triangles at the bottom of the figure.
• But also ? EAC ? DBC by the first congruence
  established before.
• Hence ? BAC ? EAC - ? EAB ? DBC - ? DBA ?
  ABC (Common Notion 3). QED

19
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 ?ABC and
  ?ABD, 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.

20
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.

21
Next, a sequence of bread-and-butter
constructions

• Proposition 9. To bisect a given angle.
• Construction
• Given the angle at A, determine on the sides two
  points B,C with AB AC
• Construct the equilateral triangle ?BCD
  (Proposition 1)
• Then AD bisects the angle.
• (Note This is probably slightly different from
  the way you learned to do this. Can you see
  why?)

22
Angle Bisection The Construction

23
Angle Bisection The Proof

• Proof AB AC by construction.
• BD CD since the triangle ?BCD is equilateral
• AD is common to the two triangles ?ABD and
• ?ACD
• Therefore, ?ABD and ?ACD are congruent
  (Proposition 8).
• Hence ltADB ltADC, and we have done what we set
  out to do the angle at A is bisected. QEF

24
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 ?ABC
• Using Proposition 9, bisect the angle at C
• Let D be the intersection of the angle bisector
  and AB. Then D bisects AB.

25
Line Segment Bisection Proof

• Proof
• AC BC since ?ABC is equilateral
• ltACD ltBCD since CD bisects ltACB
• The side CD is in both triangles ?ACD and ?BCD
• Therefore, ?ACD and ?BCD are congruent by
  Proposition 4.
• Hence AD BD since the corresponding parts of
  congruent triangles are equal. QEF

26
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 ?BCD
  (Proposition 1)
• Then DA is perpendicular to the line at A.

27
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 ?ADB and ?ADC are congruent by
  Proposition 8 (SSS). Hence ltADB ltADC are
  right angles. QEF

28
A group of 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.
  
• Proof Say the ray starts at point B on the
  line, P,Q are on the line on opposite sides of B
  and A is on the ray.
• If ltPBA ltQBA then the two angles are right
  angles (given as a Definition by Euclid).
• Otherwise, use Proposition 11 to erect a
  perpendicular to the line at B, and take C on the
  perpendicular.
• Then reasoning with Common Notions 1 and 2, ltPBA
  ltQBA ltPBC lt QBC so equal to two right
  angles. QED

29
A group of propositions about angles, continued

• 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 the equivalent for him was the
  angle described by two right angles together
  not a huge difference, of course, but it affected
  the way a number of statements were stated and
  proved.

30
A group of propositions about angles, continued

• Proposition 15. Vertical angles are equal.
• Note these are the opposite angles formed by
  the intersections of two lines like ltCPD and
  ltAPB

31
A group of propositions about angles, continued

• Proof ltBPC ltCPD is the same as two right
  angles by Proposition 13. Similarly for ltAPB
  ltBPC. Hence ltCPD ltBPC ltAPB lt BPC by
  Postulate 4. Therefore, ltCPD ltAPB by Common
  Notion 3. QED

32
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 ltDCB is greater than ltCAB,
  ltCBA

33
Proof of Proposition 16

• Euclid's proof is clever! To show ltDBA is
  greater than ltBAC
• 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 ltAEB ltFEC by Proposition 15.

34
Proof of Proposition 16, concluded

• Therefore ?AEB and ?CEF are congruent
  (Proposition 4 SAS). Hence ltBAE ltECF.
• But ltECF is a part of the exterior angle ltDCA.
  So the exterior angle is larger (Common Notion
  5). QED
• A similar argument shows ltDCA is larger than lt
  ABC.

35
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(!)

36
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.

37
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.

38
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.

39
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

40
Proposition 27, continued

• The claim is that if ltAEF ltDFE, then the lines
  AB and CD, even if extended indefinitely, never
  intersect.
• Proof Suppose they did intersect at some point
  G

41
Proposition 27, concluded

• Then the exterior angle ltAEF is equal to the
  opposite interior angle ltEFG in the triangle
  ?EFG.
• But that contradicts Proposition 16. Therefore
  there can be no such point G. QED

42
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.

43
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.

44
Parallel criteria
Proof (a) Suppose for instance that ltGEB
ltGFD. By Proposition 15, ltGEB ltAEH. So ltGFD
ltAEH (Common Notion 1). Hence AB and CD are
parallel by Proposition 27.
45
Parallel criteria
Proof (b) Now suppose for instance that ltHEB
ltGFD 2 right angles. We also have ltHEB lt HEA
2 right angles by Proposition 13. Hence ltGFD
ltHEA (Common Notion 3). Therefore AB and CD are
parallel by Proposition 27. QED
46
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.
47
Familiar facts about parallels
Proof of (c) The claim is that, for instance,
if AB and CD are parallel, then ltBEHltDFG 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 ltBEH ltGEA and lt DFG ltHFC (Proposition
15), while ltGEA ltAEH 2 right angles ltHFC
CFG (Proposition 13), then ltAEH ltCFG is less
than 2 right angles, and Postulate 5 implies the
lines meet on the other side. QED
48
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
• 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
• The other parts of Proposition 29 are proved
  similarly.

49
Further properties of parallels

• Proposition 30. If two lines are parallel to the
  same line then they are parallel to one another.
  
• 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 see discussion on pages
  171 and 172 of BJB if you are interested in
  seeing how this works.

50
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 ltGFE ltBEF (Proposition
  23)
• Proof Then FG and AB, produced indefinitely,
  are parallel lines (Proposition 27).
• 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.

51
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 ?ABC, extend AB to D and
  construct BE parallel to AC (Proposition 31).

52
The angle sum theorem, proof

• Proof (a) The exterior angle ltDBC is equal to
  ltDBE ltEBC. But ltDBE ltBAC and ltEBC lt ACB by
  Proposition 29 parts (a) and (b).
• (b) ltABC ltDBC 2 right angles by Proposition
  13. Therefore, using part (a), the sum of the
  three angles in the triangle equals 2 right
  angles. QED

53
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)

54
Parallelograms

• Proof We have ltBCA ltDCB by Proposition 29
  (a). Hence ?ABC and ?DCB are congruent
  (Proposition 4 SAS). Therefore BD AC and
  ltBCA ltDBC. But then BD and AC are also
  parallel by Proposition 27. QED

55
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.

56
Comparing areas

• The next group of propositions establishes facts
  about areas of triangles and parallelograms.
• Proposition 35. Two parallelograms with the same
  base and lying between the same parallel lines
  are equal in area.

57
Comparing areas

• The next group of propositions establishes facts
  about areas of triangles and parallelograms.
• Proposition 35. Two parallelograms with the same
  base and lying between the same parallel lines
  are equal in area.

58
Comparing areas

• Proof (another clever one!) Let ABDC and
  ABFE be the parallelograms and construct CF.
  ?ACE and ?BFD are congruent since EC FD, AC
  BD and ltACE lt BDF (Proposition 29 and
  Proposition 4). Starting with area of ?ACE,
  subtract ?GCF and add ?ABG to get the area of
  ABFE. But the same subtraction and addition also
  gives the area of ABDC(!) QED

59
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!)

60
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.

61
Corresponding facts for triangles

• Proposition 37. Two triangles with equal bases
  and lying between the same two parallel lines are
  equal in area.
• ?ABC and ?ABD below have the same area if CD is
  parallel to AB

62
Corresponding facts for triangles

• Proof comes by constructing parallelograms with
  diagonals BC and AD, then applying Propositions
  34 and 35.

63
Corresponding facts for triangles

• Proposition 38. Two triangles with congruent
  bases and lying between the same two parallel
  lines are equal in area.
• Follows from Proposition 37 by the same
  construction used to deduce Proposition 36 from
  Proposition 35.
• The next two Propositions 39 and 40 are
  corollaries of 37 and 38. Not needed for our
  purposes, so omitted.
• Proposition 41. If a parallelogram and a
  triangle have the same base and lie between two
  parallel lines, then the parallelogram has double
  the area of the triangle.
• This follows from the previous statements and
  Proposition 34.

64
Getting close!

• Propositions 42 through 45 are also not needed
  for our purposes.
• Proposition 46. To construct a square on a given
  line segment.
• Construction Let AB be the given line segment.
  Erect a perpendicular AC at A (Proposition 11)
  and find D on AC with AC AB (Proposition 3 or
  just use Postulate 3)
• Construct a parallel line to AB passing through D
  (Proposition 31)
• Construct a parallel line to AB passing through D
  (Proposition 31)
• Let E be the intersection of the parallels. Then
  ABED is a square.

65
The goal we have worked toward

• Proposition 47. In a right triangle, the square
  on the hypotenuse is equal to the sum of the
  squares on the two other sides.
• The Theorem of Pythagoras, but stated in terms
  of areas (not as an algebraic identity!)
• The proof Euclid gives has to rank as one of the
  masterpieces of all of mathematics, although it
  is far from the simplest possible proof (as we
  have seen already and will discuss shortly).
• The thing that is truly remarkable is the way the
  proof uses just what has been developed so far in
  Book I of the Elements.

66
Euclid's proof, construction

• Let the right triangle be ?ABC with right angle
  at A
• Construct the squares on the three sides
  (Proposition 46) and draw a line through A
  parallel to BD (Proposition 31)

67
Euclid's proof, step 1

• ltBAG ltBAC 2 right angles, so G,A,C are all
  on one line (Proposition 14), and that line is
  parallel to FB (Proposition 28)
• Consider ?FBG

68
Euclid's proof, step 2

• Proposition 37 implies areas of ?FBG and ?FBC
  are equal

69
Euclid's proof, step 3

• AB FB and BC BD since ABFG and BCED are
  squares
• ltABD ltFBC since each is a right angle plus ltABC
• Hence ?BFC and ?ABD are congruent (Proposition 4
  SAS)

70
Euclid's proof, step 4

• Proposition 37 again implies ?BDA and ?BDM have
  the same area.
• Hence square ABFG and rectangle BLKD have the
  same area (twice the corresponding triangles
  Proposition 41)

71
Euclid's proof, conclusion

• A similar argument on the other side shows
  ACKH and CLME have the same area
• Therefore BCDE ABFG ACKH. QED