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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.

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.

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.

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(!)

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.

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

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(!)

A QuestionPropositions 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

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.

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(!)

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 ΔABC 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)

Start of proof of Proposition 5

- By the construction CD = CE, and CA = CB by hypothesis. Moreover ᐸ ACE = ᐸ BCD.
- Hence ΔDCB ≅ ΔECA (Proposition 4)
- It follows that BD = AE, ᐸ CBD = ᐸCAE, and ᐸ CDB = ᐸ CEA (corresponding parts of congruent triangles)
- Then since CE = CD and CA = CB, we also have AD = BE (Common Notion 3)
- Therefore, ΔAEB ≅ ΔBDA (Proposition 4 – note the angle at D is the same as the angle at E from the third line above.)

Conclusion of proof of Proposition 5

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

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.

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.”

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 ΔBCD (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?)

Angle Bisection – The Proof

- Proof: AB = AC by construction.
- BD = CD since the triangle ΔBCD is equilateral
- AD is common to the two triangles ΔABD and
- ΔACD
- Therefore, ΔABD and ΔACD are congruent (Proposition 8).
- Hence <ADB = <ADC, and we have done what we set out to do – the angle at A is bisected. QEF

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.

Line Segment Bisection – Proof

- Proof:
- AC = BC since ΔABC is equilateral
- <ACD = <BCD since CD bisects <ACB
- The side CD is in both triangles ΔACD and ΔBCD
- Therefore, ΔACD and ΔBCD are congruent by Proposition 4.
- Hence AD = BD since the corresponding parts of congruent triangles are equal. QEF

“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.

“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

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 <PBA = <QBA 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, <PBA + <QBA = <PBC + < QBC so equal to two right angles. QED

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.

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 <CPD and <APB:

A group of propositions about angles, continued

- 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

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:

Proof of Proposition 16

- Euclid\'s proof is clever! To show <DBA 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.

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.

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(!)

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.

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.

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.

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

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

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

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.

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.

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.

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

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.

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

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.

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.

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).
- 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.

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).

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

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)

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

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.

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.

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.

Comparing areas

- Proof: (another “clever” one!) Let ABDC and ABFE be the parallelograms and construct CF. ᐃACE and ᐃBFD are congruent since EC = FD, AC = BD and <ACE = < BDF (Proposition 29 and Proposition 4). Starting with area of ᐃACE, subtract ᐃGCF and add ᐃABG to get the area of ABFE. But the same subtraction and addition also gives the area of ABDC(!) QED

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!)

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.

Corresponding facts for triangles

- Proposition 37. Two triangles with equal bases and lying between the same two parallel lines are equal in area.
- ᐃABC and ᐃABD below have the same area if CD is parallel to AB:

Corresponding facts for triangles

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

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.

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.

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.

Euclid\'s proof, construction

- Let the right triangle be ᐃABC 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)

Euclid\'s proof, step 1

- <BAG + <BAC = 2 right angles, so G,A,C are all on one line (Proposition 14), and that line is parallel to FB (Proposition 28)
- Consider ᐃFBG

Euclid\'s proof, step 2

- Proposition 37 implies areas of ᐃFBG and ᐃFBC are equal

Euclid\'s proof, step 3

- AB = FB and BC = BD since ABFG and BCED are squares
- <ABD = <FBC since each is a right angle plus <ABC
- Hence ᐃBFC and ᐃABD are congruent (Proposition 4 – SAS)

Euclid\'s proof, step 4

- Proposition 37 again implies ᐃBDA and ᐃBDM have the same area.
- Hence square ABFG and rectangle BLKD have the same area (twice the corresponding triangles – Proposition 41)

Euclid\'s proof, conclusion

- A similar argument “on the other side” shows ACKH and CLME have the same area
- Therefore BCDE = ABFG + ACKH. QED

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