Transcript Lecture notes

The Elements, Book I – Highlights and
Comments
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.
The First Four Postulates
1. (It is possible) to draw a straight line from
any point to any point.
2. (It is possible) to produce any finite straight
line continuously in a straight line.
3. (It is possible) to describe a circle with any
center and distance.
4. 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.
The situation in Postulate 5
After Producing the lines sufficiently far,
…
Comments
1.
2.
3.
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
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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 Accompanying Figure
The Proof
1.
AC = AB, since radii of a circle are all equal (this
was stipulated in the definition of a circle).
2.
Similarly BC = AB.
3.
Therefore, AC = BC (Common Notion 1)
4.
Hence ∆ABC is equilateral (again, this was given
as a definition previously). QEF
A Question
How do we know that such a point C exists?
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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
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later either(!)
Propositions 2 and 3
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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
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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
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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”
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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.
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Construction:
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Let the triangle be ΔABC with CA = CB
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Extend CA to D (Postulate 2)
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On the extension of CB, lay off a line segment CE
with CE = CD (Proposition 3)
Draw BD and AE (Postulate 1)
Figure for Proposition 5
Start of proof of Proposition 5
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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
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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
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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
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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
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Proposition 9. To bisect a given angle.
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Construction:
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Given the angle at A, determine on the sides two points
B,C with AB = AC
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Construct the equilateral triangle ΔBCD (Proposition 1)
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Then AD bisects the angle.
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(Note: This is probably slightly different from the way
you learned to do this. Can you see why?)
Angle Bisection – The Construction
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Angle Bisection – The Proof
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Proof: AB = AC by construction.
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BD = CD since the triangle ΔBCD is equilateral
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AD is common to the two triangles ΔABD and
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ΔACD
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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
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Proposition 10. To bisect a given line segment.
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Construction: Let AB be the given line segment
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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
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Proof:
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AC = BC since ΔABC is equilateral
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<ACD = <BCD since CD bisects <ACB
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The side CD is in both triangles ΔACD and ΔBCD
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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
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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.
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Construct an equilateral triangle ΔBCD (Proposition 1)
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Then DA is perpendicular to the line at A.
“Dropping” a perpendicular
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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.
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Let D be the midpoint of BC (Proposition 10)
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Then DA is perpendicular to the line at D.
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Proof: ΔADB and ΔADC are congruent by Proposition 8
(“SSS”). Hence <ADB = <ADC are right angles. QEF
A group of propositions about angles
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
a) One side and the two adjacent angles of one triangle are equal
to one side and the two adjacent angles of the other triangle
b) 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.
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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
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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
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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
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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
a)
Two corresponding angles are congruent, or
b)
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
a)
Two corresponding angles are congruent, or
b)
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
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This is the first use of Postulate 5 in Book I of the Elements
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It is almost as if Euclid wanted to delay using it as long as possible
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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
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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
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Proposition 31. To construct a line parallel to a given line and
passing through a given point not on that line.
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Construction: Say AB is the line and F is the given point.
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Pick any point E on AB and construct EF (Postulate 1)
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Construct FG so that <GFE = <BEF (Proposition 23)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Proof comes by constructing parallelograms with diagonals BC
and AD, then applying Propositions 34 and 35.
Corresponding facts for triangles
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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!
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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)
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Construct a parallel line to AB passing through D (Proposition 31)
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Construct a parallel line to AB passing through D (Proposition 31)
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Let E be the intersection of the parallels. Then ABED is a square.
The goal we have worked toward
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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
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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
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<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
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Proposition 37 implies areas of ᐃFBG and ᐃFBC are equal
Euclid's proof, step 3
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AB = FB and BC = BD since ABFG and BCED are squares
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<ABD = <FBC since each is a right angle plus <ABC
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Hence ᐃBFC and ᐃABD are congruent (Proposition 4 – SAS)
Euclid's proof, step 4
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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
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A similar argument “on the other side” shows ACKH and CLME
have the same area
Therefore BCDE = ABFG + ACKH. QED