Download Euclid`s Five Postulates Some of Euclid`s Book 1 Definitions

1
Euclid's Five Postulates
These are the axioms of standard Euclidean Geometry.
They appear at the start of Book I of The Elements by Euclid.
Note that while these are the only axioms that Euclid explicitly uses, he implicitly uses others
such as Pasch's Axiom.
Postulate 1
A straight line segment can be drawn joining any two points.
Postulate 2
Any straight line segment can be extended indefinitely to form a straight line.
Postulate 3
Given any line segment, a circle can be drawn using the segment as the radius
with one endpoint as the center.
Postulate 4
All right angles are congruent.
Postulate 5
If a straight line falling on two straight lines make the interior 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.
Some of Euclid's Book 1 Definitions
1. A point is that which has no part.
This can be interpreted to mean that a point is something that cannot be divided into
anything smaller.
2. A line is breadthless length.
A line is a construct that has no thickness. It can be considered as a continuous
succession of points.
4. A straight line is a line which lies evenly with the points on itself.
8. A plane angle is the inclination to one another of two lines in a plane which meet one
another and do not lie in a straight line.
Thus, the amount of rotation about the intersection required to bring one line into
correspondence with the other is the angle between the lines.
2
10. R
Right Angle / Perpendicuular Lines:
W
When
a straig
ght line set up
u on a straigght line makkes the adjaceent angles eqqual to one
annother, each
h of the equaal angles is right,
r
and thhe straight linne standing on
o the other is
caalled a perpendicular too that on whhich it standss.
Inn the diagram
m, the line CD
C has been constructedd so as to be a
peerpendicula
ar to the line AB.
b one line
15. A circle is a plane figuree contained by
such that alll the straight line [segmeents]
falling upon
n it from one point among those
lying within
n the figure are
a equal to one
o
another.
16. A
And the poin
nt is called thhe center off the
circle.
Euclid
d's Comm
mon Notioons
matic statemeents that apppear at the staart of Book I of The Elem
ments by Euuclid.
This is a set of axiom
1. Thhings which
h are equal too the same thhing are alsoo equal to eacch other.
2. Iff equals are added
a
to equuals, the whooles are equaal.
3. Iff equals are subtracted
s
frrom equals, the
t remaindeers are equall.
4. Thhings which
h coincide wiith one anothher are equal to one anotther.
5. Thhe whole is greater than the part.
3
24 Eucclidean Propositio
P
ons
1. Consstruction of
o Equilateeral Trianggle
Theorem
m
On a giveen straight liine segment,, it is possiblle to
constructt an equilaterral triangle.
Construcction
Let AB be the given straight
s
line segment.
We consttruct a circlee BCD with center
c
A andd
radius AB
B.
We consttruct a circlee ACE with center
c
B andd radius
AB.
From C, where the ciircles interseect, we draw
w a line segm
ment to A andd to B to form
m the straighht
line segm
ments AC and
d BC.
Then ABC
A
is the eq
quilateral triaangle requireed.
Proof
fr
Book I Definition
D
155: Circle thaat AC = AB.
As A is thhe center of circle BCD, it follows from
As B is thhe center of circle ACE, it follows frrom Book I Definition
D
155: Circle thaat BC = AB.
So, as AC
C = AB and BC = AB, itt follows froom Common Notion 1 that AC = BC
C.
Therefore AB = AC = BC.
Therefore ABC is equilateral.
■
Historicaal Note
This is Prroposition 1 of Book I of
o Euclid's Thhe Elements.
2. Triaangle Side--Angle-Sidde Equalityy Implies Congruenc
C
ce
Theorem
m
If two triiangles have:
• tw
wo sides equ
ual to two siddes respectivvely;
• thhe angles con
ntained by thhe equal straaight lines eqqual
they will also have:
des equal;
• thheir third sid
4
•
thhe remaining
g two angles equal to theeir respectivee remaining angles, nam
mely, those which
w
thhe equal sidees subtend.
Proof
BC and DEF
D
be twoo
Let AB
triangles having sides
AB = DE
E and AC = DF , and wiith
BAC = EDF.
If ABC
C is placed on
o
DEF
such thatt:
the pointt A is placed
d on point D,
D
and the line
l AB is placed on linee
DE thenn the point B will also
coincide with point E because
AB = DE
E.
So, with AB coincid
ding with DE
E , the line AC
A will coiincide with the
t line DF because
BAC = EDF.
Hence thhe point C will
w also coinncide with thhe point F , because AC
C = DF.
But B also coincided with E.
Hence thhe line BC will coincide with
w line EF
F.
(Otherwiise, when B coincides with
w
E and C with F, the line BC
C will not cooincide with line
EF and two
t straight lines will ennclose a spacce which is impossible.)
i
Therefore BC will coincide
c
withh EF and bee equal to it.
Thus the whole AB
BC will coinncide with thhe whole DEF and thuus ABC = DEF.
The remaaining angles on ABC
C will coincidde with the remaining anngles on DEF
D and be equal
e
to them.
■
Historicaal Note
This is Prroposition 4 of Book I of
o Euclid's Thhe Elements.
3. Isosceles Trianngles havee Two Equal Angles
Theorem
m
In isosceles triangless, the angles at the base are
a equal to each
e
other.
t equal straaight lines arre extended, the angles under
u
the
Also, if the
base willl also be equ
ual to each otther.
Proof
Let ABC be an isossceles trianglle whose sidde AB equalss side AC.
We extennd the straigh
ht lines AB and
a AC to D and E respeectively.
Let F bee a point on BD.
5
We cut off
o from AE a length AG equal to AF.
We draw
w line segmen
nts FC andd GB.
Since AF
F = AG and AB = AC , the two sidees FA and AC
A are equaal to GA annd AB
respectivvely.
They conntain a comm
mon angle, thhat is, FAG
G.
So by
, AFC = AGB.
C = GB, A
ACF = ABG
G and AF
FC = AGB
B.
Thus FC
Since AF
F = AG and AB = AC, then
t
BF = CG.
C
But FC = GB, so thee two sides BF,
B FC are equal
e
to the two sides CG
G, GB respeectively.
Then B
BFC = CG
GB while CB
B is commoon to both.
Therefore by Trianglle Side-Anglle-Side Equaality, BFC
C = CGB.
Therefore FBC = GCB andd BCF = CBG.
So since ACF = A
ABG, and inn these BC
CF = CBG,, then ABC
C = ACB.
But AB
BC and AC
CB are at thhe base of ABC.
Also, we have alread
dy proved thaat FBC = GCB, andd these are thhe angles under the base of
ABC.
Hence thhe result.
■
Historicaal Note
This is Prroposition 5 of Book I of
o Euclid's Thhe Elements.
4. Triaangle Side--Side-Side Equality Implies
I
Coongruence
Theorem
m
Let two triangles
t
hav
ve all three sides equal.
Then they also have all three anggles equal.
w
sides are
a all equal are themselvves equal.
Thus twoo triangles whose
Proof
BC and DE
EF be two
Let AB
triangles such that:
AB = DE
E;
AC = DF
F;
BC = EF
F.
Suppose ABC werre
superimpposed over DEF so
that point B is placeed on
a the line BC
B on EF.
point E and
Then C will coincid
de with F, ass BC = EF and
a so BC cooincides withh EF.
6
Now supppose BA do
oes not coinccide with ED
D and AC does not coiincide with DF.
Then they will fall ass, for example, EG and GF.
G
Thus therre will be tw
wo pairs of sttraight line segments
s
connstructed on the same linne segment, on
the same side as it, meeting
m
at diffferent pointts, in contraddiction to Tw
wo Lines Meet at Uniquee
Point.
Therefore BA coinccides with ED and AC coincides with
w DF.
Therefore BAC coiincides with EDF and is equal to it.
i
The samee argument can
c be applieed to the othher two sidess, and thus we
w show that all
corresponnding angless are equal.
■
Historicaal Note
This is Prroposition 8 of Book I of
o Euclid's Thhe Elements.
5. Biseection of ann Angle
Construcction
AC be the giv
ven angle to be bisected..
Let BA
Take anyy point D on
n AB.
We cut off
o from AC a length AE
E equal to AD.
A
We draw
w the line seg
gment DE.
We consttruct an equiilateral triangle DEF on AB.
We draw
w the line seg
gment AF.
Then the angle BA
AC has been bisected by the straight
ment AF.
line segm
Proof
We have:
AD = AE
E;
AF is coommon;
DF = EF
F.
Thus triaangles AD
DF and
Thus D
DAF = EA
AF.
AE
EF are equall.
Hence BAC has beeen bisectedd by AF.
■
Historicaal Note
This is Prroposition 9 of Book I of
o Euclid's Thhe Elements.
d easier constructions off a bisection, but this parrticular one uses
u only ressults
There aree quicker and
previouslly demonstraated.
7
6. Biseection of a Straight Line
L
Theorem
m
It is possible to bisecct a straight line
l segmentt.
Construcction
s
line segment.
Let AB be the given straight
We consttruct an equiilateral triangle ABC on AB.
We bisecct the angle ACB by thhe straight linne segment
CD.
Then AB has been bisected at thee point D.
Proof
As AB
BC is an equilateral trianngle, it follow
ws that AC = CB.
The two triangles ACD and BCD have side CD in
i common, and side AC
C of ACD
D
equals sidde BC of BCD.
The angle ACD su
ubtended by lines AC annd CD equuals the anglee BCD subbtended by lines
l
BC and CD, as A
ACB was biseected.
So trianggles ACD and BCD
D are equal.
Therefore AD = DB..
So AB haas been biseccted at the pooint D.
■
Historicaal Note
This is Prroposition 10 of Book I of Euclid's The
T Elementts.
7. Connstruction of a Perpeendicular through
t
a Given
G
Poin
nt
m
Theorem
Given ann infinite straaight line, annd a given pooint not on thhat straight line,
l
it is posssible to draw
wa
perpendicular to the given
g
straighht line.
Construcction
Let AB be two pointts on the givven infinite straight
s
line.
Let C be the given point
p
not on it.
Let D be some poin
nt not on AB
B on the otheer side
of it from
m C.
8
We consttruct a circlee EFG withh center C and
a radius CD.
C
We bisecct the straigh
ht line EG at
a the point H.
H
We draw
w line segmen
nts from C to each of G,
G H and E too form the sttraight line segments
s
CG
G,
CH and CH.
Then the line CH is perpendicula
p
ar to the giveen infinite sttraight line AB
A through thhe given poiint C.
Proof
As C is the center of
o circle BCD
D, it followss from Book I Definition 15: Circle that GC = CE.
C
As EG has
h been bissected, GH = HE.
Thus, as GC = CE an
nd GH = HE
E, and CH is
i common, by
b
,
CHG = EHG.
Therefore CHG = CHE.
So CH is
i a straight line set up on
o a straight line making the adjacennt angles equual to one
another. Thus it follo
ows from Boook I Definittion 10: Righht Angle thatt each of CHG and
CHE are
a right anglles.
So the strraight line CH
C has beenn drawn at riight angles to the given infinite
i
straight line AB
through the
t given point C.
■
Historicaal Note
This is Prroposition 12 of Book I of Euclid's The
T Elementts
8. Verttical Anglee Theorem
m
Theorem
m
If two strraight lines cut
c each otheer, they makke the opposiite angles eqqual each othher.
Proof
Let AB and CD bee two straight lines that cut
c each otheer at the poinnt E.
Since thee straight line AE stands on the
straight line CD, the angles AE
ED and
m
two rig
ght angles.
AEC make
Since thee straight line DE standds on the
straight line AB, the angles AE
ED and
m
two rig
ght angles.
BED make
But AE
ED and A
AEC also make
m
two
right anggles.
So by Coommon Notio
on 1 and the fact that all
right anggles are cong
gruent,
AED + AEC = AED + B
BED .
Let AE
ED be subtrracted from each.
e
Then by Common No
otion 3 it follows that AEC = BE
ED .
Similarlyy it can be sh
hown that BEC = AE
ED.
■
9
Historicaal Note
This is Prroposition 15 of Book I of Euclid's The
T Elementts.
Internaal/Externall Angle Deefinitions
■ Intern
nal Angle
The interrnal angle (or interior angle)
a
of a vertex
v
is the size of the angle
a
betweeen the sides
forming that
t vertex, as
a measuredd inside the polygon.
p
■ Extern
nal Angle
Surprisinngly, the exteernal angle (or exteriorr angle) of a
vertex is not the size of the anglee between the sides form
ming
that verteex, as measu
ured outside the polygon.
It is in faact an angle formed
f
by one side of a polygon andd a
line prodduced from an
a adjacent side.
While AFE is the internal
i
angle of vertex F , the exterrnal
angle of this
t vertex iss EFG.
Note: it doesn't
d
matteer which adjacent side yoou use, sincee
they are equal
e
by thee Vertical Anngle Theorem
m.
9. Exteernal Anglle of Trianngle Greateer than Intternal Oppposite Anggles
m
Theorem
The exterrnal angle off a triangle is greater thaan either of thhe opposite internal anggles.
Proof
Let AB
BC be a trian
ngle.
Let the siide BC be extended
e
to D.
Let AC be bisected at E.
Let BE be joined an
nd extended to F.
Let EF be made equ
ual to BE.
(Techniccally we reallly need to exxtend BE too a point
beyond F and then crimp
c
off a length
l
EF.)
Let CF be joined.
Let AC be extended
d to G.
We have AEB = CEF from Two Straighht Lines
make Equual Oppositee Angles.
Since AE
E = EC and BE = EF , from
f
Trianggle Side-Anglle-Side Equaality we havve
ABE = CFE.
Thus AB
B = CF and BAE = ECF.
But EC
CD is greateer than EC
CF.
10
Therefore
ACD iss greater thann
BAE.
Similarlyy, if BC werre bisected, BCG, whhich is equal to
Equal Oppposite Anglles, would bee shown to be
b greater than
ACD by Two Straaight Lines make
m
ABC as
a well.
Hence thhe result.
■
Historicaal Note
This is Prroposition 16 of Book I of Euclid's The
T Elementts.
10. Tw
wo Angles of Trianglle Less thaan Two Rigght Angless
Theorem
m
In any triiangle, two angles
a
taken together in any manner are less thann two right angles.
a
Proof
Let AB
BC be a triaangle.
Let the siide BC be ex
xtended to D.
D
Since thee angle AC
CD is an extternal angle of
ABC, it
i follows thaat it is greateer than both
BAC annd ABC.
We add ACB to bo
oth, so that
ACD + ACB is greater thann
ABC + ACB.
But AC
CD + ACB
B is equal too two right angles.
a
Therefore ABC + ACB is leess than two right angles.
w show that the same appplies to thee other two pairs
p
of internnal angles of
In a similar manner we
ABC.
■
Historicaal Note
This is Prroposition 17 of Book I of Euclid's The
T Elementts.
11. Grreater Sidee of Trianggle Subtendds Greaterr Angle
Theorem
m
In any triiangle, the greater side subtends the greater anglle.
Proof
BC be a triaangle such thhat AC is
Let AB
greater thhan AB.
Let AD be made equ
ual to AB.
Let BD be joined.
11
Then A
ADB is an exterior
e
angle of the trianngle BCD.
Therefore from AD
DB is greateer than AC
CB.
As AD = AB, the triaangle ABD
D is isosceles.
From Isoosceles Trian
ngles have Two
T Equal Anngles, AD
DB = ABD.
Therefore ABD iss greater thann ACB.
Therefore, as ABC
C = ABD + DBC, it follows
f
that ABC is greater than
ACB.
Hence thhe result.
■
Historicaal Note
This is Prroposition 18 of Book I of Euclid's The
T Elementts.
This theoorem is the converse
c
of Proposition
P
1 Greater Angle
19:
A
of Triiangle Subteended by Greeater
Side.
12. Grreater Anggle of Trianngle Subteended by Greater
G
Sid
de
Theorem
m
In any triiangle, the greater angle is subtendedd by the greaater side.
Proof
Let AB
BC be a trian
ngle such thaat
BCA.
ABC is greater thann
Suppose AC is not greater
g
than AB.
If AC were
w equal to
o AB, then byy Isosceles Triangles
T
haave Two
Equal Anngles, ABC
C = BCA, but they're not
n so it isn'tt.
If AC were
w less than
n AB, then by
b Greater Side
S of Trianngle
Subtendss Greater An
ngle it wouldd follow that ABC is leess than
BCA, but
b it's not so
o it isn't.
So AC must
m be greateer than AB
Hence thhe result.
■
Historicaal Note
This is Prroposition 19 of Book I of Euclid's The
T Elementts.
This theoorem is the converse
c
of Proposition
P
1 Greater Side of Triangle Subtendds Greater
18:
Angle.
12
13. Suum of Two Sides of Triangle
T
Grreater than
n Third Siide
Theorem
m
Given a triangle
t
ABC
C, the sum off the lengthss of any two sides of the triangle is greater
g
than the
t
length off the third sid
de.
Proof
We can extend
e
BA past
p A into a straight linee.
There exists a point D such thatt DA = CA.
Therefore, ADC = ACD beccause isoscelles triangle have
h
two equaal angles.
Thus, B
BCD > BD
DC by Eucliid's fifth com
mmon notion.
Since DCB is a triiangle havinng BCD greater
g
than
BDC, this means that
t
BD > BC.
B
But BD = BA + AD , and AD = AC.
Thus, BA
A + AC > BC
C.
A similarr argument shows
s
that AC
A + BC > BA
B and
BA + BC
C > AC.
■
Historicaal Note
This is Prroposition 20 of Book I of Euclid's The
T Elementts.
It is a geoometric interrpretation off the trianglee inequality.
14. Coonstructionn of Trianggle from Given
G
Lenggths
Theorem
m
Given thrree straight lines
l
such thhat the sum of
o the lengthhs of any twoo of the liness is greater thhan
the lengthh of the third
d line, it is possible
p
to coonstruct a triiangle having the lengthss of these linnes
as its sidee lengths.
Construcction
Let the thhree straightt lines
from whiich we are going to
constructt the trianglee be a, b,
and c.
Let D annd E be any
y distinct
points. Construct
C
DE
E and
extend it beyond E.
We cut off
o a length DF
D on
DE equaal to a.
Similarlyy, we cut offf
FG = b and
a
GH = c on DE.
Now we can construcct a circle
centered at F with radius
r
DF.
13
Similarly, we can construct a circle centered at G with radius GH.
Call one of the intersections of the two circles K, without loss of generality, let K to be the top
point of intersection in the accompanying diagram.
Finally, we can construct the segment FK.
FGK is the required triangle.
Proof
Since F is the center of the circle with radius FD, it follows from Book I Definition 15: Circle
that DF = KF, so a = KF by Euclid's first common notion.
Since G is the center of the circle with radius GH, it follows from Book I Definition 15: Circle
that GH = GK, so c = GK by Euclid's first common notion.
FG = b by construction.
Therefore the lines FK, FG, and GK are, respectively, equal to the lines a, b, and c, so FGK
is indeed the required triangle.
■
Historical Note
This is Proposition 22 of Book I of Euclid's The Elements.
Note that the condition required of the lengths of the segments is the equality shown in
Proposition 20: Sum of Two Sides of Triangle Greater than Third Side. Thus, this is a necessary
condition for the construction of a triangle.
When Euclid first wrote the poof of this proposition in The Elements, he neglected to prove that
the two circles described in the construction actually do intersect, just as he did in Proposition 1:
Construction of Equilateral Triangle.
15. Triangle Angle-Side-Angle and Side-Angle-Angle Equality Implies
Congruence
Theorem
Part 1
If two triangles have
• Two angles equal to two angles, respectively;
• The sides between the two angles equal
Then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and the included sides are equal, then the triangles are equal.
Part 2
If two triangles have
• Two angles equal to two angles, respectively;
• The sides opposite one pair of equal angles equal
Then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and a pair of opposite sides are equal, then the triangles are
equal.
14
Proof
Part 1
BC = DEF
F, BCA = EFD, andd
Let AB
BC = EF
F.
Assume AB ≠ DE. If
I this is the case, one off the
two mustt be greater. Without losss, we let AB
B > DE.
We consttruct a pointt G on AB such that
BG = ED
D, and then we
w construct the segmentt CG.
Now, sinnce we have BG
B = ED, GBC = D
DEF,
and BC = EF, from Triangle
T
Sidde-Angle-Sidde
Equality we have G
GCB = DF
FE.
But from
m Euclid's fift
fth common notion
n
DF
FE = ACB > GCB, a contradictioon.
Therefore, AB = DE
E, so from Trriangle Side-Angle-Side Equality, we
w have AB
BC =
DEF
F.
■
Part 2
Let AB
BC = DEF,, BCA = EFD, and
AB = DE
E.
Assume BC ≠ EF. If
I this is the case,
c
one off
the two must
m be greaater. Withoutt loss of
generalityy, we let BC
C > EF.
We consttruct a pointt H on BC such that
BH = EF
F, and then we
w construct the segmentt
AH.
Now, sinnce we have BH
B = EF,
ABH = DEF, and
d AB = DE, from
f
Trianggle
Side-Anggle-Side Equality we havve
BHA = EFD.
But from
m External Angle
A
of Triaangle Greateer than Internnal Oppositee,
we have BHA > HCA = E
EFD, a contrradiction.
■
Historicaal Note
This is Prroposition 26 of Book I of Euclid's The
T Elementts.
16. Eqqual Alternnate Interiior Angles Implies Paarallel
Theoreem
Given tw
wo infinite strraight lines which
w
are cuut by a transvversal, if thee alternate innterior angless are
equal, theen the lines are
a parallel.
15
Proof
Let AB and CD bee two straight lines,
and let EF
E be a tran
nsversal that cuts
them. Let the at leastt one pair of
alternate interior ang
gles, without loss of
generalityy, let AEF
F and EF
FD be
equal.
Assume that
t the liness are not parrallel.
Then the meet at som
me point G which withoout loss of geenerality is on the samee side as B and
a
D.
e
anglle of GEF
F, from Exterrnal Angle of
o Triangle Greater
G
thann
Since AEF is an exterior
Internal Opposite, AEF > E
EFG, a contraadiction.
ot meet on thhe side of A and C.
Similarlyy, they canno
Therefore, by definittion, they aree parallel.
■
Historicaal Note
This is Prroposition 27 of Book I of Euclid's The
T Elementts.
This theoorem is the converse
c
of the
t first part of Propositiion 29.
17. Eqqual Correesponding Angles
A
or Supplemeentary Inteerior Anglees
Im
mplies Para
allel
m
Theorem
Part 1
Given tw
wo infinite strraight lines which
w
are cuut by a
transverssal, if the corrresponding angles are equal, then thhe
lines are parallel.
Part 2
wo infinite strraight lines which
w
are cuut by a
Given tw
transverssal, if the inteerior angles on the samee side of the
transverssal are suppleementary, thhen the lines are parallel..
Proof
Part 1
Let AB and CD bee infinite straaight lines, annd let EF be
b a transverrsal that cuts them. Let at
least one pair of corresponding angles,
a
Withoout loss, let EGB andd GHD, bee equal.
Then GHD = EGB = AGH
H by the Verrtical Angle Theorem.
Thus AB
B CD by Equal
E
Alternaate Interior Angles
A
Impllies Parallel..
■
Part 2
Let AB annd CD be in
nfinite straighht lines, and let EF be a transversall that cuts thhem. Let at leeast
one pair of interior an
ngles on the same side of
o the transveersal, Withouut loss, let BGH and
16
DHG be supplemeentary, so byy definition DHG + BGH equaals two rightt angles.
AGH + BGH eq
quals two rigght angles.
Then from
m Euclid's first
fi and thirdd common notion
n
and Euclid's
E
fourtth postulate,,
AGH = DHG.
Finally, AB CD by
y Equal Alteernate Interioor Angles Im
mplies Paralllel.
■
Historicaal Note
This is Prroposition 28 of Book I of Euclid's The
T Elementts.
This theoorem is the converse
c
of the
t second annd third partts of Proposiition 29.
18. Paarallel Impplies Equall Alternatee Interior Angles,
A
Coorrespondiing Angles,
an
nd Supplem
mentary In
nterior Angles
Part 1
Given tw
wo infinite strraight lines which
w
are cuut by a transvversal, if thee lines are paarallel, then the
t
alternate interior ang
gles are equal.
Part 2
Given tw
wo infinite strraight lines which
w
are cuut by a transvversal, if thee lines are paarallel, then the
t
corresponnding angless are equal.
Part 3
Given tw
wo infinite strraight lines which
w
are cuut by a transvversal, if thee lines are paarallel, then the
t
interior angles
a
on thee same side of
o the transvversal are suppplementaryy.
Proof
Let AB and CD bee parallel infiinite straightt lines,
and let EF
E be a tran
nsversal that cuts them.
Part 1
t alternatee interior anggles are not equal.
e
Assume the
Then onee of the pair AGH andd GHD must
m be
greater. Without
W
losss of generalitty, let AG
GH be
greater.
Now A
AGH + BG
GH equal two
t right anngles,
so
GHD + BGH is lesss than two rigght
angles.
Lines exttended infiniitely from anngles less thaan two rightt angles mustt meet due too Euclid's fiffth
postulatee. But the lin
nes are paralllel, so by deefinition the lines do not intersect. Thhis is a
contradicction.
Thus, thee alternate in
nterior angles must be eqqual.
■
17
Part 2
From part 1,
AGH = DHG.
So
EGB = AGH = DHG due to the Vertical Angle Theorem: If two straight lines
cut each other, they make the opposite angles equal each other.
■
Part 3
From part 2 and Euclid's second common notion,
EGB + BGH = DHG + BGH.
Further, EGB + BGH equal two right angles,
so by definition BGH and DHG are supplementary.
I.e., when BGH and DHG are set next to each other, they form a straight angle.
■
Note: This is Proposition 29 of Book I of Euclid's The Elements. Proposition 29 is the first
proposition to make use of Euclid's fifth postulate.
19. Construction of a Parallel
Theorem
Given an infinite straight line, and a given point not on that straight line, it is possible to draw a
parallel to the given straight line.
Construction
Let A be the point, and let BC
be the infinite straight line.
We take a point D at random
on BC, and construct the
segment AD.
We construct DAE equal to
ADC on AD at point A.
We extend AE into an infinite straight line.
Then the line AE is parallel to the given infinite straight line BC through the given point A.
Proof
Since the transversal AD cuts the lines BC and AE and makes
that EA BC.
■
Historical Note
This is Proposition 31 of Book I of Euclid's The Elements.
DAE = ADC, it follows
18
20. Suum of Anglles of Trianngle Equalls Two Rigght Angless
Theorem
m
In a trianngle, the sum
m of the threee interior anggles equals two
t right anggles.
Proof
BC be a trian
ngle, and let BC be exteended to a pooint D. Construct CE thhrough the point
p
Let AB
C paralleel to the straaight line AB
B.
Since AB
B CE and AC is a trannsversal thatt cuts them,
it followss that BAC
C = ACE.
Similarlyy, since AB CE and BD
B is a transsversal that
cuts them
m, it follows that ECD
D = ABC.
Thus by Euclid's
E
Seccond Commoon Notion,
ACD = ABC + BAC.
Again byy by Euclid'ss Second Com
mmon Notioon,
ACB + ACD = ABC + B
BAC + ACB
CB.
But AC
CB + AC
CD equals tw
wo right anglles, so by
Euclid's First
F
Commo
on Notion
ABC + BAC + ACB equaals two rightt angles.
■
Historicaal Note
This is Prroposition 32 of Book I of Euclid's The
T Elementts.
Euclid's proposition
p
32
3 actually consists
c
of tw
wo parts, thee first of whiich is that inn a triangle, if
i one
of the siddes is extend
ded, then the exterior anggle equals thhe sum of thee two opposiite interior
angles. This
T is proved in the courrse of provinng the more useful and widely
w
know
wn part of thee
propositiion, that which is given here.
h
21. Rad
dius at Rig
ght Angle to
t Tangentt
Theorem
m
If a straigght line toucches a circle at only one point,
p
and iff a straight liine is joined from the cennter
to the point of contacct, then the straight
s
line so
s formed will
w be perpenndicular to thhe tangent.
Proof
Let DE be tangent to
t the circle ABC at C.
Let F bee the center of ABC.
Let FC be the radiu
us in questionn.
ot perpendicuular to DE.
Suppose FC were no
Instead, suppose
s
FG were drawnn perpendicuular to
DE.
19
Since FGC is a right angle, then from Two Angles of Triangle Less than Two Right Angles it
follows that FCG is acute.
From Greater Angle of Triangle Subtended by Greater Side it follows that FC > FG.
But FC = FB and so FB > FG, which is impossible.
So FG cannot be perpendicular to DE.
Similarly it can be proved that no other straight line except FC can be perpendicular to DE.
Therefore FC is perpendicular to DE.
■
Historical Note
This is Proposition 18 of Book III of Euclid's The Elements.
22. Angles Inscribed in Semicircles are Right
Theorem
In a circle the angle in a semicircle is right.
Proof
Let ABCD be a circle whose diameter is BC and
whose center is E.
Join AB, AC, and AE.
Let BA be produced to F.
Since BE = EA , from Isosceles Triangles have Two
Equal Angles it follows that
ABE = BAE.
Since CE = EA , from Isosceles Triangles have
Two Equal Angles it follows that
ACE = CAE.
So from
BAC = ABE + ACE = ABC + ACB.
But from Sum of Angles of Triangle Equals Two Right Angles FAC = ABC + ACB.
So BAC = FAC, and so from Book I Definition 10 each one is a right angle.
So the angle in the semicircle BAC is a right angle.
■
Historical Note
This is a part of Proposition 31 of Book III of Euclid's The Elements.
20
23. Pytthagorean Theorem
Theorem
m
Given anny right trian
ngle
ABC with
w
c as thhe hypotenuuse, we have a 2 + b 2 = c 2 .
Proof
Let ABC be a right trriangle whosse angle BAC
C
is a right angle.
Construcct squares BD
DEC on BC
C, ABFG on
AB and ACKH on AC.
Construcct AL paralllel to BD (oor CE).
Since BAC and BAG are booth right
angles, frrom Two Angles makingg Two Right
Angles make
m
a Straig
ght Line it foollows that
CA is in a straight lin
ne with AG.
For the same reason BA is in a straight
s
line
with AH..
We have that DBC
C = FBA, because
b
bothh
are right angles.
We add ABC to eaach one to make
m
FBC
C
and DB
DBA.
By comm
mon notion 2,
2 FBC = DBA.
By Trianngle Side-Ang
gle-Side Equuality,
ABD = FBC.
We have that the parallelogram BDLM
B
is onn the same base
b
BD andd between thhe same paraallels
BD and AL as the trriangle ABD.
A
So, by Paarallelogram
m on Same Base
B
as Trianngle has Twiice its Area, the paralleloogram BDL
LM is
twice thee area of ABD.
A
Similarlyy, we have th
hat the paralllelogram AB
BFG (whichh happens allso to be a sqquare) is on the
t
same basse FB and between
b
the same paralleels FB and GC as the triangle FBC.
F
So, by Paarallelogram
m on Same Base
B
as Trianngle has Twiice its Area, the paralleloogram ABF
FG is
twice thee area of FBC.
F
So BDL
LM = 2 ⋅ ABD
A
= 2 ⋅ FBC = AB
BFG.
By the saame construcction, we havve that CEL
LM = 2 ⋅ ACE
A
= 2 ⋅ KBC
K
= ACK
KH.
But BDL
LM + CELM
M is the whoole of the sqquare BDEC
C.
Therefore the area off the square BDEC is eqqual to the coombined area of the squaares ABFG and
ACKH.
■
Historicaal Note
This is Prroposition 47 of Book I of Euclid's The
T Elementts.
21
24. Eqquiangularr Triangless are Simillar
Theorem
m
Let two triangles
t
hav
ve the same correspondin
c
ng angles.
Then their correspon
nding sides arre proportionnal.
Thus, by definition, such
s
trianglees are similaar.
As Euclidd defined it:
In equianngular triang
gles the sidees about the equal
e
angless are
proportioonal, and tho
ose are corresponding siides which
subtend the
t equal ang
gles.
(The Elem
ments: Book
k VI: Proposition 4)
Proof
Let AB
BC, DCE be equianguular triangless such
that:
ABC = DCE
BAC = CDE
ACB = CED
Let BC be placed in
n a straight liine with CE
E.
From Tw
wo Angles off Triangle Leess than Twoo Right Anglles ABC + ACB is less
l than twoo
right anggles.
As AC
CB = DEC
C, it follows that ABC
C + DEC is also less than
t
two righht angles.
So from the
t Parallel Postulate, BA
B and ED
D , when prodduced, will meet.
m
Let this happen
h
at F.
F
We have that ABC
C = DCE.
So from Equal
E
Correesponding Angles Impliees Parallel, BF CD.
Again, we
w have that ACB = CED.
Again froom Equal Co
orresponding Angles Imp
mplies Parallel, AC FE..
Therefore by definitiion □FACD
D is a parallelogram.
Therefore from Oppo
osite Sides and
a Angles off Paralleloggram are Equual FA = DC
C and AC =
FD.
C FE, it folllows from Parallel
P
Linee in Trianglee Cuts Sides Proportionaally that
Since AC
BA : AF = BC : CE .
But AF = CD so
BA : CD = BC : CE
( That is,
( or
BA
BC
=
AF
CE
).
BC
BA
B
=
C
CD
CE
).
From Prroportional Magnitudes
M
are Proporttional Alternnately,
AB : BC = DC : CE
( or
DC
AB
A
=
B
BC
CE
).
Since CD
D BF , fro
om Parallel Line
L in Trianngle Cuts Siddes Proportiionally,
22
But FD = AC so
BC : CE = FD : DE
( or
BC
FD
=
CE
DE
) .
BC : CE = AC : DE
( or
BC
AC
=
CE
DE
) .
So from Proportional Magnitudes are Proportional Alternately,
BC : CA = CE : ED
( or
BC
CE
=
CA
DE
) .
BA
CD
=
AC
DE
) .
It then follows from Equality of Ratios Ex Aequali that
BA : AC = CD : DE
( or
■
Historical Note
This is Proposition 4 of Book VI of Euclid's The Elements.