Euclid's Five Postulatesmypages.iit.edu/~maslanka/16Prop.pdf · Euclid's Five Postulates These are the axioms of standard Euclidean Geometry. They appear at the start of Book I of

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.

10. RWanca

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15. A 16. A

Euclid

This is a

1. Th

2. If

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4. Th

5. Th

Right Angle /When a straignother, eachalled a perp

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24 Euc 1. Cons TheoremOn a giveconstruct Construc Let AB b We constradius ABWe constAB. From C, line segmThen A Proof As A is thAs B is th So, as ACThereforTherefor■ HistoricaThis is Pr 2. Tria TheoremIf two tri

• tw• th

they will• th

clidean P

struction o

m en straight lit an equilater

ction

e the given s

truct a circleB. truct a circle

where the ciments AC andABC is the eq

he center of he center of

C = AB and e AB = AC =e ABC is

al Note roposition 1

angle Side-

m iangles havewo sides equhe angles con also have:

heir third sid

Propositio

of Equilate

ine segment,ral triangle.

straight line

e BCD with c

e ACE with c

ircles intersed BC. quilateral tria

circle BCD,circle ACE,

BC = AB, it= BC. equilateral.

of Book I o

-Angle-Sid

: ual to two sidntained by th

des equal;

ons

eral Triang

, it is possibl

segment.

center A and

center B and

ect, we draw

angle require

it follows frit follows fr

t follows fro

of Euclid's Th

de Equality

des respectivhe equal stra

gle

le to

d

d radius

w a line segm

ed.

from Book I Drom Book I D

om Common

he Elements

y Implies C

vely; aight lines eq

ment to A and

Definition 15Definition 15

Notion 1 th

.

Congruenc

qual

d to B to form

5: Circle tha5: Circle tha

at AC = BC

ce

m the straigh

at AC = AB.at BC = AB.

C.

3

ht

• thth

Proof Let ABtriangles AB = DE BAC =

If ABCsuch thatthe pointand the lDE thencoincide AB = DESo, with BAC =Hence thBut B aHence th(OtherwiEF and tThereforThus the The remato them. ■ HistoricaThis is Pr 3. Isos Theorem

In isosceother. Also, if tbase will Proof Let ABWe exten Let F be

he remaininghe equal side

BC and Dhaving side

E and AC = = EDF. C is placed ot: t A is placedline AB is p

n the point Bwith point E

E. AB coincid

= EDF. he point C wlso coincide

he line BC wise, when B two straight e BC will cwhole AB

aining angle


celes Trian

m

les triangles

the equal stral also be equ

C be an isosnd the straigh

e a point on

g two angleses subtend.

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on DEF

d on point Dlaced on line

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ding with DE

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coincide withBC will coins on ABC

of Book I o

ngles have

s, the angles

aight lines arual to each ot

sceles trianglht lines AB a

BD.

equal to the

o

ith

D, e

E , the line A

ncide with th

with line EFwith E and nclose a spach EF and bencide with th

C will coincid

of Euclid's Th

e Two Equ

at the base a

re extended,ther.

le whose sidand AC to D

eir respective

AC will coi

he point F ,

F. C with F, ce which is ie equal to it. he whole de with the r

he Elements

al Angles

are equal to e

the angles u

de AB equalsD and E respe

e remaining

incide with t

because AC

the line BCimpossible.) DEF and thuemaining an

.

each

under the

s side AC. ectively.

angles, nam

the line DF

C = DF.

C will not co

us ABC =ngles on D

mely, those w

because

oincide with

DEF. DEF and be e

4

which

line

equal

We cut oWe draw Since AFrespectivThey conSo by Thus FC Since AFBut FC =Then BThereforTherefor So since But ABAlso, we

ABC. Hence th■ HistoricaThis is Pr 4. Tria TheoremLet two tThen the Thus two Proof Let ABtriangles AB = DEAC = DFBC = EFSuppose superimpthat poinpoint E aThen C

off from AE w line segmen

F = AG and vely. ntain a comm

-C = GB, A

F = AG and = GB, so theBFC = CGe by Triangle FBC =

ACF = ABC and AC have alread

he result.


angle Side-

m triangles havy also have

o triangles w

BC and DEsuch that:

E; F; F.

ABC werposed over t B is place

and the line Bwill coincid

a length Ants FC and

AB = AC ,

mon angle, th-

ACF = ABG

AB = AC, te two sides BGB while CBle Side-Angl GCB and

ABG, and inCB are at th

dy proved tha

of Book I o

-Side-Side

ve all three sall three ang

whose sides a

EF be two

re DEF so

ed on BC on EF. de with F, as

G equal to d GB.

the two side

hat is, FAG,

G and AF

then BF = CBF, FC are eB is commole-Side Equad BCF =

n these BChe base of at FBC =

of Euclid's Th

Equality I

ides equal. gles equal.

are all equal

s BC = EF a

AF.

es FA and A

G. AFC = A

FC = AGB

CG. equal to the

on to both. ality, BFC CBG.

CF = CBG,ABC. GCB, and

he Elements

Implies Co

are themselv

and so BC co

AC are equa

GB. B.

two sides CG

C = CGB.

, then ABC

d these are th

.

ongruence

ves equal.

oincides with

al to GA an

G, GB respe

C = ACB.

he angles un

h EF.

nd AB

ectively.

der the base

5

of

Now supThen theThus therthe same Point. ThereforThereforThe samecorrespon■ HistoricaThis is Pr 5. Bise Construc Let BA Take anyWe cut oWe drawWe constWe draw Then theline segm Proof We haveAD = AEAF is coDF = EFThus triaThus D Hence ■ HistoricaThis is Pr There arepreviousl

ppose BA doy will fall asre will be tw side as it, m

e BA coince BAC coie argument cnding angles


ection of an

ction

AC be the giv

y point D onoff from AC w the line seg

truct an equiw the line seg

angle BAment AF.

: E; ommon; F. angles ADDAF = EA

BAC has be


e quicker andly demonstra

oes not coincs, for examp

wo pairs of stmeeting at dif

cides with EDincides with can be applies are equal.

of Book I o

n Angle

ven angle to

n AB. a length AE

gment DE. ilateral trian

gment AF.

AC has been

DF and AEAF.

een bisected

of Book I o

d easier conated.

cide with EDle, EG and Gtraight line sfferent point

ED and AC EDF and

ed to the oth

of Euclid's Th

be bisected.

E equal to A

gle DEF

bisected by

EF are equal

d by AF.

of Euclid's Th

structions of

D and AC GF. segments conts, in contrad

coincides wis equal to i

her two sides

he Elements

.

AD.

on AB.

the straight

l.

he Elements

f a bisection

does not coi

nstructed ondiction to Tw

with DF. it. s, and thus w

.

.

, but this par

incide with

the same linwo Lines Me

we show that

rticular one u

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ne segment, et at Unique

all

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6

on e

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6. Bise TheoremIt is poss Construc Let AB b We constWe bisecCD. Then AB Proof As ABThe two equals sidThe anglBC and So triangTherefor So AB ha■ HistoricaThis is Pr 7. Con TheoremGiven anperpendi Construc Let AB line. Let C b Let D bof it from

ection of a

m ible to bisec

ction

e the given s

truct an equict the angle

has been bi

BC is an equtriangles de BC of e ACD suCD, as A

gles ACD e AD = DB.

as been bisec


nstruction

m n infinite stracular to the g

ction

be two point

e the given p

e some poinm C.

Straight L

ct a straight l

straight line

ilateral trian ACB by th

sected at the

ilateral trianACD and BCD.

ubtended by ACB was bise

and BCD.

cted at the po

0 of Book I

of a Perpe

aight line, angiven straigh

ts on the giv

point not on

nt not on AB

Line

line segment

segment.

gle ABC he straight lin

e point D.

ngle, it followBCD hav

lines AC anected. D are equal.

oint D.

of Euclid's T

endicular t

nd a given poht line.

ven infinite s

it.

B on the othe

t.

on AB. ne segment

ws that AC =e side CD i

nd CD equ

The Element

through a G

oint not on th

straight

er side

= CB. in common,

uals the angle

ts.

Given Poin

hat straight l

and side AC

e BCD sub

nt

line, it is pos

C of ACD

btended by l

ssible to draw

7

D

lines

w a

We constWe bisecWe drawCH and Then the Proof As C is As EG hThus, as

CHG =ThereforSo CH ianother. CHE a

So the strthrough t■ HistoricaThis is Pr 8. Vert TheoremIf two str Proof Let AB Since thestraight l AEC m

Since thestraight l BED mBut AEright angSo by Coright ang AED +

Let AEThen by Similarly■

truct a circlect the straigh

w line segmenCH.

line CH is p

the center ohas been bis GC = CE an

= EHG. e CHG =is a straight Thus it follo

are right anglraight line Cthe given po


tical Angle

m raight lines c

and CD bee straight linine CD, the

make two rige straight linine AB, the make two rigED and A

gles. ommon Notiogles are cong+ AEC = ED be subtrCommon No

y it can be sh

e EFG withht line EG ants from C

perpendicula

of circle BCDsected, GH =nd GH = HE

CHE. line set up oows from Boles. CH has beenint C.

2 of Book I

e Theorem

cut each othe

e two straighe AE standangles AE

ght angles. e DE standangles AE

ght angles. AEC also m

on 1 and the gruent, AED + B

racted from eotion 3 it folhown that

h center C aat the point Hto each of G

ar to the give

D, it follows= HE. E, and CH i

on a straight ook I Definit

n drawn at ri

of Euclid's T

m

er, they mak

t lines that cs on the ED and

ds on the ED and

make two

fact that all

BED . each. lows that BEC = AE

and radius CH.

G, H and E to

en infinite st

s from Book

is common, b

line makingtion 10: Righ

ight angles t

The Element

ke the opposi

cut each othe

AEC = BEED.

CD.

o form the st

traight line A

I Definition

by

the adjacenht Angle that

o the given i

ts

ite angles eq

er at the poin

ED .

traight line s

AB through th

15: Circle

- -

nt angles equt each of

infinite strai

qual each oth

nt E.

segments CG

he given poi

that GC = C

ual to one CHG and

ght line AB

her.

8

G,

int C.

CE.

,

HistoricaThis is Pr Interna ■ InternThe interforming t ■ ExternSurprisinvertex is that verteIt is in faline prod While angle of tNote: it dthey are e 9. Exte TheoremThe exter Proof Let ABLet the siLet AC Let BE Let EF (Technicbeyond FLet CF Let AC We have make EquSince AE

ABE =Thus ABBut EC


al/External

nal Angle rnal angle (that vertex, a

nal Angle ngly, the exte

not the size ex, as measuact an angle fduced from a

AFE is the ithis vertex isdoesn't matteequal by the

ernal Angl

m rnal angle of

BC be a trianide BC be ebe bisected be joined anbe made equ

cally we reallF and then cbe joined. be extended

AEB = ual OppositeE = EC and = CFE. B = CF and CD is greate

5 of Book I

l Angle De

or interior aas measured

ernal angle of the angle

ured outside formed by o

an adjacent s

internal angls EFG. er which adje Vertical An

le of Trian

f a triangle i

ngle. extended to at E.

nd extended ual to BE. ly need to excrimp off a l

d to G.

CEF from e Angles. BE = EF , f

BAE = er than EC

of Euclid's T

efinitions

angle) of a vd inside the p

(or exteriore between ththe polygonne side of a ide.

e of vertex

acent side yongle Theorem

ngle Greate

s greater tha

D.

to F.

xtend BE tolength EF.)

Two Straigh

from Triang

ECF. CF.

The Element

vertex is the polygon.

r angle) of ae sides form. polygon and

F , the exter

ou use, sincem.

er than Int

an either of th

o a point

ht Lines

gle Side-Angl

ts.

size of the a

a ming

d a

rnal

e

ternal Opp

he opposite

le-Side Equa

angle betwee

posite Ang

internal ang

ality we hav

en the sides

gles

gles.

ve

9

Therefor SimilarlyEqual Op Hence th■ HistoricaThis is Pr 10. Tw TheoremIn any tri Proof Let ABLet the siSince the

ABC, i BAC an

We add ACD + ABC +

But ACTherefor In a simi

ABC. ■ HistoricaThis is Pr 11. Gr TheoremIn any tri Proof Let ABgreater thLet AD Let BD

e ACD is

y, if BC werpposite Angl

he result.


wo Angles

m iangle, two a

BC be a triaide BC be exe angle ACit follows thand ABC. ACB to bo

+ ACB is + ACB. CD + ACBe ABC +

lar manner w


reater Side

m iangle, the g

BC be a triahan AB. be made eqube joined.

s greater than

re bisected, les, would be

6 of Book I

of Triangl

angles taken

angle. xtended to DCD is an extat it is greate

oth, so that greater than

B is equal to ACB is le

we show tha

7 of Book I

e of Triang

reater side s

angle such th

ual to AB.

n BAE.

BCG, whe shown to b

of Euclid's T

le Less tha

together in

D. ternal angle er than both

n

o two right aess than two

t the same ap

of Euclid's T

gle Subtend

ubtends the

hat AC is

hich is equal be greater th

The Element

an Two Rig

any manner

of

angles. right angles

pplies to the

The Element

ds Greater

greater angl

to ACD an ABC a

ts.

ght Angles

are less than

.

e other two p

ts.

r Angle

le.

by Two Straas well.

s

n two right a

pairs of intern

aight Lines m

angles.

nal angles o

10

make

f

Then AThereforAs AD =From IsoThereforTherefor Hence th■ HistoricaThis is PrThis theoSide. 12. Gr TheoremIn any tri Proof Let AB BCA.

Suppose If AC wEqual AnIf AC wSubtends BCA, b

So AC m Hence th■ HistoricaThis is PrThis theoAngle.

ADB is an ee from AD

= AB, the triaosceles Triane ABD ise, as ABC

he result.


orem is the c

reater Ang

m iangle, the g

BC be a trian

AC is not gwere equal tongles, ABC

were less thans Greater Anbut it's not so

must be greate

he result.


orem is the c

exterior anglDB is greateangle ABDngles have Ts greater thanC = ABD +

8 of Book I converse of P

gle of Trian

reater angle

ngle such tha

greater than o AB, then byC = BCA, n AB, then bngle it wouldo it isn't.

er than AB

9 of Book I converse of P

e of the trianer than ACD is isoscel

Two Equal Ann ACB.

+ DBC, it f

of Euclid's TProposition 1

ngle Subte

is subtended

at ABC is

AB. y Isosceles Tbut they're n

by Greater Sd follow that

of Euclid's TProposition 1

ngle BCD. CB. es. ngles, AD

follows that

The Element19: Greater A

ended by G

d by the grea

greater than

Triangles hanot so it isn'tSide of Trian ABC is le

The Element18: Greater

DB = ABD

ABC is g

ts. Angle of Tri

Greater Sid

ater side.

n

ave Two t.

ngle ess than

ts. Side of Tria

.

reater than

iangle Subte

de

ngle Subtend

ACB.

ended by Gre

ds Greater

11

eater

13. Su TheoremGiven a tlength of Proof We can eThere exTherefortwo equaThus, BSince BDC, But BD Thus, BAA similarBA + BC■ HistoricaThis is PrIt is a geo 14. CoTheoremGiven thrthe lengthas its side Construc Let the thfrom whiconstructand c. Let D anpoints. Cextend it We cut oDE equaSimilarlyFG = b aNow we centered

um of Two

m triangle ABCf the third sid

extend BA pists a point e, ADC =

al angles. BCD > BDDCB is a trithis means t= BA + AD ,A + AC > BCr argument s

C > AC.

al Note roposition 2ometric inter

onstructionm ree straight lh of the thirde lengths.

ction

hree straightich we are gt the triangle

nd E be anyConstruct DE

beyond E. off a length Dal to a. y, we cut off and GH = ccan construcat F with r

Sides of T

C, the sum ofde.

past A into aD such that ACD bec

DC by Eucliiangle havinthat BD > B, and AD = C. shows that A

0 of Book I rpretation of

n of Triang

lines such thd line, it is p

t lines oing to

e be a, b,

y distinct E and

DF on

f on DE.

ct a circle radius DF.

Triangle Gr

f the lengths

a straight linet DA = CA. cause isoscel

id's fifth comng BCD gBC.

AC.

AC + BC > B

of Euclid's Tf the triangle

gle from G

hat the sum opossible to co

reater than

s of any two

e. les triangle h

mmon notiongreater than

BA and

The Elemente inequality.

Given Leng

of the lengthonstruct a tri

n Third Si

sides of the

have

.

ts.

gths

hs of any twoiangle havin

ide

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o of the linesg the lengths

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s is greater ths of these lin

12

the

han nes

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.

Proof Part 1 Let ABBC = EFAssume two mustWe constBG = EDNow, sinand BC =Equality But fromTherefor■ Part 2 Let ABAB = DEAssume the two mgeneralityWe constBH = EFAH. Now, sin ABH =

Side-Ang BHA =

But fromwe have ■ HistoricaThis is Pr 16. Eq TheoreGiven twequal, the

BC = DEFF .

AB ≠ DE. It be greater. truct a point

D, and then wnce we have B= EF, from Twe have G

m Euclid's fifte, AB = DE

BC = DEF,E.

BC ≠ EF. Imust be greay, we let BCtruct a point

F, and then w

nce we have B= DEF, andgle-Side Equ= EFD. m External A

BHA >


qual Altern

em wo infinite stren the lines a

F, BCA =

If this is the Without los

t G on AB we construct BG = ED, Triangle SidGCB = DFfth common nE, so from Tr

, BCA =

If this is the cater. WithoutC > EF. t H on BC we construct

BH = EF, d AB = DE, fality we hav

Angle of Tria HCA = E

6 of Book I

nate Interi

raight lines ware parallel.

EFD, and

case, one ofss, we let AB

such that the segment GBC = D

de-Angle-SidFE. notion DFriangle Side

EFD, and

case, one of t loss of

such that the segment

from Triangve

angle GreateEFD, a contr

of Euclid's T

ior Angles

which are cu

d

f the B > DE.

t CG. DEF, de

FE = ACB -Angle-Side

f

t

gle

er than Internradiction.

The Element

Implies Pa

ut by a transv

> GCB, aEquality, w

nal Opposite

ts.

arallel

versal, if the

a contradictiowe have AB

e,

e alternate in

on. BC = DEF

nterior angles

14

F.

s are

Proof Let AB and let Ethem. Lealternate generalityequal. Assume tThen theD. Since Internal SimilarlyTherefor■ HistoricaThis is PrThis theo 17. Eq Im Theorem Part 1 Given twtransverslines are Part 2 Given twtransverstransvers Proof Part 1 Let AB least oneThen Thus AB■ Part 2 Let AB anone pair

and CD beEF be a trant the at leastinterior angy, let AEF

that the lines meet at som

AEF is an eOpposite,

y, they cannoe, by definit


orem is the c

qual Corremplies Para

m

wo infinite strsal, if the corparallel.

wo infinite strsal, if the intesal are supple

and CD be pair of corrGHD = E

B CD by E

nd CD be inof interior an

e two straighnsversal that t one pair of

gles, without F and EF

s are not parme point G

exterior angl AEF > E

ot meet on thtion, they are

7 of Book I converse of t

esponding Aallel

raight lines wrresponding

raight lines werior angles ementary, th

e infinite straesponding aGB = AGH

Equal Alterna

nfinite straighngles on the

t lines, cuts

loss of FD be

rallel. which witho

le of GEFEFG, a contrahe side of Ae parallel.

of Euclid's Tthe first part

Angles or

which are cuangles are e

which are cuon the same

hen the lines

aight lines, anangles, WithoH by the Verate Interior A

ht lines, and same side o

out loss of ge

F, from Exteradiction. and C.

The Elementof Propositi

Suppleme

ut by a qual, then th

ut by a e side of the

are parallel.

nd let EF bout loss, let rtical Angle Angles Impl

let EF be aof the transve

enerality is

rnal Angle o

ts. ion 29.

entary Inte

he

.

be a transver EGB and

Theorem. lies Parallel.

a transversalersal, Withou

on the same

of Triangle G

erior Angle

rsal that cuts d GHD, be

.

l that cuts thut loss, let

e side as B a

Greater than

es

them. Let ae equal.

hem. Let at le BGH and

15

and

n

t

east

DHG AGH +

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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 DAE = ADC, it follows that EA BC. ■ Historical Note This is Proposition 31 of Book I of Euclid's The Elements.

20. Su TheoremIn a trian Proof Let ABC paralleSince ABit followsSimilarlycuts themThus by E ACD =

Again by ACB +But ACEuclid's F ABC ■ HistoricaThis is Pr Euclid's pof the sidangles. Tpropositi 21. Rad TheoremIf a straigto the po Proof Let DE Let F beLet FC Suppose Instead, sDE.

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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.

23. Pyt TheoremGiven an Proof Let ABC is a right ConstrucAB and ConstrucSince angles, frAngles mCA is in For the swith AH. We have are right We add and DBBy commBy Trian

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ally that

21

o

=

22

BC : CE = FD : DE ( or DEFD

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BC : CA = CE : ED ( or DECE

CABC = ) .

It then follows from Equality of Ratios Ex Aequali that BA : AC = CD : DE ( or

DECD

ACBA = ) .

■ Historical Note This is Proposition 4 of Book VI of Euclid's The Elements.

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Euclid's Five Postulatesmypages.iit.edu/~maslanka/16Prop.pdf · Euclid's Five Postulates These are the axioms of standard Euclidean Geometry. They appear at the start of Book I of