Euclid.

# Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Font size formed, equals the given triangle ABC.

(/) The sum of the squares on GH, KE, and FD will be equal
to six times the square on the hypotenuse.

{g) The difference of the squares on AB, A C, is equal to the
difference of the squares on AD, AE.

161. The area of any two parallelograms described on the two
sides of a triangle, is equal to that of a parallelogram on the base,
whose side is equal and parallel to the line drawn from the vertex of
the triangle, to the intersection of the two sides of the former paral-
lelograms produced to meet.

162. If one angle of a triangle be a right angle, and another
equal to two-thirds of a right angle, prove from the First Book of
Euclid, that the equilateral triangle described on the hypotenuse,
is equal to the sum of the equilateral triangles described upon the
sides which contain the right angle.

BOOK 11.

DEFINITIONS.

I.

Every right-angled parallelogram is called a rectangle, and is said
to be contained by any two of the straight lines which contain one of
the right angles.

n.

In every parallelogram, any of the parallelograms about a diameter
together with the two complements, is called a gnomon.

A E D

F y^

B G

" Thus the parallelogram HG together with the complements ^F, FC,
is the gnomon, which is more briefly expressed by the letters AGK, or
EHC, which are at the opposite angles of the parallelograms which make
the gnomon."

PROPOSITION I. THEOREM.

If there he two straight lines, one of which is divided into any number
of parts ; the rectangle contained by the two straight lines, is equal to the
rectangles contained by the undivided line^ and the several parts of the
divided line.

Let A and BChe two straight lines ;
and let BChe divided into any parts BD, DE, EC, in the points B, E.
Then the rectangle contained by the straight lines A and BC, shall
be equal to the rectangle contained by A and BD, together with that
contained by A and I)E, and that contained by A and EC.

86

EUCLID S ELEMENTS.
B DEC

K L H

From the point B, draw JBF at right angles to BC, (l. 11.)
and make BG equal to A ; (l. 3.)
through G draw 6^7/ parallel to JBC, (l. 31.)
and through I), E, C, draw DK, EL, CjET parallel to BG, meeting

GHmK,L,H.

Then the rectangle BHis equal to the rectangles BK, DL, EH.

And BSis contained by A and BC,

for it is contained by GB, B C, and GB is equal to A :

and the rectangle BK is contained by A, BD,

for it is contained by GB, BD, of which GB is equal to A :

also JDL is contained by A, DE,

because DK, that is, BG, (l. 34.) is equal to A ;

and in like manner the rectangle EH is contained by Af EC:

therefore the rectangle contained by A, BC, is equal to the several

rectangles contained by A, BD, and by A, DE, and hy A, EC.

Wherefore, if there be two straight lines, &c. Q. e.d.

PROPOSITION II. THEOREM.

If a straight line be divided into any two parts, the rectangles contained
hy the ichole and each of the parts, are together equal to the square on the
whole line.

Let the straight line AB be divided into any two parts in the point C.
Then the rectangle contained by AB, BC, together with that con-
tained by AB, A C, shall be equal to the square on AB.

A C B

D F E

Upon AB describe the square A DEB, (l. 46.) and through Cdrr.w
Ci^ paraUel to AD or BE, (L 31.) meeting DE in F.

Then AE is equal to the rectangles AF, CE.

And AE is the square on AB ;

and ^i^is the rectangle contained by BA, A C;

for it is contained by DA, A C, of which DA is equal to AB :

and CE is contained by AB, BC,

for BE is equal to AB:

therefore the rectangle contained by AB, A C, together with the

rectangle AB, BC is equal to the square on AB.

If therefore a straight line, &c. Q. E. D.

BOOK II. PROP. Ill, IV.

PROPOSITION III. THEOREM.

87

If a straight line he divided into any txoo parts, the rectangle contained by
the xohole and one of the parts, is egtial to the rectangle contained by the two
parts, together with the square on the aforesaid part.

Let the straight line ABhe divided into any "^wo parts in the point C.

KThen the rectangle AB, BC, shall be equil to the rectangle
AC, CB, together with the square on BC.

U|xrn ^C describe the square CDEB, (i. 46.) and produce ED to F,

trough A draw ^i^parallel to CD or BE, (l. 31.) meeting j^i^'in F.

Then the rectangle AE is equal to the rectangles AD, CE.

And AE is the rectangle contained by AB, BC,

for it is contained by A B, BE, of which BE is equal to BCi

and AD is contained by A C, CB, for CD is equal to CB :

and CE is the square on BC:

therefore the rectangle AB, BC, is equal to the rectangle AC, CB,

together with the square on BC.

If therefore a straight line be divided, &c. Q. E. D.

PROPOSITION IV. THEOREM.

If a straight line be divided into any two parts, the square on the whole
line is equal to the squares on the two parts, together with twice the rectangle
contai7ied by the parts.

Let the straight line ^i? be divided into any two parts in C.
Then the square on AB shall be equal to the squares on AC, and
CB, together with twice the rectangle contained by A C, CB.

A C B

G

/

D F E

Upon AB describe the square ADEB, (l. 46.) join BD,
through C draw CGF parallel to AD or BE, (i. 31.) meeting BD

in G and DE in F ',
and through G draw JTG^jK' parallel to ^ji5 or DE, meeting AD in

H, and BE in K;
Then, because CF is parallel to AD and BD falls upon them,
therefore the exterior angle BGC is equal to the interior and opposite
angle BDA ; (i. 29.)

but the angle BDA is equal to the angle DBA, (l. 5.)
because BA is equal to AD, being sides of a square ;

88 Euclid's elements.

wherefore the angle BGCh equal to the angle DBA or GBC)

and therefore the side BCi^ equal to the side CG; (l. 6.)

but ^Cis equal also to GK, and CG to BK] (l. 34.)

wherefore the figure CGKB is equilateral.

It is likewise rectangular ;

for, since CG is pai'allel to BK, and ^C meets them,

therefore the angles J^J5(7, BCG are equal to two right angles ; (l. 29.)

but the angle XBCis a right angle ; (def. 30. constr.)

wherefore B CG is a right angle :

and therefore also the angles CGX, GXB, opposite to these, are right

angles ; (I. 34.)

wherefore CGKB is rectangular :

but it is also equilateral, as was demonstrated ;

wherefore it is a square, and it is upon the side CB.

For the same reason HF is a square,

and it is upon the side HG, which is equal to AC. (l. 34.)

Therefore the figures HF, CK, are the squares on A C, CB.

And because the complement AG h equal to the complement GE,

(I. 43.)^

and that AG is the rectangle contained by A C, CB,

for GCh equal to CB-,

therefore GE is also equal to the rectangle A C, CB ;

wherefore A G, GE are equal to twice the rectangle A C, CB ;

and HF, CK are the squares on A C, CB ;
wherefore the four figures HF, CK, A G, GE, are equal to the

squares on A C, CB, and twice the rectangle A C, CB :
hut HF, CK, AG, GE make up the whole figure ABEB, which

is the square on ^^ ;
therefore the square on AB is equal to the squares on A C, CB, and

twice the rectangle A C, CB.
Wherefore, if a stmight line be divided, &c. Q. E. D.

CoE. From the demonstration, it is manifest, that the parallelo-
grams about the diameter of a square, are likewise squares.

PROPOSITION V. THEOREM.

If a straight line be divided into tioo equal parts, and also into two
ttneq^ial parts ; the rectangle contained by the unequal parts, together with
the square on the line between the points of section, is equal to the square on
half the line.

Let the straight line AB be divided into two equal parts in the
point C, and into two unequal parts in the point D.

Then the rectangle AD, DB, together with the square on CD, shall
be equal to the square on CB.

A C D B

E G F

BOOK II. PROP. VI. 89

thi'ough D draw DHG parallel to CE or BF, (i. 31.) meeting DE

in JT, and^i^in G,
and through ^ draw XX Jf parallel to CB or EF, meeting CE
in Z, and BF in Jf;
also through A draw ^X parallel to CL or JSili", meeting ilfXX in X.
Then because the complement CH is equal to the complement HF,
(i. 43.) to each of these equals add D3I;

therefore the whole CMh equal to the whole DF\

but because the line AC is equal to CB,

therefore AL is equal to CM, (i. 36.)

therefore also ^Z is equal to EF;

to each of these equals add CH,

and therefore the whole All is equal to EFand CH:

but ASis the rectangle contained by AD, DB, for DHis equal to DB-,

and DF together with CH is the gnomon CMG ;

therefore the gnomon CMG is equal to the rectangle AID, DB :

to each of these equals add ZG, which is equal to the square on

CD; (II. 4. Cor.)
therefore the gnomon CMG, together with LG, is equal to the

rectangle AD, DB, together with the square on CD :
but the gnomon CMG and LG make up the whole figure CEFB,

which is the square on CB ;
therefore the rectangle AD, DB, together with the square on CD
is equal to the square on CB.

Wlierefore, if a straight line, &c. Q. E. D.
Cor. Prom this proposition it is manifest, that the difference of
the squares on two unequal lines A C, CD, is equal to the rectangle
contained by their sum AD and their difiference DB.

PROPOSITION VI. THEOREM.

t If a straight line he bisected, and produced to any point ; the rectangle
contained by the whole line thus produced, a^id the part of it produced,
together with the square on half the li?ie bisected, is equal to the square on
the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point J).
Then the rectangle AD, DB, together with the square on CB, shalJ
be equal to the square on CD.

A c B D

L

Hi/

/

E G F

Upon CD describe the square CEFD, (i. 46.) and join DE,
through B draw BHG parallel to CE or DF, (l. 31.) meeting DE

inH,andEFin G;
through H draw XZJf parallel to AD or EF, meeting DF in
If, and CE in Z;
and through A draw ^Z" parallel to CZ or DM, meeting MZK in K.

90 Euclid's elements.

Then because the line ^ C is equal to CB, ]

therefore the rectangle AL is equal to the rectangle CII, (l. 36.)
but CH is equal to HF-, (l 43.)

therefore ^X is equal to HF',

to each of these equals add CM\

therefore the whole ^If is equal to the gnomon CMG:

but A3I is the rectangle contained by AD, DB,

for DMi^ equal to DB : (n. 4. Cor.)

therefore the gnomon CMG is equal to the rectangle AD, DB :

to each of these equals add L G which is equal to the square on CB ;

therefore the rectangle AD, DB, together with the square on CB, is

equal to the gnomon CMG, and the figure LG',
but the gnomon CMG and LG make up the whole figure CEFD,

which is the square on CD ;
therefore the rectangle AD, DB, together with the square on CB,
is equal to the square on CD.

Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION VII. THEOREM.

If a straight line he divided into any two parts, the squares on the whole
line, and on one of the parts, are equal to twice the rectangle contained by
the whole and that part, together with the square on the other part.

Let the straight line AB be divided into any two paiis in the point C
Then the squares on AB, i^C shall be equal to twice the rectangle
AB, BC, together with the square on AC.
A c B

Upon AB describe the square ADEB, (i. 46.) and join BD ;
through Cdi-aw CjP parallel to AD or BE (l. 31.) meeting BD'm

G, and DE in F-,
through G draw ^6^^ parallel to ^jB or DE, meeting ADmH,
and BE in K.

Then because ^ G^ is equal to GE, (l. 43.)
add to each of them CK',
therefore the whole AKi^ equal to the whole CE',
and therefore AK, CE, are double of AK:
but AK, CE, are the gnomon AKF and the square CK;
therefore the gnomon AKF and the square CK are double of AK:
but twice the rectangle AB, BC, is double of AK,
for BK is equal to BC; (IL 4. Cor.)
therefore the gnomon AKF and the square CK, are equal to twice the
rectangle AB, BC;
to each of these equals add HF, which is equal to the square on A C,
therefore the gnomon AKF, and the squares CK, HF, are equal to
twice the rectangle AB, BC, and the square on AC;
but the gnomon AKF, together with the squares CK, HF, make

BOOK II. PROP. VIII. 91

up the whole figure ADJEB and CK, which are the squares on AB
aud JBC;

therefore the squares on AJB and J? Care equal to twice the rectangle

kAB, J3C, together with the square on AC.
Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION VIII. THEOREM.

If a straight line he divided into any tioo parts, four times the rectangle
contained by the whole Ime, and one of the parts, together with the square on
the other part, is equal to the square on the straight line, which is made up
of the xchole and that part.

Let the straight line ^^ be divided into any tw- o parts in the point C.

Then four times the rectangle AB, BC, together with the square on
AC, shall be equal to the square on the sti^aight line made up oi AB
and ^C together,

A C B D

M ÂŁS2^N

o

G

k!/!

P

/

R

/

E li L F

Produce AB to D, so that BB be equal to CB, (i. 3.)

upon AT> describe the square AEFD, (l. 46.) and join DE,

through B, C, draw BL, CtZT parallel to AE or DF, and cutting DE

in the points K, P respectively, and meeting EE in L, H-,

through K, P, draw MGKN, XPRO parallel to AD or EF.

Then because CB is equal to BD, CB to GK, and BD to KNi

therefore GK\% equal to KN',

for the same reason, PR is equal to RO;

and because CB is equal to BD, and GK to KN^

therefore tiie rectangle CK is equal to BN, and GR to RN-, (l. 36.)

but CK is equal to RN, (i. 43. )

because they are the , complements of the parallelogram CO;

therefore also BN is equal to GR ;

and the four rectangles BN, CK, GR, RN, are equal to one another,

and so are quadruple of one of them CK.

Again, because CB is equal to BD, and BD to BK, that is, to C^*";

and because CB is equal to GK, that is, to GP ;

therefore CG is equal to GP.

And because CG is equal to GP, and PR to R O,

therefore the rectangle AG i^ eqjial to 3ÂŁP, and PL to RF]

but the rectangle MP is equal to PL, (l. 43.)
because they are the complements of the parallelogram ML :
wherefore also AG h equal to RF:
therefore the four rectangles A G, MP, PL, RF, are equal to one
another, and so are quadruple of one of them A G.
And it was demonstrated, that the four CK, BN, GR, and RN, are

92 EUCLID'S ELEMENTS.

therefore the eight rectangles which contain the gnomon A OH, are

And because ^^is the rectangle contained by AB, BC,

for^Xisequalto J?C;

therefore four times the rectangle AB, BC is quadruple ofAK:

but the gnomon A OH was demonstrated to be quadruple of ^^;

therefore four times therectangle ^^, B Cis equal to the gnomon^ OH;

to each of these equals add XH, which is equal to the square on A C;

therefore four times the rectangle AB, BC, together with the square

on AC, is equal to the gnomon A OH and the square XH;
but the gnomon A OH and XH make up the figure AJEFD, which is
therefore four times the rectangle AB, -B C together with the square
on A C, is equal to the square on AD, that is, on AB and BC added
together in one straight line.

Wherefore, if a straight line, &c, q.e.d.

PROPOSITION IX. THEOREM.

If a straight line be divided into two equal, and also into two unequal
parts ; the squares on the two unequal parts are together double of the square
on half the line, and of the square on the line between the points of section.

Let the straight line AB be divided into two equal parts in the point
C, and into two unequal parts in the point D.

Then the squares on AD, DB together, shall be double of the
squares on A C, CD.

E

From the point C draw CE at right angles to AB, (l. 11.)

make CE equal to ^ Cor CB, (l. 3.) and join EA, EB ;

through D draw Z)P parallel to CE, meeting EB in F, (l. 31.)

through F draw EG parallel to BA, and join AF.

Then, because ^ C is equal to CE,

therefore the angle AEC is equal to the angle EA C; (l. 5.)

and because A CE is a right angle,

therefore the two other angles AEC, EA C of the triangle are together

equal to a right angle ; (l. 32.)

and since they are equal to one another ;

therefore each of them is half a right angle.

For the same reason, each of the angles CEB, EB Cis half a right angle;

and therefore the whole AEB is a right angle.

And because the angle GEE is half a right angle,

and EGF a right angle,

for it is equal to the interior and opposite angle ECB, (l. 29.)

therefore the remaining angle EFG is half a right angle ;

wherefore the angle GEF is equal to the angle EFG,

and the side 6^i^ equal to the side EG. (i. 6.)

BOOK II. PROP. X,

93

Again, because the angle at B is half a right angle,

and FDB a right angle,

it is equal to the interior and opposite angle JECJB, (l. 29.)

therefore the remaining angle BFD is half a right angle ;

wherefore the angle at B is equal to the angle BFl),

and the side UF equal to the side BB. (l. 6.)

And because ACh equal to CF,

the square on ^Cis equal to the square on CF;

lerefore the squares on A C, CF are double of the square on AC;

but the square on AF is equal to the squares on A C, CF, (l. 47.)

because A CF is a right angle ;

therefore the square on AF is double of the square on A CI

Again, because FG is equal to GF,

the square on FG is equal to the square on GF;

therefore the squares on FG, GFare double of the square on GF;

but the square on FFis equal to the squares on FG, GF; (i. 47.)

therefore the square on FF is double of the square on GF;

and GFh equal to CD; (i. 34.)

therefore the square on FF is double of the square on CF ;

but the square on AF is double' of the square on A C;

therefore the squares on AF, FF are double of the squares on A C, CF ;

but the square on AF is equal to the squares on AF, FF,

because AFFis a right angle : (l. 47.)

therefore the square on AFis double of the squares on AC, C :

but the squares on AF, FF are equal to the square on AF;

because the angle AFFis a right angle ; (i. 47.)

therefore the squares on AF, FFave double of the squares on A C, CF -,

and FF is equal to FB ;

therefore the squares on AF, FB are double of the squares on A C, CF.

If therefore a straight line be divided, &c. q.e.d.

PROPOSITION X. THEOREM.

If a straight line be bisected, and produced to any point, the square on

the whole line thus produced, and the square on the part of it produced, re

together double of the square on half the line bisected, and of the squ-are on
the line made up of the half and the part produced.

Let the straight line AB he bisected in C, and produced to the
point F.

Then the squares on AF, FB, shaU be double of the squares on
AC, CD,

E F

From the point Cdraw CF at right angles to AB, (l. 11.)

make CF equal to ^ C or CB, (l. 3.) and join AF, FB ;

through F draw J?jP parallel to AB, (l. 31.)

and through D draw X>i^ parallel to CF^ meeting ^i^in F.

94

Then because the straight line JEF meets the parallels CE, FDi
therefore the angles CEF, EFD are equal to two right angles ; (l. 29.)
and therefore the angles BEF, EFD are less than two right angles.
But straight lines, which with another straight line make the in-
terior angles upon the same side of a line, less than two right angles,
will meet if produced far enough; (I. ax. 12.)

therefore EB, FD will meet, if produced towards B, D ;

let them be produced and meet in G, and join AG.

Then, because ^ C is equal to CE,

therefore the angle CEA is equal to the angle EAC; (l. 5.)

and the angle A CE is a right angle ;

therefore each of the angles CEA, EA C is half a right angle. (l. 32.)

For the same reason,

each of the angles CEB, EBCis half a right angle;

therefore the whole AEB is a right angle.

And because EB C is half a right angle,

therefore DBG is also half a right angle, (l. 15.)

for they are vertically opposite ;

but BDG is a right angle,

because it is equal to the alternate angle DCE; (i. 29.)

therefore the remaining angle DGB is half a right angle;

and is therefore equal to the angle DBG;

wherefore also the side BD is equal to the side DG. (i. 6.)

Again, because EGF is half a right angle, and the angle at jP is a

right angle, being equal to the opposite angle ECD, (l. 34.)

therefore the remaining angle FEG is half a right angle,

and therefore equal to the angle EGF;

wherefore also the side GF is equal to the side FE. (i. 6.)

And because ECh equal to CA ;

the square on EC is equal to the square on CA ;

therefore the squares on EC, CA are double of the square on CA ;

but the square on EA is equal to the squares on EC, CA ; (l. 47.)

therefore the square on EA is double of the square on A C.

Again, because GF is equal to FE,

the square on GF is equal to the square on FE ;

therefore the squares on GF, FE are double of the square on FE;

but the square on EG is equal to the squares on GF, FE; (i. 47.)

therefore the square on EG is double of the square on FE;

and FE is equal to CD ; (i. 34.)
wherefore the square on EG is double of the square on CD ;
but it was demonstrated,

that the square on EA is double of the square on AC;

therefore the squares on EA, EG are double of the squares on A C, CD ;

but the square on ^6^ is equal to the squares on EA, EG; (l. 47.)

therefore the square on ^G^ is double of the squares on AC, CD-.

but the squares on AD, DG are equal to the square on AG;

therefore the squares on A D,DG are double of the squares on A C, CD ;

but DG is equal to DB ;

therefore the squares on AD, DB are double of the squares on A C, CD.

Wherefore, if a straight line, &c. Q. E. d.

r

BOOK II. PEOP. XI.

PROPOSITION XI. PROBLEM.

95

^^

!
1

To divide a given straight line itito tioo parts ^ so that the rectangle con-
tained by the whole and one of the parts, shall be equal to the square on
the other part.

Let AB be the given straight line.
It is required to divide AB into two parts, so that the rectangle
contained by the whole line and one of the parts, shall be equal
to the square on the other part.

F G

C K D

Upon AB describe the square ACDB; (i. 46.)

bisect AC in JS, (l. 10.) and join BJE,

produce CA to F, and make FF equal to FB, (t. 3.)

upon AF describe the square FGIIA. (i. 46.)

Then AB shall be divided in S, so that the rectangle AB, BH is

equal to the square on AH.

Produce G^^to meet CD in K.
Then because the straight line ^ Cis bisected in F, and produced to F,
therefore the rectangle CF, FA together with the square on AE,
is equal to the square on EF', (ll. 6.)

but EF is equal to EB ;
therefore the rectangle CF, FA together with the square on AE, is
equal to the square on EB ;
but the squares on BA, AE are equal to the square on EB, (l. 47.)
because the angle FAB is a right angle ;
therefore the rectangle CF, FA, together with the square on AE,
is equal to the squares on BA, AE',
take away the square on AE, which is common to both ;
therefore the rectangle contained by CF, FA is equal to the square
on BA.
But the figure FK is the rectangle contained by CF, FA,
for FA is equal to FG ;
and AD IB the square on AB ;
therefore the figure FK is equal to AD ;
take away the common part AK,
therefore the remainder FH is equal to the remainder HD ;
but HD is the rectangle contained by AB, BH,
for ^i? is equal to BD ;
and FHis the square on AH;
therefore the rectangle AB, BH, is equal to the square on AH.
Wherefore the straight line AB is divided in H, so that the
rectangle AB, BH is equal to the square on AH. q.e.f.

i

96 eucud's elements.

PROPOSITION XII. THEOREM.
In ohtuse-angled triangles, if a perpendicular he drawn from either of
the acute angles to the opposite side produced, the sqtiare on the side sub-
tending the obtuse angle, is greater than the squares on the sides containing
the obtuse angle, by twice the rectangle contained by the side upon which,
when produced, the perpendicular falls, and the straight line i?itercepted
without the triangle between the perpendicular and the obtuse angle.

liBt AJBChe an obtuse-angled triangle, having the obtuse angle
A CJB, and from the point A, let AD be di-awn perpendicular to BC
produced.

Then the square on AB shall be greater than the squares on A C,
CB, by twice the rectangle BC, CD.

A

Because the straight line BDis divided into two parts in the point C,
therefore the square on BD is equal to the squares on BC, CD,
and twice the rectangle BC, CD ; (ii. 4.)

to each of these equals add the square on DA ;