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

. **(page 11 of 38)**

Online Library → 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 text (page 11 of 38)

Font size

therefore the squares on BD, DA are equal to the squares on B C,

CD, DA, and twice the rectangle BC, CD;

but the square on BA is equal to the squares on BD, DA, (l. 47.)

because the angle at D is a right angle ;

and the square on CA is equal to the squares on CD, DA ;

therefore the square on BA is equal to the squares on BC, CA, and

twice the rectangle B C, CD ;

tJiat is, the square on BA is greater than the squares on BC, CA, by

twice the rectangle BC, CD.

Therefore in obtuse-angled triangles, &c. q.e.d.

PROPOSITION XIII. THEOREM.

Tn every triangle, the square on the side subtending either of the acut'

angles, is less than the squares on the sides containing that angle, by twic

the rectangle contained by either of these sides, and the straight line inter

cepted beticeeii the acute angle and the perpendicular let fall upon it from

the opposite angle.

Let ABC be any triangle, and the angle at B one of its acute

angles, and upon BC, one of the sides containing it, let fail the

perpendicular AD from the opposite angle. (l. 12.)

Then the square on ^C opposite to the angle B, shall be less than

the squares on CB, BA, by twice the rectangle CB, BD.

BOOK 11. PROP. XIII.

97

First, let AD fall within the triangle ABC,

Then because the straight line CB is divided into two parts in D,

the squares on CB, BD are equal to twice the rectangle contained by

CB, BD, and the square on DC; (ll. 7.)

to each of these equals add the square on AD ;

therefore the squares on CB, BD, DA, are equal to twice the

rectangle CB, BD, and the squares on AD, DC;

but the square on ^jB is equal to the squares on BD, DA, (l. 47.)

because the angle BDA is a right angle ;

and the square on ^ C is equal to the squares on AD, DC',

therefore the squares on CB, BA are equal to the square on A C,

and twice the rectangle CB, BD :

that is, the square on ^C alone is less than the squares on CB, BA,

by twice the rectangle CB, BD.

Secondly, let AD fall without the triangle ABC,

B CD

Then, because the angle at -D is a right angle,

the angle ACB is greater than a right angle ; (l. 16.)

and therefore the square on ^^ is equal to the squares on AC, CB,

and twice the rectangle J? C CD; (il. 12.)

to each of these equals add the square on BC;

therefore the squares on AB, BC are equal to the square on AC,

twice the square on BC, and twice the rectangle BC, CD;

but because BD is divided into two parts in C,

therefore the rectangle DB, BC is equal to the rectangle BC, CD,

and the square on BC; (il. 3.)

and the doubles of these are equal ;

that is, twice the rectangle DB, BC is equal to twice the rectangle

BC, CD and twice the square on BC:

tiierefore the squares on AB, J5Care equal to the square on AC,

and twice the rectangle DB, BC:

wherefore the square on ^ C alone is less than the squares on AB,BC;

by twice the rectangle DB, BC.

Lastly, let the side AC he perpendicular to BC.

A

B c

Then BC is the straight line between the perpendicular and the

acute angle at B ;

and it is manifest, that the squares on AB, BC,_ are equal to the

square on A C, and twice the square on BC. (l. 47.)

.^^ Therefore in any triangle, &c. q.e.d.

98 Euclid's elemeni'i

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure.

It is required to describe a square that shall be equal to A.

H

Describe the rectangular parallelogram BCDE equal to the rect

lineal figure A. (i. 45.)

Then, if the sides of it, BE, ED, are equal to one another,

it is a square, and what was required is now done.

But if BE, EL), are not equal,

produce one of them BE to F, and make EF equal to ED^

bisect ^i^ in G; (i. 10.)

from the center Gj at the distance GB, or GF, describe the semicircle

BHF,

and produce DE to meet the ch'cumference in H,

The square described upon EH shall be equal to the given recti-

lineal figure A.

Join GH.

Then because the straight line BF is divided into two equal pai

in the point G, and into two unequal parts in the point E',

therefore the rectangle BE, EF, together with the square on EQ

is equal to the square on GF', (ii. 5.)

but GF'i^ equal to GH; (def. 15.)

therefore the rectangle BE, EF, together with the square on EG, i

equal to the square on GH;

but the squares on HE, EG are equal to the square on GH; (l. 47j

therefore the rectangle BE, EF, together with the square on EC^^

is equal to the squares on HE, EG ; '

take away the square on EG, which is common to both ;

therefore the rectangle BE, EF is equal to the square on HE.

But the rectangle contained by BE, EF is the parallelogram BD,

because EF is equal to ED ;

therefore BD is equal to the square on EH;

but BD is equal to the rectilineal figure A ; (constr.)

therefore the square on EH is equal to the rectilineal figure A.

Wherefore a square has been made equal to the given rectilineal

figure Af namely, the square described upon EH. Q. e.f.

NOTES TO BOOK IL

In Book T, Geometrical magnitudes of the same kind, lines, angles

and surfaces, more particularly triangles and parallelograms, are com-

pared, either as being absolutely equal, or unequal to one another.

In Book II, the properties of right-angled parallelograms, but without

reference to their magnitudes, are demonstrated, and an important

extension is made of Euc. i. 47, to acute-angled and obtuse- angled

triangles. Euclid has given no definition of a rectangular parallelogram

or rectangle : probably, because the Greek expression -rrapaXk^Xoy pajxixov

SpdoywvLov, or opdoyuviov simply, is a definition of the figure. In English,

the term rectangle, formed from rectus angulus, ought to be defined before

its pi'operties are demonstrated. A rectangle may be defined to be a

parallelogram having one angle a right-angle, or a right angled paral-

lelogram ; and a square is a rectangle having all its sides equal.

As the squares in Euclid's demonstrations are squares described or

supposed to be described on straight lines, the expression 'â– '"the square

on AB," is a more appropriate abbreviation for *' the square described on

the line AB," than " the square of AB." The latter expression more

fitly expresses the arithmetical or algebraical equivalent for the square

on the line AB.

In Euc. I. 35, it may be seen that there may be an indefinite number

of parallelograms on the same base and between the same parallels whose

areas are always equal to one another ; but that one of them has all its

angles right angles, and the length of its boundary less than the boundary

of any other parallelogram upon the same base and between the same

parallels. The area of this rectangular parallelogram is therefore de-

termined by the two lines which contain one of its right angles. Hence

it is stated in Def. 1, that every right-angled parallelogram is said to be

contained by any two of the straight lines which contain one of the right

angles. No distinction is made in Book ii, between equality and identity,

as the rectangle may be said to be contained by two lines which are

equal respectively to the two which contain one right angle of the figure.

It may be remarked that the rectangle itself is bounded by four straight

lines.

It is of primary importance to discriminate the Geometrical conception

of a rectangle from the Arithmetical or Algebraical representation of it.

The subject of Geometry is magnitxide not number, and therefore it would

be a departure from strict reasoning on space, to substitute in Geometrical

demonstrations, the Arithmetical or Algebraical representation of a rect-

angle for the rectangle itself. It is however, absolutely necessary that

the connexion of number and magnitude be clearly understood, as far

as regards the representation of lines and areas.

All lines are measured by lines, and all surfaces by surfaces. Some

one line of definite length is arbitrarily assumed as the linear unit, and

the length of every other line is represented by the number of linear

units, contained in it. The square is the figure assumed for the measure

of surfaces. The square unit or the unit of area is assumed to be that

square, the side of which is one unit in length, and the magnitude of

every surface is represented by the number of square units contained

in it. But here it may be remarked, that the properties of rectangles

and squares in the Second Book of Euclid are proved independently

f2

R

100 ErCLlD's

of the consideration, whether the sides of the rectangles can be repre-

sented by any multiples of the same linear unit. If, however, the

sides of rectangles are supposed to be divisible into an exact number

of linear units, a numerical representation for the area of a rectangle

may be deduced.

On two lines at right angles to each other, take AB equal to 4, and

AD equal to 3 linear units.

Complete the rectangle ABCD, and through the points of division of

AB, AD, draw EL, FM, GN parallel to AD ; and HP, KQ parallel to

AB respectively.

A E F G B

I

:p

-Q

I

D L M N C

Then the whole rectangle AC is divided into squares, all equal to each

other.

And AC is equal to the sum of the rectangles AL, EM, FN, GC ; (ii. 1.)

also these rectangles are equal to one another, (i. 36.)

therefore the whole ^C is equal to four times one of them AL.

Again, the rectangle AL is equal to the rectangles EH, HR, RD,

and these rectangles, by construction, are squares described upon the

equal lines AH, HK, KD, and are equal to one another.

Therefore the rectangle AL \q equal to 3 times the square on AH,

but the whole rectangle AC is equal to 4 times the rectangle AL,

therefore the rectangle ^C is 4 x 3 times the square on AH, or 12

square units :

that is, the product of the two numbers which express the number of

linear xmits in the two sides, will give the number of square units in the

rectangle, and therefore will be an arithmetical representation of its area.

And generally, if AB, AD, instead of 4 and 3, consisted of a and b

linear units respectively, it may be shewn in a similar manner, that the

area of the rectangle A C would contain ab square units ; and therefore the

product ab is a proper representation for the area of the rectangle AC. j

Hence, it follows, that the term rectangle in Geometry corresponds t(^

the term product in Arithmetic and Algebra, and that a similar com-

parison may be made between the products of the two numbers which

reiu'csent the sides of rectangles, as between the areas of the rectangleÂ«i'

themselves. This forms the basis of what are called Arithmetical or,

Algebraical proofs of Geometrical properties.

If the two sides of the rectangle be equal, or if b be equal to a,^

the figure is a square, and the area is represented by aa or a*.

Also, since a triangle is equal to the half of a parallelogram of the

same base and altitude ;

Therefore the area of a triangle will be represented by half the rect-

angle which has the same base and altitude as the triangle : in other

words, if the length of the base be a units, and the altitude be b units ;

Then the area of the triangle is algebraically represented by hab.

The demonstrations of the first eight propositions, exemplify th(

obvious axiom, that, *' the whole area of every figure in each case, is

equal to all the parts of it taken together."

Def. 2. The parallelogram Â£iiC together with the complements AF

NOTES TO BOOK II. 101

(7, is also a gnomon, as well as the parallelogram HG together with the

same complements.

Prop. I. For the sake of brevity of expression, ** the rectangle con-

tained by the straight lines JB, BC," is called "the rectangle AB, BC;"

and sometimes " the rectangle ABC."

To this proposition may be added the corollary : If two straight lines

be divided into any number of parts, the rectangle contained by the two

straight lines, is equal to the rectangles contained by the several parts of

one line and the several parts of the other respectively.

The method of reasoning on the properties of rectangles, by means of

the products Avhich indicate the number of square units contained in their

arcas,is foreign to Euclid's ideas of rectangles, as discussed in his Second

Book, which have no reference to any particular unit of length or measure

of surface.

Prop. T. The figures J5//, BK, DL, EH are rectangles, as may

readily be shewn. For, by the parallels, the angle CEL is equal to EDK ;

and the angle EDK is equal to BDG (Euc. i. 29.). But BDG is a right

angle. Hence one of the angles in each of the figures BH, BK, DL, EH

is a right angle, and therefore (Euc. i. 46, Cor.) these figures are

rectangular.

Prop. I. Algebraically, (fig. Prop, i.)

liCt the line BC contain a linear units, and the line A, h linear units ol

the same length.

Also suppose the parts BD, DE, EC to contain m, n, p linear units

respectively.

Then a â€” m -\- n â– \- p,

multiply these equals by h,

therefore ah â€” hni + bji + hp.

That is, the product of two numbers, one of which is divided into any

number of parts, is equal to the sum of the products of the undivided

number, -and the several parts of the other ;

or, if the Geometrical interpretation of the products be restored.

The number of square units expressed by the product ab, is equal

to the number of square units expressed by the sum of the products 6m,

bUf hp.

Prop. II. Algebraically, (fig. Prop, ii.)

Let AB contain a linear units, and AC, CB, m and w linear units

respectively.

Then in + n = a,

multiply these equals by a,

therefore am + an = a^.

That is, if a number be divided into any two parts, the sum of the

products of the whole and each of the parts is equal to the square of the

whole number

Prop. III. Algebraically, (fig. Prop, iii.)

Let AB contain a linear units, and let BC contain m, and AC, n linear

units.

Then a = m + n,

multiply these equals by m,

therefore ma = m^ + mn.

That is, if a number be divided into any two parts, the product of

the whole number and one of the parts, is equal to the square of that

part, and the product of the two parts.

10^ Euclid's elements.

Prop. IV. might have been deduced from the two preceding propo-

sitions ; but Euclid has preferred the method of exhibiting, in the de-

monstrations of the second book, the equality of the spaces compared.

In the corollary to Prop. xlvi. Book I, it is stated that a parallelogram

which has one right angle, has all its angles right angles. By applying

this corollary, the demonstration of Prop. iv. may be considerably

shortened.

If the two parts of the line be equal, then the square on the whole

line is equal to four times the square on half the line.

Also, if a line be divided into any three parts, the square on the whole

line is equal to the squares on the three parts, and twice the rectangles

contained by every two parts.

Prop. IV. Algebraically, (fig. Prop, iv.)

Let the line AB contain a linear units, and the parts of it ^Cand BC^

in and n linear units respectively.

Then a = m + 7i,

squaring these equals, .'. a^ = (w -f n)-,

or a* = vi^ + Iran + w*.

That is, if a number be divided into any two parts, the square of the

number is equal to the squares of the two parts together with twice the

product of the two parts.

From Euc. ii. 4, may be deduced a proof of Euc. i, 47. In the fig.

take BL on BE^ and FM on EB, each equal to Â£C, and join CH, HL,

LM, MC. Then the figure HLMC is a square, and the four triangles

CAH, HBL, LEM, MB Care equal to one another, and together are equal

'to the two rectangles AG, GE.

Now AG, GE, FH, CK are together equal to the whole figure ABEB ;

and HLMC, with the four triangles CAH, HDL, LEB^ MBC also make

up the whole figure ADEB ; â€¢

Hence ^G, GE, FH, CK are equal to i7I<xVC together with the four

triangles ;

but AG, GE are equal to the four triangles.

wherefore FH, CK are equal to HLMC,

that is, the squares on AC, AH are together equal to the square on CH.

Prop. V. It must be kept in mind, that the sum of two straight lines

in Geometry, means the straight line formed by joining the two lines

together, so that both may be in the same straight line.

The following simple properties respecting the equal and tmequal

division of a line are worthy of being remembered.

I. Since AB = 2BC = 2 {BB + BC) = 2BD + 2BC. (fig. Prop, v.)

and AB = AB + BB;

.\ 2CB + 2BB = AB + BB,

and by subtracting 2BB from these equals,

.-. 2CB = AB - DB,

and CD = i {AD - BB).

That is, if a line ^B be divided into two equal parts in C, and into two

unequal parts in D, the part CD of the line between the points of section

is equal to half the difference of the unequal parts AB and DB.

II. Here AD = AC + CB, the sum of the unequal parts, (fig. Prop, v.)

and BB = AC - CD their difference.

KOTES TO BOOK II. 103

Hence by adding these equals together,

.*. AD + DB = 2 AC,

the sum and difference of two lines AC, CD, are together equal to

twice the greater line.

And the halves of these equals are equal,

.-. i,AD +^.DB== AC,

or, half the sum of two unequal lines AC, CD added to half their dijffe-

rence is equal to the greater line A C.

III. Again, since AD = AC + CD, and DB = AC -^ CD,

by subtracting these equals,

/. AD - DB= 2CD,

or, the difference between the sum and difference of two unequal lines is

equal to twice the less line.

And the halves of these equals are equal,

.-. -l-.AD-^.DB =CD,

or, half the difference of two lines subtracted from half their sum is equal

to the less of the two lines.

IV. Since AC - CD = DB the difference,

.-. AC= CD + DB,

and adding CD the less to each of these equals,

.-. JC + CD = 2CD + DB,

or, the sum of two unequal lines is equal to twice the less line together

with the difference between the lines.

Prop. V. Algebraically.

Let AB contain 2a linear units,

its half BC will contain a linear units.

And let CD the line between the points of section contains linear units.

Then AD the greater of the two unequal parts, contains a + m linear units ;

and DB the less contains a â€” m units.

Also m is half the difference of a + m and a â€” m-,

.'. (a + m) (a â€” m) = a^ â€” m^,

to each of these equals add m^ ;

.'. (a + m) (a â€” m) + m* = a*.

That is, if a number be divided into two equal parts, and also into two

unequal parts, the product of the unequal parts together with the square

of half their difference, is equal to the square of half the number.

Bearing in mind that AC, CD are respectively half the sum and half

the difference of the two lines AD, DB ; the corollary to this proposition

may be expressed in the following form : "The rectangle contained by

two straight lines is equal to the difference on the squares of half their

sum and half their difference."

The rectangle contained by ^D and DB, and the square on BC are

each bounded by the same extent of line, but the spaces enclosed differ

by the square on CD.

A given straightline is said to he produced when it has its length increased

in cither direction, and the increase it receives, is called the part produced.

If a point be taken in a line or in a line produced, the line is said to

be divided internally or externally, and the distances of the point from

104 Euclid's elements.

the ends of the line are called the internal or external segments of the

line, according as the point of section is in the line or the line produced.

Prop. VI. Algebraically.

Let AB contain 2a linear units, then its half BC contains a units ; and

let BD contain m units.

Then AD contains 2a + m units,

and .-. (2a + m) m = 2am + m^ ;

to each of these equals add a^,

.*. (2a + m) m + a^ = a- + 2am + Â«i'.

But a^ + 2am + n^ = (a + w)^

.â€¢. (2a + m)m + a^ = (a -\- my.

That is, If a number be divided into two equal numbers, and anothe:

number be added to the whole and to one of the parts ; the product o:

the whole number thus increased and the other number, together wi:h th(

square of half the given number, is equal to the square of the numbe]

which is made up of half the given number increased.

The algebraical results of Prop. v. and Prop. vi. are identical, as it if

obvious that the difference of a + m and a â€” m in Prop. v. is equal to th<

difference of 2a + m and m in Prop, vi, and one algebraical result ex-

presses the truth of both propositions.

This arises from the two ways in which the diflference between two^

unequal lines may be represented geometrically, when they are in the

same direction.

In the diagram (fig. to Prop, v.), the difference DjBof the two unequa

lines AC and CD is exhibited by producing the less line CD, and making

CB equal to ^C the greater.

Then the part produced DB is the difference between AC and CD^

for AC is equal to CB^ and taking CD from each,

the difference of ^C and CD is equal to the difference of CB and CD.

In the diagram (fig. to Prop, vi.), the difference DB of the two un-

equal lines CD and CA is exhibited by cutting off from CD the greater,

a part CB equal to CA the less.

Prop. VII. Either of the two parts ACy CB of the line AB may b

taken : and it is equally true, that the squares on AB and AC are equ

to twice the rectangle AB, AC, together with the square on BC.

Prop. vii. Algebraically.

Let AB contain a linear units, and let the parts AC and CB contain m

and n linear units respectively.

Then a = m + n ;

squaring these equals,

.*. a^ = m^ + 2mn + w^,

add 11^ to each of these equals,

.*. a* + n^ = w'* + 2mn + 2w*.

But 2mn + 2w* = 2 (m + n) n = 2a w,

.*. a' + 71^ = m* + 2an.

That is. If a number be divided into any two parts, the squares of the

whole number and of one of the parts,are equal to twice the product of the

whole number and that part, together with the square of the other part.

Prop. VIII. As in Prop. vii. eitlier part of the line may be talcen,

and it is also true in this Proposition, that four times the rectangle con-

^B JJOTES TO BOOK II. 105

^Bkned by JB, ^C together with the square on BC, is equal to the square

' on the straight line made up oi AB and AC together.

The truth of this proposition may be deduced from Euc. ii. 4 and 7.

For the square on AD (fig. Prop. 8.) is equal to the squares on AB^

BB, and twice the rectangle ^S, BD\ (Euc. ii. 4.) or the squares on

AB, BC, and twice the rectangle AB, BC, because BC is equal to BD:

and the squares on AB, BCare equal to twice the rectangle AB, BC with

the square on ^C: (Euc. ii. 7.) therefore the square on AD is equal to

four times the rectangle AB, BC together with the square on AC.

Prop. viir. Algebraically.

Let the whole line AB contain a linear units of which the parts AC,

CB contain m, n units respectively.

Then m + w = Â«,

and subtracting or taking n from each,

.'.m â€” a â€” Hy

squaring these equals,

/, ni' â€” a^ â€” 2an 4- yi'^,

and adding 4a7i to each of these equals,

.'. 4rt7i + m^ = a* + 2an + n^.

But a^ + 2an + w* = (a + ny,

.'. ian + m^ = (a + Â«)^.

That is, If a number be divided into any two parts, four times the pro-

duct of the whole number and one of the parts, together with the square

of the other part, is equal to the square of the number made of the whole

and the part first taken.

Prop. Till, may be put under the following form : The square on the

sum of two lines exceeds the square on their difference, by four times the

rectangle contained by the lines.

Prop. IX. The demonstration of this proposition may be deduced

from Euc. ii. 4 and 7.

For (Euc. II. 4.) the square on AD is equal to the squares on AC, CD

and twice the rectangle ^C, CD; (fig. Prop. 9.) and adding the square

on DB to each, therefore the squares on /4D, DB are equal to the squares

on AC, CD and twice the rectangle AC, CD together with the square on

DB ; or to the squares on BC, CD and twice the rectangle BC, CD with

the square on DB, because J?C is equal to AC.

But the squares on BC, CD are equal to twice the rectangle BC, CD,

with the square on DB. (Euc. ii. 7.)

Wherefore the squares on AD, DB are equal to twice the squares on

BC and CD.

Prop. IX. Algebraically.

Let AB contain 2a linear units, its half ^C or BC will contain a units ;

and let CD the line between the points of section contain m units.

Also AD the greater of the two unequal parts contains a + m units,

and DB the less contains a â€” m units.

Then {a + my = a^ + 2am + m%

and (a - my = a^ â€” 2am + m^.

Hence by adding these equals,

/. (a + my + (a - tny = 2a- + 2mÂ».

f5

106 Euclid's elements.

That is, If a number be divided into two equal parts, and also into two

unequal parts, the sum of the squares of the two unequal parts is equal

to twice the square of half the number itself, and twice the square of

CD, DA, and twice the rectangle BC, CD;

but the square on BA is equal to the squares on BD, DA, (l. 47.)

because the angle at D is a right angle ;

and the square on CA is equal to the squares on CD, DA ;

therefore the square on BA is equal to the squares on BC, CA, and

twice the rectangle B C, CD ;

tJiat is, the square on BA is greater than the squares on BC, CA, by

twice the rectangle BC, CD.

Therefore in obtuse-angled triangles, &c. q.e.d.

PROPOSITION XIII. THEOREM.

Tn every triangle, the square on the side subtending either of the acut'

angles, is less than the squares on the sides containing that angle, by twic

the rectangle contained by either of these sides, and the straight line inter

cepted beticeeii the acute angle and the perpendicular let fall upon it from

the opposite angle.

Let ABC be any triangle, and the angle at B one of its acute

angles, and upon BC, one of the sides containing it, let fail the

perpendicular AD from the opposite angle. (l. 12.)

Then the square on ^C opposite to the angle B, shall be less than

the squares on CB, BA, by twice the rectangle CB, BD.

BOOK 11. PROP. XIII.

97

First, let AD fall within the triangle ABC,

Then because the straight line CB is divided into two parts in D,

the squares on CB, BD are equal to twice the rectangle contained by

CB, BD, and the square on DC; (ll. 7.)

to each of these equals add the square on AD ;

therefore the squares on CB, BD, DA, are equal to twice the

rectangle CB, BD, and the squares on AD, DC;

but the square on ^jB is equal to the squares on BD, DA, (l. 47.)

because the angle BDA is a right angle ;

and the square on ^ C is equal to the squares on AD, DC',

therefore the squares on CB, BA are equal to the square on A C,

and twice the rectangle CB, BD :

that is, the square on ^C alone is less than the squares on CB, BA,

by twice the rectangle CB, BD.

Secondly, let AD fall without the triangle ABC,

B CD

Then, because the angle at -D is a right angle,

the angle ACB is greater than a right angle ; (l. 16.)

and therefore the square on ^^ is equal to the squares on AC, CB,

and twice the rectangle J? C CD; (il. 12.)

to each of these equals add the square on BC;

therefore the squares on AB, BC are equal to the square on AC,

twice the square on BC, and twice the rectangle BC, CD;

but because BD is divided into two parts in C,

therefore the rectangle DB, BC is equal to the rectangle BC, CD,

and the square on BC; (il. 3.)

and the doubles of these are equal ;

that is, twice the rectangle DB, BC is equal to twice the rectangle

BC, CD and twice the square on BC:

tiierefore the squares on AB, J5Care equal to the square on AC,

and twice the rectangle DB, BC:

wherefore the square on ^ C alone is less than the squares on AB,BC;

by twice the rectangle DB, BC.

Lastly, let the side AC he perpendicular to BC.

A

B c

Then BC is the straight line between the perpendicular and the

acute angle at B ;

and it is manifest, that the squares on AB, BC,_ are equal to the

square on A C, and twice the square on BC. (l. 47.)

.^^ Therefore in any triangle, &c. q.e.d.

98 Euclid's elemeni'i

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure.

It is required to describe a square that shall be equal to A.

H

Describe the rectangular parallelogram BCDE equal to the rect

lineal figure A. (i. 45.)

Then, if the sides of it, BE, ED, are equal to one another,

it is a square, and what was required is now done.

But if BE, EL), are not equal,

produce one of them BE to F, and make EF equal to ED^

bisect ^i^ in G; (i. 10.)

from the center Gj at the distance GB, or GF, describe the semicircle

BHF,

and produce DE to meet the ch'cumference in H,

The square described upon EH shall be equal to the given recti-

lineal figure A.

Join GH.

Then because the straight line BF is divided into two equal pai

in the point G, and into two unequal parts in the point E',

therefore the rectangle BE, EF, together with the square on EQ

is equal to the square on GF', (ii. 5.)

but GF'i^ equal to GH; (def. 15.)

therefore the rectangle BE, EF, together with the square on EG, i

equal to the square on GH;

but the squares on HE, EG are equal to the square on GH; (l. 47j

therefore the rectangle BE, EF, together with the square on EC^^

is equal to the squares on HE, EG ; '

take away the square on EG, which is common to both ;

therefore the rectangle BE, EF is equal to the square on HE.

But the rectangle contained by BE, EF is the parallelogram BD,

because EF is equal to ED ;

therefore BD is equal to the square on EH;

but BD is equal to the rectilineal figure A ; (constr.)

therefore the square on EH is equal to the rectilineal figure A.

Wherefore a square has been made equal to the given rectilineal

figure Af namely, the square described upon EH. Q. e.f.

NOTES TO BOOK IL

In Book T, Geometrical magnitudes of the same kind, lines, angles

and surfaces, more particularly triangles and parallelograms, are com-

pared, either as being absolutely equal, or unequal to one another.

In Book II, the properties of right-angled parallelograms, but without

reference to their magnitudes, are demonstrated, and an important

extension is made of Euc. i. 47, to acute-angled and obtuse- angled

triangles. Euclid has given no definition of a rectangular parallelogram

or rectangle : probably, because the Greek expression -rrapaXk^Xoy pajxixov

SpdoywvLov, or opdoyuviov simply, is a definition of the figure. In English,

the term rectangle, formed from rectus angulus, ought to be defined before

its pi'operties are demonstrated. A rectangle may be defined to be a

parallelogram having one angle a right-angle, or a right angled paral-

lelogram ; and a square is a rectangle having all its sides equal.

As the squares in Euclid's demonstrations are squares described or

supposed to be described on straight lines, the expression 'â– '"the square

on AB," is a more appropriate abbreviation for *' the square described on

the line AB," than " the square of AB." The latter expression more

fitly expresses the arithmetical or algebraical equivalent for the square

on the line AB.

In Euc. I. 35, it may be seen that there may be an indefinite number

of parallelograms on the same base and between the same parallels whose

areas are always equal to one another ; but that one of them has all its

angles right angles, and the length of its boundary less than the boundary

of any other parallelogram upon the same base and between the same

parallels. The area of this rectangular parallelogram is therefore de-

termined by the two lines which contain one of its right angles. Hence

it is stated in Def. 1, that every right-angled parallelogram is said to be

contained by any two of the straight lines which contain one of the right

angles. No distinction is made in Book ii, between equality and identity,

as the rectangle may be said to be contained by two lines which are

equal respectively to the two which contain one right angle of the figure.

It may be remarked that the rectangle itself is bounded by four straight

lines.

It is of primary importance to discriminate the Geometrical conception

of a rectangle from the Arithmetical or Algebraical representation of it.

The subject of Geometry is magnitxide not number, and therefore it would

be a departure from strict reasoning on space, to substitute in Geometrical

demonstrations, the Arithmetical or Algebraical representation of a rect-

angle for the rectangle itself. It is however, absolutely necessary that

the connexion of number and magnitude be clearly understood, as far

as regards the representation of lines and areas.

All lines are measured by lines, and all surfaces by surfaces. Some

one line of definite length is arbitrarily assumed as the linear unit, and

the length of every other line is represented by the number of linear

units, contained in it. The square is the figure assumed for the measure

of surfaces. The square unit or the unit of area is assumed to be that

square, the side of which is one unit in length, and the magnitude of

every surface is represented by the number of square units contained

in it. But here it may be remarked, that the properties of rectangles

and squares in the Second Book of Euclid are proved independently

f2

R

100 ErCLlD's

of the consideration, whether the sides of the rectangles can be repre-

sented by any multiples of the same linear unit. If, however, the

sides of rectangles are supposed to be divisible into an exact number

of linear units, a numerical representation for the area of a rectangle

may be deduced.

On two lines at right angles to each other, take AB equal to 4, and

AD equal to 3 linear units.

Complete the rectangle ABCD, and through the points of division of

AB, AD, draw EL, FM, GN parallel to AD ; and HP, KQ parallel to

AB respectively.

A E F G B

I

:p

-Q

I

D L M N C

Then the whole rectangle AC is divided into squares, all equal to each

other.

And AC is equal to the sum of the rectangles AL, EM, FN, GC ; (ii. 1.)

also these rectangles are equal to one another, (i. 36.)

therefore the whole ^C is equal to four times one of them AL.

Again, the rectangle AL is equal to the rectangles EH, HR, RD,

and these rectangles, by construction, are squares described upon the

equal lines AH, HK, KD, and are equal to one another.

Therefore the rectangle AL \q equal to 3 times the square on AH,

but the whole rectangle AC is equal to 4 times the rectangle AL,

therefore the rectangle ^C is 4 x 3 times the square on AH, or 12

square units :

that is, the product of the two numbers which express the number of

linear xmits in the two sides, will give the number of square units in the

rectangle, and therefore will be an arithmetical representation of its area.

And generally, if AB, AD, instead of 4 and 3, consisted of a and b

linear units respectively, it may be shewn in a similar manner, that the

area of the rectangle A C would contain ab square units ; and therefore the

product ab is a proper representation for the area of the rectangle AC. j

Hence, it follows, that the term rectangle in Geometry corresponds t(^

the term product in Arithmetic and Algebra, and that a similar com-

parison may be made between the products of the two numbers which

reiu'csent the sides of rectangles, as between the areas of the rectangleÂ«i'

themselves. This forms the basis of what are called Arithmetical or,

Algebraical proofs of Geometrical properties.

If the two sides of the rectangle be equal, or if b be equal to a,^

the figure is a square, and the area is represented by aa or a*.

Also, since a triangle is equal to the half of a parallelogram of the

same base and altitude ;

Therefore the area of a triangle will be represented by half the rect-

angle which has the same base and altitude as the triangle : in other

words, if the length of the base be a units, and the altitude be b units ;

Then the area of the triangle is algebraically represented by hab.

The demonstrations of the first eight propositions, exemplify th(

obvious axiom, that, *' the whole area of every figure in each case, is

equal to all the parts of it taken together."

Def. 2. The parallelogram Â£iiC together with the complements AF

NOTES TO BOOK II. 101

(7, is also a gnomon, as well as the parallelogram HG together with the

same complements.

Prop. I. For the sake of brevity of expression, ** the rectangle con-

tained by the straight lines JB, BC," is called "the rectangle AB, BC;"

and sometimes " the rectangle ABC."

To this proposition may be added the corollary : If two straight lines

be divided into any number of parts, the rectangle contained by the two

straight lines, is equal to the rectangles contained by the several parts of

one line and the several parts of the other respectively.

The method of reasoning on the properties of rectangles, by means of

the products Avhich indicate the number of square units contained in their

arcas,is foreign to Euclid's ideas of rectangles, as discussed in his Second

Book, which have no reference to any particular unit of length or measure

of surface.

Prop. T. The figures J5//, BK, DL, EH are rectangles, as may

readily be shewn. For, by the parallels, the angle CEL is equal to EDK ;

and the angle EDK is equal to BDG (Euc. i. 29.). But BDG is a right

angle. Hence one of the angles in each of the figures BH, BK, DL, EH

is a right angle, and therefore (Euc. i. 46, Cor.) these figures are

rectangular.

Prop. I. Algebraically, (fig. Prop, i.)

liCt the line BC contain a linear units, and the line A, h linear units ol

the same length.

Also suppose the parts BD, DE, EC to contain m, n, p linear units

respectively.

Then a â€” m -\- n â– \- p,

multiply these equals by h,

therefore ah â€” hni + bji + hp.

That is, the product of two numbers, one of which is divided into any

number of parts, is equal to the sum of the products of the undivided

number, -and the several parts of the other ;

or, if the Geometrical interpretation of the products be restored.

The number of square units expressed by the product ab, is equal

to the number of square units expressed by the sum of the products 6m,

bUf hp.

Prop. II. Algebraically, (fig. Prop, ii.)

Let AB contain a linear units, and AC, CB, m and w linear units

respectively.

Then in + n = a,

multiply these equals by a,

therefore am + an = a^.

That is, if a number be divided into any two parts, the sum of the

products of the whole and each of the parts is equal to the square of the

whole number

Prop. III. Algebraically, (fig. Prop, iii.)

Let AB contain a linear units, and let BC contain m, and AC, n linear

units.

Then a = m + n,

multiply these equals by m,

therefore ma = m^ + mn.

That is, if a number be divided into any two parts, the product of

the whole number and one of the parts, is equal to the square of that

part, and the product of the two parts.

10^ Euclid's elements.

Prop. IV. might have been deduced from the two preceding propo-

sitions ; but Euclid has preferred the method of exhibiting, in the de-

monstrations of the second book, the equality of the spaces compared.

In the corollary to Prop. xlvi. Book I, it is stated that a parallelogram

which has one right angle, has all its angles right angles. By applying

this corollary, the demonstration of Prop. iv. may be considerably

shortened.

If the two parts of the line be equal, then the square on the whole

line is equal to four times the square on half the line.

Also, if a line be divided into any three parts, the square on the whole

line is equal to the squares on the three parts, and twice the rectangles

contained by every two parts.

Prop. IV. Algebraically, (fig. Prop, iv.)

Let the line AB contain a linear units, and the parts of it ^Cand BC^

in and n linear units respectively.

Then a = m + 7i,

squaring these equals, .'. a^ = (w -f n)-,

or a* = vi^ + Iran + w*.

That is, if a number be divided into any two parts, the square of the

number is equal to the squares of the two parts together with twice the

product of the two parts.

From Euc. ii. 4, may be deduced a proof of Euc. i, 47. In the fig.

take BL on BE^ and FM on EB, each equal to Â£C, and join CH, HL,

LM, MC. Then the figure HLMC is a square, and the four triangles

CAH, HBL, LEM, MB Care equal to one another, and together are equal

'to the two rectangles AG, GE.

Now AG, GE, FH, CK are together equal to the whole figure ABEB ;

and HLMC, with the four triangles CAH, HDL, LEB^ MBC also make

up the whole figure ADEB ; â€¢

Hence ^G, GE, FH, CK are equal to i7I<xVC together with the four

triangles ;

but AG, GE are equal to the four triangles.

wherefore FH, CK are equal to HLMC,

that is, the squares on AC, AH are together equal to the square on CH.

Prop. V. It must be kept in mind, that the sum of two straight lines

in Geometry, means the straight line formed by joining the two lines

together, so that both may be in the same straight line.

The following simple properties respecting the equal and tmequal

division of a line are worthy of being remembered.

I. Since AB = 2BC = 2 {BB + BC) = 2BD + 2BC. (fig. Prop, v.)

and AB = AB + BB;

.\ 2CB + 2BB = AB + BB,

and by subtracting 2BB from these equals,

.-. 2CB = AB - DB,

and CD = i {AD - BB).

That is, if a line ^B be divided into two equal parts in C, and into two

unequal parts in D, the part CD of the line between the points of section

is equal to half the difference of the unequal parts AB and DB.

II. Here AD = AC + CB, the sum of the unequal parts, (fig. Prop, v.)

and BB = AC - CD their difference.

KOTES TO BOOK II. 103

Hence by adding these equals together,

.*. AD + DB = 2 AC,

the sum and difference of two lines AC, CD, are together equal to

twice the greater line.

And the halves of these equals are equal,

.-. i,AD +^.DB== AC,

or, half the sum of two unequal lines AC, CD added to half their dijffe-

rence is equal to the greater line A C.

III. Again, since AD = AC + CD, and DB = AC -^ CD,

by subtracting these equals,

/. AD - DB= 2CD,

or, the difference between the sum and difference of two unequal lines is

equal to twice the less line.

And the halves of these equals are equal,

.-. -l-.AD-^.DB =CD,

or, half the difference of two lines subtracted from half their sum is equal

to the less of the two lines.

IV. Since AC - CD = DB the difference,

.-. AC= CD + DB,

and adding CD the less to each of these equals,

.-. JC + CD = 2CD + DB,

or, the sum of two unequal lines is equal to twice the less line together

with the difference between the lines.

Prop. V. Algebraically.

Let AB contain 2a linear units,

its half BC will contain a linear units.

And let CD the line between the points of section contains linear units.

Then AD the greater of the two unequal parts, contains a + m linear units ;

and DB the less contains a â€” m units.

Also m is half the difference of a + m and a â€” m-,

.'. (a + m) (a â€” m) = a^ â€” m^,

to each of these equals add m^ ;

.'. (a + m) (a â€” m) + m* = a*.

That is, if a number be divided into two equal parts, and also into two

unequal parts, the product of the unequal parts together with the square

of half their difference, is equal to the square of half the number.

Bearing in mind that AC, CD are respectively half the sum and half

the difference of the two lines AD, DB ; the corollary to this proposition

may be expressed in the following form : "The rectangle contained by

two straight lines is equal to the difference on the squares of half their

sum and half their difference."

The rectangle contained by ^D and DB, and the square on BC are

each bounded by the same extent of line, but the spaces enclosed differ

by the square on CD.

A given straightline is said to he produced when it has its length increased

in cither direction, and the increase it receives, is called the part produced.

If a point be taken in a line or in a line produced, the line is said to

be divided internally or externally, and the distances of the point from

104 Euclid's elements.

the ends of the line are called the internal or external segments of the

line, according as the point of section is in the line or the line produced.

Prop. VI. Algebraically.

Let AB contain 2a linear units, then its half BC contains a units ; and

let BD contain m units.

Then AD contains 2a + m units,

and .-. (2a + m) m = 2am + m^ ;

to each of these equals add a^,

.*. (2a + m) m + a^ = a- + 2am + Â«i'.

But a^ + 2am + n^ = (a + w)^

.â€¢. (2a + m)m + a^ = (a -\- my.

That is, If a number be divided into two equal numbers, and anothe:

number be added to the whole and to one of the parts ; the product o:

the whole number thus increased and the other number, together wi:h th(

square of half the given number, is equal to the square of the numbe]

which is made up of half the given number increased.

The algebraical results of Prop. v. and Prop. vi. are identical, as it if

obvious that the difference of a + m and a â€” m in Prop. v. is equal to th<

difference of 2a + m and m in Prop, vi, and one algebraical result ex-

presses the truth of both propositions.

This arises from the two ways in which the diflference between two^

unequal lines may be represented geometrically, when they are in the

same direction.

In the diagram (fig. to Prop, v.), the difference DjBof the two unequa

lines AC and CD is exhibited by producing the less line CD, and making

CB equal to ^C the greater.

Then the part produced DB is the difference between AC and CD^

for AC is equal to CB^ and taking CD from each,

the difference of ^C and CD is equal to the difference of CB and CD.

In the diagram (fig. to Prop, vi.), the difference DB of the two un-

equal lines CD and CA is exhibited by cutting off from CD the greater,

a part CB equal to CA the less.

Prop. VII. Either of the two parts ACy CB of the line AB may b

taken : and it is equally true, that the squares on AB and AC are equ

to twice the rectangle AB, AC, together with the square on BC.

Prop. vii. Algebraically.

Let AB contain a linear units, and let the parts AC and CB contain m

and n linear units respectively.

Then a = m + n ;

squaring these equals,

.*. a^ = m^ + 2mn + w^,

add 11^ to each of these equals,

.*. a* + n^ = w'* + 2mn + 2w*.

But 2mn + 2w* = 2 (m + n) n = 2a w,

.*. a' + 71^ = m* + 2an.

That is. If a number be divided into any two parts, the squares of the

whole number and of one of the parts,are equal to twice the product of the

whole number and that part, together with the square of the other part.

Prop. VIII. As in Prop. vii. eitlier part of the line may be talcen,

and it is also true in this Proposition, that four times the rectangle con-

^B JJOTES TO BOOK II. 105

^Bkned by JB, ^C together with the square on BC, is equal to the square

' on the straight line made up oi AB and AC together.

The truth of this proposition may be deduced from Euc. ii. 4 and 7.

For the square on AD (fig. Prop. 8.) is equal to the squares on AB^

BB, and twice the rectangle ^S, BD\ (Euc. ii. 4.) or the squares on

AB, BC, and twice the rectangle AB, BC, because BC is equal to BD:

and the squares on AB, BCare equal to twice the rectangle AB, BC with

the square on ^C: (Euc. ii. 7.) therefore the square on AD is equal to

four times the rectangle AB, BC together with the square on AC.

Prop. viir. Algebraically.

Let the whole line AB contain a linear units of which the parts AC,

CB contain m, n units respectively.

Then m + w = Â«,

and subtracting or taking n from each,

.'.m â€” a â€” Hy

squaring these equals,

/, ni' â€” a^ â€” 2an 4- yi'^,

and adding 4a7i to each of these equals,

.'. 4rt7i + m^ = a* + 2an + n^.

But a^ + 2an + w* = (a + ny,

.'. ian + m^ = (a + Â«)^.

That is, If a number be divided into any two parts, four times the pro-

duct of the whole number and one of the parts, together with the square

of the other part, is equal to the square of the number made of the whole

and the part first taken.

Prop. Till, may be put under the following form : The square on the

sum of two lines exceeds the square on their difference, by four times the

rectangle contained by the lines.

Prop. IX. The demonstration of this proposition may be deduced

from Euc. ii. 4 and 7.

For (Euc. II. 4.) the square on AD is equal to the squares on AC, CD

and twice the rectangle ^C, CD; (fig. Prop. 9.) and adding the square

on DB to each, therefore the squares on /4D, DB are equal to the squares

on AC, CD and twice the rectangle AC, CD together with the square on

DB ; or to the squares on BC, CD and twice the rectangle BC, CD with

the square on DB, because J?C is equal to AC.

But the squares on BC, CD are equal to twice the rectangle BC, CD,

with the square on DB. (Euc. ii. 7.)

Wherefore the squares on AD, DB are equal to twice the squares on

BC and CD.

Prop. IX. Algebraically.

Let AB contain 2a linear units, its half ^C or BC will contain a units ;

and let CD the line between the points of section contain m units.

Also AD the greater of the two unequal parts contains a + m units,

and DB the less contains a â€” m units.

Then {a + my = a^ + 2am + m%

and (a - my = a^ â€” 2am + m^.

Hence by adding these equals,

/. (a + my + (a - tny = 2a- + 2mÂ».

f5

106 Euclid's elements.

That is, If a number be divided into two equal parts, and also into two

unequal parts, the sum of the squares of the two unequal parts is equal

to twice the square of half the number itself, and twice the square of

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38