Lawrence S. (Lawrence Sluter) Benson. # Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson online

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Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 5 of 21)

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together with the rectangle AC.CB. If, therefore, etc.

8cho. Otherwise : Take a line D equal to CB. Then (II. 1)

the rectangle AB.D is equivalent to the

-^ ^ ^ rectangles BCD and AC.D ; that is (const.

and L def 15), the rectangle AB.BC is

^ equivalent to the square of BC togctlier with

the rectangle AC.CB.

Prop. IV.â€” Tiieor.â€” 7/" a straight line he divided into

any tico j^^-^^ts, the square of the v-hole line is equivalent to

the squares of the tioo i)arts\ together with twice their red

angle.

Let the straight line AB be divided into any two parts in C ;

the square of AB is equivalent to the squares of AC and CB,

together with twice the rectangle under AC and CB.

On AB describe (L 23) the square of AE, and join BD;

through C draw (I. 18) CGF parallel to AD or BE; and

through G draw UK parallel to AB or DE. Then, l)ecause

G

/

BOOK II.] EUCLID AND LEGENDEE. 51

CF is parallel to AD, aucl BD falls upon them, the exterior

angle CGB is equivalent (I. 16) to the inte-

rior and remote angle ADB ; but ADB is ^ ^ ?

â€¢equal (I. l) to ABD, because BA and AD

are equal, being sides of a square; wherefore

(I. ax. 1) the angle CGB is equal to GBC;

iiad therefore the side BC is equal (I. 1, cor.)

to the side CG. But (const.) the figure CK Â» ^ B

is a parallelogram ; and since CBK is a right

angle, and BC equal to CG, CK (I. def 15) is a square, and it

is upon the side CB. For the like reason HF also is a square,

and it is upon the side HG, which is equal (I. 15, cor. 1) to AC;

therefore HF, CK are the squares of AC, CB. And because

(I. 15, cor. 8) the complements AG, GEare equivalent, and that

AG is the rectangle contained by AC, CB, for CG has been

proved to be equal to CB ; therefore GE is also equivalent to

the rectangle AC.CB; wherefore AG, GE are equivalent to

twice the rectangle AC.CB. The four figures, therefore, HF,

-CK, AG, GE are equivalent to the squares of AC, CB, and

twice the rectangle AC.CB. But HF, CK, AG, GE make up

the Avhole figure AE, which is the square of AB ; therefore the

square of AB is equivalent to the squares of AC and CB, and

twice the rectangle AC.CB. Wherefore, etc.

Otherwise: AB^ = AB.AC + AB.CB (H. 2). But (H. 3)

A.B. AC =AC=+ AC.CB, and AB.CB=zCB-+AC.BC. Hence

<I. ax. 2) AB.AC + AB.BC, or AB-=AC^+CB-+2AC.CB.

Cor. 1. It follows from this demonstration, that the parallel-

ograms about the diagonal of a square are likewise squares.

Cor. 2. Hence the square of a straight line â€¢ is equivalent to

four times the square of its half; for the straight line being

iDisected, the rectangle of the parts is equivalent to the square

of one of them.

Pkop. V. â€” Theor. â€” If a straight line be divided into two

' equal parts, and also into tioo unequal parts ; the rectangle con-

tained by the unequal parts., together with the square of the line

hettoeen the points of section, is equivalent to the square of half

the line.

Let the straio;ht line AB be bisected in C, and divided un-

H

/

K L

/

52 THE ELEMENTS OF [bOOK II.

equally in D ; the rectangle AD.DB, together with the square

of CD, is equivalent to the square of CB.

On CB describe (I. 23) the square CF, join BE, and through

D draw (I. 18) DHG parallel to CE or BF ; also through H

draw KLM parallel to CB or EF ; and

â– ^ c D B through A draw AK parallel to CL or

BM. Then (I. 15, cor. 5) AL and CM

^^ are equal, because AC is equal to CB ;

and (I. 15, cor. 8) the complements CH

E G F and HF are equivalent. Therefore (I.

ax. 2) AL and CH together are equal to

CM and HF together ; that is, AH is equivalent to the gnomon

CMC To each of those add LG, and (I. ax. 2) the gnomon

CMG, together with LG, is equivalent to AH together with

LG. But the gnomon CMG, and LG make up the figure

CEFB, which is the square of CB ; also AH is the rectangle

imder AD and DB, because DB is equal (IL 4, cor. 1) to DH ;

and LG is the square of CD. Thei-efore the rectangle AD.DB

and the square of CD are equivalent to the square of CB.

Wherefore, if a straight line, etc.

Otherwise : Since, as is easily seen from the proof in the

above, DF is equal to AL, take these separately from the entire

figure, and there remain AH and LG equivalent to the square

CF, as before. The proof may also be as follows:

AD.DB = CD.DB + AC.DB (H. 1) or AD.DB =CD.DB+

CB.DB, because CB=:AC. Hence (H. 3) AD.DB =CD.DB +

CD.DB+DB^=i:2CD.DB + DBl To each of these add CD=;

then AD.DB + CD^=2CD.DB+DB^+CD^ or (H. 4) AD.DB

+ CD'=CB\

Cor. 1. Hence the ditference of the squares of CB and CD is

equivalent to the rectangle under AD and DB. But since AC

is equal to CB, AD is equivalent to the sum of CB and CD, and

DB is the difterence of these lines. Hence the difference of the

squares of two straight lines is equivalent to the rectangle under

their sum and difference.

Cor. 2. Since the square of CB, or, which is the same, the

rectangle AC.CB, is greater than the rectangle AD.DB by the

equare of CD, it follows, that to divide a straight line into two

BOOK n.] EUCLID AND LEGENDKE. 53

parts, the rectangle of wliich may be the greatest possible, or,

as is termed, a tnaxlmiim, tlie line is to be bisected.

Cor. 3. Hence also tlie sum of the squares of tlie two parts

into which a straight line is divided, is the least possible, or is,

as it is termed, a ininimuni, Avlien the line is bisected. For (II.

4) the square of tlie line is equivalent to the squares of the

parts and twice their rectangle ; and therefore the greater the

rectangle is, the less are the squares of the j^arts; but, by the

foregoins; corollarv, the rectangle is a maximum when the line

is bisected.

C<yr. 4. Since (I. 24, cor. 5) the difference of the squares of

the sides of a triangle is equivalent to the difference of the

squares of the segments of the base, it follows, from the first

corollary above, that the rectangle under the sum and differ-

ence of the sides of a triangle is equivalent to the rectano-le

under the sum and difference of the segments, intercepted be-

tween the extremities of the base and the jjoint in which the

perpendicular cuts the base, or tlie base produced.

Cor. 5. Hence, also, if a straight line be drawn from the ver-

tex of an isosceles triangle to any point in the base, or its con-

tinuation, the difference of the squares of that line and either of

the equal sides is equivalent to the rectangle under the seoments

intercepted between the extremities of the base and the jDoint.

Cor. 6. Since (I. 24, cor. 4) the square of one of the legs of a

right-angled triangle is equivalent to the difference of the

squares of the hypothenuse and the other leg, it follows (IT. 5,

cor. 1) that the square of one leg of a right-angled triangle is

equivalent to the rectangle under the sum and difference of the

hypothenuse and the other.

^cho. And a parallelogram can be constructed equivalent

to a ffiven triangle, or anv given rectilinear fiorure havinor an

angle equal to a given angle, or applied to a given straight line

â€” that is, having that straight line for one of its sides, when

the jjarallelogram shall be equivalent to a given triangle or

given rectilinear figure, and have one of its angles equal to a

given angle, by applying a parallelogram equivalent to the

given triangle with an equal angle, to a given straight line, and

then constructing an equal triangle to the given triangle (I. 15

and 15, cor. 4).

A C

B D

K

H

/

L

/

54 THE ELEMENTS OF [bOOK II,

Prop. VI. â€” Theok. â€” If a straight line he bisected, and be

produced to any point, the rectangle contained by the whole

line thus produced, and the part of it produced, together with

the square of half the line bisected, is equivalent to the square

of the straight line which is made %ip of the half and the p>art

produced.

Let the straight line AB be bisected in C, and produced to

D ; the rectangle AD.DB, and the square of CB, are equiva-

lent to the square of CD.

Upon CD describe (I. 23) the square CF, and join DE ;

through B draw (I. 18) BHG parallel to CE or DF ; through

H draw KLM parallel to AD or EF, and

through A draw AK parallel to CL or

jj DM. Then because AC is equal to CB,

the rectangle AL is equal (I. 15, cor. 5) to

CH ; but (1.15, cor. 8) the complements CH,

J, ^j J, HF are equivalent ; therefore, also, AL is

equal to HF, To each of these add CM

and LG ; therefore AM and LG are equivalent to the whole

square CEFD, But AM is the rectangle under AD and DB,

because (11. 4, cor. 1) DB is equal to DM; also, LG is the

square of CB, and CEFD the square of CD. Therefore the

rectangle AD.DB and the square of CB are equivalent to the

square of CD. Wherefore, if a straight line, etc.

Otherwise : Produce CA to N, and make CN equal to CD.

To these add the equals CB and CA ;

^ ^ ^ ^ Â° therefore NB is equal (L ax. 2) to AD.

But (IL 5) the rectangle NB.BD, oi-

AD.BD, together with the square of CB, is equivalent to the

square of CD ; which was to be proved.

Prop. VIL â€” Theor. â€” If a straight line be divided into a?2y

two parts, the squares of the whole line and of one of the p>arts

are equivalent to ticice the rectangle contained by the whole and

that part, together tcith the square of the other part.

Let the straight line AB be divided into any two parts in the

point C ; the squares of AB, BC are equivalent to twice the

rectangle AB.BC, together with the square of AC.

BOOK II.] EUCLID AND LEGENDRE. 65

Upon AB describe (I. 23) the square AE, and construct the

figure as in the preceding propositions. Then,

because (I. 15, cor. 8) the complements CII, ^ c B

FK are equivalent, add to each of thorn CK;

the whole AK is therefore equal to the whole ^

CE ; therefore AK, CE are together double

of AK. But AK, CE are the gnomon AKF,

K

together with the square CK ; therefore the i> F E

gnomon AKF and the square CK are double

of AK, or double of the rectangle AB.BC, because BC is equal

(II. 4, cor. 1) to BK. To each of these equals add HF, which

is equal (II. 4, cor. 1 ) to the square of AC ; therefore the gno-

mon AKF, and the squares CK, HF are equivalent to twice the

rectangle AB.BC and the square of AC. But the gnomon

AKF, and the squares CK, HF make up the whole figure AE,

together with CK; and these are the squares of AB and BC ;

therefore the squares of AB and BC are equivalent to twice the

rectangle AB.BC, together with the square of AC ; wherefore,

if a straight line, etc.

Otherwise: Because (II. 4) AB^ = AC=+BC-+2AC.BC, add

BC= to both; then AB=+BC==AC= + 2BC=+2AC.BC. But

(II. 3) BC=+AC.BC= AB.BC, and therefore 2BC=+2AC.BC

=2AB.BC; wherefore AB-+BC^=AC=+2AB.BC.

Cor. 1. Since AC is the diflierence of AB and BC, it follows

that the square of the difterence of two straight lines is equiva-

lent to the sum of their squares, wanting twice their rectangle.

Cor. 2. Since (II. 4) the square of the sum of two lines ex-

ceeds the sum of their squares by twice their rectangle, and

since, by the foregoing corollary, the square of their difference

is less than the sum of their squares by twice their rectangle, it

follows that the square of the sum of two lines is equivalent to

the square of their difference, together with four times their

rectangle.

Pkop. VIII. â€” TiiEOR. â€” If a straight line he divided into two

equal, and also into tvw imequal parts, the squares of the un-

equal parts are together double of the square of half the linCj

and of the square of the line hetioeen the points of section.

Let the straight line AB be divided equally in C, and un-

56

THE ELICMENTS OF

[book n.

E

equally in D ; the square? of AD, DB are together double of

the squares of AC, CD.

From C draAV (I. 7) CE j)ei-|ie!Klicular to AB, and make it

equal to AC or CB, and join EA, EB ; through D draw (I. 18)

DF jjarallel to CE, and tlirougli V draw FG parallel to AB ;

and join AF. Then because (const.) the triangles ACE, BCE

are right-angled and isosceles, tlie angles CAE, AEC, CEB,

O O 7 O 7 3 7

EBC are (I. 20, cor. 4) each half a riglit angle. So also (I. 16,

part 2) are EEC, BED, because FG is parallel to AB, and ED

to EC ; and for the same reason EGF, ADF are right angles.

The angle AEB is also a right angle, its parts being each half

a right angle. Hence (I. 1) EG is equal to GF or CD, and ED

to DB. Again (I. 24, cor. 1) : the square

of AE is equivalent to the squares of AC,

CE, or to tAvice the square of AC, since

AC and CE are equal. In like manner,

the square of EF is equivalent to twice

the square of GF or CD. Now (I. 24,

^ cor. 1), the squares of AD and DF, or

of AD and DB, are equivalent to the square of AF ; and the

squares of AE, EF, that is, twice the square of AC and twice

the square of CD, are also equivalent to the square of AF;

therefore (I. ax. 1) the squares of AD, DB are equivalent to

twice the square of AC and twice the square of CD. If, there-

fore, a straight line, etc.

Otherwise: DB^+2BC.CD=BC=+CD^ (II. 7), or DB=+

2AC.CD=:AC= + CD^-; also (11. 4) AD==:AC^+CD=+2AC.

CD. Add these equals togethei-, and from the sums take

2AC.CD; then AD-+DB^=2AC^+2CD^

Prop. IX. â€” ^Theor. â€” If a straight line he bisected, and pro-

duced to any point, the squares of the whole line thus produced,

and of the part of it produced, are together double of the square

of half the lifie bisected, and of the square of the line made up

of the half and the part produced.

Let the straight line AB be bisected in C, and produced to

D ; the squares of AD, DB arc double of the squares of AC,

CD.

From C draw (I. 7) CE perpendicular to AB ; and make it

BOOK II.]

EUCLID AND LEGENDRE.

57

A

equal to AC or CB ; join AE, EB, and tln-ough E and D draw

(I. 1 8) EF parallel to AB, and

DF parallel to CE. Then be-

cause the straiu'lit line EF

meets the parallels EC, FD,

the ano-les CEF, EFD are

equivalent (I. 9) to two right

anixles : and therefore the an-

gles BEF, EFD are less than

two right angles; therefore (I. 19) EB, FD will meet, if pro-

duced toward B, D ; let them meet in G, and join AG. Then

it would be proved, as in the last proposition, that the angles

CAE, xVEC, CEB, EBC are each half a right angle, and AEB

a right angle. BDG is also a right angle, being equal (I. 16)

to ECB, since (const.) EC, EG are parallel ; DGB, DBG are

each half a right angle, being equal (I. 10 and I. 11) to CEB,

CBE, each to each ; and FEG is half a right angle, being (I.

16) equal to CBE. It would also be proved, as in the last

proposition, that the squai-e of AE is twice the square of AC,

and the square of EG twice the square of EF or CD. Now

(I. 24, cor. l),the squares of AD, DG, or of AD, DB, are equiv-

alent to tlie sipiare of AG; and the squares of AE, EG, or twice

the square of AC and twice the square of CD, are also equiva-

lent to the square of AG. Therefore (I. ax. 1) the squares of

AD, DB are equivalent to twice the square of AC and twice

the square of CD. If, therefore, a straight line, etc.

Otherwise: Produce CA, making CH equal to CD. To

these add CB, CA ; therefore IIB, AD are

equal. Then (II. 8) IIB - +BD^ or AD=+

BD^ = 2CD=-f2AC-.

&eho. The nine foregoing propositions may all be proved

very easily by means of algebra, in connection with the princi-

ples of mensuration, already established in the corollaries to

the 23d proposition of the first book. Thus, to pi'ove the fourth

proposition, let AC=:a, CB = 5, and, consequently, AB=a+5.

Now, the area of the square described on AB will be found (I.

23, cor. 4) by multiplying a-\-h by itself. This product is

found, by performing tlie actual operation, to be Â«'+2aJ+^';

an expression, the first and third parts of which are, by the

n A

B D

58 THE ELEMENTS OF [bOOK H.

same corollary, the areas of the squares of AC and CB, and the

second is twice the rectangle of those lines.

In like manner, to prove the eighth, adopting the same nota-

tion, Ave have the line which is made up of the whole and CB

= a + 2$y and, multiplying this by itself, we get for the area ot

the square of that line, cr+iab+ib", or a'-\-4{a-]-b)b, the

first part of which is the area of the square of AC, and the

second four times the area of the rectangle under AB and CB.

It will be a useful exercise for the student to prove the other

propositions in a similar manner. He will also find it easy to

investigate various other relations of lines and their parts by

means of algebra.

All the properties delivered in these propositions hold also

respecting numbers, if products be substituted for rectangles.

Thus, 7 being equal to the sum of 5 and 2, the square, or sec-

ond power of 7, is equal to the squares of 5 and 2 and twice

their product ; that is, 49 = 25 + 4+20.

Pkop. X. â€” Peob. â€” To divide a given straight line into txoo

parts, so that the rectangle contained by the \chole and one of

the parts may be equivalent to the square of the other part.

Let AB be the given straight line ; it is required to divide it

into two parts, so that the rectangle under the whole and one

of the parts may be equivalent to the square of the other.

Upon AB describe (I. 23) the square AD ; bisect (I. 6) AC

in E, and join E with B, the remote extremity of AB ; produce

CA to F, making EF equal to EB, and cut off AH equal to AF ;

AB is divided in H, so that the rectangle AB.BH is equivalent

to the square of AH.

Complete the parallelogram AG, and produce GH to K.

Then, since BAC is a right angle, FAH is also (I. 9) a right

angle; and (I. def 15) AG is a square, AF, AH being equal

by construction. Because the straight line AC is bisected in

E, and produced to F, the rectangle CF.FA and the square of

AE are together equivalent (II. 6) to the square of EF or of EB,

since (const.) EB, EF are equal. But the squares of BA, AE

are equivalent (I. 24, cor. 1) to the square of EB, because the

angle EAB is a right angle; therefore the rectangle CF.FA

and the square of AE are equivalent (I. ax. 1) to the squares of

BOOK n.]

EUCLID AND LEGENDRE.

59

BA, AE. Take away the square of AE, -u-hich is common to

both; therefore the remainino- rectano-le CF.FA

F

G

E

H

-"

B

D

is equivalent (I. ax. 3) to the square of AB.

But the figure FK is the rectangle contained

by CF, FA, for AF is equal to FG ; and AD

is the square of AB ; therefore FK is equal

to AD. Take away the common part AK,

and (I. ax. 3) the remainders FH and HD are

equivalent. But HD is the rectangle AB.BH,

for AB is equal to BD ; and FH is the square

of AH. Therefore the rectangle AB.BH is equivalent to the

square of AH ; Avherefore the straight line AB is divided in H,

so that the rectangle AB.BH is equal to the square of AH ;,

which was to be done.

Sc/w. 1. In the practical construction in this proposition, and

in the 2d cor. to 9 of the fifth book, which is virtually the same, it

is sufficient to draw AE perpendicular to AB, making it equal

to the half of AB, and producing it through A ; then, to make

EF equal to the distance from E to B, and AH equal to AF.

It is plain that BD might be bisected instead of AC, and that

in this way another point of section would be obtained.

While the enunciation in the text serves for ordinary pur-

poses, it is too limited in a geometrical sense, as it comprehends

only one case, excluding another. The following include&

both :

In a given straight line^ or its continuatio7i, to find ai^ointy

such that the rectangle contained hy the given line^ and the seg-

ment between one of its extremities and the required point, may

be equal to the square of the segment between its other extremity

and the same point.

The point in the continuation of BA will be found by cutting

ofiT a line on EC and its continuation, equal to EB, and describ-

ing on the line composed of that line and AE a square lying on

the opposite side of AC from AD ; as the angular j^oint of that

square in the continuation of BA is the point required. The

proof is the same as that given above, except that a rectangle

corresponding to AK is to be added instead of being sub-

tracted.

Scho. 2. The line CF is equivalent to BA and AH ; and since

60 THE ELEMENTS OF [bOOK H,

it has been shown that the rectangle CF.FA is eqiiivalent to

the square of BA or CA, it follows that if any straight line AB

(see the next diagram) be divided according to this proposition

in C, AC being the greater part, and if AD be made equal to

AB, DC is similarly divided in A. So also if DE be made

F

E D A C B

equal to DC, and EF to EA, EA is divided similarly in D, and

FD in E ; and the like additions may be continued as far as we

please.

Conversely, if any straight line FD be divided according to

this proposition in E, and if EA be made equal to EF, DC to

DE, etc., EA is similarly divided in D, DC in A, etc. It fol-

lows also, that the greater segment of a line so divided will be

itself similarly divided, if a part be cut oif from it equal to the

less ; and that by adding to the whole line its greater segment,

another line will be obtained, which is similarly divided.

Prop. XI. â€” Theok. â€” In an oUuse-angkd triangle^ the

square of the greatest side exceeds the squares of the other tioo^

hy tioice the rectangle contained by either of the last-mentioned

sides, and its continuation to meet a perpendicular drawn to it

from the opposite angle.

Let ABC be a triangle, having the angle ACB obtuse ; and

let AD be perpendicular to BC produced ; the square of AB is

equivalent to the squares of AC and CB, and twice the rectan-

gle BC.CD.

Because the straight line BD is divided into two parts in the

point C, the square of BD is equivalent (II. 4) to the squares

of BC and CD, and tAvice the rectangle

BC.CD. To each of these equivalents add

the square of DA ; and the squares of DB,

DA are equivalent to the squares of BC,

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

But, because the angle D is a right angle,

the square of BA is equivalent (I. 24, cor. 1) to the squares of

BD, DA, and tlie square of CA is equivalent to the squares of

BOOK II.] EUCLID AND LEGENDRE. 61

CD, DA ; therefore the square of BA is equivalent to the

squares of BC, CA, and twice the rectangle BC.CD. Therefore,

in an obtuse-angled triangle, etc.

Pkop. XII. â€” TiiEOR. â€” I/i any triangle^ the square of a side

subtending an acute a?igle is less than the squares of the other

sides, by twice the rectangle contained by either of those sides,

and the straight line hitercej^ted between the acute angle and

the perpendicular drawn to that side from tJie opposite angle.

Let ABC (see this figure and that of the foregoing proposi-

tion) be any triangle, having the angle B acute ; and let AD

be perpendicular to BC, one of the sides containing that angle;

the square of AC is less than the squares of AB, BC, by twice

the rectangle CB.BD.

The squares of CB, BD are equivalent (II. 1) to twice the

rectangle contained by CB, BD, and the square of DC. To

each of these equals add the square of

AD ; therefore the squares of CB, BD,

DA are equivalent to twice the rect-

angle CB.BD, and the squares of AD,

DC. But, because AD is perpendicu-

lar to BC, the square of AB is equiva-

lent (I. 24, cor. 1) to the squares of

BD, DA, and the square of x\.C to the squares of .VD, DC ;

therefore the squares of CB, BA are equivalent to the square of

AC, and twice the rectangle CB.BD ; tliat is, tlie square of AC

alone is less than the squares of CB, BA, by twice the rectangle

CB.BD.

If the side AC be perpendicular to BC, then BC is the

straight line between the perpendicular and the acute angle at

B ; and it is manifest that the squares of AB, BC are equiva-

lent (I. 24, cor. 1) to the square of AC and twice the square of

BC. Therefore, in any triangle, etc.

Scho. By means of this or the foregoing proposition, the area

of a triangle may be computed, if the sides be given in num-

bers. Thus, let AB = 17, BC = 28, and AC = 25. From AB=+

BC take AC; that is, from l7'4-2SUake 25=; the remainder

448 is twice CB.BD. Dividing this by 56, twice BC, the quo-

tient 8 is BD, Hence, from either of the triangles ABD,

62

THE ELEMENTS OF [bOOK II.

ACD, we find the perpendicular AD to be 15 ; and thence the

area is found, by taking half tlie product of BC and AD, to be

210.

The segments of the base are more easily found by means -of

tJie 4th corollary to the fifth proposition of this book, in connec-

tion with the pi-inciple, that if half the difference of two mag-

nitudes le added to half their sum ^ the resvlt is the greater; and

if half the difference he taken from half the sum, the remainder

8cho. Otherwise : Take a line D equal to CB. Then (II. 1)

the rectangle AB.D is equivalent to the

-^ ^ ^ rectangles BCD and AC.D ; that is (const.

and L def 15), the rectangle AB.BC is

^ equivalent to the square of BC togctlier with

the rectangle AC.CB.

Prop. IV.â€” Tiieor.â€” 7/" a straight line he divided into

any tico j^^-^^ts, the square of the v-hole line is equivalent to

the squares of the tioo i)arts\ together with twice their red

angle.

Let the straight line AB be divided into any two parts in C ;

the square of AB is equivalent to the squares of AC and CB,

together with twice the rectangle under AC and CB.

On AB describe (L 23) the square of AE, and join BD;

through C draw (I. 18) CGF parallel to AD or BE; and

through G draw UK parallel to AB or DE. Then, l)ecause

G

/

BOOK II.] EUCLID AND LEGENDEE. 51

CF is parallel to AD, aucl BD falls upon them, the exterior

angle CGB is equivalent (I. 16) to the inte-

rior and remote angle ADB ; but ADB is ^ ^ ?

â€¢equal (I. l) to ABD, because BA and AD

are equal, being sides of a square; wherefore

(I. ax. 1) the angle CGB is equal to GBC;

iiad therefore the side BC is equal (I. 1, cor.)

to the side CG. But (const.) the figure CK Â» ^ B

is a parallelogram ; and since CBK is a right

angle, and BC equal to CG, CK (I. def 15) is a square, and it

is upon the side CB. For the like reason HF also is a square,

and it is upon the side HG, which is equal (I. 15, cor. 1) to AC;

therefore HF, CK are the squares of AC, CB. And because

(I. 15, cor. 8) the complements AG, GEare equivalent, and that

AG is the rectangle contained by AC, CB, for CG has been

proved to be equal to CB ; therefore GE is also equivalent to

the rectangle AC.CB; wherefore AG, GE are equivalent to

twice the rectangle AC.CB. The four figures, therefore, HF,

-CK, AG, GE are equivalent to the squares of AC, CB, and

twice the rectangle AC.CB. But HF, CK, AG, GE make up

the Avhole figure AE, which is the square of AB ; therefore the

square of AB is equivalent to the squares of AC and CB, and

twice the rectangle AC.CB. Wherefore, etc.

Otherwise: AB^ = AB.AC + AB.CB (H. 2). But (H. 3)

A.B. AC =AC=+ AC.CB, and AB.CB=zCB-+AC.BC. Hence

<I. ax. 2) AB.AC + AB.BC, or AB-=AC^+CB-+2AC.CB.

Cor. 1. It follows from this demonstration, that the parallel-

ograms about the diagonal of a square are likewise squares.

Cor. 2. Hence the square of a straight line â€¢ is equivalent to

four times the square of its half; for the straight line being

iDisected, the rectangle of the parts is equivalent to the square

of one of them.

Pkop. V. â€” Theor. â€” If a straight line be divided into two

' equal parts, and also into tioo unequal parts ; the rectangle con-

tained by the unequal parts., together with the square of the line

hettoeen the points of section, is equivalent to the square of half

the line.

Let the straio;ht line AB be bisected in C, and divided un-

H

/

K L

/

52 THE ELEMENTS OF [bOOK II.

equally in D ; the rectangle AD.DB, together with the square

of CD, is equivalent to the square of CB.

On CB describe (I. 23) the square CF, join BE, and through

D draw (I. 18) DHG parallel to CE or BF ; also through H

draw KLM parallel to CB or EF ; and

â– ^ c D B through A draw AK parallel to CL or

BM. Then (I. 15, cor. 5) AL and CM

^^ are equal, because AC is equal to CB ;

and (I. 15, cor. 8) the complements CH

E G F and HF are equivalent. Therefore (I.

ax. 2) AL and CH together are equal to

CM and HF together ; that is, AH is equivalent to the gnomon

CMC To each of those add LG, and (I. ax. 2) the gnomon

CMG, together with LG, is equivalent to AH together with

LG. But the gnomon CMG, and LG make up the figure

CEFB, which is the square of CB ; also AH is the rectangle

imder AD and DB, because DB is equal (IL 4, cor. 1) to DH ;

and LG is the square of CD. Thei-efore the rectangle AD.DB

and the square of CD are equivalent to the square of CB.

Wherefore, if a straight line, etc.

Otherwise : Since, as is easily seen from the proof in the

above, DF is equal to AL, take these separately from the entire

figure, and there remain AH and LG equivalent to the square

CF, as before. The proof may also be as follows:

AD.DB = CD.DB + AC.DB (H. 1) or AD.DB =CD.DB+

CB.DB, because CB=:AC. Hence (H. 3) AD.DB =CD.DB +

CD.DB+DB^=i:2CD.DB + DBl To each of these add CD=;

then AD.DB + CD^=2CD.DB+DB^+CD^ or (H. 4) AD.DB

+ CD'=CB\

Cor. 1. Hence the ditference of the squares of CB and CD is

equivalent to the rectangle under AD and DB. But since AC

is equal to CB, AD is equivalent to the sum of CB and CD, and

DB is the difterence of these lines. Hence the difference of the

squares of two straight lines is equivalent to the rectangle under

their sum and difference.

Cor. 2. Since the square of CB, or, which is the same, the

rectangle AC.CB, is greater than the rectangle AD.DB by the

equare of CD, it follows, that to divide a straight line into two

BOOK n.] EUCLID AND LEGENDKE. 53

parts, the rectangle of wliich may be the greatest possible, or,

as is termed, a tnaxlmiim, tlie line is to be bisected.

Cor. 3. Hence also tlie sum of the squares of tlie two parts

into which a straight line is divided, is the least possible, or is,

as it is termed, a ininimuni, Avlien the line is bisected. For (II.

4) the square of tlie line is equivalent to the squares of the

parts and twice their rectangle ; and therefore the greater the

rectangle is, the less are the squares of the j^arts; but, by the

foregoins; corollarv, the rectangle is a maximum when the line

is bisected.

C<yr. 4. Since (I. 24, cor. 5) the difference of the squares of

the sides of a triangle is equivalent to the difference of the

squares of the segments of the base, it follows, from the first

corollary above, that the rectangle under the sum and differ-

ence of the sides of a triangle is equivalent to the rectano-le

under the sum and difference of the segments, intercepted be-

tween the extremities of the base and the jjoint in which the

perpendicular cuts the base, or tlie base produced.

Cor. 5. Hence, also, if a straight line be drawn from the ver-

tex of an isosceles triangle to any point in the base, or its con-

tinuation, the difference of the squares of that line and either of

the equal sides is equivalent to the rectangle under the seoments

intercepted between the extremities of the base and the jDoint.

Cor. 6. Since (I. 24, cor. 4) the square of one of the legs of a

right-angled triangle is equivalent to the difference of the

squares of the hypothenuse and the other leg, it follows (IT. 5,

cor. 1) that the square of one leg of a right-angled triangle is

equivalent to the rectangle under the sum and difference of the

hypothenuse and the other.

^cho. And a parallelogram can be constructed equivalent

to a ffiven triangle, or anv given rectilinear fiorure havinor an

angle equal to a given angle, or applied to a given straight line

â€” that is, having that straight line for one of its sides, when

the jjarallelogram shall be equivalent to a given triangle or

given rectilinear figure, and have one of its angles equal to a

given angle, by applying a parallelogram equivalent to the

given triangle with an equal angle, to a given straight line, and

then constructing an equal triangle to the given triangle (I. 15

and 15, cor. 4).

A C

B D

K

H

/

L

/

54 THE ELEMENTS OF [bOOK II,

Prop. VI. â€” Theok. â€” If a straight line he bisected, and be

produced to any point, the rectangle contained by the whole

line thus produced, and the part of it produced, together with

the square of half the line bisected, is equivalent to the square

of the straight line which is made %ip of the half and the p>art

produced.

Let the straight line AB be bisected in C, and produced to

D ; the rectangle AD.DB, and the square of CB, are equiva-

lent to the square of CD.

Upon CD describe (I. 23) the square CF, and join DE ;

through B draw (I. 18) BHG parallel to CE or DF ; through

H draw KLM parallel to AD or EF, and

through A draw AK parallel to CL or

jj DM. Then because AC is equal to CB,

the rectangle AL is equal (I. 15, cor. 5) to

CH ; but (1.15, cor. 8) the complements CH,

J, ^j J, HF are equivalent ; therefore, also, AL is

equal to HF, To each of these add CM

and LG ; therefore AM and LG are equivalent to the whole

square CEFD, But AM is the rectangle under AD and DB,

because (11. 4, cor. 1) DB is equal to DM; also, LG is the

square of CB, and CEFD the square of CD. Therefore the

rectangle AD.DB and the square of CB are equivalent to the

square of CD. Wherefore, if a straight line, etc.

Otherwise : Produce CA to N, and make CN equal to CD.

To these add the equals CB and CA ;

^ ^ ^ ^ Â° therefore NB is equal (L ax. 2) to AD.

But (IL 5) the rectangle NB.BD, oi-

AD.BD, together with the square of CB, is equivalent to the

square of CD ; which was to be proved.

Prop. VIL â€” Theor. â€” If a straight line be divided into a?2y

two parts, the squares of the whole line and of one of the p>arts

are equivalent to ticice the rectangle contained by the whole and

that part, together tcith the square of the other part.

Let the straight line AB be divided into any two parts in the

point C ; the squares of AB, BC are equivalent to twice the

rectangle AB.BC, together with the square of AC.

BOOK II.] EUCLID AND LEGENDRE. 65

Upon AB describe (I. 23) the square AE, and construct the

figure as in the preceding propositions. Then,

because (I. 15, cor. 8) the complements CII, ^ c B

FK are equivalent, add to each of thorn CK;

the whole AK is therefore equal to the whole ^

CE ; therefore AK, CE are together double

of AK. But AK, CE are the gnomon AKF,

K

together with the square CK ; therefore the i> F E

gnomon AKF and the square CK are double

of AK, or double of the rectangle AB.BC, because BC is equal

(II. 4, cor. 1) to BK. To each of these equals add HF, which

is equal (II. 4, cor. 1 ) to the square of AC ; therefore the gno-

mon AKF, and the squares CK, HF are equivalent to twice the

rectangle AB.BC and the square of AC. But the gnomon

AKF, and the squares CK, HF make up the whole figure AE,

together with CK; and these are the squares of AB and BC ;

therefore the squares of AB and BC are equivalent to twice the

rectangle AB.BC, together with the square of AC ; wherefore,

if a straight line, etc.

Otherwise: Because (II. 4) AB^ = AC=+BC-+2AC.BC, add

BC= to both; then AB=+BC==AC= + 2BC=+2AC.BC. But

(II. 3) BC=+AC.BC= AB.BC, and therefore 2BC=+2AC.BC

=2AB.BC; wherefore AB-+BC^=AC=+2AB.BC.

Cor. 1. Since AC is the diflierence of AB and BC, it follows

that the square of the difterence of two straight lines is equiva-

lent to the sum of their squares, wanting twice their rectangle.

Cor. 2. Since (II. 4) the square of the sum of two lines ex-

ceeds the sum of their squares by twice their rectangle, and

since, by the foregoing corollary, the square of their difference

is less than the sum of their squares by twice their rectangle, it

follows that the square of the sum of two lines is equivalent to

the square of their difference, together with four times their

rectangle.

Pkop. VIII. â€” TiiEOR. â€” If a straight line he divided into two

equal, and also into tvw imequal parts, the squares of the un-

equal parts are together double of the square of half the linCj

and of the square of the line hetioeen the points of section.

Let the straight line AB be divided equally in C, and un-

56

THE ELICMENTS OF

[book n.

E

equally in D ; the square? of AD, DB are together double of

the squares of AC, CD.

From C draAV (I. 7) CE j)ei-|ie!Klicular to AB, and make it

equal to AC or CB, and join EA, EB ; through D draw (I. 18)

DF jjarallel to CE, and tlirougli V draw FG parallel to AB ;

and join AF. Then because (const.) the triangles ACE, BCE

are right-angled and isosceles, tlie angles CAE, AEC, CEB,

O O 7 O 7 3 7

EBC are (I. 20, cor. 4) each half a riglit angle. So also (I. 16,

part 2) are EEC, BED, because FG is parallel to AB, and ED

to EC ; and for the same reason EGF, ADF are right angles.

The angle AEB is also a right angle, its parts being each half

a right angle. Hence (I. 1) EG is equal to GF or CD, and ED

to DB. Again (I. 24, cor. 1) : the square

of AE is equivalent to the squares of AC,

CE, or to tAvice the square of AC, since

AC and CE are equal. In like manner,

the square of EF is equivalent to twice

the square of GF or CD. Now (I. 24,

^ cor. 1), the squares of AD and DF, or

of AD and DB, are equivalent to the square of AF ; and the

squares of AE, EF, that is, twice the square of AC and twice

the square of CD, are also equivalent to the square of AF;

therefore (I. ax. 1) the squares of AD, DB are equivalent to

twice the square of AC and twice the square of CD. If, there-

fore, a straight line, etc.

Otherwise: DB^+2BC.CD=BC=+CD^ (II. 7), or DB=+

2AC.CD=:AC= + CD^-; also (11. 4) AD==:AC^+CD=+2AC.

CD. Add these equals togethei-, and from the sums take

2AC.CD; then AD-+DB^=2AC^+2CD^

Prop. IX. â€” ^Theor. â€” If a straight line he bisected, and pro-

duced to any point, the squares of the whole line thus produced,

and of the part of it produced, are together double of the square

of half the lifie bisected, and of the square of the line made up

of the half and the part produced.

Let the straight line AB be bisected in C, and produced to

D ; the squares of AD, DB arc double of the squares of AC,

CD.

From C draw (I. 7) CE perpendicular to AB ; and make it

BOOK II.]

EUCLID AND LEGENDRE.

57

A

equal to AC or CB ; join AE, EB, and tln-ough E and D draw

(I. 1 8) EF parallel to AB, and

DF parallel to CE. Then be-

cause the straiu'lit line EF

meets the parallels EC, FD,

the ano-les CEF, EFD are

equivalent (I. 9) to two right

anixles : and therefore the an-

gles BEF, EFD are less than

two right angles; therefore (I. 19) EB, FD will meet, if pro-

duced toward B, D ; let them meet in G, and join AG. Then

it would be proved, as in the last proposition, that the angles

CAE, xVEC, CEB, EBC are each half a right angle, and AEB

a right angle. BDG is also a right angle, being equal (I. 16)

to ECB, since (const.) EC, EG are parallel ; DGB, DBG are

each half a right angle, being equal (I. 10 and I. 11) to CEB,

CBE, each to each ; and FEG is half a right angle, being (I.

16) equal to CBE. It would also be proved, as in the last

proposition, that the squai-e of AE is twice the square of AC,

and the square of EG twice the square of EF or CD. Now

(I. 24, cor. l),the squares of AD, DG, or of AD, DB, are equiv-

alent to tlie sipiare of AG; and the squares of AE, EG, or twice

the square of AC and twice the square of CD, are also equiva-

lent to the square of AG. Therefore (I. ax. 1) the squares of

AD, DB are equivalent to twice the square of AC and twice

the square of CD. If, therefore, a straight line, etc.

Otherwise: Produce CA, making CH equal to CD. To

these add CB, CA ; therefore IIB, AD are

equal. Then (II. 8) IIB - +BD^ or AD=+

BD^ = 2CD=-f2AC-.

&eho. The nine foregoing propositions may all be proved

very easily by means of algebra, in connection with the princi-

ples of mensuration, already established in the corollaries to

the 23d proposition of the first book. Thus, to pi'ove the fourth

proposition, let AC=:a, CB = 5, and, consequently, AB=a+5.

Now, the area of the square described on AB will be found (I.

23, cor. 4) by multiplying a-\-h by itself. This product is

found, by performing tlie actual operation, to be Â«'+2aJ+^';

an expression, the first and third parts of which are, by the

n A

B D

58 THE ELEMENTS OF [bOOK H.

same corollary, the areas of the squares of AC and CB, and the

second is twice the rectangle of those lines.

In like manner, to prove the eighth, adopting the same nota-

tion, Ave have the line which is made up of the whole and CB

= a + 2$y and, multiplying this by itself, we get for the area ot

the square of that line, cr+iab+ib", or a'-\-4{a-]-b)b, the

first part of which is the area of the square of AC, and the

second four times the area of the rectangle under AB and CB.

It will be a useful exercise for the student to prove the other

propositions in a similar manner. He will also find it easy to

investigate various other relations of lines and their parts by

means of algebra.

All the properties delivered in these propositions hold also

respecting numbers, if products be substituted for rectangles.

Thus, 7 being equal to the sum of 5 and 2, the square, or sec-

ond power of 7, is equal to the squares of 5 and 2 and twice

their product ; that is, 49 = 25 + 4+20.

Pkop. X. â€” Peob. â€” To divide a given straight line into txoo

parts, so that the rectangle contained by the \chole and one of

the parts may be equivalent to the square of the other part.

Let AB be the given straight line ; it is required to divide it

into two parts, so that the rectangle under the whole and one

of the parts may be equivalent to the square of the other.

Upon AB describe (I. 23) the square AD ; bisect (I. 6) AC

in E, and join E with B, the remote extremity of AB ; produce

CA to F, making EF equal to EB, and cut off AH equal to AF ;

AB is divided in H, so that the rectangle AB.BH is equivalent

to the square of AH.

Complete the parallelogram AG, and produce GH to K.

Then, since BAC is a right angle, FAH is also (I. 9) a right

angle; and (I. def 15) AG is a square, AF, AH being equal

by construction. Because the straight line AC is bisected in

E, and produced to F, the rectangle CF.FA and the square of

AE are together equivalent (II. 6) to the square of EF or of EB,

since (const.) EB, EF are equal. But the squares of BA, AE

are equivalent (I. 24, cor. 1) to the square of EB, because the

angle EAB is a right angle; therefore the rectangle CF.FA

and the square of AE are equivalent (I. ax. 1) to the squares of

BOOK n.]

EUCLID AND LEGENDRE.

59

BA, AE. Take away the square of AE, -u-hich is common to

both; therefore the remainino- rectano-le CF.FA

F

G

E

H

-"

B

D

is equivalent (I. ax. 3) to the square of AB.

But the figure FK is the rectangle contained

by CF, FA, for AF is equal to FG ; and AD

is the square of AB ; therefore FK is equal

to AD. Take away the common part AK,

and (I. ax. 3) the remainders FH and HD are

equivalent. But HD is the rectangle AB.BH,

for AB is equal to BD ; and FH is the square

of AH. Therefore the rectangle AB.BH is equivalent to the

square of AH ; Avherefore the straight line AB is divided in H,

so that the rectangle AB.BH is equal to the square of AH ;,

which was to be done.

Sc/w. 1. In the practical construction in this proposition, and

in the 2d cor. to 9 of the fifth book, which is virtually the same, it

is sufficient to draw AE perpendicular to AB, making it equal

to the half of AB, and producing it through A ; then, to make

EF equal to the distance from E to B, and AH equal to AF.

It is plain that BD might be bisected instead of AC, and that

in this way another point of section would be obtained.

While the enunciation in the text serves for ordinary pur-

poses, it is too limited in a geometrical sense, as it comprehends

only one case, excluding another. The following include&

both :

In a given straight line^ or its continuatio7i, to find ai^ointy

such that the rectangle contained hy the given line^ and the seg-

ment between one of its extremities and the required point, may

be equal to the square of the segment between its other extremity

and the same point.

The point in the continuation of BA will be found by cutting

ofiT a line on EC and its continuation, equal to EB, and describ-

ing on the line composed of that line and AE a square lying on

the opposite side of AC from AD ; as the angular j^oint of that

square in the continuation of BA is the point required. The

proof is the same as that given above, except that a rectangle

corresponding to AK is to be added instead of being sub-

tracted.

Scho. 2. The line CF is equivalent to BA and AH ; and since

60 THE ELEMENTS OF [bOOK H,

it has been shown that the rectangle CF.FA is eqiiivalent to

the square of BA or CA, it follows that if any straight line AB

(see the next diagram) be divided according to this proposition

in C, AC being the greater part, and if AD be made equal to

AB, DC is similarly divided in A. So also if DE be made

F

E D A C B

equal to DC, and EF to EA, EA is divided similarly in D, and

FD in E ; and the like additions may be continued as far as we

please.

Conversely, if any straight line FD be divided according to

this proposition in E, and if EA be made equal to EF, DC to

DE, etc., EA is similarly divided in D, DC in A, etc. It fol-

lows also, that the greater segment of a line so divided will be

itself similarly divided, if a part be cut oif from it equal to the

less ; and that by adding to the whole line its greater segment,

another line will be obtained, which is similarly divided.

Prop. XI. â€” Theok. â€” In an oUuse-angkd triangle^ the

square of the greatest side exceeds the squares of the other tioo^

hy tioice the rectangle contained by either of the last-mentioned

sides, and its continuation to meet a perpendicular drawn to it

from the opposite angle.

Let ABC be a triangle, having the angle ACB obtuse ; and

let AD be perpendicular to BC produced ; the square of AB is

equivalent to the squares of AC and CB, and twice the rectan-

gle BC.CD.

Because the straight line BD is divided into two parts in the

point C, the square of BD is equivalent (II. 4) to the squares

of BC and CD, and tAvice the rectangle

BC.CD. To each of these equivalents add

the square of DA ; and the squares of DB,

DA are equivalent to the squares of BC,

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

But, because the angle D is a right angle,

the square of BA is equivalent (I. 24, cor. 1) to the squares of

BD, DA, and tlie square of CA is equivalent to the squares of

BOOK II.] EUCLID AND LEGENDRE. 61

CD, DA ; therefore the square of BA is equivalent to the

squares of BC, CA, and twice the rectangle BC.CD. Therefore,

in an obtuse-angled triangle, etc.

Pkop. XII. â€” TiiEOR. â€” I/i any triangle^ the square of a side

subtending an acute a?igle is less than the squares of the other

sides, by twice the rectangle contained by either of those sides,

and the straight line hitercej^ted between the acute angle and

the perpendicular drawn to that side from tJie opposite angle.

Let ABC (see this figure and that of the foregoing proposi-

tion) be any triangle, having the angle B acute ; and let AD

be perpendicular to BC, one of the sides containing that angle;

the square of AC is less than the squares of AB, BC, by twice

the rectangle CB.BD.

The squares of CB, BD are equivalent (II. 1) to twice the

rectangle contained by CB, BD, and the square of DC. To

each of these equals add the square of

AD ; therefore the squares of CB, BD,

DA are equivalent to twice the rect-

angle CB.BD, and the squares of AD,

DC. But, because AD is perpendicu-

lar to BC, the square of AB is equiva-

lent (I. 24, cor. 1) to the squares of

BD, DA, and the square of x\.C to the squares of .VD, DC ;

therefore the squares of CB, BA are equivalent to the square of

AC, and twice the rectangle CB.BD ; tliat is, tlie square of AC

alone is less than the squares of CB, BA, by twice the rectangle

CB.BD.

If the side AC be perpendicular to BC, then BC is the

straight line between the perpendicular and the acute angle at

B ; and it is manifest that the squares of AB, BC are equiva-

lent (I. 24, cor. 1) to the square of AC and twice the square of

BC. Therefore, in any triangle, etc.

Scho. By means of this or the foregoing proposition, the area

of a triangle may be computed, if the sides be given in num-

bers. Thus, let AB = 17, BC = 28, and AC = 25. From AB=+

BC take AC; that is, from l7'4-2SUake 25=; the remainder

448 is twice CB.BD. Dividing this by 56, twice BC, the quo-

tient 8 is BD, Hence, from either of the triangles ABD,

62

THE ELEMENTS OF [bOOK II.

ACD, we find the perpendicular AD to be 15 ; and thence the

area is found, by taking half tlie product of BC and AD, to be

210.

The segments of the base are more easily found by means -of

tJie 4th corollary to the fifth proposition of this book, in connec-

tion with the pi-inciple, that if half the difference of two mag-

nitudes le added to half their sum ^ the resvlt is the greater; and

if half the difference he taken from half the sum, the remainder

Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 5 of 21)