Book II.

THE

ELEMENTS

E

OF

EUCLID.

BOOK II.

DEFINITIONS,

I.

VERY right angled parallelogram is faid to be contained
by any two of the straight lines which contain one of the
right angles.

II.

In every parallelogram, any of the parallelograms about a diame

'posite angles of the parallelograms which make the gnomon."

PROP. I. THEOR.

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

Book II.

Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle contained by the

d. 34. 1.

a. 46. 1.

b. 31. 1.

From the point B draw a BFG at right angles to BC, and make BG equal to A; and thro' GF draw GH parallel to BC; and

KLH
A

thro' D, E, C draw • DK, EL, CH parallel to BG. then the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is d BG, is equal to A; and in like manner the rectangle EH is contained by A, EC. therefore the rectangle contained by A, BC is equal to the several rectangles contained by A, BD, and by A, DE, and also by A, EC. Wherefore if there be two straight lines, &c. Q. E. D.

I

PROP. II. THEOR.

a straight line be divided into any two parts, the rectangles contained by the whole and eash of the parts, are together equal to the square of the whole line.

Let the straight line AB be divided into A
any two parts in the point C; the rectangle
contained by AB, BC together with the
rectangle * AB, AC shall be equal to the
square of AB.

Upon AB describe the square ADEB,
and thro' C draw & CF parallel to AD or
BE. then AE is equal to the rectangles AF,
CE; and AE is the square of AB; and AFD

CB

FE

* N. B. To avoid repeating the word Contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.

is the rectangle contained by BA, AC; for it is contained by Book II. DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB. therefore the rectangle contained by AB, AC together with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D.

IF

PROP. III. THEOR.

a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the forefaid part.

Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC.

Upon BC describe the square A C CDEB, and produce ED to F, and thro' A draw & AF parallel to CD or BE. then the rectangle AE is equal to the rectangles AD, CE;

a. 46. I.

B

b. 31. 1.

and AE is the rectangle contained by AB, BC, for it is contained

:

by AB, BE, of which BE is equal
to BC; and AD is contained by F D

E

AC, CB, for CD is equal to CB; and DB is the square of BC. therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. If therefore a straight line, &c. Q. E. D.

IF

PROP. IV. THEOR.

F a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB and to twice the rectangle contained by AC, CB.

1

Book II.

a. 46. 1.

b. 31. 1.

C. 29. I.

d. 5. 1.

e. 6. 1.

f. 34. 1.

g. 43. I.

Upon AB describe a the square ADEB, and join BD, and thro' C draw CGF parallel to AD or BE, and thro' G draw HK parallel to AB or DE. and because CF is parallel to AD; and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB; but ADB is equal d to the angle ABD, because BA is equal to AD, being fides of a square; wherefore the

angle CGB is equal to the angle GBC,

and therefore the fide BC is equal to

A

D

CB

Gy

K

FE

the fide CG. but CB is equal also f to
GK, and CG to BK; wherefore the H
figure CGKB is equilateral. it is like-
wise rectangular; for CG is parallel to
BK, and CB meets them, the angles
KBC, GCB are therefore equal to two
right angles; and KBC is a right an-
gle, wherefore GCB is a right angle; and therefore also the an-
gles of CGK, GKB oppofite to these are right angles, and CGKB
is rectangular. but it is also equilateral, as was demonstrated;
wherefore it is a square, and it is upon the side CB. for the fame
reason HF also is a square, and it is upon the fide HG which is
equal to AC. therefore HE, CK-are the squares of AC, CB. and
because the complement AG is equal & to the complement GE,
and that AG is the rectangle contained by AC, CB, for GC is
equal to CB; therefore GE is also equal to the rectangle AC,
CB; wherefore AG, GE are equal to twice the rectangle AC,
CB. and HF, CK are the squares of AC, CB; wherefore the four
figures HF, CK, AG, GE are equal to the squares of AC, CB
and to twice the rectangle AC, CB. but HF, CK, AG, GE make
up the whole figure ADEB which is the square of AB. therefore
the square of AB is equal to the squares of AC, CB and twice
the rectangle AC, CB. Wherefore if a straight line, &c.
Q. E. D.

COR. From the demonstration it is manifest, that the parallelograms about the diameter of a square are likewise squares.

IF

PROP. V. THEOR.

Fa straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB together with the square of CD, is equal to the square of CB.

Book II.

b. 31. I.

Upon CB describe the square CEFB, join BE, and thro' D a. 46. 1. draw DHG parallel to CE or BF; and thro' H draw KLM parallel to CB or EF; and also thro' A draw AK parallel to CL or BM. and because the complement CH is equal to the comple- c. 43. 1.

AH is the rectangle contained by AD, DB, for DH is equal to e. Cor. 4. 2. DB; and DF together with CH is the gnomon CMG; therefore

the gnomon CMG is equal to the rectangle AD, DB. to each of these add LG, which is equal to the square of CD, therefore the gnomon CMG together with LG is equal to the rectangle AD, DB together with the square of CD. but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB. therefore the rectangle AD, DB together with the fquare of CD is equal to the square of CB. Wherefore if a ftraight line, &c.

E. D.