GB, HC; and thro' A draw b AL parallel to BD or CE, and join Book I. AD, FC./then becaufe each of the angles BAC, BAG is a right

angle, the two straight lines AC, AG upon the oppofite fides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the fame ftraight line with AG. for the fame reafon, AB and AH are in the fame ftraight line., and because the angle DBC is equal to the angle FBA, each

of them being a right angle,

add to each the angle ABC,

D

E

L

and the whole angle DBA is

equal to the whole FBC./and because the two fides AB, BD are e. 2. Ax. equal to the two FB, BC, each to each, and the angle DBA equal

to the angle FBC; therefore the base AD is equal to the bafe FC, f. 4. 1. and the triangle ABD to the triangle FBC./now the parallelogram

BL is double of the triangle ABD, because they are upon the fame g. 41. I. bafe BD, and between the fame parallels ED, AL; and the fquare GB is double of the triangle FBC, becaufe thefe alfo are upon the fame base FB, and between the fame parallels FB, GC. but the doubles of equals are equal to one another. therefore the paralle- h. 6. Ax. logram BL is equal to the fquare GB. and in the fame manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole fquare BDEC is equal to the two fquares GB, HC. and the fquare BDEC is described upon the straight line BC, and the fquares GB, HC upon BA, AC. wherefore the fquare upon the fide BC is equal to the squares upon the fides BA, AC. Therefore in any right angled triangle, &c. Q. E. D.

PROP. XLVIII. THEOR.

IF the fquare defcribed upon one of the fides of a triangle, be equal to the fquares described upon the other two fides of it; the angle contained by these two fides is a right angle.

Book I.

*. II. I.

If the fquare defcribed upon BC one of the fides of the triangl ABC be equal to the fquares upon the other fides BA, AC; the angle BAC is a right angle.

Ꭰ

From the point A draw AD at right angles to AC, and make AD equal to BA, and join DC. then because DA is equal to AB, the square of DA is equal to the fquare of AB; to each of these add the square of AC, therefore the fquares of DA, AC are equal to the fquares of BA, AC. but the

5. 47. 1. fquare of DC is equal to the fquares of DA, AC, because DAC is a right angle; and the

1

fquare of BC, by Hypothefis, is equal to the
fquares of BA, AC; therefore the fquare of
DC is equal to the fquare of BC; and there- B

A

C

fore alfo the fide DC is equal to the fide BC. and because the fide DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the bafe DC is equal to the base BC; therefore the angle DAC is ec. 8. 1. qual to the angle BAC. but DAC is a right angle, therefore alfo BAC is a right angle. Therefore if the fquare, &c. Q. E. D.'

THE

Book II.

THE

ELEMENTS

OF

EUCLID.

EVE

BOOK IL

DEFINITION S.

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 diameter,

together with the two complements, is called a Gnomon. Thus the parallelo'gram HG together with 'the complements AF, FC 'is the gnomon, which is 'more briefly expressed by 'the letters AGK, or EHC 'which are at the oppofite

angles of the parallelograms which make the gnomon.'

PROP, I. THEOR.

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 feveral parts of the divided line.

Let

Book II.

2. II. I.

b. 3. 1.

c. 31. 1.

B

DEC

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 straight lines A, BC is equal to the rectangle contained by A, BD; and to that contained by A, DE; and alfo to that contained by A, EC.

From the point B draw a BF at right angles to BC, and make BG equal to A; and thro' G draw c GH parallel to BC; and thro' D, E, C draw DK, EL, CH paral

G

F

KLH

A

lel 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 d. 34. 1. DL is contained by A, DE, because DK, that is 4 BG, is equal to

2. 46. 1,

b. 31. 1.

A; and in like manner the rectangle EH is contained by A, EC. therefore the rectangle contained by A, BC is equal to the feveral rectangles contained by A, BD, and by A, DE, and alfo by A, EC. Wherefore if there be two ftraight lines, &c. Q. E. D.

IF

PROP. II. THEOR.

Fa ftraight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the fquare 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 fhall be equal to the fquare of AB.

Upon AB defcribe the fquare ADEB, and thro' C draw b CF parallel to AD or BE. then AE is equal to the rectangles AF, CE; and AE is the fquare of AB; and AF is the

D

C B

FE

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

rectangle

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

IF

PROP. III. THEOR.

Fa ftraight 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 fquare 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 fquare of BC.

a

Upon BC defcribe the fquare

CDEB, and produce ED to F, and A C

thro' A draw b AF parallel to CD or

BE. then the rectangle AE is equal to
the rectangles AD, CE; and AE is
the rectangle contained by AB, BC,
for it is contained by AB, BE, of which
BE is equal to BC; and AD is con-
tained by AC, CB, for CD is equal to F
CB; and DB is the fquare of BC.

therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the fquare of BC.

Q. E. D.

If therefore a straight line, &c.

PROP. IV. THEOR.

a ftraight line be divided into any two parts, the fquare of the whole line is equal to the fquares 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 fquare of AB is equal to the fquares of AC, CB and to twice the rectangle contained by AC, CB,

Upon