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c 8. I.

Book I. therefore also 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, and the base DC likewise equal to the base BC, the angle DAC is equal 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.

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ELE

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EVERY right angled parallelogram, or rectangle, is faid

to be contained by any two of the straight lines which are about one of the right angles.

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66

“Thus, the right angled parallelogram AC is called the rectangle contained by AD and DC, or by AD and AB, "&c. For the fake of brevity, instead of the rectangle con"tained by AD and DC, we fhall fimply fay the rectangle AD.DC, placing a point between the two fides of the rectangle. Alfo, inftead of the square of a line, for inftance of "AD, we may fometimes in what follows write AD2.” "The fign+placed between the names of two magnitudes, "fignifies that thofe magnitudes are to be added together; and "the fign -placed between them, fignifies that the latter is to "be taken away from the former."

66

"The fignfignifies, that the things between which it is placed are equal to one another."

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N.

Book II.

II.

E

D

In every parallelogram, any of the parallelograms about a
diameter, together with the
two complements, is called A
a Gnomon., Thus the pa-
rallelogram HG, toge-

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⚫ther with the complements
AF, FC, is the gnomon
of the parallelogram AC.
This gnomon may also,
for the fake of brevity,
be called the gnomon B
AGK, or EHC.'

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a II. 1.

I'

PROP. I. THEOR.

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

B

D E C

Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle A.BC is equal to the feveral rectangles A.BD, A.DE, A.EC. From the point B draw a BF at right angles to BC, and b 3. 1. make BG equal to A; and c31. 1. through G draw c GH parallel to BC; and through D, E, C, draw c DK, EL, CH parallel to BG; then BH, BK, DL and EH are rectangles, and BH=BK+DL+

EH.

G

F

K L

H

A

But BHBG.BCA.BC, because BG=A: Alfo BK= BG.BD A.BD, because BGA; and DL DK.DE= 34.1. A.DE, becaufed DKBG=A. In like manner, EH A.EC. Therefore A.BCA.BD+A.DE+ A.EC; that is, the rectangle A.BC is equal to the feveral rectangles

A.BD,

A.BD, A.DE, A.EC. Therefore, if there be two ftraight Book II. lines, &c. Q. E. D.

PROP. II. THEOR.

Fa ftraight line be divided into any two parts,

of the parts, are together equal to the fquare of the whole line.

Let the ftraight line AB be divided into any two parts in the point C; the rectangle AB.BC, together with the rectangle AB.AC, is equal to the fquare of AB; or AB.AC+AB.BCĦAB2

On AB defcribe a the fquare ADEB, and through C draw CF b parallel to AD or BE; then AF+ CE AE. But AF=AD.AC= AB.AC, because AD AB; CE BE.BC AB.BC; and AE

AB2. D

Therefore AB.AC+AB.BCAB2.

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

C

B

T

E

a 46. I.

b 31. I.

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PROP. III. THEOR.

a 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 ftraight line AB be divided into two parts in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2.

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52

Book II.

31. L.

Upon BC describe a the

a 46. 1. fquare CDEB, and produce ED to F, and through A draw b AF parallel to CD or BE; then AE-AD+CE. But AE-AB.BE=AB.BC, because BE=BC. So alfo AD AC.CD AC.CB;

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and CE BC2; therefore F

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AB.BC AC.CB+BC2. Therefore, if a straight line, &c.

Q. E.D.

a 46. I.

€ 29. I. d 5.1.

IF

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 ftraight 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, that is, AB AC2+CB2+2AC.CB.

A

C B

G

K

Upon AB defcribe a the fquare ADEB, and join BD, and b31. 1. through C draw b CGF parallel to AD or BE, and through 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 oppofite angle ADB; but ADB is equal d to the angle ABD, because BA is H equal to AD, being fides of a fquare; wherefore the angle CGB is equal to the angle GBC; and therefore the fide BC is equal e to the fide CG: but CB is equal alfo to GK, and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other g Cor. 46.1. angles of the parallelogram CGKB are alfo right angles 8.

e 6. I. £ 34. I.

f

D

F E

Wherefore

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