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

angle, the two straight lines

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AC, AG upon the oppofite fides of AB, make with it at

the point A the adjacent angles F equal to two right angles; therefore CA is in the fame ftraight line with AG. for the fame reason, AB and AH are in the fame straight 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,

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and the whole angle DBA is

- equal to the whole FBC. and because the two fides AB, BD are c. 2. Ax. equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the bafe AD is equal f 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 g. 41. 1. fame base BD, and between the fame parallels BD, AL; and the fquare GB is double of the triangle FBC, because these also 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 h. 6. Ax. parallelogram BL is equal to to the fquare GB. and in the fame manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the fquare HC. Therefore the whole square BDEC is equal to the two fquares GB, HC. and the square BDEC is described upon the ftraight 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.

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F the fquare described 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 thefe two fides is a right angle.

Book I.

2. II. I.

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

a

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

47.1. fquare of DC is equal to the fquares of A

c. 8. I.

b

DA, AC, because DAC is a right angle;
and the fquare of BC, by Hypothesis, is
equal to the fquares of BA, AC; therefore
the fquare of DC is equal to the fquare of B

D

BC; and therefore 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 bafe BC; therefore 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.

c

Book II.

THE

ELEMENTS

O F

EUCLI D.

E

BOOK II.

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.

A,

E

D

In every parallelogram, any of the parallelograms about a diame-
ter, together with the two
complements, is called a
Gnomon. Thus the pa-
rallelogram HG together
• with the complements AF,
FC is the gnomon, which

F

H

K

is more briefly expreffed

by the letters AGK, or B

C

G

EHC which are at the op

IF

pofite angles of the parallelograms which make the gnomon.'

PROP. I. THEO R.

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.

Book II.

3. II. I.

b. 3. 1.

6. 31. I.

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DEC

Let A and BC be two ftraight lines; and let BC be divided into any parts in the points D, E; the rectangle contained by the ftraight lines A, BC is equal to B 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 BF
at right angles to BC, and make

BG equal b to A; and thro' GF
draw GH parallel to BC; and

c

C

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 d. 34. 1. by A, DE, because DK, that is 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 feveral rectangles contained by A, BD, and by A, DE, and alfo by A, EC. Wherefore if there be two straight lines, &c. Q. E. D.

a. 46. 1.

b. 31. I.

IF

PROP. II. THEO R.

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.

*

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C B

FE

N. B. To avoid repeating the word Contained too frequently, the rectangle contained by two ftraight lines AB, AC is fometimes fimply 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 fquare of AB. If therefore a straight line, &c. Q. E. D.

IF

PROP. II. THEO R.

F 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 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 A C

B

2. 46. T

b. 31. I

CDEB, and produce ED to F, and
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 contain-
ed by AB, BC, for it is contained
by AB, BE, of which BE is equal
to BC, and AD is contained by
AC, CB, for CD is equal to CB; and DB is the fquare 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.

FD

E

PROP. IV. THEO R.

IF a straight line be divided into any two parts, the square 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.

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