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c. 30. Des.

С

GB, HC; and thro' A draw 6 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

b. 31.1. AC, AG upon the opposite

G fides of AB, make with it at

H the point A the adjacent angles ecual to two right angles; therefore CA is in the same

K straight lined with AG. for

d. 14. I. the fame reason, AB and AH ale in the same 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,

E and the whole angle DBA is equal o to the whole FBC./and because the two sides 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 base AD is equal to the base FC, f. 4. I. and the triangle ABD to the triangle FBC. now the parallelogram BL is double 8 of the triangle ABD, because they are upon the fame g. 41.1. base BD, and between the fame parallels BD, AL; and the square GB is double of the triangle FBC, because thicse also are upon the fame base FB, and between the fame parallels TB, GC. but the doubles of cquals are equal h to one another. therefore the paralle- h. 6. Ax. logram BL is equal to the square GB. and in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC. and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC. wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore in any right angled triangle, &c. R. E. D.

PROP. XLVIII. THEOR.
If the square described upon one of the fides of a tri-

angle, be equal to the squares described upon the other two sides of it; the angle contained by these two fides is a right angle.

1. II.1.

Book 1. If the square described upon BC one of he sides of the triangle mABC be equal to the squares upon the other sides BA, AC; the

angle BAC is a right angle.

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 D A is equal to the square

D
of AB; to each of these add the square of
AC, therefore the squares of DA, AC are

equal to the squares of BA, AC. but the 5. 47. 1. square of DC is equalb to the squares of DA,

А
AC, because DAC is a right angle; and the
square of BC, by Hypothesis, is equal to the
squares of BA, AC; therefore the square of
DC is equal to the square of BC; and there-

B

С fore also the side DC is equal to the side BC. and because the side DA is equal to AB, ani AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the

base DC is equal to the base BC; therefore the angle DAC is e6, 8.1. qual to the angle BAC. but DAC is a right angle, therefore also

BAC is a right angle. Therefore if the square, &c. Q. E. D.

Τ Η Σ

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DEFINITION S.

I.
VERY right angled parallelogram is said 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 com-
plements, is called a Gno-

E
A

$

D mon. “Thus the parallelo

gram HG together with
the complements AF, FC
‘is the gnomon, which is

F
-H

K K
'more briefly expressed by
the letters AGK, or EHC B
which are at the opposite
angles of the parallelograms which make the gnomon.'

IF

1

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 several parts of the divided

Let

2. JI. I.

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 straight lines A, BC is equal to the rect

B D E C angle contained by A, BD; and to that contained by A, DE; and also to that contained by A, EC.

From the point B draw * BF at right angles to BC, and make BG

G b. 3. 1. equal 6 to A; and thro’G draw c

K L H c. 31. 1. GH parallel to BC; and thro' D,

F
E, C draw C DK, EL, CH paral-

A
lel to BG. then the rectangle B H 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. 31.1. DL is contained by A, DE, because DK, that is a BG, is equal to

A; and in like manner the rectangle E H 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.

PROP. II. THEOR.

F a straight line be divided into any two parts, the

rectangles contained by the whole and each of the parts, are together equal to the square of the whole line.

IT

Q. 46. I, b. 31. 3.

Let the straight line AB be divided into

A CB
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 a the square ADEB,
and thro’C draw 6 CF parallel to AD or BE.
then AE is equal to the rectangles AF, CE;
and AE is the square of AB; and AF is the D

* 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.

rectangle

F E

EUCLID. 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 BEM 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.

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IF

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 square of the foresaid part. .

a. 46. s.

b. 31. I.

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
CDEB, and produce ED to F, and A O

В
thro’ A draw 6 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 D
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.

PRO P. IV. THE OR.

IF

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,

Upon

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