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Book I. GB, HC ; and through A drawb AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is m

b 31. 1. a right angle, the two

G

< 30. del. Atraight lines AC, AG upon the opposite Gides of AB,

H make with it at the point A the adjacent angles equal to

F two right angles; therefore

K CA is in the same straight

d 14. 11 lined with AG; for the same

B reason, AB and AH are in

С 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, D

E and the whole angle DBA is

L equal to the whole FBC ; and because the two sides AB, BD e 2. Ax. are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equals to the base FC, and the triangle ABD to che triangle

f4.1. FBC: Now the parallelogram BL is double 6 of the triangle 8 41... ABD, because they are upon the same base BD, and between the same parallels, BD, AL; and the square GB is double of the triangle FBC, because these allo are upon the fame base FB, and between the fame parallels FB, GC. But the doubles of equals are equal h to one another: Therefore the parallelo. h 6. AX, gram BL is equal to the square GB: And in the same manner, by joining AE, BK, it is demonstrated that the parallelogranı 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 itraight line BO, and the aquares GB, HC upon BA, AC: Wherefore the square upon the Gde BC is equal to the squares upon the sides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D.

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the O

F the square described upon one of the sides of a tri

upon ther two sides of it; the angle contained by these two fides is a right angle.

Book 1.

a 11.1.

b 47. I.

If the square described upon BC, one of the fides of the triangle ABC, be equal to the squares upon the other fides 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 DA is equal
to the square of AB: To each of these

D
add the square of AC; therefore the squares
of DA, AC, are equal to che squares of
BA, AC: But the square of DC is equal A

to the squares of DA, AC, because DAC
is a right angle; and the square of BC, by
hypothelis, is equal to the squares of BA,
AC; therefore the square of DC is equal
to the square of BC; and therefore allo B
the side DC is equal to the fide BC. And
because the side 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 base DC is equal to the base BC; there-
fore the angle DAC is equal to the angle BAC: But DAC
is a right angle; therefore also BAC is a right angle. There-
fore, if the square, &c. Q. E. D.

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I.
VERY right angled parallelogram is said to be contained

by any two of the straight lines which contain one of the right angles.

11. In every parallelogram, any of the parallelograms about a dia.

meter, together with the kwo complements, is called A

D a Gnomon. « Thus the

pa: i ' rallelogram HG, toge

ther with the comple"ments AF, FC, is the gno

F • mon, which is more brief. H

K ly exprefled by the letters «ÁGK, or EHĆ which are B

G at the opposite angles of • the parallelograms which make the gnomon.'

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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.

Let

D 2

Book II.

a II. I

b 3. I.
с 3І. І.

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,

B

D E C BD, together with that contain: ed by A, DE, and that contained by A, EC.

From the point B draw • BF at right angles to BC, and makeG BG equal to A; and through

K L Η G draw GH parallel to BC; and through D, E, C draw DKF Α 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, a 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 leveral rectangles contained by A, BD, and by A, DE; and also by A, EC. Wherefore, if there be two straight lines, &c. 0. E, D,

d 34. 1.

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IF

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.

CB

Let the straight line AB be divided into

A
any two parts in the point C; the rect-
angle 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 through C drawb CF, paralld) to AD,
or BE ; then AE is equal to the rectangles
AF, CE; and ĄL is the lquare of AB;D

a 46. I. bo 31. 1.

F E

• N. B. To avoid repeating the word contained too frequently, the rectang contained by two Itraight lines AB, AC is sometimes simply called the rectang AB, AC,

and AF is the rectangle contained by BA, AC; for it is con- Book II. tained by 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.

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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.

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 O

B

a 46. I. CDEB, and produce ED to F, and

b 31. 1. through A drawb AF parallel to CD or BE, then the rectangle AE is e. qual 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 contained by AC, CB, for CD is equal to CB ; and DB is the F D square of BC; therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the {quare of BC. If therefore a straight line, &c. Q. E. D.

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IF
F a straight line be divided into any two parts, thie

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. D3

Upon

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