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F

G

Book I.

b 31. I.

€ 30. def.

H

K

d 14. 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
the two
a right angle, the
ftraight lines AC, AG up-
on the opposite sides 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 e-
qual to the angle FBA, each
of them being a right angle,
add to each the angle ABC,

B

D

and the whole angle DBA is

L

E

£ 4. x.

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equal to the whole FBC; and because the two fides 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 bafe AD is equal to the bafe FC, and the triangle ABD to the triangle FBC: Now the parallelogram BL is double of the triangle g 41. 1. ABD, because they are upon the fame bafe BD, and between the fame parallels, BD, AL; and the fquare 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 to one another: Therefore the parallelo- h 6. Ax, gram BL is equal to the fquare GB: And in the fame manner, by joining AE, BK, it is demonftrated that the parallelogramı CL is equal to the fquare HC: Therefore the whole fquare BDEC is equal to the two fquares GB, HC; and the fquare BDEC is defcribed upon the ftraight line BC, and the fquares GB, HC upon BA, AC: Wherefore the fquare upon the fide BC is equal to the fquares 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 defcribed upon the other two fides of it; the angle contained by these two fides is a right angle.

Book I.

a II. I.

b 47. I.

c 8.

If the fquare defcribed 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.

D

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 fquare of DA is equal to the fquare of AB: To each of thefe add the fquare of AC; therefore the squares of DA, AC, are equal to the squares of BA, AC: But the fquare of DC is equal A to the fquares of DA, AC, because DAC is a right angle; and the square 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 therefore alfo B 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. There fore, if the fquare, &c. Q. E. D.

THE

THE

ELEMENTS

O F

EU CLI D.

BOOK II.

DEFINITIONS.

I.

VERY right angled parallelogram is faid to be contained

right angles.

II.

E

D

In every parallelogram, any of the parallelograms about a dia.
meter, together with the
two complements, is called A
a Gnomon. Thus the pa-
'rallelogram HG, toge-
ther with the comple-
ments AF, FC, is the gno-
mon, which is more brief-
ly expreffed by the letters
AGK, or EHC which are

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at the oppofite angles of

F

H

HK

B

G

'the parallelograms which make the gnomon.'

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F there be two straight 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.

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Book II.

Book II.

a II. I'

b 3. I.

C 31. I.

d 34. x.

a 46. I.

b 31. 1.

B

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 the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC.

с

с

KLH

A

From the point B draw BF at right angles to BC, and make BG equal to A; and through G draw GH parallel to BC; and through D, E, C draw DKF 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, becaufe DK, that is, d 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 also by A, EC. Wherefore, if there be two ftraight lines, &c. Q. E, D.

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.

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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.

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 fquare of AB. If therefore a straight line, &c. Q. E. D.

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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 square 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 CDEB, and produce ED to F, and through A drawb 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 contained by AC, CB, for CD is equal to CB; and DB is the F fquare of BC; therefore the rectangle

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AB, BC is equal to the rectangle AC, CB together with the fquare of BC. If therefore a ftraight line, &c. Q. E. D.

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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.

D 3

Upon

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