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1. EVERY right angled parallelogram, or rectangle, is Book II. said to be contained by any two of the straight lines which are about one of the right angles.

“ Thus, the right angled parallelogram AC* is called the rectangle contained by AD and DC, or by AD and AB, “ &c. For the sake of brevity, instead of the rectangle con"tained by AD and DC, we shall simply say the rectangle “ AD.DC, placing a point between the two sides of the rect“ angle. Also, instead of the square of a line, as AD, we n. may sometimes write AD2." “ The sign + placed between the names of two magnitudes, signifies that those magnitudes are to be added together; " and the sign --- placed between them, signifies that the latter is to be taken away from the former.”

“The sign = signifies that the things between which it is “placed are equal to one another."

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* See the next figure.

Book II.

II. In every parallelogram any of the parallelograms about a diameter, together with the

E

D
two complements, are called A
a gnomon. Thus, the

pa-
rallelogram HG, together
with the complements AF,
· FC, are the gnomon of
the parallelogram AC. H

K
· This gnomon may also,
• for the sake of brevity,
called the

gnomon

B G
AGK, or EHC.'

F

• be

PROP. I. THEOR.

IF 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 line.

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"a 11. 1.

b 3. 1.
€ 31. 1.

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 several rectangles A.BD, A.DE, A.EC. From the point B drawa.

B

*

DE
BF at right angles to BC, and
make BG equalb to A; and
through G drawe GH paral-
lel to BC, and through D,
E, C, draw DK, EL, CH
parallel to BG; then BH,

K L H
BK, DL, and EH are rect-
angles, and BH=BK+DL+
EH. But BH=BG.BC F
A:BC, because BG=A; and
BK=BG.BD=A.BD, and DL-DK.DE=A.DE, becaused
DK=BG=A. In like manner, EH=A.EC. Therefore

d 34 1.

A.BC=A.BD+A.DE+A.EC; that is, the rectangle A.BC Book II. is equal to the several rectangles. A.BD, A.DE, A.EC. Therefore, if there be two straight lines, &c. Q. E. D.

PROP. II. THEOR.

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.

Let the straight line AB be divided into any two parts in
the point C; the rectangle AB.BC,
together with the rectangle AB.AC,

A c B В
are equal to the square of AB; or
AB.AC+AB.BC=AB2.
On AB describea the

square ADEB, and through C drawb CF 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.BC=AB?. Therefore, if a straight line, &c. Q. E. D.

a 46. 1. b 31. 1.

PROP. III. THEOR.

IF 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 fore.

said part.

Let the straight line AB be divided into two parts in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC%.

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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, that is,

AB2=AC2+CB2+2AC.CB. a 46. 1.

Upon AB describe the square ADEB, and join BD, and b 31. 1. through C drawb CGF parallel to AD or BE, and through

G draw HK parallel to AB or DE. Because CF is parallel to AD, and BD falls upon

A А.

c B them, the exterior angle BGC is c 29. 1.

equal to the interior and opposite d 5.1. angle ADB; but ADB is equald

Hi

K to the angle ABD, because BA is equal to AD; wherefore the angle

CGB is equal to the angle GBC; e 6.1. therefore the side BC is equale f 34. 1. to the side CG; but CB is equale

D F . also to GK, and CG to BK; wherefore the figuré CGKB is equilateral. It is likewise rectan

gular; for the angle CBK being a right angle, the other & Cor.46.1. angles of the parallelogram CGŘB are also right angless.

Wherefore CGKB is a square, and it is upon the side CB. Book II. For the same reason HF also is a square, and it is upon

the side HG, which is equal to AC; therefore HF, CK are the squares of AC, CB. The complement AG is equalh to the h 43. 1. complement GE, and AG=AC.CG=AC.CB; therefore GE=AC.CB, and AG+GE=2AC.CB. Now HF=AC, and CK=CB: therefore HF+CK+AG+GE=AC2+CB2 +2AC.CB. But HF+CK+AG+GE=the figure AE, or AB; therefore AB2=AC2+CB*+2AC.CB. Wherefore, if a straight line be divided, &c. Q. E. D.

CoR. From the demonstration it is manifest that the parallelograms about the diameter of a square are likewise squares.

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IF a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, are equal to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D; the rectangle AD.DB, together with the square of CD, are equal to the square of CB, or AD.DB+CD:=CB.

Upon CB describe the square CEFB, join BE, and a 46. 1.
through D drawb DHG parallel to CE or BF; and through b 31. 1.
H draw KLM parallel to
CB or EF; and also

С D
А,

B
through A draw AK pa-
rallel to CL or BM. Be-
K

M cause CH=HF, if DM

H be added to both, CM= DF. But AL = CM,

c 36. 1. therefore AL=DF, and

I G F CH being added to both, AH=gnomon CMG. But AH=AD.DH=AD.DB, because DH=DBd; therefore gnomon CMG=AD.DB. To d Cor. 4. 2.

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