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Book II. Upon AB describe the square ADEB, and join BD, and

med through C drawb CGF parallel to AD or BE, and through G 1 46. 1.

draw HK parallel to AB or DE: and because CF is parallel to b 31. 1.

AD, and BD falls upon them, the exterior angle BGC is equal c 29. 1.

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

to the angle ABD, because BA is equal to AD, being sides of a square ; wherefore the angle CGB A C B

is equal to the angle GBC; and there. e 6 1.

fore the side BC is equal e to the side f34. 1. CG: but CB is equall also to GK,

H

K and CG to BK; wherefore the figure CGKB is equilateral; it is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right

D

F angles; and KBC is a right angle; wherefore GCB is a right angle: and therefore also the angles fCGK, GKB opposite to these, are right angles, and CGKB is rectangular: but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB: 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; and because the complement AG is equals to the complement GE, ảnd that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB : and HF, CK are the

squares

of AC, CB: wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB; but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c. Q. E. D.

Cor. From the demonstration it is manifest, that the paralJelograms about the diameter of a square are likewise squares:..

g 43. 1,

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Book II.

PROP. V. THEOR.

с

If a straight line be divided into two cqual parts, and also into two unequal parts; the rectang!. contained by the unequal parts, together with the square of the line between the points of section, is 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 at the point D; the rectangle AD, DB, together with the square of CD, is equal to the square of CB.

Upon CB describe a the square CEFB, join BE, and a 46. ư. through D draw. DHG parallel to CE or BF; and through b 31. i. H draw KLM parallel to CB or EF; and also, through A draw AK parallel to CL or BM: and because the complement CH is equal c to the complement HF, to each of these c 43. : add DM; therefore the whole CM is equal to the A

DB whole DF; but CM is equald!

d 36. 1. to AL, because AC is equal

L. il

к to CB; therefore also AL is

M M equal to DF. To each of these add , CH, and the whole AH is equal to DF and CH: but AH is the

E G F rectangle contained by AD, DB, for DH, is equale to DB; e Cor. 4. and DF together with CH is the gnomon CMG; therefore the gnomon CHG is equal to the rectangle AD, DB: to each of these add LG, which is equal to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG makes up the whole figure CEFB, which is the square of CB; therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, &c. Q. E. D.

From this proposition it is manifest, thay the difference of two unequal lines AC, CD, is equal to the rectangle con. tained by their sum and difference.

Book IL

PROP. VI. THEOR.

If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together withi. the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. * Let the straight line AB be bisected in C, and produced to the point D; the rectangle AD, DB, together with the

square of CB, is equal to the square of CD. a 46. 1.

Upon CD describe a the square CEFD, join DE, and b 31. 1.

through B drawb BHG parallel to CE or DF, and through
H draw KLM parallel to AD or EF, and also through A
draw AK" parallel to CL or
DM: and because AC is A

С

B D equal to CB, the rectangle c 43. 1. AL is equal c to CH; but

L
d 36. 1.
CH is equal to HF; there-
K

M м
fore also AL is equal to
HF: to each of these add
CM; therefore the whole
AM is equal to the gnomon
CMG: and DM is the rect-

E.

G angle contained by AD, e Cor. 4. 2. DB, for DM is equale to DB: therefore the gnomon CMG

is equal to the rectangle AD, DB: add to each of these LG, which is equal to the square of CB; therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG and the figure LG:- but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D.

PROP. VII. THEOR.

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

of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

Let the straight line AB be divided into any two parts in

the point C; the squares of AB, BC are equal to twice the Book 11. . rectangle AB, BC, together with the square of AC.

Upon AB describe a the square ADEB, and construct the a 46. 1. figure as in the preceding propositions : and because AG is equalb to GE, add to each of them CK; the whole AK is b 43. 1. therefore equal to the whole CE; therefore AK, CE are double of A с B AK : but AK, CE are the gnomon • AKF together with the squale

CK; therefore the gnomon AKF, toge

G ther with the square CK, is double H н

K of AK: but twice the rectangle AB, BC is double of AK, for BK is equal

c Cor. 4.2. cto BC: therefore the gnomon AKF, together with the square CK, is equal

D to twice the rectangle AB, BC: to

F E each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC: there, fore the squares of AB and BC are equal to twice the rect. angle AB, BC, together

with the square of AC. Wherefore, if a straight line, &c. Q. E. D.

PROP. VIII. THEOR.

IF a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

Let the straight line AB be divided into any two parts in the point C; four times the rectangle AB, BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together.

Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal to GK, and BD to KN; therefore GK is , 34. A

H

Book II. equal KN; for the same reason, PR is equal to RO; and

i because CB is equal to BD, and GK to KN, the rectangle b 36. 1. CK is equal b to BN, and GR to RN: but CK is equal to c 43. 1. RN, because they are the complements of the parallelogram

CO; therefore also BN is equal to GR; and the four rect-
angles BN, CK, GR, RN are therefore equal to one another,
and so are quadruple of one of them CK: again, because CB
is equal to BD, and that BD is
equal to BK, that is, to CG;

CB a Cor. 4. 2. and CB equal to GK, that d is, to A

D
GP; therefore CG is equal to

GK
GP: and because CG is equal to M

N
GP, and PR to RO, the rectangle

P

R
AG is equal to MP, and PL to X

O e 43. 1. RF: but MP is equal e to PL,

because they are the complements
of the parallelogram ML; where-

E
fore AG is equal also to RF:
therefore the four rectangles

H L. AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG. And it was demonstrated, that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which contain the gnomon AOH are quadruple of AK: and because AK is the rectangle contained by AB, BC, for BK is equal to BC, four times the rectangle AB, BC is quadruple of AK : but the gnomon AOH was demonstrated to be quadruple of AK;

therefore four times the rectangle AB, BC is equal to the d Cor. 4. 2. gnomon AOH. To each of these add XH, which is equal a

to the square of AC: therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the square XH: but the' gnomon AOH and XH make up the figure AEFD, which is the square of AD : therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D.

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