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b 36. 1. C 43. I.

b

Book II. equal to KN: For the fame reason, PR is equal to RO; and becaufe CB is equal to BD, and GK to KN, the rectangle CK is equal to BN, and GR to RN: But CK is equal equal to RN, because they are the complements of the parallelogram CO; therefore alío BN is equal to GR; and the four rectangles BN, CK, GR, RN, are therefore equal to one another, and fo are quadruple of one of them CK: Again, because CB is equal to BD, and that BD is

d Cor. 4. 2. equal d to BK, that is, to CG;

C 43. I.

CBD

GK
PRO

N

H L

LF

and CB equal to GK, that is, to A GP; therefore CG is equal to GP: And because CG is equal to M GP, and PR to RO, the rectangle AG is equal to MP, and PL to X RF: But MP is equal to PL, because they are the complements of the parallelogram ML; wherefore AG alfo is equal to RF: Therefore the four rectangles E AG, MP, PL, RF are equal to one another, and fo are quadruple of one of them AG. And it was demonstrated, that the four CK, BN, GR, 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 demonftrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH. Cor. 4. 2. To each of thefe add XH, which is equal to the fquare of AC: Therefore four times the rectangle AB, BC, together with the fquare of AC, is equal to the gnomon AOH and the fquare XH: But the gnomon AOH and XH make up the figure AEFD which is the fquare of AD: Therefore four times. the rectangle AB, BC, together with the fquare of AC, is equal to the fquare of AD, that is, of AB and BC added together in one straight line. Wherefore, if a ftraight line, &c. Q. E. D.

d

PROP

Book II.

PROP. IX. THEOR.

IF a ftraight line be divided into two equal, and also into two unequal parts; the fquares of the two unequal parts are together double of the fquare of half the line, and of the fquare of the line between the points of fection.

Let the straight line AB be divided at the point Cinto two equal, and at Ď into two unequal parts: The fquares of AD, DB are together double of the fquares of AC, CD.

E

G

F

CDB € 29. 1.

From the point C draw CE at right angles to AB, and a 11. 1. make it equal to AC or CB, and join EA, EB; thro' D draw DF parallel to CE, and through F draw FG parallel to AB; b 31. 1. and join AF: Then, because AC is equal to CE, the angle EAC is equal to the angle AEC; and because the angles. 1. ACE is a right angle, the two others AEC, EAC together make one right angled; and they are equal to one another; d 32. t. each of them therefore is half of a right angle. For the fame reafon each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle: And because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and oppofite angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the fide EG equal to the fide GF: Again, because the angle at B f 6. I. is half a right angle, and FDB a right angle, for it is equal to the interior and oppofite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the fide DF to the fide DB: And be caufe AC is equal to CE, the fquare of AC is equal to the fquare of CE; therefore the fquares of AC, CE are double of the fquare of AC: But the fquare of EA is equal to the 8 47. 1. fquares of AC, CE, becaufe ACE is a right angle; therefore the fquare of EA is double of the fquare of AC: Again, be caufe EG is equal to GF, the fquare of EG is equal to the fquare of GF; therefore the fquares of EG, GF are double of

A

the

i 47. I.

h

Book II. the fquare of GF; but the fquare of EF is equal to the fquares n of EG, GF; therefore the fquare of EF is double of the fquare h 34. I. GF: And GF is equal to CD; therefore the fquare of EF is double of the fquare of CD: But the fquare of AE is likewife double of the fquare of AC; therefore the fquares of AE, EF are double of the fquares of AC, CD: And the fquare of AF is equal to the fquares of AE, EF, because AEF is a right angle; therefore the fquare of AF is double of the fquares of AC, CD: But the fquares of AD, DF are equal to the fquare of AF, because the angle ADF is a right angle; therefore the fquares of AD, DF are double of the fquares of AC, CD: And DF is equal to DB; therefore the fquares of AD, DB are double of the fquares of AC, CD. If therefore a straight line, &c. Q. E. D.

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a 11. I.

b 31. I.

€ 29. I.

a straight line be bifected, and produced to any point, the fquare of the whole line thus produced, and the fquare of the part of it produced, are together double of the fquare of half the line bifected, and of the fquare of the line made up of the half and the part produced.

Let the ftraight line AB be bifected in C, and produced to the point D; the fquares of AD, DB are double of the fquares of AC, CD.

From the point C draw a CE at right angles to AB: And make it equal to AC or CB, and join AE, EB; through E draw EF parallel to AB, and through D draw DF parallel to CE : And because the ftraight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles; and therefore the angles BEF, EFD are lefs than two right angles: But straight lines which with another ftraight line make the interior angles 12. Ax. upon the fame fide lefs than two right angles, do meet if produced far enough: Therefore EB, FD fhall meet, if produced, toward B, D: Let them meet in G, and join AG: Then, becaufe AC is equal to CE, the angle CEA is equal to the angle EAC and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a right angle: For the fame reafon,

e 5.1.

f $2. I.

:

each

c

each of the angles CEB, EBC is half a right angle; therefore Book II. AEB is a right angle: And because EBC is half a right angle, DBG is alfof half a right angle, for they are vertically oppo- f 15. 1. fite; but BDG is a right angle, because it is equal to the al- c 29. I. ternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore alfo the fide BD is equal to the fide DG: Again, g 6. I.

becaufe EGF is half a

right angle, and that the angle at F is a right angle, because it is equal to the oppofite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF;

[blocks in formation]

wherefore alfo the fide GF is equal to the fide FE. And becaufe EC is equal to CA, the fquare of EC is equal to the fquare of CA; therefore the fquares of EC, CA are double of the fquare of CA: But the fquare of EA is equal to the fquares of EC, CA; therefore the fquare of EA is double of the fquare of AC: Again, because GF is equal to FE; the fquare of GF is equal to the fquare of FE; and therefore the fquares of GF, FE are double of the fquare of EF: But the fquare of EG is e-quali to the fquares of GF, FE; therefore the fquare of EG is double of the fquare of EF: And EF is equal to CD; wherefore the fquare of EG is double of the fquare of CD: But it was demonftrated, that the fquare of EA is double of the fquare of AC; therefore the fquares of AE, EG are double of the fquares of AC, CD: And the fquare of AG is equal to the fquares of AE, EG; therefore the fquare of AG is double of the fquares of AC, CD: But the fquares of AD, DG are equal to the fquare of AG; therefore the fquares of AD, DG are double of the fquares of AC, CD: But DG is equal to DB; therefore the fquares of AD, DB are double of the fquares of AC, CD: Wherefore, if a straight line, &c. Q. E. D.

i 47. I.

PROP

Book II.

a 46. I. b 10. I. C 3. I.

d 6.2.

$47.1.

PROP. XI. PROB.

TO divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, fhall be equal to the fquare of the other part.

Let AB be the given ftraight line; it is required to divide it into two parts, fo that the rectangle contained by the whole, and one of the parts, fhall be equal to the fquare of the other 'part.

Upon AB defcribe the fquare ABDC; bife&b AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF defcribe the fquare FGHA; AB is divided in H fo, that the rectangle AB, BH is equal to the fquare of AH.

a

G

Produce GH to K: Because the straight line AC is bifected in E, and produced to the point F, the rectangle CF, FA, together with the fquare of AE, is equal d to the fquare of EF: But EF is equal to EB; therefore the rectangle CF, FA, together with the fquare of AE, is equal to the fquare of EB: And the fquares of BA, AE are equal to the F fquare of EB, because the angle EAB is a right angle; therefore the rectangle CF, FA, together with the fquare of AE, is equal to the fquares of BA, AE: Take away the fquare of AE, A which is common to both, therefore the remaining rectangle CF, FA is equal to the fquare of AB: And the fi- E gure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the fquare of AB; therefore FK is equal to AD: Take away the common part AK, and the remainder C FH is equal to the remainder HD:

H R

K D

And HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the fquare of AH: Therefore the rectangle AB, BH is equal to the fquare of AH: Wherefore the ftraight line AB is divided in H, fo that the rectangle AB, BH is equal to the fquare of AH. Which was to be done.

PROP.

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