Book II.

3 46. I. b 31. I.

© 36. I. ₫ 43. I.

PROP. VI. THEOR.

F a ftraight line be bifected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the fquare of half the line bifested, is equal to the fquare of the ftraight line which is 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 rectangle AD, DB, together with the fquare of CB, is equal to the fquare of CD..

Upon CD defcribe a the fquare CEFD, join DE, and through B draw b BIG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and alfo through A draw AK paral

AM is the rectangle con

Cor. 4. 2. tained by AD, DB, for DM is equal e to DB: Therefore the gnomon CMG is equal to the the rectangle AD, DB: Add to each of thefe LG, which is equal to the fquare of CB, therefore the rectangle AD, DB, together with the fquare 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 fquare of CD; therefore the rectangle AD, DB together with the fquare of CB, is equal to the fquare of CD. Wherefore, if a straight line, &c. Q. E. D.

IF

PROP. VII. PRO B.

Fa ftraight line be divided into any two parts, the fquares 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 fquare of the other part.

Let the ftraight line AB he divided into any two parts in

the

the point C; the fquares of AB, BC are equal to twice the Book II. rectangle AB, BC together with the fquare of AC.

Upon AB defcribe a the fquare ADEB, and construct the a 46. 1. figure as in the preceding propofitions: and because AG is

A

C

B

G

K

equal b to GE, add to each of them CK; the whole AK is b 43. I. therefore equal to the whole CE; therefore AK, CE, are double of AK: But AK, CE are the gnomon AKF together with the fquare CK; therefore the gnomon AKF, toge-H ther with the fquare CK, is double of AK: But twice the rectangle AB BC is double of AK, for BK is equal to BC: Therefore the gnomon AKF, together with the fquare CK, is equal to twice the rectangle D AB, BC: To each of thefe equals

€ Cor. 4. 2.

F

E

add HF, which is equal to the fquare of AC; therefore the gnomon AKF, together with the fquares CK, HF, is equal to twice the rectangle AB, BC and the fquare of AC: But the gnomon AKF, together with the fquares CK, HF, make up the whole figure ADEB and CK, which are the fquares of AB and BC therefore the fquares of AB and BC are equal to twice the rectangle AB, BC, together with the fquare of AC. Wherefore if a straight line &c. Q. E. D.

IF

PROP. VIII. THEOR,

Fa ftraight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the fquare of the other part, is equal to the fquare of the ftraight 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 fquare of AC, is equal to the fquare of the straight line made up of AB and BC together.

Produce AB to D, fo that BD be equal to CB, and upon AD describe the fquare AEFD; and conftruct two figures fuch as in the preceding. Becaufe CB is equal to BD, and that CB is equal a to GK, and BD to KN; therefore GK is a 34. 1.

equal

Book II.

b 36. I. C 43. I.

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 b to BN, and GR to RN: But CK is equal to RN, because they are the complements of the parallelogram CO; therefore alfo BN is equal to GR; and the four recangles 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;

€ 43. I.

X

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
RF: But MP is equal e to PL,
because they are the complements
of the parallelogram ML; where-
fore AG is equal alfo to RF:
Therefore the four rectangles E
AG, MP, PL, RF are equal
to one another and fo are qua-

GK

CBD

N

PRO

HL F

druple 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 demonftrated to be quadruple of AK; therefore four times the rectangle AB, BC, is equal to the gnomon AOH. To Cor. 4. 2. each of these add XH, which is equal f 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 ftraight line. Wherefore, if a straight line, &c. Q. E. D.

PROP.

1

PROP. IX. THEOR.

[Fa ftraight line be divided into two equal, and alfo

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 ftraight line AB be divided at the point C into two equal, and at D into two unequal parts: The fquares of AD, DB are together double of the fquares of AC, CD.

Book II.

From the point C draw a CE at right angles to AB, and a 1ì. 1, make it equal to AC or CB, and join EA, EB; through D draw

E

G

F

CD B

b 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 becaufe the angle c 5. 1. ACE is a right angle, the two others AEC, EAC together make one right angle d; and they are equal to one another; d 32. I, 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 A equal e to the interior and oppofite angle ECB, the re- e 29. 1. maning angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the fide EG equal f to the fide GF: Again, because the angle at B is f 6. 1. half a right angle and FDB a right angle, for it is equal e 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 f the fide DB: And becaufe 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 g to the fquares of AC, CE, because ACE is a right angle; therefore the fquare of EA is double of the fquare of AC: Again, becaufe EG is equal to GF, the fquare of EG is equal to the square of GF; therefore the fquares of FG, GF are double of

the

g 47. I.

b 34. I.

i 47. I.

Book II. the fquare of GF; but the fquare of EF is equal to the fquares of EG, GF; therefore the fquare of EF is double of the fquare GF; and GF is equal h to CD; therefore the square of EF is double of the fquare of CD: But the fquare of AE is likewise double of the fquare of AC; therefore the fquares of AE, EF are double of the fquares of AC, CD: And the square 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 square 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 ftraight line, &c. Q. E. D.

a II. I.

b 31. 1.

€ 29. I.

IF

PROP. X. THEOR.

Fa ftraight line be bifected, and produced to any point, the square of the whole line thus porduced, 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 squares 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 b EF parallel to AB, and through D draw DF parallel to CE: And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles; and therefore the angles BEF, EFD are less than two right angles: but straight lines which with another ftraight line make the interior angles d 12. Ax. upon the fame fide lefs than two right angles, do meet d if produced far enough: Therefore EB, FD fhall meet, if produced towards B, D: Let them meet in G, and join AG: Then, because 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 f; For the fame reason,

5. I.

f 32. I.

each