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a 46. 1.

b 43. 1.

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

Upon A B describe a the square ADEB, and construct the figure as in the preceding propositions; and because AG equal to GE, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE, are double AK: But AK, CE, are thegnomon A с B AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, iš H

G K double AK: But twice the rect

angle AB, BC is double AK, c Cor. 4. 2. for BK is equal to BC: There

fore the gnomon AKF, together
with the square CK, is equal to D

F

E twice the rectangle AB, BC: To 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: therefore the squares of AB and BC are equal to twice the rectangle 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 a to GK, and BD to KN: therefore GK is equal to KN: For the same reason, PR is equal to RO; and Book II. because CB is equal to BD, and GK to KN, the rectangle CK is equal o to BN, and GR to RN: But CK is equal o to b 36. l. RN, because they are the complements of the parallelogram < 43. 1. CO; therefore also BN is equal to GR; and the four rectangles 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 d to BK, that is, to CG;

a 34. 1.

d Cor. 4. 2. and CB equal to GK, that is, to GP; therefore CG is equal

A
CB

D
to GP, and because CG is equal M M GK N
to GP, and PR to RO, the rect-
angle AG is equal to MP, and X

PVR PL to RF: But MP is equal e

e 43. 1. to PL, because they are the complements of the parallelogram ML; wherefore AG is equal also to RF: Therefore the four rect- E

HL

F angles 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 gnomon AOH. To each of these add XH, which is equal to the square of AC: f Cor. 4. 2. 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.

D

Book II.

PROP. IX. THEOR.

If a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided at the point C into two equal, and at V into two unequal parts : The squares of AD,

DB are together double of the squares of AC, CD. a ll. 1.

From the point C draw a CE at right angles to AB, and

make it equal to AC or CB, and join EA, EB; through D draw b 31. 1. • DF parallel to CE, and through F draw FG parallel to AB;

and join AF: Then, because AC is equal to CE, the angle c 5. 1.

EAC is equal to the angle AEC; and because the angle

ACE is a right angle, the two others AEC, EAC together d 32. ). make one right angled; and they are equal to one another;

each of them therefore is half
a right angle. For the same

E
reason each of the angles CEB,
EBC is half a right angle; and
therefore the whole AEB is a

G
right angle: And because the
angle GEF is half a right angle,
and EGF a right angle, for it is

А. C D B e 29. 1. equal e to the interior and op

posite angle ECB, the remain

ing angle EFG is half a right angle ; therefore the angle f 6. 1. GEF is equal to the angle EFG, and the side EG equal

to the side GF: Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to the side DB: And because AČ is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE, are double the square of AC: But the square of EA is equal to the squares of AC, CE, because ACE is a right angle; therefore the square of EA is double the square of AC: Again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF are double the square of GF; but the square of EF is equal to the squares of EG,

g 47. 1.

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GF; therefore the square of EF is double the square GF; Book II. and GF is equal to CD; therefore the square of EF is double the

square of CD: But the square of AE is likewise double h 34. 1. the square of AC; therefore the squares of AE, EF are double the squares of AC, CD: And the square of AF is equal' to the squares of AE, EF, because AEF is a right i 47. 1. angle; therefore the square of AF is double the squares of AC, CD: But the squares of AD, DF are equal to the square of AF, because the angle ADF is a right angle; therefore the squares of AD, DF are double the squares of AC, CD: And DF is equal to DB; therefore the squares of AD, DB are double the squares of AC, CD. If, therefore, a straight line," &c. Q. E. D.

PROP. X. THEOR.

If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line 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 squares of AD, DB are double of the squares of AC, CD. .

From the point C draw a CE at right angles to AB: And a 11. 1. make it equal to AC or CB, and join A E, EB; through E draw • EF parallel to AB, and through D draw DF parallel to CE: b 31. 1. And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles: c 29. 1. and therefore the angles BEF, EFD are less than two right angles; but straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet d if produced far enough: Therefore d 12. Ax. EB, FD shall meet, if produced towards BD: 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 e 5. 1. is a right angle; therefore each of the angles CEA, EAC is half a right angle :' For the same reason each of the angles f 32. 1. CEB, EBC is half a right angle; therefore AEB is a right angle: And because EBC is half a right angle, DBG is alsof f 15. 1. half a right angle, for they are vertically opposite; but BDG

g 6. 1.

С

B D

Book II. is a right angle because it is equal to the alternate angle

w DCE; therefore the remaining angle DGB is half a right anc 29. 1. gle, and is therefore equal to the angle DBG; wherefore also

the side BD is equal 8 to the side DG: Again, because EGF
is half a right angle, and
that the angle at F is a

E

F right angle, because it is h 34. 1. equal" to the opposite an

gle ECD, the remaining
angle FEG is half a right
angle, and equal to the an- A
gle EGF; wherefore also
the side GF is equal & to
the side FE. And be-

G cause EC is equal to CA, the square of EC is equal to the

square of CA; therefore the squares of EC, CA are double i 47. 1. the square of CA: but the square of EA is equal to the

squares of EC, CA; therefore the square of EA is double the square of AC: Again, because GF is equal to FE, the square of GF is equal to the square of FE: and therefore the squares of GF, FE are double the square of EF: But the square of EG is equal to the squares of GF, FE; therefore the square of EG is double the square of EF: and EF is equal to CD; wherefore the square of EG is double the square of CD: But it was demonstrated, that the square of EA is double the square of AC; therefore the squares of AE, EG are double the squares AC, CD: and the square of AG is equal' to the squares of AE, EG; therefore the square of AG is double the squares of AC, CD: But the squares of AD, GD are equal i to the square of AG; therefore the squares of AD, DG are double the squares of AC, CD: But DG is equal to DB; therefore the squares of AD, DB are double the squares of AC, CD. Wherefore, “ if a straight line,&c. Q. E. D.

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, shall be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

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