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Book II.

a. 46. I. b 31. I. C 29. I. ds.I.

e 6. I. f34. I.

Upon AB describe a the square ADEB, and join BD, and through C drawbCGF parallel to AD or BE, and through Gdraw HK parallel to AB or DE: And because CF is parallel to AD, and BD falts upon them, the exterior angle BĠC is equal to the interior and opposite angle ADB ; but ADB is equal to the angle ABD, because BA is equal to AD, being lides of a {quare ; wherefore the angle CGB A С B is equal to the angle GBC; and there. fore the side BC is equal to the lide CG: Bat CB is equal f also to

G

K GK, and CG to BK; wherefore H the figure CGKB is equilateral : It is likewile rectangular ; for CG is pa'rallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right angles ; and KBC D F

E is a right angle; wherefore GCB is a right angle; and therefore also the angles #CGK, 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 fide CB: For the same reason HF also is a square, and it is upon the fide HG which is equal to AC: Therefore HF, CK are the squares of AC, CB; and becaule the complement AG is equal 8 to the complement GE, and that AG is the rectangle contain: ed 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 manifeft, that the parallelograms about the diameter of a square are likewite squares.

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PROP.

Book II.

PRO P. V.

THEOR.

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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 fe&tion, 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 & the square CEFB, join BE, and through a 46. f. D draw DHG parallel to CE or BF; and through H draw b 31. 1. KLM parallel to CB or EF; and also through A draw AK Patallel to CL or BM: And because the complement CH is equal to the complement HF, to each of these add DM; C 43. I. iherefore the whole CM is equal to the whole DF;

А but CM is equal to Al.,

С D B d 36.1. because AC is equal to

L H CB, therefore also AL is K

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 rectangle contained by

E G F AD, DB, for DH is equal to DB; and DF together with CH is the gnomon CMG ; c Cor. 4. therefore the gnomon CMG 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 make up the whole figure CEFB, which is the fquare 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 manifeft, that the difference of the [quares of two unequal lines AC, CU, is equal to the rectangle contained by their lum and difference.

2.

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I

a 46. 1.

Book 11.

PRO P. VI. THE O R. [F a straight line be bisected, and produced to any

t; produced, and the part of it produced, together with the square of half of the line bifected, 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.

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

B draw b 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

С
A

B D is equal to CB, the rectangle C 36. I.

AL is equal to CH; but
d 43. I.
CH is equal to HF; there

H Н
fore allo AL is equal to
K

M
HF: To each of these add
CM; therefore the whole
AM is equal to the gno-
mon CMG: And AM is
the rectangle contained by

E G F AD, DB, for DM is equal e Cor. 4. 2. c to DB : Therefore the gnomon CMG is equal to the re&t.

angle 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. 0. E. D.

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

squares 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

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

Upon AB describe the square ADEB, and construct the a 46. I. figure as in the preceding propositions : And because AG is equal 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

A С AK : But AK, CE are the gnomon

B AKF together with the square CK ; therefore the gnomon AKF, toge-H G K ther with the square CK, is double of AK : But twice the rectangle AB, BC is double of AK, for BK is e. qual to BC: Therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle D F

E 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 iwice the rectangle AB, BC and the square of AC: But the gnomon AKF, together with the squares CK, HF, make up the whole figure A DEB 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.

c Cor. 1223

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F 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 inade 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 Iquare 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 a 34. 1.

equal

Book Il equal to KN: For the fame reason, PR is equal to RO; añid

because CB is equal to BD, and GK to KN, the rectangle 136. I.

CK is equal to BN, and GR to RN: But CK is equal to 6 43. 1.

AN, because they are the complements of the parallelogram 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
2 Cor.4. 2. equal to BK, that is, to CG;
and CB equal to GK, that d is, to A

CBD
GP; therefore CG is equal to
GP: And because CG is equal to M

GK

N
GP, and PR to RO, the rectangle
AG is equal to MP, and PL to

PIRO

x
RF: But MP is equal to PL,
because they are the complements
of the parallelogram ML; where-
fore AG is cqual also to RF:
Therefore the four rectangles

E

H LF AG, MP, PL, RF are equal to one another, and so are quadruple of one of the 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 re angle contain: ed 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. 1 Cor. 4. 2. To each of these add XH, which is equal 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 fquare 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 togeiber in one straight line. Wherefore, if a straight line, &c. Q. E. D.

PROP

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