BOOK II. rectangle AB, BC, is equal to the fquare of AB. If therefore a ftraight line, &c. Q. E. D.

a 46. 1.

PROP. III. THEOR.

Fa ftraight line be divided into any two parts, the rectangle contained by the whole, and one of the parts, is equal to the rectangle contained by the two parts, together with the fquare of the forefaid part.

Let the ftraight line AB be divided into any two parts in the point C; the rectangle AB, AC is equal to the rectangle AC, CB, together with the fquare of CA.

Upon AC describe à the fquare B CDEA, and produce ED to F, and b 31. 1. through B draw BF parallel to CD or AE; then the rectangle BE is equal to the rectangles BD, CE; and BE is the rectangle contained by AB, AC, for it is contained by AB, AE, of which AE is equal to AC; and BD

a 46. I. 31. I.

C 29. 1.

d 5. 1.

D

C

E

is contained by AC, CB, for CD is equal to CA; and DA is the fquare of AC; therefore the rectangle AB, AC is equal to the rectangle AC, CB, together with the fquare of AĈ. If, therefore, a ftraight line, &c. Q. E. D.

PROP. IV. THEOR.

[F a ftraight line be divided into any two parts, the fquare of the whole line is equal to the fquares of the two parts, together with twice the rectangle contained by the parts.

Let the ftraight line AB be divided into any two parts in C; the fquare of AB is equal to the fquares of AC, CB, and to twice the rectangle contained by AC, CB.

d

Upon AB defcribe a the fquare ADEB, and join BD, and through C draw b CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: And becaufe CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal © to the interior and oppofite angle ADB ; but ADB is equal to the angle ABD, becaufe BA is equal to AD, being fides of a fquare; wherefore the angle CGB is equal to the angle GBC; and therefore the fide BC is equal to the fide CG: But CB is f 34. 1. equal alfo to GK, and CG to BK; wherefore the figure CGKB is equilateral: It is likewife rectangular; for CG is pa

e 6. I.

f

e

rallel

f

A

C B

C 29. I. f 34. I.

G

K

rallel to BK, and CB meets them; the angles KBC, GCB are Book. II. therefore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle; and therefore alfo the angles CGK, GKB oppofite to these are right angles, and CGKB is rectangular: But it is alfo H equilateral, as was demonftrated; wherefore it is a fquare, and it is upon the fide CB For the fame reafon, HF alfo is a fquare, and it is upon the fide HG, which is equal to AC: Therefore HF, D

g

F

E

CK are the fquares of AC, CB; and because the complement AG is equal to the complement GE, and that AG is the rect- 43. angle contained by AC, CB, for GC is equal to CB; therefore GE is alfo equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB: And HF, CK are the fquares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the fquares 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 fquares of AC, CB and twice the rectangle AC, CB. Wherefore, if a ftraight line, &c. Q. E. D.

COR. I. From the demonftration, it is manifeft, that the parallelograms about the diameter of a fquare, are likewife fquares.

COR. 2. Hence it is manifeft, that the fquares of two lines AC, CB, and twice the rectangle contained by them, are together equal to the fquare of their sum AB.

COR. 3. And if AC be equal to CB or CG, the four figures AG, GE, HF, CK are fquares, and are equal to one another; therefore AE is quadruple of CK. Confequently the fquare of a line is four times the fquare of the half of that line.

IF

PROP. V. THEOR.

Fa ftraight line be divided into two equal parts, and alfo into two unequal parts; the rectangle contained by the unequal parts, together with the fquare of the line between the points of fection, is equal to the fquare of half the line.

Let the ftraight 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 fquare of CD, is equal to the fquare of CB.

a

Upon CB defcribe the fquare CEFB, join BE, and through a 46. ±. D drawb DHG parallel to CE or BF; and through H draw b 31. 1. KLM

G

C43. 1.

I.

d 36. 1.

d

C

D R

H

M

E

BOOK II. KLM parallel to CB or EF; and alfo through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of these add DM; therefore the whole CM is equal to the whole DF; but CM is equal to AL, because AC is equal to CB; therefore alfo AL is equal to DF: To each of thefe add CH, and the whole AH is equal to DF and CH: But AH is the rectangle contained eCor. 4.2. by AD, DB, for DH is equal to DB; and DF, together with CH, is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equal to the fquare of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the fquare 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 fquare of CD, is equal to the fquare of CB. Wherefore, if a straight line, &c. Q. E. D.

a 46, 1.

e

COR. From this propofition it is manifeft, that the difference of the fquares of two unequal lines AC, CD is equal to the rectangle contained by their fum AD and difference DB.

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 of the line bifected, 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.

c

d

Upon CD defcribe the fquare CEFD, join DE, and through b 31. 1. B draw ↳ 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 becaufe AC is equal to CB, the rectangle 36. I. AL is equal to CH; but CH is equal to HF; therefore allo d 43. 1. AL is equal to HF: To each of these add CM; therefore the whole AM is equal to the gnomon CMG: And AM is the e Cor. 4.2. rectangle contained by AD, DB, for DM is equal to DB: Therefore the gnomon CMG is equal to the rectangle AD,

e

DB:

C

В

D

H

M

L

DB: Add to each of thefe LG, which is equal to the fquare of Book II. CB; therefore the rectangle AD, A DB, together with the fquare of CB, is equal to the gnomon CMG and the figure LG: But the K 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

F

G F

fquare of CD. Wherefore, if a ftraight line, &c. Q. E. D.

IF

PROP. VII. THEOR.

F a 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 be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the fquare of AC.

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

b

C

K

equal 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 AK: But AK, CE are the gnomon AKF together with the fquare CK; therefore the gno- H mon AKF, together with the fquare CK, is double of AK: But twice the rectangle AB, BC is double of AK, for BK is equal c to BC: Therefore the gnomon AKF, together with the fquare CK, is D equal to 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 fquares CK, HF, is equal to twice the rectangle AB, BC and the square of AC: But the gnomon AKF, together with the fquares CK, HF, make up the whole figures 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 ftraight line, &c. Q. E. D.

COR. Hence it is manifeft, that the fquares of two lines AB, BC are equal to twice the rectangle AB, BC contained by thefe lines, together with the fquare of their difference AC.

G 2

PROP.

c Cor. 4.1.

BOOK II,

See N.

IF

PROP. VIII. THEOR.

F a 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 ftraight 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 ftraight line made up of AB and BC together.

b

Produce AB to D, fo that BD be equal to CB, and upon AD defcribe the fquare AEFD; and conftruct two figures fuch as in 34.1 the preceding. Becaufe GK is equal to CB, and CB to BD; Cor. 4.2. GK is equal to BD: and GR, BN are fquares defcribed upon them; therefore GR is equal to BN, and the fide KR to the fide BK and because BK is equal to KR, the rectangle AK is 36. 1. equal to MR: but AK is equal to KF; therefore MR is equal to KF: And because MP is equal

C

:

d 43.1. d to PL, for they are the complements

d

X

A

C

B D

GK

P

R

Q

of the parallelogram ML; and GR is equal to BN, the whole MR is equal to PL, BN together: and AK, KF are each of them equal to MR; therefore PL, BN together, and the three AK, KF, MR, are all equal, and fo are quadruple of one of them AK: but PL, BN, together with AK, KF, MR, make up the gnomon AOH; therefore the gnomon AOH is quadruple of AK: and becaufe 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 ftraight line. Wherefore, if a straight line, &c. Q. E. D.

d

COR,