Sidebilder
PDF
ePub

and AF is the rectangle contained by BA, AC; for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the fquare of AB. If therefore a ftraight line, &c. Q. E. D.

[blocks in formation]

IF a 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, BC is equal to the rectangle AC, CB, together with the fquare of BC.

A C

Upon BC defcribe the fquare CDEB, and produce ED to F, and through A drawb AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC, CB, for CD is equal to CB; and DB is the F fquare of BC; therefore the rectangle

D

Book II.

B

a 46. I.

b 31. 1.

E

AB, BC is equal to the rectangle AC, CB together with the fquare of BC. If therefore a ftraight line, &c. Q. E. D.

IF

PROP. IV. THEOR.

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..

Ꭰ 3

Upon

[ocr errors]

Book II.

a 46. I. b 31. 1. c 29. I. d 5.1.

e 6. I.

£ 34. 1.

8 43. T.

c

Upon AB defcribe the fquare ADEB, and join BD, and thro' C draw b CGF parallel to AD or BE, and thro' 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, because BA is equal to AD, being sides of a fquare; wherefore the angle CGB A

e

is equal to the angle GBC; and there-
fore the fide BC is equal to the
fide CG: But CB is equal f alfo to
GK, and CG to BK; wherefore H
the figure CGKB is equilateral: It is
likewife rectangular; for CG is
rallel to BK, and CB meets them ;
the angles KBC, GCB are therefore
equal to two right angles; and KBC D

C B

[blocks in formation]

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 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, CK are the fquares of AC, CB; and because the complement AG is equal 8 to the complement GE, and that AG is the rectangle 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 squares 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 fquare 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. QE. D.

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

PROP.

Book II.

IF

PROP. V. THEOR.

F a 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.

2

Upon CB defcribe the fquare CEFB, join BE, and thro' a 46. 1. D draw DHG parallel to CE or BF; and thro' H draw b 31. I. KLM parallel to CB or EF; and alfo thro' 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; c 43. I therefore the whole CM

A

is equal to the whole DF;
but CM is equal to AL,
because AC is equal to
CB; therefore alfo AL is K
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 by
AD, DB, for DH is equal

[ocr errors]
[blocks in formation]

e

EGF

c

2.

to DB; and DF together with CH is the gnomon CMG; 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 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 square of CB. Wherefore, if a straight line, &c. Q. E. D.

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 and difference.

[blocks in formation]

Book II.

a 46. . b 31. 1.

c 36. I.

d 43. I.

[blocks in formation]

IF 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 square of CB, is equal to the fquare of CD.

Upon CD defcribe the fquare CEFD, join DE, and thro B draw BHG parallel to CE or DF, and thro' H draw KLM parallel to AD or EF, and also thro' A draw AK parallel to CL or DM: And because AC is cqual to CB, the rectangle AL is equal to CH; but CH is equal to HF; therefore alfa AL is equal to K

c

HF: To each of these add

CM; therefore the whole

A

C

B

D

L

H

M

[blocks in formation]

AD, DB, for DM is equal

[blocks in formation]

c Cor. 4. 2. to DB: Therefore the gnomon CMG is equal to 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 ftraight line, &c. Q. E. D.

PRO P. VII. THEOR.

IF 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

[ocr errors]

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 the fquare ADEB, and conftruct the a 46. I. figure as in the preceeding propofitions: And because AG is

A

C

B

G

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 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

c Cor. 4.2.

F

E

add HF, which is equal to the fquare of AC; therefore the gromon 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 squares 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.

PROP. VIII. THEOR.

IF 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.

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 the preceeding. Becaufe CB is equal to BD, and that CB is equal to GK, and BD to KN; therefore GK is a 4.7. equal

[ocr errors]
« ForrigeFortsett »