Sidebilder
PDF
ePub

Book II. each add LG=CD, then gnomon CMG+LG=AD.DB+

CD. But CMG+LG=BC, therefore AD.DB+CD=
BC%. Wherefore, if a straight line, &c. Q. E. D.

“ Cor. From this proposition it is manifest that the difference of the squares of two unequal lines AC, CD is equal to the rectangle contained by their sum and difference, or that ACP-CD-=AC+CD. ACCD.”

[ocr errors][ocr errors][merged small]

a 46. 1. b 31. 1.

IF a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square of half the line bisected, are 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, are equal to the square of CD.

Upon CD describea the square CEFD, join DE, and through B drawb 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. Because AC is equal to CB, the rectangle AL is equal to CH; but A

c

B D CH is equald to HF; therefore also AL is equal to HF. To each of these K L

HM add CM; therefore the whole AM is equal to the gnomon CMG. Now AM =AD.DM=AD.DB, because DM=DB. There

E

G F fore gnomon CMG=AD.DB, and CMG+LG=AD.DB+ CB2. But CMG+LG=CF=CD2 ; therefore AD.DB+CB2 =CD. Therefore, if a straight line, &c. Q. E. D.

c 36. 1. d 43. 1.

PROP. VII. THEOR. ;

Book II.

IF 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.

[ocr errors]

Let the straight 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 square of AC, or AB'+ BCP=2A B.BC+AC2.

Upon AB describe the square ADEB, and construct the a 46. 1. figure as in the preceding propositions. Because AG=GE, AG+CK=GECK, that is, AKO

A

B в
CE, and therefore AK+CE=2AK.
But AK+CE = gnomon AKF+
CK; therefore AKF+CK=2AK
2AB.BK=2AB.BC, because BK

K =b BC; therefore AKF+CK+HF

b Cor. 4. 2. =2AB.BC+HF; and because AKF+HF=AE=AB, ABP+CK -2AB.BC+HF, that is (since CK

D F E =CB?, and HF=AC?), ABP+CB2 2AB.BC+AC%. Wherefore, if a straight line, &c? Q. E. D.

[ocr errors]

Otherwise.

“ Because AB=AC2+BC2+2AC.BC, if BC be added to a 4.2. both, AB2+BC=AC? +2BCP+ 2AC.BC. But BC2+AC:BC=

С

AB AB.BCb, and therefore 2BC?+

b 3. 2. 2AC.BC=2AB.BC; therefore AB2+BC:=AC?+2AB.BC.”

Book 11.

“ Cor. Hence, the sum of the squares of any two lines is equal to twice the rectangle contained by the lines together “ with the square of the difference of the lines.”

, ic /

2 acti

2

2

te

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, are equal to the square of the straight line which is made up of the whole and the firstmentioned 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, are 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 a 34. 1.

such as in the preceding. Because GK is equal to CB, and
CB to BD, and BD to KN, GK is equal to KN, For the
same reason PR is equal to RO; and because CB is equal to

BD, and GK to KN, the rectangles CK and BN are equal, b 43. L. as also the rectangles GR and RN. But CK is equal to

RN, because they are the complements of the parallelogram
CO; therefore also BN is equal to GR; and the four rect-
angles BN, CK, GR, RN are therefore equal to one another,

and so CK+BN+GR+RN=4CK. Again, because CB is c Cor. 4. 2. equal to BD, and BD equal to

А C B D
BK, that is, to CG; and CB e-
qual to GK, that is, to GP;

GI KI
therefore CG is equal to GP; M

N and because CG is equal to GP,

P R
and PR to RO, the rectangle AG X

0
is equal to MP, and PL to RF.
But MP is equalb to PL, because
they are the complements of the
parallelogram ML; wherefore
AG is equal also to RF. There E

H L F

fore the four rectangles AG, MP, PL, RF, are equal to one Book II. another; therefore AG+MP+PL+RF-4AG. It has been demonstrated that CK+BN+GR+RN=4CK; wherefore, if equals be added to equals, the whole gnomon AOH=4AK. Now AK=AB.BK=AB.BC, and 4AK=4AB.BC; therefore gnomon AOH=4AB.BC; and if XH, or AC?, be c Cor. 4. 2. added to both, gnomon AOH+XH=4AB.BC+AC?. But AOH+XH=AF-AD2; therefore AD-4AB.BC+AC?. Now AD is the line that is made up of AB and BC. Wherefore, if a straight line, &c. Q. E. D.

“ CoR. 1. Hence, because AD is the sum, and AC the “ difference of the lines AB and BC, four times the recta “ angle contained by any two lines, together with the square “ of their difference, are equal to the square of the sum of “ the lines.”

0 Cor. 2. From the demonstration it is manifest that, " since the square of CD is quadruple of the square of CB, “the square of any line is quadruple of the square of half “ that line.”

Otherwise.

“ AD=AC+CD+2CD.ACa, and CD'=CB+BD + 4 4 2. 2CB.BD. But CD=2CB, therefore CD’=4CB?, and also 2CD.AC=4CB.AC; therefore ADP=AC+4BC24BC.AC. Now BCR + BC.AC = AB.BCb; therefore AD =AC + b3.2. 4AB.BC. Q. E. D.”

Book II.

PROP. IX. THEOR.

11

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.

a 11. 1.

b 31. 1.

c 5.1.

d 32. 1.

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

From the point C drawa CE at right angles to AB, and make it equal to AC or CB, and join EA, EB ; through D drawb DF parallel to CE, and through F draw FG parallel to AB, and join AF. Then, because AC is equal to CE, the angle EAC is equal to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angled; therefore each of them is half of a right angle.

For the same reason each of the angles CEB, EBC is half a right angle. Hence the whole AEB is a right angle. Because the angle GEF is half a right an

C D B gle, and EGF a right angle, for it is equale to the interior and opposite angle ECB, the remaining angle EFG is half a right angle ; therefore the angle GEF is equal to the angle EFG ; therefore the side EG is equal to the side GF. Again, because the angle at B is half a right angle, and FDB a right angle, for it is equalto 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 BPD; therefore the side DF is equalf to the side DB. Now, because AC=CE, ACP=CE', and ACP+

€ 29.1.

f 6. 1.

« ForrigeFortsett »