Book II.

= CD2, then gnomon CMG + LG AD.DB + CD2 But CMG+LG BC, therefore AD.DB+CD2=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 dif"ference, or that AC2CD2=(AC+CD) (AC—CD)."

ǎ 46. 1.

b 31. 1.

PROP. VI. THEOR.

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, 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 through 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 be cause AC is equal to JCB, the rectangle AL

с

c 36. 1. is equal to CH; but

с

d 43. 1.

d

A

CH is equal to HF;
therefore also AL is K
equal to HF: To each
of these add CM; there-
fore the whole AM is
equal to the gnomon

CMG. Now, AM

=

C

B

D

AD.DM = AD.DB, because DM DB.

G F

Therefore

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

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.

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 AB2+ BC2 = 2AB.BC + AC2.

1

Upon AB describe the square ADEB, and construct a 46. 1. the figure as in the preceding propositions: Because AG

=

GĚ, AG + CK — GE+ CK,

that is, AKCE, and therefore" AK + CE2AK. But AK + CE gnomon AKF + CK; and therefore, AKF + CK-2AK-H 2AB.BK-2AB.BC, because BK

BC. Since then, AKF+ CK -2AB.BC, AKF+ CK + HF ≈ 2AB.BC + HF; and because AKF+ HF AE AB2, AB D

4 CK2AB.BC+ HF, that is, (since CKCB2, and HF = AC2), AB2 + CB2 - 2AB.BC+AC Wherefore, if a straight fine, &c. Q. E. Dos of upo

"Because AB2=AC2+BC2+2AC.BC, adding BC2 a 4. 2. "to both, AB2+ BC2 AC2+2BC2 + 2AC.BC. But

BC2+ AC.BC = - AB.BC; and therefore, 2BC2 + b 3. 2. "2AC.BC=2AB.BC; and therefore AB2+ BC2= AC' + 2AB.BC.".

66

Book II.

"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."

ǎ 34. 1.

If a

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 the first-mentioned part.

Let the straight line AB be divided into any two parts in the point, C; four times the rectangle AB.BC, toge ther with the square of AC, is equal to the square of the straight line made up of AB and BC together.

a

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 GK is equal a 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 Snood because CB is equal to BD, and GK to KN, the rectangles CK and BN are equal, as also the rectangles GR b 43. 1. 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 rectangles BN, CGK GR, RN are therefore equal to one another, and so CK + BN + GR + RN = 4CK. Again, because CB is M

à Cor. 4. 2. equal to BD, and BD equal

d

A

to BK, that is, to CG; and X
CB equal to GK, that is, to
GP; therefore CG is equal to
GP; and because CG is equal
to GP, and PR to RO, the

rectangle AG is equal to MP, E

e four

C B D

G K

N

P

R

and PL to RF: But MP is equal to PL, because they are the complements of the parallelogram ML; wherefore

=

AG is equal also to RF: Therefore the four rectangles Book I. AG, MP, PL, RF, are equal to one another, and so AGMP+PL + RF 4AG. And it was domonstrated, that CK + BN + GR + RN-4CK; wherefore adding equals to equals, the whole gnomon AOH= 4AK. Now AK AB.BKAB.BC, and 4AK=4AB.BC; therefore, gnomon AOH=4AB.BC; and adding XH, ord AC, to both, gnomon AOH+XH=4AB.BC + d Cor. 4. d. AC. But AOH+XHAFAD; therefore AD*= 4AB.BC+AC. Now AD is the line that is made up of AB and BC, added together into one line: 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 "rectangle contained by any two lines together with the "square of their difference, is equal to the square of the "sum of the lines."

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:

a 4. 2.

"Because AD is divided any how in Ca, AD2: "AC+ CD +2CD.AC. But CD=2CB: and there"fore CD2CB2+ BD2+2CB.BDa = 4CB2, and also "2CD.AC-4CB.AC; therefore, AD2 AC + 4BC2 "+4BC.AC. Now BC+BC.ACAB.BC; and b 3. 2. "therefore ADAC + 4AB.BC. Q. E. D.

=

a 11. 1.

b 31. 1.

c 5. 1.

d 32. 1.

e 29. 1.

f 6. 1.

g 47. 1.

PROP. IX. THEOR..

If a straight line be divided into two equal, and
also into treo unequal parts; the squares of the
two unequal parts are together double of the
two unequal parts
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 D into two unequal parts: The squares of AD, DB are together double of the squares of AC, CD.

a

с

From the point C draw CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D draw 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 an gle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angled; and they are equal to one another; each of them therefore is half of a right angle. For the same rea son each of the angles CEB, EBC is half a right angle: and therefore, the whole AEB is a right angle: And because the angle GEFA is half a right angle, and

e

E

D

EGF a right angle, for it is equal 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, 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 BFĎ, and the side DF to the side DB. Now, because ACCE, AC2-CE2, and AC2 + CE2 = 2AC2. But AE AC2 + CE; therefore AE = 2AC AC2. Again, because EG=GF, EG2 = GF2, and

f