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Book II.

PROP. V. THEOR.

IF 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 section, 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 a the square CEFB, join BE, and through a 46. 1. D draw b 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 parallel to CL or BM: and because the complement CH is equal c to the complement HF, to each of these add DM; c 43. 1. therefore the whole CM is equal to the whole DF ; A

C DB but CM is equal d to AL,

d 36. 1. because AC is equal to

L
H

M M
CB; therefore also AL is K
equal to Dr. To each of
these add CH, and the
whole AH is equal to DF
and CH: but AH is the

E rectangle contained by AD, DB, for DH is equal é to DB ; and Df together with CH is the e Cor.4.2. gnomon CMG ; therefore the gnomon CMG is equal to the rectangle AD, DB : to each of these add LG, which is equal e 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 square 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 manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference.

Book II.

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. a 46. 1. Upon CD describe a the square CEFD, join DE, and through b 31. 1. B draw b BHG parallel to CE or DF, and through H draw KLM

parallel to ADor EF, and also through A draw AK parallel to CL or DM: and because AC is А

С

B D equal to CB, the rectangle c 36. 1. AL is equal. c to CH; but

L

H d 43. 1. CH is equal d to HF; there

K

M
fore also 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 rect-

E
angle contained by AD, DB,

G F e Cor.4.2. for DM is equal é to DB: therefore the gnomon CMG is equal

to the rectangle 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: byt 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. Q. E. D.

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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 rect- Book II. angle AB, BC, together with the square of AC.

Upon AB describe a the square ADEB, and construct the a 46. 1. figure as in the preceding propositions : and because AG is equal b 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 A C B AK : but AK, CE are the gnomon AKF together with the square CK; therefore the gnomon AKF, toge

G ther with the square CK, is double H н

K of AK: but twice the rectangle AB, BC is double of AK, for BK is equal e to BC : therefore the gnomon AKF,

c eor. 4.2. iogether with the square CK, is equal to twice the rectangle AB, BC: to

D

F E 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 twice the rectangle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up the whole figure . ADEB 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,

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, is equal to the square of the straight line which is made 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 square 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 BI and that CB is equal a to GK, and BD to KN ; therefore GK is a 34. 1.

H

Book II. equal to KN; for the same reason, PR is equal to RO; and

because CB is equal to BD, and GK to KN, the rectangle b 36. 1. CK is equal to BN, and GR to RN: but CK is equal e to c 43. 1. RN, 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 d Cor.4.2. equal d to BK, that is, to CG;

CB and CB equal to GK, that d is, to A

D
GP; therefore CG is equal to

G K
GP: and because CG is equal to M м

N
GP, and PR to RO, the rectangle

P. R
AG is equal to MP, and PL to X

0 e 43. 1. RF: but MP is equal e to PL,

because they are the complements
of the parallelogram ML; where-
fore AG is equal also to RF: E

F therefore the four rectangles

H L AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG. And it was demonstrated, that the four CK, BN, GR, and RN are quad. ruple of CK: therefore the eight rectangles which contain the gnomon AOH are quadruple of AK : and because 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 demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH. To each of these add XH, which is equal d 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 square 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 together in one straight line. Wherefore, if a straight line, &c. Q. E. D.

Book II.

PROP. IX. THEOR.

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.

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 draw a CE at right angles to AB, and a 11. 1. 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; b 31. 1. and join AF: then, because AC is equal to CE, the angle EAC is equal c to the angle AEC; and because the angle c 5. 1. ACE is a right angle, the two others, AEC, EAC together make one right angle d; and they are equal to one another; d 32. 1. each of them therefore is half

E of a right angle.

For the same reason each of the angles CEB, EBC is half a right angle; and

F therefore the whole AEB is a right angle: and because the angle GEF is half a right angle, and EGF a right angle, for it is А

C D B equal e to the interior and oppo

e 29. 1. site 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 f to the side GF: again, because the angle at Bf 6. 1. is half a right angle, and FDB a right angle, for it is equal e 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 BFD, and the side DF to f the side DB : and because AC is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE are double of the square of AC: but the square of EA is equal & to the g 47. 1. squares of AC, CE, because ACE is a right angle; therefore the square of EA is double of the square of AC: again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of FG, GF are double of

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