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G d 1. 6.

Let BC, CG be placed in a straight line ; therefore DC and Book VI. CE are also in a straight linea; and complete the parallelogram DG; and, taking any straight line K, make b as BC to CG, a 14. 1. so K to L, and as DC to CE, so make b L to M: therefore b 12. 6 the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE. But the ratio of K to M is that which is said to be compounded < of the c A.def.5: ratios of K to L, and L to M: wherefore also K has to M the ratio compounded of the ratios of the A D

H sides; and because as BC to CG, so is the parallelogram AC to the parallelogram CHd; but as BC to CG, so is K to L; therefore K is to L, as the pa- B

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e 11. 5. rallelogram AC to the parallelogram CH: again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; bat as DC to CE, so is L to M; wherefore L is e to M, as the parallelogram CH to the parallelogram

K L M E F CF: therefore, since it has been proved that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; ex æqualif, K is to M, as the parallelogram AC to the f 22. 5.. parallelogram CF: but K has to M the ratio which is compounded of the ratios of the sides ; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelq. grams, &c. Q. E. D.

PROP. XXIV. THEOR.

THE parallelograms about the diameter of any See n. parallelogram, are similar to the whole, and to one another.

Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter: the parallelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another.

Because DC, GF are parallels, the angle ADC is equal a to a 29. 1. the angle AGF: for the same reason, because BC, EF are pa

G

H

Book VI. rallels, the angle ABC is equal to the angle AEF : and each

w of the angles BCD, EFG is equal to the opposite angle DAB , b 34. 1. and therefore are equal to one another; wherefore the paral

lelograms ABCD, AEFG are equiangular : and because the angle ABC is equal to the angle AEF, and the angle BAC

common to the two triangles BAC, EAF, they are equianguC4. 6. lar to one another; therefore c as AB

А E

B to BC, so is AE to EF: and because

the opposite sides of parallelograms d 7.5.

F
are equal to one another b, AB is d to
AD, as AE to AG; and DC to CB,
as GF to FE; and also CD to DA,
as FG to GA: therefore the sides of
the parallelograms ABCD, AEFG
about the equal angles are proportion- D K

с als ; and they are therefore similar to el. def. 6. one another e: for the same reason, the parallelogram ABCD

is similar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is similar to DB: but rectilineal figures

which are similar to the same rectilineal figure, are also similar f 21. 6. to one anotherf; therefore the parallelogram GE is similar to

KH. Wherefore, the parallelograms, &c. Q. E. D.

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See N.

TO describe a rectilineal figure which shall be si. milar to one, and equal to another given rectilineal figure.

1

Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar

to ABC, and equal to D. a Cor.45. Upon the straight line BC describe a the parallelogram BE

equal to the figure ABC ; also upon CE describe a the paral

lelogram CM equal to D, and having the angle FCE equal S 29. 1. to the angle CBL : therefore BC and CF are in a straight

2 14. 1. line b, as also LE and EM: between BC and CF find c a mean c 13. 6. proportional GH; and upon GH described the rectilineal fid 18. 6. gure KGH similar and similarly situated to the figure ABC : e 2. Cor. and because BC is GH as GH to CF, and if three straight 20. 6. lines be proportionals, as the first is to the third, so ise the

b

figure upon the first to the similar and similarly described figure Book VI. upon the second; therefore as BC to CF, so is the rectilineal figure ABC to KGH: but as BC to CF, so is f the parallelogram f 1.6. BE to the parallelogram EF: therefore as the rectilineal figure ABC is to GH, so is the parallelogram BE to the parallelogram EFs: and the rectilineal figure ABC is equal to the parallelo- g 11. 5.

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gram BE; therefore the rectilineal figure KGH is equal h to the h 14. 5. parallelogram EF: but EF is equal to the figure D; wherefore also KGH is equal to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Which was to be done.

PROP. XXVI. THEOR.

IF two similar parallelograms have a common an. gle, and be similarly situated; they are about the same diameter.

Let the parallelograms ABCD, AEFG be similar and simi-
larly situated and have the angle DAB common. ABCD and
AEFG are about the same diameter.
For, if not, let, if possible, the A G

D
parallelogram BD bave its diame-
ter AHC in a different straight K

H line from AF, the diameter of the parallelogram EG, and let GF E

F meet AHC in H; and through H draw HK parallel to AD or BC: therefore the parallelograms

B

с ABCD, AKHG being about the same diameter, they are similar to one another, a : wherefore, as a 24. 6. DA to AB, so isb'GA to AK: but because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE; b 1.def. 6. Book VI. therefore as GA to AE, so GA to AK; wherefore GA has the

same ratio to each of the straight lines AE, AK; and consec 11. 5. quently AK is equald to AE, the less to the greater, which d 9. 5. is impossible : therefore ABCD and AKHG are not about the

same diameter; wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c. Q. E. D.

• To understand the three following propositions more easily, « tt is to be observed,

• 1. That a parallelogram is said to be applied to a straight • line, when it is described upon it as one of its sides. Ex. gr. 'the parallelogram AC is said to be applied to the straight line * AB.

• 2. But a parallelogram AE is said to be applied to a straight ' line AB, deficient by a parallelogram, when AD the base of " AE is less than AB, and therefore • AE is less than the parallelogram

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G • AC described upon AB in the same

angle, and between the same pa• rallels, by the parallelogram DC; • and DC is therefore called the de"fect of AE.

A D B

F 3. And a parallelogram AG is said to be applied to a straight line AB, exceeding by a parallelogram,

when AF the base of AG is greater than AB, and there. "fore AG exceeds AC the parallelogram described upon AB in • the same angle, and between the same parallels, by the parallelograin BG.

PROP. XXVII. THEOR.

See N.

OF all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is described upon the half of the line; that which is applied to the half, and is similar to its defect, is the greatest.

Let AB be a straight line divided into two equal parts in C; and let the parallelogram AD be applied to the half AC; which is therefore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon the other half CB: of all the parallelograms applied to any other parts of

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AB, and deficient by parallelograms that are similar, and simi- Book VI. larly situated to CE, AD is the greatest.

Let AF be any parallelogram applied to AK, any other part of
AB than the half, so as to be deficient from the parallelogram upon
the whole line AB by the parallelogram KH similar, and similar-
ly situated to CE ; AD is greater than AF.

First, let AK the base of AF, be greater than AC the half of
AB; and because CE is similar to the

D L E
parallelogram KH, they are about the
same diametera: draw their diameter

a 26. 6. DB, and complete the scheme: be

G cause the parallelogram CF is equal b

H b 43. 1. to FE, add KH to both, therefore the whole CH is equal to the whole KE: but CH is equal to CG, because the

c 36. 1. base AC is equal to the base CB ;

A ск B
therefore CG is equal to KE: to each
of these add CF; then the whole AF is equal to the gnomon
CHL: therefore CE, or the parallelogram AD, is greater than
the parallelogram AF.

Next, let AK the base of AF, be less G F M H
than AC, and, the same construction be-
ing made, the parallelogram DH is equal
to DG«, for HM is equal to MG 4, be-

L

d 34.1. cause BC is equal to CA; wherefore DH is greater than LG: but DH is equal b to DK ; therefore DK is greater than LG: to each of these add AL ; then the whole AD is greater than the whole AF. Therefore, of all parallelograms applied, &c. Q. E. D.

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