four straight lines be proportionals, those straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to GH, and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly described; and upon EF, GH the similar rectilineal figures MF, NH in like manner: the rectilineal figure KAB is to LCD, as MF to NH.

To AB, CD take a third proportional (11. 6.) X; and to EF, GH a third proportional O: and because AB is to CD, as EF to GH, and that CD is (11. 5.) to X, as GH to O; wherefore, ex æquali (22. 5.), as AB to X, so is EF to O: but as AB to X, so is (2 Cor. 20. 6.) the rectilineal KAB to the rectllineal LCD, and as EF to O, so is (2 Cor. 20. 6.) the rectilineal MF to the rectilineal NH: therefore, as KAB to LCD, so (11. 5.) is MF to NH.

And if the rectilineal KAB be to LCD, as MF to NH; the straight line AB is to CD, as EF to GH.

Make (12. 6.) as AB to CD so EF to PR, and upon PR describe (18. 6.) the rectilineal figure SR similar and similarly situated to

either of the figures MF, NH: then, because as AB to CD, so is EF to PR, and that upon AB, CD are described the similar and similarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR; KAB is to LCD, as MF to SR; but by the hypothesis, KAB is to LCD, as MF to NH: and therefore the rectilineal MF having the same ratio to each of the two NH, SR, these are equal (9. 5.) to one another: they are also similar, and similarly situated; therefore GH is equal to PR: and because as AB to CD, so is EF to PR, and that PR is equal to GH; AB is to CD, as EF to GH. If, therefore, four straight lines, &c. Q. E. D.

PROP. XXIII. THEOR.

EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides.*

Let AC, CF be equiangular parallelograms, having the angle

* See Note.

BCD equal to the angle ECG: the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides.

B

A

D H

وح

G

Let BG, CG, be placed in a straight line; therefore DC and CE are also in a straight line (14. 1.); and complete the parallelogram DG; and, taking any straight line K make (12. 6.) as BC to CG, so K to L; and as DC to CE, so make (12. 6.) L to M: therefore 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 (A. def. 5.) of the ratios of K to L, and L to M: wherefore also K has to M the ratio compounded of the ratios of the sides; and because as BC to CG, so is the parallelogram AC to the parallelogram CH (1. 6.); but as BC to CG, so is K to L; therefore K is (11. 5.) to L, as the parallelogram AC to the parallelogram GH: again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but as DC to CE, so is L to M; wherefore Lis (11. 5.) to M, as the parallelogram CH to the parallelogram 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 æquali (22. 5.), K is to M, as the parallelogram AC to the 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 parallelograms, &c. Q. E. D.

PROP. XXIV. THEOR.

KLM E F

THE parallelograms about the diameter of any 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 (29. 1.) to the angle AGF: for the same reason, because BC, EF are parallels, the angle ABC is equal to the angle AEF: and each of the angles BCD, EFG is equal to the opposite angle DAB (34. 1.), and therefore are equal to one another; wherefore the parallelograms 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 equiangular to one another; therefore (4. 6.) as AB to BC, so is AE to EF: and because the

• See Note.

A

E

B

F

G

H

opposite sides of parallelograms are equal to one another (34. 1.), AB is (7. 5.) 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 proportionals; and they are therefore similar to one another (1. def. 6.): 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 to one another (21. 6.); therefore the parallelogram GE is similar to KH. Wherefore, the parallelograms, &c. Q. E. D.

D K

C

PROP. XXV. PROB.

To describe a given rectilineal figure which shall be similar to one, and equal to another given rectilineal figure.*

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.

Upon the straight line BC describe (Cor. 45. 1.) the parallelogram BE equal to the figure ABC; also upon CE describe (Cor. 45. 1.) the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL; therefore BC and CF are in a straight line (29. 1. 14. 1.), as also LE and EM: between BC and CF find (13. 6.) a mean proportional GH, and upon GH describe (18. 6.) the rectilineal figure KGH similar and similarly situated to the figure ABC; and because BC is to GH as GH to CF, and if three straight lines be proportionals, as the first is to the third, so is (2. Cor. 20. 6.) the figure upon the first to the similar and similarly described figure upon the second; therefore as BC to CF, so is the rectilineal figure ABC to KGH; but as BC to CF, so is (1. 6.) the parallelogram BE to the parallelogram EF: therefore as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF (11. 5.): and the A

rectilineal figure ABC is equal to the parallelogram BE; therefore the

* See Note.

BOOK VI.

THE ELEMENTS OF EUCLID.

rectilineal figure KGH is equal (14. 5.) to the 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. XXVII. THEOR.

Ir two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.

Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common: ABCD and AEFG are about the same diameter.

A

G

D

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F

For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF the diame- K ter of the parallelogram EG, and let GF meet AHC in H; and through H draw HK E parallel to AD or BC; therefore the parallelograms ABCD, AKHG being about the same diameter, they are similar to one another (24. 6.): wherefore as DA to AB, so is (1. def. 6.) GA to AK: but because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE; therefore (11. 5.) as GA to AE, so GA to AK; wherefore GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal (9. 5.) to AE, the less to the greater, which 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.

B

C

To understand the three following propositions more easily, it 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.

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 AC described upon AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE.

E C

G

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 therefore AG exceeds AC the parallelogram described upon AB in the same angle, and between the same parallels, by the same parallelogram BG.'

19

PROP. XXVII. THEOR.

Or 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 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 paral·lelogram CL upon the other half CB: of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar and similarly 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 similarly situated to CE. AD is greater than AF.

G

F

H

A

C K

B

First, let AK the base of AF, be greater than AC the half of AB; and because CE is similar to the paralD LE lelogram KH, they are about the same diameter (26. 6.): draw their diameter DB, and complete the scheme: because the parallelogram CF is equal (43. 1.) to FE, add KH to both, therefore the whole CH is equal to the whole KE: but CH is equal (36. 1.) to CG, because the base AC is equal to the base CB: therefore CG is equal to KE: to each of these add CF; then the whole AF is equal to the gnomon CHL: thereforce CE, or the parallelogram AD is greater, than the parallelogram AF. Next, let AK the base of AF be less than AC, and, the same construction being made, the parallelogram DH is equal to DG (36. 1.), for HM is equal to MG (34. 1.) because BC is equal to CA; wherefore DH is greater than LG: but DH is equal (43. 1.) 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.

G F M

H

L

D

E