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PROP. XXI. THEOR. RECTILINEAL figures which are similar to the same rectilineal figure, are also similar to one another.

Let each of the rectilineal figures A, B be similar to the rectilineal figure C: The figure A is similar to the figure B. .

Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportionala. Again, '1 Def. 6. because B is similar to C, they are equiangular, and have their sides about the equal angles proportionalsa. Therefore the figures A, B, are each of them equiangular to C, and have the sides about the equal'angles of each of them and of B proportionals. Wherefore the rectilineal figures A and C are equiangular b, and have their b 1 Ax. 1. sides about the equal angles proportionals. Therefore A c 11. 5. is similar a to B. Q. E. D.

PROP. XXII. THEOR. If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals; and if the similar reçtilineal figures similarly described upon 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 proportionala X; and to EF, 11. 6. GH a third proportional O: And because AB is to CD as

EF to GH, and that CD is by to X as GH to 0; wherefore, " 11.5. . ex æqualic, as AB to X, so EF to 0: But as AB to X, so o 22. 5.

20.6.

Book VI. is the rectilineal KAB to the rectilineal LCD, and as EF

m to 0, so is d the rectilineal MF to the rectilineal NH: d 2 Cor.

05: Therefore, as KAB to LCD, sob is MF to NH. b 11.5. And if the rectilineal KAB be to LCD, as MF to NH;

the straight line AB is to CD, as EF to GH. • 12. 6. Make é as AB to CD, so EF to PR, and upon PR de. { 18. 6. scribef the rectilineal figure SK similar and similarly situ

ated 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 59. 5. of the two NH, SR, these are equals 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. XXII. THEOR. See N. EQUIANGULAR parallelograms have to one ano

ther the ratio which is compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms, having the angle 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 12. 6.

11.6.

e 11. 5.

11. 5.

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Let BC, CG be placed in a straight line : therefore DC Book VI. and CE are also in a straight linea; and complete the pa.

ha 14.1. rallelogram DG; and taking any straight line K, makebi as BC to CG, so K to L; and as DC to CE, so makeb 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 compoundedc of the ratios of K to L,' and L to M: A. Def. 5. 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 parallelogtam CHd; but as BC to 4

D H CG, so is K to L; therefore K is e to L, as the parallelogram AC to the parallelogram CH: Again, i 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 L ise 1 to M, as the parallelogram CH I to the parallelogram CF: There

ť fore since it has been proved. kla that as K to L, so is the parallelogram AC to tlie parallel. ogram CH; and as L to M, so the parallelogram CH to the parallelogram CF: ex æqualis, K is to M, as the pa- f 92. 5. rallelogram 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.

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PROP. XXIV. THEOR. ' Tue parallelograms about the diameter of any See X. parallelogram, aie siinilar 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 parallelogramas EG, HK are similar both to the whole parallelogram ABCD, and to one another.

· Because DC, GF are parallels, the angle ADC is equala a 29. 1. to the angle AGF: For the same reason, because BC, EF

Book VI. are parallels, the angle ABC is equal to the angle AEF: un and each of the angles BCD, EFG is equal to the opposite $ 34. 1. angle DABb, 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, * 4. 6. EAF, they are equiangular to one anothere, therefore as AB

to BC, so is AE to EF: And
because the opposite sides of

parallelograms are equal to one
- 07.5. another b, ABd is to AD, as AE G

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 D

are proportionals; and they are therefore similar to one ei Def. 6. anothere: 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 rec

tilineal figures which are similar to the same rectilineal 191. 6. figure are also similar to one anotherf; therefore the paral

lelogram GE is similar to KH. Wherefore the parallelograms, &c. Q. E. D.

.: PROP. XXV. PROB. See N. To describe a rectilineal figure which shall be si

milar 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. * Cor. 45.1. Upon the straight line BC describe a the parallelogram

BE equal to the figure ABC; also upon CE describe a the parallelogran CM equal to D, and having the angle FCE

equal to the angle CBL: Therefore BC and CF are in a us 29. 1. straight line b, as also LE and EM: Between BC and CF

14. 1: findc a mean proportional GH, and upon GH described the d 18. örectilineal 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

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third, so ise the figure upon the first to the similar and si- Book VI., milarly described figure upon the second ; therefore, as BC to CF, so is the rectilineal figure ABC to KGH: But as BC 20. to CF, so isf the parallelogram BE to the parallelogram 11. 6. EF: Therefore as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EFs: And11. 5.

Cor. 20. 6.

the rectilineal figure ABC is equal to the parallelogram BE; therefore the rectilineal figure KGH is equal h to the 14.5mm 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 uyas to be done.

PROP. XXVI. THEOR. If 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.

For, if not, let, if possible, the parallelogram BD have its diame- A ter AHC in a different straight line from AF, the diameter of the KF

H : parallelogram EG, and let GF E meet AHC in H; and through H draw HK parallel to AD or I BC: Therefore the parallelo- Blu grams ABCD, AKHG being in to about the same diameter, they are similar to one another a : - 24.6. Wherefore as DA to AB, so is b GA to AK: But because b 1 Def. 6.

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