« ForrigeFortsett »
PROP. XXI. THEO R.
ECTILINEAL 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 proportionals'. Again, because a. 1. Def.6, B is similar to C, they are equiangular, and have their sides about the equal angles proportio
.B nals 2. 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 C proportionals. · Wherefore the rectilineal figures A and B are equiangular 6, and have their sides about b. 1. Ax. s. the equal angles proportionals. Therefore A is similar a to B. c. 15. S. Q. E. D.
PROP. XXII. THEOR.
tilineal figures similarly described upon them shall also be proportionals. and if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals.
Let the four straight lines AB, CI), EF, GH be proportionals, yiz. 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 • X; and to EF, GH a a. 11.6. third proportional O. and because AB is to CD, as EF to GH, bist. s therefore CD is o to X, as GH to 0; wherefore ex aequali“, as AB C. 22. 5.
Book VI. to X, fo EF to 0. but as AB to X, so is d the rectilineal KAB to w the rectilineal LCD, and as EF to O, fo is the rectilineal MF to d. 2. Cor. the rectilineal NH. therefore as KAB to LCD, so bis 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 as AB to CD, fo EF to PR, and upon PR describe! f. 18.6, the rectilineal figure SR similar and fimilarly situated to either of
b. 11. 5.
c. I2. 6.
E F G H P R the figures MF, NH. then because as AB to CD, fo 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 rectilincals 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 8 to one another. they are also similar, and si: milarly situated; therefore GH is equal to PR. and because as AB to CD, fo 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.
PRO P. XXIII. THEO R.
EQUIANGULAR parallelograms have to one
have to one another the ratio which is compounded of the ratios of their fides,
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 com pounded of the ratios of their fides,
b. 12. 6.
Let BC, CG be placed in a straight line, therefore DC and CE Book VI. áre also in a straight line“; and complete the parallelogram DG, m and, taking any straight line K, make b as BC to CG, fo K to L; 2. 14. 1. and as DC to CE, fo make bL 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 Mis that which is said to be compounded of the ratios of K to L, and L to M. c.A. Def.s. wherefore also K has to M, the A 1) Н ratio coinpounded of the ratios of the sides. and because as BC to CG, fo is the parallelogram AC to
B C С the parallelogram CH 4; but as BC to CG, fo is K to L; therefore K is e to L, as the parallelogram AC to the parallelogram CH. again, because as DC to CE, so is the parallelogram CH to the parallelo KLM gram CF; but as DC to CE, fo is L to M; wherefore L is to M, as the parallelogram CH to the parallelogram CF. therefore fince it has been proved that as K to L, fo is the parallelogram AC to the parallelogram CH ; and as L to M, so the parallelogram CH to the parallelogram CF; ex aequalif, K is to M, as the parallelo- f. 22. Sv gram 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 conpounded of the ratios of the fides. Wherefore equiregular farallelograms, &c. Q. E. D.
C. II. So
PROP. XXIV. THEOR.
HE parallelograms about the diameter of any pa- sec N.
rallelogram, 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' to the a. 29. r angle AGF. for the same reason, because BC, EF are paralels, the
Book VI. angle ABC is equal to the angle AEF. and each of the angies
VBCD, EFG is equal to the oppolite angle DAB), and therefore are D. 34. 1.
cqual to one another; wherefore the parallelograms ABCI), AEFG are equiangular. and because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAX, they are equiangular to one another ;
A E C. 4. 6. therefore as AB to BC, fo is AE to
EF. and because the opposite sides of
parallelograms are equal to one anod. 7. 5.
ther 6, AB is d to AD, as AE to AG;
AEFG about the equal angles are pro6.1. Def.6. portionals; and they are therefore similar to one another for the
fame reafon, the parallelogram ABCD is fimilar to the parallelogram FHCK. Wherefore each of the parallelogiams GE, KH
is similar to DB. but rectilineal figures which are limilar to the f. 21. 6. fame rectilineal figure, are also similar to one another f, therefore
the parallelogram GE is similar to KH. Wherefore the parallelo. grams, &c. 0. E. D.
PROP. XXV. Ń ROB.
lar 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 inust be equal. It is required to describe a rectilineal figure similar to'
ABC and equal to D. a.Cor.45.1. Upon the straight line BC describe a the parallelogram BE equal
to the figure APC; also upon CE describe a the parallelogram CM
equal to D, and having the angle FCE equal to the angle CBL. 5 29.1.
therefore EC and CF are in a straight line b, as alfo LE and EM. between BC and CF find a mean proportional CH, and oron GH describe d the rectilineal figure KGH similar and fimilarly
situated to the figure ABC. and because BC is to GH, as GH to c.2. Cor. 20. CF, and if three straight lines be proportionals, as the first is to
the third, fo ise the figure upon the first to the fimilar and limilarly
14. 1. c. 13. 6. d. 13. 6.
described figure upon the second ; therefore as BC to CF, fo is the Book VI. rectilineal figure ABC 10 KGH. but as BC to CF, fo is f the parallelogram BE to the parallelogram EF. therefore as the rectilineal f. 1.6. figure ABC is to KGH, fo is the parallelogram BE to the parallelogram EF $. and the rectilineal figure ABC is equal to the pa- 8.11.50
rallelogram BE; therefore the rectilineal figure KGH is equal 1.14 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 nas been described fimilar to the figure ABC, and equal to D. Which was to be done.
PROP. XXVI. THEOR.
and be fimilarly situated; they are about the same diameter.
Let the parallelograms ABCD, AEFG be similar and similarly fituated, and have the angle DAB common. ABCD and AEFG are about the same diameter. For if not, let, if posible, the
A G parallelogram BD have its diameter AHC in a different straight line
K from AF the diameter of the paral
H lelogram EG, and let GF meet AHC in H; and thro'H draw HK parallel to AD or BC. therefore the paral
B lelograms ABCD, AKHG being about the fame diameter, they are similar to one another, where- a. 24. 6. fore as DA to AB, fo is o GA to AK. but because ABCD and 6.1. Def.6. AEFG are similar parallelograms, as DA is to AB so is GA to AE.