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If any four Right Lines A, B, C, and D, be proposed, the Ratio of the first A to the fourth D, is equal to the Ratia compounded of the Ratio of the firft A to the second B, and of the Ratio of the second B to the third C, and of the Ratio of the third C to the fourth D.
For in three Right Lines A, C, and D, the Ratio of A to D, is equal to the Ratio's compounded of the Ratio's of A to C, and of C to D; and it has been already demonstrated, that the Ratio of A to C is equal to the Ratio compounded of the Ratio's of A to B, and B to C. Therefore the Ratio of A to D is equal to the Ratio compounded of the Ratio's of A to B, of B to C, and of C to D. After the same Manner we demonstrate, in any number of Right Lines, that the Ratio of the first ta
the last is equal to the Ratio compounded of the Ratio's A B C D of the firft to the second, of the second to the third, of
the third to the fourth, and so on to the last.
This is true of any other Quantities besides Right Lines, which will be manifeft, if the same Number of Right Lines A, B, C, &c. as there are Magnitudes be asumed in the fame Ratio, viz. so that the Right Line X is to the Right Line B, as the first Magnitude is to the second, and the Right Line B to the Right Line C as the second Magnitude is to the third, and so on. It is manifeft (by 22, 5.) by Equality of Proportion, that the first Right Line A is to the last Right Line, as the first Magnitude is to the last; but the Ratio of the Right Line A to the last Right Line, is equal to the Ratio compounded of the Ratio's of A to B, B to C, and so on to the last Right Line : But (by the Hyp.) the Ratio of any one of the Right Lines to that nearesi to it, is the fame as the Ratio of a Magnitude of the fame Order to that nearest it. And therefore the Ratio of the first Magnitude to the last, is equal to the Ratio compounded of the Ratio's of the first Magnitude to the second, of the fecond to the third, and so on to the last; which was to be demonstrated.
one another that is compounded of their Sides.
having the Angle BCD equal to the Angle ECG. I say, the Parallelogram AC, to the Parallelogram CF, is in the Proportion compounded of their Sides, viz. compounded of the Proportion of BC to CG, and of D C to CE.
For let B C be placed in the same Right Line with CG. Then DC shall be * in a strait Line with CE,
14. and complete the Parallelogram DG; and then t as t 12 of tbiš. BC is to CG, fo is fome Right Line K to L; and as DC is to CE, so let L be to M.
Then the Proportions of K to L, and of L to M, are the same as the Proportions of the Sides, viz. of BC to CG, and DC to CE; but the Proportion of K to M is compounded of the Proportion of K 1 Lemning to L, and of the Proportion of L to M. Wherefore precede also K to M hath a Proportion compounded of the Sides. Then because B C is to CG as the Parallelogram AC is * to the Parallelogram CH: And since * 1 of this. BC is to CG as K is to L, it shall be f as K is to L, † 11. s. so is the Parallelogram AC, to the Parallelogram CH. Again, because DC is to CE as the Parallelogram CH is to the Parallelogram CF; and since as DC is to CE, fo is L to M. Therefore as L is to M, so shall + the Parallelogram CH be to the Parallelogram CF; and consequently since it has been proved that K is to L, as the Parallelogram AC is to the Parallelogram CH, and as L is to M, so is the Parallelogram CH to the Parallelogram CF; it shall be I by Equality of Proportion, as K is to M, fo is † 21. 5the Parallelogram AC to the Parallelogram CF; but K to M hath a Proportion compounded of the Sides: Therefore also the Parallelogram AC, to the Parallelogram CF, hath a Proportion compounded of the Sides. Wherefore equiangular Parallelograms have
the Proportion to one another that is compounded of their Sides; which was to be demonstrated.
are about the Diameter, are similar to the whole,
ter is A C; and EG, HK, be Parallelograms about the Diameter AC. I say the Parallelograms EG, HK, are similar to the whole ABCD, and also to each other.
For because EF is drawn parallel to BC, the Side 2 of ibis of the Triangle ABC, it shall be * as BE to EA,
fo is CF to FA. Again, because F G is drawn parallel to CD, the side of the Triangle ACD, it hall be as CF to FA, so is * DG to GA. But CF is to
FA, (as has been proved) as BE is to EA. There+11. 5.
fore, as BE is to E A, fo is + DG to GA; and by I 18. 5.
compounding, as BA is to AE, fo is DA to AG; and by Alternation, as B A is to AD, fo is EA to AG. Therefore the sides of the Parallelograms ABCD, EG, which are about the common Angle
BAD, are proportional. And because G F is paral29. 9. lel to DC, the Angle AGF is * equal to the Angle
ADC, and the Angle GFA equal to the Angle DCĂ; and the Angle DĄ C is common to the two Triangles ADC, AGF. Wherefore the Triangle ADC will be equiangular to the Triangle AGF. For the fame Reason, the Triangle ACB is equiangular to the Triangle AFE. Therefore the whole Parallelogram
ABCD is equiangular to the. Parallelogram EG; + 4 of this and fo AD is to DC as AG ist to GF. But DC
is to CA as GF is to FA; and AC is to CB, as AF is to FE; and moreover, CB is to BA as FE is to EA. Wherefore, since it has been proved, that DC is to CA, as GF is to FA; and AC is to CB, as AF is to FE; it fhall be, by Equality of Proportion, as DC is to CB, so is GF'to FE. Therefore the Sides that are about the equal Angles of the Parallelogramas
ABCD, EG, are proportional ; and accordingly the Parallelogram ABCD is fimilar to the Parallelogram EG. For the same Reason the Parallelogram ABCD is similar to the Parallelogram KH. Therefore, both the Parallelograms EG, HK, are similar to the Parallelogram ABCD. But Right-lined Figures that are similar to the same Right-lined Figure, are simi- * 21 of this lar to one another. Therefore the Parallelog
am EG is similar to the Parallelogram HK. And so in every Parallelogram, the Parallelograms that are about the Diameter are similar to the whole, and also to one an
which was to be demonstrated,
lined Figure which shall be given, and equal to
ET ABC be a given Right-lined Figure, to which
» On the Side B C of the given Figure ABC, *make * 44.e!! the Parallelogram B E equal to the Right-lined Figure ABC; and on the Side CE make * the Parallelogram CM equal to the Right-lined Figure D, in' the Angle F CE, equal to the Angle CBL Then BC, CF, as also LE, EM, will be + in two strait Lines. † 14. I, Find GH a mean Proportional between BC, CF, 13 of tbiq and on GH let there be described * the Right-lined + 18 of this. Figure KGH fimilar and alike situate to the Rightdined Figure ABC.
And then because B C is to GH, as GH is to CF, and since when three Right Lines are proportional, the first is to the third as the Figure described on the first is † to a similar and alike situate Figure described on t Cor. zp. the second, it shall be as BC is to CF, so is the Right of this. lined Figure ABC to the Right-lined Figure KGH. But as BC is to CF, fo is I the Parallelogram BE to f 1 of slika the Parallelogram EF. Therefore, as the Right-lined Figure ABC is to the Right-lined Figure KGH, lo is the Parallelogram BE to the Parallelogram EF.
Wherefore, (by Alternation) as the Right-lined Figure ABC is to the Parallelogram BE, so is the Rightlined Figure KGH to the Parallelogram EF. But the Right-lined Figure ABC is equal to the Parallelo grami BE. Therefore the Right-lined Figure KGH is also equal to the Parallelogram E F. But the Parallelogram E F is equal to the Right-lined Figure D. Therefore the Right-lined Figure KGH is equal to D. But KGH is fimilar to ABC. Consequently there is described the Right-lined Figure KGH fimilar to the giveri Figure ABC, and equal to the given Figure D; which was to be done.
THEOREM. If from a Parallelogram be taken away another similar to the whole, and in like manner situate, having also an Angle common with it, then is that Parallelogram about the same Diameter with the whole. ET the Parallelogram AF be taken away from the Parallelogram ABCD similar to ABCD,
, and in like manner situate, having the Angle DAB common. I say the Parallelogram ABCD is about the same Diameter with the Parallelogram AF.
For if it be not, let AHC be the Diameter of the Parallelogram BD, and let GF be produced to H; also let HK be drawn parallel to AD, or BC.
Then because the Parallelogram ABCD is about
the same Diameter as the Parallelogram KG, the Pa• 24 of this. rallelogram ABCD shall be * fimilar to the ParalleDef. 1. of logram KG; and so as DA is to AB, so is + GA to tbis.
AK. But because of the Similarity of the Parallelo
grams ABCD, EG, as D A is to AB, fo is GA to 111. S. AE. And therefore as GA is I to A E, so is GA
to AK. And since GA has the same Proportion to AK as to AE, AE is equal to AK, the less to a greater, which is abfurd. Therefore the Parallelogram ABCD is not about the same Diameter as the Parallelogram AH. And therefore it will be about the fame Diameter with the Parallelogram AF. Therefore, if from a Parallelogram be taken away another formilar