« ForrigeFortsett »
the Proportion to one another that is compounded of their Sides; which was to be demonstrated.
are about the Diameter, are similar to the
a ameter is AC; and EG, HK, be Parallelograms about the Diameter AC. I say, the Parallelograms E G, H K, are similar to the Whole ABCD, and also to each other.
For, because E F is drawn parallel to BC, the Side of the Triangle A B C, it shall be *, as B E is to EA, .2 of tbita fo is C F to F A. Again, because F G is drawn parallel to CD, the side of the Triangle ACD, it shall be as C F is to FA, fo is * DG to GA. But CF is to F A (as has been proved), as B E is to E A. Therefore, as B.E is to E A, fu is + D G to G A; +11.5. and by compounding, as B A is to A E, fo is DA $ 18. s to. AĞ; and by Alternation, as B A is to AD, so is A E to AG. Therefore, the sides of the Parallelograms ABCD, EG, which are about the common Angle B A D are proportional. And because G F is parallel to D C, the Angle AGF is * equal to the 29. 1. Angse A DC, and the Angle G F A equal to the Angle DCA; and the Angle DAC is common to the Triangles ADC, AGF. Wherefore the Triangle ADC will be equiangular to the Triangle AGF. For the same Reason, the Triangle A C B is equiangular to the Triangle AFE. Therefore, the whole Parallelogram A B C D is equiangular to the Parallelogram EG; and so A D is to D C, as AG is + tot 4 of tbisa GF; DC is to CA, as G F is to FA; and AC is 10 CB, as AF is to FE; and, moreover, CB is to BA, as F E is to E A. Wherefore, since it has been proved, that DC is to C A, as G F is to FA; and AC is to CB, as AF is to FE; it shall be, by Equality of Proportion, as DC is to CB, fo is G Fio FÉ. Therefore the Sides that are about the equal
Angles of the Parallelograms A B CD, EG, are proportional; and, accordingly, the Parallelogram ABCD is similar to the Parrallelogram E G. For the same Reason, the Parallelogram ABCD is fimilar to the Parallelogram K H. Therefore, both the Parallelograms EG, HK, are fimilar to the Paralle.
logram ABCD. But Right-lined Figures that † 21 of this. are similar to the t same Right-lined Figure, are fimi
Jar to one another. Therefore the Parallelogram EG is fimilar to the Parallelogram HK. And so, in every Parallelogram, the Parallelograms that are about the Dia ameter, åre similar to the Whole, and also to one another; which was to be demonstrated.
lined Figure which mall be given, and equal to
Figures ; it is required to describe another Fia
gure, fimilar to A B C and equal to D. 44. 3.
On the Side B C of the given Figure ABC*, make the Parallelogram B E equal to the Right-lined Figure A BC, and on the Side CE make the Parallelogram C M equal to the Right-lined Figure D,
in the Angle FC E, equal to the Angle C BL. + 14. .
Then B C, CF, as allo L E, EM, will be + in two I 13 of this. (trait Lines. Find I GH a mean Proportional be
tween B C and CF; and on GH let there be de* 38 of tbiso fcribed * the Right-lined Figure KGH, fimilar, and
alıke fituate, to the Right-lined Figure A BC.
And then, because B C is to GH, as G H is to CF; and since, when three Right Lines are propor
tional, the first is to the third, as the Figure described * Cor. 20 of on the first is t to a' fimilar and alike situate Figure
described on the second : it shall be, as B C is to CF,
fo is the Right-lined Figure A B C to the Right-lined 12, 0f this. Figure KGH. But as B C is to CF, fo is I the
Parallelogram B E to the Parallelogram EF. Therefore, as the Right-lined Figure A B C is to the Rightlined Figure R GH, so is the Parallelogram Bê to
the Parallelogram EF. Wherefore (by Alternation), as the Right lined Figure ABC is to the Parallelogram BĚ, fo is the Right-lined Figure KGH to the Parallelogram E F. But the Right-lined Figure A B C is equal to the Parallelogram B E. 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 Rightlined Figure KGH is equal to D. But KGH is fimilar to A B C. Confequently, there is described the Right-lined Figure KGH similar to the given Figure A B C, and equal to the given Figure D; which was to be done.
similar to tbe Wbole, and in like manner situate,
with the whole. LE T'he Parallelogram A F be taken away from
the Parallelogram ABCD, fimilar to ABCD, and in like manner situate, having the Angle DAB COIN:01). I say, the Parallelogram ABCD is about the faine Diameter with the Parallelogram AF.
For, if it be noi, les AHC be the Diameter of the Parallelogram BD, and let G F be produced to H; allo, let H K be drawn parallel to AD, or BC.
Then, because the Parailclogram ABCD is about the fame Diameter as the Parallelogram KG, the Parallelogram A B C D shall be * fimilar to the 24 of tbisa Parallelogram KG; and so, as D A is to AB, so is + G A 10 A K. But because of the Similarity of the Def. I. Parallelograms * ABCD, EG; as D A is to A B, Hyp. so is GĂ O A E. And therefore, as G A is $ 10 1 11. s. A E, so is G A to A K. And since GA has the fame Proportion to A K as to A E, A E is t equal to 7 9. So AK, the less to a greater, or the greater to a less which is absurd. Therefore, the Parallelogram ABCD is not about the fame Diameter as the Parallelogram A H.' And therefore it will be about the
fame Diameter with the Parallelogram AF. There-
Line, and wanting in Figure by Parallelograms
ibe balf Line, and it is similar to the Defeet. LET AB be a Right Line, bifected in the Point
C; and let the Parallelogram A D be applied to the Right Line A B, wanting in Figure the Parailelogram CE, similar and alike fruate to that described on half of the Right Line AB. I say, A D is the greatest of all Parallelograms applied to the Right Line Å B, wanting in Figure by Parallelograms similar and alike situate to CE. For, let the Parallelogram AF be applied to the Right Line AB, wanting in Figure the Parallelogram HK, similar and alike situate to the Parallelogram CE. I say, the Parallelogram A D is greater than the Parallelogram AF.
For, because the Parallelogram CE is fimilar to the 26 of tbis. Parallelogram HK, they stand * about the same Dia
meter. Let D B, their Diameter, be drawn, and the
Figure described; then, since the Parallelogram CF is +43. 1. + equal to F E, let H K, which is common, be added ;
and the Whole CH is equal to the Whole K E. But I 36. 1. CHis I equal to CG, because the Right Line AC is
equal to CB; therefore CG is equal to KE; add the common Parallelogram CF, and the Whole AF is equal to the Gnomon LNM; and fo CE, that is the Parallelogram A D, is greater than the Parallelogram AF. Therefore, of all the Parallelograms ap. plied to the same Right Line, and wanting in Figure by: Parallelograms similar and alike ftuate, to that described on the half Line, the greatest is that which is applied to the half Line, and it is similar to the Defeet, which was to be demonstrated.
equal to a Right-lined Figure given, deficient by
Parallelogram to be applied.
given Right-lined Figure, to which the Parallelogram to be applied to the Right Line A B must be equal, be C, which must not be greater than the Parallelogram applied to the half Line, the Defects being similar; and let D be the Parallelogram, to which the Defect of the Parallelogram to be applied is similar. Now it is required to apply a Parallelogram equal to the given Right lined Figure C to the given Right Line AB, deficient by a Parallelogram similar to D.
Let A B be bilected in E, and on E B describe * the * 18 of ibis. Parallelogram E B F G, fimilar and alike ficuate to D; and compleat the Parallelogram A G.
Now, A G is either equal to C, or greater than it, because of the Determination. If AG be equal to C, what was proposed will be done; for the Parallelogram A G is applied to the Right Line A B equal to the given Right-lined Figure C, deficient by the Parallelogram EF, fimilar to the Parallelogram D. But, if it be not equal, then H E is greater than C; but E F is equal to HE; therefore E F shall also be greater than c. Now make + the Parallelogram + 25 of thing, K L M N similar and alike fituate to D, and equal to the Excess, by which E F exceeds C. But D is fimilar to EF; wherefore KM fhall also be similar to EF. Therefore let the Right Line K L be homologous to G E, and L M to GF: Then because EF