« ΠροηγούμενηΣυνέχεια »
22 Th. If four straight lines be proportionals, the similar rectilineal figures described upon them shall be proportionals: and if the similar rectilineal figures described upon four straight lines be proportionals, those lines shall also be proportionals.
1. Let AB, CD, EF,GH, be four straight lines, of such kind that AB: CD :: EF : GH; and upon AB, CD, describe similar figures, ABK, CDL; also, upon EF, GH, describe similar figures FM, HN.
Then ABK: CDL::FM: HN.
To AB, CD let X be a third proportional; and to EF, GH let O be a third proportional (a). And since AB : CD :: EF : GH by hyp.
and CD: X ::GH: 0, (b); ex æquali AB : X :: EF: 0, (c). But AB : X :: ABK: CDL (d);
and EF: 0 :: FM: HN (d); therefore ABK: CDL :: FM: HN (a), as stated.
2. If ABK be to CDL as FM is to HN; then AB : CD:: EF: GH.
Let AB be to CD as EF is to PR (e); and upon PR describe the rectilineal figure RS, similar to HN, or FM (f). Then since AB : CD:: EF : PR; and upon AB, CD similar figures ABK, CDL, are described; and upon EF, PR similar figures FM, RS, are described ; therefore ABK : CDL :: FM : RS. But, by hyp. ABK: CDL:: FM: HN; therefore FM has to RS and HN the same ratio; and so, RS equals HN (g); therefore, being equal and similar, their sides about equal angles are equal: hence PR equals GH; and for this cause, and that AB : CD:: EF: PR, it is clear that AB : CD:: EF : GH. Therefore, if four, &c.
Q. E. D
(6) p. 11, 5; (c) p. 22, 5;
23 Th. Equiangular parallelograms have to each other the ratio which is composed of the ratios of their sides :the sides of the antecedent being the antecedents, and the sides of the consequent being the consequents.
Given AC, CF, two parallelograms, having equal angles at C: the ratio of AC to CF is composed of the ratios of BC to CD and of EC to CG.
Place the angles at C vertical to each other (a); then the sides adjacent to them shall be in straightG lines (6): complete the parallelogram DG.
Now, since parallelograms of the same altitude are to each other as their bases (c),
AC: DG :: BC: CG, and DG : CF:: DC: CE. And, since in compound ratios, the product of the antecedents is the antecedent, and the product of the consequents is the consequent (d);
Therefore ACXDG : DGXCF :: BC XDC: CGXCE. But magnitudes have the same ratio as their equimultiples (e); therefore AC: CF :: BCX DC : CG X CE.
Wherefore, equiangular parallelograms, &c. Q. E. D.
(e) p. 15,5
24 Th. The parallelograms about the diameter of any parallelogram, are similar to the whole and to one another.
Given the whole parallelogram BD, of which the diameter is AC; and ĒG, HK, parallelograms about the parts of AC: the three are
H Н similar to each other.
1. BD and EG are equiangular: for since DC, GF are parallels, the angles ADC, AGF are equal (a); and since BC, EF are parallels, the D K
C angles ABC, AEF are equal (a); also the common angle DAB equals each of the opposite angles BCD, EFG (b).
2. The sides of BD, EG, about equal angles, are proportionals; for, in the triangles BAC, EAF, the angles ABC, AEF are equal, and BAC is common: the triangles are therefore equiangular; and so, AB : BC :: AE : EF (c). And since the opposite sides of parallelograms are equal (6), AB : AD :: AE: AG, and DC : CB :: GF: FE, also CD: DA:: FG: GA (d).
Therefore the parallelograms BD, EG are equiangular; and the sides about their equal angles are proportional : they are therefore similar to each other, as stated (e). For the same reason, BV is similar to HK; wherefore, also, EG is similar to HK (f), Wherefore, the parallelograms, &c.
Q. E. D Recite (a) p. 29, 1; (1) p. 34, 1; (c) p. 4, 6; (d) p. 7,5; lej def. 1, 6;
(f) p. 21, 6.
25 P. To describe a rectilineal figure which shall be similar to one and equal to another given rectilineal figure.
Given the rectilineal figures ABC and DE: it is required to describe a rectilineal figure similar to ABC, and equal to DE.
Draw AO perpendicular to BC (a); produce AO to P, so that OP be equal to half AO; through P draw FT parallel to BC (6); complete the parallelogram BT; to the straight line BF apply the parallelogram EF (c), making EF equal to DE (d); bisect the straight line CE in I (e); on the centre I describe the semicircle CGE: through B draw BG at right angles to CE (f); make BH equal to BG, and upon it describe the triangle BHK equiangular to ABC (g). BHK is equal to DE.
DE is equal to EF; for they are complements about the diameter of a parallelogram (d): but BT: EF:: BC: EB (k); and because CGE is a semicircle, BG is a mean proportional between BC and EB (i); therefore BC is to EB as the triangle ABC is to a similar triangle described upon BG (k), or upon its equal BH; that is, BHK. And since BC: EB :: BT: EF, or ABC : BHK; therefore ABC : BHK :: BT: EF (1). But ABC equals BT; for they have the same base, and the triangle has two altitudes of the parallelogram (nu); therefore BHK is equal to EF (n), or DE, as stated.
Wherefore, because BHK is equiangular to ABC, it is also similar to it (o); and it also proves equal to DE. Therefore, a rectilineal figure has been described, &c.
Q. E. F.
(d) p. 43, 1; (e) p. 10, 1; (f) p. 11, 1;
26 Th. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diaineter.
Given the parallelograms BD, EG, similar, and having the angle A common, they are about the
K same diameter.
For, it BD have for its diameter AHC, and EG have AF, let GF meet AHC in H; draw KH parallel to AD, or BC: then the parallelograms BD, KG, being about the same diameter, are similar to one another (a): wherefore AB : AD :: AK : AG
(6): but because BD, EG are similar, AB : AD :: AE : AG (e); and BO, AE : AG:: AK : AG, and AE eqnals AK (d); but it is evidently greater. Therefore AHC is not the diameter of BD; nor is any other line, that is not also the diameter of EG: therefore BD and EG are about the same diameter.
Q. E. D. Recite (a) p. 24, 6; (6) p. 1, 6;
(c) p. 11, 5; (d) p. 9,5.
1. A parallelogram is exactly applied to a straight A, line, when that line makes one of its sides; as AC to AB.
2. A parallelogram is defectively applied to a straight D line, when that line makes more than one of its sides; as AE to AB, where the defect is DC.
B 3. A parallelogram is excessively applied to a straight line, when that line makes less than one of its sides; as AG to AB, where the excess is BG.
27 Th. The greatest of all defective parallelograms applied to the same straight line, is that described upon its half; provided each of the defects be similar to the paral. lelogram described upon half the line. .
Given the straight line AB bisected in C, and the figures completed, as in the margin ; the defective parallelogram AD, whose defect is similar to CE (for it is CE), described upon half the line, is greater than AF, whose defect KH, is similar to CE, because they are about the same diameter BD (a). 1. Let AK, the base of AF, exceed
DLE O FM AC; because CF equals FE (), add KH to each; therefore the whole CH is equal
HE to the whole KE. But CH is equal to CG (c); therefore CG equals KE: to each of these add CF; therefore the parallelogram AF equals the gnomon CHL, and is therefore less than CE, or
CKB AD, as stated.
2. Let AK, the base of AF, be less than AC: now DH equals DG (c); for HM equals MG, because BC equals AC: wherefore DH exceeds LG: but DH equals DK (6); therefore DK exceeds LG: to each add AL; therefore the whole AD exceeds the whole AF, as stated.
Wherefore, the greatest of all defective, &c. Q. E. D.
28 P. To a given straight line to apply a defective parallelogram equal to a given triangle, whose defect is similar to a given parallelogram; but the given triangle must not exceed the parallelogram applied to half the given line, whose defect is similar to that of the parallelogram to be applied.
Given the straight line AB; also the triangle CC, equal to the defective parallelogram to be applied to AB, but not greater than the T parallelogram applied to half AB, and D the figure to which both defects are to be similar.
Bisect AB in E (a), and upon EB describe the parallelogram EF similar to D. (b), and complete the parallelogram AG; which, if equal to CC, is the thing required: for AG is applied to AB, defective by a parallelogram similar to D.
But if AG be not equal to CC, it is greater by hypothesis ; and so, EF exceeds CC. Make (c) the parallelogram KM equal to the excess of EF above CC, and similar to D, or EF (d); and let EG be homologous to KL, and GF to LM. Therefore ÈF, being equal to KM and CC both, is greater than KM; and its sides EG, GF exceed the sides KL, LM. Make GX equal to LK, and GO to LM, and complete the parallelogram XO; which is equal and similar to KM, and similar to EF; and EF, XO are about the same diameter (e). Draw the diameter BG and complete the scheme.
Now, because EF equals KM and CC, or XO and C, the gnomon ERO equals the part CC; but EP equals RO (f): to each add RS; the sums ER, OB are equal: but ER equals ET (8); therefore ET equals OB: to each add EP; then the sum ST equals the gnomon ERO, which proved equal to CC.
Therefore ST equals CC; and it is applied to AB, defective by a parallelogram RS, similar to D; because RS and EF are about the same diameter (h): which was to be done. Recite (a) p. 10, 1; () p. 18, 6; (c) p. 25, 6;
(d) p. 21, 6; (e) p. 26, 6; (f) p. 43, 1 ;
29 P. To a given straight line to apply an excessive parallelogram equal to a given triangle: the excess to be similar to a given parallelogram.