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AE, the leses AE, AK, as the same cato AÉ, SO

Book VI. ABCD and AEFG are similar parallelograms, as DA is to m AB, so is GA to AE; therefore c as GA to AE, so GA to

110 3. AK; wherefore GA has the same ratio to each of the 1 9. 5. straight lines AE, AK; and consequently AK is equald to

AE, the less to the greater, which is impossible: There. fore ABCD and AKHG are not about the same diameter: wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c. Q.E.D.

• To understand the three following propositions more easily, it is to be observed,

"1. That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. Ex. gr. the parallelogram AC is said to be applied to the • straight line AB.

(2. But a parallelogram AE is said to be applied to a • straight line AB, deficient by a parallelogram, when AD

the base of AE is less than “AB, and therefore AE is

less than the parallelogram "AC described upon AB in

the same angle, and be*tween the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE.

"3. And a parallelogram AG is said to be applied to a 6 straight line AB, exceeding by a parallelogram, when AF I the base of AG is greater than AB, and therefore AG ex

ceeds AC the parallelogram described upon AB in the same angle, and between the same parallels, by the paralI lelogram BG.

PROP. XXVII. THEOR. See N. Of all parallelograms applied to the same straight

line, and deficient by parallelograms, similar and similarly situated to that which is described upon the half of the line; that which is applied to the half, and is similar to its defect, is the greatest.

Let AB be a straight line divided into two equal parts in C, and let the parallelogram AD be applied to the half AC, which is therefore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon

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the other half CB: of all the parallelograms applied to any Book VI. other parts of AB, and deficient by parallelograms that are similar and similarly situated to CE, AD is the greatest.

Let AF be any parallelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH similar and similarly situated to CE: AD is greater than AF.

First, let AK the base of AF be greater than AC the half of AB; and because CE is si

DL E milar to the parallelogram KH, they are about the same diametera:

1 26.6. Draw their diameter DB, and complete the scheme : Because the parallelogram CF is equalb to FE, add

1 6.43. 1. KH to both, therefore the whole CH is equal to the whole KE: But CH is equal to CG, because the A CK B C36. 1. base AC is equal to the base CB; therefore CG is equal to KE: To each of these add CF; then the whole AF is equal to the goomon CHL: Therefore CE, or the parallelogram AD, is greater than the parallelogram AF.

Next, Let AK the base of AF, od G EM HI be less than AC, and, the same construction being made, the pa-'mi rallelogram DH is equal to DGC,

LVD E for HM is equal to MG, because ! BC is equal to CA; wherefore DH . is greater than LG: But DH is all equalb to DK; therefore DK is greater than LG; To each of dit late these add AL; then the whole AD is greater than the whole AF. Therefore, of all parallelograms

ami A KC applied, &c. Q. E. D.

llese aud AU; then the whole

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tilineal figou be equal, wied to the line similae the paral

PROP. XXVIII. PROB. See N. To a given straight line to apply a parallelogram

equal to a given rectilineal figure, and deficient by à parallelogram similar to a given parallelogram: But the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied : that is, to the given parallelogram.

Let AB be the given straight line, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater than the parallelogram applied to the half of the line, having its defeet from that upon the whole line similar to the defect of that which is to be applied, and let D be the parallelogram to which this defect is required to be similar. It is

required to apply a
- parallelogram to the

straight line AB, which
shall be equal to the
figure C, and be defi-
cient from the parallel-

ogram upon the whole
· line by a parallelogram
similar to D.

Divide AB into two • 10. 1. equal partsa in the point

E, and upon EB de-
scribe the parallelogram

sed Khand? • 18.6. EBFG similar and similarly situated to D, and complete

the parallelogram AG, which must either be equal to C, or greater than it, by the determination: And if AG be equal to C, then what was required is already done: For, upon the straight line AB, the parallelogram AG is applied equal to the figure C, and deficient by the parallelogram EF similar to D: But, if AG be not equal to C, it is greater than

it; and EF is equal to AG; therefore EF also is greater • 25. 6. than C. Makec the parallelogram KLMN equal to the

excess of EF above C, and similar and similarly situated to

UND

D; but Das KL be there is equal to the strai

D; but D is similar to EF, therefore d also KM is similar Book VI. to EF: Let KL be the homologous side to EG, and LM to GF; and because EF is equal to C and KM together, ***

21.6. EF is greater than KM; therefore the straight line EG is greater than KL, and GF than LM: Make GX equal to LK, and GO equal to LM, and complete the parallelogram XGOP : Therefore XO is equal and similar to KM, but KM is similar to EF; wherefore also XO is similar to EF, and therefore XO and EF are about the same diametere: Let' 26. 6. GPB be their diameter, and complete the scheme: Then, because EF is equal to C and KM together, and XO a part of the one is equal to KM a part of the other, the remainder, viz, the gnomon ERO is equal to the remainder C: and because OR is equalf to XS, by adding SR to each, the r 34. 1. whole OB is equal to the whole XB: But XB is equals to 5 36. 1. TE, because the base AE is equal to the base EB; wherefore also TE is equal to OB; Add XS to each, then the whole TS is equal to the whole, viz. to the gnomon ERO: But it has been proved that the gnomon ERO is equal to C, and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB deficient by the parallelogram SR similar to the given one D, because SR is similar to EFh. Which was to be done.

194. 6.

PROP. XXIX. PROB. To a given straight line to apply a parallelogram See N. equal to a given rectilineal figure, exceeding by a parallelogram similar to another given.

Let AB be the given straight line, and the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given Ime is required to be similar. It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D.

Divide AB into two equal parts in the point E, and upona EB describe a the parallelogram EL similar and similarly 18, 6,

b95.

Book VI. situated to D: And makeb the parallelogram GH equal to

EL and C together, and similar and similarly situated to 21.6D; wherefore GH is similar to ELC: Let KH be the side

homologous to FL, and KG to FE; And because the pa-
rallelogram GH is greater than EL, therefore the side KH
is greater than FL, and KG than FE: Produce FL and
FÈ, and make FLM equal to KH, and FEN to KG, and
complete the parallelogram MN. MN is therefore equal
and similar to GH:
But GH is similar
to EL; wherefore
MN is similar to

· EL, and conse" quently EL and 3 : MN are about the a 26. 6. same diameterd:

Draw their diame-
ter FX, and com-
plete the scheme.

VB 10
DOS A EL
Therefore since
GH is equal to EL
and C together,

Ν Ρ Χ and that GH is equal to MN; MN is equal to EL and C: Take away the common part EL; then the remainder, viz.

the gnomon NOL, is equal to C. And because AE is equal • 36. 1. to EB, the parallelogram AN is equal e to the parallelogram $43. 1. NB, that is, to BMf, Add NO to each; therefore the whole,

viz. the parallelogram AX, is equal to the gnomon NOL. But the gnomon NOL is equal to C; therefore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal

C, exceeding by the parallelogram PO, which is similar to 3 24.6. D, because PO is similar to EL.. Which was to be done.

PROP. XXX. PROB. To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line, it is required to cut it in extreme and mean ratio.

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