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c 11. 5.

Book VI. rallelograms, as DA is to AB, so is GA to AE; therefore ‹ as GA to AE so GA to AK; wherefore GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal to AE, the less to the greater, which is impossible therefore 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.

à 9.5.

See Note.

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 AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE.

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3. And a parallelogram AG is said to be applied to a 'straight line AB, exceeding by a parallelogram, when AF the base of AG is greater than AB, and therefore AG exceeds AC the parallelogram described upon AB in the same angle, and between the same parallels, by the parallelogram BG.

PROP. XXVII. THEOR.

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 the other half CB of all the parallelograms applied to any other parts of

AB, and deficient by parallelograms that are similar, and si- Book VI. milarly situated to CE; AD is the greatest.

D L

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 similar to the E parallelogram KH, they are about the same diameter: draw their diameter DB, and complete the scheme: because the parallelogram CF is equal G to FE, add KH to both, therefore the whole CH is equal to the whole KE: but CH is equal to CG, because the base AC is equal to the basc CB: therefore CG is equal to KE: to each of these add CF; then the whole AF is equal to the gnomon CHL: therefore CE, or the parallelogram AD, is greater than the parallelogram AF.

Next, let AK the base of AF, be less than AC, and, the same construction being made, the parallelogram DH is equal to DG, for HM is equal to MG, because BC is equal to CA; wherefore DH is greater than LG: but DH is equal to DK; therefore DK is greater than LG: to each of these add AL; then the whole AD is greater than the whole AF. Therefore of all parallelograms applied, &c. Q. E. D.

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PROP. XXVIII. PROB.

To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a 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 defect 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 deficient from the parallelogram upon the whole line by a parallelogram similar to D.

Divide AB into two equal parts in the point E, and upon EB describe the parallelogram EBFG similarb and similarly situated to D, and complete the parallelogram AG, which must either be equal to C,4

or

greater than it, by

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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 than
C. Make the parallelogram KLMN equal to the excess of
EF above C, and similar and similarly situated to D; but D is
similar to EF, therefore also KM is similar to EF: let KL

be the homologous side to EG, and LM to GF: and because Book VI, EF is equal to C and KM together, 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 GPB be their diameter, and complete the e 26.6. 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 equal to XS, by adding SR to each, the f 34. 1. whole OB is equal to the whole XB: but XB is equals to TE, g 36. 1. 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. h 24. 6.

PROP. XXIX. PROB.

TO a given straight line to apply a parallelogram See Note. equal to a given rectilineal figure, exceeding by a parallelogram similar to another given.

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, and D the parallelogram to which the excess of the one to be applied above that upon the given line 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 upon EB describe the parallelogram EL similar and similarly situa- a 18. 6.

b 25. 6. c' 21, 6.

Book VI. ted to D: and make b the parallelogram GH equal to EL and. C together, and similar and similarly situated to D; wherefore GH is similar to EL; let KH be the side homologous to FL, and KG to FE: and because the parallelogram GH is greater than EL, therefore the side KH is greater than FL, and KG than FE: produce FL and FE, 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 consequently EL

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and MN are about
the same diameter:
draw their diameter
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the scheme. There-
fore, since GH is e-
qual to EL and C to-

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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 be-
cause AE is equal to EB, the parallelogram AN, is equale to
the parallelogram NB, that is, to BM. 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 rec-
tilineal C, exceeding by the parallelogram PO, which is simi-
lar to D, because PO is similar to ELs. 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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