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figure upon the first to the similar and similarly described Book VI. figure upon the second ; therefore, as BC to CF, so is the rectilineal figure ABC to KGH: But as BC to CF, so isf f 1. 6. the parallelogram BE to the parallelogram EF: Therefore, as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF 8: And the rectilineal g 11.5.
figure ABC is equal to the parallelogram BE: therefore the rectilineal figure KGH is equal to the parallelogram EF: h 14. 5. But EF is equal to the figure D: wherefore also KGH is equal to D; and it is similar to ABC. Therefore, the rectilineal figure KGH has been described, similar to the figure ABC, and equal to D. Which was to be done.
PROP. XXVI. THEOR.
If two similar parallelograms have a common angle, and be similarly situated ; they are about the same diameter.
Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common. ABCD and AEFG are about the same diameter.
For, if not, let, if possible, the parallelogram BD have its diame- A G D ter AHC in a different straight line from AF, the diameter of the K
С the same diameter, are similar
a 24. 6. to one another 4: Wherefore as DA to AB, so is 6 GA to AK: b 1.def.6.
Book vi. But because ABCD and AEFG are similar parallelograms,
as DA is to AB, so is GA to AE; therefore c as GA to AE, cll. 5. so is GA to AK; wherefore GA has the same ratio to each of d 9. 5. the straight lines A E, AK; and consequently AK is equal
to A E, the less to the greater, which is impossible: therefore
• 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
• 2. But a parallelogram AE is said to be applied to a
* 3. And a parallelogram AG is said to be applied to a straight
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 Book VI. 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 similar to the parallelogram KH, they are
DL E about the same diameter a : Draw
a 26. 6. their diameter DB and complete the
F scheme : because the parallelogram G
Н. CF is equal to FE, add KH to both,
b 43. 1. therefore the whole CH is equal to the whole KE: But CH is equal
c 36. 1. to CG, because the base AC is equal
A to the base CB; therefore CG is
CK B 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 G F M
H construction being made, the parallelogram DH is equal to DG , for HM is equal to MG 4, 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 A KO B applied,” &c. Q. E. D.
E d 34. 1.
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 the 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 rectili-
H G OF
A E SB
L M M b 18. 6. lelogram EBFG similar 6, and similarly situated to D,
DI and complete the parallelo
K N gram 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 than c 25. 6. C. Make c the parallelogram KLMN equal to the excess of
EF above C, and similar and similarly situated to D; but D is d. 21.6. similar to EF, therefore d 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 diameter e Let GPB be their diameter, and com- e 26. 6. plete 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 equal f to XS, by adding f34. 1. SR to each, the whole OB is equal to the whole XB: But XB is equal to TE, because the base AE is equal to the base EB; 8 36. 1. 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 EF h, h 24. 6. Which was to be done.
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 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 a the parallelogram EL similar, and similarly si. a 18. 6.