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AB, and deficient by parallelograms that are similar, and simi- Book VI. larly situated to CE, AD is the greatest.
Let AF be any parallelogram applied to AK, any other part of
First, let AK the base of AF, be greater than AC the half of
D L E
a 26. 6. DB, and complete the scheme: be
G cause the parallelogram CF is equal b
H b 43. 1. to FE, add KH to both, therefore the whole CH is equal to the whole KE: but CH is equal to CG, because the
c 36. 1. base AC is equal to the base CB ;
A ск B
Next, let AK the base of AF, be less G F M H
d 34.1. cause BC is equal to CA; wherefore DH is greater than LG: but DH is equal b 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.
А к с B
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
HG G OF
А b 18. 6. lelogram EBFG similar b and
E SB similarly situated to D, and
K к N 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 complete the scheme : e 26. 6. 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 SR to each, the whole OB is f 34. 1. equal to the whole XB : but XB is equal 8 to TE, because the g 36, 1.. 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 gno, mon 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. Which was to be done.
h 24. 6.
PROP. XXIX. PROB.
TO a given straight line to apply a parallelogram Şee 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 tlie 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 situa- a 18, 6:
d 26. 6. same
Book VI. ted to D: and make b the parallelogram GH equal to EL and a C together, and similar and similarly situated to D; wherefore b 25. 6. GH is similar to EL<: let KH be the side homologous to FL, € 21. 6. and KG 10 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
N P X
then the remainder, viz. the gnomon NOL, is equal to C. And e 36. 1. because AE is equal to EB, the parallelogram AN is equal e to f 43. 1. the parallelogram NB, that is, to BMf. Add NO to each : there
fore 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 D, be8 24. 6. cause PO is similar to EL %. Which was to be done.
TO cut a given straight line in extreme and mcan ratio.
Let AB be the given straight line ; it is required to cut it in extreme and mean ratio.
Upon AB describe a the square BC, and to AC apply the Book VI: parallelogram CD equal to BC, exceeding by the figure AD si. milar to BC b; but BC is a square,
a 46. 1.
ID therefore also AD is a square; and be
b 29. 6. cause BC is equal to CD, by taking the common part CE from each, the re
A mainder BF is equal to the remainder
B AD: and these figures are equiangular, therefore their sides about the equal
c 14. 6. angles are reciprocally proportional c : wherefore, as FE to ED, so AE to EB :
d 34. 1. but FE is equal to AC4, that is, to AB; and ED is equal to AE: therefore, as
F BA to AE, so is AE 10 EB: but AB is greater than AE; wherefore AE is greater than EB •; therefore e 14. 5. the straight line AB is cut in extreme and mean ratio in E f. f 3. def. Which was to be done.
Let AB be the given straight line; it is required to cut it in extreme and mean ratio.
Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC&. Then,
g 11. 2. because the rectangle AB, BC is equal to the
А C B square of AC, as BA to AC, so is AC to CBh:
h 17. 6. therefore AB is cut in extreme and mean ratio in Cf. Which was to be done.
PROP. XXXI. THEOR.
IN right angled triangles, the rectilineal figure des- See N: cribed upon the side opposite to the right angle is equal to the similar and similarly described figures upon the sides containing the right angle.
Let ABC be a right angled triangle, having the right angle BAC: the rectilineal figure described upon BC is equal to the similar and similarly described figures upon BA, AC.
Draw the perpendicular AD; therefore, because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one another, and a 8. 6.