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Book VI.

PROP. XXVIII.

PROB.

See N.

To

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10 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 hav. ing 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

H G OF
a parallelogram to the straight
line AB, which shall be equal
to the figure C, and be deficient
from the parallelogram upon

X P

R
the whole line by a parallelo-
gram similar to D.
Divide AB into two equal

A E SB
parts a in the point E, and upon

L M
EB describe the parallelogram
EBFG similar b and similarly

D situated to D, and complete the

KY N 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 c. 25. 6. AG, therefore EF also is greater than C. Make c the parallelogram

KLMN equal to the excess of EF above C, and similar and fimid. 25. 6. larly situated to D; but D is limilar to EF, therefore d also KM is

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b. 18. 6.

fimilar to EF. let KL be the homologous fide to EG, and LM to Book VI. GF. and because 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 fimilar to KM; but KM is similar to EF; wherefore also XO is similar to EF, and therefore XO and EF are about the fame diameter 4. let GPB be their diameter, and complete the scheme. then € 26. 6. 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 equal to the whole XB. but XB is equal 8 to TE, because the base AE

g. 36. 1. 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 EF h. Which was to be done.

f. 34. I.

h. 24. 6.

PROP. XXIX. PROB. T° O 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 K

H given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelo

C

T

L M gram to which the ex

D cess of the one to be applied above that up

E on the given line is required to be similar. It is required to apply

N P x

JB

A

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Book VI. 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 a. 18. 6. describe the parallelogram EL similar and similarly situated to D. b. 25.6. and make the parallelogram GH equal to EL and C together, and C. 21. 6. fimilar and similarly situated to D; wherefore GH is similar to ELC.

let KH be the fide 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 parallel-
ogram MN. MN is
therefore equal and

K

H
fimilar to GH; but
GH is similar to EL ;
wherefore MN is fi-
milar to EL, and con-
sequently EL and MN

C

L M
are about the same dia-
meter d. draw their dia
ameter FX, and com-

E
A

BO
plete the scheme. there-
fore fince GH is equal
to EL and C together,

N P X 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 to EB, the parallele. 36. r. ogram AN is equal to the parallelogram NB, that is to BM f.

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 fimilar to D, because PO is similar to EL 6. Which was to be done.

d. 26. 6.

A

f. 43. 1.

g. 24. 6.

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To

10 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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Upon AB describe the square BC, and to AC apply the paral- Book VI. lelogram CD equal to BC exceeding by the figure AD similar to

a. 46. 1. BC b. but BC is a square, therefore also AD is a square. and because BC is equal

D to CD, by taking the common part CE from each, the remainder BF is equal to

E the remainder AD. and these figures are

B cquiangular, therefore their fides about the equal angles are reciprocally proportional c. wherefore as FE to ED, fo AE

e. 14. 6. to EB. but FE is equal to AC d, that is

d. 34. .. to AB; and ED is equal to AE. therefore as BA to AE, fo is AE to EB. but

F AB is greater than AE; wherefore AE is greater than EBC, there. c. 14. 5. fore the straight line AB is cut in extreme and mean ratio in E f, f. 3. Def. 8. Which was to be done.

Otherwise, 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 8. then

8. II. 2 because the rectangle AB, BC is equal to the

A

B square of AC, as BA to AC, fo is AC to CB h, therefore AB is cut in extreme and mean ratio in Cf. Which

hi 37.6. was to be done.

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N right angled triangles the rectilineal figure defcrib- See N.

ed upon the side opposite to the right angle, is equal to the similar, and similarly defcribed 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 because the triangle a. 8. 6.

20. 6.

Book VI. ABC is similar to ABD, as CB to BA, fo is BA to BD b, and be

cause these three straight lines are proportionals, as the first to the b. 4.6.

third, so is the figure upon the first to the similar, and similarly

described figure upon the f. 2. Cor. second c. therefore as CB to

BD, fo is the figure upon CB
to the fimilar and similarly
described figure upon BA,
and, inversely d, as DB to
BC, so is the figure upon B

D

C BA to that upon BC. for the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore as BD and DC together to BC, so are the figures upon BA, AC to that upon BC & but BD and DC together are equal to BC. Therefore the figure described on BC is equal f to the fimilar and similarly describ ed figures on BA, AC. Wherefore in right angled triangles, &c. Q. E, D,

d. B. $.

f. A. S.

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PROP. XXXII. THEOR. See N. TF two triangles which have two sides of the one pro,

portional to two sides of the other, be joined at one angle so as to have their homologous fides parallel to one another ; the remaining sides thall be in a straight line,

Let ABC, DCE be two triangles which have the two sides BA,
AC proportional to the two CD, DE, viz. BA to AC, as CD to
DE; and let AB be parallel to DC, and AC to DE. BC and CE
are in a straight line.

Because AB is parallel to
DC, and the straight line
AC meets them, the alter-

nate angles BAC, ACD are 18. 29. 1. equal a; for the same reason,

the angle CDE is equal to.
the angle ACD; wherefore
also BAC is equal to CDE...

E
and because the triangles
ABC, DCE have one angle at A equal to one at D, and the sides
about these angles proportionals, viz. BA to AC, as CD to DE,

D

B В

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