Book VI. therefore as GA to AE, fo GA to AK; wherefore GA has the fame ratio to each of the ftraight lines AE, AK; and confequently' AK is equal to AE, the lefs to the greater, which is impoffible. therefore ABCD and AKHG are not about the fame diameter; wherefore ABCD and AEFG must be about the fame diameter. Therefore if two fimilar, &c. Q. E. D.

c. 11. 5. d. 9.5.

Sec N.

To understand the three following propofitions more eafily, it is to be obferved,

1. That a parallelogram is faid to be applied to a straight line, when it is defcribed upon it as one of its fides. Ex. gr. the pas rallelogram AC is faid to be applied to the straight line AB.

2. But a parallelogram AE is faid 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 lefs than the parallelogram AC
defcribed upon AB in the fame
'angle, and between the fame pa-
rallels, by the parallelogram DC;
and DC is therefore called the A

defect of AE.

E

C

G

DB

F

3. And a parallelogram AG is faid 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 defcribed upon AB in the fame angle, and between the fame parallels, by the parallelogram BG.'

ΟΡ

PROP. XXVII. THEOR.

Fall parallelograms applied to the fame ftraight line, and deficient by parallelograms fimilar and fimilarly fituated to that which is described upon the half of the line; that which is applied to the half, and is fimilar 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 fimilar and fimilarly fituated to CE, AD is the greateft. Book VI. Let AF be any parallelogram applied to AK any other part of AB than the half, fo as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH fimilar and fimilarly fituated to CE; AD is greater than AF.

G

DL E

H

a. 26. 6.

b. 43. I.

First, Let AK the bafe of AF be greater than AC the half of AB; and becaufe CE is fimilar to the parallelogram KH, they are about the fame diameter 2. draw their diameter DB, and complete the scheme. because the parallelogram CF is equal to FE, add KH to both, therefore the whole CH is equal to the whole KE. but CH is equal to CG, because the bafe AC is equal to the bafe 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.

A

CK B

Next, Let AK the bafe of AF be lefs than AC, and, the fame conftruction being made, the paral

lelogram DH is equal to DG, for HM is equal to MG, becaufe 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.

GFM

H

d. 34. I

D

E

AKC

B

Book VI.

See N.

a. Io. I.

b. 18. 6.

то

PROP. XXVIII. PROB.

a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram fimilar to a given parallelogram. but the given rectilineal figure to which the parallelogram to be applied is to be equal, muft not be greater than the parallelogram applied to half of the given line having its defect fimilar to the defect of that which is to be applied; that is, to the given parallelogram.

Let AB be the given ftraight line, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure muft not be greater than the parallelogram applied to the half of the line having its defect from that upon the whole line fimilar to the defect of that which is to be applied; and let D be the parallelogram to which this defect is required to be fimilar. 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 fimilar to D.

Divide AB into two equal parts in the point E, and upon EB defcribe the parallelogram EBFG fimilar and fimilarly fituated to D, and complete the parallelogram AG, which must

H

G OF

X

P

T

R

A E

SB

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 fimilar to D. but if AG be not equal to C, it is greater than it; and EF is equal to AG, therefore EF alfo is greater than C. Make the parallelogram KLMN equal to the excefs of EF above C, and fimilar and fimid. 21. 6. larly fituated to D; but D is fimilar to EF, therefore alfo KM is

c. 25. 6.

fimilar to EF. let KL be the homologous fide to EG, and LM to Book VI. GF. and becaufe 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 fimilar to EF; wherefore allo XO is fimilar to EF, and therefore XO and EF are about the fame diameter . let GPB be their diameter, and complete the fcheme. then e. 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 becaufe OR is equal to XS, by adding SR to each, the whole OB is equal to f. 34. 1. the whole XB. but XB is equal to TE, because the bate AE g. 36. í. is equal to the bafe EB; wherefore alfo 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 alfo TS is equal to C. Wherefore the parallelogram TS equal to the given rectilineal figure C, is applied to the given ftraight line AB deficient by the parallelogam SR is fimilar to the given one D, becaufe SR is fimilar to EF . Which was to be done.

T°

PROP. XXIX. PRO B.

h. 24. 6.

O a given straight line to apply a parallelograin See N. equal to a given rectilineal figure, excceding by

a parallelogram fimilar to another given.

Book VI. a parallelogram to the given ftraight line AB which fhall be equal to W the figure C, exceeding by a parallelogram fimilar to D.

a. 18. 6.

b. 25. 6.

C. 21. 6.

Divide AB into two equal parts in the point E, and upon EB defcribe the parallelogram EL fimilar and fimilarly fituated to D. and make the parallelogram GH equal to EL and C together, and fimilar and fimilarly fituated to D; wherefore GH is fimilar to EL. let KH be the fide homologous to FL, and KG to FE. and because the parallelogram GH is greater than EL, therefore the fide 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

c. 36. 1.

f. 43. 1.

8. 24. 6.

e

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 parallelogram 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 alfo 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 fimilar to EL. Which was to be done.

PRO P. XXX. PRO B.

cut a given straight line in extreme and mean

ratio.

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