c. 11. 5. d. 9. 5.

Book VI. therefore as GA to AE, fo GA to AK; wherefore GA has the fame ratio to each of the straight 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. fore if two fimilar, &c. Q. E. D.

There

'To understand the three following propofitions more easily, it is to be obfervéd,

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 parallelogram 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 lefs than AB, and therefore AE is lefs

than the parallelogram AC de

fcribed upon AB in the fame

E C

G

See N.

'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.'

PROP. XXVII. THEOR.

OF all parallelograms applied to the same straight 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 others parts of AB and deficient by parallelograms

that

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.

DL E

a. 26. Ċ. Hb. 43.1.

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

G

but CH is equal to CG, because the

bafe AC is equal to the bafe CB; there

A

c 36. 1.

CK B

fore CG is equal to KE. to each of thefe add CF; then the whole AF is equal to the gnomon CHL. therefore CE or the parallelogram AD is greater than the parallelogram AF.

G F M H

L D

d. 34. f.

E

Next, Let AK the bafe of AF be less than AC, and, the fame conftruction being made, the parallellogram 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 b to DK; therefore DK is greater than LG. to each of thefe add AL; then the whole AD is greater than the whole AF. Therefore of all parallelograms applied, &c. Q. E. D.

Book VI.

See N.

PROP. XXVIII. PROB.

TO 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 ap plied 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 fi gure, to which the parallelogram to be applied is required to be equal, which figure muft not be greater than the parallelogram ap. plied 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 Jet D be the parallelogram to which this defect is required to be fimilar. It is required to apply a parallelogram to the ftraight line AB, which fhall be equal to the figure C, and be deficient from the parallelogram upon the whole line by a parallelogram fimilar to D.

H

G OF

X

P

T

PR

A

E SB

Divide AB into two equal .ro. r. parts in the point E, and upon EB defcribe the parallelogram

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 ftraight 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 4. 5. 6. AG, therefore EF alfo is greater than C. Make the parallelogram

KLMN equal to the excefs of EF above C, and fimilar and fimid...larly fituated to D; but D is fimilar to EF, therefore alfo KM is

fimilag

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 ftraight 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 also XO is fimilar to EF, and therefore XO and EF are about the fame diameter. let GPB be their diameter, and complete the scheme. then e. 26. 6. becaufe 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 bafe AE is g. 36. s. equal to the bafe EB; wherefore alfo T'E 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 parallelogram SR fimilar to the given one D, because SR is fimilar to EFh. Which was to be h. 24. 6. done.

To

PROP. XXIX. PROB.

O a given straight line to apply a parallelogram equal See N. to a given rectilineal figure, exceeding by a paral

felogram fimilar to another given.

Let AB be the given Atraight line, and C the given reti.ineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excefs of the one to be applied above that upon the given line is required to be fimilar. It is required

K

H

G

C

L M

D

B

N P

Book VI. 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.

C. 21. 6.

a

Divide AB into two equal parts in the point E, and upon EB a 18 6. defcribe the parallelogram EL fimilar and fimilarly fituated to D. b. 25. 6. and make the parallelogram GH equal to EL and C together, and fmilar and fimilarly fituated to D; wherefore GH is fimilar to EL. et KH be the fide homologous to FL, and KG to FE. and becaufe the parallelogram GH is greater than EL, therefore the side KH is greater that 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 fi-
milar to GH; but GH
is fimilar to EL;
wherefore MN is fi-
milar to EL, and con-
fequently EL and MN

are about the fame

d. 26. 6. diameter, draw their

diameter FX, and

complete the fcheme.

therefore fince GI is

K

H

N

P X

e. 36. 1.

equal to EL and C together, 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 becaufe AE is equal to EB, the parallelogram AN is equal to the parallelogram NB, that is to f. 43. 1. BMf. 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 furaight line AB there is applied the parallelogram AX equal to the given rectilineal C, exceeding by the parallelogram PO, which is g. 24. 6. fimilar to D, becaufe PO is fimilar to EL". Which was to be done.

то

PROP. XXX.

PROB.

cut a given ftraight line in extreme and mean

ratio.

Let AB be the given straight line; it is required to cut it in exfrome and mean ratio,

Upon