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 consequently C. 11.5. AK is equal to AE, the less to the greater, which is impossible, d. 9. s. therefore ABCD and AKHG are not about the same diameter; where fore ABCD and AEFG must be about the fame diameter. There fore if two similar, &c. Q. E, D. 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 line, ' when it is described upon it as one of its sides. Ex. gr, the parallelogram AC is said to be applied to the straight line AB. ' 2. But a parallelogram AE is said 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 less * than the parallelogram AC de scribed upon AB in the same 'angle, and between the same pa rallels, by the parallelogram DC; and DC is therefore called the de. A D B F • feet of AE. ‘3. And a parallelogram AG is said 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 described upon AB in the same angle, and between the fame parallels, by the parallelogram BG.' See N. PROP. XXVII. THEOR. and deficient by parallelograms fimilar 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 there fore 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 similar and similarly situated to CE, AD is the greatest, Book VI. 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 D£ E parallelogram KH, they are about the fame diameter a draw their diameter a. 26. c. DB, and complete the scheme. be G H cause the parallelogram CF is equal b b. 43. : to FE, add KH to both, therefore the whole CH is equal to the whole K E. but CH i3 equal to CG, because the c 36. 1. base AC is equal to the base CB; there A CK B Next, Let AK the base of AF be less than AC, and, the same d. 34. t. is equal to CA; wherefore DH is L D E B Book VI. PROP. XXVIII. PROB. See N. O 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 sigure, 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 X P R A E SB M L K N if AG be not equal to C, it is greater than it; and EF is equal to 6.15.5. AG, therefore EF also is greater than C. Make the parallelogram KLMN equal to the excess of EF above C, and similar and simid. 31.6. larly situated to D; but D is similar to EF, therefore d also KM is similar upon С D similar to EF. let KL be the homologous, side to EG, and LM to Book VI. GP. 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 similar to KM; but KM is fimilar to EF; wherefore also XO is fimilar to EF, and therefore XO and EF are about the fame diamèter let GPB be their diameter, and complete the scheme. then c. 26. 6. because EF is equal to C and KM together, and X0 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 to XS, by adding SR to each, the whole OB is equal to f. 34. s. the whole XB. but XB is equal 8 to TE, because the base AE is g. 36. s. equal to the base EB; wherefore also 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 also TS is equal to 6. Wherefore the parallelograin 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 EFh. Which was to be h. 24. 6. done. PROP. XXIX. PROB, to a given rectilineal figure, exceeding by a paralfelogram limilar to another given. Let AB be the given Itraight line, and the K K given reti.incal figure to which the parallelogi am to be applied is required to be G equal, and D the pa C L M F rallelogram to which the excess of the one to be applied above that upon the giren line is required to be N X {imilar. It is required box k VI. to apply a parallelogram to the given straight line AB which Mall 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 b the parallelogram GH equal to EL and C together, and c. 21. 6. ímilar and similarly situated to D; wherefore GH is similar to EL'. !: KH be the fide homologous to FL, and KG to FE. and because G L M F are about the same D d. 20.6. cameterd. draw their diameter FX, and B o N Р Х e jual to EL and C to gether, 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 C. 36. 1. parallelogram AN is equal to the parallelogram NB, that is to f. 43. 1. DM, add NO to each ; therefore the whole, viz, the parallelo gram 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 firaight line AB there is applied the parallelogram AX equal to the given rectilineal C, exceeding by the parallelogram PO, which is g. 24.6. similar to D, becauie PO is funilar to EL. Which was to be done. PROP. XXX. PROB. ratio. Let AB be the give? Atright line; it is reqaired to cut it in exicine and mean ratio, Upon |