Book VI. therefore c as GA to AE, fo GA to AK; wherefore GA has the vlame ratio to each of the straight lines AE, AK; and consequently cji.s. AK is equal to AE, the less to the greater, which is impoflibie. therefore ABCD and AKHG are not about the same diameter ; wherefore ABCD and AEFG must be about the same diameter. Therefore if two similar, &c. 0. E. D. d. 9.5 • To understand the three following propofitions more eally, 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 pa• rallelogram 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 Е С G described upon AB in the same angle, and between the fame parallels, by the parallelogram DC; and DC is therefore called the A D B F defect 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 pa6 rallelogram described upon AB in the fame angle, and between the fame parallels, by the parallelogram BG. PRO P. XXVII. THEO R. Scc N. O Fall parallelograms applied to the same straight line, and deficient by parallelograms similar and simnilarly 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 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 a. 26. 6. c. 36. ,, 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, fo as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH fimilar 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 DL E parallelogram KH, they are about the fame diameter 2 draw their diameter DB, and complete the scheme. because G H the parallelogram CF is equal to FE, b. 43. 1. add KH to both, therefore the whole CH is equal to the whole KE. but CH is equal to CG, because the base AC А. CK B 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. Next, Let AK the base of AF be less than AC, and, the same construction being made, the parallelogram DH is equal to DG“, for G F M Н. HM is equal to MG4, because BC is equal to CA ; wherefore DH is LI VD greater than LG. but DUI is equal 5 E 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. 0. E. D. A KO B d. 34 Book VI. PROP. XXVIIT. PROB. See N. To 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 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 wliole 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 G OF P R A E SB M c D fituated to D, and complete the K 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. 15. 6. AG, therefore EF also is greater than C. Make the parallelogram KLMN equal to the excess of EF above C, and similar and fimid. 21.6. larly situated to D; but D is similar to EF, therefore d allo KM is a 10. I. b. 16. 6. similar 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 fimilar to EF; wherefore alo XO is similar to EF, and therefore XO and EF are about the same diameter let GPB be their diameter, and complete the scheme. 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 because OR is equal f to XS, by adding SR to each, the vihole OB is equal to f. 34. 1. the whole Xb. but XB is equal to TE, because the bate AE 8:36. í. is equal to the base EB; wherefore also TE is equal to OB. adu XS to cach, 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 allo TS is equal to C. Wherefore the parallelogram TS equal to the giren rectilineal figure C, is applied to the given straight line AB deficient by the parallelogʻam SR is similar to the given one D, because SR is similar to EF). Which was to be done. h. 24. 6. To. PROP. XXIX. PROB. equal to a given rectilincal 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 L M C gram to which the excess of the cne to be applied above that upon the given line is required to be similar. It is required to apply N I 고 D a. 18. 6. b. 25. 6. C. 21. 6. Book VI. a parallclogram 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 the parallelogram EL similar and similarly situated to D. and make the parallelogram GH equal to EL and C together, and fimilar and funilarly situated to D; wherefore GH is similar to EL, let KH be the side homologous to FL, and KG to FE. and because the parallelogram GH is greater than EL, therefore the Gde 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 K H с L M are about the fame diad. 26. 6. meter d. draw their diameter FX, and com E А 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 parallelc. 36. I. ogram AN is equal o to the parallelogram NB, that is to BM f. f. 43. 1. 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 similar to D, because PO is similar to EL. Which was to be done. 8. 14. 6. PRO P. XXX. P R O B. "O cut a given straight linc in extreme and mean ratio. Let AB be the given straight line ; it is required to cut it in extreme and mean ratio. То |