f 1. 6. figure upon the first to the fimilar and fimilarly defcribed fi- Book VI. gure upon the fecond; therefore as BC to CF, fo is the rectilineal figure ABC to KGH: But as BC to CF, fo is f the parallelogram BE to the parallelogram EF: Therefore as the rectilineal figure ABC is to KGH, fo is the parallelogram BE to the parallelogram EF: And the rectilineal figure ABC is equal & 11. 5. to the parallelogram BE; therefore the rectilineal figure KGH is equal to the parallelogram EF: But EF is equal to the fi- h 14. 5. gure D; wherefore alfo KGH is equal to D; and it is fimilar to ABC. Therefore the rectilineal figure KGH has been defcribed fimilar to the figure ABC, and equal to D. Which was to be done. F two fimilar parallelograms have a common angle, and be fimilarly fituated; they are about the fame diameter. Let the parallelograms ABCD, AEFG be fimilar and fimilarly fituated, and have the angle DAB common. AEFG are about the fame diameter. For, if not, let, if poffible, the parallelogram BD have its diameter AHC in a different straight line from AF the diameter of the parallelogram EG, and let GF meet AHC in H; and through H draw HK parallel to AD or BC: Therefore the parallelograms ABCD, AKHG being about the B fame diameter, they are fimilar ABCD and A G D H K E to one another: Wherefore as DA to AB, fo is GA to AK: & 24 But because ABCD and AEFG are fimilar parallelograms, M4 bl. as Book VI. C II. 5. d 9. 5. as DA is to AB, fo is GA to AE; 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. To understand the three following propofitions more eafily, it is to be observed, 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 there 'fore AE is less than the paral- E C G D B 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.' See N. PROP. XXVII. THEOR. all parallelograms applied to the fame straight line, and deficient by parallelograms fimilar and fimilarly fituated to that which is defcribed 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, AB, and deficient by parallelograms that are similar, and fimi- Book VI. larly fituated to CE, AD is the greatest. A DL E F CK B a 26. 6. H b 43. I. C 36. I. 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. First, Let AK, the bafe 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 feheme: Because the parallelogram CF is equal G b to FE, add KH to both, therefore the whole CH is equal to the whole KE: But CH is equal to CG, because the base AC is equal to the bafe CB; therefore 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. Next, Let AK the bafe of AF, be lefs than AC, and, the fame conftruction being made, the parallelogram DH is equal to DG, for HM is equal to MG4, because 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 thete add AL; then the whole AD is greater than the whole AF. Therefore of all parallelograms applied, &c. Q. E. D. G F M H L D Ed 34. 1. I. АКС В PROP. Book VI. See N. a Io. I. b 18. G. C25.6. d 21. 6. T PROP. XXVIII. P R O B. A O a given ftraight 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 flraight line, and C the given rectili neal 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 pa rallelogram to the ftraight 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. H T G OF X P R A E SB L C D K 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 muft either be e qual 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 AG; therefore EF alfo is greater C. Make the parallelogram KLMN equal to the excefs of EF above C, and fimilar and fimilarly fituated to D; but D is fimilar to EF, therefore d alfo KM is fimilar to EF: Let KL than be Book VI. n be the homologous fide to EG, and LM to GF: And because 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 ĠX 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 alfo XO is fimilar to EF, and therefore XO and EF are about the fame diameter: Let GPB be their diameter, and complete the € 26. 6. scheme: Then 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 to XS, by adding SR to each, f 34. I. the whole OB is equal to the whole XB: But XB is equal to 8 36. I. TE, because the bafe AE 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 straight line AB deficient by the parallelogram SR, fimilar to the gi ven one D, becaufe SR is fimilar to EF. Which was to be h 24, 6) done. To PRO P. XXIX. PROB. O a given straight line to apply a parallelogram e- see N. qual to a given rectilineal figure, exceeding by a parallelogram fimilar to another given. Let AB be the given ftraight line, and C the given rectili. neal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be fimilar. It is required to apply a parallelogram to the given ftraight line AB which shall be equal to the figure C, exceeding by a parallelogram fimilar to D, Divide AB into two equal parts in the point E, and upon EB describe the parallelogram EL fimilar and fimilarly fitua- 2 18. 6, a ted |