Book VI. See N. 8. 10. I. b. 18. 6. To equal PROP. XXVIII. PROB. a given straight line to apply a parallelogram 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 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 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 H a parallelogram to the straight a 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 G OF 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 c. 25. 6. AG, therefore EF alfo is greater than C. Make the parallelogram KLMN equal to the excess of EF above C, and fimilar and fimi d. 21. 6. larly fituated to D; but D is fimilar to EF, therefore alfo KM is fimilar 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 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 alfo 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. 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 whole OB is equal to the whole XB. but XB is equal to TE, becaufe the base AE 8. 36. I, is equal to the base 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 also 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 given one D, because SR is fimilar to EF h. Which was to be done. f. 34. I. g. h. 24. 6. O a given straight line to apply a parallelogram See N. equal to a given rectilineal figure, exceeding by a parallelogram fimilar to another given. 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, 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 to apply K H C L M F D A E BO N P X Book VI. a parallelogram to the given straight line AB which shall be equal to 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 ELc. 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 e. 36. I. f. 43. I. 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. то 10 cut a given ftraight line in extreme and mean ratio. Let AB be the given straight line; it is required to cut it in extreme and mean ratio. a. 46. 1. b. 29. 6. D E B Upon AB describe the square BC, and to AC apply the paral- Book VI. lelogram CD equal to BC exceeding by the figure AD fimilar to BC . but BC is a fquare, therefore also AD is a fquare. and because BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD. and these figures are A equiangular, therefore their fides about the equal angles are reciprocally proportional. wherefore as FE to ED, fo AE to EB. but FE is equal to AC d, that is to AB; and ED is equal to AE. therefore as BA to AE, fo is AE to EB. but F e. 14. 6. d. 34. I. AB is greater than AE; wherefore AE is greater than EB. there- e. 14. 5. fore the straight line AB is cut in extreme and mean ratio in E f. f. 3. Def. 6. Which was to be done. Otherwise, Let AB be the given straight line; it is required to cut it in extreme and mean ratio. A g. 11. 2. с в Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC 8. then because the rectangle AB, BC is equal to the fquare of AC, as BA to AC, fo is AC to CBh. therefore AB is cut in extreme and mean ratio in Cf. Which ha 17. 6. was to be done. PROP. XXXI. THEOR. 37.6. N right angled triangles the rectilineal figure defcrib- See N. ed upon the fide oppofite to the right angle, is equal to the fimilar, and fimilarly defcribed figures upon the fides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC. the rectilineal figure defcribed upon BC is equal to the fimilar and fimilarly defcribed figures upon BA, AC. Draw the perpendicular AD; therefore because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another, and because the triangle a. 8. 6 Book VI. b. 4. 6. ABC is fimilar to ABD, as CB to BA, fo is BA to BD b. and be cause these three straight lines are proportionals, as the first to the third, fo is the figure upon the first to the fimilar, and fimilarly defcribed figure upon the c. 2. Cor. fecond. therefore as CB to 20. 6. BD, fo is the figure upon CB to the fimilar and fimilarly defcribed figure upon BA. d. B. 5. and, inversely, as DB to e. 24. 5. f. A. 5. See N. BC, fo is the figure upon B BA to that upon BC. for the fame reason, as DC to CB, fo is the figure upon CA to that upon CB. Wherefore as BD and DC together to BC, fo are the figures upon BA, AC to that upon BC. but BD and DC together, are equal to BC. Therefore the figure described on BC is equal f to the fimilar and fimilarly defcrib ed figures on BA, AC. Wherefore in right angled triangles, &c. Q. E. D. I two triangles which have two fides of the one pro. portional to two fides of the other, be joined at one angle fo as to have their homologous fides parallel to one another; the remaining fides fhall be in a straight line. Let ABC, DCE be two triangles which have the two fides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE. BC and CE are in a straight line. ABC, DCE have one angle at A equal to one at D, and the fides about thefe angles proportionals, viz. BA to AC, as CD to DE, |