Book VI. See N. a 10. I. b 18. 6. C25.6. d 21. 6. T PROP. XXVIII. PROB. O a given ftraight line to apply a parallelogram e. qual 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, must 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 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 a pa rallelogram to the straight 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 R A E SB L M C Divide AB into two equal parts a in the point E, and upon EB describe the parallelogram EBFG fimilar b and fimilarly fituated to D, and complete the 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 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 than C. Make the parallelogram KLMN equal to the excefs of EF above C, and fimilar and fimilarly fituated to D; but Dis fimilar to EF, therefore d alfo KM is fimilar to EF: Let KL be be the homologous fide to EG, and LM to GF: And becaufe Book VI. 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 alfo XO is fimilar to EF, and therefore XO and EF are about the fame diameter: Let GPB be their diameter, and complete the e 26. 6. fcheme: 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 becaufe OR is equal to XS, by adding SR to each, the f 34. T. whole OB is equal to the whole XB: But XB is equal g to TE, 8 36. I. 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 given one D, because SR is fimilar to EF h. Which was to be done. h 24. 6. PROP. XXIX. PRO B. T "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 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 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 fhall 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 defcribe a the parallelogram EL fimilar, and fimilarly fitua a 18. 6. ted Book VI. b 25. 6. € 21.6. ¿ 26. 6. K H ted to D: And make b'the parallelogram GH equal to EL and gether, and that GH G C L M D E B N PX € 36. I. f 43. I. g24. 6. is equal to MN; MN T PROP. XXX. PROB. cut a given ftraight line in extreme and mean ratio.. Let AB be the given ftraight line; it is required to cut it in extreme and mean ratio. Upon A D EB a 46. I. b 29. 6. Upon AB defcribe a the fquare BC, and to AC apply the Book VI. parallelogram CD equal to BC, exceeding by the figure AD fimilar to BC: But BC is a fquare, therefore also AD is a fquare; and becaufe 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 equiangular, therefore their fides about the equal angles are reciprocally proportional c: 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 C BA to AE, fo is AE to EB: But AB is greater than AE; wherefore AE is C 14. 6. d 34. I. F greater than EB: Therefore the straight line AB is cut in ex- e 14. 5. treme and mean ratio in E f. Which was to be done. Otherwise, Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, fo that the rectangle contained by AB, BC be equal to the fquare of ACg: Then, because the rectangle AB, BC is e- A qual to the fquare of AC, as BA to AC, fo Св f 3. def. 6. g II. 2. is AC to CB h: Therefore AB is cut in extreme and mean ra- h 17. 6 tio in Cf. Which was to be done. PROP. XXXI. THEOR. N right angled triangles, the rectilineal figure defcri- see N. bed 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 bafe BC, the triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another 2, and a 8. 6. because b 4. 6. Book VI because the triangle ABC is fimilar to ADB, as CB to BA, fo is BA to BD b; and because these three straight lines are proportionals, as the first to the third, fo is the figure upon the firft to the fimilar, and fimilarly defcribed figure upon the fecond c: Therefore as CB to € 2. Cor. 20. 6. BD, fo is the figure upon CB to the fimilar and fimilarly defcribed figure upon BA: And, inversely d,as DB to BC, fo is the figure upon 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 e: But BD and DC together are equal to BC. Therefore the figure defcribed on BC is equal f to the fimilar and fimilarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D. PROP. XXXII. THEOR. F two triangles which have two fides of the one proportional 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 AG to DE, BC and CE are in a straight line. Because AB is parallel to |