AB, and deficient by parallelograms that are similar, and similarly situated to CE; AD is the greatest. 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. G D LE H First, let AK the base of AF, be greater than AC the half of AB; and because CE is similar to the parallelogram KH, they are about the same diameter (26. 6.): draw their diameter DB, and complete the scheme: because the parallelogram CF is equal (43. 1.) to FE, add KH to both, therefore the whole CH is equal to the whole KE: but CH is equal (36. 1.) to CG, because the base AC is equal to the base 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 (36. 1.), for HM is equal to MG (34. 1.) because BC is equal to CA; wherefore DH is greater than LG: but DH is equal (43. 1.) 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. Q. E. D. A CK. B G F M H L D E A KC B PROP. XXVIII. PROB. To 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 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 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 a parallelogram to the straight line AB, which shall be equal to the figure C, and be deficient from the parallelogram upon the whole line by a parallelogram similar to D. H GO F T A X R S E Σ C K N Divide AB in two equal parts (10. 1.) in the point E, and upon EB describe the parallelogram EBFG similar (18. 6.) and similarly situated 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 similar to D: but, if AG be not equal to C, it is greater than it and EF is equal to AG; therefore EF also is greater than C. Make (25. 6.) the parallelogram KLMN equal to the excess of EF above C, and similar and similarly situated to D; but D is similar to EF, therefore (21. 6.) also KM is similar to EF; let KL * See Note. be the homologous side to EG, and LM, to 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 similar to KM; but KM is similar to EF; wherefore also XO is similar to EF, and therefore XO and EF are about the same diameter (26. 6.): let GPB be their diameter, and complete the 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 (34. 1.) to XS, by adding SR to each, the whole OB is equal to the whole XB: but XB is equal (36. 1.) to TE, because the base AE is equal to the base EB; wherefore also 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, similar to the given one D, because SR is similar to EF (24. 6.). Which was to be done. PROP. XXIX. PROB. To a given straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar 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 similar. It is required to apply a parallelogram 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 (18. 6.) the parallelogram EL similar and similarly si * See Note. A a PROP. XXVIII. PROB. To 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 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 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 a parallelogram to the straight line AB, which shall be equal to the figure C, and be deficient from the parallelogram upon the whole line by a parallelogram similar to D. Divide AB in two equal parts (10. 1.) in the point E, and upon EB describe the parallelogram EBFG similar (18. 6.) and similarly situated to D, and complete the parallelogram AG, which must either be equal to C or greater than it, by the determination: and if H GO F T A X P R 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 AG; therefore EF also is greater than C. Make (25. 6.) the parallelogram KLMN equal to the excess of EF above C, and similar and similarly situated to D; but D is similar to EF, therefore (21. 6.) also KM is similar to EF; let KL * See Note. be the homologous side to EG, and LM, to 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 'similar to KM; but KM is similar to EF; wherefore also XOis similar to EF, and therefore XO and EF are about the same diameter (26. 6.): let GPB be their diameter, and complete the 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 (34. 1.) to XS, by adding SR to each, the whole OB is equal to the whole XB: but XB is equal (36. 1.) to TE, because the base AE is equal to the base EB; wherefore also 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, similar to the given one D, because SR is similar to EF (24. 6.). Which was to be done. PROP. XXIX. PROB. To a given straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar 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 similar. It is required to apply a parallelogram 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 (18. 6.) the parallelogram EL similar and similarly si See Note. A a |