H opposite sides of parallelograms are A B equal to one another (34. 1.), AB is (7. 5.) to AD, as AE to AG; and DC to CB, as GF to FE; and also CD to G DA, as FG to GA; therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportionals; and they are therefore similar to one another (1. def. 6.): for the same reason, the parallelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is similar to DB; but rectilineal figures which are similar to the same rectilineal figure are also similar to one another (21. 6.); therefore the parallelogram GE is similar to KH. Wherefore, the parallelograms, &c. Q. E. D. PROP. XXV. PROB. To describe a given rectilineal figure which shall be similar to one, and equal to another given rectilineal figure.* Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar to ABC, and equal to D. Upon the straight line BC describe (Cor. 45. 1.) the parallelogram BE equal to the figure ABC; also upon CE describe (Cor. 45. 1.) the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL; therefore BC and CF are in a straight line (29. 1. 14. 1.), as also LE and EM: between BC and CF find (13. 6.) a mean proportional GH, and upon GH describe (18. 6.) the rectilineal figure KGH similar and similarly situated to the figure ABC; and because BC is to GH as GH to CF, and if three straight lines be proportionals, as the first is to the third, so is (2. Cor. 20. 6.) the figure upon the first to the similar and similarly described figure upon the second; therefore as BC to CF, so is the rectilineal figure ABC to KGH; but as BC to CF, so is (1. 6.) the parallelogram BE to the parallelogram EF: therefore as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF (11. 5.): and the Ем rectilineal figure ABC is equal to the parallelogram BE; therefore the * Sce Note. rectilineal figure KGH is equal (14. 5.) to the parallelogram EF: but EF is equal to the figure D; wherefore also KGH is equal to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Which was to be done. PROP. XXVII. THEOR. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter. В Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common : ABCD and AEFG are about the same diameter. For, if not, let, if possible, the paral- A G lelogram BD have its diameter AHC in a different straight line from AF the diame. K ter of the parallelogram EG, and let GF meet AHC in H; and through H draw HK E parallel to AD or BC; therefore the parallelograms ABCD, AKHG being about the same diameter, they are similar to one another (24. 6.): wherefore as DA to AB, so is (1. def. 6.) GA to AK : but because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE; therefore (11. 5.) as GA to AE, so GA to AK; wherefore GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal (9. 5.) to AE, the less to the greater, which is impossible: 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. Q. E. D. • To understand the three following propositions more easily, 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 parallelogram AC is said to be applied to the straight line AB. • 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 par E C G allelogram AC described upon AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE. A D B F •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 parallelogram described upon AB in the same angle, and between the same parallels, by the same parallelogram BG.' 19 PROP. XXVII. THEOR. Of all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly 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 equal parts in C; and let the parallelogram AD be applied to the half AC, which is therefore defi cient from the parallelogram upon the whole line AB by the paral· lelogram CL upon the other half CB: of all the parallelograms applied to any other parts of 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 situat. ed 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 paral. D L E Jelogram KH, they are about the same diameter, (26. 6.): draw their diameter DB, IF 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 A c K B. each of these add CF; then the whole AF is equal to the gnomon CHL: thereforce CE, or the parallelogram AD is greater, than the parallelogram AF. Next, let AK the base of AF be less than G F M H 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 JL/VD 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. 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 HG OF 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 E SB (18. 6.) and similarly situated L par M to D, and complete the parallelogram AG, which must either be equal to C or greater than it, by the determination: Κ Ν 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 be the homologous side to EĞ, 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 * See Note. 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 defi. cient 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 situated to D: and make (25. 6.) the parallelogram GH equal to EL and C together, and similar and similarly situated to D; wherefore GH is similar to EL (21. 6.); let KH be the side homologous to FL, and KG to FE; and because the parallelogram GH is greater than EL, therefore the side 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 similar to GH, but GH is K Η similar to EL; wherefore MN is similar to EL and consequently EL and MN are about the same diameter (26. 6.): draw their diameter FX, and complete the scheme. Therefore, since GH is equal to EL and C together, and that GH is equal to MN; MN is equal * See Note. |