c A B DH G d. 1. 6. é. II. g. Let BC, CG be placed in a straight line, therefore DC and CE Book VI. are also in a straight line *; and complete the parallelogram DG, a. 14. 1. and, taking any straight line K, make as BC to CG, fo K to Lib 12.6. and as DC to CE, fo make b L to M. therefore the ratios of K to L, and L to M are the fame with the ratios of the fides, viz. of BC to CG, and DC to CE. But the ratio of K to M is that which is faid to be compounded of the ratios of K to L, and L to M. c. A.Def. 51 wherefore alfo K has to M, the ratio compounded of the ratios of the fides. and because as BC to CG, fo is the parallelogram AC to the parallelogram CH; but as BC to CG, fo is K to L; therefore K is e to L, as the parallelogram AC to the parallelogram CH. again, because as DC to CE, fo is the parallelogram CH to the parallelo gram CF; but as DC to CE, fo is L to M; wherefore L is to M, as the parallelogram CH to the parallelogram CF. therefore fince it has been proved that as K to L, fo is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; ex aequali f, K is to M, as the parallelo- f. 22. gə gram AC to the parallelogram CF. but K has to M the ratio which is compounded of the ratios of the fides; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the fides. rallelograms, &c. Q. E. D. KLM E F Wherefore equiangular pa E TH HE parallelograms about the diameter of any pa- See N rallelogram, are fimilar to the whole, and to one another. Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter. the parallelograms EG, HK are fimilar both to the whole parallelogram ABCD, and to one another. Becaufe DC, GF are parallels, the angle ADC is equal a to the a. 29. Të angle AGF. for the fame reason, because BC, EF are parallels, the b. 34. 1. Book VI. angle ABC is equal to the angle AEF. and each of the angles BCD, EFG is equal to the oppofite angle DAB, and therefore are equal to one another; wherefore the parallelograms ABCD, AEFG are equiangular. and because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore as AB to BC, fo is AE to EF. and because the oppofite fides of c. 4. 6. d. 7. 5. parallelograms are equal to one ano- G ther, AB is to AD, as AE to AG; D A E B H e. 1.Def. 6. portionals; and they are therefore fimilar to one another. for the fame reason, the parallelogram ABCD is fimilar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is fimilar to DB. but rectilineal figures which are fimilar to the fame rectilineal figure, are alfo fimilar to one another, therefore the parallelogram GE is fimilar to KH. Wherefore the parallelograms, &c. Q. E. D. f. 21. 6. T O describe a rectilineal figure which fhall be fimilar to one, and equal to another given rectilineal figure. Let ABC be the given rectilineal figure, to which the figure to be defcribed is required to be fimilar, and D that to which it must be equal. It is required to defcribe a rectilineal figure fimilar to ABC and equal to D. Upon the straight line BC describe a the parallelogram BE equal to the figure ABC; also upon CE describe a 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, as also LE and EM. between BC and CF find a mean proportional GH, and upon GH defcribe the rectilineal figure KGH fimilar and fimilarly fituated 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, fo is the figure upon the first to the fimilar and fimilarly described figure upon the fecond; therefore as BC to CF, fo is the Book VI. rectilineal figure ABC to KGH. but as BC to CF, so is the pa- f. 1. 6. rallelogram BE to the parallelogram EF. therefore as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EF . and the rectilineal figure ABC is equal to the pa- 5. II. 5. rallelogram BE; therefore the rectilineal figure KGH is equal h. 14. 5. to the parallelogram EF. but EF is equal to the figure 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. IF F two fimilar parallelograms have a common angle, and be fimilarly fituated; they are about the fame diameter. Let the parallelograms ABCD, AEFG be similar and similarly fituated, and have the angle DAB common. ABCD and AEFG are about the fame diameter. For if not, let, if poffible, the parallelogram BD have its diameter AHC in a different ftraight line from AF the diameter of the parallelogram EG, and let GF meet AHC in H; and thro' H draw HK parallel to AD or BC. therefore the parallelograms ABCD, AKHG being K E A G D H F B about the fame diameter, they are fimilar to one another, where- a. 24. 6. fore as DA to AB, fo is b GA to AK. but becaufe ABCD and b. 1. Def. 6. AEFG are similar parallelograms, as DA is to AB fo is GA to AE. c. ii. 5. d. 9. 5. d Book VI. therefore as GA to AE, fo GA to AK; wherefore GA has the fame ratio to each of the straight lines AE, AK; and consequently AK is equal to AE, the lefs to the greater, which is impossible. 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. See N. To understand the three following propofitions more easily, it is to be observed, 1. That a parallelogram is faid to be applied to a straight line, when it is defëribed upon it as one of its fides. Ex. gr. the pa rallelogram AC is faid to be applied to the ftraight 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 therefore AE is lefs than the parallelogram AC defcribed upon AB in the fame angle, and between the fame parallels, by the parallelogram DC; and DC is therefore called the A • defect of AE. E C G F DB 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 pa⚫rallelogram defcribed upon AB in the fame angle, and between the fame parallels, by the parallelogram BG.' OF PROP. XXVII. THEÖR. all parallelograms applied to the fame ftraight 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 greateft. 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 CÈ upon the other half CB. of all the parallelograms applied to any other parts of AB and deficient by parallelograms that are fimilar and fimilarly fituated to CE, AD is the greatest. Book VI. 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. DL E First, Let AK the base 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 scheme. because the parallelogram CF is equal 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 C A CK B H is equal to the bafe CB; therefore CG a. 26. 6. b. 43. I c. 36. I. G F M H d. 34. Ta D E Next, Let AK the base of AF be less than AC, and, the fame construction being made, the parallelogram DH is equal to DG c, for HM is equal to MG 4, becaufe 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 these add AL; then the whole AD is greater than the whole AF. Therefore of all parallelograms applied, &c. Q. E. D. AKC B M |