a. 14. Let BC, CG be placed in a straight line, therefore DC and CE Book VL are also in a straight line *; and complete the parallelogram DG, and, taking any straight line K, make bas BC to CG, so K to L; 6 126. 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 said to be compounded of the ratios of K to L, and L to M. C. A. Def.si wherefore also K has to M, the A DH ratio compounded of the ratios of the fides. and because as BC to CG, so is the parallelogram AC to the parallelogram CHd, but as BC B d. 1. 6. to CG, so is K to L; therefore K is e to L, as the parallelogram AC €. II. $. to the parallelogram CH. again, because as DC to CE, fo is the patallelogram CH to the parallelo KLM E F gram CF; but as DC to CE, fo is L to M; wherefore L is e to M, as the parallelogram CH to the parallelogram CF. therefore fince it has been proved that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; ex aequali', K is to M, as the parallelo- f. 42. go 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 sides. Wherefore equiangular parallelograms, &c. Q. E. D. A E THE parallelograms about the diameter of any pa. Šée N. rallelogram, are similar to the whole, and to one another. THE Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter. the parallelograms EG, HK are similar 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. I. angle AGF. for the same reason, because BC, EF are parallels, the G d. 7. 5. Book VI, angle ABC is equal to the angle AEF. and each of the angles BCD, EFG is equal to the opposite angle DAB , and therefore are b. 34. 1. 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, EAY, they are equiangular to one another; C. 4. 6. therefore as AB to BC, fo is AE to Α. Ε B EF. and because the opposite fides of parallelograms are equal to one ano F H ther , AB is to AD, as AE to AG; and DC to CB, as GF to FE; and also CD to DA, as FG to GA. therefore the sides of the parallelograms ABCD, AEFG about the equal angles are pro DK e. 1. Def. 6. portionals; and they are therefore similar to one another e. for the same reason, the parallelogram ABCD is similar to the parallelo FHCK. Wherefore each of the parallelograms GE, KH is fimilar to DB. but rectilineal figures which are similar to the f. 21. 6. fame rectilineal figure, are also similar to one another f, therefore the parallelogram GE is similar to KH. Wherefore the parallelo Q. E. D. gram FHCK. grams, &c. PROP. XXV. PROB. See N. 10 describe a rectilineal figure which shall be fimi lar 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. a. Cor.45.1. 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. 114. 1. therefore BC and CF are in a straight line 6, as also LE and EM. C. 13. 6. between BC and CF find ca mean proportional GH, and upon GH described the rectilineal figure KGH similar and fimilarly situated to the figure ABC. and because BC is to GH, as GH to e. 2.Cor.20. CF, and if three straight lines be proportionals, as the firft is to the third, so is the figure upon the first to the similar and fimilarly b. Ş 29. 1. d. 18. 6. 6. f. 1. 6. described figure upon the second; therefore as BC to CF, so is the Book VI. rectilineal figure ABC to KGH. but as BC to CF, fo is f 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 s, and the rectilineal figure ABC is equal to the pa- 5. 11. So А D К. B G H M rallelogram BE; therefore the rectilineal figure KGH is equal 1 h. 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. F two similar parallelograms have a common angle, and be similarly situated; they are about the same diameter. 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 G parallelogram BD have its diame D ter AHC in a different straight line from AF the diameter of the paral K к H E lelogram EG, and let GF meet AHC FI in H; and thro' H draw HK parallel to AD or BC. therefore the parallelograms ABCD, AKHG being B! c about the same diameter, they are similar to one another 2. where- 2. 24. 6. fore as DA to AB, so is o GA to AK. but because ABCD and b. I. Def. 6. ALFG are similar parallelograms, as DA is to AB so is GA to AE. Book VI. therefore e as GA to AE, fo GA to AK; wherefore GA has the Kapitel fame ratio to each of the straight lines AE, AK; and consequently c. ii. s. AK is equal d to AE, the less to the greater, which is impossible. therefore ABCD and AKHG are riot about the fame diameter ; wherefore ABCD and AEFG must be about the same diameter. Therefore if two fimilar, &c. Q. E. D. gr: the pas • To understand the three following propofitions more easily, it * 1. That a parallelogram is said to be applied to a straight line, • 2. But a parallelogram AE is said to be applied to a straight E C G described upon AB in the same • angle, and between the same pa rallels, by the parallelogram DC; 6 and DC is therefore called the A D B F 6 defect of AE. 3. And a parallelograr 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 described upon AB in the fame angle, and between the same parallels, by the parallelogram BG.' PROP. XXVII. THÉO Ř. See No F all parallelograms applied to the fame straight line, and deficient by parallelograms fimilar and similarly situated to that which is described upon the half of the line ; that which is applied to the half, and is fimilar to its defea, 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 similar and similarly situated to CE, AD is the greatest. Book VI. 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 fimilarly situated 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 D L E parallelogram KH, they are about the fame diameter draw their diameter a. 26. 6. DB, and complete the scheme. because G H н the parallelogram CF is equal to FE, b. 43. It 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. 36. I А CK B 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 C, for G F M H HM is equal to MG 4, because BC d. 34. 11 is equal to CA; wherefore DH is L D greater than LG. but DH is equal b E 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 KC B M |