logram CF, is the same with the ratio which is compounded of the ratios of their sides. Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line (14. 1.); and complete the parallelogram DG; and, taking any straight line K, make (12. 6.) as BC to CG, so K to L; and as DC to CE, so make (12. 6.) L to M: Therefore, the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE. But the ratio of K to M is that which is said to be compounded (A def. 5.) of the ratios of K to L, and L to M: Wherefore also K has to M the ratio compounded of the ratios of the sides: And because as BC to CG, so is the parallelogram AC to the parallelogram CH (1. 6.) but as BC to CG, so is K to L; therefore K is (11, 5.) to L, as the parallelogram AC to the parallelogram CH: Again, be PROP. XXIV. KL M D H G E F cause as DC to CE, so is the paralle logram CH to the parallelogram CF but as DC to CE, so is L to M; wherefore L is (11. 5.) to M, as the parallelo- B gram CH to the parallelogram CF : Therefore, since 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 æquali, (22. 5.) K is to M, as the parallelogram AC to the parallelogram CF: but K has to M the ratio which is compounded of the ratios of the sides: 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. THEOR. The parallelograms about the diameter of any parallelogram, are similar 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 similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the angle ADC is equal (29. 1.) to the angle AGF: For the same reason, because BC, EF are parallels, the angle ABC is equal to the angle AEF. And each of the angles BCD, EFG is equal to the opposite angle DAB, (34.1.) 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 (4. 6.) as AB to BC, so is AE to EF: And because the oppo AE F G 73 H site sides of parallelograms are 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 DA, as FG to GA: Therefore the sides of the pa rallelograms 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 the 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. D K C PROP. XXV. PROB. To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. Let ABC be the giver 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 ĈE 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. 30. 6.) the figure upon the first to the similar and similarly described figure PROP. XXVI. THEOR. If two similar parallelograms have a common angle, and be similarly situated; they are about the same diameter. 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 ABD 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. 2. 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 parallelogram AC described upon AB 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 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 CE 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 situated to CE; AD is greater than AF. DLE 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: G (26. 6.) Draw their diameter DB, and complete the scheme: CKB PROP. XXVIII. PROB. To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficieni 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 paral lelogram. Divide AB into 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 (26. 1.) also KM is similar to EF: Let KL 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.) 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, be cause SR is similar to EF. (24. 6.) Which was to be done. PROP. XXIX. PROBLEM. 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.) 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 similar to EL; wherefore MN is similar to EL, and consequently EL and MN are about the same diameter: since GH is equal to EL and C together, and that GH is equal to MN; MN is equal to EL and C: Take away the common part EL; then the remainder, viz. the gnomon NOL, is equal to C. And because AE is equal to EB, the parallelogram AN ic equal (36. 1.) to the parallelogram NB, that is to BM. (43. 1.) Add NO to each; therefore the whole, viz. the parallelogram AX, is equal to the gnomon NOL. But the gnomon NOL is equal to C; therefore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal C, exceeding by the parallelogram PO, which is similar to D, because PO is similar to EL (24. 6.) Which was to be done. PROP. XXX. PROBLEM. To cut a given straight line in extremc and mean ratio. Let AB be the given straight line: it is required to cut it in extreme aud mean ratio. Upon AB describe (46. 1.) the square BC, and to AC apply the parallelogram CD equal to BC, exceed ing by the figure AD similar to BC: (29. 6.) But BC is a square, therefore also AD is a square; and because BC is equal, to CD, by taking the common part CE from each, the remainder BF is equal to the remainder |