G M KLH N en; wherefore (9. dat.) the ratio of BE to EC, as also (cor. 6. dat.) the ratio of EC to CB is given: and the ratio of BC to CD is given (7. dat.), because the ratio of BD to DC is given; therefore (9. dat.) the ratio of EC to CD is given, and consequently (7. dat.) the ratio of DE to EC: and the ratio of EC to EA was shown to be given, therefore (9. dat.) the ratio of DE to EA is given; and ADE is a right angle, wherefore (46. dat.) the triangle ADE is given in species, and the angle AED given: and the ratio of CE to EA is given, therefore (44. dat.) the triangle AEC is given in species, and consequently the angle ACE is given, as also the adjacent angle ACB. In the same manner, because the ratio of BE to EA is given, the triangle BEA is given in species, and the angle ABE is therefore given: and the angle ACB is given; wherefore the triangle ABC is given (43. dat.) in species. But the ratio of the greater side BA to the other AC must be less than the ratio of the greater segment BD to DC: because the square of BA is to the square of AC, as the squares of BD, DA to the squares of DC, DA; and the squares of BD, DA have to the squares of DC, DA a less ratio than the square of BD has to the square of DC, because the square of BD is greater than the square of DC; therefore the square of BA has to the square of AC a less ratio than the square of BD has to that of DC: and consequently the ratio of BA to AC is less than the ratio of BD to DC. This being premised, a triangle which shall have the things mentioned to be given in the proposition, and to which the triangle ABC is similar, may be found thus: take a straight line GH given in position and magnitude, and divide it in K, so that the ratio of GK to KH may be the same with the given ratio of BA to AC: divide also GH in L, so that the ratio of GL to LH may be the same with the given ratio of BD to DC, and draw If A be greater than B, and C any third magnitude; then A and C together have to B and C together a less ratio than A has to B. Let A be to B as C to D, and because A is greater than B, C is greater than D: but as A is to B, so A and C to B and D; and A and C have to B and C a less ratio than A and C have to B and D, because C is greater than D, therefore A and C have to B and C a less ratio than A to B. LM at right angles to GH: and because the ratio of the side of a triangle is less than the ratio of the segments of the base, as has been shown, the ratio of GK to KH is less than the ratio of GL to LH; wherefore the point L must fall between K and H : also make as GK to KH; so GN to NK, and so shall (19. 5.) NK be to NH. And from the centre N, at the distance NK, describe a circle, and let its circumference meet LM in O, and join OG, OH; then OGH is the triangle which was to be described because GN is to NK, or NO, as NO to NH, the triangle OGN is equiangular to HON; therefore as OG to GN, so is HO to ON, and by permutation, as GO to OH, so is GN to NO, or NK, that is, as GK to KH, that is, in the given ratio of the sides; and by the construction, GL, LH have to one another the given ratio of the segments of the base. Ir a parallelogram given in species and magnitude be increased or diminished by a gnomon given in magnitude, the sides of the gnomon are given in magnitude. First, let the parallelogram AB given in species and magnitude be increased by the given gnomon ECBDFG, each of the straight lines CE, DF is given. GA Because AB is given in species and magnitude, and that the gnomon ECBDFG is given, therefore the whole space AG is given in magnitude: but AG is also given in species, because it is similar (2. def. 2. and 24. 6.) to AB; therefore the sides of AG are given (60. dat.): each of the straight lines AE, AF is therefore given; and each of the straight lines CA, AD is given (60. dat.), therefore each of the remainders EC, DF is given (4. dat.). Next, let the parallelogram AG given in species and magnitude, be diminished by the given gnomon ECBDFG, each of the straight lines ČE, DF is given. E B C FD A H Because the parallelogram AG is given, as also its gnomon ECBDFG, the remaining space AB is given in magnitude: but it is also given in species: because it is similar (2. def. 2. and 24. 6.) to AG; therefore (60. dat.) its sides CA, AD are given, and each of the straight lines EA, AF is given; therefore EC, DF are each of them given. The gnomon and its sides CE, DF may be found thus in the first case. Let H be the given space to which the gnomon must be made equal, and find (25. 6.) a parallelogram similar to AB and equal to the figures AB and H together, and place its sides AE, AF from the point A, upon the straight lines AC, AD, and complete the parallelogram AG which is about the same diameter (26. 6.) with AB; because therefore AG is equal to both AB and H, take away the common part AB, the remaining gnomon ECBDFG is equal to the remaining figure H; therefore a gnomon equal to H, and its sides CE, DF are found and in like manner they may be found in the other case, in which the given figure H must be less than the figure FE from which it is to be taken. If a parallelogram equal to a given space be applied to a given straight line, deficient by a parallelogram given in species, the sides of the defect are given. Let the parallelogram AC equal to a given space be applied to the given straight line AB, deficient by the parallelogram BDCL given in species; each of the straight lines CD, DB are given. G HF K CL Bisect AB in E; therefore EB is given in magnitude: upon EB describe (18. 6.) the parallelogram EF similar to DL and similarly placed; therefore EF is given in species, and is about the same diameter (26. 6.) with DL; let BCG be the diameter, and construct the figure; therefore, because the figure EF given in species is described upon the given straight line EB, EF is given (56. A dat.) in magnitude, and the gnomon ELH is equal (36. and 43. 1.) to the given figure AC: therefore (82. dat.) since EF is diminished by the given gnomon ELH, the sides EK, FH of the gnomon are given; but EK is equal to DC, and FH to DB; wherefore CD, DB are each of them given. ED B This demonstration is the analysis of the problem in the 28th prop. of book 6, the construction and demonstration of which proposition is the composition of the analysis; and because the given space AC or its equal the gnomon ELH is to be taken from the figure EF described upon the half of AB similar to BC, therefore AC must not be greater than EF, as is shown in the 27th prop. B. 6. : 59. PROP. LXXXIV. Ir a parallelogram equal to a given space be applied to a given straight line, exceeding by a parallelogram given in species; the sides of the excess are given. Let the parallelogram AC equal to a given space be applied to the given straight line AB, exceeding by the parallelogram BDCL given in species; each of the straight lines CD, DB are given. G FH Bisect AB in E; therefore EB is given in magnitude: upon EB describe (18. 6.) the parallelogram EF similar to LD, and similarly placed; therefore EF is given in species, and is about the same diameter (26. 6.) with LD. Let CBG be the diameter, and construct the figure therefore, because the figure EF given in species is described upon the given A straight line EB, EF is given in magnitude, (56. dat.) and the gnomon ELH is equal to B D KL C the given figure (36. dat. 43. 1.) AC; wherefore, since EF is increased by the given gnomon ELH, its sides EK, FH are given (82. dat.); but EK is equal to CD, and FH to BD; therefore CD, DB are each of them given. This demonstration is the analysis of the problem in the 29th prop. book 6, the construction and demonstration of which is the composition of the analysis. COR. If a parallelogram given in species be applied to a given straight line, exceeding by a parallelogram equal to a given space; the sides of the parallelogram are given. Let the parallelogram ADCE given in species be applied to the given straight line AB, exceeding by the parallelogram BDCG equal to a given space; the sides AD, DC of the parallelogram are given. E G C F K B D Draw the diameter DE of the parallelogram AC, and construct the figure. Because the parallelogram AK is equal (43. 1.) to BC which is given, therefore AK is given; and BK is similar (24. 6.) to AC, therefore BK is given in species. And since the parallelogram AK given in magnitude is applied to the given straight line AB, exceeding by the parallelogram BK given in species, therefore, by this proposition, BD, DK the sides of the excess are given, and the straight line AB is given; therefore the whole AD, as also DC, to which it has a given ratio, is given. PROBLEM. H A To apply a parallelogram similar to a given one to a given straight line AB, exceeding by a parallelogram equal to a given space. To the given straight line AB apply (29. 6.) the parallelogram AK equal to the given space, exceeding by the parallelogram BK similar to the one given. Draw DF, the diameter of BK, and through the point A draw AE parallel to BF, meeting DF produced in E, and complete the parallelogram AC. The parallelogram BC is equal (43. 1.) to AK, that is, to the given space; and the parallelogram AC is similar (24. 6.) to BK; therefore the parallelogram AC is applied to the straight line AB similar to the one given, and exceeding by the parallelogram BC which is equal to the given space. If two straight lines contain a parallelogram given in magnitude, in a given angle; if the difference of the straight lines be given, they shall each of them be given. Let AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let the excess of BC above AB be given; each of the straight lines AB, BC is given. Let DC be the given excess of BC above BA, therefore the remainder BD is equal to BA. Complete the parallelogram AD; and because AB is equal to BD, the ratio of AB to BD is given; and the angle ABD is given, therefore the parallelogram AD is given in species; and A E B D C |