DC, and draw LM at right angles to GH : and because the ratio of the sides of a triangle is less than the ratio of the segments of the base, as has been shown, the ratio GK to KH is less than the ratio of GL to LH: wherefore the point L inust fall betwixt K. and H: also make as GK to KH, so GN h 19. 5 . to NK, and so shall b 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 (GH 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. IF 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. a 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. 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 spieces, be2.def. cause it is similara to AB: therefore the sides of AG are gi2. and venb; each of the straight lines AE, AF G E 24. 6. is therefore given ; and each of the straight b 60. dat. lines CA, AD is given b, therefore each of C B F D A Because the parallelogram AG is given, H in magnitude: but it is also given in species : because it is simi- 52.def. Jar a t0 AG ; therefore bits sides CA, AD are given, and each of a 24.06. the straight lines EA, AF is given ; therefore EC, DF are each b 60. dat. 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 a parallelogram similar to AB. and a 25. 6. 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 ce 26.8. 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. PROP. LXXXIII. 58. 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. Bisect AB in E; therefore EB is given in magnitude : upon EB describe a the parallelogram EF similar to DL and similarly a 18. 6. placed; therefore EF is given in species, and G H F is about the same diameterb with DL; let 6 26. 6. BCG be the diameter, and construct the K figure; therefore, because the figure EF given in species is described upon the giver straight line EB, EF is given in magnitude, A c 56. dat. and the gnomon ELH is equal d to the given figure AC : therefore e since EF is diminished by the given d 36. and gnomon ELH, the sides EK, FH of the gnomon are given ; 43. 1. but EK is equal to DC, and FH to DB; wherefore CD, DB e 82. dat: are each of them given, 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. IF 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. Bisect AB in E; therefore EB is given in magnitude ; upon a 18.6. EB describe a the parallelogram EF similar to LD, and similar ly placed ; therefore EF is given in species, and is about the b 26. 6. same diameter b with LD. Let CBG be G FH the diameter and construct the figure : therefore, because the figure EF given B in species is described upon the given D straight line EB, EF is given in magnic 56. dat. lude c, and the gnomon ELH is equal K L C d 36. and to the given figured AC; wherefore, 43. 1. since EF is encreased by the given gnomon ELH, its sides EK e 82. dar. FH are given e; but EK is equal to CD, and FH to BD; there fore 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. Draw the diameter DE of the parallelogram AC, and construct the figure. Because the parallelogram AK is equal a to a 43. 1. BC which is given, therefore AK is · E G C b 24. 6. given ; and BK is similar to AC, therefore BK is given in species. And since the parallelogram AK given in magni I HL tude is applied to the given straight line K AB, exceeding by the parallelogram BK given in species, therefore by this pro A B D position, 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. PROB. 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 - the parallelogram AKC 29.6. 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 a to AK, that is, to the given space ; and the parallelogram AC is similar b 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 А E 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 B D C is given, therefore the parallelogram AD'is given in specics; and because the given parallelogram AC is в с. EUCLID'S applied to the given straight line DC, exceeding by the paral lelogram AD given in species, the sides of the excess are gia 84. dat. ven a : therefore BD is given ; and DC is given, wherefore the whole BC is given: and AB is given, therefore AB, BC are each of them given. 85. PROP. LXXXVI. IF two straight lines contain a parallelogram given in magnitude, in a given angle ; if both of them together be given, they shall each of them be given. Let the two straight lines AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let AB, BC together be given ; each of the straight lines AB, BC is given. Produce CB, and make BD equal to AB, and complete the parallelogram ABDE. Because DB is equal to BA, and the angle ABD given, because the adjacent an E A ven straight line DC, deficient by the parallelogram AD giren a 83. dat, in species, the sides AB, BD of the defect are given a ; and DC is given, wherefore the remainder BC is given ; and each of the straight lines AB, BC is therefore given. D ᄑ 87. PROP. LXXXVII. IF two straight lines contain a parallelogram given in magnitude, in a given angle; if the excess of the square of the greater above the square of the lesser be given, each of the straight lines shall be given. Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC; if the excess of the square of BC above the square of BA be given; AB and BC are each of them given. Let the given excess of the square of BC above the square of BA be the rectangle CB, BD : take this from the square |