a b 26. 6. 43. 1. Bisect AB in E; therefore EB is given in magnitude ; a 18.6. upon EB describe the parallelogram EF similar to LD, and . similarly placed; therefore EF is given in species, and is G F H the figure EF, given in species, is de E B şcribed upon the given straight line A D e 56. dat. EB, EF is given in magnitude, and the gnomon ELH is equal to the K L C d 36. and given figured AC; wherefore, since EF is increased by the given gnomon ELH, its sides EK, e 82. dat. FH are given; 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 wbich 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 cona 43. I. struct the figure: Because the parallelogram AK is equala to BC, which is given, therefore AK is b b 24. 6. given; and BK is similar b. to AC, therefore BK is given in species. And E С G F K rallelogram BK, given in species, therefore by this proposition, BD, DK the A B D 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 c AK equal to the given space, exceeding by the parallelogram c 29. 6. a 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 a 43. 1. given space; and the parallelogram AC is similar to BK; b 24. 6. 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 is given, each of them is 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. Com E plete the parallelogram AD; and because A 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 because the given paral- D B C lelogram AC is applied to the given straight line DC, exceeding by the parallelogram AD given in species, the sides of the excess are givena, 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. a 84. dat. If two straight lines contain a parallelogram given in magnitude, in a given angle ; if both of them together are given, each of them is 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 BA, and complete the parallelogram ABDE. Because DB is equal to BA, and the A E the parallelogram AD, given in species, the sides AB, BD of a 83. dat. the defect are given a ; and DC is given, wherefore the re mainder BC is given: and each of the straight lines AB, BC is therefore given. 87. a 2. 2. 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 less is given, each of the straight lines is 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 of BC, the remainder, which is a the rectangle BC, CD is equal to the square of AB; and because the angle ABC of the parallelogram AC is given, the ratio of the rectangle of b 62. dat. the sides AB, BC to the parallelogram AC is given ; and AC is given, therefore the rectangle AB, BC is given; and the rectangle CB, BD is given; therefore the ratio of the rectcl. 6. angle CB, BD to the rectangle AB, BC, that is “, the ratio of d 54. dat. the straight line DB to BA is given; therefore d the ratio of the square of DB to the square of BA is А. В. PD с e 7. dat. square of BD; and, by composition, the ratio of four times . the rectangle BC, CD together with the square of BD to the square of BD is given : But four times the rectangle BC, CD. f 8. 2. together with the square of BD, is equal to the square of the straight lines BC, CD taken together: therefore the ratio EUCLID'S Data. of the squares of BC, CD together to the square of BD, is The preceding demonstration is the analysis of this problem, viz. A parallelogram AC, which has a given angle ABC, being given in magnitude, and the excess of the square of BC one of its sides above the square of the other BA being given ; to find the sides : And the composition is as follows. Let EFG be the given angle, to which the angle ABC is required to be equal, and from any point E in FE, draw EG perpendicular to FG; let the rectangle EG, GH be the given space M HN Because, as CB to BD, so is OG to GL, the square of CB d 8. 2. But the square of BC and CD together, is equal d to four times the rectangle BC, CD together with the square of BD; therefore four times the rectangle BC, CD together with the square of BD is to the square of BD, as the square of ML to the square of LG: And, by division, four times the rectangle BC, CD is to the square of BD, as the square of MG to the square of GL; wherefore the rectangle BC, CD is to the square of BD as (the square of KG the half of MG to the square of GL, that is, as) the square of AB to the square of M A K B PDC F G L O HN BD, because as LG to GK, so DB was made to BA: Thereb 14. 5. fore, the rectangle BC, CD is equal to the square of AB. To each of these add the rectangle CB, BD, and the square N. PROP. LXXXVIII. If two straight lines contain a parallelogram, given in magnitude, in a given angle; if the sum of the squares of its sides are given, each of the sides is given. Let the two straight lines AB, BC contain the parallelogram ABCD, given in magnitude, in the given angle ABC, and let |