The Elements of Euclid

GH F

milarly 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. 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.

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.

K L C

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 straight line EB, EF is given in magnitude, (56. dat.) and the gnomon ELH is equal to 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

Draw the diameter DE of the parallelogram AC, and construct

G C

་

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.

B D

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.

Ir 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.

D C

Let DC be the given excess of BC above BA, A 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 because the given parallelogram AC is applied to the given straight line DC, exceeding by the parallelogram AD given in species, the sides of the excess are given (84. dat.): 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.

Ir 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 angle ABC is given, the parallelogram AD is given in species; and because AB, BC together are given, and AB is equal to BD; therefore DC is given; and because the given parallelogram AC is applied to the given straight line DC, deficient by the parallelogram AD given in species, the sides AB, BD of the defect are given (83. dat.); and DC is given, wherefore the remainder BC is given; and each of the straight lines AB, BC is therefore given.

PROP. LXXXVII.

B C

87.

Ir 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 of BC; the remainder, which is (2. 2.) 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 the sides AB, BC to the parallelogram AC is given (62. dat.); and AC is given, therefore the rectangle AB, BC is given; and the rectangle CB BD is given; therefore the ratio of the rectangle CB, BD to the rectangle AB, BC, that is (1. 6.), the ratio of the straight line DB to BA is given; therefore (54. dat.) the ratio of the square of DB to the square of BA is given: and the square of BA is equal to the rectangle BC, CD: wherefore the ratio of the rectangle BC, CD to the square of BD is given, as also the ratio of four times the rectangle BC, CD to the square of BD; and, by composition (7. dat.), the ratio of four times the rectangle BC, CD, together with the square of BC to the square of BD is given: but four times the rectangle BC, CD, together with the square of BD is equal (8. 2.) to the square of the straight lines BC, CD taken together: therefore the ratio of the square of BC, CD together to the square of BD is given; wherefore (58. dat.) the ratio of the straight line BC, together with CD to BD, is given: and, by composition, the ratio of BC, together with CD and DB, that is, the ratio of twice BC to BD, is given; therefore the ratio of BC to BD is given, as also (1. 6.) the ratio of the square of BC to the rectangle CB, BD: but the

B PD

rectangle CB, BD is given, being the given excess of the square of BC, BA; therefore the square of BC, and the straight line BC, is given and the ratio of BC to BD, as also of BD to BA, has been shown to be given; therefore (9. dat.) the ratio of BC to BA is given; and BC is given, wherefore BA is given.

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 EF, draw EG perpendicular to FG; let the rectangle EG, GH be the given space to which the parallelogram AC is to be made equal; and the rectangle HG, GL be the given excess of the squares of BC, BA.

F G

Take, in the straight line GE, GK equal to FE, and make GM double of GK: join ML, and in GL produced, take LN equal to LM: bisect GN in O, and between GH, GO find a mean proportional BC: as OG to GL, so make CB to BD; and make the angle CBA equal to GFE, and as LG to GK, so make DB to BA; and complete the parallelogram AC: AC is equal to the rectangle EG, GH, and the excess of the squares of CB, BA is equal to the rectangle HG, GL.

Because as CB to BD, so is OG to GL, the square of CB is to the rectangle CB, BD as (1. 6.) the rectangle HG, GO to the rectangle HG, GL: and the square of CB is equal to the rectangle HG, GO, because GO, BC, GH are proportionals; therefore the rectangle CB, BD is equal (14. 5.) to HG, GL. And because as CB to BD, so is OG to GL; twice CB is to BD, as twice OG, that is GN, to GL: and, by division, as BC together with CD is to BD, so is NL, that is, LM, to LG: therefore (22. 6.) the square of BC together with CD is to the square of BD, as the square of ML to the square of LG: but the square of BC. and CD together is equal (8. 2.) 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 BD, because as LG to GK, so DB was made to BA: therefore (14. 5.) the rectangle BC, CD is equal to the square of AB. To each of these add the rectangle CB, BD, and the square of BC becomes equal to the square of AB together with the rectangle CB, BD; therefore this rectangle, that is, the given rectangle KG, GL, is the excess of the squares of BC, AB. From the point A, draw AP perpendicular to BC, and because the angle ABP is equal to

the angle EFG, the triangle ABP is equiangular to EFG: and DB was made to BA, as LG to GK; therefore as the rectangle CB, BD to CB, BA, so is the rectangle HG, GL to HG, GK; and as the rectangle CB, BA to AP, BC, so is (the straight M

B PD C

F G L O HN

line BA to AP, and so is FE or GK to EG, and so is) the rectangle HG, GK to HG, GE; therefore ex æquali, as the rectangle CB, BD to AP, BC, so is the rectangle HG, GL to EG, GH: and the rectangle CB, BD is equal to HG, GL: therefore the rectangle AP, BC, that is, the parallelogram AC, is equal to the given rectangle EG, GH.

Ir two straight lines contain a parallelogram given in magnitude, in a given angle; if the sum of the squares of its sides be given, the sides shall each of them be given.

Let the two straight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle ABC, and let the sum of the squares of AB, BC be given; AB, BC are each of them given. First, let ABC be a right angle; and because twice the rectangle contained by two equal straight lines is equal to both their squares; but if two straight lines are unequal, twice the rect- Al angle contained by them is less than the sum of their squares, as is evident from the 7th prop. book 2, Elem.; therefore twice the given space, to which space the rectangle of which the sides are to be found is equal, must not be greater than the given sum of the squares of the sides: and if twice that space be equal to the given sum of the squares, the sides of the rectangle must necessarily be equal to one another; therefore in this case describe a square ABCD equal to the given rectangle, and its sides AB, BC are those which were to be found: for the rectangle AC is equal to the given space, and the sum of the squares of its sides AB, BC is equal to twice the rectangle AC, that is, by the hypothesis, to the given space to which the sum of the squares was required to be equal

But if twice the given rectangle be not equal to the given sum of the squares of the sides, it must be less than it, as has been shown. Let ABCD be the rectangle; join AC and draw BE perpendicular to it, and complete the rectangle AEBF, and describe the circle ABC about the triangle ABC; AC is its diameter (Cor. 5. 4.): and because the triangle ABC is similar (8. 6.) to AEB, as AC to CB, so is AB to BE; therefore the rectangle

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The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...