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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 rectangle contained by them A D is less than the sum of their squares, as is evident from the 7th Prop. B. 2, Elem. ; there- B С fore 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 are 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 a : And because the triangle ABC is simi- a Cor. 5.4. lar 6 to AEB, as AC to CB, so is AB to BE; therefore, the b 8. 6. rectangle AC, BE, is equal to AB, BC; and the rectangle AB, BC is given, wherefore AC, BE is given: And because the sum of the squares of AB, BC is given, the square of AC which is equal < to that sum is given; and AC itself is therefore given c 47. 1. in magnitude : Let AC be likewise given in position, and the point A ; therefore AF is given in po

d 32. dat. sition : And the rectangle AC, BE is

А

D given, as has been shown, and AC is

E given, wherefore · BE is given in mag

e 61. dat. nitude, as also AF which is equal to it; F

B and AF is also given in position, and the point A is given, wherefore the

f 30. dat. point F is given; and the straight line

G FB in position 8: And the circumfe

K HL

8 31. dat, rence ABC is given in position, wherefore h the point B is h 28. dat. given ; and the points A, C are given ; therefore, the straight lines AB, BC are giveni in position and magnitude.

i 29. dat. The sides AB, BC of the rectangle may be found thus: Let the rectangle GH, GK be the given space to which the rect

.

angle AB, BC is equal; and let GH, GL be the given rect

angle, to which the sum of the squares of AB, BC is equal ; k 14. 2. Find k a square equal to the rectangle GH, GL; and let its

side AC be given in position ; upon AC as a diameter, describe the semicircle ABC, and as AC to GH, so make GK to AF,

and from the point A place AF at right angles to AC; there1 16. 6. fore the rectangle CA, AF is equal to GH, GK; and, by

the hypothesis, twice the rectangle GH, GK is less than GH, GL, that is, than the square of AC; wherefore, twice the rectangle CA, AF is less than the square of AC, and the rectangle CA, AF itself less than half the square of AC, that is, than the rectangle contained by the diameter AC and its half: wherefore, AF is less than the semidiameter of the circle, and consequently the straight line drawn through the point F, parallel to AC, must meet the circumference in two points. Let B be either of them, and join AB, BC, and complete the rect

angle ABCD, ABCD is the rectangle which was to be found: m 34. 1. Draw BE perpendicular to AC; therefore BE is equal" to AF,

and because the angle ABC in a semicircle is a right angle, the b 8. 6. rectangle AB, BC is equal to AC, BE, that is, to the rect

angle CA, AF, which is equal to the given rectangle GH, GK: c 47. 1. And the squares of AB, BC are together equal to the square

of AC, that is, to the given rectangle GH, GL.

But if the given angle ABC of the parallelogram AC be not à right angle, in this case, because ABC is a given angle, the

ratio of the rectangle contained by the sides AB, BC to the n 62. dat. parallelogram AC, is given "; and AC is given, therefore the

rectangle AB, BC is given; and the sum of the squares of AB, BC is given; therefore the sides AB, BC are given, by the preceding case.

The sides AB, BC, and the parallelogram AC, may be found thus : Let EFG be the given angle of the parallelogram, and from any point E in

А

D
FE draw EG perpendicular to FG; and
let the rectangle EG, FH be the given
space, to which the parallelogram is to be
made equal, and let EF, FK be the given B L С
rectangle to which the sum of the squares
of the sides is to be equal. And, by the

E
preceding case, find the sides of a rect-
angle which is equal to the given rect-
angle EF, FH, and the squares of the
sides of which are together equal to the
given rectangle EF, FK; therefore, as

F HG K
was shown in that case, twice the rect-
angle EF, FH must not be greater than the rectangle EF,

86.

FK; let it be so, and let AB, BC be the sides of the rectangle joined in the angle ABC equal to the given angle EFG, and complete the parallelogram ABCD, which will be that which was to be found: Draw AL perpendicular to BC, and because the angle ABL is equal to EFG, the triangle ABL is equiangular to EFG; and the parallelogram AC, that is, the rectangle AL, BC is to the rectangle AB, BC as (the straight line AL to AB, that is, as EG to EF, that is, as) the rectangle EG, FH to EF, FH; and, by the construction, the rectangle AB, BC is equal to EF, FH, therefore the rectangle AL, BC, or, its equal, the parallelogram AC, is equal to the given rectangle EG, FH; and the squares of AB, BC are together equal, by construction, to the given rectangle EF, FK.

PROP. LXXXIX. If two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of them above a given space has a given ratio to the square of the other ; each of the straight lines is given.

Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the excess of the square of BC above a given space have a given ratio to the square of AB, each of the straight lines AB, BC is given.

Because the excess of the square of BC above a given space bas a given ratio to the square of BA, let the rectangle CB, BD be the given space; take this from the square of BC, the remainder, viz. the rectangle BC, CD has a given ratio to a 2. 2. the square of BA: Draw A E perpendicular to BC, and let the square of BF be equal to the rectangle BC, CD, then, because the angle A BC, as also BEA is given, the

T triangle ABE is given b in species, and the ratio of AE to A B given: And because А the ratio of the rectangle BC, CD, that is, of the square of BF to the square of BA, is given, the ratio of the straight line B.ED с BF to BA is giveno; and the ratio of AE to AB is given; wherefored the ratio of AE to BF is given; d 9. dat. as also the ratio of the rectangle AE, BC, that is, of the e 35. 1. parallelogram AC to the rectangle FB, BC, and AC is given, wherefore the rectangle FB, BC is given. The excess of the square of BC above the square of BF, that is, above the rectangle BC, CD, is given, for it is equal a to the given rectangle CB, BD: therefore, because the rectangle contained by the

b 43. dat.

c 58. dat.

straight lines FB, BC is given, and also the excess of the

square of BC above the square of BF; FB, BC are each of f 87. dat. them given'; and the ratio of FB to BA is given; therefore

AB, BC are given.

G

А

The Composition is as follows :
Let GHK be the given angle to which the angle of the
parallelogram is to be made equal, and from any point G in
HG, draw GK perpendicular to HK; let GK, HL be the rect-
angle to which the parallelogram is to be

N
made equal, and let LH, HM be the rect-
angle equal to the given space which is to
be taken from the square of one of the
sides; and let the ratio of the remainder

HKM L
to the square of the other side be the same
with the ratio of the square of the given straight line NH to
the square of the given straight line HG.

By help of the 87th dat. find two straight lines BC, BF, which contain a rectangle equal to the given rectangle NH, HL, and such that the excess of the square

I
of BC above the square of BF may be equal
to the given rectangle LH, HM; and join
CB, BF, in the angle FBC, equal to the
given angle GHK: and as NH to HG, so

BED с
make FB to BA, and complete the paral-
lelogram AC, and draw A E perpendicular to BC; then AC
is equal to the rectangle GK, HL; and if from the square
of BC the given rectangle LH, HM be taken, the remainder
shall have to the square of B A the same ratio which the square
of NH has to the

square of HG. Because, by the construction, the square of BC is equal to the square of BF together with the rectangle LH, HM; if

from the square of BC there be taken the rectangle LH, HM, 8 22. 6. there remains the square of BF, which has 8 to the square of

BA the same ratio which the square of NH has to the square of HG, because as NH to HG, so FB was made to BA; but as HG to GK, so is BA to A E, because the triangle GHK is

equiangular to ABE; therefore, ex æquali, as NH to GK, so hl. 6. is FB to AE; wherefore the rectangle NH, HL is to the

rectangle GK, HL, as the rectangle FB, BC to AE, BC;

but by the construction, the rectangle NH, HL is equal to i 14. 5. FB, BC; therefore the rectangle GK, HL is equal to the

rectangle AE, BC, that is, to the parallelogram AC.

The analysis of this problem might have been made as in the 86th Prop. in the Greek, and the composition of it may be made as that which is in Prop. 87. of this edition.

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If two straight lines contain a given parallelogram in a given angle, and if the square of one of them together with the space which has a given ratio to the square of the other be 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, and let the square of BC together with the space which has a given ratio to the square of AB be given, AB, BC are each of them given.

Let the square of BD be the space which has the given ratio to the square of AB; therefore, by the hypothesis, the square of BC together with the square of BD is given. From the point A, draw A E perpendicular to BC; and because the angles A BE, BEA are given, the triangle ABE is given a in a 43. dat. species; therefore the ratio of BA to AE is given : And because the ratio of the square of BD to the square of BA is given, the ratio of the straight line BD to BA is given b; and b 58. dat. the ratio of BA to AE is given'; therefore the ratio of AE c 9. dat. to BD is given, as also the ratio of the rectangle AE, BC, that is, of the parallelogram AC to the rectangle DB, BC; and AC is given, therefore the rectangle DB, BC is given ; and the square of BC together with the square of BD is given :

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BE С GH K L therefored because the rectangle contained by the two straight d 88. dat. lines DB, BC is given, and the sum of their squares is given: The straight lines DB, BC are each of them given; and the ratio of DB to BA is given; therefore AB, BC are given.

The Composition is as follows :

Let FGH be the given angle to which the angle of the parallelogram is to be made equal, and from any point Fin GF, draw FH perpendicular to GH, and let the rectangle FH, GK be that to which the parallelogram is to be made equal ; and let the rectangle KG, GL be the space to which the square

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