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Dat.

8 43. Dat.

the ratio of CE to EA is given, therefore f the triangie AEC is f. given in species, and consequently the angle ACE is given, as also the adjacent angle ACB. in the fame manner, becaule the ratio of BE to E A 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 : in species.

But the ratio of the greater (de BA to the other AC must be less than the ratio of the greater segment BD to DC. because the fqoare 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 t, because the square of ED 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 firaight line GH given in position and magnitude, and divide it in K so that the ratio of GK to KH may be the fame with the given ratio of BA to AC; divide alfo GH in L so that the ratio of GLto LH may be the same with the given ratio of BD to DC, aod draw LM at right angles to GH. and because the ratio of the sides of a triangle is lefs than the ratio of the segments of the base, as has been shewn, the ratio of GK 10 KH is less than the ratio of GL to LH, wherefore the point L must fall betwixt K and II. also make as GK to KH, fo CN to NK, and fo Mall 1 NK be to NH. and from the center N, at the distance h. 19. 5. NK describe a circle, and let its circumference mect LM in 0, 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 CG to GN, fo 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

| If A be greater than B, and Cany third magnitnde ; 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 aud C have to Band C
a less ratio than A to B.

construction, GL, LH have to one another the given ratio of the segments of the base.

60.

PROP. LXXXII.
Fa parallelogram given in fpecies and magnitude be

encreased, or diminished by a gnomon given in magnitude; the sides of the gnomon are given in magnitude.

IF

2.Der.

First, let the parallelogram AB given in species and magnitude be encreased by the given gnomon ECBDFG; each of the itraight lines CE, DF is given.

Because AB is given in species and magnitude, and that the gno2. and mon ECBDFG is given, therefore the whole space AG is given in

24.6. magnitude. but AG is also given in species, because it is fimilar? b.60. Dat. to AB; therefore the sides of AG are givenb. each of the Araight lines AE, AF is therefore given ; and each of

G the straight lines CA, AD is given, therefore c. 4. Dat. each of the remainders EC, DF is given

c Next, let the parallelogram AG given in species and magnitude be diminished by the given TD

А gnomon ECBDPG ; each of the straight lines CE, DF is given.

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 a to AG; therefore bits fides 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 d. 25. 6.

made equal, and find d a parallelogram similar to AB and equal to the figures AB and H together, and place its fides AE, AF

from the point A, upon the straight lines AC, AD, and complete e. 26. 6. the parallelogram AG, which is about the fame diameter e 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.

58.

IF

PRO P. LXXXIII.
F 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.

Bifeet AB in E; therefore EB is given in magnitude. upoa EB defcribe the parallelogram EF fimilar to DL and similarly placed; 2. 19.6. therefore EF is given in species, and is about the same diameter 6 with DL ; let BCG

G HT b. 16.6. be the diameter, and construct the figure. therefore because the figure EF given in

K к species is described upon the given straight line EB, EF is given in magnitude. and

А.

d. 36. and the gnomon ELH is equal to the given figure AC, therefore e since EF is diminished by the given gno- e. 82. Dat. mon ELH, the sides EK, FH of the gnomon are given. bur EK is equal to DC, and FH to DB; wlierefore CD, DB are each of them given.

This Demonstration is the Analysis of the Problem in the 28. 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 shewn in the 27. Prop. B.O.

c. 30. Dat.

43. 1.

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PROP. LXXXIV.

59 IF a parallelogram equal to a given space

be applied to a given straight line, exceeding by a parallelogram given in fpecics; 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.

Birect AB in E; therefore EB is given in magnitude. upon EB describe the parallelogram EF similar to LD, and funilar!, pleced; 2. 19. 6. therefore EF is given in fpecies, and is about the fame diameter o b. 36. 6.

Ee

43. I.

with LD. let CLG be the diameter, and
construct the figure. therefore becaute the

TH figure EF given in fpecies is described up

E on the given ftraight line EB, EF is given A

D C. 56. Da:. in magnitude . and the gnomon ELH is d. 36. and equal to the given figure d AC; where

K L c fore fince EF is encreased by the given e. 82. Dat. gnomon ELH, its fides EK, FH are giveno. but EK is equal to

CD, and TH to BD; therefore CD, DB are each of them given.

This Demonstration is the Analysis of the Problem in the 29. 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 excceding 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 to BC which is given, therefore AK is given. and BK is

similar b to AC, therefore PK is given in E G C b 24. 6.

species, and since the parallelogram AK
given in magnitude is applied to the given
Straight line AB, exceeding by the paral-

H

K lelogram BK given in species, therefore, by this Proposition, BD, DK the sides of A the excess are given. and the straight line AB is given, therefore the whole AD, as alfo DC to which it has a given ratio is given.

PRO B. To apply a parallelogram similar to a given one to a giren straight line AB, exceeding by a parallelogram equal to a given space.

To the given straight line AB apply 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 thro' the point A draw AE parallel to BF meeting DF produced in E, and complete the parallelogram AC.

The parallelogram BC is equal - to AK, that is to the given space; and the parallelogram AC is fimilar b to BK. therefore the parallelogram AC is applied to the straight line AB similar to the

2. 43. I.

C 29. 6.

345

IF

öne given and exceeding by the parallelogram BC which is equal
to the given space.

PRO P. LXXXV.
Ftwo straight lines contain a parallelogram given in

magnitude, in a given angle; if the difference of the straight lines be given, they ihall 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 gia ven; each of the straight lines AB, BC is given.

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 fpecies, and because the given parallelogram AC is applied to the given straight line DC, exceeding by the parallelograin AD given in fpecies, the sides of the exce's are given'; therefore BD is given. and DC is given, wherefore a. 2.4. Dat. the whole DC is given. and AB is given, therefore AB, BC are each of them given.

PROP. LXXXVI.
Ftwo straight lines contain a parallelogrım given in

magnitude, in a given angle; if both of them ingether be given, they fall each of them be givea.

Let the two straight lines AB, BC contain the parallelogram AC given in maynitude, in the given angle ABC, and let AD, BC together be given ; each of the straight lines AB, DC is given.

Produce CD and make BD equal to BA, and complete the parallelogram ABDE. becaule DB is equal to BA, and the angle ABD given, becaule the adjacent angle ABC is E A given; the parallelogiam AD is given in species. and becaute AB, BC together are given, and AB is equal to BD ; therefore DC is given. and because the given parallelogram AC is ap-D plied to the given straight line DC, deficient by the partielo ram AD given in species, the sides AB, BD of the defect are given. 2. 9; 1*t. and DC is given, wherefore the remainder BC is given ; and each of the straight sides AB, BC is therefore given.

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