the angle ACE is given, as also the adjacent angle ACB. In the same manner, because the ratio of BE to EA 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 side BA to the other AC must be less than the ratio of the greater segment BD to DC; because the square 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*, because the square of BD 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.

g 43. dat.

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 straight line GH given in position and magnitude, and divide it in K, so that the ratio of GK to KH may be the same with the given ratio of BA to AC: Divide also GH in L, so that the ratio of GL to LH may be the same with the given ratio of BD to 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 of GK to KH is less than the ratio of GL to LH: wherefore the point L must fall betwixt K and H: Also as GK to KH, so make GN to NK, and so shall NK be to NH. And from the centre N, h 19. 5. at the distance NK, describe a circle, and let its circumference meet LM in O, 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 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 be greater than B, and C any third magnitude; then A and C together have to B and C together a less ratio than A bas 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 and C have to B and C a less ratio than A to B.

a

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.

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 species, beS2. def. 2. and cause it is similar to AB; therefore the 24. 6. sides of AG are given: Each of the straight lines AE, AF is therefore given; and each of the straight lines CA, AD is given, therefore each of the remainders

b 60. dat.

c 4. dat. EC, DF is given.

a

Next, Let the parallelogram AG, given in species and magnitude, be diminished by the given gnomon ECBDFG, each of the straight lines CE, DF is given.

Ꮐ

E

C

B

Because the parallelogram AG is given, as also its gnomon (2. def. ECBDFG, the remaining space AB is given in magnitude: a) 2. and But it is also given in species: because it is similar a to AG: 24. 6. therefore its sides CA, AD are given, and each of the straight lines EA, AF is given; therefore EC, DF are each of them given.

b 60. dat.

c 25. 6.

d 26. 6.

d

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 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 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.

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.

58.

GHF

b 26. 6.

K

L

ED B

c 56. dat.

Bisect AB in E; therefore EB is given in magnitude: upon EB describe the parallelogram EF similar to DL, and a 18. 6. similarly placed; therefore EF is given in species, and is about the same diameter b 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 in magnitude, and the gnomon ELH is equal to the given figure d 36. dat. AC; therefore 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.

с

с

A

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 simi. lar to BC, therefore, AC must not be greater than EF, as is shown in the 27th Prop. B. 6.

43. 1.

e 82. dat.

PROP. LXXXIV.

ap

If a parallelogram equal to a given space be plied 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.

59.

a 18. 6.

b 26. 6.

e 56. dat.

d 36. and

43. 1.

e 82. dat.

a 43. 1.

b 24. 6.

Bisect AB in E; therefore EB is given in magnitude; upon EB describe the parallelogram EF similar to LD, and similarly placed; therefore EF is given in species, and is about the same diameter with LD.

A

G F H

E

B

D

KLC

Let CBG be the diameter, and con-
struct the figure: Therefore, because
the figure EF, given in species, is de-
scribed upon the given straight line
EB, EF is given in magnitude, and
the gnomon ELH is equal to the
given figured AC; wherefore, since
EF is increased by the given gnomon ELH, its sides EK,
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 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.

E

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 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 pa- H rallelogram BK, given in species, there

G C

F

K

fore by this proposition, BD, DK the

sides of the excess are given, and the

c 29. 6.

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.

с

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 through the point A draw AE parallel to BF,
meeting DF produced in E, and complete the parallelogram
AC.

a

The parallelogram BC is equal 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.

PROP. LXXXV.

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.

E A

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. 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 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 given, 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.

PROP. LXXXVI.

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

81.

a 84. dat.

85.