square of PQ, greater than the rectangle GMH, that is, than the square of ML, and the straight line PQ is therefore greater than ML. Draw LR parallel to GP, and from P draw PR at right angles to GP. Because PQ is greater than ML, or PR, the circle described from the centre P, at the distance PQ, must necessarily cut LR in two points; let these be O, S, and join OG, OH; SG, SH: each of the triangles OGH, SGH, have the things mentioned to be given in the proposition: Join OP, SP; and because as GP to PQ, or PO, so is PO to PH, the triangle OGP is equiangular to HOP; as therefore OG to GP, so is HO to OP; and, by permutation, as GO to OH, so is GP to PO, or PQ; and so is GQ to QH: Therefore the triangle OGH has the ratio of its sides GO, OH, the same with the given ratio of GQ to QH; and the perpendicular has to the base the given ratio of K to GH, because the perpendicular is equal to LM, or K: The like may be shown in the same way of the triangle SGH.

This construction by which the triangle OGH is found, is shorter than that which would be deduced from the demonstration of the datum, by reason that the base GH is given in position and magnitude, which was not supposed in the demonstration: The same thing is to be observed in the next proposition.

If the sides about an angle of a triangle be unequal, and have a given ratio to one another, and if the perpendicular from that angle to the base divides it into segments that have a given ratio to one another, the triangle is given in species.

Let ABC be a triangle, the sides of which about the angle BAC are unequal, and have a given ratio to one another, and let the perpendicular AD to the base BC divide it into the segments BD, DC, which have a given ratio to one another, the triangle ABC is given in species.

Let AB be greater than AC, and make the angle CAE equal to the angle ABC; and because the angle AEB is 4. 6. common to the triangles ABE, CAE, they are a equiangular to one another: Therefore as AB to BE, so is CA

to AE, and, by permutation, as AB to AC, so is BE to EA, and so is EA to EC: But the ratio of BA to AC is given, therefore the ratio of BE to EA, as also the ratio of EA to EC, is given; wherefore the ratio of BE to EC, as also the ratio of EC to CB, is given: And the ratio of BC to CD is givend, because the ratio of BD to DC is given therefored the ratio of EC to CD is given, and consequentlyd

the ratio of DE to EC: And the ratio of EC to EA was shown to be given, therefore the ratio of DE to EA is

• Cor. 6. Dat. d 7 Dat.

given: And ADE is a right angle, wherefore e the triangle. 46 Dat. ADE is given in species, and the angle AED given; And the ratio of CE to EA is given, thereforef the triangle £44 Dat. AEC is given in species, and consequently 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 givens 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 DCt, 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.

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

+ 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 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 and C have to B and C a less ratio than A to B.

$ 43 Dat

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: 19. 5. Also make as GK to KH, so GN to NK, and so shall NK be to NH. And from the centre N, 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.

b

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 2 Def. AG is given in magnitude: But AG is also given in species, 2 and because it is similar a to AB; therefore the sides of AG are given: Each of the straight lines AE, AF, is therefore given; and each of the straight G

24.6.

60 Dat.

lines CA, AD, is given, therefore each of

4 Dat. the remainders EC, DF, is given.

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.

Because the parallelogram AG is given, as

FD

H

also its gnomon ECBDFG, the remaining space AB is given in

2 Def. 2. and

24.6.

magnitude: But it is also given in species; because it is similar to AG; therefore its sides CA, AD, are given, and each of the straight lines EA, AF, is given; therefore 60 Dat EC, DF are each of them given.

The gnomon and its sides CE, DF, may be found thus Fin the first case. Let H be the given space to which the gnomon must be made equal, and find a parallelogram si-a 25. 6. milar 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 diametere with AB; because 26. 6. 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.

Ir 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

Bisect AB in E; therefore EB is given in magnitude; upon EB describe the parallelogram EF similar to DL * 18. 6. and similarly placed; therefore EF is given in species, and is about the same diameter 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 mag- A nitude, and the gnomon ELH is equal d to the given figure 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 BD; wherefore CD, DB are each of them given."

с

EDB

B 56 Dat.

d 36. and

43. 1.

e 82 Dat.

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.

GFH

Bisect AB in E; therefore EB is given in magnitude: a 18. 6. Upon BE describe the parallelogram EF similar to LD, and similarly placed; therefore EF is given in species, and 26. 6. is about the same diameter 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 € 56 Dat. EB, EF is given in magnitude, and the gnomon ELH is equal to the given

43. 1.

A E

B

KLC

36. and figured AC; wherefore, since EF is increased by the given 82 Dat, 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..

Draw the diameter DE of the parallelogram AC, and