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would then fall on L; por can PO be less than LM, therefore it is greater; and consequently the rectangle GPH is greater than the rectangle GMH, and the straight line GP greater than GM: Therefore the ratio of GM to MH is greater than the ratio of GP to PH, and the ratio of the square of GM to the square of ML is therefore' greater than the ratio of i 2. Cor. the square of GP to the square of PO, and the ratio of the 20. 6. straight line GM to ML greater than the ratio of GP to PO. But as GM to ML, so is GL to LH; and as GP to PO, so is GO to OH; therefore the ratio of GL to LH is greater than the ratio of GO to OH; wherefore the ratio of GL to LH is the greatest of all others; and consequently the given ratio of the greater side to the less must not be greater than this ratio. . But if the ratio of the sides be not the same with this greatest ratio of GM to ML, it must necessarily be less than it: Let any less ratio be given, and the same things being supposed, viz. that GH is the base, and K equal to the altitude of the triangle, it may be found as follows. Divide GH in the point Q, so that the ratio of GQ to QH may be the same with the given ratio of the sides; and as GQ to QH, so make GP to PQ, and so will' PQ be to PH; wherefore the square f 19. 5. of GP is to the square of PQ, as the straight line GP to

i PH: And because GM, ML, MH, are proportionals, the square of GM is to the square of ML, as the straight line GM to MH: But the ratio of GQ to QH, that is, the ratio of GP to PQ, is less than the ratio of GM to ML; and therefore the ratio of the square of GP to the square of PQ is less than the ratio of the square of GM to that of ML; and consequently the ratio of the straight line GP to PH is less than the ratio of GM to MH; and, by division, the ratio of GH to HP is less than that of GH to HM; wherefore k the k 10. 5. straight line HP is greater than HM, and the rectangle GPH, that is, the 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

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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.

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a

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 coma 4. 6. mon 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 BE

A
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

B DC E

E b 9. dat. of EA to EC is given ; wherefore the

M c Cor. 6. ratio of BE to EC, as also the ratio of

EC to CB is given : And the ratio of d 7. dat. BC to CD is given“, because the ratio

of BD to DC is given ; therefore b the
ratio of EC to CD is given; and conse-

G KLH N
quently d 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 given: And ADE is a right angle, e 46. dat. wherefore e the triangle ADE is given in species, and the

angle AED given: And the ratio of CE to EA is given, theref 44. dat. fore the triangle AEC is given in species, and consequently

с

dat.

e

a

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 8 in species.

g 43. dat. 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.

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 bas 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 0, and join OG, OH; then OG H 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.

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a

• 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 ihan A to B.

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If a parallelogram, given in species and magnitude, be increased or diminished by a gnomon, given in mag. nitude, the sides of the gnomon are given in magni. tude.

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b 60. dat.

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 (2. def. is given in magnitude : But AG is also given in species, be2. and cause it is similar a to AB; therefore the (24. 6. sides of AG are given : Each of the

G E straight lines AE, AF is therefore given; and each of the straight lines CA, AD is

с

B given b, therefore each of the remainders c 4. dat. EC, DF is given Next, Let the parallelogram AG, given

F D A in species and magnitude, be diminished by the given gnomon ECBDFG, each of the

H straight lines CE, DF is given.

Because the parallelogram AG is given, as also its gnomon (2. def. ECBDFG, the remaining space AB is given in magnitude:

2. and But it is also given in species : because it is similar to AG: 224, 6. therefore bits sides 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 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 d with A'B; 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 É must be less than the figure FE from which it is to be taken.

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b 60. dat.

c 25. 6.

d 26. 6.

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If a parallelogram equal to a given space be ap

. plied 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.

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 dia

G H F meter with DL: Let BCG be the dia

b 26. 6. meter, and construct the figure; there

K

L fore, because the figure EF, given in species, is described upon the given straight line EB, EF is given in mag

А E D B

c 56. dat. nitude, and the gnomon ELH is equal d to the given figure d 36. dat. AC; therefore e since EF is diminished by the given gnomon

43. 1. 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.

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.

e

e 82. dat.

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If 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.

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