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base BC into the segments BF, FC, which have to one an

other the ratio of the sides BA, CA, because BE, EA, or 19. 5. EF, and EC, were shown to be proportiopals, therefore

BF is to FC, as BE to EF or EA, that is, as BA to AC; and AE cannot be less than the altitude of the triangle ABC, but it may be equal to it, which if it be, the triangle, in this case, as also the ratio of the sides, may be thus found: Having given the ratio of the perpendicular to the base, take the straight line GH, given in position and magnitude, for the base of the triangle to be found; and let the given ratio of the perpendicular to the base be that of the straight line K to GH, that is, let K be equal to the perpendicular; and suppose GLH to be the triangle which is to be found, therefore having made the angle #LM equal to LGH, it is required that LM be perpendicular to GM, and equal to K; and because GM, ML, MH, are proportionals, as was shown of BE, EA, EC, the rectangle GMH is equal to the

square of ML. Add the coinmon square of NH (having 56.2. bisected GH in N), and the square of NM is equals to the

squares of the given straight lines NH and ML, or K; therefore the square of NM, and its side NM, is given, as also the point M, viz. by taking the straight line NM, the square of which is equal to the squares of NH, ML. Draw ML equal to K, at right angles to GM; and because ML is given in position and magnitude, therefore the point L is given: Join LG, LH; then the triangle LGH is that which was to be found; for the square of NM is equal to

of NH and ML, and taking away the common
square of NH, the rect-
angle GMH is equals K

LR
to the square of ML;
therefore as GM to ML,
3.

HO so is ML to MH, and

To 1993 16.6. the triangle LGM is h

therefore equiangular to
HLM, and the angle
HLM equal to the an-
gle LGM, and the straight line LM, drawn from the vertex
of the triangle, making the angle HLM equal to LGH, is
perpendicular to the base, and equal to the given straight
line K, as was required ; and the ratio of the sides GL, LH,
is the same with the ratio of GM to ML, that is, with the
ratio of the straight line which is made up of GN, the half
of the given base, and of NM, the square of which is equal
to the squares of GN and K, to the straight line K.

the squares

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G NQ H M PRO

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And whether this ratio of GM to ML is greater or less than the ratio of the sides of any other triangle upon the base GH, and of which the altitude is equal to the straight line K, that is, the vertex of which is in the parallel to GH drawn through the point L, may be thus found. Let OGH be any such triangle, and draw OP, making the angle HOP equal to the anglę OGH; therefore, as before, GP, PO, PH, are proportionals, and PO cannot be equal to LM, because the rectangle GPH would be equal to the rectangle GMH, which is impossible; for the point P cannot fall upon M, because O would then fall on L; nor 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 thereforei greater than the ratio of the square of GP 2 Cor. to the square of PO, and the ratio of the straight line GM 20. 6. 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; * 19. 5. wherefore the square of GP is to the square of PQ, as i the straight line GP to PH : Anıl because GM, ML, MH, are proportionals, the square of GM is to the square of ML, asi 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; whereforek the straight line HP10.5. 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 0, 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 saine 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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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 & equi

angular to one another: Therefore as AB to BE, so is CA

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• Cor. 6.

to AE, and, by permutation, as AB to AC, so is BE to EA, and so is EA

A 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, B DC is given ; wherefore the ratio of BE M.

69 Dat. to EC, as also the ratio of EC to CB,

Dat. is given : And the ratio of BC to CD

d7 Dat. is givend, because the ratio of BD to DC is given : therefored the ratio of Ć KLH EC to CD is given, and consequently the ratio of DE to EC: And the ratio of EC to EA was shown to be given, thereforeb the ratio of DE to EA is 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, therefore 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 given in species.

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

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. 3. Also make as GK to KH, so GN to NK, and so shallb NK

be to NH. And from the centre N, at the distance NK, describe a circle, and let its circumference meet 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 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 rativ of the segments of the base.

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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 224,6. given": Each of the straight lines AE, AF,

G is therefore given; and each of the straight

E lines CA, AD, is given", therefore each of

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

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

FD given gnomon ECBDFG, each of the straight lines CE, DF, is given.

H Because the parallelogram AG is given, as also its gnomon ECBDFG, theremainingspace AB is givenia

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