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2. S. 4.

Describe the circle BAC about the triangle, and from its cen

ter E draw EA, EB, EC, ED. because the angle BAC is given, b. 20. 3. the angle BEC at the center, which is the double b of it, is given.

1 and the ratio of BE to EC is given, because they are equal to one c. 44. Dat. another; therefore the triangle BEC is given in species, and the d. 7. Dat. ratio of EB to BC given. alfo the ratio of CB to BD is given ", be

cause the ratio of BD to DC is given; therefore the ratio of EB to c. 9. Dat. BD is given and the angle EBC is given, wherefore the triangle

EBD is given in species, and the ratio of EB, that is of EA to ED

is therefore given. and the angle EDA is given, because each of the f. 47. Dat, angles BDE, BDA is given. therefore the triangle AED is given i in species, and the angle AED given; allo

A
the angle DEC is given, because each of the
angles BED, BEC is given ; therefore the
angle AEC is given. and the ratio of EA to
EC, which are equal, is given; and the tri-

B
angle AEC is therefore given in species,
and the angle ECA given. and the angle

ECB is given, wherefore the angle ACB is given. and the angle 3:43. Dat. BAC is also given; therefore the triangle ABC is given in species.

A triangle similar to ABC may be found, by taking a straight line given in position and magnitude, and dividing it in the given ra- ; tio which the segments BD, DC are required to have to one another ; then if upon that straight line a segment of a circle be defcribed containing an angle equal to the given angle BAC, and a straight line be drawn from the point of division in an angle equal to the given angle ADB, and from the point where it meets the circumference, straight lines be drawn to the extremity of the first line, these together with the first line shall contain a triangle similar to ABC, as may easily be thewn.

The Demonstration may be also made in the manner of that of the 77. Prop. and that of the 77. may be made in the manner of this.

L.

PRO P. LXXX, the sides about an angle of a triangle have a given ra

tio to one another, and if the perpendicular drawn from that angle to the base has a given ratio to the base; the triangle is given in species.

Let

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b. 44. Dat,

Let the fides BA, AC, about the angle BAC of the triangle ABC have a given ratio to one another, and let the perpendicular AD have a given ratio to the base BC; the triangle ABC is given in species.

First, let the sides AB, AC be equal to one another, therefore the perpendicular AD bileets the base BC. and the

A ratio of AD to BC, and therefore to its half DB is given; and the angle ADB is given. wherefore 1 the triangle * ABD and consequently the triangle

*.43. Dat, ABC is given b in species.

B D C But let the sides be unequal, and BA be greater than AC; and make the angle CAE equal to the angle ABC. because the angle AEB is common to the triangles AEB, CEA, they are similar ; therefore as AB to BE, fo is CA to AE, and, by permutation, as BA to AC, fo is BE to EA, and fo is EA to EC. and the ratio of BA to AC is given, therefore the ratio of BE to EA, and the ratio of EA to EC, as also the ratio of BE to EC is given"; wherefore c.9. Dat. the ratio of EB to BC is given d. and the ra

d. 6. Dat.

A tio of AD to BC is given by the Hypothesis,

2. therefore the ratio of AD to BE is given; and the ratio of BE to EA was shewn to be given ; wherefore the ratio of AD to AE is given. and B FC E D ADE is a right angle, therefore the triangle ADE is given e in fpe- c.46. Dat. cies, and the angle AEB given; the ratio of BE to EA is likewise given, therefore b the triangle ABE is given in species, and consequently the angle EAB, as also the angle ABE, that is the angle CAE is given; therefore the angle BAC is given, and the angle ABC being also given, the triangle ABC is given f in species.

Det. How to find a triangle which shall have the things which are mentioned to be given in the Proposition, is evident in the first case. and to find it the more easily in the other case, it is to be observed that if the straight line EF equal to EA be placed in EB towards B, the point F divides the base BC into the segments BF, FC which have to one another the ratio of the sides BA, AC. because BE, EA, or EF, and EC were shewn to be proportionals, therefore * BF is • 19. Ş. 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

f.

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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 fuppose GLH to be the triangle which is to be found. therefore having made the angle HLM 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 shewn of BE, EA, EC, the rectangle GMH

is equal to the square of ML. add the common square of NH, 8. 6. 2. (having bisected GH in N) and the square of NM is equals to the

squares of the given straight lines NH and MI., 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 the
squares of NH and ML, and taking away the common squarc of
NH, the rectangle GMH is
cqual 8 to the square of

K

LR
0

S
ML; therefore as GM to
ML, fo is ML to MH, and

3. h. 6.6. the triangle LGM is there

fore equiangular to HLM,
and the angle HLM equal

G
to the angle 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 fidei
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.

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 thro' the point L, may be thus found. Let OGH be any such triangle, and draw OP making the angle HOP equal to the angle 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

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20. 6.

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the fall on L; nor can PO be less than LM, therefore it is greater ; and confequently 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 GP to PH, and the ratio of the square of GM to the square of ML is therefore i greater

i. 2. Cor. than the ratio of the square of GP to the square of PO, and the ratio of the 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 lefs 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 f PQ_be to PH; f.19. s. wherefore the square of GP is io the square of PQ, as i the straight line GP to PH, and because GM, ML, MH are proportionals, the square of GM is to the square of ML, as i the straight line GM to BH, 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 {quare 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 Hine CP 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 k. 10. 5. the 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 righť angles to GP. because PQ is greater than ML, or PR, the circle described from the center 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.

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 shewn 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 Propofition.

M.

IF

PRO P. LXXXI.
F 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 fpecies.

Let AB be greater than AC, and make the angle CAE equal to the angle ABC, and because the angle AEB is common to the tri6. angles ABE, CAE, they are a equiangular to one another, there

fore as AB to BE, fo is CA to AE, and, by
permutation, as AB to AC, fo BE to EA,
and so is EA to EC. but the ratio of BA to
AC is given, therefore the ratio of BE to B DC E
EA, as also the ratio of EA to EC is given ;

M,
b. 9. Dat. wherefore b the ratio of BE to EC, as also o
c. Cor. 6.

the ratio of EC to CB is given. and the ra-
Dat.
d. 7. Dat.

tio of BC to CD is given , because the ra-
tio of BD to DC is given; therefore b the

KLHN ratio of EC to CD is given, and consequently d the ratio of DE to EC. and the ratio of EC to EA was shewn to be given, therefore b

the ratio of DE to EA is given. and ADE is a right angle, where6. 46. Dat. fore the triangle ADE is given in species, and the angle AED gi

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