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the segment of a circle EGF containing an angle equal to the given angle BAC, draw GH bisecting EF at right angles, and join EG, GF: then, since the angle EGF is equal to the angle BAC, and that EGF is an isosceles triangle, and ABC is not, the angle FEG is not equal to the angle CBA: draw EL making the angle FEL equal to the angle CBA; join FL, and draw LM perpendicular to EF; then, because the triangles ELF, BAC are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB; and as LE to EF, so is AB to BC; wherefore, ex æquali, as LM to EF, so is AD to BC; and because the ratio of AD to BC is given, therefore the ratio b 2. dat. of LM to EF is given; and EF is given, wherefore LM also is given. Complete the parallelogram LMFK; and because LM is given, FK is given in magnitude; it is also given in position, and the point F is given, and consequently the point K; and because through K the straight line KL is drawn parallel to EF which is given in position, therefored KL is given in position:

c 30. dat.

d 31. dat.

BRD C

G

N

E OHM F

e 28. dat. f 29. dat.

g 42. dat.

and the circumference ELF is given in position; therefore the point L is given. And because the points L, E, F, are given, the straight lines LE, EF, FL, are givenf in magnitude; therefore the triangle LEF is give in species; and the triangle ABC is similar to LEF, wherefore also ABC is given in species. Because LM is less than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be less than the ratio of GH to EF, which the straight line, in a segment of a circle containing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base.

COR. 1. If two triangles, ABC, LEF have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the base BC, as the perpendicular LM to the base EF, the triangles ABC, LEF are similar.

Describe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicular to EF; because the angles ENF, ELF are equal, and that

the angle EFN is equal to the alternate angle FNL, that is, to
the angle FEL in the same segment; therefore the triangle
NEF is similar to LEF; and in the segment EGF there can be
no other triangle upon the base EF, which has the ratio of its
perpendicular to that base the same with the ratio of LM or NO
to EF, because the perpendicular must be greater or less than
LM or NO; but, as has been shown in the preceding demon-
stration, a triangle similar to ABC can be described in the seg-
'ment EGF upon the base EF, and the ratio of its perpendicular
to the base is the same, as was there shewn, with the ratio of
AD to BC, that is, of LM to EF; therefore that triangle must
be either LEF, or NEF, which therefore are similar to the tri-
angle ABC.

COR. 2. If a triangle ABC have a given angle BAC, and if the straight line AR drawn from the given angle to the opposite side BC, in a given angle ARC, have a given ratio to BC, the triangle ABC is given in species.

Draw AD perpendicular to BC; therefore the triangle ARD is given in species; wherefore the ratio of AD to AR is given: and the ratio of AR to BC is given, and consequentlyh the ratio h 9. dat. of AD to BC is given; and the triangle ABC is therefore given in species i.

i77. dat.

COR. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if straight lines drawn from these angles to the bases, making with them given and equal angles, have the same ratio to the bases, each to each; then the triangles are similar; for having drawn perpendiculars to the bases from the equal angles, as one perpendicular is to its base, so is the other to its basek; wherefore, by Cor. 1, the triangles are k similar.

A triangle similar to ABC may be found thus: having described the segment EGF, and drawn the straight line GH, as was directed in the proposition, find FK, which has to EF the given ratio of AD to BC; and place FK at right angles to EF from the point F; then because, as has been shown, the ratio of AD to BC, that is of FK to EF, must be less than the ratio of GH to EF; therefore FK is less than GH; and consequently the parallel to EF, drawn through the point K, must meet the circumference of the segment in two points: let L be either of them, and join EL, LF, and draw LM perpendicular to EF: then, because the angle BAC is equal to the angle ELF, and that AD is to BC, as KF, that is LM, to EF, the triangle ABC is similar to the triangle LEF, by Cor. 1.

$4.6.

22. 5.

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PROP. LXXVIII.

IF a triangle have one angle given, and if the ratio of the rectangle of the sides which contain the given angle to the square of the third side be given, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the square of BC be given; the triangle ABC is given in species.

From the point A, draw AD perpendicular to BC, the rectangle AD, BC has a given ratio to its halfa, the triangle ABC; and because the angle BAC is given, the ratio of the triangle ABC to the rectangle BA, AC is given b; and by the hypothesis, the ratio of the rectangle BA, AC to the square of BC is given; therefore the ratio of the rectangle AD, BC to the square of BC, that is, the ratio of the straight line AD to BC is given; wherefore the triangle ABC is given in speciese.

A triangle similar to ABC may be found thus: take a straight line EF given in position and magnitude, and make the angle FEG equal to the given angle BAC, and draw FH perpendicular to EG, and BK perpendicular to AC; therefore the triangles ABK, EFH

are similar, and the rect-
angle AD, BC, or the
rectangle BK, AC which
is equal to it, is to the
rectangle BA, AC, as the
straight line BK to BA,

A

M O

that is, as FH to FE. Let BD N
the given ratio of the rect-

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angle BA, AC to the square of BC be the same with the ratio of the straight line EF to FL; therefore, ex æquali, the ratio of the rectangle AD, BC to the square of BC, that is, the ratio of the straight line AD to BC, is the same with the ratio of HF to FL; and because AD is not greater than the straight line MN in the segment of the circle described about the triangle ABC, which bisects BC at right angles; the ratio of AD to BC, that is, of HF to FL, must not be greater than the ratio of MN to BC: · let it be so, and, by the 77th dat. find a triangle OPQ which has one of its angles POQ equal to the given angle BAC, and the ratio of the perpendicular OR, drawn from that angle to the base PQ the same with the ratio of HF to FL; then the triangle ABC is similar to

OPQ: because, as has been shown, the ratio of AD to BC is the same with the ratio of (HF to FL, that is, by the construc tion, with the ratio of) OR to PQ; and the angle BAC is equal to the angle POQ; therefore the triangle ABC is similar ff1. Cor. to the triangle POQ.

Otherwise,

Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the square of BC be given; the triangle ABC is given in species.

Because the angle BAC is given, the excess of the square

of both the sides BA, AC together above the square of the

77. dat.

third side BC has a given a ratio to the triangle ABC. Let the a 76. dat. figure D be equal to this excess; therefore the ratio of D to the triangle ABC is given: and the ratio of the triangle ABC

to the rectangle BA, AC is given ↳, because BAC is a given b Cor. 62.

angle; and the rectangle BA, AC has

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A

C

dat.

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is given; but D together with the square of BC is equal to the square of both BA and AC together; therefore the ratio of the square of BA, AC together to the square of BC is given; and the ratio of BA, AC together to BC is therefore given e; and the e 59. dat. angle BAC is given, wherefore the triangle ABC is given in f48. dat. species.

The composition of this, which depends upon those of the 76th and 48th propositions, is more complex than the preceding composition, which depends upon that of prop. 77, which is easy.

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IF a triangle have a given angle, and if the straight See Note. line drawn from that angle to the base, making a given angle with it, divide the base into segments which have a given ratio to one another; the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the straight line AD drawn to the base BC making the given angle ADB, divide BC into the segments BD, DC which have

19. 5.

g 6. 2.

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 shown to be proportionals, 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 tri-
angle to be found; and let the given ratio of the perpendicu-
lar 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
HLM equal to LGH, it is required that LM be perpendicular to
GM, and equal to K; and because GM, ML, MH are propor-
tionals, as was shewn of BE, EA, EC, the rectangle GMH is
equal to the square of ML. Add the common square of NH,
(having bisected GH in N), and the square of NM is equals to
the squares of the given straight lines NH and ML or K; there-
fore 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 square of
NH, the rectangle GMH is
equal to the square of
ML: therefore as GM to

3

K

L R

S

h 6. 6.

ML, so is ML to MH, and

++
NQ H

M P

the triangle LGM ish there-
fore equiangular to HLM,
and the angle HLM equal to G
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 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.

And whether this ratio of GM to ML be 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,

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