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tion and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join KH, and draw GL perpendicular to it: then the ratio of HK to the half of GL is the same with the ratio of the rect. angle DC, CE to the triangle ABC: because the angles HGK, DAC at the vertices of the isosceles triangles GHK, ADC are equal to one another, these triangles are similar ; and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so is he (DC to AF, and so is) the rectangle DC, CE to the rect. angle AF, CE; but as GL to its half, so is the rectangle AF, CE to its half, which is the triangle ACE, or the triangle ABC; therefore, ex æquali, HK is to the half of the straight line GL, as the rectangle DC, CE is to the triangle ABC.

Cor. And if a triangle have a given angle, the space by which the square of the straight line which is the difference of the sides which contain the given angle is less than the square of the third side, shall have a given ratio to the triangle. This is demonstrated the same way as the preceding proposition, by help of the second case of the lemma.

22. S.

PROP. LXXVII.

L.

1

IF the perpendicular drawn from a given angle of See Note. a triangle to the opposite side, or base, have a given ratio to the base, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC have a given ratio to it, the triangle ABC is given in species. If ABC be an isosceles triangle, it is evident that if any a 5. & 32:

1.

K

A

B RDC

E

H MF one of its angles be given, the rest are also given ; and therefore the triangle is given in species, without the consideration of the ratio of the perpendicular to the base, which in this case is given by prop. 50.

But when ABC is not an isosceles triangle, take any straight line EF given in position and magnitude, and upon it describe

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 make ing 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 b LM also is given. Complete the parallelogram LMFK; and, because LM

is given, FK is given in magnitude ; it is also given in position, c 30. dat. and the point F is given, and consequently the point K; and be

cause through K the straight line KL is drawn parallel to EF I d-31. dat, which is given in position, therefore d KL is given in position :

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and the circumference ELF is given in position; therefore the e 28. dat. point L is given And because the points L, E, F, are given, f 29. dat. the straight lines LE, EF, FL, are given fin magnitude ; there42. dat. fore the triangle LEF is given 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 contain: ing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base.

Cor. i. 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 perpendicu. far 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 tri-
angle 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 consequently be the ratio h 9. dat. of AD to BC is given ; and the triangle ABC is therefore given in species

i 77. 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 base k; wherefore, by Cor. 1, the triangles are k$4.6. similar.

22 5. A triangle similar to ABC may be found thus: having de. scribed the segment EGF, and drawo 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 La 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 si-. milar to the triangle LEF, by Cor. 1.

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80.

PROP. LXXVIII.

IF a triangle have one angle given, and if the ratiò 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 gi. ven; the triangle ABC is given in species.

From the point A, draw AD perpendicular to BC, the recta 41. 1.

angle AD, BC has a given ratio to its half, the triangle ABC;

and because the angle BAC is given, the ratio of the triangle b Cor. 62. ABC to the 'rectangle BA, AC is given b; and by the hypodat.

thesis, the ratio of the rectangle BA, AC to the square of BC is c 9. dat.

given ; therefore e the ratio of the rectangle AD, BC to the d 1.6.

square of BC, that is, the ratio of the straight line AD to BC e 77. dat. is given; wherefore the triangle ABC is given in species .

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

M 0
are similar, and the rect. . A
angle AD, BC, or the

E L
rectangle BK, AC which
is equal to it, is to the
rectangle BA, AC, as the
straight line BK to BA,
that is, ás FH to FE. Let BD N С F

G
the given ratio of the rect-
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 ise 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

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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 construction, with the ratio of) OR to PQ; and the angle BAC is equal to the angle POQ; therefore the triangle ABC is similar f f 1 Cor. to the triangle POQ.

77, date,

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 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 by because BAC is a given b Cor. 62. angle ; and the rectangle BA, AC has

dat. a given ratio to the square of BC: wherefore the ratio of D to the square

c 10. dat. of BC is given ; and by composition a,

d 7. dat, the ratio of the space D together with the square of BC to the square of BC B 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 f the triangle ABC is given in f 48. 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 augle ADB, divide BC into the segments BD, DC which have

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