80.

a 41. 1.

b Cor. 62.

dat.

c 9. dat.

d 1. 6.

e 77. dat.

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 half, the triangle ABC; and because the angle BAC is given, the ratio of the triangle ABC to the rectangle BA, AC is given; 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 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 are similar, and

the rectangle AD, BC,

or the rectangle BK,

A

M O

AC, which is equal to
it, is to the rectangle

C F

G

AB, AC as the straight BD N
line BK to BA, that is,
as FH to FE. Let the given ratio of the rectangle 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 DA, 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 construction, with the ratio of) OR to PQ; and the angle BAC is equal to the angle POQ. Therefore the triangle ABC is similar 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

fl. Cor.

77. dat.

third side BC has a givena 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

A

to the rectangle BA, AC is given, because BAC is a given b Cor. 62. angle; and the rectangle BA, AC has a given ratio to the square of BC; wherefore the ratio of D to the square of BC is given; and, by composition, the ratio of the space D, B together with the square of BC to

dat.

D

c 10. dat.

d 7. dat.

C

the square of BC 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 ; and the angle BAC is given, wherefore fe 59. dat. the triangle ABC is given in 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.

PROP. LXXIX.

f 48. dat.

K.

If a triangle has a given angle, and if the straight See N. line drawn from that angle to the base, making a given angle with it, divides 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 CB into the segments BD, DC which

a 5. 4.

b 20. 3.

c 44. dat.

have a given ratio to one another; the triangle ABC is given in species.

Describe the circle ABC about the triangle, and from its centre E, draw EA, EB, EC, ED: because the angle BAC is given, the angle BEC at the centre, which is the double of it, is given And the ratio of BE to EC is given, because they are equal to one another, therefore the triangle BEC is given in species, and the ratio of EB to BC is given; also the ratio of CB to BD is given 4, because the ratio of BD to DC is given; therefore the ratio of EB to 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 angles BDE, BDA is given; therefore the triangle AED is f 47. dat. given in species, and the angle AED is

d 7. dat.

e 9. dat.

B

C

D

given: Also 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 triangle 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 g 43. dat. angle BAC is also given; therefore the triangle ABC is given in species.

L.

C

A triangle similar to ABC may be found by taking a straight line given in position and magnitude, and dividing it in the given ratio which the segments BD, DC are required to have to one another; then, if upon that straight line a segment of a circle be described 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 shown.

The demonstration may be also made in the manner of that of the 77th Prop. and that of the 77th may be made in the manner of this,

PROP. LXXX.

If the sides about an angle of a triangle have a given ratio 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 the sides 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.

a 26. 1.

A

⚫ 43. dat. b 44. dat.

First, Let the sides AB, AC be equal to one another, therefore the perpendicular AD bisects the base BC; and the ratio of AD to BC, and therefore to its half DB is given; and the angle ADB is given; wherefore the triangle * ABD, and consequently the triangle ABC is given' in species.

2

1

BD C

A d 6. dat.

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, so is CA to AE, and, by permutation, as BA to AC, so is BE to EA, and so 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 the ratio of EB to c 9. dat. BC is given; and the ratio of AD to BC is given by the hypothesis, therefore the ratio of AD to BE is given; and the ratio of BE to EA was shown to be given: wherefore the ratio of AD to AE is given, and ADE is a right angle, therefore the triangle ADE is given in species, and the angle AEB given, e 46. dat. the ratio of BE to EA is likewise given, therefore, 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 in species.

BFCED

f 43. dat.

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 shown to be proportionals, therefore* BF is to FC, as 19. 5. 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

g 6. 2.

h 6. 6.

angle 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 HLM equal to LGH, it is required that LM be per-
pendicular 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. And the com-
mon square of NH (having bisected GH in N), and the square
of NM is equal 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
the squares of NH and ML, and taking away the common
square of NH, the rect-
angle GMH is equals to
the square of ML; there-
fore as GM to ML, so is
ML to MH, and the tri-
angle LGM is therefore
equiangular to HLM,
and the angle HLM
equal to the angle LGM
and the straight line LM,

3

K

LR

GN Q H MP

S

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 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 ÓP, 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