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47. 1. equal to the squares of AE, ED, that is, to b the square of AD, together with the rectangle BD, DC; the other case is shown in the same way by 6. 2. Elem.

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Ir a triangle have a given angle, the excess of the square of the straight line which is equal to the two sides that contain the given angle, above the square of the third side, shall have a given ratio to the triangle.

Let the triangle ABC have the given angle BAC, the excess of the square of the straight line which is equal to - BA, AC, together above the square of BC, shall have a given ratio to the triangle ABC.

Produce BA, and take AD equal to AC, join DC, and
produce it to E, and through the point B draw BE parallel
to AC: join AE, and draw AF perpendicular to DC; and
because AD is equal to AC, BD is equal to BE; and BC
is drawn from the vertex B of the isosceles triangle DBE:
therefore, by the Lemma, the square of BD, that is, of BA
and AC together, is equal to the rectangle DC, CE, toge-
ther with the square of BC: and therefore the square of
BA, AC, together, that is, of BD,
is greater than the square of BC by
the rectangle DC, CE; and this
rectangle has a given ratio to the
triangle ABC, because the angle
BAC is given, the adjacent angle

CAD is given; and each of the
angles ADC, DCA, is given, for

5. & 32.1. each of them is the halfa of the

B

C

H

L

E

K

given angle BAC; therefore the triangle ADC is given in 43 Dat. species; and AF is drawn from its vertex to the base in a

given angle; wherefore the ratio of AF to the base CD is 50 Dat. given; and as CD to AF, so isd the rectangle DC, CE, to 1. 6. the rectangle AF, CE; and the ratio of the rectangle AF, 41. 1. CE, to its halfe, the triangle ACE, is given; therefore the ratio of the rectangle DC, CE, to the triangle ACE, that 137. 1. is, to the triangle ABC, is givens; and the rectangle DC, 9 Dat. CÉ, is the excess of the square of BA, AC, together, above the square of BC: Therefore the ratio of this excess to the triangle ABC is given.

The ratio which the rectangle DC, CE, has to the trian

gle ABC is found thus: Take the straight line HG given
in position 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 rectangle DC, CE, to the triangle
ABC: Because the angles HGK, DAC, at the vertices of
the isosceles triangles GHK, ADC, are equal to one an-
other, these triangles are similar; and because GL, AF, are
perpendicular to the bases HK, DC, as HK to GL, so ish-
(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 tri-
angle 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 in the preceding proposition, by help of the second case of the Lemma.

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4.6. 22. 5.

Ir the perpendicular drawn from a given angle of See N. a triangle to the opposite side, or base, has 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 *5. & 39. 1.

B RDC

N

E 0

H M

К

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, 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 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; and the point F is given, and conse30 Dat. quently the point K; and because through K the straight line KL is drawn parallel to EF, which is given in position, 31 Dat. therefored KL is given in position: and the circumference

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A

N

K

B RDC

E O H M

• 28 Dat. ELF is given in position; therefore the point L is given. And because the points L, E, F, are given, the straight lines f 29 Dat. LE, EF, FL, are givenf in magnitude; therefore the tri42 Dat. angle LEF is given in speciess; 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 per

pendicular 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 demonstration, a triangle, similar to ABC, can be described in the segment EGF upon the base EF, and the ratio of its perpendicular to the base is the same, as was there shown, 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 triangle ABC.

COR. 2. If a triangle ABC has 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, has 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 consequently the ratio of AD to BC is given; and the triangle ↳ 9 Dat. ABC is therefore given in speciesi.

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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; wherefore, by Cor. 4. 6. 1. the triangles are 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.

22.5.

80.

PROP. LXXVIILY

Ir 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 * 41. 1. rectangle AD, BC, has a given ratio to its halfa the triangle ABC; and because the angle BAC is given, the ratio of the Cor. 62. triangle BAC to the rectangle BA, AC is given ; and by

Dat.

the hypothesis, the ratio of the rectangle BA, AC, to the 9 Dat. square of BC is given; therefore the ratio of the rectan1. 6. gle AD, BC, to the square of BC, that is, the ratio of the straight line AD to BC, is given; wherefore the triangle 77 Dat. 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 si

A

M

K

milar, and the rect

angle AD, BC, or the

E

H

rectangle BK, AC,
which is equal to it,

is to the rectangle B D N

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BA, AC, as the straight 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 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 OK, drawn from that angle to the base PQ, the same with the ratio of HF to FL; then the triangle ABC is si

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