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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

per-
pendicular must be greater or less than LM or NO; but,
as has been shown in the preceding demonstration, a trian-
gle, 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 ÅR is given: and the ratio of AR to BC is given, and consequently the ratio of AD to BC is given ; and the triangle h 9 Dat. ABC is therefore given in species i

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. «

S4. 6.

222. 5. 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.

i 77 Dat. 80.

Dat.

PROP. LXXVIII IT a triangle have one angle given, and if the ra. tio 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 A D perpendicular to BC, the *41. 1. rectangle AD, BC, has a given ratio to its balfa 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 b; and by

the hypothesis, the ratio of the rectangle BA, AC, to the “9 Dat. square of BC is given ; therefore c the ratio of the rectan. 4 1. 6. gle AD, BC, to the square of BC, that is d, 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 M. ABK, EFH, are si

K milar, and the rect

E angle AD, BC, or the

JI rectangle BK, AC,

R which is equal to it, is to the rectangle B D N BA, AC, as the straight line BK to BA, that is, as FH 10 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 rectungle 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 FI.; 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 si

A

milar 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.

"I Cor.

77 Dat.

Otherwise,

Dat.

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 givena ratio to the triangle ABC. Let: 76 Dat. the 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 • Cor. 62 BAC is a given angle; and the

А rectangle BA, AC, has a given ratio to the square of BC; where

D forec the ratio of D to the square

10 Dat. of BC is given; and, by compo- B sition, the ratio of the space

27 Dat. D, together with the square of BC, to 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 FC is therefore givene; and the angle BAC is given, 59 Dat wherefore the triangle ABC is given in species.

? 48 Dat. 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,

K.

PROP. LXXIX. Ir a triangle have a given angle, and if the straight Sce X. line drawn from that angle to the base, inaking a given angle with it, divides the base into segments which have a given ratio to one another; the tria angle 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 have a given ratio to one another; the triangle ABC

is given in species. * 5, 4.

Describe a the circle BAC 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 b 20. 3. double b of it, is given. And the ratio of BE to EC is given, 44 Dat. because they are equal to one another; therefore the c tri

angle BEC is given in species, and the ratio of EB to BC # 7 Dat is given ; also the ratio of CB to BD is given, because the ratio of BD to DC is given;

therefore the ratio of EB to e 9 Dat. BD is givene, 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 * 47 Dat. given ; therefore the triangle AED is given f in species,

and the angle AED given: also, the
angle DEČ is given, because each of
the angles BĚD, BEC, is given;
therefore the angle AEC is given,
and the ratio of EA to EC, which
are equal, is given ; and the triangle B
AEC is therefore givene in species,
and the angle ECA is given; and the

angle ECB is given; wherefore the angle ACB is given; 6 43 Date and the angle 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 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,

D

* 26. I.

PROP. LXXX.

L. Ir 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.

First, let the sides AB, AC, be equal to one another, therefore the perpendicular AD bisects a the base BC; and the ratio of AD to BC, and therefore to its half DB, is given ; and the

1 angle ADB is given; wherefore the triangle*

* 43 Dat. ABD and consequently the triangle ABC is B D с given b in species.

b 44 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 E Ato 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 giveno; wherefore the ra-c9 Dat. tio of EB. to BC is givend; and the ratio of AD to BC is given by the 2. hypothesis, therefore the ratio of AD to BE is given; and the ratio of BE to EA was shown to be given; where- B

C D fore the ratio of AD to EA is given ; and ADE is a right angle, therefore the triangle ADE is givene in species, and 46 Dat. the angle AEB given; 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.

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 Eą be placed in EB towards B, the point F divides the

A 16 Dat

143 Dat.

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