Sidebilder
PDF
ePub

the segment of a circle EGF containing all 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 62. 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 Fis given, and consequently e the point K ; and be

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

[merged small][merged small][merged small][merged small][merged small][ocr errors]

g 42. dat.

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 giveni, f 29.dat. the straight lines LE, EF, FL, are given fin magnitude ; there

fore the triangle LEF is given in species 8; 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 segnient 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 LF, 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 thie 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 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 h the ratio h 9. dar, of AD to BC is given ; and the triangle ABC is therefore given in speciesi.

i 77. das 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, cach to each ; then the trian. gles 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

$ 22. 5. simila"

triangle similar to ABC may be found thus: having deEuribed 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 ihat AD is to BC, as KF, that is LM, to EF, the triangle ABC is similar to the triangle LEF, by Cor. 1.

24. 6.

80s

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.

A

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 recta 41. 1. angle AD, BC has a given ratio to its half a 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 c9. dat. given; therefore c the ratio of the rectangle AD, BC to the d 1.6. square of BC, that is d, the ratio of the straight line AD to BC e 77. dat. is given ; wherefore the triangle ABC is given in species e.

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

мо
are similar, and the rect-
angle AD, BC, or the

E L
rectangle BK, AC which

P
is equal to it, is to the
rectangle BA, AC, as the
straight line BK to BA,
that is, as FH to FE. Let BD N C 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 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 an-
gle 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

R

shown, the ratio of AD to BC is the same with the ratio of (HF 10 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 similarf to the triangle POQ.

f 1. Cor.

77. dai. 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 by Cor. 62. angle ; and the rectangle BA, AC has

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

D

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

d 7. dat. the ratio of the space D together with ihe square of BC to the square of BC B

с 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 43. 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.

[blocks in formation]

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, 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 BC into the segments BD, DC which have

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

species. a 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 b 20. 3. given, the angle BEC at the centre, which is the double b of it,

is given. And the ratio of BE to EC is given, because they c 44. dat. are equal to one another; therefore the triangle BEC is

given in species, and the ratio of EB to BC is given ; also the d 7. dat. ratio of CB to BD is givend, because the ratio of BD to DC e 9. dat. is given ; therefore the ratio of EB 10 BD is given, and the

angle EBC is given, wherefore the triangle EBD is given e 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 f in species, and the angle AED gi

ven ; also the angle DEC is given, because
each of the anglos BED, BEC is given ;
therefore the angle AEC is given, and the
ratio of EA to EC, which are equal, is gi-

B

C ven; and the triangle AEC is therefore

D given in species, and the angle ECA gi

ven; and the angle ECB is given, wherefore the angle ACB is g 43. dat. given, and the angle BAC is also given ; therefore 5 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.

« ForrigeFortsett »