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I.

b 47. fquares of AE, ED, that is, to b the fquare of AD, together with the rectangle BD, DC; the other cafe is fhewn in the fame way by 6.2. Elem.

67.

PROP. LXXVI.

IF a triangle have a given angle, the excess of the fquare of the ftraight line which is equal to the two fides that contain the given angle, above the fquare of the third fide, fhall have a given ratio to the triangle.

Let the triangle ABC have the given angle BAC, the excess of the fquare of the ftraight line which is equal to BA, AC together above the fquare of BC, fhall 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 becaufe AD is equal to AC, BD. is equal to BE; and BC is drawn from the vertex B of the ifofceles triangle DBE, therefore, by the Lemma, the fquare of BD, that is, of BA and AC together, is equal to the rectangle DC, CE together with the fquare of BC; and therefore, the fquare of BA, AC to. gether, that is, of BD, is greater than the fquare of BC by the rectangle DC, CE; and this rectangle has a given ratio to the triangle ABC: Becaufe the angle BAC is given, the adjacent angle CAD is given; and each of the angles ADC, DCA is given, for cach a 5.&32.1. of them is the half a of the given angle BAC; therefore the triangle ADC is

b 43. dat. given in fpecies; and AF is drawn

D

FC

B

E

from its vertex to the bafe in a given angle; wherefore the ratio 50. dat. of AF to the bafe CD is given ; and as CD to AF, fo is d the

d 1. 6.

C 41. I.

rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its halfe the triangle ACE is given; therefore the ratio of the rectangle DC, CE to the triangle f 37. 1. ACE, that is f, to the triangle ABC, is given 8; and the rectangle 9. dat. DC, CE is the excefs of the fquare of BA, AC together above the fquare of BC; therefore the ratio of this excess to the triangle ABC is given.

The ratio which the rectangle DC, CE has to the triangle ABC is found thus: Take the ftraight line GH given in pofi

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 fame with the ratio of
the rectangle DC, CE to the triangle ABC: Because the angles
HGK, DAC at the vertices of the ifofceles triangles GHK,
ADC are equal to one another, thefe triangles are fimilar; and
because GL, AF are perpendicular to the bafes HK, DC, as
HK to GL, fo is h (DC to AF, and fo is) the rectangle DC, h
CE to the rectangle AF, CE; but as GL to its half, fo is the 2
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 ftraight 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 fquare of the ftraight line which is the difference of the fides which contain the given angle is less than the fquare of the third fide, fhall have a given ratio to the triangle. This is demonftrated the fame way as the preceding propofition, by help of the fecond cafe of the Lemma.

PROP. LXXVII.

4. 6. 22. 5.

I.

IF the perpendicular drawn from a given angle of a see N. triangle to the oppofite fide, or bafe, has a given ra

tio to the bafe; the triangle is given in fpecies.

Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the bafe BC, have a given ratio to it; the triangle ABC is given in fpecies.

If ABC be an ifofceles triangle, it is evident that if any as. & 32. 2.

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one of its angles be given, the reft are alfo given; and therefore the triangle is given in fpecies, without the confideration of the ratio of the perpendicular to the bafe, which in this cafe is given by prop. 50.

But when ABC is not an ifofceles triangle, take any ftraight line EF given in pofition and magnitude, and upon it defcribe

the

the fegment of a circle EGF containing an angle equal to the given angle BAC; draw GH bifecting EF at right angles, and join EG, GF: Then, fince the angle EGF is equal to the angle BAC, and that EGF is an ifofceles 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 alfo are the triangles MLĒ, DAB, as ML to LE, fo is DA to AB; and as LE to EF, fo is AB to BC; wherefore, ex æquali, as LM to EF, fo is AD to BC; and because the ratio of AD to BC is given, therefore the ratio of LM b 2. dat. to EF is given; and EF is given, wherefore b LM allo is given. Complete the parallelogram LMFK; and because LM is given, FK is given in magnitude; it is alfo given in pofition, and c 30. dat. the point F is given, and confequently c the point K; and becaufe through K the ftraight line KL is drawn parallel to EF d 31. dat. which is given in pofition, therefore d KL is given in pofition;

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and the circumference ELF is given in position; therefore the e 28. dat. point L is given e. And because the points L, E, F are given, fag. dat. the ftraight lines LE, EF, FL are given fin magnitude; there8 43, dat. fore the triangle LEF is given in fpecies g; and the triangle ABC is fimilar to LEF, wherefore alfo ABC is given in fpecies.

Because LM is lefs than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be lefs than the ratio of GII to EF, which the ftraight line, in a fegment of a circle containing an angle equal to the given angle, that bifects the bafe of the fegment 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 bafe BC, as the perpendicular LM to the base EF; the triangles ABC, LEF are fimilar.

Defcribe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicular to EF; becaufe the angles ENF, ELF are equal, and that

the

the angle EFN is equal to the alternate angle FNL, that is, to the angle FEL in the fame fegment; therefore the triangle NEF is fimilar to LEF; and in the fegment EGF there can he no other triangle upon the base EF, which has the ratio of its perpendicular to that bafe the fame 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 fhewn in the preceding demonstration, a triangle fimilar to ABC can be described in the segment EGF upon the base EF, and the ratio of its perpendicular to the base is the fame, as was there fhewn, 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 fimilar to the triangle ABC.

COR. 2. If a triangle ABC has a given angle BAC, and if the ftraight line AR drawn from the given angle to the oppofite fide BC, in a given angle ARC, has a given ratio to BC; the triangle ABC is given in fpecies.

Draw AD perpendicular to BC; therefore the triangle ARD is given in fpecies; wherefore the ratio of AD to AR is given; and the ratio of AR to BC is given, and confequently h the ra-h tio of AD to BC is given; and the triangle ABC is therefore given in fpecies i.

9. dat.

i 77. dat.

COR. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if ftraight lines drawn from these angles to the bases, making with them given and equal angles, have the fame ratio to the bafes, each to each; then the triangles are fimilar; for having drawn perpendiculars to the bafes from the equal angles, as one perpendicular is to its bafe, fo is the other to its bafe k; wherefore, by Cor. 1. the triangles are f4. 6. fimilar.

A triangle fimilar to ABC may be found thus; having defcribed the fegment EGF and drawn the ftraight line GH as was directed in the propofition, 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 fhewn, 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 lefs than GH; and confequently the parallel to EF drawn through the point K, must meet the circumference of the fegment in two points: Let L be ei ther 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 fimilar to the triangle LEF, by Cor. 1.

PROP.

k

122. 3.

80.

2 41. 1.

b Cor. 62.

PROP. LXXVIII.

IF a triangle have one angle given, and if the ratio of the rectangle of the fides which contain the given angle to the fquare of the third fide be given; the triangle is given in fpecies.

Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the fquare of BC be given; the triangle ABC is given in fpecies.

From the point A, draw AD perpendicular to BC; the rectangle 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 ABC to the rectangle BA, AC is given b; and, by the hypothefis, the ratio of the rectangle BA, AC to the fquare of BC is given; therefore the ratio of the rectangle AD, BC to the fquare of BC, that is, d the ratio of the ftraight line AD to BC, e 77. dat. is given; wherefore the triangle ABC is given in fpecies e

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d 1. 6.

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A triangle fimilar to ABC may be found thus; take a ftraight line EF given in pofition 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 fimilar, and the rect. angle AD, BC, or the rectangle BK, AC, which is equal to it, is to the rectangle BA, AC, as the ftraight line BK to BA, that is, ás FH to FE. Let the given ratio of the rectangle BA, AC to the fquare of BC be the fame with the ratio of the ftraight line EF to FL; therefore, ex æquali, the ratio of the rectangle AD, BC to the fquare of BC, that is, the ratio of the ftraight line AD to BC, is the fame with the ratio of HF to FL; and becaufe AD is not greater than the ftraight line MN in the fegment of the circle defcribed about the triangle ABC, which bifects 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 fo, 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 bafe PQ the fame with the ratio of HF to FL; then the triangle ABC is fimilar to

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