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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 tri-
angle ABC. because the angles HGK, DAC at the vertices of the
Ifofceles triangles GHK, ADC are equal to one another, these tri-
angles 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 rect- h.
angle DC, CE to the rectangle AF, CE; but as GL to its half, fo
is the rectangle AF, CE to its half which is the triangle ACE, or
the triangle ABC; therefore, ex aequali, 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 fquare of the straight line which is the difference of the s which contain the given angle is lefs than the fquare of the third fide, fhall have a given ratio to the triangle. this is demonftrated the fame way as the preceeding Propofition, by help of the fecond cafe of the Lemma.

54.6. 222.51

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IF the perpendicular drawn from a given angle of a tri- See N. angle to the oppofite fide, or bafe, has a given ratio 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 base BC, have a given ratio to it; the triangle ABC is given in fpecies.

a

If ABC be an Ifofceles triangle, it is evident that if any one of a. 5.& 32.1.

G

N

K

B RDC

ЕОНМ F

its angles be given, the reft are alfo given; and therefore the tiangle 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 firaight line EF given in pofition and magnitude, and upon it defcribe the fegDd 3

ment

ment 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 also are the triangles MLE, DAB, as ML to LE, fo is DA to AB; and as LE to EF, fo is AB to BC; wherefore, ex aequali, as LM to EF, fo is AD to BC. and because the ratio of AD to BC is given, therefore b. 3. Dat. the ratio of LM to EF is given; and EF is given, wherefore ↳ LM alfo is given. complete the parallelogram LMFK, and because LM

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is given, FK is given in magnitude; it is alfo given in pofition, and c. 30. Dat. the point F is given, and confequently the point K; and because thro' K the straight line KL is drawn parallel to EF which is given in d. 31. Dat. pofition, therefore KL is given in pofition; and the circumference

G

N

B RDC

K

EOHM F

e. 18. Dat. ELF is given in pofition, therefore the point L is given . and be: caufe the points L, E, F are given, the straight lines LE, EF, FL f. 29. Dat. are given fin magnitude; therefore the triangle LEF is given in 8. 41. Dat. fpecies . and the triangle ABC is fimilar to LEF, wherefore also 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 less than the ratio of GH 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 bafe.

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, 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. because the angles ENF, ELF are equal, and that the angle EFN is

equal

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 be no other triangle upon the bafe 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 muft be greater or lefs than LM or NO. but, as has been fhewn in the preceeding demonstration, a triangle fimilar to ABC can be described in the feg- ment EGF upon the bafe EF, and the ratio of its perpendicular to the bafe 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 fpecics; wherefore the ratio of AD to AR is given; and the ratio of AR to BC is given, and confequently h the ratio of AD to h. 9. Dat. BC is given; and the triangle ABC is therefore given in fpecies i.

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 fame ratio to the bases, 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. where- k. 54.6. fore, by Cor. 1. the triangles are fimilar.

A triangle fimilar to ABC may be found thus; having deferibed 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 becaufe, as has been fhewn, the ratio of AD to BC, that is of FK to EF, muft be lefs than the ratio of GH to EF; therefore FK is lefs than GH; and confequently the parallel to EF drawn thro' the point K muit meet the circumference of the fegment 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 fimilar to triangle LEF. by Cor 1.

222.5.

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80.

IF a

PROP. LXXVIII.

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 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 ABC to p. Cor. 62. 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 square of BC, that is the ratio of the straight line AD to BC is given. wherefore the trie. 77. Dat.angle ABC is given in fpecies.

Dat.

c. 9. Dat. d. 1. 6.

A

E

H

BD N

C F

G

A triangle fimilar to ABC may be found thus; take a straight 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 rectMO 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 as 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 straight line EF to FL; therefore, ex aequali, 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 because 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 77. 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

OPQ

OPQ. because, as has been shewn, the ratio of AD to BC is the fame 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 fimilar f 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 fquare of BC be given; the triangle ABC is given in species.

Because the angle BAC is given, the excefs of the fquare of both the fides BA, AC together above the square of the third fide BC has

f. I. Cor.

77. Dat.

a given a ratio to the triangle ABC. let the figure D be equal to a. 76. Dat. this excefs; therefore the ratio of D to the triangle ABC is given; and the ratio of the triangle ABC to the rectangle B A,

AC is given b, because BAC is a given angle; and the rectangle BA, b. Cor. 62.

AC has a given ratio to the fquare of
BC; wherefore the ratio of D to the

c

fquare of BC is given. and, by compofition, the ratio of the space D together with the fquare of BC to the fquare B

Dat.

A

c. 10. Dat.

D

C

of BC is given. but D together with the fquare of BC is equal to the fquare of both BA and AC together; therefore the ratio of the fquare of BA, AC together to the fquare of BC is given;

d. 7. Dat.

and the ratio of BA, AC together to BC is therefore given . and e. 59. Dat. the angle BAC is given, wherefore f the triangle ABC is given in f. 48. Dat. fpecies.

The compofition of this which depends upon thofe of the 76. and 48. Propofitions is more complex than the preceeding compofition which depends upon that of Prop. 77. which is easy.

IF

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F a triangle have a given angle, and if the straight line See N. drawn from that angle to the bafe, making a given angle with it, divides the bafe into fegments which have a given ratio to one another; the triangle is given in fpecies.

Let the triangle ABC have the given angle BAC, and let the ftraight line AD drawn to the bafe BC making the given angle ADB, divide BC into the fegments BD, DC which have a given ratio to one another; the triangle ABC is given in fpecies.

Defcribe

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