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 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 shewn in the preceding demonstration, a triangle fimilar to ABC can be defcribed in the fegment EGF upon the bafe 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 similar 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 species.

Draw AD perpendicular to BC; therefore the triangle ARD is given in fpecies; wherefore the ratio of AD to AR is given; and the

h

ratio of AR to BC is given, and consequently ↳ the ratio of AD h. 9. Dat. to 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 ftraight lines drawn from these angles to the bafes, 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 base, so is the other to its base *. wherefore, by Cor. 1. the triangles are fimilar.

A triangle fimilar to ABC may be found thus; haying defcribed 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 fhewn, the ratio of AD to BC, that is of FK to EE, 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 fimilar to triangle LEF. by Cor. 1.

k.

$4.6.

222954

F a triangle have one angle given, and if the ratio of

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 reca. 41. 1. tangle AD, BC has a given ratio to its half the triangle ABC. and because the angle BAC is given, the ratio of the triangle ABC b. Cor. 62. to the rectangle BA, AC is given b; and, by the hypothesis, the ratio of the rectangle BA, AC to the fquare of BC is given. therec. 9. Dat. fore 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. wheree. 77. Dat. fore the triangle ABC is given in fpecies.

Dat.

d. 1. 6.

A

A triangle fimilar 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 fimilar. and the rectangle AD, BC, or the rectangle BK, AC, which is equal to it, is to the rectangle BA, AC, as the straight line BK to BA, that is as FH to FE;. let the given ratio of the BD

M

E

L

P

H

R

N

C F

rectangle BA, AC to the square of BC be the fame with the ra tio 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 same with the ratio of HF to FL. and because AD is not greater than the flraight line MN in the fegment of the circle described 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 base PQ_ the fame with the ratio of HF to FL. then the triangle ABC is fimilar to

OPQ, because, as has been fhewn, 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. f. 1. Cor. 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.

77. Dat.

Because the angle BAC is given, the excefs of the square of both the fides BA, AC together above the square of the third fide BC has a given a ratio to the triangle ABC. let the figure D a. 76. Dat. 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

Dat.

A

c. 10. Dat.

D

d. 7. Dat.

BA, AC is given ", because BAC is a given angle; and the rec- b. Cor. 62, tangle BA, AC has a given ratio to the fquare of BC; wherefore the ratio of D to the square of BC is given. and, by composition, the ratio of the space D together with the fquare of BC to theB

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

The compofition of this which depends upon those of the 76. and 48. Propofitions is more complex than the preceding compo fition which depends upon that of Prop. 77. which is easy.

IF

F a triangle have a given angle, and if the ftraight See N. line drawn from that angle to the bafe, making a given angle with it, divides the base 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 base 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,

a. 5. 4.

b

Defcribe the circle BAC about the triangle, and from its cen ter E draw EA, EB, EC, ED. because the angle BAC is given, the b. 20. 3. angle BEC at the center, which is the double of it, is given. and the ratio of BE to EC is given, because they are equal to one anoc. 44. Dat. ther; therefore the triangle BEC is given in species, and the ratio d. 7. Dat. of EB to BC 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 given . and the angle EBC is given, wherefore the triangle EBD is given in fpecies, and the ratio of EB, that is of EA to ED is therefore given. and the angle EDA is given, because each of the f. 47. Dat. angles BDE, BDA is given. therefore the triangle AED is given £ in species, and the angle AED given; also the angle DEC is given, because each of the angles BED, BEC is given; therefore the angle AEC is given. and the ratio of EA to EC, which are equal, is given; and the triangle AEC is therefore given in fpecies, and the angle ECA given. and the angle

B

D

ECB is given, wherefore the angle ACB is given. and the angle g. 43. Dat. BAC is also given; therefore & the triangle ABC is given in fpecies.

L.

A triangle fimilar to ABC may be found, by taking a straight line given in pofition and magnitude, and dividing it in the given ratio which the fegments 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 ftraight line be drawn from the point of divifion 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 fhall contain a triangle fimilar to ABC, as may eafily be fhewn.

The Demonstration may be also made in the manner of that of the 77. Prop. and that of the 77. may be made in the manner of this.

IF

PROP. LXXX.

F the fides 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 fides 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 fpecies. First, let the fides AB, AC be equal to one another, therefore

the perpendicular AD bifects the bafe BC. and
the ratio of AD to BC, and therefore to its half
DB is given; and the angle ADB is given. where- 1
fore the triangle * ABD and confequently the tri-
angle ABC is given in species.

b

2

A

BD

A

d. 6. Dat.

But let the fides 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, fo is CA to AE, and, by permutation, as BA to AC, fo is BE to EA, and fo is EA to 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 alfo the ratio of BE to EC is given ; wherefore c. 9. Date the ratio of EB to BC is given d. and the ratio of AD to BC is given by the Hypothefis, therefore the ratio of AD to BE is given; and the ratio of BE to EA was fhewn to be given;, wherefore the fatio of AD to AE is given. andB F C E D ADE is a right angle, therefore the triangle ADE is given in fpecies, and the angle AEB given; the ratio of BE to EA is likewife given, therefore the triangle ABE is given in fpecies, and confequently the angle EAB, as alfo 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 fpecies.

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 cafe, it is to be observed that if the straight line EF equal to EA be placed in EB towards B, the point F divides the base BC into the fegments BF, FC which have to one another the ratio of the fides BA, AC, because BE, EA, or £F, and EC were fhewn to be proportionals, therefore * BF is to FC, as BE to EF, or EA, that is as BA to AC. and AE cannot be less than the altitude of the triangle ABC, but it may be equal to it; which if it be, the triangle, in this case, as also the ratio of the fides, may be thus found, having given the ratio of the perpendi cular to the base. take the straight line GH given in pofition and magnitude, for the bafe of the triangle to be found; and let the

e. 46. Dat.

f. 43, Dat.

* 19. Sa