PROP. XLVI.

If the sides of a right angled triangle about one of the acute angles have a given ratio to one another; the triangle is given in species.

Let the sides AB, BC about the acute angle ABC of the triangle ABC which has a right angle at A, have a given ratio to one another; the triangle ABC is given in species.

b

43.

Take a straight line DE given in position and magnitude; and because the ratio of AB to BC is given, as AB to BC, so make DE to EF; and because DE has a given ratio to EF, and DE is given, therefore a EF is given; and because as AB a 2. dat. to BC, so is DE to EF; and AB is less than BC; therefore b 19. 1. DE is less than EF. From the point D draw DG at right an- c A. 5. gles to DE, and from the centre E, at the distance EF, describe a circle which

shall meet DG in two points; let G be either of them, and join EG; therefore the circumference of

e

A

D

F

G

the circle is given in position; and the straight line DG is d 6. def. given in position, because it is drawn to the given point D e 32. dat. in DE given in position, in a given angle; therefore f the f28. dat. point Gis given: and the points D, E are given, wherefore DE, EG, GD are given & in magnitude, and the triangle DEG in g 29. dat. species h. And because the triangles ABC, DEG have the h 42. dat. angle BAC equal to the angle EDG, and the sides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD less than a right angle; the triangle ABC is equiangular and similar to the triangle DEG: But DEG i 7. 6. is given in species; therefore the triangle ABC is given in species: And in the same manner, the triangle made by drawing a straight line from E to the other point in which the circle meets DG is given in species.

See N.

a 32. 1.

44.

b 43. dat.

PROP. XLVII.

If a triangle has one of its angles which is not a right angle given, and if the sides about another angle have a given ratio to one another; the triangle is given in species.

Let the triangle ABC have one of its angles ABC a given angle, but not a right angle, and let the sides BA, AC about another angle BAC have a given ratio to one another; the triangle ABC is given in species.

First, Let the given ratio be the ratio of equality, that is, let the sides BA, AC, and consequently the angles ABC, ACB be equal; and because the angle ABC is given, the angle ACB, and also the remaining a angle BAC is given; therefore the triangle ABC is given bin species; and it is evident, that, in this case, the given angle ABC must be

acute.

Next, Let the given ratio be the ratio of a less to a greater, that is, let the side AB adja

B

C

cent to the given angle be less than the side AC: Take a straight line DE, given in position and magnitude, and make the angle

с

e 32. dat. DEF equal to the given angle ABC; therefore EF is given in position; and because the ratio of BA to AC is given, as BA to AC, so make ED to DG; and because the ratio of ED to DG is given, and ED is given, the straight line DG is

e A. 5.

f 6. def.

B

A

C

D

E

F

d 2. dat. given, and BA is less than AC, therefore ED is less than DG. From the centre D, at the distance DG, describe the circle GF, meeting EF in F, and join DF: And because the circle is given in position, as also the straight line EF, the g 28. dat. point Fis given; and the points D, E are given; wherefore the straight lines DE, EF, FD are given in magnitude, and the triangle DEF, in species. And because BA is less than AC, the angle ACB is less than the angle ABC, and therefore ACB is less than a right angle.

h 29. dat.

i 42. dat.

k 18. 1. 1 17. 1.

h

G

1

In the same manner, because ED is less than DG or DF; the angle DFE is less than a right angle: And because the triangles ABC, DEF have the angle ABC equal to the angle DEF, and the sides about the angles BAC, EDF proportionals, and each of the other angles ACB, DFE less than a right angle; the triangles ABC, DEF are similar, and DEF is given in m 7. 6. species, wherefore the triangle ABC is also given in species.

с

m

A

c 32. dat.

Thirdly, Let the given ratio be the ratio of a greater to a less, that is, let the side AB adjacent to the given angle be greater than AC; and as in the last case, take a straight line DE, given in position and magnitude, and make the angle DEF equal to the given angle ABC; therefore EF is given in position: Also draw DG perpendicular to EF; therefore if the ratio of BA to AC be the same with the ratio of ED B to the perpendicular DG, the triangles ABC, DEG are similar m, because the angles ABC, DEG are equal, and DGE is a right angle: Therefore the angle ACB is a right angle, and the E triangle ABC is given in species.

e

G F

b 43. dat.

A

p 10 5.

M

L

C

e A. 5.

Ꭰ

But if, in this last case, the given ratio of BA to AC be not the same with the ratio of ED to DG, that is, with the ratio of BA to the perpendicular AM drawn from A to BC; the ratio of BA to AC must be less than the ratio of BA to AM, o 8. 5. because AC is greater than AM. As BA to AC, so make ED to DH; therefore the ratio of ED to DH is less than the ratio of (BA to AM, that is, than the ratio of) ED to DG; and consequently DH is greater P than DG; and because BA is greater than AC, ED is greater than DH. From the centre D, at the distance DH, describe the circle KHF, which necessarily meets the straight line EF in two points, because DH is greater than DG, and less than DE. Let the circle meet EF in the points F, K which are given, as was shown in the preceding case; and DF, DK being joined, the triangles DEF, DEK are given in species, as was there shown. From the centre A, at the distance AC, describe a circle meeting BC again in L: And

B

E K

F

H

m 7. 6.

B L

A

M

C

D

if the angle ACB be less than a right angle, ALB must be
greater than a right angle; and on the contrary. In the same
manner, if the angle DFE be less than a right angle, DKE
must be greater than a right angle; and on the contrary. Let
each of the angles ACB, DFE be
either less or greater than a right angle;
and because in the triangles ABC,
DEF the angles ABC, DEF are equal,
and the sides BA, AC, and ED, DF,
about two of the other angles propor-
tionals, the triangle ABC is similar m
to the triangle DEF. In the same
manner, the triangle ABL is similar to
DEK. And the triangles DEF, DEK
are given in species: therefore also the
triangles ABC, ABL are given in
species. And from this it is evident,
that, in this third case, there are always two triangles of a dif-
ferent species, to which the things mentioned as given in the
proposition can agree.

E K

F

H

PROP. XLVIII.

a 9 1. b 3. 6.

45.

c 12. 5.

If a triangle has one angle given, and if both the sides together about that angle have a given ratio to the remaining side; the triangle is given in species.

Let the triangle ABC have the angle BAC given, and let the sides BA, AC together about that angle have a given ratio to BC; the triangle ABC is given in species.

A

Bisect the angle BAC by the straight line AD; therefore the angle BAD is given. And because as BA to AC, so is b BD to DC, by permutation, as AB to BD, so is as AC to CD; and as BA and AC together to BC, so is AB to BD. But the ratio of BA and AC together to BC is given, wherefore the ratio of AB to BD is given, and the B

с

D

C

d 47. dat. angle BAD is given; therefore the triangle ABD is given. in species, and the angle ABD is therefore given; the angle BAC is also given, wherefore the triangle ABC is given in

e 43. dat. species.

A triangle, which shall have the things that are mentioned in the proposition to be given, can be found in the following

manner. Let EFG be the given angle, and let the ratio of H to K be the given ratio which the two sides about the angle EFG must have to the third side of the triangle; therefore, because two sides of a triangle are greater than the third side, the ratio of H to K must be the ratio of a greater to a less. Bisect the angle EFG by the straight line FL; and by the a 9. 1. 47th proposition, find a triangle of which EFL is one of the angles, and in which the ratio of the sides about the angle opposite to FL is the same with the ratio of H to K: To do which, take FE given in position and magnitude, and draw EL perpendicular to FL: Then, if the ratio of H to K be the same with the ratio of FE to EL, produce EL, and let it meet FG in P; the triangle FEP is that which was to be found: for it has the given angle EFG; and because this angle is bisected by FL, the sides EF, FP together are to EP, as b FE to EL, that is, as H to K.

But, if the ratio of H to K be not the same with the ratio of FE to EL, it must be less than it, as was shown in Prop.

H

E

K

G

P

has the given angle EFL, and

b 3. 6.

the ratio of the sides about the angle opposite to FL the same with the ratio of H to K. By Prop. 47. find these triangles EFM, EFN, each of which has the angle EFL for one of its angles, and the ratio of the side FE to EM or EN the same with the ratio of H to K; and let the angle EMF be greater, and ENF less than a right angle. And because H is greater than K, EF is greater than EN, and therefore the angle EFN, that is, the angle NFG, is less than the angle ENF. To f 18. 1. each of these add the angles NEF, EFN; therefore the angles NEF, EFG are less than the angles NEF, EFN, FNE, that is, than two right angles; therefore the straight lines EN, FG must meet together when produced; let them meet in O, and produce EM to G. Each of the triangles EFG, EFO has the things mentioned to be given in the proposition: for each of them has the given angle EFG; and because this angle is bisected by the straight line FMN, the sides EF, FG together have to EG the third side the ratio of FE to EM, that is, of H to K. In like manner, the sides EF, FO together have to EO the ratio which H has to K.