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IF a triangle has one of its angles which is not a right angle given, and if the sides about another angie 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.
qual; and because the angle ABC is given, a S2. 1.
the angle ACB, and also the remaining a
angle BAC is given ; therefore the trianh 43. dat. gle ABC is given b in species : and it is B с
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 adjacent to the given angle be less than the side AC: take a straight line DE given in position and
magnitude, and make the angle DEF equal to the given angle c 32. dat. ABC; therefore EF is given c in position; and because the
ratio of B A 10 AC is given, as BA
given, and ED is given, the di 2. dat. straight line DG is givend, and BA is less than AC, therefore ED
C e A. 5. is lesse than DG. From the centre D at the distance DG de
D scribe the circle GF meeting EF
in F, and join DF; and because * 6. def. the circle is givenf in position, as
also the straight line EF, the point E $ 23. dat. F is giveng ; and the points D, E
are giren ; wherefore the straight 11 29. dat. lines DE, EF, FD are givenh in G i 49. dat. magnitude, and the triangle D:F k 18. 1. in speciesi, and because BA is less than AC, the angle ACB is I 1. 7. 1. lessk than the angle ABC, and therefore ACB is less than a right angle. 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 m similar, and m 7. 6. DEF is given in species, wherefore the triangle ABC is also given in species.
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
A case, take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC; therefore FE is given c in posi
& 32. dat. tion: also draw DG perpendicular to B EF; therefore if the ratio of BA to AC be the same with the ratio of ED
D to the perpendicular DG, the triangles ABC, DEG are similarm, because the angles ABC, DEG are equal, and DGE is a right angle: therefore the angle E.
G F ACB is a right angle, and the triangle
b 43. dat, ABC is given inb species.
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 thano the ratio of BA to AM, because AC is greater than AM. Make as BA to ACO 8.5. so 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 greaterp than DG ; and because BA is great
p 10.5. er than AC, ED is greatere than DH.
e A. 5. 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
E K EF in the points F, K which are given,
H 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 if the angle ACB be less than a right angle, ALM must
be greater than a right angle ; and on the contrary. In the same
F in this third case, there are always two
H triangles of a different species, to the things mentioned as given in the proposition can agree.
m 7. 6.
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.
Bisect a the angle BAC by the straight line AD; therefore a 9.1. b 3, 0.
the angle BAD is given. And because as BA to AC, so is b
A c 12. 5.
gether to BC, so is C AB to BD. But the
is given, and the angle BAD is given; B D d 47. dat. therefored the triangle ABD is given in
species, and the angle ABD is therefore given ; the angle BAC e 43. dat. is also given, wherefore the triangle ABC is given in species e.
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
a 9. 1.
G are to EP, as b FE to EL, that
b 3. 6. is, as H to K.
But if the ratio of H to K
IF a triangle has one angle given, and if the sides about another angle, both together, have a given ra. tio to the third side; the triangle is given in species.
Let the triangle ABC have one angle ABC given, and let the two sides BA, AC about another angle BAC have a given ratio to BC; the triangle ABC is given in species.
Suppose the angle BAC to be bisected by the straight line AD; BA and AC together are to BC, as AB to BD, as was shown in the preceding proposition. But the ratio of BA and
AC together to BC is given, therefore also the ratio of AB to . a 44. dat. BD is given. And the angle ABD is given, wherefore a the
triangle ABD is given in species : and consequently the angle BAD, and its double the angle BAC
A are given ; and the angle ABC is gi
Therefore the triangle ABC is b 43. dat. given in speciesb. A triangle which shall have the
к tio of H to K the given ratio ; and
M by prop. 44, find the triangle EFL, which has the angle EFG for one of its angles, and the ratio of the sides F
LG EF, FL about this angle the same with the ratio of H to K; and make the angle LEM equal to the angle FEL. And because the ratio of H to K is the ratio which two sides of a triangle have to the third, H must be greater than K; and because EF is to FL, as H to K, therefore EF is greater than FL, and the angle FEL, that is, LEM, is therefore less than the angle ELF. Wherefore the angles LFE, FEM are less than two right angles, as was shown in the foregoing proposition, and the straight lines FL, EM must meet if produced; let them meet in G, EFG is the triangle which was to be found; for EFG is one of its angles, and because the angle FFG is bisected by EL, the two sides FE EG together have to the third side FG the ratio of EF to FL that is, the given ratio of H to K.