(8. 1.) to the angle BAC; therefore, because the angle EDF which is equal to the angle BAC, has been found, the angle BAC is given (1. def.); in like manner the angles at B, C are given. And because the sides AB, BC, CA are given, their ratios to one another are given (1. dat.), therefore the triangle ABC is given (3. def.) in species.

Ir each of the angles of a triangle be given in magnitude, the triangle is given in species.

Let each of the angles of the triangle ABC be given in magnitude, the triangle ABC is given in species.

B

A

D

CE F

Take a straight line DE given in position and magnitude, and at the points D, E make (23. 1.) the angle EDF equal to the angle BAC; and the angle DEF equal to ABC; therefore the other angles EFD, BCA are equal, and each of the angles at the points A, B, C is given; wherefore each of those at the points D, E, F is given: and because the straight line FD is drawn to the given point D in DE, which is given in position, making the given angle EDF; therefore DF is given in position (32. dat.). In like manner EF also is given in position; wherefore the point F is given: and the points D, E are given; therefore each of the straight lines DE, EF, FD is given (29. dat.) in magnitude; wherefore the triangle DEF is given in species (42. dat.); and it is similar (4. 6. 1. def. 6.) to the triangle ABC: which is therefore given in species.

Ir one of the triangles of a triangle be given, and if the sides about it have a given ratio to one another; the triangle is given in species.

Let the triangle ABC have one of its angles BAC given, and let the sides BA, AC about it have a given ratio to one another; the triangle ABC is given in species.

Take a straight line DE given in position and magnitude, and at the point D, in the given straight line DE, make the angle EDF equal to the given angle BAC; wherefore the angle EDF is given; and because the straight line FD is drawn to the given point D in ED which is given in position, making the given angle EDF; therefore FD is given in position (32. dat.). And because the ratio of BA to AC is given, make the

A

D

ratio of ED to DF the same with it,

and join EF; and because the ratio of

ED to DF is given, and ED is given,

C

E

F

therefore (2. dat.) DF is given in mag- B
nitude: and it is given also in position, and the point D is given,

1

wherefore the point F is given (30. dat.); and the points D, E are given, wherefore DE, EF, FD are given (29. dat.) in magnitude; and the triangle DEF is therefore given (42. dat.) in species; and because the triangles ABC, DEF have one angle BAC equal to one angle EDF, and the sides about these angles proportionals; the triangles are (6. 6.) similar; but the triangle DEF is given in species, and therefore also the triangle ABC.

Ir the sides of a triangle have to one another given ratios, the triangle is given in species.

Let the sides of the triangle ABC have given ratios to one another, the triangle ABC is given in species.

A

DE F

C

G

Take a straight line D given in magnitude; and because the ratio of AB to BC is given, make the ratio of D to E the same with fit; and D is given, therefore (2. dat.) E is given. And because the ratio of BC to CA is given, to this make the ratio of E to F the same; and E is given, and therefore (2. dat.) F; and because as AB to BC, so is D to E; by composition AB and BC together are to BC, as D and E to F; but as BC to CA, so is E to F; therefore, ex æquali, (22. 5.) as AB and BC are to CA, so are D and E to F, and AB and BC are greater (20. 1.) than CA; therefore D and E are greater (A. 5.) than F. In B the same manner any two of the three D, E, F are greater than the third. Make (22. 1.) the triangle GHK whose sides are equal to D, E, F, so that GH be equal to D, HK to E, and KG to F; and because D, E, F, are each of them given, therefore GH, HK, KG are each of them given in magnitude; therefore the triangle GHK is given (42. dat.) in species; but as AB to BC, so is (D to E, that is) GH to HK; and as BC to CA, so is (E to F, that is) HK to KG; therefore, ex æquali, as AD to AC, so is GH to GK. Wherefore (5. 6.) the triangle ABC is equiangular and similar to the triangle GHK; and the triangle GHK is given in species; therefore also the triangle ABC is given in species.

H

K

COR. If a triangle is required to be made; the sides of which shall have the same ratios which three given straight lines D, E, F have to one another; it is necessary that every two of them be greater than the third.

Ir 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.

Take a straight line DE given in position and magnitude; and because the ratio of AB to BC is given, make as AB to BC, so DE to EF; and because DE has a given ratio to EF, and DE is given, therefore (2. dat.) EF is given; and because as AB to BC, so is DE to EF; and AB is less (19. 1.) than BC, therefore DE is less (A. 5.) than EF. From the point D draw DG at right angles 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 the circle is given (6. def.) in position; and straight line DG is given (32.

A

D F

C

B

E

dat.) in position, because it is drawn to the given point D in DE given in position, in a given angle; therefore (28. dat.) the point G is given; and the points D, E are given: wherefore DE, EG, GD are given (29. dat.) in magnitude, and the triangle DEG in species (42. dat.). And because the triangles ABC, DEG have the 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 (7. 6.) and similar to the triangle DEG: but DEG 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.

Ir 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 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 (32. 1.) angle BAC is given; therefore the triangle ABC is given (43. dat.) in species; and it is evident that in this case the given angle ABC must be acute.

A

B

C

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

A

Ꭰ

the side AC; 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 (32. dat.) 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 given (2. dat.), and BA is less than AC, therefore ED is less (A. 5.) than DG. B 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 (6. def.) in position, as also the straight line EF, the point F is given (28. dat.); and the points D, E are given; wherefore the straight lines DE, EF, FD are given (29. dat.) in magnitude, and the triangle DEF in species (42. dat.). And because G

E

F

BA is less than AC, the angle ACB is less (18. 1.) than the angle ABC, and therefore ABC is less (1. 7. 1.) 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 (7. 6.) similar, and DEF is given in species, wherefore the triangle ABC is also given in species.

A

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 (32. dat.) in position: also draw DG perpendicular to EF; therefore if the ratio of BA to AC be the same with the ratio of ED to the perpendicular DG, the triangles ABC, DEG are similar (7. 6.), because the angles ABC, DEG are equal, and DGE is a right angle: therefore the angle ACB is a right angle, and the triangle ABC is given in (43. dat.) E 'species.

B

с

G F

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 (8. 5.) the ratio of BA to AM, because AC is greater than AM. Make as BA to AC, 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 greater (10. 5.) than DG; and because BA is great

er than AC, ED is greater (A. 5.) 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 B L 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 cir

A

C

cle meeting BC again in L: and if the angle ACB E K 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 DGF be less than a right angle, DKE must be greater than one; 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 proportionals, the triangle ABC is similar (7. 6.) to the triangle DEF. In the same man- B L ner, the triangle ABL is similar to DEK. 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 different species, to which the things mentioned as given in the proposition can agree.

F

H

A

C

And

E K

Ir 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 (9. 1.) the angle BAC by the straight line AD; therefore the angle BAD is given. And because as BA to AC, so is (3. 6) BD to DC; by permutation, as AB to BD, so is AC to CD: and as BA and AC together to BC, so is (12. 5.) 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 angle BAD is given therefore (47. dat.) the triangle ABD is given in species, and the angle ABD is therefore given: the angle D C BAC is also given: wherefore the triangle ABC is given in species (43. dat.).

B