ACD, and that magnitudes have the same ratio which their equimultiples have (15. 5.); as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; and because it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore as the base BC is to the base CD, so is (11. 5.) the parallelogram EC to the parallelogram CF. Wherefore triangles, &c. Q. E. D.

COR. From this it is plain, that triangles and parallelograms that have equal altitudes, are one to another as their bases.

Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are (33. 1), because the perpendiculars are both equal and parallel to one another: then, if the same construction be made as in the sition, the demonstration will be the same.

PROP. II. THEOR.

propo

Ir a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle.*

Let DE be drawn parallel to BC, one of the sides of the triangle ABC; BD is to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal to the triangle CDE (37. 1.), because they are on the same base DE, and between the same parallels DE, BC; ADE is another triangle, and equal magnitudes, have to the same the same ratio (7. 5.); therefore, as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is (1.6.) BD to DA, because having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA. Therefore, as BD to DA, so is CE to EA (11. 5.).

• See Note.

Next, let the sides AB, AC of the triangle ABC, or these

produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; DE is parallel to BC.

The same construction being made; because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE (1. 6.); and as CE to EA, so is the triangle CDE. to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore (9. 5.) the triangle BDE is equal to the triangle CDE; and they are on the same base DE; but equal triangles on the same base are between the same parallels (39. 1.) therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PROP. III. THEOR.

If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another: and if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles.*

Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: BD is to DC, as BA to AC.

See Note:

E

Through the point C draw CE parallel (31. 1.) to DA, and let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD (29. 1.); but CAD, by the hypothesis is equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC: but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal B

A

D C

to the side (6. 1.) AC; and because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE (2. 6.): but AE is equal to AC; therefore, as BD to DC, so is BA to AC (7. 5.).

Let now BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line AD.

The same construction being made; because, as BD to DC, so is BA to AC: and as BD to DC, so is BA to AE (2. 6.), because AD is parallel to EC; therefore BA is to AC, as BA to AE (11. 5.): consequently AC is equal to AE (9. 5.), and the angle AEC is therefore equal to the angle ACE (5. 1.): but the angle AEC is equal to the outward and opposite angle BAD: and the angle ACE is equal to the alternate angle CAD (29. 1.): wherefore also the angle BAD is equal to the angle CAD: therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D.

PROP. A. THEOR.

If the outward angle of a triangle made by producing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced: the segments between the dividing line and the extremities of the base have the same ratio which the other sides of the triangle have to one another: and if the segments of the base produced, have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section divides the outward angle of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be divided into two equal angles by the straight line AD which meets the base produced in D; BD is to DC, as BA to AC.

E

Through Cdraw CF parallel to AD (31. 1.): and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD (29. 1.): but CAD is equal to the angle DAE (Hyp.); therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and opposite angle CFA: but the angle ACF has been proved equal to the angle DAE; therefore also the angle ACF is equal to the angle CFA, and consequently the side AF is B

C

D

equal to the side AC (6. 1.): and because AD is parallel to FC, a side of the triangle BCF, BD is to DC, as BA to AF (2. 6.): but AF is equal to AC; as therefore BD is to DC, so is BA to AC.

Let now BD be to DC, as BA to AC, and join AD; the angle CAD is equal to the angle DAE.

The same construction being made, because BD is to DC, as BA to AC; and that BD is also to DC, as BA to AF (11. 5.): therefore BA is to AC, as BA to AF (9. 5.); wherefore AC is equal to AF (5. 1.), and the angle AFC equal (5. 1.) to the

angle ACF; but the angle AFC is equal to the outward angle EAD, and the angle ACF to the alternate angle CAD: therefore also EAD is equal to the angle CAD. Wherefore, if the outward, &c. Q. E. D.

PROP. IV. THEOR.

The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently (32. 1.) the angle BAC equal to the angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals: and those are the homologous sides which are opposite to the equal angles.

B

D

C

E

Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it: and because the angles ABC, ACB are together less than two right angles (17. 1.) ABC, and DEC, F which is equal to ACB, are also less than two right angles; wherefore A BA, ED produced shall meet (12. Ax. 1.); let them be produced and meet in the point F; and because the angle ABC is equal to the angle DCE, BF is parallel (28. 1.) to CD. Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE (28. 1.): therefore FACD is a parallelogram: and consequently AF is equal to CD, and AC to FD (34. 1.): and because AC is parallel to FE, one of the sides of the triangle FBE, BA is to AF; as BC to CE (2. 6.): but AF is equal to CD; therefore (7. 5.), as BA to CD, so is BC to CE; and alternately, as AB to BC, so is DC to CE (17.1.): again, because CD is parallel to BF, as BC to CE, so is FD to DE (2. 6.): but FD is equal to AC: therefore, as BC to CE, so is AC to DE: and alternately, as BC to CA, so CE, to DE; therefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so CE to ED, ex æquali, (22. 5.) BA is to AC, as CD to DE. Therefore the sides, &c. Q. E. D.