as the base BC, is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

EAF

Produce BD both ways to the points H, L, and take any number of straight lines BG, GH, each equal to the base BC; and DK, KL, any number of them, each equal to the base CD; and join AG, AH, AK, AL. Then, because CB, BG, GH are all equal, the triangles AHG, AGB, ABC are all equal (38.1.); Therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC. For the same reason, whatever the base LLC is of the base CD, the same multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC (38. 1.): and if the base HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC; and if less, less. Therefore, since there

HGBCDKL

are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC, and the triangle AHC; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC; and since it has been shewn, that if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; and if less, less; Therefore (def. 5. 5.), as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle ABC (41. 1.), and the parallelogram CF double of the triangle ACD, and because 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 shewn, 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.

COR. From this it is plain, that triangles and parallelograms that have equal altitudes, are to one 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 proposition, the demonstration will be the same.

1

PROP. II. THEOR.

If a straight line be drawn parallel to one of the sides of a triangle, it will cut the other sides, or the other sides produced, proportionally: And if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section will 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: but 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.).

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

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.

PROP. III. THEOR.

If the angle of a triangle be bisected by a straight line which also cuts the buse; 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, bisects the vertical angle.

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.

E

A

Through the point C draw CE parallel (Prop. 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 exterior angle BAD is equal to the interior 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 to the side (6. 1.) AC. And because AD is drawn parallel to one of the sides of

B

the triangle BCE, viz. to EC, BD is

D

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.).

Next, let 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 AB is to AC, as AB 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

A

E

AEC is equal to the exterior and op

angle BAD is equal to the angle

CAD: Therefore the angle BAC is cut into two equal angles by the straight

PROP. A. THEOR.

I' the exterior angle of a triangle be bisected by a straight line which also cuts the base produced; the segments between the bisecting line and the extremities of the base have the same ratio which the other sides of the triangles have to T one another; And if the segments of the base produced have the same ratio which the other sides of the triangles have, the straight line, drawn from the vertex to the point of section, bisects the exterior angle of the triangle.

Let the exterior angle CAE, of any triangle ABC, be bisected by the straight line AD which meets the base produced in D; BD is to DC, as BA to AC.

Through C draw CF parallel to AD (Prop. 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 exterior angle DAE is equal to the interior and opposite angle CFA; But the angle ACF has

been proved to be equal to the an

gle DAE; therefore also the angle ACF is equal to the angle CFA, and consequently the side AF is 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; therefore as BD is to DC, so is BA to AC.

Now let 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 also BD to DC, BA to AF (2.6.); therefore BA is to AC, as BA to AF (11. 5.), wherefore AC is equal to AF (9.5.), and the angle AFC equal (5. 1.) to the angle ACF: but the angle AFC is equal to the exterior angle EAD, and the angle ACF to the alternate angle CAD; therefore also EAD is equal to the angle CAD

PROP. IV. THEOR.

The sides about the equal angles of equiangular triangles are proportionals; ana 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 (4. Cor. 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 an gles.

F

A

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, which is equal to ACB, are also less than two right angles: wherefore BA, ED produced shall meet (1 Cr. 29.1.); 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,

B

C

D

E

one of the sides of the triangle FBE, BA:AF:: BC: CE (2.6.): but AF is equal to CD; therefore (7. 5.) BA: CD :: BC: CE; and alternately, BA: BC :: DC: CE (16.5.): Again, because CD is parallel to BF, BC: CE :: FD: DE (2.6.); but FD is equal to AC; therefore BC : CE :: AC: DE; and alternately, BC: CA :: CE: ED. Therefore because it has been proved that AB : BC :: DC: CE; and RC CA

CE: ED, ex æquali, BA : AC :: CD: DE.

PROP. V. THEOR.

If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides.

Let the triangles ABC, DEF have their sides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and consequently ex æquali, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC being equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

At the points E, F, in the straight line EF, make (Prop. 23.1.) the angle FEG equal to the angle ABC, and the angle EFG equal to BCA, wherefore the remaining angle BAC is equal to the remaining angle EGF (4. Cor. 32. 1.), and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they have their sides opposite to the B equal angles proportionals (4. 6.).

Wherefore,

D

A

E

F

AB: BC:: GE: EF; but by supposition,

AB: BC:: DE: EF, therefore,

CG

DE: EF :: GE: EF. Therefore (11.5.) DE and GE have