Book VI.

See N.

a 38. I.

TRI

PROP. I. THEOR.

RIANGLES and parallelograms of the fame altitude are one to another as their bases.

Let the triangles ABC, ACD, and the parallelograms EC, CF have the fame altitude, viz. the perpendicular drawn from the point A to BD: Then, as the base BC is to the base CD, fo is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

Produce BD both ways to the points H, L, and take any
number of straight lines BG, GH, each equal to the bafe BC;
and DK, KL, any number of them, each equal to the bafe CD;
and join AG, AH, AK, AL: Then, because CB, BG, GH
are all equal, the triangles AHG, AGB, ABC are all equal a :
Therefore, whatever multiple the bafe HC is of the bafe BC,
the fame multiple is the triangle AHC of the triangle ABC :
For the fame reason, whatever multiple the base LC is of the
bafe CD, the fame mul-
tiple is the triangle ALC
of the triangle ADC :
And if the base HC be
equal to the base CL, the
triangle AHC is alfo
equal to the triangle
ALC; and if the base

HC be greater than the HG BC
bafe CL, likewife the

E A F

D K

L

triangle AHC is greater than the triangle ALC; and if lefs, lefs Therefore, fince there are four magnitudes, viz. the two bafes BC, CD, and the two triangles ABC, ACD; and of the bafe BC and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC and triangle AHC; and of the base CD and triangle ACD, the fecond and fourth, have been taken any equimultiples whatever, viz. the bafe CL and triangle ALC; and that it has been fhown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; b5. def. 5. and if lefs, lefs: Therefore b, as the bafe BC is to the bafe CD, fo is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle

ABC

d 15.5.

ABC,and the parallelogram CF double of the triangle ACD, Book VI. and that magnitudes have the fame ratio which their equimul tiples have d; as the triangle ABC is to the triangle ACD, fo 4.1. 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 is to the triangle ACD, fo is the parallelogram EC to the parallelogram CF; therefore, as the bafe BC is to the base CD, fo is 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 bafes.

e II.5.

Let the figures be placed fo as to have their bases in the fame ftraight line; and having drawn perpendiculars from the vertices of the triangles to the bafes, the straight line which Joins the vertices is parallel to that in which their bafes are f, be- f 33. 1. cause the perpendiculars are both equal and parallel to one another. Then, if the fame construction be made as in the propofition, the demonftration will be the fame.

IF

PROP. II. THEOR.

Fa ftraight line be drawn parallel to one of the fides See N. of a triangle, it fhall cut the other fides, or those produced, proportionally: And if the fides, or the fides produced, be cut proportionally, the straight line which joins the points of fection shall be parallel to the remaining fide of the triangle.

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

:

a 37. I.

Join BE, CD; then the triangle BDE is equal to the triangle CDEa, because they are on the fame base DE, and between the same parallels DE, BC: ADE is another triangle, and equal magnitudes have to the fame, the fame ratio b; there- b 7. 5. fore, 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, fo is BD to DA, because having the fame c 1. 6. 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

Book VI.

d 11. 5.

e I. 6.

£9.5.

€39. 1.

See N.

as the triangle CDE to the triangle ADE, fo is CE to EA. Therefore, as BD to DA, fo is CE to EA d.

Next, Let the fides AB, AC of the triangle ABC, or thefe

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

The fame conftruction being made, Because as BD to DA, fo is CE to EA; and as BD to DA, fo is the triangle BDE to the triangle ADE; and as CE to EA, fo 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 fame ratio to the triangle ADE; and therefore f the triangle BDE is equal to the triangle CDE: and they are on the fame bafe DE; but equal triangles on the fame base are between the fame parallels g; therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PRO P. III. THEOR.

F the angle of a triangle be divided into two equal

a

the fegments of the base fhall have the fame ratio which the other fides of the triangle have to one another: And if the fegments of the base have the fame ratio which the other fides of the triangle have to one another, the ftraight line drawn from the vertex to the point of fection, divides the vertical angle into two equal angles.

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

Through

4

Through the point C draw CE parallel a to DA, and let BA Book VI. produced meet CE in E. Because the straight line AC meets

the parallels AD, EC, the angle ACE is equal to the alternate a 31. I. angle CAD b: But CAD, by the hypothefis, is equal to the b 29. I. angle BAD; wherefore BAD is equal to the angle ACE. A

gain, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and oppofite angle AEC: But the angle ACE has been proved equal to the angle BAD; therefore alfo ACE is equal to the

E

A

angle AEC, and confequently B

the fide AE is equal to the

fide AC: And because AD is drawn parallel to one of the c 6. 1. fides of the triangle BCE, viz. to EC, BD is to DC, as BA to AEd; but AE is equal to AC; therefore, as BD DC, so is d 2. 6. BA to AC c.

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.

€ 7.5.

The fame construction being made; because, as BD to DC, fo is BA to AC; and as BD to DC, fo is BA to AE d, because AD is parallel to EC; therefore BA is to AC, as BA to AE f: f 11. 5. Confequently AC is equal to AE g, and the angle AEC is there- g 9. 5. fore equal to the angle ACE h: But the angle AEC is equal to h 5. I. the outward and oppofite angle BAD; and the angle ACE is equal to the alternate angle CAD b: Wherefore also the angle BAD is equal to the angle CAD: Therefore the angle BAC is cut into two equal angles by the ftraight line AD. Therefore, if the angle, &c. Q. E. D.

PROP.

Book VI. 13

a 31. I.

b 29. I. ⚫ Hyp.

d 6. 1. € 2.6.

f11. 5.

g 9.5.

h 5. I.

PROP. A. THEOR.

IF the outward angle of a triangle made by producing one of its fides, be divided into two equal angles, by a ftraight line which also cuts the base produced; the fegments betweeen the dividing line and the extremities of the bafe have the fame ratio which the other fides of the triangle have to one another: And if the fegments of the bafe produced, have the fame ratio which the other fides of the triangle have, the ftraight line drawn from the vertex to the point of fection 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 ftraight line AD which meets the base produced in D: BD is to DC, as BA to AC.

E

Through Cdraw CF parallel to AD a; and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD : But CAD is equal to the angle DAE; therefore alfo DAE is equal to the angle ACF. Again, because the ftraight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and oppofite angle CFA: But the angle ACF has been proved equal to the angle DAE; therefore alfo the angle ACF is equal to the angle CFA, and confequently the fide AF is equal to the fide

A

F

B

C

D

ACd: And because AD is parallel to FC, a fide of the triangle BCF, BD is to DC, as BA to AF e; but AF is equal to AC; as therefore BD is to DC, fo 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 fame conftruction being made, becaufe BD is to DC, as BA to AC; and that BD is alfo to DC, as BA to AFf; therefore BA is to AC, as BA to AF g; wherefore AC is equal to AF h, and the angle AFC equal h to the angle ACF: But

the