Two fides of one figure are faid to be reciprocally proportional to two fides of another, when one of the fides of the firft is to one of the fides of the other, as the remaining fide of the other is to the remaining fide of the first.

III.

A ftraight line is faid to be cut in extreme and mean ratio, when the whole is to the greater fegment, as the greater fegment is to the lefs.

IV.

The altitude of any figure is the straight line drawn from its vertex perpendicular to the base.

1

PROP.

Book VI.

PROP. I. THEOR.

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

TR

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.

E A F

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 a 38. 1. equal a: Therefore, whatever multiple the base HC is of the bafe BC, the fame multiple is the triangle AHC of the triangle ABC. For the fame reafon, whatever multiple the bafe LC is of the bafe CD, the same multiple is the triangle ALC of the triangle ADC : And if the base HC be equal to the bafe CL, the triangle AHC is alfo equal to the triangle ALC; and if the bafe HC be greater than the base CL, likewife the trian le AHC is greater than the triangle ALC; and if lefs, lefs. Therefore, fince there are four magni tudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the bafe BC and the triangle ABC, the first and third, any equimult ples whatever have been taken, viz. the bafe 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 fince it has been thown, that if the base HC be greater than the bafe CL, the triangle AHC is greater than the triangle

H G B C

D K

L

angle ALC; and if equal, equal; and if lefs, lefs: There- Book VI. foreb, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.

b 5. def. 5.

c

15. 5.

And because the parallelogram CE is double of the triangle ABC, and the parallelogram CF double of the triangle 41. 1. ACD, and that magnitudes have the fame ratio which their equimultiples haved; as the triangle ABC is to the triangle d ACD, fo is the parallelogram EC to the parallelogram CF. And because it has been shown, that, as the base BC is to the bafe CD, fo is the triangle ABC to the triangle ACD; and as the triangle ABC to the triangle ACD, fo is the parallelogram EC to the parallelogram (F; therefore, as the base BC is to the base CD, fo is e the parallelogram EC to the paral- e 11. 5. lelogram 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.

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 ftraight line which joins the vertices is parallel to that in which their bases areƒ, f 33. 1. because the perpendiculars are both equal and parallel to one another. Then, if the fame conftruction be made as in the propofition, the demonftration will be the fame.

PRO P. II. THE O R.

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

Join BE, CD; then the triangle BDE is equal to the triangle CDE, because they are on the fame base DE, and between a 37. r. the fame parallels DE, BC: but ADE is another triangle,

and

b 7.5. C 1.6.

Book VI. and equal magnitudes have to the fame the fame ratio b; therefore, as the triangle BDE to the triangle ADE, fo is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, fo is c BD to DA, because having the fame altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the fame reason, as the triangle CDE to the triangle ADE, fo is 11.5 CE to EA. Therefore, as BD to DA, fo is CE to EAd. Next, let the fides AB, AC of the triangle ABC, or these

ex. 6.

fides 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 e; 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 f9. 5. triangle ADE; and therefore f the triangle BDE is equal to the triangle CDE: And they are on the same base DE; but equal triangles on the fame base are between the fame paralg39. 1. lels g; therefore DE is parallel to BC. Wherefore, if a ftraight line, &c. Q. E. D.

PROP.

Book VI.

PROP. III. THEOR.

[F the angle of a triangle be bifected by a ftraight line which also cuts the base; the segments of the base shall have the fame ratio which the other fides of the triangle have to one another: And if the segments 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, bifects the vertical angle.

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.

E

Through the point C draw CE parallel a to DA, and let a 31. 1. BA produced meet CE in E. Because the ftraight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CADb: But CAD, by the hypothefis, is b 29. £. 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 oppofite angle AEC: But the angle ACE has been proved equal to the angle BAD; therefore alfo ACE is equal to the angle AEC, and confequently the fide AE is B equal to the fide c AC:

A

C

c6.15

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

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