bafe BC is to the bafe CD, fo is the triangle ABC to the triangle ACD.

And because [by 41. 1.] the parallelogram EC is double to the triangle ABC, and the parallelogram CF is double to the triangle AC D. But [by 15. 5.] parts have the fame ratio as their equimultiples; the triangle ABC will be to the triangle ACD, as the parallelogram E C is to the parallelogram FC. Therefore because it has been proved, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, but as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC: Therefore [by 11. 5.] as the base BC is to the base CD, fo will the parallelogram EC be to the parallelogram Fc.

Therefore triangles and parallelograms, which have the fame altitude, will have the fame ratio to one another as their bases. Which was to be demonftrated.

d The demonstration of this propofition is one very natural and eafy confequence or cafe of Euclid's fifth definition of his fifth book in triangles and parallelograms of the fame altitude; fhewing the exceeding eafy application and ufefulness of that definition.

if

PROP. II. THEOR.

If a right line be drawn parallel to one fide of a triangle, this will cut the fides of the triangle proportionally and if the fides of a triangle be cut proportionally, the right line which joins the points of fection will be parallel to the uncut fide of the triangle.

D

For let the right line p£ be drawn parallel to one fide B C of a triangle ABC: I fay as B D is to DA, fo is CE

to EA.

For join BE, CD.

Then [by 37. 1] the triangle BDE is equal to the triangle CDE; for they ftand upon the fame bafe DE, and are between the fame parallels DE, BC. But ADE is another triangle; and [by 7. 5. equal magnitudes have the fame R 2

E

ratio

Book VI. ratio to the fame magnitude: Therefore as the triangle BDE is to the triangle A DE, fo is the triangle C D E, to the triangle ADE. But as the triangle BDE is to the triangle A DE, fo is BD to D A: For fince they have the fame altitude, viz. the perpendicular drawn from the point E to A B, they are [by 1. 6.] to one another as their bafes. And for the fame reafon the triangle cDE is to the triangle ADE, as CE is to EA: Therefore as BD is to DA, fo is [by 11. 5.] CE to EA.

Again, let the fides A B, A C of the triangle A B C be cut proportionally in the points D, E, fo that B D be to DA, as CE is to EA: and join DE: I fay DE is parallel

to BC.

For the fame construction remaining, because as BD is to DA, fo is CE to EA: But as B D is to D A, fò [by 1. 6.] is the triangle B D E to the triangle A D E; and as CE is to EA, fo is the triangle CDE to the triangle ADE: it will be [by 11. 5.] as the triangle BDE is to the triangle ADE, fo is the triangle CDE to the triangle ADE, Therefore the triangles BDE, CDE have each the fame ratio to the triangle ADE: Wherefore [by 9. 5.] the triangle BDE is equal to the triangle CDE. But they ftand both upon the fame base D E. And equal triangles that ftand upon the fame bafe [by 39. 1.] are alfo between the fame parallels. Wherefore D E is parallel to B C..

If therefore a right line be drawn parallel to one fide of a triangle, cutting the other two fides; this line will cut the fides proportionally and if the fides of a triangle be cut proportionally, a right line joining the points of section, will be parallel to the uncut fide of the triangle. Which was to be demonftrated.

If one angle of a triangle be bifected, and the bifecting right line cuts the bafe, the fegments of the bafe will have the fame ratio to one another that the other two fides of the triangle have to one another and if the fegments of the base have the fame ratio to one another, as the other two fides of the triangle bave to one another: then will the right line drawn from the point of fection to the

vertex of the triangle bifect the angle at the

vertex.

Let there be a triangle A B C, and let the angle B A C be cut into two equal parts by the right line AD; I fay, as BD is to DC, fo is BA to AC.

For [by 31. 1.] draw CE thro' c parallel to DA, and let BA continued out meet the fame in E,

Then because the right line AC falls upon the two parallels AD, EC; the angle ACE [by 29. 1.] will be equal to the angle CAD. But the angle CAD is fuppofed to be equal to the angle BAD: Therefore BAD will also be equal to the angle ACE. Again, because the right line BAE falls upon the parallel right lines A D, E C, the outward angle B A D [by 29, 1.] is equal to the inward angle AEC. And it has been alfo proved that the angle ACE is equal to the angle BAD Therefore ACE will also be

equal to the angle A E C; and fo [byB

A

6. 1.] the fide A E is equal to the fide A c. And because AD is drawn parallel to one fide, viz. EC of the triangle BCE; [by 2. 6.] it will be as B D is to D C, fo is B A to

AE.

But AE is equal to AC: Therefore [by 7. 5.] as BD

is to DC, fo is BA to AC.

But now let BD be to DC, as BA is to AC: and join AD: I fay the angle BAC is bifected by the right line AD.

For the fame conftruction remaining, because it is as BD to DC, fo is BA to AC; and [by 2. 6.] as B D is to DC, fo is BA to AE (for AD is drawn parallel to one fide EC of the triangle BCE) it will be as BA to A ç, so is BA to AE: Therefore [by 9. 5.] AC is equal to AE, and fo [by 5. 1.] the angle A E C is equal to the angle a CE. But the angle A EC [by 29. 1.] is equal to the outward angle B A D, and the angle A CE equal to the alternate angle CAD: Therefore the angle BAD will be equal to the angle CAD: Wherefore the angle B A C is bifected by the right line AD.

Therefore if one angle of a triangle be bifected, and the bifecting line alfo cuts the base; the fegments of the bafe will have the fame ratio, as the other two fides of the triangle. And if the segments of the base have the fame

R 3

ratio

Book VI. ratio as the other fides of the triangle, the right line drawn - from the vertex to the point of fection, will bifect the angle at the vertex of the triangle. Which was to be demonftrated.

PROP. IV. THEOR.

The fides about the equal angles of equiangular triangles, are proportional: and the fides oppofite to the equal angles are homologous or co-rationals.

Let A B C, DCE be equiangular triangles, having the angle A B C equal to the angle DCE, and the angle A C B equal to the angle D E C, and moreover the angle B A C equal to the angle CDE: I fay the fides about the equal angles of the triangles A B C, DC E are proportional; and the fides oppofite to the equal angles are homologous or co-rationals.

For let BC, CE be both put in the fame right line: and becaufe [by 17 1.] the angles A B C, A CB are lefs than two right angles. But the angle A C B is equal to DEC; the angles A B C, DEC will be lefs than two right angles : Wherefore [by II. ax.] BA, ED being produced will meet one another. Let them be produced and meet in F.

F

A

B

Then because the angle DCE is equal to the angle ABC; [by 28. 1.] BF will be parallel to c D. Again, because the angle ACB is equal to the angle DEC; A C will be parallel to FE: Therefore F A C D is a parallelogram: And fo [by 34. 1.] FA Eis equal to CD, and AC to FD. And because A c is drawn parallel to one fide FE of the triangle F BE; [by 2. 6.] it will be as BA is to A F, fo is B C to CE. But AF is equal to CD: Therefore as BA is to C D, fo is [by 7. 5.] BC to CE: and alternately [by 1 6. 5.] as A B is to BC, fo is D C to CE. Again, because c D is parallel to B F, it will be as B C is to CE, fo is F D to D E. But DE is equal to AC; Therefore as BC is to cE, fo is AC to E D. And alternately as B C is to CA, fo is CE to E D. And so because it has been proved that as AB is to B C, fo is DC to c E, and

as

as B C is to CA, fo is CE to ED; it shall be by equality [by 22. 5.] as B A is to A C, fo is CD to D E.

Therefore the fides about the equal angles of equiangular triangles, are proportional; and the fides oppofite to the equal angles are homologous or co-rationals. Which was to be demonstrated.

PROP. V. THEOR.

If two triangles have their fides proportional; thofe triangles will be equiangular; thofe angles being equal which are oppofite to the homologous or corational fides.

Let there be two triangles A B C, DE F, having proportional fides: and let A B be to BC, as DE is to E F: alfo as B C is to CA, fo is E F to F D, and moreover as B A is to A C, fo is E D to DF: I fay the triangles A B C, D E F are equiangular, those angles being equal which are oppofite to the homologous fides, viz. the angle ABC equal to the angle D E F, the angle BCA equal to the angle E F D, and the angle BAC equal to the angle EDF.

For [by 23. 1.] at the right line EF, and at the points E, F in it, make the angle FEG equal to the angle ABC, and the angle EFG, equal to the angle BCA; then [by 32. 1.] the remaining angle B Ac is equal to the remaining angle EGF.

Therefore the triangles ABC, EGF are equiangular; and fo the fides about the equal angles of the triangles ABC, EGF are [by 4. 6.] proportional: and the fides oppofite to the equal angles are homologous: Wherefore as AB is to BC, fo is GE to

EF. But as A B is to BC, fo is DE to EF: There. fo

[by 11. 5.] is GE to EF: Wherefore DE, GE have each the same ratio to EF:

D

And fo [by 9. 5.] DE will be equal to GE. By the fame reafon D F will also be equal to GF. Then because DE is equal to E G, and EF is common, the two fides DE, EF, are equal to the fides G E, E F, and the base DF is equal to the bafe GF: Therefore [by 8. 1.] the angle DEF is equal

R 4