to the Triangle ADE, fo is CE to EA: And therefore as BD is to DA, fo* is CE to EA.

And if the Sides AB, AC, of the Triangle ABC, be cut proportionally, that is, fo that BD be to DA, as CE is to EA; and if DE be joined, I fay, DE is parallel to BC.

* II. 5.

For the fame Conftruction remaining, because BD is to DA, as CE is to EA; and BD is + to DA, as † 1 of this. the Triangle BDE is to the Triangle ADE; and CE is to EA, as the Triangle CDE is to the Triangle ADE: It fhall be as the Triangle BDE, is to the Triangle ADE, fo is the Triangle CDE to the Triangle A.DE. And fince the Triangles BDE, CDE, have the fame Proportion to the Triangle ADE, the Triangle BDE, fhall be † equal to the t 9. 5. Triangle CDE; and they have the fame Base DE: But equal Triangles being upon the fame Base, are 39. 1. + between the fame Parallels; therefore DE is parallel to BC. Wherefore, if a Right Line be drawn parallel to one of the Sides of a Triangle, it shall cut the Sides of the Triangle proportionally; and if the Sides of the Triangle be cut proportionally, then a Right Line joining the Points of Section, shall be parallel to the other Side of the Triangle; which was to be demonstrated.

PROPOSITION III.

THEOREM.

If one Angle of a Triangle be bifected, and the Right Line that bijects the Angle, cuts the Bafe alfo; then the Segments of the Bafe will have the fame Proportion, as the other Sides of the Triangle. And if the Segments of the Base have the fame Proportion that the other Sides of the Triangle bave; then a Right Line drawn from the Vertex, to the Point of Section of the Bafe, will biJect the Angle of the Triangle.

L

ET there be a Triangle ABC, and let its Angle

*

BAC, be bifected by the Right Line AD. I* 9. 1. fay, as BD is to DC, fo is BA to AC.

** 31. 1.

+29, I,

For thro' C draw* CE parallel to DA, and produce B A till it meets CE in the Point E.

Then because the Right Line AC, falls on the Parallels AD, EC, the Angle ACE, will be fequal to the Angle CAD: But the Angle CAD (by the Hypothefis) is equal to the Angle BAD. Therefore the Angle BAD, will be equal to the Angle ACE. Again, because the Right Line BAE, falls on the Parallels AD, EC, the outward Angle BAD, is tequal to the inward Angle AEC; but the Anglè ACE, has been proved equal to the Angle BAD: Therefore ACE fhall be equal to AEC; and fo the Side AE is equal to the Side AC. And becaufe the Line AD is drawn parallel to CE, the Side of the 2 of thi Triangle BCE, it fhall be* as BD is to DC, fo is BA to AE; but AE is equal to AC. Therefore as BD is to DC, fo is + BA to AC.

$6.1.

29. I.

And if BD be to DC, as BA is to AC; and the Right Line AD be joined, then, I say, the Angle BAC, is bifected by the Right Line AD.

*

For the fame Conftruction remaining, becaufe BD is to DC, as BA is to AC; and as BD is to DC, so 2 of this. is BA to AE; for AD is drawn parallel to one Side EC of the Triangle BCE, it fhall be as BA is to AC, fo is BA to AE. Therefore AC is equal to AE; and accordingly the Angle AEC, is equal to the Angle ECA: But the Angle AE C, is equal *to the outward Angle BAD; and the Angle ACE, equal to the alternate Angle CAD. Wherefore the Angle BAD is alfo equal to the Angle CAD; and fo the Angle BAC is bifected by the Right Line AD. Therefore, if the Angle of a Triangle be bifected, and the Right Line that bifects the Angle, cuts the Bafe alfo; then the Segments of the Bafe will have the fame Proportion as the other Sides of the Triangle. And if the Segments of the Bafe have the fame Proportion that the other fides of the Triangle have; then a Right Line drawn from the Vertex, to the Point of Section of the Bafe, will bifect the Angle of the Triangle; which was to be demonftrated.

PRO

PROPOSITION IV.

THEORE M.

The Sides about the equal Angles of equiangular Triangles, are proportional; and the Sides which are fubtended under the equal Angles, are bomologous, or of like Ratio.

LET, ABC, DCE; be equiangular Triangles, having the Angle ABC equal to the Angle DCE; the Angle ACB equal to the Angle DEC, and the Angle BAC equal to the Angle CDE, I fay, the Sides that are about the equal Angles of the Triangles ABC, DCE, are proportional; and the Sides that are fubtended under the equal Angles, are homologous, or of like Ratio.

*

*

Set the Side BC, in the fame Right Line with the Side CE; and because the Angles ABC, ACB, are *lefs than two Right Angles, and the Angle A CB 17. 1. is equal to the Angle DEC, the Angles ABC, DEC, are less than two Right Angles, And fo BA, ED, produced, will meet † each other; let them be pro- † Ax. 12, duced, and meet in the Point F. Then because the Angle DCE, is equal to the Angle ABC, BF fhall be parallel to DC. Again, because the Angle ACB 28. 1. is equal to the Angle DEC, the Side AC will be t parallel to the Side FE; therefore FACD is a Parallelogram; and confequently FA is equal to DC, 34. I« and AC to FD; and because AC is drawn parallel to FE, the Side of the Triangle F BE, it fhall + be † 2 of this, as BA is to AF, fo is BC to CE; and (by Alternation) as BA is to BC, fo is CD to CE. Again; because CD is parallel to BF, it fhall be † as BC is to CE, fo is FD to DE; but FD is equal to A C. Therefore as BC is to CE, fo is AC to DE: And ‡ 7.5 fo (by Alternation) as BC is to CA, fo is CE to ED. Wherefore because it is demonftrated that AB is to BC, as DC is to CE; and as BC is to CA, fo is CE to ED; it fhall be by equality, as BA is to 21. x, AC, fo is CD to DE. Therefore, the Sides about the equal Angles of equiangular Triangles, are proportional; and the Sides, which are fubtended under the

*

L 4

equal

*23. I.

equal Angles, are homologous, or of like Ratio; which was to be demonftrated.

PROPOSITION V.

THE ORE M.

If the Sides of two Triangles are proportional, the Triangles fhall be equiangular; and their Angles, under which the homologous Sides are fubtended, are equal.

I

ET there be two Triangles, ABC, DEF, having their Sides proportional, that is, let AB be to BC, as DE is to EF; and as BC to CA, so is EF to FD. And alfo as BA to CA, fo ED to DF. I fay, the Triangle ABC is equiangular to the Triangle DEF; and the Angles equal, under which the homologous Sides are fubtended, viz. the Angle ABC, equal to the Angle DEF; and the Angle BCA equal to the Angle EFD, and the Angle BAC equal to the Angle EDF.

For at the Points E and F, with the Line EF, make *the Angle FEG, equal to the Angle ABC; and the Angle EFG, equal to the Angle BCA: Then +Cor. 32. 1. the remaining Angle BAC, is † equal to the remaining Angle EGF.

*II. 5.

† 9.5.

And fo the Triangle ABC is equiangular to the Triangle EGF; and confequently the Sides that are fubtended under the equal Angles, are proportional. 14 of this. Therefore as AB is to BC, fo is GE to EF; but as AB is to BC, fo is DE to EF: Therefore as DE is to EF, fo is * GE to EF. And fince DE, EG, have the fame Proportion to EF, DE fhall be † equal to E G. For the fame Reafon, DF is equal to FG; but EF is common. Then because the two Sides DE, EF, are equal to the two Sides GE, EF, and the Bafe DF is equal to the Bafe F G, the Angle DEF is equal to the Angle GEF; and the Triangle DEF equal to the Triangle GEF; and the other Angles of the one, equal to the other Angles of the other, which are fubtended by the equal Sides. Therefore the Angle DEF is equal to the Angle GEF, and the Angle EDF equal to the Angle EGF;

8. 1.

EGF; and because the Angle DEF is equal to the Angle GEF; and the Angle GEF equal to the Angle ABC; therefore the Angle ABC fhall be alfo equal to the Angle FED; For the fame Reason, the Angle ACB fhall be equal to the Angle DF E; as alfo the Angle A equal to the Angle D; therefore the Triangle ABC will be equiangular to the Triangle DEF. Wherefore, if the Sides of two Triangles are proportional, the Triangles fhall be equiangular; and their Angles, under which the homologous Sides are fubtended, are equal; which was to be demonftrated.

PROPOSITION VI.

THEOR E M.

If two Triangles have one Angle, of the one equal to one Angle of the other; and if the Sides about the equal Angles be proportional, then the Triangles are equiangular, and have thofe Angles equal, under which are fubtended the homologous Sides.

ET there be two Triangles ABC, DEF, having

EDF of the other; and let the Sides about the equal Angles be proportional, viz. let AB be to AC, as ED is to DF. I fay, the Triangle ABC is equiangular to the Triangle DEF; and the Angle ABC equal to the Angle DEF; and the Angle ACB equal to the Angle DF E.

For at the Points D and F, with the Right Line DF, make the Angle FDG equal to either of the* 23. 1. Angles BAC, EDF; and the Angle DFG equal to the Angle ACB.

Then the other Angle B, is + equal to the other † Cor. 32. 1. Angle G; and fo the Triangle ABC, is equiangular to the Triangle DGF; and confequently, as BA is to AC, fo is GD to DF: But (by the Hyp.) ast 4 of this. BA is to AC, fo is ED to DF. Therefore as ED is to DF, fo is GD to DF; whence ED is † equal 11. 5. to DG, and DF is common; therefore the two Sides † 9. 5. ED, DF, are equal to the two Sides GD, DF; and the Angle EDF, equal to the Angle GDF:

Con

* II.