fame Altitude, viz. a Perpendicular drawn from the Point E to A B, they are to each other as their Bafes. And for the fame Reafon, as the Triangle CDE, is to the Triangle ADE, fo is CE to EA: And therefore as BD is to D'A, fo* is CE to E A.

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; CA; and if DE be joined, I fay, DE is parallel to BC.

For the fame Construction remaining, because BD

11. 5.

is to DA, as CE is to EA; and BD is t 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 ADE. And fince the Triangles ADE, CDE, have the fame Proportion to the Triangle ADE, the Triangle BDE, fhall be equal to the † 9.5. Triangle CDE; and they have the fame Bafe DE: But equal Triangles being upon the fame Bafe,, 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 ef the Triangle proportionally; and if the Sides of the Triangle be cut proportionally, then a Right Line joining the Points of Section, fhall 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 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 Seg-
ments of the Bafe have the fame Proportion that the
other Sides 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.

LE ET there be a Triangle ABC, and let its Angle
ABC, be bifected by the Right Line AD. I* 9. 1.
fay, as BD is to DC, fo is BA to AC.

Lo 4

For

* 31. 1.

+ 29.3.

‡6. 1.

*2 of this.

17.5.

* 29. 1..

For thro' C draw* CE parallel to DA, and produce BA 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 † equal
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 Angle
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 because the
Line AD is drawn parallel to CE, the Side of the
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.
t

*

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

*

*

For the fame Conftruction remaining, because BD is to DC, as BA is to AC; and as BD is to DC, fo 2 of this. is BA to AE; for A D is drawn parallel to one Side EC of the Triangle BCE, it shall be as BA is to AC, fo is BA to AE. Therefore AC is equal to A E; and accordingly the Angle AEC, is equal to the Angle ECA: But the Angle A EC, 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 Sides 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.

THEOREM.

The Sides about the equal Angles of equiangular Triangles, are proportional; and the Sides which are fubtended under the equal Angles, are homologous, 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.

34. I.

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 ACB * 17. 1. is equal to the Angle DEC, the Angles ABC, DEC, are less than two Right Angles. And fo BÁ, ED, produced, will meet † each other; let them be pro- † 12 Ax. 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 is equal to the Angle DEC, the Side AC will be + + 28. 1. parallel to the Side FE; therefore FACD is a Parallelogram; and confequently FA is equal to CE, * and AC to FD; and because AC is drawn parallel to FE, the Side of the Triangle FBE, it fhall be t + 2 of this. as BA is to AF, fo is BC to ČE; and (by Alternation) as BA is to B C, 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 AC. Therefore as BC is to CÉ, 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 AC, * 21. 1. fo is CD to DE. Therefore the Sides about the equal Angles of equiangular Triangles, are proportional; and

the

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

PROPOSITION V.

3

THEOREM.

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.

LE

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, fo 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 the remaining Angle BAC, is † equal to the remaining

† Cor. 32. Angle EGF.

1.

11. 5.

+ 9.5.

*

And fo the Triangle ABC is equiangular to the Triungle EGF; and confequently the Sides that are fubtended under the equal Angles, are proportional. #4 of this. Therefore as AB is to BC, fo is GE to EF; but as A B 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 EG. 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

8. 1.

EGF;

EGF; and because the Angle DEF is equal to the Angle GEF; and the Angle GEF equal to the Angle A B C; therefore the Angle ABC fhall be alfo equal to the Angle FED: For the fame Reafon, the Angle ACB fhall be equal to the Angle DFE; as alfo the Angle A equal to the Angle D; there fore 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 homologus Sides are fubtended, are equal; which was to be demonstrated.

PROPOSITION VI.

THORE 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.

LET there be two Triangles ABC, DEF, having

one Angle BAC of the one equal to the Angle 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 DFE.

For at the Points D and F, with the Right Line DF, make the Angle F D G 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. Angle G; and fo the Triangle ABC, is equiangular 1. to the Triangle DGF; and confequently, as BA is

*

11. 5.

to A C, fo is GD to DF: But (by the Hyp.) as ‡ 4 of this. BA is to AC, fo is ED to DF. Therefore as ED isto DF, fo is GD to DF; whence ED is † equal to DG, and D F 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