*

And because the Parallelogram EC is + double to† 41.1. the Triangle ABC; and the Parallelogram FC, double to the Triangle ACD; and Parts have the fame Proportion as their like Multiples: There. 15. 1. fore, as the Triangle ABC is to the Triangle A CD, fo is the Parallelogram EC to the Parallelogram CF. And fo, fince it has been proved that the Bafe BC is to the Bale CD, as the Triangle ABC is to the Triangle ACD; and the Triangle A B C is to the Triangle A CD, as the Parallelogram EC is to the Parallelogramı CF, it fhall be ‡, as the Bafe BC is ‡ 11. 5. to the Bafe CD, fo is the Parallelogram EC to the Parallelogram FC. Waerefore, Triangles and Parallelograms, that have the fame Altitude, are to each other as their Bafes; which was to be demonstrated.

PROPOSITION II.

THEOREM.

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.

LET

ET DE be drawn parallel to BC, a Side of the
Triangle ABC. I fay, DB is to DA, as CE is

to E A.

For, let BE, CD be joined.

*

Then the Triangle BDE is equal to the Tri-* 37. L angle CDE; for they ftand upon the fame Bafe DE, and are between the fame Parallels DE and BC, and ADE is fome other Triangle. But equal Magnitudes have + the fame Proportion to one and † 7.5 the fame Magnitude. Therefore, as the Triangle BDE is to the Triangle ADE, fo is the Triangle CDE to the Triangle ADE.

But as the Triangle BDE is to the Triangle t 1 of this. ADE, fo is BD to DA; for fince they have the fame Altitude, viz. a Perpendicular drawn from the Point E to A B, they are to each other as their Bafes, And, for the fam: Reafon, as the Triangle C D E is to

# 11.3.

+ 9.5.

the Triangle A DE, fo is CE to E A: And therefore as BD is to D A, fo is CE to EA.

And if the Sides A B, A C, 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.

For, the fame Conftruction remaining, becaufe BD t1 of this. is to DA, as CE is to EA; and BD is † to DA, as the Triangle BDE is to the Triangle ADE; and CE is to E A, 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 Triangle B DE, CDE, have the fame Proportion to the Triangle ADE, the Triangle B D E fhall be equal to the Triangle CDE; and they have the fame Bafe DE: But equal Triangles, being upon the fame Bafe, I are between the fame Parallels; therefore DE is parallel to BC. Wherefore, if e 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 demonftrated.

$39. I.

9. 1.

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 Bale 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 bife the Angle of the Triangle.

LET there be a Triangle ABC, and let its Angle BAC be bifected by the Right Line A D. Į fay, as BD is to DC, fo is B A to A C.

For,

For, thro' C draw CE parallel to DA, and pro-* 31. 1. duce B A, till it meets CE in the Point E.

Then, because the Right Line AC falls on the Parallels AD, E C, the Angle ACE will be equal ‡ 29. 1. 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 A CE. Again, because the Right Line BAE falls on the Parallels A D, E C, the outward Angle B A D is + equal † 7. 5. to the inward Angle A EC; but the Angle ACE has been proved equal to the Angle BAD: Therefore ACE fhall be equal to A EC; and fo the Side A E is equal to the Side A C. And because the 16. Line AD is drawn parallel to CE, the Side of the Triangle BCE, it fhall be, as BD is to DC, fo is * 2 of thin BA to AE; but AE is equal to A C. Therefore, as BD is to DC, fo is + BA to A C.

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

29. 1.

For, the fame Conftruction remaining, because B D is to DC, as BA is to AC; and as BD is to DC, fo is BA to AE; for AD is drawn parallel to one of this. Side EC of the Triangle BCE; it fhall be, as BA is to A C, fo is BA to A E. Therefore A C is equal to AE+; and, accordingly, the Angle A EC is equal † 9. 4. to the Angle ECA: But the Angle A E 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 B A C is bifected by the Right Line AD. Therefore, if one Angle of a Triangle be bifeted, and the Right Line, that bifects the Angle, cuts the Bafe alfo; then the Segments of the Base 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

17. I.

$28. I.

34. I.

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, DEC, be equiangular Triangles,

having the Angle ABC equal to the Angle DCE, the Angle, A C B equal to the Angle D EC, and the Angle B A C equal to the Angle CDE. I fay, the Sides that are about the equal Angles of the Triangles A B C, D C E, are proportional; and the Sides that are fubtended under the equal Angles, are homologous, or of like Ratio.

*

Set the Side B C 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 is equal to the Angle DEC, the Angles A B C, DEC, are less than two Right Angles. And fo d. 12. BA, ED, produced, will meet each other; let them be produced, and meet in the Point F. Then, because the Angle DCE is equal to the Angle ABC, BF fhall be parallel to DC. Again, becaufe the Angle ACB is cqual to the Angle DEC, the Side AC will be parallel to the Side FE; therefore FA CD is a Parallelogram, and confe quently FA is equal to DC, and AC to FD; and because A C is drawn parallel to F E, the Side of the +2 of bis. Triangle FB E, it fhall + be, as BA is to A F, fo is BC to CE: But CD is equal to A F, and (by Alternation) as BA is to BC. fo is CD to CE. Again, becaufe CD is parallel to B F, it fhall be † as BC is to CE, fo is FD to DE, but F D is qual to AC. Therefore as BC is to CE, to is AC to DE: And fo by Alternarion, as BC is to CA, fo is CE to ED. Wherefore, becaufe it is demonftrated, that A B 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, fo is CD to DE. Therefore, the Sides about the equal Angles of equiangular Triangles are proportional; and the Side, which are fubtended under

17.5.

#21.5.

the

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

PROPOSITION V.

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

E T there be two Triangles A B C, DEF, have ing their Sides proportional; that is, let A B 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 are 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; * 23. 1. and the Angle EFG equal to the Angle BCA: Then the remaining Angle B A C is + is equal to the + Cor. 32. remaining Angle EG F.

And fo the Triangle ABC is equiangular to the Triangle EGF; and, confequently, the Sides that are fubtended under the equal Angles, are proportional. Therefore, as A B is to BC, fo is GE tot 4 of this. EF; but (by the Hyp.) as AB is to BC, fo is DE to E F Therefore, as D E is to EF, fo is GE to* 11. 5.

.

are

EF. And fince DE, EG, have the fame Proportion to EF, DE fhall be + equal to EG. For thet 2. 5. fame Reafon D F is equal to FG; but EF is common. Then, because the two Sides DE, E F, equal to the two Sides G E, EF, and the Bafe DF is equal to the Bafe F G, the Angle D E F is equal to 18. the Angle GEF; and the Triangle D E F 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 DFE is equal to the Angle GFE, and the AngleEDF equal to the Angle EGF. And because the

Angle