* 17. 1.

28. 1.

34.3.

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.

LE

ET ABC, D E C, be equiangular Triangles, having the Angle ABC equal to the Angle DCE, the Angle AC B equal to the Angle DEC, and the Angle BAC equal to the Angle CDE. Í fay, the Sides that are about the equal Angles of the Triangles A B C, DCE, 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 becaufe the Angles A B C, ACB, are lefs then two Right Angles, and the Angle A CB is equal to the Angle DEC, the Angles ABC, DEC, are lefs than two Right Angles. And fo † Ax. 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 D C. Again. be cause the Angle A C B is equal to the Angle D E C, the Side AC will be parallel to the Side FE; therefore F ACD is a Parallelogram, and confe quently F A is equal to D C, and A C to F D ; and because A C is drawn parallel to F E, the Side of the 2 of this. Triangle F B 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 B C, fo is C D to CE. Again, because C D is parallel to B F, it fhall be as + BC is to CE, fo is F D to D E, but FD is equal to A C. Therefore, as B C is to CE, fo is AC to DE: And fo by Alternation, as BC is to C A, fo is CE to ED. Wherefore, because, it is demonftrated, that A B is to B C, as DC is to CE; and as BC is to CA, fo is CE to E D'; it fhall be, by Equality, as BA is to A C, fo is CD to D E. Therefore, the Sides about the equal Angles of equiangular Triangles are proportional; and the Sides, which are fubtended under

$7.5.

21. 5.

the

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

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.

L

ET there be two Triangles, A B C, DEF, hav. ing their Sides proportional; that is, let A B be to BC, as DE is to EF; and as B C to CA, fo is EF to F D: And, allo, as B A to C A, fo E D to DF. I fay, the Triangle A B C is equiangular to the Triangle DEF; and the Angles are equal, under which the homologous Sides are fubtended, viz. the Angle A BC, equal to the Angle DEF; and the Angle B C A equal to the Angle EDF.

*

For, at the Points E and F, with the Line E F, make the Angle FEG equal to the Angle ABC; 23. x. and the Angle EF G equal to the Angle B CA:

Then the remaining Angle B A C is + equal to the + Cor. 32. 1. remaining Angle EGF.

And fo the Triangle A B C 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 B C, fo is ‡ G E to‡ 4 ofthis. EF; but (by the Hyp.) as A B is to BC, fo is DE to E F: Therefore as DE is, to E F,, fo is * G E. to* 11. 5. EF. And fince DE, E G, have the fame Proportion to E F, DE fhall be † equal to E G. For the† 2. 5. fame Reafon D F is equal to FG; but E F is common. Then, because the two Sides DE, EF, are equal to the two Sides GE, EF, and the Base DF is equal to the Bafe F G, the Angle D E F is equal to s. 1. 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 DFE is equal to the Angle GFE, and the Angle EDF equal to the Angle EGF. And because the Angle

23. I.

+ Cor. 32.1.

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 DEF. For the fame Reafon the Angle A C B fhall be equal to the Angle D F 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.
THEOREM.

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

ET there be two Triangles A BC, DEF, having one Angle BA C, of the one, equal to the Angle E D F of the other; and let the Sides about the equal Angles be proportional, viz. let A B be to AC, as ED is to DF. I fay, the Triangle A B C is equiangular to the Triangle DEF; and the Angle A B C equal to the Angle DE F; and the Angle ACB equal to the Angle D F E.

*

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

Then the other Angle B is + equal to the other Angle G; and fo the Triangle A B C is equiangular to the Triangle DGF; and, confequently, as BA is 14 of this. to A C, fo is G D to DF: But (by the Hyp.) as BA is to AC, fo is ED to D F. Therefore, as ED is to D F, fo is GD to DF; whence ED is + equal to D G, and D F is common; therefore the two Sides E D, D F, are equal to the two Sides GD, DF; and the Angle EDF equal to the Angle GDF:

11. 5.

9.5.

*

GDF: Confequently the Bafe E F is equal to the * 4. 1. Bafe FG, and the Triangle DEF equal to the Triangle D G F; and the other Angles of the one equal to the other Angles of the other, each to each; under which the equal Sides are fubtended. Therefore the Angle DFG is equal to the Angle D F E, and the Angle G equal to the Angle E; but the Angle DFG is equal to the Angle A C B: Wherefore the Angle ACB is equal to the Angle DFE; but the Angle BAC is is also equal to the Angle EDF: Therefore, the other Angle at B is equal to the other Angle at E; and fo the Triangle ABC is equiangular to the Triangle D E F. Therefore, 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 theje Angles equal, under which are fubtended the homer logous Sides; which was to be demonstrated.

PROPOSITION VII.
THEOREM.

If there are two Triangles, having one Angle of the one equal to one Angle of the other, and the Sides about the other Angles proportional; and if the remaining third Angles are either both lefs, or both not lefs, than Right Angles; then fhall the Triangles be equiangular, and have thofe Angles equal, about which are the proportional Sides.

By H

+ 520

LE ET two Triangles A DC, D E F, have one Angle of the one equal to one Angle of the other, viz. the Angle BAC equal to the Angle EDF; and let the Sides about the other Angles A B C, DEF, be proportional, viz. as D E is to E F, fo let A B be to BC and let the other Angles at C and F be both lefs, or both not lefs, than Right Angles. I fay, the Triangle ABC is equiangular to the Triangle DEF; and the Angle ABC is equal to the Angle DEF; as all the other Angle at C equal to the other Angle at F.

For,

23.1.

4 of this. * By Hyp.

11.

† 9. 5. 5. I.

5.

For, if the Angle ABC be not equal to the Angle DEF, one of them will be the greater, which let be ABC. Then at the Point B with the Right Line A B, make the Angle A B G equal to the Angle DEF.

Now, because the Angle A is equal to the Angle D, and the Angle A B G equal to the Angle D E Cor. 32.1, the remaining Angle A G B is † equal to the remaining Angle DFE: And therefore the Triangle A B G is equiangular to the Triangle D E F ; and fo, as A B is to BG, fo is ‡ DE to EF; but as DE is to E F, so is * A B to BC. Therefore, as A B is to BC, fo is A B to BG; and fince A B has the fame Proportion to BC, that it has to B G, BC fhall be + equal to BG; and, confequently, the Angle at C equal to the Angle BG C. Wherefore each of the Angles BCG, or BGC, is less than a Right Angle; and confequently A G B is greater than a Right Angle. But the Angle AGB has been proved equal to the Angle at F; therefore the Angle at F is greater than a Right Angle: But (by the Hyp.) it is not greater, fince C is not greater than a Right Angle, which is abfurd. Wherefore the Angle ABC is not unequal to the Angle DEF; and fo it must be equal to the remaining Angle at F; and confequently the Triangle ABC is equiangular to the Triangle DEF. Therefore, if there are two Triangles having one Angle of the one equal to one Angle of the other, and the Sides about the other Angles proportional; and if the remaining third Angles are either both lefs, or both not lefs, than Right Angles; then shall the Triangles be equiangular, and have thofe Angles equal, about which are the proportional Sides ; which was to be demonftrated.

PRO