PROPOSITION IV. THEOREM. Triangles are proportional ; and ibe Sides, are bomologous, or of like Ratio. LE ÉT ABC, DEC, be equiangular Triangles, having the Angle A B C equal to the Angle DCE, the Angle A C B equal to the Angle D EC, and the Angle BAC equal to the Angle C D'E. I fay, the Sides that are about the equal Angles of the Triangles A B C, DC.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 same Right Line with the Side CE; and because the Angles A BC, ACB, are less then two Right Angles, and the Angle ACB is equal to the Angle DEC, the Angles ABC, DEC, ale less than two Right Angles. And fo + Ax. 12. BA, ED, produced, will meet t each other”; let them be produced, and meet in the Point F. Then, because the Angle DCE is equal to the Angle 1 28. 1. ABC, B F shall be f parallel to D C. Again. because the Angle AC B is equal to tbe Angle D) EC, the Side A C will be I parallel to the Side FE; therefore FAC D is a Parallelogram, and conse quently F A is * equal to DC, and A C to FD; and 34• because AC is drawn parallel to F E, the side of the + 2 of bis. Triangle F B E, it shall + be, as B A is to A F, fois B C to CE: But C D is equal to A F, and (by (Alternation) as B A is to B C, fo is C D to CE. Again, because C D is parallel to B F, it shall be as + B C is to Ç E, fo is F D to D Е, but F D is equal to A C. Therefore, as B C is to CE, so is I AC to DE: 17.5. And so by Alternation, as B C is to C A, fo is C Eto the . 3• • 21. 5. the equal Angles are homologous, or of like Ratio; which was to be demonstrated. PROPOSITION V. THEOREM. Triangles shall be equiangular ; and their Angles, ing their Sides proportional; that is, let A B be to BC, as D E is to EF; and as BC to CA, so is E F to FD: And, allo, as B A to C A, fo E D to DF. I say, 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 B C, equal to the Angle DEF; and the Angle B C A equal to the Angle E DF. For, at the Points E and F, with the Line EF, make * the Angle F E G equal to the Angle A B C ;* 23. 1. and the Angle E F G equal to the Angle BCA: Then the remaining Angle BAC is † equal to the + Cor. 32. 1. remaining Angle EGF. And so the Triangle A B C is equiangular to the Triangle E GF; and, consequently, the sides that are subtended under the equal Angles, are proportional. Therefore, as A B is to BC, so is IG É cof 4 of tbis. EF; but (by the Hyp.) as A B is to BC, lo is DE to E F: Therefore as D E is to E F, Co. is * G E.co* 11. 5. EF. And fince D E, E G, have the same Proportion to EF, D E fhall be + equal to E G. For the + 2, s. same Reason D F is equal to F G; but E F is com Then, because the two Sides DE, EF, are equal to the two Sides G E, EF, and the Base D F is equal to the Base F G, the Angle D E F is I equal to 1 3. 1. 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 subtended by the equal Sides. Therefore the. Angle DFE is equal to the Angle GFE, and the Angle E D F equal to the Angle EGF. And because the Angle mon. Angle D E F is equal to the Angle G EF; and the Angle GEF equal to the Angle ABC ; therefore the Angle ABC shall be also equal to the Angle DEF. For the fame Reason the Angle ACB shall be equal to the Angle D FE; as also the Angle A equal to the Angle D: Therefore the Triangle ABC will be equiangular to the Triangle D EF. Wherefore, if the sides of two Triangles are proportional, the Triangles Mall be equiangular ; and their Angles, under which the homologous Sides are subtended, are equal; which was to be demonstrated. PROPOSITION VI. THEOREM. 10 ome Angle of the other, and if the Sides Angles, equal, under which are subtended the ing one Angle BAC, 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 E D is to DF. I say, the Triangle a BC is equiangular to the Triangle DEF; and the Angle A B C equal to the Angle DEF; and the Angle ACB equal to the Angle D FE. For, at the Points D and F, with the Right Line * 23. 1. DF, make the * Angle F D G equal to either of the Angles B AC, 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 so the Triangle A B C is equiangular to the Triangle D GF; and, consequently, as B A is 1 4 of tbis. to A C, so is I G D to DF: But (by the Hyp.) as B A is to A C, fo is E D to DF. Therefore, as ED is * to DF, so is G D to DF; whence E D is + equal to DG, and D F is common; therefore the two Sides E D, DF, are equal to the two Sides GD, DF; and the Angle E D F equal to the Angle GDF: + Cor. 32.I. • 11. 5 GDF: Consequently the Base E F is * equal to the *4. 1 Base F G, and the Triangle D E F equal to the Triangle DGF; and the other Angles of the one equal to the other Angles of the other, each to each ; under which the equal Sides are subtended. Therefore the Angle DFG is equal to the Angle D FE, and the Angle G equal to the Angle E; but the Angle DĚ G is equal to the Angle ACB: Wherefore the Angle ACB is equal to the Angle DFE; but the Angle BAC is I is also equal to the Angle EDF: 1 By Hide Therefore, the other Angle at B is + equal to the other † 52. 10 Angle at E ; and so the Triangle A B C is equiangular to the Triangle DEF. Therefore, if iwo 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 bomar logous Sides; which was to be demonstrated. PROPOSITION VII. THEOREM. the one equal to one Angle of the other, and tbe LET wo Triangles ADC, DE F, have one Ana gle 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 ABC, DEF, be proportional, viz, as D E is 1o E F, fo let A B be to B C; and let the other Angles at Cand F be both less, or both not less than Right Angles. I say, the Triangle A B C is equiangular to the Triangle D EF; and ine Angle A B C is equal to the Angle DEF as also the other Angle at C equal to the other Angle at F. For, For, if the Angle A B C 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 23.8 A B, make * the Angle A B G equal to the Angle Now, because the Angle A is equal to the Angle D, and the Angle A B G equal to the Angle DEF; Cor. 32.1, the remaining Angle A G B ist equal to the remain ing Angle DFE. And therefore the Triangle A BG is equiangular to the Triangle DEF; and so, as AB 14 of tbis. is to BG, so is D E to E F; but as D E is to E F, • By Hyp. fo is * AB to BC. Therefore, as A B is to B C, fo is A B I to BG; and fince A B has the same Propor tion to B C, that it has to BG, B C shall be + equal 5. I. 10 BG; and, consequently, the Angle at C * equal to the Angle BGC. Wherefore each of the Angles BCG, or BGC, is less than a Right Angle; and consequently A G B is greater than a Right Angle. But the Angle A G B 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, since C is not greater than a Right Angle, which is absurd. Wherefore the Angle ABC is not unequal to the Angle DEF; and so it must be equal to the remaining Angle at F; and consequently the Trio angle A B C is equiangular to the Triangle D E F. 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 less, or both not less, than Right Angles; then fall the Triangles be equiangular, and have those Angles equal, about which are the proportional Sides, which was to be demonstrated. |