PROPOSITION I. TH É O R E M. ET the Triangles ABC, ACD, and the viz. the Perpendicular drawn from the Point A to BD. I say, as the Base BC, is to the Base CD, so is the Triangle ABC, to the Triangle ACD, and so is the Parallelogram EC to the Parallelogram CF. For produce BD both ways to the Points H and L, and take GB, GH, any Number of Times equal to the Base BC; and DK, KL, any Number of Times equal to the Base ÇD, and join AG, AH, AK, AL. Then because CB, BG, GH, are equal to one another, the Triangles AHG; AGB, Å BC, also will be * equal to one another : Thejefore the same * 38. Xá Multiple that the Base HC is of BC, shall the Triangle AHC be of the Triangle ABC. By the fame Reason, the fame Multiple that the Base LC is of the Base CD, fhäll the Triangle ALC be of the Triangle ACD. And if HC, be equal to the Base CL, the Triangle AHC is also * equal to the Triangle ALC: And if the Base HC, exceeds the Base CL, then the Triangle AHC, will exceed the Triangle ALC. And if Å C be less, then the Triangle AħC will be less. Therefore since there are four Magnitudes, viz, the two Bases BC, CD, and the two Triangles ABC, ACD; and since the Base HC, and the Triangle AHC, are Equimultiples of the Base B C, and the Triangle ABC: And the Base CL, and the Triangle ALC, are Equimultiples of the Base CD, and the Triangle ADC. And it has been proved, that if the Base HC, exceeds the Base CL, the Triangle AHC, will exceed the Triangle ALC; and if equal, equal; if lefs, less. Then as the Base BC, is to the Base CD, fo fis the Trian- + Def, si si gle AB. C, to the Triangle ACD, + 41. 1. And because the Parallelogram EC, is † double to the Triangle ABC; and the Parallelogram FC, double † to the Triangle A CD; and Parts have the same Proportion as their like Multiples. Therefore as the Triangle A B C is to the Triangle ACD, so is the Parallelogram EC to the Parallelogram CF. And so since it has been proved, that the Base B C is to the Base CD, as the Triangle A B C, is to the Triangle ACD; and the Triangle ABC is to the Triangle ACD, as the Parallelogram EC is to the Parallelogram CF; it shall be | as the Base B C is! to the Base CD, so is the Parallelogram EC to the Parallelogram FC. Wherefore Triangles, and Parallelograms, that have the fame Altitude, are to each other as their Bases; which was to be demonstrated | 11. 5. PROPOSITION II. THEOREM. Sides of a Triangle, it shall cut the sides of the Triangle be cut proportionally, then a Right Line the Triangle ABC. I say, DB is to DA, as CE is to E A. For let B E, CD, be joined. + 37. I. Then the Triangle BDE is * equal to the Triangle CDE, for they stand upon the fame Base DE, and are between the same Parallels DE and BC; and ADE is some other Triangle. But equal Magni+ 7. 5. tudes have t the same Proportion to one and the fame Magnitude. Therefore as the Triangle BDE is to the Triangle ADE, so is the Triangle CDE to the Triangle ADE. But as the Triangle BDE, is to the Triangle 1 1 of this. ADE, fo I is BD to DA; for since they have the fame Altitude, viz. a Perpendicular drawn from the Point E to AB, they are to each other as their Bases. And for the same Reason, as the Triangle CD E, is * II. 5. to the Triàngle ADE, so is CE to EA: And therefore as B D is to DA, fo* is CE to E A. And if the Sides AB, AC, of the Triangle ABC, be cut proportionally, that is, so that BD be to DA, as CE is to EA; and if D E be joined, I say, DE is parallel to BC. For the fame Construction remaining, because BD is to DA, as CE is to EA; and B D is 7 to DA, as t 1 of this. the Triangle BD E is to the Triangle ADE; and CE is to EA, as the Triangle CDE is to the Triangle ADE: It shall be as the Triangle BDE, is to the Triangle 'ADE, fo is * the Triangle CDE to the Triangle ADE. And since the Triangles BDE, CDE, have the same Proportion to the Triangle ADE, the Triangle BDE, shall be + equal to the t 9.5. Triangle CDE; and they have the same Base DE: But equal Triangles being upon the same Base, I are 39. 1. between the same 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. Line that bijeets the Angle, cuts the Base allo; ET there be a Triangle ABC, and let its Angle I* 9. 1. say, as B D is to DC, so is B A to AC. L 3 For 31. ?. For thro' C draw * CE parallel to DA, and pro. duce B A till it meets CE in the Point E. Then because the Right Line AC, falls on the +29, I, Parallels AD, EC, the Angle ACE, will be f equal to the Angle CAD: But the Angle CAD (by the Hypothesis) is equal to the Angle BAD. Therefore the Angle B AD, will be equal to the Angle ACE. Again, becaufe the Right Line BAE, falls on the Parallels AD, EC, the outward Angle BAD, is tequal to the inward Angle A EC; but the Anglè ACE, has been proved equal to the Angle BAD: Therefore ACE shall be equal to AEC; and so the $ 6. I. Side A E'is equal to the Side AC." And becaufe the Line AD is drawn parallel to CE, the side of the 2 of this. Triangle BCE, it shall be * as B D is to DC, so is B A to AE; but AE is equal to AC. Therefore as tr. 5 BD is to DC, fo is + B A to AC. And if B D be to DC, as B A is to AC; and the Right Line AD be joined, then, I say, the Angle BAC, is bisected by the Right Line AD. For the same Construction remaining, becaufe BD is to DC, as BA is to AC; and as BD is to DC, so 12 of this. is İ BA to AE; for AD is drawn parallel to one Side EC of the Triangle BCE, it shall be as BA is to AC, so is B A to AE. Therefore A C is equal to AE; and accordingly the Angle AEC, is equal to the Angle ECA: But the Angle AEC, is equal * to the outward Angle BAD; and the Angle ACE, equal * to the alternate Angle CAD. Wherefore the Angle BAD is also equal to the Angle CAD, and so the Angle BAC is bisected by the Right Line AD. Therefore, if the Angle of a Triangle be bifected, and the Right Line that biseɛts the Angle, cuts the Base also; then the Segments of the Base will have the fame Proportion as the other sides of the Triangle. And if the Segments of the Base have the same Proportion that the other sides of the Triangle have ; then a Right Line drawn from the Vertex, to the Point of Section of the Base, will bisect the Angle of the Triangle; which was to be demonstrated, 29• I. PRO PROPOSITION IV. THEOREM. The Sides about the equal Angles of equiangular Triangles, are proportional, and the sides which are subtended under the equal Angles, are bomologous, or of like Ratio. ET ABC, DCE, be equiangular Triangles, having the Angle ABC equal to the Angle DCE; the Anglę AÇß equal to the Angle DEC, and the Angle BAC equal to the Angle CDE, I say, the Sides that are about the equal Angles of the Triangles ABC, DCE, are proportional, and the Sides that are subtended under the equal Angles, arę homolo gous, 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 * less than two Right Angles, and the Angle ACB* 19. 1. is equal to the Angle DEC, the Angles ABC, DEC, are less than two Right Angles, And so BA, ED, produced, will meet + each other; let them be pro- † Ax. 124 duced, and meet in the Point F. Then because the Angle DCE, is equal to the Angle ABC, BF shall be parallel to DC. Again, because the Angle ACB 1 28. 1. is equal to the Angle DEC, the Side AC will be 1 parallel to the Side FE; therefore FACD is a Parallelogram; and consequently FA is * equal to DC, * 34. In and AC to FD; and because AC is drawn parallel to FE, the side of the Triangle F BE, it shall + be † 2 of this. as BA is to AF, fo is BC to CE; and (by Alternation) as BA is to B C, fo is CD to CE. Again ; because CD is parallel to BF, it shall be t as B C is to ÇE, fo is FD to DE; but FD is equal to AC, Therefore as B C is to CE, fo is P A C to DE: And 1 7. 54 so (by Aļternation) as BC is to CÀ, so is CE to ED. Wherefore because it is demonstrated that AB is to BC, as DC is to CE; and as B C is to CA, fo is CE to ED; it shall be * by equality, as B A is to * 21. In AC, so is CD to D E. Therefore, the Sides about equal Angles of equiangular Triangles, are propora timal; and the Sides, which are subtended under the L4 equal the |