tio's ; as if AB: BG:: EB: BC, then the Antecedents are AB and EB, one of which is in each Figure, and the Confequents are BG, BC; one of which is likewife in each Figure. III. A right Line AC is faid to be divided into extreme and mean Proportion, when the whole AC is to the greater Segment AB, as the greater Segment AC is to the leffer one CB. that is AC: AB :: AB: CB. B Ratio's multiplied new Ratio. As the Ratio of IV. The Altitude of any Figure ABC, is a Perpendicular Line AD drawn from the Vertex A to the Bafe BC. V. A Ratio is faid to be compounded of Ratio's, when the Quantities of the into each other, beget fome A to C is compounded of the Ratio's of A to B, and of B to C, for A Ba A' X AB PROP. I. Triangles ABC, FCD, and Parallelograms BCAE, CDFA, that have the fame altitude, are to one another as their Bafes BC, CD. H G a C D Take any Number of Lines on BC, as BG, HG, each equal to BC, alfo DI= CD, and join AG, AH, FI. The Triangles ACB, ABG, AGH are equal; alfo the Triang. FCD FDI. Therefore the Triangle ACH is the fame Multiple of the Triangle ACB, as the Base HC is of the Base BC; and the Triangle FCI is the fame Multiple of the Triangle FCD, as the Base CI is of the Bafe CD. But if HC be,, or than CI, in like manner fhall the Triang. AHC be,=, or FCI; and therefore BC: CD:: Triang. ABC Triang. FCD:: Pgr. CE: CF. Q. E.D. с d SCHOL SCHOL. Hence the Triangles ABC, HKM, and the Parallelograms AGBC, DKHM, that have equal BaJes BC, KM, are to one another as their Altitudes AI, HF. ИИ BL CIKE MF b a Take ILCB, and EF = KM; and join LA, LG, ED, EH; it is manifeft that the Triang. ABC: KHM :: ALI: HEF :: AI: HF:: Pgr. AGBC: DKHM. B d PRO P. II. a с d If any right Line DE be drawn parallel to one Side BC of a Triangle ABC; this fhall cut or divide the Sides of that Triang. proportionally, viz. AD: BD:: AE: EC. And if the Sides of a Triangle be cut or divided proportionally, viz. AD: BD:: C AE EC. a right Line DE joining the Points D, E, of Di- . D E vifions, will be parallel to the other Side BC of the Triangle. Draw CD, BE. e 1 Hyp. Because the Triang. DEB DEC; then fhall the Triang. ADE: DBE :: ADE: ECD; but the Triang. ADE: DBE:: AD: K 3 3. 1. 7.5. 1. 6. 41. I. & 15.5.. DB. and the Triang. ADE: DEC:: AE": 2 Hyp. Because AD: DB:: AE: EC. that is the Triang. ADE: DBE :: ADE: ECD; the Triang. DBE fhall be ECD. Therefore DE, BC, are parallel. Q. E. D. d If feveral right Lines DE, FG, be parallel to one Side BC, all the Segments of the Side's fhall be proportional. : For DF: FA:: EG : GA; and by compounding and inverting, FA DA :: GA: EA; and DA:DB:: EA: EC. therefore by Equality DF: DB:: EG: EC. Q.E.D. COROL. e If DF: DB:: EG: EC; then shall BC, DE, FG be parallel. And if the Segments of the Bafe have the fame Proportion as the other Sides of the Triangle, (BD: DC:: AB: AC) then a right Line AD drawn from from the Vertex A to D the Point of Section, will bifelt the Angle BAC of the Triangle. Continue out BA, and make AE= AC, and join CE. 1 Hyp. Because AE BAC E are Parallel; whence DC Q. E. D. f 2 Hyp. Because BA: AC: (AE) :: BD: DC, therefore shall DA, CE be parallel: whence the Ang. BADE, and the Ang. DAC = ACE E. therefore" the Ang. BAD And fo the Ang. BAC is bife&ed. Q. E. D. DAC: 5. I. h The Sides of equiangular Triangles ABC, DCE, that are about the equal Angles, B, DCE, are Proportional, viz. AB: BC:: DC: CE, &c. and the Sides AB, DC, &c. that fubtend the equal Angles ACB, É, &c, are Homologous or of like Ratio's. Set the Side BC in the same right Line with the Side CE, and continue out BA, ED, till they meet one another. i k k Because the Ang. BECD, the Sides BF, CD are parallel; alfo because the Ang. BCA CED, the Lines CA, EF are m therefore the Figure CAFD is a Parallelogram, whence " AFCD, and AC" FD. There- m 34. 1. that AB: AF (: CD) : : n 2.6. K 4 BC: fore it is manifeft |