a 4. 6. C 22. 5. b AB: BH :: CD: DF; and AG: GH2:: CE: EF. alfo AG: AH3:: CE: CF. and AH: AB:: CF: CD. Whence by equality AG: AB:: CE: CD. In like manner, GH: HB:: EF: FD. therefore the Polygons ABHG, CDFE 6 def. 6. are fimilar and fimilarly fituate. Q.E.D. C becaufe AB: DF (:: BC: EF):: EF: BG. and the Ang. BF. therefore fhall the Triang. ABG be DEF. But the Triang. ABC: BC i BC & twice; Hence if three Lines BC, EF, BG are Proportional, as the first is to the third, fo is a Triangle defcrib'd upon the firft BC, to a Triangle described upon the fecond EF, being fimilar and alike fituate to the other. Or fo is a Triangle defcrib'd upon the fecond EF to a Triangle described upon the third, fimilar and fimilarly fituate to it. PROP. PROP. XX. Similar Polygons ABCDE, FGHIK are divided into fimilar Triangles ABC, FGH; and ACD, FHI and ADE, FIK, equal in Number and Homologous to the holes (ABC: FGH:: ABCDE : FGHIK :: ACD: FHI :: ADE: FIK) and the Polygons ABCDE, FGHIK, are in that proportion duplicated, that one Homologous Side BC has to another GH. B E are D H b b 1. For the Ang. BG; and AB: BC:: a byp. FG GH. therefore the Triangles ABC, FGH equiangular. After the fame manner we prove that the Triangles AED, FKI, are fimilar. Therefore fince the Ang. BCA GHF, and the Ang. ADE FIK; and the whole Ang. BCDGHI; and the whole CDE, HIK are equal; there will remain the Ang. ACD = FHI, and the Ang. ADE = FIH. Whence © al- €32. 1. fo the Ang.. CAD HFI. therefore the Triangles ACD, FHI are fimilar. Whence, &c. 2. Therefore because the Triangles BCA, с IKF IK 16.5. fince BC: GH:: CD: HI:: DE: IK : there-hyp. fore the Triang. BCA: GHF: CAD: HFI:: cor. 32. DEA 5. L 2 twice. Whence f a DEA: IKF :: polyg. ABCDE: FGHIK :: 2. 12. 5. I. Hence if there are three right Lines proportional, as the firft is to the third, fo is a Polygon described upon the first to a Polygon defcribed upon the fecond, fimilar and alike fituate, or so shall the Polygon described upon the fecond, be to a Polygon defcribed upon the third, fimilar and alike fituate. From hence we have a way of augmenting or diminishing any given right-lined Figure in a given Ratio. As fuppofe you would make a Pentagon five times greater than a given one whofe * 18. 6. Side is CD; between AB and five times AB find a Mean Proportional, upon which * make a Pentagon fimilar to the given one, which will be five times the given one. II. Hence, likewife, if you know the Homologous Sides of fimilar Figures; you may likewife have the Proportion of the Figures; viz. by finding a third Proportional. PROP. XXI. Right-lined Figures ABC, DIE that are fimilar to the fame right-lined Figure HFG, are fimilar to one another. дал A D G E For the Ang. A= HD. and the Ang. * & def. 6. C = GE; and the Ang. BFi alfo alfo AB 6. AC :: HF: HG:: DI: DE. and ↳ 1 def. b AC: CB:: HG : GF :: DE : EI. and AB: BC :: HF: FG :: DI: IE. Therefore ABC, PROP. XXII. If four right Lines be proportional, viz. AB: CD:: EF: GH; the right-lined Figures defcribed upon them fimilarly, and fimilarly fituate, will be also proportional (ABI: CDK :: EM: GO). · And if the fimilar right-lined Figures fimilarly defcribed upon the Lines be proportional, (ABI : CDK :: EM: GO) then the right Lines fhall also be proportional (AB: CD:: EF: GH). да a I H 'ABI b Therefore ABI: CDK :: EM: byp GO. Q.E.D. 1. Hyp. EM GO® AB 2 Hyp. CD EF C GH Q.E.D. twice. Therefore d AB: CD:: EF: GH. d cor. 23. SCHOL. Hence is deduced and demonftrated the way of multiplying furd Quantities. For Example, lets be to be multiply'd by 3; I fay the Product will be 15. For by the Definition of Multiplication, it is as 1:3:: V5 to the Product. Whence (by this Prop.) as the Square of I is to the Square of 3, fo is the Square of √5 to the Square of the Product; that is, 135 to the Square of the Product. Whence the Square of the Product is 15; and fo 15 is the Product of √3 into √5. Q.E. D. If a right Line AB be any how divided in D, then fhall the Rectangle under the farts AD, DB, be a Mean Proportional between the Squares of them, and the Rectangle under the whole AB and one part AD, or DB, is a Mean Proportional between the Square of AB and the Square of the faid part AD or DB. Describe a Semicircle upon the Diameter AB, and raise a Perpendicular from D to meet the Periphery in E, and draw AE, EB. 2 2 Then it is manifeft that AD: DE:: DE: DB. Therefore AD: DE:: DE : DB2, that is, AD: ADB:: DB. Q. E. D. 00 Again BA AEa:: AE: AD. Therefore d BA2: AE AE: AD. That is BA2 BAD :: BAD: AD. In like manner AB": ABD:: ABD: BD. Q. E. D... 2 :: : C PROP. |