any Parallelepipedon is produced by multiplying the Altitude into the Bale, as is manifest from the 31ft Prop. of this Book. Moreover, fince the whole Parallelepipedon is produced by Multiplying the Altitude into the whole Base, the half thereof, that is the Triangular Prifm, is produced by multiplying "the Altitude into half the Base. PROP. I. Lib. X. being necessary to demonftrate fome Things in this Book, Itherefore shall infert it here. H I If there be two B C unequal Magnitudes AB and C D ...! propos'd and F G E from the greater part AH greater than its Half; and if from HB what remains, there be taken again a part HI greater than half this remainder; and if this be done continually, there will at last remain fome Magnitude IB, that shall be less than the propofed leffer Magnitude. Take C fuch a Number of times, that DE a Multiple of it fhall nearest exceed AB; and let DF FG GE = C. Now take AH from AB greater than its half, and HI from the remainder HB, more than its half, and fo on until the Parts AH, HI, IB are the fame in Number with the Parts DF, FG, GE. Now it is manifeft, that FE, which is not less than half DE, is greater than HB, which is lefs than half AB DE. In like manner GE, which is not lefs than half FE, is greater than IB half HB. Therefore C, or GEIB. Q. E. D. The fame is demonftrated, if from AB you take AH its half, and from the Remainder HB, again its half HI, and so on. EU. 1 EUCLID's ELEMENTS. S B BOOK XII. DEFINITIONS. PROP. I. Imilar Polygons ABCDE, FGHIK de- Draw AC, BL, FH, GM. Because the Ang: ABC FGH, and A AB: BC:: FG: GH1 def. 6. O 2 Therefore b 6.6. 21.3 C d 31.3. Therefore the Ang. ACB (ALB) fhall be FHG (FMG). But the Angles ABL, FGM ‘, are right Angles, and fo equal to each other. Therefore the Triangles ABL,FGM are Equicor. 4. 6. angular, whence 'AB: FG:: AL: FM. Therefores ABCDE: FGMIK :: AL': FM2. · 32. 3. f 8 22.6. CORO L. Becaufe AB: FG:: AL: FM :: BC: GH, &c. therefore the Area's of fimilar Figures de1.12. fcribed in a Circle, are in the Ratio of the h PROP. II. Circles ABT, EFN, have the fame Proportion to one another as the Squares of their Diameters, AC, EG. For firft, if poffible, let I be less than the Circle EFN, and let K be the Excefs or Difference. Infcribe the Square EFGH in the C a d e 41. 1. Circle EFN, which will be equal to one Schol. 7. half of the eircumfcribed Square, and fo 4 greater than the Semicircle. Bifect. the Arches" 30. 3. EF, FG, GH, HE, and at the Points of Divifion join the right Lines EL, LF, &c. At L draw the Tangent PQ, and continue out HEP, GFQ (which will be parallel to EF). Schol. 27. Then the Triangle ELF will be half of the Parallelogram EPQF, and fo greater than the half of the Segment ELF. And after the fame manner, the other Triangles do exceed the halves of the other Segments. And if the Arches EL, LF, FM, &c. be again bifected, and right Lines drawn, the Triangles will likewife exceed the halves of the Segments. Whence if the Square EFGH be taken from the Circle EFN, and the Triangles from the other Segments, and this be done continually, there willat laft remain fome Magnitude lefs 1. 10. than K. Let us have gone fo far, viz. to the Segments EL, LF, FM, &c. taken together. Therefore I (Circ. EFNK) Polyg. hyp. E ELFMGNHO (Circ. EFN-Segm. EL+LF, 1 ax. c.) infcrib'd in the Circle ABT fuppos'd 30. 3.. fimilar to the Polygon ARBSCTDV, and fo & post. I. fince ARBSCTDV: ELFMGNHO: AC EG: Circ. ABT Space I. And the Polyg. byp. ARBSCTDV Circ. ABT. the Polyg. 19 ax. 1. ELFMGNHO fhall be I. But we had m 14. 5. before I ELFMGNHO, which is abfurd. Again, (2.) if poffible, let I 2 Circ. EFN, e : h k 1. 12. :n hyp. then because AC EG" :: Circ ABT:n K, but EG: AC: Circ. EFN: K. P 11. 5 which is contradictory, as has been already demonftrated. 3 There I. |