Clavius, Christoph
,
Geometria practica
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GEOMETR. PRACT.
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KAO; </
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<
s
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xml:space
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">ac proinde multo mai{us}, quam dimidium trianguli mixti KAO, cui{us} vnum la-
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t{us} eſt arc{us} A O. </
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<
s
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xml:space
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">Eadem ratione erit K O X, mai{us}, quam dimidium trianguli mixti
<
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KOB, cui{us} vnum lat{us} eſt arc{us} O B. </
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<
s
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xml:space
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">Auferendo ergo triangulum KVX, ex figura mi-
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ſtilinea K A B, in qua vnum lat{us} eſt arc{us} A O B, ablatum erit pl{us} quam dimidium.
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</
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<
s
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xml:space
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">Subtractis igitur quatuor eiuſmodi triangulis K V X, L Y a, M b c, I d e, ablatum erit
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plus, quam dimidium ex quatuor reſiduis extra circulum, & </
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<
s
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xml:space
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">ſic deinceps. </
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<
s
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xml:space
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">Ponantur
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igituriam octo triãgula mixta reſidua, quorũ baſes ſunt arcus AO, OB, BP, &</
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<
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">c. </
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minora magnitudine z. </
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<
s
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xml:space
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">Cum ergo circulus cum z, æqualis poſitus ſit triangulo
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F G H, erit circulus cum illis octo reſiduis, hoc eſt, figura Octogona V X Y,
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a b c d e V, minor eodem triangulo F G H. </
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<
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xml:space
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">quod eſt ab ſurdum, cum maius fit: </
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<
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<
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quippe cum perpẽdicularis EO, æqualis ſit lateri F G, & </
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<
s
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xml:space
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">ambitus Octogini ma-
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ior circumferentia circuli, hoc eſt, recta GH. </
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<
s
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xml:space
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">Hinc enim fit, triangulum rectan-
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gulum, cuius latus F G, æquale perpendiculari EO, & </
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<
s
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xml:space
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">alterum latus æquale am-
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bitui Octogoni, maius videlicet, quam GH, maius eſſe triangulo FGH. </
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<
s
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xml:space
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">Cum er-
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go illud triangulum ſit, ex ſcholio propoſ. </
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<
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">41. </
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<
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">lib. </
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<
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xml:space
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">æquale rectangulo ſub
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FG, & </
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<
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">ſemiſſe ambitus Octogoni comprehenſo; </
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<
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">hoc autem rectangulum O-
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ctogono æquale, ex propoſ. </
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<
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">de Iſoperimetris: </
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">erit quoq; </
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<
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xml:space
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">Octogonum
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maius triangulo FGH. </
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<
s
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xml:space
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">Nõ ergo minus eſſe poteſt, ac proinde circulus ABCD,
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minor non eſt triangulo FGH: </
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<
s
xml:id
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xml:space
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">Sed neque maior eſt, vt demonſtrauimus. </
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<
s
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xml:space
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tur æqualis eſt, quod erat demonſtrandum.</
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<
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Scaliger, vel quia vim huius demonſtrationis non perpendit,
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vel quia ſuæ circuli quadrandi rationi vidit eſſe contrariam, non eſt veritus Ar-
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chimedem hoc loco falſitatis arguere: </
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<
s
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">conaturque oſtendere, non rectè ab eo
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demonſtratum, circulum æqualem eſſe triangulo rectangulo, cuius vnum latus
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ſemidiametro, & </
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<
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">alterum circumferentiæ circuli eſt æquale. </
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<
s
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xml:space
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">Nam, ait, ſi de-
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monſtratio Archimedis bona eſt, demonſtrabitur eodem modo, circulũ æqua-
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lem eſſe triangulo rectangulo, cuius vnum latus circa angulum rectum ſemidi-
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ametro æquale eſt, & </
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<
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">alterum peripheria circuli maius. </
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<
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">Sit enim in triangulo
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lmn, latus quidem lm, trianguli ſemidiametro circuli E A, æquale, at mn, peri-
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pheria maius. </
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<
s
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xml:space
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">Concedit ergo Scaliger, circulum non eſſe maiorem triangulo
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FGH, rectè eſſe ab Archimede demonſtratum, hoc eſt, triangulum F G H, cuius
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latus GH, peripheriæ eſt æquale, non eſſe minus circulo, ac proinde neque tri-
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angulum lmn, cuius latus m n, maius eſt peripheria, circulo minus eſſe. </
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<
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xml:space
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cedititem, rectè probatum eſſe, circulũ non eſſe minorem triangulo FGH, ſi la-
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tus GH, peripheriæ ſit ęquale, hoc eſt, triangulum FGH, non eſſe maius circu-
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lo. </
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<
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">Sed negat, ex hoc ſequi, triangulum FGH, eſſe æquale circulo. </
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<
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<
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inquit, eodem modo, ſi baſis m n, maior eſt peripheria, ſed minor circumſcripti
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polygoni ambitu, (hoc enim contingere, ait, nihil prohibet) polygonum erit
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quidem maius triangulo l m n, quod ambitus polygoni maior ſit recta m n, & </
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<
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ſemidiameter EA, rectæ l m, æqualis. </
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<
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xml:space
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">Sed reſectis portionibus, ſequeretur, idẽ
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polygonum eſſe triangulo l m n, minus, quod eſt ineptum. </
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<
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liger? </
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<
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<
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xml:space
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">Nam poſito latere m n, ma-
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iore, quam peripheria; </
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<
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">quando eo peruentum erit, polygonum eſſe minus tri-
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angulo l m n, (ſi nimirumrelictæ portiones minores fuerint magnitudine z,) </
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