Clavius, Christoph
,
Geometria practica
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LIBER SEXTVS.
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<
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his puto ſatis ſtudio ſum Lectorem intelligere, quo pacto in alijs e-
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xemplis ſe gerere debeat. </
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<
s
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xml:space
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">Nam ſi verbi gratia ex hoc propoſito rectilineo irre-
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gulariſsimo per lineam lateri AM, parallelam abſcindenda ſit portio æqualis al-
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teri cuipiam rectilineo minori, producemus MA, vſ-
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178
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275-01
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que ad N. </
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<
s
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xml:space
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">Et ſi quidem deprehenſum fuerit trian-
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gulum A B N, eſſe æquale dato rectilineo minori,
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(quod ſcietur, ſi quadratum triangulo æquale con-
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ſtructum, fuerit æquale quadrato, quod dato recti-
<
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lineo minori conſtruitur æquale) recta AN, proble-
<
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ma efficiet. </
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<
s
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echoid-s11414
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xml:space
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">Siverò maius, conſtruemus ſuper AN,
<
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verſus B, trapezium per parallelam ipſi A N, æquale exceſſui: </
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<
s
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echoid-s11415
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xml:space
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">At ſi minus, du-
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cemus L O, parallelam. </
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<
s
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xml:space
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">Nam ſi fuerit deprehenſum rectilineum NL, æquale
<
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defectui, problema efficiet parallela L O: </
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<
s
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echoid-s11417
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xml:space
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">Si verò maius, conſtituemus ſuper
<
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L O, verſus MN, per parallelam ipſi MN, trapezium exceſſui æquale. </
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<
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xml:space
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">Ea enim
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parallela problema ſoluet: </
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<
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xml:space
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">At ſi minus, producemus OL, ad P: </
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<
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">Etſi quidem
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triangulum KLP, fuerit æquale defectui, tota parallela O P, quæſtioni ſatisfa-
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ciet: </
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<
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xml:space
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">Si verò maius, conſtituemus in angulo K, triangulum ſimile
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note-275-01a
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">25. ſexti.</
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KLP, & </
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<
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">exceſſuiæquale; </
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<
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xml:space
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">ita vt hoc triangulum vna cum rectilineo per paralle-
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lam L O, abſciſſo ſit dato rectilineo minori æquale. </
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<
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xml:space
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">Ex quo colliges, proble-
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ma in hoc caſu ſolui non poſſe, cum duæ parallelæ, nimirum L O, & </
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<
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xml:space
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">illa, quæ
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triangulum ipſi KLP, ſimile aufert, reſecent ex toto rectilineo BG, partem dato
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rectilineo minori æqualem. </
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<
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xml:space
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">At ſi triangulum KLP, fuerit minus defectu præ-
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dicto, ita vt hoc triangulum vna cum rectilineo per parallelam L O, abſciſſo ſit
<
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minus dato rectilineo minore, ducemus per D, parallelam Q R. </
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<
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xml:space
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">Et ſi quidem
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rectilineum P R, æquale fuerit defectui, quo figura KPLMABNO, à dato re-
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ctilineo minore deficit, factum erit per parallelam QR, quod iubetur: </
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<
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">Siverò
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maius, parallela, quæ cum QR, verſus OP, auferet rectilineum huic exceſſuiæ-
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quale, ſatisfaciet problemati: </
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<
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xml:space
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">At ſi rectilineum PR, fuerit minus prædicto de-
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fectu, & </
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<
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">triangulum C D R, inuentũ fuerit vltimo huic defectui, quo rectiline-
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um PR, à prædicto defectu deficit, æquale, parallela DQ@ quæſtionem diſſoluet:
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</
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<
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xml:space
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atur triangulum exceſſui æquale, & </
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<
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ni duæ parallelæ, videlicet D Q. </
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<
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">& </
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<
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">baſis prædictitrianguli conſtituti; </
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<
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hoc caſu per vnicam parallelam ſatisfieri problemati nequit: </
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<
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gulum CDR, minus extiterit eo dem illo vltimo defectu, ducemus parallelam
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IS. </
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<
s
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xml:space
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">Et ſi quidẽ rectilineum DI, æquale fuerit illi, quo triangulum CDR, minus
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eſt vltimo illo defectu, erit totum rectilineum ISDCBAMLKI, dato minori re-
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ctilineo æquale: </
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<
s
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">Si autem rectilineum DI, inæquale fuerit, progrediemur vlte-
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rius, vt iam ſæpius dictum eſt, donec rectilineum inueniamus dato minori recti-
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lineo æquale; </
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<
s
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xml:space
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">Inuenietur autem omnino vnum æquale, cum totũ rectilineum
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BG, maius ponatur. </
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<
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">Vides igitur, facilè conijci poſſe, quando problema per v-
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nicam parallelam ſolui poſsit, & </
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<
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">quando non, ſed per duas: </
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<
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">Quotieſcunque
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enimincidemus in eiuſmo ditriangulum in ipſa conſtructione, qualia fu-
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erunt K L P, & </
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<
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æquale exceſſui alicui, ſolui problema nequit, niſi per
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duas parallelas.</
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