Clavius, Christoph
,
Geometria practica
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LIBER OCTAVVS.
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">SI per idem punctum diametri in rectangulo duæ lineæ ducantur late-
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ribus parallelæ: </
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<
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henſum æquale duobus rectangulis ſub ſegmentis duorum laterum
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comprehenſis.</
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rectangulo BD, per E, punctum diametri AC, ductæ ſint FG, HI, lateri-
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bus parallelæ. </
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<
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FB, & </
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<
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AE, EC, vna cum rectangulo bis ſub AE, EC. </
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"> Suntautem quadrata ex AB, BC, quadrato ex AC, æqualia; </
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<
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">erunt quoque duo quadrata ex AE, EC, vna
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cum rectangulo bis ſub AE, EC, æqualia quadratis ex
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http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/385-01
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AB, BC. </
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<
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FB; </
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<
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">Et quadratum ex BC, æquale eſt duobus quadra-
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tis ex BI, IC, vna cum rectangulo bis ſub BI, IC. </
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<
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tur duo quadrata ex AE, EC, vna cum rectangulo bis
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ſub AE, EC, æqualia erunt quatu or quadratis ex AF,
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FB, BI, IC, vna cum rectangulis bis ſub AF, FB, & </
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<
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dratum ex AE, quadratis ex AF, FE, hoc eſt, ex AF, BI: </
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">& </
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<
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">quadratum ex EC,
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quadratis ex EI, IC, hoc eſt ex FB, IC: </
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<
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">æquale ſit; </
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ex AF, BI, FB, IC, vna cum rectangulo bis ſub AE, EC, æqualia quatuor qua-
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dratis ex AF, FB, BI, IC, vna cum rectangulis bis ſub AF, FB, & </
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<
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latiſque vtrobique prædictis quatuor quadratis communibus, eritreliquum re-
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ctangulum bis ſub AE, EC, reliquis rectangulis bis ſub AF, FB, & </
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<
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quale. </
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">Ideo que rectangulum ſemel ſub AE, EC, rectangulis ſemel ſub AF.
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in quadrato, vt in poſteriori figura, rectangulum ſub ſegmentis
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diametri AE, EC, comprehenſum æquale eſt duobus complementis DE, BE,
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quippe cum complementa ſub ſegmentis laterum comprehendantur.</
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<
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<
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bus punctis ad eaſdem partes axis, vel centri, per quæ tranſire dicatur
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Ellipſis: </
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<
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problema conicum eſt, & </
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<
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<
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dium alicuius Ellipſis in linea axis maioris BC, quantacunq;</
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<
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eadem Ellipſi D, E, verſus eandem partem, hoc eſt, ſiue ſupra centrum A, ſiue
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infra; </
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<
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