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Apollonii Pergaei quae graece exstant cum ... - Wilbourhall.org

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CONICORUM LIBER I. 165<br />

et a ad AE perpendicularis ducatur @E, per E<br />

autem rectae B& parallela EA, et ab A ad EA<br />

perpendicularis<br />

^J<br />

ducatur AA^ EA autem in K in duas<br />

A^L<br />

perpendicularis<br />

partes aequales secetur,<br />

et a Jf ad EA<br />

ducatur<br />

iiTMproducaturque<br />

ad Z, H, et sit<br />

AKxKM=^AA^,<br />

datis autem duabus<br />

rectis<br />

AK, KM, quarum<br />

KA positione<br />

data est ad K terminata,<br />

KM autem magnitudine,<br />

et angulo recto describatur parabola, cuius<br />

diametrus sit KAj uertex autem K, et latus rectum KM,<br />

ita ut supra demonstratum est [prop. LII]; per A igitur<br />

ueniet, quia AKx KM== AA^ [prop. XI], et EA<br />

sectionem continget, quia EK = KA [prop. XXXIII].<br />

et 0^ rectae EKA parallela est; itaque 0AB diametrus<br />

sectionis est,<br />

et rectae a sectione ad eam ductae<br />

rectae AE parallelae ab ^J5 in binas partes aequales<br />

secabuntur [prop. XLVI]. ducentur autem in angulo<br />

0AE [Eucl. I, 29]. et quoniam est L AE0 = L AHZ,<br />

communis autem angulus ad A positus, erit<br />

A®E(yr)AHZ.<br />

quare [EucL YI, 4] ®A: EA = ZAiAH itaque<br />

2A&:2AE=ZA:AH [Eucl. V, 15]. est autem<br />

rA = 2@A', itaque ZA : AH = TA : 2AE. ergo<br />

propter ea, <strong>quae</strong> in propositione XLIX demonstrata<br />

sunt, r^ latus rectum est.

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