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

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aliquod<br />

CONICORUM LIBER I. 179<br />

H sumatur, per id autem rectae ^E parallela<br />

ducatur HA^ <strong>quae</strong> <strong>cum</strong> AB producta in K concurrat.<br />

producatur igitur ZO et <strong>cum</strong> HK in A concurrat.<br />

quoniam igitur arcus A^ arcui z/5 aequalis est,<br />

erit [Eucl. III, 27] LABA^LAZB. et quoniam<br />

est [Eucl. I, 32] L EZA = ZJA + ZAJ, et<br />

LZA/i = LZB^,<br />

L ZAA ^ LZBA [Eucl. III, 27], erit etiam<br />

L EZA = L ^BA = L BZ^,<br />

uerum etiam JE parallela est rectae AH. quare<br />

L EZA = L ZH®, L AZB = L Z®H [Eucl. I, 29].<br />

quare etiam L Z if0 = L Z H et [Eucl. 1 , 6] Zif= Z(9.<br />

describatur igitur cir<strong>cum</strong> ®H circulus H®N ad<br />

triangulum @HZ perpendicularis, et fingatur conus,<br />

cuius basis sit H®N circulus, uertex autem Z punctum;<br />

conus igitur rectus erit, quia HZ = Z0 [def. 3]. et<br />

quoniam circulus H@N ad planum ®HZ perpendicularis<br />

est, uerum etiam planum subiacens ad planum<br />

rectarum if0, ®Z perpendiculare est, etiam communis<br />

eorum sectio ad planum rectarum if0, ®Z perpendicularis<br />

erit [Eucl. XI, 19]. KM igitur communis<br />

eorum sectio sit. itaque KM ad utramque AK, KH<br />

perpendicularis est [Eucl. XI def. 3]. et quoniam<br />

conus, cuius basis est H®N circulus, uertex autem Z<br />

punctum, plano per axem sectus est, quod sectionem<br />

efficit triangulum HSZ, uerum etiam alio plano<br />

rectarum AK^ KMj quod est planum subiacens, sectus<br />

est secundum rectam KM ad HK perpendicularem,<br />

et<br />

hoc planum <strong>cum</strong> ZH, Z® lateribus coni concurrit,<br />

sectio orta ellipsis est, cuius diametrus est AB,<br />

ordinate ductae autem in recto angulo ducentur<br />

12*

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