Z=A+E and A>E; he lets also X=A-E. From these relations he obtains identities which, in modern notation, are ¼Z²-AE=(½Z-E)²=¼X². Now, if we know Z and AE, we can find ½X. Then ½(Z+X)=A, and ½(Z-X)=E, and A=½Z+√(¼Z²-AE). Having established these preliminaries, he proceeds thus: Datis igitur linea inaequaliter secta Z (10), & rectangulo sub segmentis AE (21) qui gnomon est: datur semidifferentia segmentorum ½X: & per consequens ipsa segmenta. Nam ponatur alterutrum segmentum A: alterum erit Z-A: Rectangulum auctem est ZA-A_q=AE. Et quia dantur Z & AE: estque ¼Z_q-AE=¼X_q: & per 5c. 18, ½Z+½X=A: & ½Z-½X=E: Aequatio sic resoluetur: ½Z±√_q:¼Z_q-AE:=A {maius segment/minus segment. Itaque proposita equatione, in qua sunt tres species aequaliter in ordine tabellae adscendentes, altissima autem species ponitur negata: Magnitudo data coefficiens mediam speciem est linea bisecanda: & magnitudo absoluta data, ad quam sit aequatio, est rectangulum sub segmentis inaequalibus, sine gnomon: vt ZA-A_q=AE: in numeris autem 10l-l_q=21: Estque A, vel 1l, alterutrum segmentum inaequale. Inuenitur autem sic: Dimidiata coefficiens median speciem est Z/2 (5); cuius quadratum est Z_q/4 (25): ex hoc tolle AE (21) absolutum: eritque Z_q/4-AE (4) quadratum semidifferentiae segmentorum: latus huius quadratum (2) est semidifferentia: quam si addas ad Z/2 (5) semissem coefficientis, sive lineae bisecandae, erit maius segment.; sin detrahas, erit minus segment: Dico Z/2±√_q:Z_q/4-AE:=A {maius segmentum/minus segmentum. We translate the Latin passage, using the modern exponential notation and parentheses, as follows: Given therefore an unequally divided line Z (10), and a rectangle beneath the segments AE (21) which is a gnomon. Half the difference of the segments ½X is given, and consequently the segment itself. For, if one of the two segments is placed equal to A, the other will be Z-A. Moreover, the rectangle is ZA-A²=AE. And because Z and AE are given, and there is ¼Z²-AE=¼X², and by 5c.18, ½Z+½X=A, and ½Z-½X=E, the equation will be solved thus: ½Z±√(¼Z²-AE)=A {major segment/minor segment. And so an equation having been proposed in which three species (terms) are in equally ascending powers, the highest species, moreover, being negative, the given magnitude which constitutes the middle species is the line to be bisected. And the given absolute magnitude to which it is equal is the rectangle beneath the unequal segments, without gnomon. As ZA-A²=AE, or in numbers, 10x-x²=21. And A or x is one of the two unequal segments. It may be found thus: