Apollonius <Pergaeus> - Apollonii Pergaei Conicorvm Lib. V. VI. VII. paraphraste Abalphato Asphahanensi : nunc primum editi ; additvs in calce Archimedis assvmptorvm liber, ex codibvs arabicis mss Abrahamus Ecchellensis Maronita latinos reddidit, Jo. Alfonsvs Borellvs curam in geometricis versione contulit & [et] notas vberiores in vniuersum opus adiecit

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228190Apollonij Pergæi ctangulum V a E ad quadratum a A eandem proportionem habebit, quàm axis
tranſuerſus ad eius erectum, ſeu quàm axis tranſuerſus alterius ſectionis C D
ad eius erectum: ſed in eadem proportione eſt rectangulum γ c F ad quadratũ
1137. lib. 1. c C; igitur in duobus triangulis A V E, & C γ F rectæ A a, & C c cũ baſibus
angulos æquales a, & c, nempe rectos efficiunt, cum ordinatim applicatæ ſint ad
axes; atque duo anguli verticales V A E, & γ C F æquales ſint inter ſe, cum
propter parallelas æquales ſint angulis O, & M æqualibus in ſegmentis ſimilibus;
22Propoſ. 7.
præmiſſ. igitur duo triangula A E V, & C F γ æquiangula, & ſimilia ſunt inter ſe: &
proptered V A ad A E erit, vt γ C ad C F, & c.

Ponamus iam P A ad duplam A E, vt Q C ad duplam C F: ergo ex
33e æqualitate A T diameter ad A P erectum eius, & c. In hoc textu nonnulla
videntur deficere, eiuſq; ſenſus talis erit. Quia veluti ſupra dictum eſt, triã-
gula R A L, & S C N ſimilia ſunt inter ſe, habebit R A ad A L eandem pro-
portionem, quàm S C ad C N: Ponamus iam P A ad duplam A E, vt R A ad
A L, & Q C ad duplam C F, vt S C ad C N, erunt A P, & C Q latera re-
cta diametrorum A M, & O C; ſed earundem diametrorum figuræ oſtenſæ ſunt
4450 lib. 1.
Lem. 8. ſimiles; igitur latus tranſuerſum A T ad A P erectum eius eſt, vt latus tran-
uerſum X C ad C Q erectum eius. Et quia vt latus tranſuerſum ad rectum

256[Figure 256] ita eſt rectangulum T M A ad quadratum M G, & ſimiliter rectangulum X O
5521. lib. 1. C ad quadratum O I eandem proportionem habebit, quàm latus tranſuerſum ad
rectum, ſcilicet eandem, quàm habent latera figurarũ earundẽ diametrorũ; igi-
tur rectangulum T M A ad quadratum M G eandem proportionẽ habebit, quàm
rectangulum X O C ad quadratum O I; habet verò M G ad M A eandem pro-
portionem, quàm I O ad O C propter ſimilitudinem ſegmentorum; ergo quadra-
tum G M ad quadratum M A erit vt quadratum I O ad quadratum O C: &
propterea ex æquali ordinata rectangulum T M A ad quadratum M A, ſeu T

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