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# Title: [Problema Austriacum Plus Ultra Quadratura Circuli] Problema Avstriacvm Plvs Vltra Qvadratvra Circvli (shelfmark Leibn. Marg. 230, Stück 1)
# Folio 650, page id 1815201908:0707
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AYtede i
eROPOSFLLD. CCXXVII.
TIE AB diametertran suersa hyperbo.
D lae BC D: ductisque ordinarim CF,
AanE G F ad F GE, VEC F ad FC.
Dico BGC esse ad eandem hyperbo.
lam cujus eransuertum lacus est AB. p
Demonstratio.
Voniam ex suppositione AB est diameter
Leransuerla hyperbola BCD, igitur ve est
LC ad E D, quadracum, ica esrectangulum
AFB, ad AF B rectanigulum. ergo ve quadea
cum CF adFIF. quadracum, ica quoque est
AFB rectangulum ad rectangulum AF B. ivi-
T hyperbola est cuius latus cranluer-
uC AB. Duod demonstrandum fuit.
Demonstratio.
Filac recta OK xequalis ON: e
Ponaneur GL, TL ParallelaCD.
quemadmodum ctiam LML _ LMI
quoniam BP L ponitur lixperbola
cuius cransuersa diameter AB, crit
AMB ad AMB, rectangulum ve est
quadratum P M ad F MT quadratum,
hoc est ve quadracum G L ad TL qua-
dratum. est auTeM C F ad OB, vE
CG ad OK; igitur eAt OT ad OK,
DE OM ad OB: _E OK ad RL, ve
OB ad BMI; _ NR adR L, veAB
ad BMI; denique _ N L ad L R, vE
ATT, K aNNLK FCCAMS MIAMI. GUALL CAMI SC rectanoulum AMB ad A NIB, re-
atangulum ve clt quadracum EM 2d FMI, guadracum, hoc est GL quadrarcum ad
quadratum IE, criam eric rectangulum NYL R ad NY L. K, rectangulum ve quadra.
cum OL. ad TF quadracum. igicur manifestum cst K C P eNc ad candem hyperbo.
lam cuius diameter cransuerla cA NK. Quodtuic demonctrandum.
CUF B
 t  1 11.
ane aquidistantes C F. DCICL LdC VE CE ad HIF, ita CG ad FTI.
DicoTCC este ad candem hypetoblam.
ROPOSITIO COXXVIII.
VInE AB, C. D diamerri conius acx hyperbolX BE F _ diameerO AB
AM ad MB. quare rectangulum AME B 2d R M B, rectanoulum est vE rectangulum
enet3tti benata
AIEhyperbola AB B, cuiusaxis AD: ductisq; ad eandem ordinatim
rcao LLD aguato MCa L, C.
Do ACF elle ad eandem hiyperbolam cuius axis est AD.
lemon
u   u te  s t  e tst te tete te s e te te t s i tes 1t est  e ste e   t . e   s ite ete    tet  t ete e a  t  ete t