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# Title: [Problema Austriacum Plus Ultra Quadratura Circuli] Problema Avstriacvm Plvs Vltra Qvadratvra Circvli (shelfmark Leibn. Marg. 230, Stück 1)
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4
2X8. D.
tribva
bLhmins:
AIYPERBOL A.
Demonstratio.
Voniam C D aequales sunt rectis A F, qua.
Ldracum DC aequale est quadraris AC,
CF. Fiacigicur recta AF, aequalis recto lareri
Parabola AF F; erir icaque a rectangulu F AC
xquale quadrato C P. isicur rectangulu FC A,
xquale est quadrato AC, vna cum rectangulo
F A C hoc eAt quadrato C F. vnde FC A rC.
ctangulo aequale est quadracum A F, hoc cAt
CD quadeacum. Igitur rectangulum FCA,
ad F C A rectangulum, candem obcince ratio-
nem, quam quadearum CD ad CD quadea-
cum. Ergo ADD cAbhyperbola cuius diame.
cer est AB. lacus cransuersum rero AF. Quod
fuic demonstrandum.
IBer id iptum aliter demonstrare. sic parabolx ABB axis AC, d Or.
dinarim poliex ad eandem . Udne verorectis AB aequales linea
Dico ADD fore ad hsperbolam cuius aris AC C.
Demonstratio.
CIE A F, lacerirecto parabolx AB B ae qualis,
ee parallela C B; ducacurgi AG qua rectum
angulum F. AC bitariam partiacur, cui aequi-
distet recta F II ducta per P. Pric itaguere.
ctangulum C AF, xquale C B quadraco, hoc
est eric quadrato CB aequale rectangulum
COTHCUM OCREC A, irEm A F, C L(Xqua-
lessine; ee consequenter rect angulo C CIIx.
qualc quadracum CD, cum illud xquale posi-
cum (E quadraco AB, hoc cs quadraris C B e
OAhoccCC C. igicur cum sIC CC FI rocta
sulum ad C CFT rectangulum, ve quadrarum
CDad CD quadracum, paECE AD D hyper.
bolam este, de rectas C. D ordinarim esse politas
ad axem AC. Ouod fuir demonstrandum.
PROPOSITTO CEKVIII
IE AB diamerer lecundaria parabole
ad quam ordinarim pOnanEuEC D;
quadraris aurem AC, C D RanE Xqu2.
lia quadrata C F..
Dico AEF esse hyperbolam, cuius
lacus cranluersum (ic lacus rectum did-
merri AB in parabola.
Demonstratio.
FiTaC AF aequale laceri recto diameeri AB in
Parabola ADD. cric iraque quadracu CDC
xequale FAC rectangulo, e conlequenter
rectanxulus F. C A xquale quadraris AC, C D.
hoc
PROPOSITTO CCXV