Valerio, Luca - De centro gravitatis solidorum

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1ad triangulum FBG, hoc eſt vt AF ad FG, ita eſt
triangulum AFC ad triangulum FCG; triangulum er
go FBG triangulo FCG æquale erit, & baſis BG ba
ſi GC æqualis. Quoniam igitur & AE eſt æqualis
EC, ſimiliter vt ante, oſtenderemus, triangulum BCF,
triangulo ACF, eademque ratione triangulum ABF,
triangulo BCF æquale eſſe: igitur vnumquodque trian
gulorum ABF, ACF, BCF, tertia pars eſt trianguli
ABC: ſed vt triangulum ABC, ad triangulum BCF,
ita eſt AG, ad GF; tripla igitur eſt AG ipſius GF,
ac proinde AF, ipſius FG dupla. Eadem ratione
BE, ipſius FE, & CF, ipſius FD, dupla concludetur.

Sed ſint ſi fieri poteſt, trianguli ABC duo centra qua
lia diximus D, E: & ab ipſis ad ſingulos angulos du
cantur binæ rectæ lineæ:
& eadat D in aliquo trian
gulo BEC. Quoniam
igitur D eſt centrum trian
guli ABC erit triangu
lum BDC tertia pars
trianguli ABC. Eadem
ratione triangulum BEC
tertia pars erit trianguli
ABC; triangulum ergo
DBC æquale erit trian
gulo BEC pars toti, quod
fieri non poteſt, atqui idem

8[Figure 8]
abſurdum ſequitur, ſi punctum D cadat in aliquo latere
triangulorum, quorum vertex E; Manifeſtum eſt igitur
propoſitum.

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