DelMonte, Guidubaldo
,
In duos Archimedis aequeponderantium libros Paraphrasis : scholijs illustrata
Text
Text Image
Image
XML
Thumbnail overview
Document information
None
Concordance
Figures
Thumbnails
Page concordance
<
1 - 30
31 - 60
61 - 90
91 - 120
121 - 150
151 - 180
181 - 207
>
Scan
Original
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
<
1 - 30
31 - 60
61 - 90
91 - 120
121 - 150
151 - 180
181 - 207
>
page
|<
<
of 207
>
>|
<
archimedes
>
<
text
>
<
body
>
<
chap
id
="
N10019
">
<
p
id
="
N176FE
"
type
="
main
">
<
s
id
="
N17700
">
<
pb
xlink:href
="
077/01/195.jpg
"
pagenum
="
191
"/>
què AB ipſam DG in H. Quoniam enim parabole
<
arrow.to.target
n
="
marg376
"/>
<
lb
/>
ſeſquitertia eſt trianguli ABC, itidemquè parabole DBE
<
lb
/>
trianguli DBE ſeſquitertia exiſtit, erit parabole ABC ad trian
<
lb
/>
gulum ABC, vt parabole DBE ad triangulum DBE. &
<
arrow.to.target
n
="
marg377
"/>
per
<
lb
/>
mutando parabole ABC ad parabolen DBE, vt triangulum
<
lb
/>
ABC ad triangulum DBE. Quoniam autem AC ordina
<
lb
/>
tim eſt applicata, vnde AF ipſi FC eſt æqualis, ac per conſe
<
lb
/>
〈que〉ns AF eſt ipſius AC dimidia. </
s
>
<
s
id
="
N17721
">erit triangulum ABF dimi
<
lb
/>
dium trianguli ABC. cùm vtraquè ſub eadem ſint
<
arrow.to.target
n
="
marg378
"/>
<
lb
/>
eademquè ratione triangulum DBG trianguli DBE dimi
<
lb
/>
dium exiſtit. </
s
>
<
s
id
="
N1772C
">quare vt triangulum ABF ad triangulum
<
lb
/>
DBG, ita eſt triangulum ABC ad DBE triangulum, ac pro
<
lb
/>
pterea triangulum ABF ad DBG triangulum eſt, vt parabo
<
lb
/>
le ABC ad parabolen DBE. Cùm autem ſit HG æquidiſtans
<
lb
/>
ipſi AF, ſiquidem ſunt ordinatim applicatæ, ob
<
expan
abbr
="
triangulorũ
">triangulorum</
expan
>
<
arrow.to.target
n
="
marg379
"/>
<
lb
/>
ſimilitudinem ABF HBG, ita erit FB ad BG, vt AF ad HG
<
lb
/>
vt autem FB ad BG, ita quadratum ex AF ad quadratum
<
arrow.to.target
n
="
marg380
"/>
<
lb
/>
DG, erit igitur quadratum ex AF ad quadratum ex DG, vt AF
<
lb
/>
ad HG. quare lineę AF DG HG ſunt proportionales. </
s
>
<
s
id
="
N17748
">Pro
<
lb
/>
ducatur FB, ducaturquè à puncto D ipſi AB æquidiſtans
<
lb
/>
DK, erit vtiquè triangulorum ABF DKG anguli ABF
<
lb
/>
DHG æquales, & angulus AFB angulo DGK eſt æqualis, erit
<
lb
/>
igitur, & reliquus BAF reliquo KDG æqualis, ac propterea
<
lb
/>
triangulum ABF eſt triangulo DKG ſimile. </
s
>
<
s
id
="
N17754
">quare triangu
<
lb
/>
lum ABF ad triangulum DKG eam habet proportionem,
<
lb
/>
quàm AF ad DG duplicatam, hoc eſt quàm AF ad HG, quę
<
lb
/>
eſt ea, quàm habet FB ad BG. atqui triangulum ABF ad
<
lb
/>
DKG eam quo〈que〉 habet proportionem, quam FB ad GK
<
lb
/>
duplicatam. </
s
>
<
s
id
="
N17760
">tres igitur lineę FB GK GB ſunt proportiona
<
lb
/>
les. </
s
>
<
s
id
="
N17764
">ex quibus ſequiturita eſſe FB ad GK, vt AF ad DG; &
<
lb
/>
GK ad GB, vt DG ad GH. ſed quoniam triangulum GDK
<
lb
/>
ad GDB (cùm ſint ſub eadem altitudine) ita eſt, vt KG
<
arrow.to.target
n
="
marg381
"/>
<
lb
/>
BG, ſi igitur fiat HG ad L, vt KG ad BG, erit triangulum
<
lb
/>
GDK ad triangulum GDB, vt HG ad L. Cùm autem ſit
<
expan
abbr
="
triã
">triam</
expan
>
<
lb
/>
gulum ABF ad DKG, vt AF ad HG, eſtquè
<
expan
abbr
="
triangulũ
">triangulum</
expan
>
DKG
<
lb
/>
ad DBG, vt HG ad L, erit ex ęquali triangulum ABF ad
<
lb
/>
triangulum DBG, vt AF ad L. ac propterea parabole ABC </
s
>
</
p
>
</
chap
>
</
body
>
</
text
>
</
archimedes
>