Gravesande, Willem Jacob 's
,
An essay on perspective
Text
Text Image
Image
XML
Thumbnail overview
Document information
None
Concordance
Notes
Handwritten
Figures
Content
Thumbnails
Table of Notes
<
1 - 3
>
[Note]
Page: 20
[Note]
Page: 21
[Note]
Page: 21
[Note]
Page: 21
[Note]
Page: 21
[Note]
Page: 21
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 22
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 23
[Note]
Page: 24
[Note]
Page: 25
[Note]
Page: 25
[Note]
Page: 30
[Note]
Page: 31
[Note]
Page: 31
[Note]
Page: 32
[Note]
Page: 33
[Note]
Page: 33
[Note]
Page: 37
<
1 - 3
>
page
|<
<
(20)
of 237
>
>|
<
echo
version
="
1.0RC
">
<
text
xml:lang
="
en
"
type
="
free
">
<
div
xml:id
="
echoid-div69
"
type
="
section
"
level
="
1
"
n
="
37
">
<
pb
o
="
20
"
file
="
0042
"
n
="
44
"
rhead
="
An ESSAY
"/>
<
p
>
<
s
xml:id
="
echoid-s578
"
xml:space
="
preserve
">Now, it is evident, that to prove the
<
note
symbol
="
@
"
position
="
left
"
xlink:label
="
note-0042-01
"
xlink:href
="
note-0042-01a
"
xml:space
="
preserve
">27.</
note
>
pearance of A is in the Line C H, we need but
<
lb
/>
demonſtrate that O D is parallel to A E; </
s
>
<
s
xml:id
="
echoid-s579
"
xml:space
="
preserve
">which
<
lb
/>
may be done thus:</
s
>
<
s
xml:id
="
echoid-s580
"
xml:space
="
preserve
"/>
</
p
>
<
p
>
<
s
xml:id
="
echoid-s581
"
xml:space
="
preserve
">Becauſe the Triangles O G V, and A B F, are
<
lb
/>
ſimilar.
<
lb
/>
</
s
>
<
s
xml:id
="
echoid-s582
"
xml:space
="
preserve
">A F: </
s
>
<
s
xml:id
="
echoid-s583
"
xml:space
="
preserve
">A B:</
s
>
<
s
xml:id
="
echoid-s584
"
xml:space
="
preserve
">: O G: </
s
>
<
s
xml:id
="
echoid-s585
"
xml:space
="
preserve
">O V: </
s
>
<
s
xml:id
="
echoid-s586
"
xml:space
="
preserve
">
<
lb
/>
altern. </
s
>
<
s
xml:id
="
echoid-s587
"
xml:space
="
preserve
">
<
lb
/>
A F: </
s
>
<
s
xml:id
="
echoid-s588
"
xml:space
="
preserve
">O G:</
s
>
<
s
xml:id
="
echoid-s589
"
xml:space
="
preserve
">: A B: </
s
>
<
s
xml:id
="
echoid-s590
"
xml:space
="
preserve
">O V: </
s
>
<
s
xml:id
="
echoid-s591
"
xml:space
="
preserve
">
<
lb
/>
Divid. </
s
>
<
s
xml:id
="
echoid-s592
"
xml:space
="
preserve
">and altern. </
s
>
<
s
xml:id
="
echoid-s593
"
xml:space
="
preserve
">the firſt
<
lb
/>
Proportion. </
s
>
<
s
xml:id
="
echoid-s594
"
xml:space
="
preserve
">
<
lb
/>
AF—AB (=CF): </
s
>
<
s
xml:id
="
echoid-s595
"
xml:space
="
preserve
">O G—O V=HG:</
s
>
<
s
xml:id
="
echoid-s596
"
xml:space
="
preserve
">:AB:</
s
>
<
s
xml:id
="
echoid-s597
"
xml:space
="
preserve
">OV.</
s
>
<
s
xml:id
="
echoid-s598
"
xml:space
="
preserve
"/>
</
p
>
<
p
>
<
s
xml:id
="
echoid-s599
"
xml:space
="
preserve
">But becauſe the Triangles E C F, H G D are
<
lb
/>
ſimilar.
<
lb
/>
</
s
>
<
s
xml:id
="
echoid-s600
"
xml:space
="
preserve
">C F: </
s
>
<
s
xml:id
="
echoid-s601
"
xml:space
="
preserve
">H G :</
s
>
<
s
xml:id
="
echoid-s602
"
xml:space
="
preserve
">: E F : </
s
>
<
s
xml:id
="
echoid-s603
"
xml:space
="
preserve
">G D.</
s
>
<
s
xml:id
="
echoid-s604
"
xml:space
="
preserve
"/>
</
p
>
<
p
>
<
s
xml:id
="
echoid-s605
"
xml:space
="
preserve
">Now, by obſerving the two laſt Proportions of
<
lb
/>
the other two Triangles,
<
lb
/>
E F: </
s
>
<
s
xml:id
="
echoid-s606
"
xml:space
="
preserve
">G D:</
s
>
<
s
xml:id
="
echoid-s607
"
xml:space
="
preserve
">: A F: </
s
>
<
s
xml:id
="
echoid-s608
"
xml:space
="
preserve
">O G,
<
lb
/>
And the Angle A F E, being equal to the Angle
<
lb
/>
O G D, the Triangles A E F and O D G are
<
lb
/>
ſimilar; </
s
>
<
s
xml:id
="
echoid-s609
"
xml:space
="
preserve
">and therefore A E is parallel to O D:
<
lb
/>
</
s
>
<
s
xml:id
="
echoid-s610
"
xml:space
="
preserve
">Which was to be demonſtrated.</
s
>
<
s
xml:id
="
echoid-s611
"
xml:space
="
preserve
"/>
</
p
>
<
p
>
<
s
xml:id
="
echoid-s612
"
xml:space
="
preserve
">After the ſame manner we prove, that the
<
lb
/>
Appearance of the Point A is in the Line L I,
<
lb
/>
and conſequently is in the Interſection of this
<
lb
/>
Line and HC.</
s
>
<
s
xml:id
="
echoid-s613
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
<
div
xml:id
="
echoid-div71
"
type
="
section
"
level
="
1
"
n
="
38
">
<
head
xml:id
="
echoid-head40
"
xml:space
="
preserve
">
<
emph
style
="
sc
">Remark</
emph
>
.</
head
>
<
p
>
<
s
xml:id
="
echoid-s614
"
xml:space
="
preserve
">Altho’ this Method appears more difficult than
<
lb
/>
the precedent one, as to the Geometrical Conſi-
<
lb
/>
deration thereof, yet the Operation is eaſier, if
<
lb
/>
the Points are not too far diſtant from the Baſe
<
lb
/>
Line: </
s
>
<
s
xml:id
="
echoid-s615
"
xml:space
="
preserve
">For Lines may well enough be drawn by
<
lb
/>
Gueſs, or Sight only, to touch Circles, and Cir-
<
lb
/>
cles to touch Lines.</
s
>
<
s
xml:id
="
echoid-s616
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
</
text
>
</
echo
>