Gravesande, Willem Jacob 's, An essay on perspective

Table of figures

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[11] Plate 4.Page 24.Fig. 10.f F O G g V D N L a P E H I M A
Page: 53
[12] Plate 5.page 26.Fig. 11.O Y b X a E Z A B
Page: 58
[13] Fig. 12.M O Y F S C L D X a E B Z A
Page: 58
[14] Plate 6.page 28.Fig. 13.O D c b a g E G B A C
Page: 65
[15] Fig. 14.O b 1 2 3 a c 1 2 3 g D A C 3 1 2 2 1 3 B G
Page: 65
[16] page 28.Plate. 7Fig. 16Fig. 15O G F I Vl d c e m n b a h B A H M N C E P D L
Page: 68
[17] page 36.Plate 8Fig. 17O G F c d b a A B D C
Page: 77
[18] Page 36.Plate 9Fig. 18.G F C S V I E B A
Page: 80
[19] Fig. 19.O i M X L D @ b a
Page: 80
[20] Fig. 20.S x G n H S V D l R f Q m P t
Page: 80
[21] Fig. 21.I X f T L B N A C l M E F
Page: 80
[22] page 38Plate 10.Fig. 22.V F I N a G H M P D E B C L A
Page: 83
[23] Fig. 23.O F I H a G D E B C L A M
Page: 83
[24] Fig. 24.@ o f X a e A
Page: 83
[25] page 42Plate 11.Fig. 25.S F V M I N P H a L D E G C A B
Page: 92
[26] Fig. 26.Fig. 27.S V P Q R n l g h G H B N I A C M L
Page: 92
[27] page 46Plate 12.Fig. 28.
Page: 99
[28] Fig. 29.F S V q q q E L p p p I G H q D P n n n T R m m m C B Q A
Page: 99
[29] Fig. 30.O X E L N M G Z Y D
Page: 99
[30] Fig. 31.f 3 c l n m g 4
Page: 99
[31] page 48.Plate 13.Fig. 32.V S R L P B D Q T M I F A E Y C G O H Z N
Page: 104
[32] page 52.Plate. 14.Fig. 34D C F G A B H L E
Page: 111
[33] Fig. 33S X 8 1 h 6 g 3 z q 9 m 2 4 m a 5 Y
Page: 111
[34] Fig. 35O M D P T Q R m p q B A S C
Page: 111
[35] page 56.Plate. 15F H O Z D G
Page: 118
[36] Fig. 36T N M L a R Q E I A C P B
Page: 118
[37] Fig. 37F S V T I E M A N X P C B
Page: 118
[38] page 58.Plate. 16Fig. 39Fig. 38F Q O p l r s 1 2 3 4 G
Page: 123
[39] page 60.Plate. 17F V
Page: 128
[40] Fig. 40c θ b e a F G H I K L A B E C D
Page: 128
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page |< < (20) of 237 > >|
4420An ESSAY
Now, it is evident, that to prove the 1127. pearance of A is in the Line C H, we need but
demonſtrate that O D is parallel to A E; which
may be done thus:
Becauſe the Triangles O G V, and A B F, are
ſimilar.
A F: A B: : O G: O V:
altern.
A F: O G: : A B: O V:
Divid. and altern. the firſt
Proportion.
AF—AB (=CF): O G—O V=HG: :AB: OV.
But becauſe the Triangles E C F, H G D are
ſimilar.
C F: H G : : E F : G D.
Now, by obſerving the two laſt Proportions of
the other two Triangles,
E F: G D: : A F: O G,
And the Angle A F E, being equal to the Angle
O G D, the Triangles A E F and O D G are
ſimilar; and therefore A E is parallel to O D:
Which was to be demonſtrated.
After the ſame manner we prove, that the
Appearance of the Point A is in the Line L I,
and conſequently is in the Interſection of this
Line and HC.
Remark.
Altho’ this Method appears more difficult than
the precedent one, as to the Geometrical Conſi-
deration thereof, yet the Operation is eaſier, if
the Points are not too far diſtant from the Baſe
Line: For Lines may well enough be drawn by
Gueſs, or Sight only, to touch Circles, and Cir-
cles to touch Lines.

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