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:
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.
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.
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.
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.
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.
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.