Now, it is evident, that to prove the Ap- 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.

38. 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.