Full text: Pergaeus, Apollonius: The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus

From A draw two perpendiculars to the right lines DB, ZC; viz. ADF and
AZX; and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z,
equal to the Radius of the given circle: and through F [?] and X draw lines pa-
rallel to DB, ZC; viz. FG, XH; and then by the preceding Problem draw
a circle which ſhall paſs through the given center A and touch the two lines
FG, XH; and E the center of this circle will alſo be the center of the circle
required, as appears by ſubtracting equals from equals in Figure 1: and by adding
equals to equals in Figure 2.

21. LEMMA II.

If the two circles CEB and CED cut one another C, then I ſay a line drawn
from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-
ments from thoſe circles.

1ſt. Suppose CB to be the Diameter of one of them: then draw to the
other point of ſection E the line CE, and joining EB, ED, the angle CEB will
be a right one, and the angle CED either greater or leſs than a right one, and
conſequently CD cannot be a Diameter of the other.

2dly. Suppose CBD not to paſs through the center of either: then through
C draw a Diameter CAG, and continue it to meet the other circle in F, and
join BG, DF: then the angle CBG is a right one, and the angle CDF is either
greater or leſs than a right one: and therefore the lines BG and DF are not
parallel: let H be the center of the other circle, and let a Diameter CHI be
drawn: draw DI and continue it meet to meet CG in K: then DIK will be pa-
rallel to BG: hence CB: CD: : CG: CK. But CI and CK are unequal,
(being both applied from the ſame point in a right angle) and therefore it cannot
be 2s CB: CD: : CG: CI: and hence it appears that the Segments CB and
CD are diſſimilar.

22. LEMMA III.

If through the legs of any triangle EDF (ſee Figure to Problem 10.) a line
BI be [?] drawn parallel to the baſe DF, ſo that there be conſtituted two ſimilar
triangles about the ſame vertex; and a circle be circumſcribed about each of
theſe triangles; theſe circles will touch one another in the common vertex E.

It is plain that they will either touch or cut each other in the point E: if
they cut each other, then by the preceding Lemma the Segments BE and DE
would be diſſimilar; but they are ſimilar, and they muſt therefore touch each
other.

23. PROBLEM X.

Having a point A, and alſo a right line BC, given in poſition; together
with a circle whoſe center is G given both in m [...] de and [?] poſition; to deſcribe

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