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