LEMMA.
If two Circles touch one another within, the innermoſt of which
toucheth ſome Right Line, and the exteriour one cutteth it,
three Lines produced from the Contact of the Circles unto
three points of the Tangent Right-Line, that is, to the Con
tact of the interiour Circle, and to the Sections of the exte
riour ſhall contain equall Angles in the Contact of the
Circles.
toucheth ſome Right Line, and the exteriour one cutteth it,
three Lines produced from the Contact of the Circles unto
three points of the Tangent Right-Line, that is, to the Con
tact of the interiour Circle, and to the Sections of the exte
riour ſhall contain equall Angles in the Contact of the
Circles.
Let two Circles touch one another in the point A, of which let the
Centers be B, that of the leſſer, and C that of the greater; and let
the interiour Circle touch any Line F G in the point H, and let the grea
ter cut it in the points F and G, and connect the three Lines A F, A H,
and A G. I ſay, that the Angles by
130[Figure 130]
them contained F A H and G A H are
equal. Produce A H untill it meeteth
the Circumference in I, and from the
Centers draw B H and C I, and thorow
the ſaid Centers let B C be drawn,
which continued forth ſhall meet with
the Contact A, and with the Circum
ferences of the Circles in O and N.
And becauſe the Angles I C N and
H O B are equal, for as much as either
of them is double to the Angle I A N,
the Lines B H and C I ſhall be Parallels: And becauſe B H drawn
from the Center to the Contact is Perpendicular to F G; C I ſhall alſo be
Perpendicular to the ſame, and the Arch F I equal to the Arch I G, and,
which followeth of conſequence, the Angle F A I to the Angle I A G:
Which was to be demonſtrated.
Centers be B, that of the leſſer, and C that of the greater; and let
the interiour Circle touch any Line F G in the point H, and let the grea
ter cut it in the points F and G, and connect the three Lines A F, A H,
and A G. I ſay, that the Angles by
130[Figure 130]them contained F A H and G A H are
equal. Produce A H untill it meeteth
the Circumference in I, and from the
Centers draw B H and C I, and thorow
the ſaid Centers let B C be drawn,
which continued forth ſhall meet with
the Contact A, and with the Circum
ferences of the Circles in O and N.
And becauſe the Angles I C N and
H O B are equal, for as much as either
of them is double to the Angle I A N,
the Lines B H and C I ſhall be Parallels: And becauſe B H drawn
from the Center to the Contact is Perpendicular to F G; C I ſhall alſo be
Perpendicular to the ſame, and the Arch F I equal to the Arch I G, and,
which followeth of conſequence, the Angle F A I to the Angle I A G:
Which was to be demonſtrated.