1ſect A E, the Angle A being acute: Let the interſection be in E,
from whence let fall a Perpendicular to A E, which produced, will
meet with A B infinitely prolonged in F. I ſay, firſt, that the
Right-lines F E, and F C are equal: ſo that drawing the Line
E C, we ſhall, in the
59[Figure 59]
two Triangles D E C,
B E C, have the two
Sides of the one, D E,
and C E, equal to the
two Sides of the other
B E, and E C; the
two Sides, D E, and
E B, being Tangents
to the Circle D B,
and the Baſes D C,
and C B, are likewiſe
equal: wherefore the
two Angles D E C,
and B E C, ſhall be
equal. And becauſe the Angle B C E wanteth of being a Right
Angle, as much as the Angle B E C; and the Angle C E F, to
make it a Right-Angle, wants the Angle C E D, thoſe Supple
ments being equal, the Angles F C E, and F E C ſhall be equal,
and ſo conſequently the Sides F E, and F C; wherefore making
the point F a Center, and at the diſtance F E, deſcribing a Circle,
it ſhall paſs by the point C. Deſcribe it, and let it be C E G. I ſay,
that this is the Circle required, by any point of the Circumfe
rence of which, any two Lines that ſhall interſect, departing from
the terms A and B, ſhall be in proportion to each other, as are the
two parts A C, and B C, which beſore did concur in the point C.
This is manifeſt in the two that concur or interſect in the point E,
that is A E, and B E; the Angle E of the Triangle A E B being
divided in the midſt by C E; ſo that as A C is to C B, ſo is A E
to B E. The ſame we prove in the two A G, and B G, determined
in the point G. Therefore being (by the Similitude of the Tri
angles A F E, and E F B) that as A F is to E F, ſo is E F to F B;
that is, as A F is to F C, ſo is C F to F B: So by Diviſion; as A C
is to C F, (that is, to F G) ſo is C B to B F; and the whole A B
is to the whole B G, as the part C B to the part B F: and by Com
poſition; as A G is to G B, ſo is C F to F B; that is, as E F to
F B, that is, as A E to E B, and A C to C B: Which was to be de
monſtrated. Again, let any other Point be taken in the Circum
ference, as H; in which the two Lines A H and B H concur. I ſay, in
like manner as before, that as A C is to C B, ſo is A H to B H.
Continue H B untill it interſect the Circumference in I, and draw
from whence let fall a Perpendicular to A E, which produced, will
meet with A B infinitely prolonged in F. I ſay, firſt, that the
Right-lines F E, and F C are equal: ſo that drawing the Line
E C, we ſhall, in the
59[Figure 59]two Triangles D E C,
B E C, have the two
Sides of the one, D E,
and C E, equal to the
two Sides of the other
B E, and E C; the
two Sides, D E, and
E B, being Tangents
to the Circle D B,
and the Baſes D C,
and C B, are likewiſe
equal: wherefore the
two Angles D E C,
and B E C, ſhall be
equal. And becauſe the Angle B C E wanteth of being a Right
Angle, as much as the Angle B E C; and the Angle C E F, to
make it a Right-Angle, wants the Angle C E D, thoſe Supple
ments being equal, the Angles F C E, and F E C ſhall be equal,
and ſo conſequently the Sides F E, and F C; wherefore making
the point F a Center, and at the diſtance F E, deſcribing a Circle,
it ſhall paſs by the point C. Deſcribe it, and let it be C E G. I ſay,
that this is the Circle required, by any point of the Circumfe
rence of which, any two Lines that ſhall interſect, departing from
the terms A and B, ſhall be in proportion to each other, as are the
two parts A C, and B C, which beſore did concur in the point C.
This is manifeſt in the two that concur or interſect in the point E,
that is A E, and B E; the Angle E of the Triangle A E B being
divided in the midſt by C E; ſo that as A C is to C B, ſo is A E
to B E. The ſame we prove in the two A G, and B G, determined
in the point G. Therefore being (by the Similitude of the Tri
angles A F E, and E F B) that as A F is to E F, ſo is E F to F B;
that is, as A F is to F C, ſo is C F to F B: So by Diviſion; as A C
is to C F, (that is, to F G) ſo is C B to B F; and the whole A B
is to the whole B G, as the part C B to the part B F: and by Com
poſition; as A G is to G B, ſo is C F to F B; that is, as E F to
F B, that is, as A E to E B, and A C to C B: Which was to be de
monſtrated. Again, let any other Point be taken in the Circum
ference, as H; in which the two Lines A H and B H concur. I ſay, in
like manner as before, that as A C is to C B, ſo is A H to B H.
Continue H B untill it interſect the Circumference in I, and draw