LEMMAI.
Let D C be Perpendicular to the Diameter B A; and from the Term
B continue forth B E D at pleaſure, and draw a Line from F to B. I
ſay, that F B is a Mean-proportional be-
136[Figure 136]
twixt D B and B E. Draw a Line from E
to F, and by B draw the Tangent B G;
which ſhall be Parallel to the former C D:
Wherefore the Angle D B G ſhall be equal
to the Angle F D B, like as the ſame G B D
is equal alſo to the Angle E F B in the al
tern Portion or Segment: Therefore the
Triangles F B D and F E B are alike: And,
as B D is to B F, ſo is F B to B E.
B continue forth B E D at pleaſure, and draw a Line from F to B. I
ſay, that F B is a Mean-proportional be-
136[Figure 136]twixt D B and B E. Draw a Line from E
to F, and by B draw the Tangent B G;
which ſhall be Parallel to the former C D:
Wherefore the Angle D B G ſhall be equal
to the Angle F D B, like as the ſame G B D
is equal alſo to the Angle E F B in the al
tern Portion or Segment: Therefore the
Triangles F B D and F E B are alike: And,
as B D is to B F, ſo is F B to B E.
LEMMA II.
Let the Line A C be greater than D F; and let A B have greater
proportion to B C, than D E hath to E F. I ſay, that A B is greater
than D E. For becauſe A B hath to B C
137[Figure 137]
greater proportion than D E hath to D F,
therefore look what proportion A B hath to
B C, the ſame ſhall D E have to a Line leſ
ſer than E F; let it have it to E G: And
becauſe A B to B C, is as D E, to E G, there
fore, by Compoſition, and by converting the Proportion, as C A is to A B,
ſo is G D to D E: But C A is greater than G D: Therefore B A ſhall
be greater than D E.
proportion to B C, than D E hath to E F. I ſay, that A B is greater
than D E. For becauſe A B hath to B C
137[Figure 137]greater proportion than D E hath to D F,
therefore look what proportion A B hath to
B C, the ſame ſhall D E have to a Line leſ
ſer than E F; let it have it to E G: And
becauſe A B to B C, is as D E, to E G, there
fore, by Compoſition, and by converting the Proportion, as C A is to A B,
ſo is G D to D E: But C A is greater than G D: Therefore B A ſhall
be greater than D E.
LEMMA III.
138[Figure 138]Let A C I B be the Quadrant of a Circle:
and to A C let B E be drawn from B Pa
rallel: And out of any Center taken in the
ſame deſcribe the Circle B O E S, touching
A B in B, and cutting the Circumference of
the Quadrant in I; and draw a Line from
C to B, and another from C to I continued
out to S. I ſay, that the Line C I is alwaies
leſſe than C O. Draw a Line from A to I;
which toucheth the Circle B O E. And if
D I be drawn it ſhall be equal to D B: And
becauſé D B toucheth the Quadrant, the ſaid
D I ſhall likewiſe touch it; and ſhall be Per-
and to A C let B E be drawn from B Pa
rallel: And out of any Center taken in the
ſame deſcribe the Circle B O E S, touching
A B in B, and cutting the Circumference of
the Quadrant in I; and draw a Line from
C to B, and another from C to I continued
out to S. I ſay, that the Line C I is alwaies
leſſe than C O. Draw a Line from A to I;
which toucheth the Circle B O E. And if
D I be drawn it ſhall be equal to D B: And
becauſé D B toucheth the Quadrant, the ſaid
D I ſhall likewiſe touch it; and ſhall be Per-
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