1be drawn by D, with which let C B prolonged meet in A; and let fall
the Perpendiculars D N and M C to M D, and B N to B D; and about
the Right-angled Triangle D B N deſcribe the Semicircle D F B N,
cutting D C in F; and let D O be a Mean-proportional betwixt C D
and D F; and A V a Mean-proportional betwixt C A and A B: And
let P S be the time in which the whole D C, or B C, ſhall be paſſed;
(for it is manifeſt that they ſhall be both paſt in the ſame Time;) And
look what proportion C D hath to D O, the ſame ſhall the Time S P
have to the Time P R: the Time P R ſhall be that in which a Movea
ble out of D will paſſe D F; and R S that in which it ſhall paſſe the re
mainder F C. And becauſe P S is alſo the Time in which the Movea
ble out of B ſhall paſſe B C; if it be ſuppoſed that as B C is to C D, ſo is
S P to P T, P T ſhall be the Time of the Deſcent out of A to C: by
reaſon D C is a Mean-proportional betwixt A C and C B, by what was
before demonſtrated: Laſt of all, as C A is to A V, ſo let T P be to
140[Figure 140]
P G: P G ſhall be the Time,
in which thé Moveable out
of A deſcendeth to B. And
becauſe of the Circle D F N
the Diameter erect to the
Horizon is D N, the Lines
D F and D B ſhall be paſ
ſed in equal Times. So that
if it ſhould be demonſtra
ted that the Moveable would
ſooner paſſe B C after the
Deſcent D B, than F C after the Lation D F; we ſhould have our in
tent. But the Moveable will with the ſame Celerity of Time paſſe B C
coming out of D along D B, as if it came out of A along A B: for that
in both the Deſcents D B and A B it acquireth equal Moments of Velo
city: Therefore it ſhall reſt to be demonſtrated that the Time is ſhorter
in which B C is paſſed after A B, than that in which F C is paſt after
D F. But it hath been demonſtrated, that the Time in which B C is
paſſed after A B is G T; and the Time of F C after D F is R S. It is
to be proved therefore, that R S is greater than G T: Which is thus
done. Becauſe as S P is to P R, ſo is C D to D O, therefore, by Conver
ſion of proportion, and by Inverſion, as R S is to S P, ſo is O C to C D:
and as S P is to P T, ſo is D C to C A: And, becauſe as T P is to PG,
ſo is C A to A V: Therefore alſo, by Converſion of the proportion, as
P T is to T G, ſo is A C to C V: therefore, ex equali, as R S is to G T,
ſo is O C to C V. But O C is greater than C V, as ſhall anon be de
monſtrated: Therefore the Time R S is greater than the Time G T:
Which it was required to demonſtrate. And becauſe C F is greater than
C B, and F D leſſe than B A, therefore C D ſhall have greater propor
tion to D F than C A to A B: And as C D is to D F, ſo is the Square
the Perpendiculars D N and M C to M D, and B N to B D; and about
the Right-angled Triangle D B N deſcribe the Semicircle D F B N,
cutting D C in F; and let D O be a Mean-proportional betwixt C D
and D F; and A V a Mean-proportional betwixt C A and A B: And
let P S be the time in which the whole D C, or B C, ſhall be paſſed;
(for it is manifeſt that they ſhall be both paſt in the ſame Time;) And
look what proportion C D hath to D O, the ſame ſhall the Time S P
have to the Time P R: the Time P R ſhall be that in which a Movea
ble out of D will paſſe D F; and R S that in which it ſhall paſſe the re
mainder F C. And becauſe P S is alſo the Time in which the Movea
ble out of B ſhall paſſe B C; if it be ſuppoſed that as B C is to C D, ſo is
S P to P T, P T ſhall be the Time of the Deſcent out of A to C: by
reaſon D C is a Mean-proportional betwixt A C and C B, by what was
before demonſtrated: Laſt of all, as C A is to A V, ſo let T P be to
140[Figure 140]P G: P G ſhall be the Time,
in which thé Moveable out
of A deſcendeth to B. And
becauſe of the Circle D F N
the Diameter erect to the
Horizon is D N, the Lines
D F and D B ſhall be paſ
ſed in equal Times. So that
if it ſhould be demonſtra
ted that the Moveable would
ſooner paſſe B C after the
Deſcent D B, than F C after the Lation D F; we ſhould have our in
tent. But the Moveable will with the ſame Celerity of Time paſſe B C
coming out of D along D B, as if it came out of A along A B: for that
in both the Deſcents D B and A B it acquireth equal Moments of Velo
city: Therefore it ſhall reſt to be demonſtrated that the Time is ſhorter
in which B C is paſſed after A B, than that in which F C is paſt after
D F. But it hath been demonſtrated, that the Time in which B C is
paſſed after A B is G T; and the Time of F C after D F is R S. It is
to be proved therefore, that R S is greater than G T: Which is thus
done. Becauſe as S P is to P R, ſo is C D to D O, therefore, by Conver
ſion of proportion, and by Inverſion, as R S is to S P, ſo is O C to C D:
and as S P is to P T, ſo is D C to C A: And, becauſe as T P is to PG,
ſo is C A to A V: Therefore alſo, by Converſion of the proportion, as
P T is to T G, ſo is A C to C V: therefore, ex equali, as R S is to G T,
ſo is O C to C V. But O C is greater than C V, as ſhall anon be de
monſtrated: Therefore the Time R S is greater than the Time G T:
Which it was required to demonſtrate. And becauſe C F is greater than
C B, and F D leſſe than B A, therefore C D ſhall have greater propor
tion to D F than C A to A B: And as C D is to D F, ſo is the Square