PROBL. XVI. PROP. XXXVIII.
Two Horizontal Planes cut by the Perpendicular
being given, to find a ſublime point in the Per
pendicular, out of which Moveables falling
and being reflected along the Horizontal
Planes may in Times equal to the Times of
the Deſcents along the ſaid Horizontal Planes,
namely, along the upper and along the lower,
paſſe Spaces that have to each other any given
proportion of the leſſer to the greater.
being given, to find a ſublime point in the Per
pendicular, out of which Moveables falling
and being reflected along the Horizontal
Planes may in Times equal to the Times of
the Deſcents along the ſaid Horizontal Planes,
namely, along the upper and along the lower,
paſſe Spaces that have to each other any given
proportion of the leſſer to the greater.
LET the Planes C D and B E be interſected by the Perpendicular
A C B, and let the given proportion of the leſſe to the greater be
N to F G. It is required in the Perpendicular A B to find a point
on high, out of which a Moveable falling, and reflected along C D may
in a Time equal to the Time of its Fall, paſſe a Space, that ſhall have
unto the Space paſſed by the other Moveable coming out of the ſame ſub
lime point in a Time equal to the Time of its Fall with a Reflex Motion
along the Plane B E the ſame proportion as the given Line N batb to
143[Figure 143]
F G. Let G H be
made equal to the
ſaid N; and as F H
is to H G, ſo let
B C be to C L. I ſay,
L is the ſublime
point required. For
taking C M double
to C L, draw L M
meeting the Plane
B E in O; B O
ſhall be double to
B L: And becauſe,
as F H is to H G, ſo is B C to C L; therefore, by Compoſition and In
verſion, as H G, that is, N is to G F, ſo is C L to L B, that is, C M to
B O: But becauſe C M is double to L C; let the Space C M be that
which by the Moveable coming from L after the Fall L C is paſſed along
the Plane C D; and by the ſame reaſon B O is that which is paſſed after
the Fall L B in a Time equal to the Time of the Fall along L B; foraſ
much as B O is double to B L: Therefore the Propoſition is manifeſt.
A C B, and let the given proportion of the leſſe to the greater be
N to F G. It is required in the Perpendicular A B to find a point
on high, out of which a Moveable falling, and reflected along C D may
in a Time equal to the Time of its Fall, paſſe a Space, that ſhall have
unto the Space paſſed by the other Moveable coming out of the ſame ſub
lime point in a Time equal to the Time of its Fall with a Reflex Motion
along the Plane B E the ſame proportion as the given Line N batb to
143[Figure 143]F G. Let G H be
made equal to the
ſaid N; and as F H
is to H G, ſo let
B C be to C L. I ſay,
L is the ſublime
point required. For
taking C M double
to C L, draw L M
meeting the Plane
B E in O; B O
ſhall be double to
B L: And becauſe,
as F H is to H G, ſo is B C to C L; therefore, by Compoſition and In
verſion, as H G, that is, N is to G F, ſo is C L to L B, that is, C M to
B O: But becauſe C M is double to L C; let the Space C M be that
which by the Moveable coming from L after the Fall L C is paſſed along
the Plane C D; and by the ſame reaſon B O is that which is paſſed after
the Fall L B in a Time equal to the Time of the Fall along L B; foraſ
much as B O is double to B L: Therefore the Propoſition is manifeſt.