1demonſtrated, and the Motions thorow them ſhall be performed in equal
Times ſeeing that they terminate in A unto the Circumference of the
Circle A G O from the higheſt point of it A.
Times ſeeing that they terminate in A unto the Circumference of the
Circle A G O from the higheſt point of it A.
PROBL. XII. PROP. XXXIII.
A Perpendicular and Plane inclined to it being
given, whoſe height is one and the ſame, as al
ſo the higheſt term, to find a point in the Per
pendicular above the common term, out of
which if a Moveable be demitted that ſhall
afterwards turn along the inclined Plane, the
ſaid Plane may be paſt in the ſame Time in
which the Perpendicular ex quiete would be
paſſed.
given, whoſe height is one and the ſame, as al
ſo the higheſt term, to find a point in the Per
pendicular above the common term, out of
which if a Moveable be demitted that ſhall
afterwards turn along the inclined Plane, the
ſaid Plane may be paſt in the ſame Time in
which the Perpendicular ex quiete would be
paſſed.
Let the Perpendicular and inclined Plane, whoſe Altitude is the
ſame, be A B and A C. It is required in the Perpendicular B A,
continued out from the point A to find a Point out of which a
Moveable deſcending may paſſe the Space A C in the ſame Time in
which it will paſſe the ſaid Perpendicular A B out of Reſt in A. Draw
D C E at Right-Angles to A C, and let C D be cut equal to A B, and
draw a Line from A to D: The Angle A D C ſhall be greater than the
Angles C A D: (for C A is greater than A B or C D:) Let the
Angle D A E be equal to the Angle A D E; and to A E let E F an in
clined Plane be Perpen-
132[Figure 132]
dicular, and let both be
ing prolonged meet in F,
and unto both A I and
A G ſuppoſe C F to be
equal, and by G draw
G H equidiſtant to the
Horizon. I ſay, that H
is the point which is
ſought. For ſuppoſing the
Time of the Fall along
the Perpendicular A B
to be A B, the Time along
A C ex quiete in A ſhall be the ſame A C. And becauſe in the Right
angled Triangle A E F, from the Right Angle E unto the Baſe A F,
E C is a Perpendicular, A E ſhall be a Mean-Proportional betwixt F A
and A C, and C E a Mean betwixt A C and C F, that is, betwixt C A
and A I: and foraſmuch as the Time of A C out of A is A C, A E
ſame, be A B and A C. It is required in the Perpendicular B A,
continued out from the point A to find a Point out of which a
Moveable deſcending may paſſe the Space A C in the ſame Time in
which it will paſſe the ſaid Perpendicular A B out of Reſt in A. Draw
D C E at Right-Angles to A C, and let C D be cut equal to A B, and
draw a Line from A to D: The Angle A D C ſhall be greater than the
Angles C A D: (for C A is greater than A B or C D:) Let the
Angle D A E be equal to the Angle A D E; and to A E let E F an in
clined Plane be Perpen-
132[Figure 132]dicular, and let both be
ing prolonged meet in F,
and unto both A I and
A G ſuppoſe C F to be
equal, and by G draw
G H equidiſtant to the
Horizon. I ſay, that H
is the point which is
ſought. For ſuppoſing the
Time of the Fall along
the Perpendicular A B
to be A B, the Time along
A C ex quiete in A ſhall be the ſame A C. And becauſe in the Right
angled Triangle A E F, from the Right Angle E unto the Baſe A F,
E C is a Perpendicular, A E ſhall be a Mean-Proportional betwixt F A
and A C, and C E a Mean betwixt A C and C F, that is, betwixt C A
and A I: and foraſmuch as the Time of A C out of A is A C, A E