1Mechanicks: That the Moment of the Weight elevated upon the Plane
according to the Line A B C, is
to its total Moment, as B E to B A;
94[Figure 94]
And that the Moment of the ſame
Weight upon the Elevation A D,
is to its total Moment, as D F to
D A or B A: Therefore the Mo
ment of the ſaid Weight upon the
Plane inclined according to D A,
is to the Moment upon the Plane
inclined according to A B C, as
the Line D F to the Line B E:
Therefore the Spaces which the
ſaid Weight ſhall paſſe in equal
Times along the Inclined Planes C A and D A, ſhall be to each other as
the Line B E to D F; by the ſecond Propoſition of the Firſt Book:
But as B E is to D F, ſo A C is demonſtrated to be to D A:
Therefore the ſame Moveable will in equal Times paſſe the Lines
C A and D A.
according to the Line A B C, is
to its total Moment, as B E to B A;
94[Figure 94]And that the Moment of the ſame
Weight upon the Elevation A D,
is to its total Moment, as D F to
D A or B A: Therefore the Mo
ment of the ſaid Weight upon the
Plane inclined according to D A,
is to the Moment upon the Plane
inclined according to A B C, as
the Line D F to the Line B E:
Therefore the Spaces which the
ſaid Weight ſhall paſſe in equal
Times along the Inclined Planes C A and D A, ſhall be to each other as
the Line B E to D F; by the ſecond Propoſition of the Firſt Book:
But as B E is to D F, ſo A C is demonſtrated to be to D A:
Therefore the ſame Moveable will in equal Times paſſe the Lines
C A and D A.
And that C A is to D A as B E is to D F, is thus demonſtrated.
Draw a Line from C to D; and by D and B draw the Lines
D G L, (cutting C A in the point I) and B H, Parallels to A F:
And the Angle A D I ſhall be equal to the Angle D C A, for that
the parts L A and A D of the Circumference ſubtending them, are
equal, and the Angle D A C common to them both: Therefore of
the equiangled Triangles C A D and D A I, the ſides about the
equal Angles ſhall be proportional: And as C A is to A D, ſo is
D A to A I, that is B A to A I, or H A to A G; that is, B E to
D F: Which was to be proved.
D G L, (cutting C A in the point I) and B H, Parallels to A F:
And the Angle A D I ſhall be equal to the Angle D C A, for that
the parts L A and A D of the Circumference ſubtending them, are
equal, and the Angle D A C common to them both: Therefore of
the equiangled Triangles C A D and D A I, the ſides about the
equal Angles ſhall be proportional: And as C A is to A D, ſo is
D A to A I, that is B A to A I, or H A to A G; that is, B E to
D F: Which was to be proved.
Or elſe the ſame ſhall be demonſtrated more ſpeedily thus.
Vnto the Horizon A B, let a Circle be erect, whoſe Diameter is
perpendicular to the Horizon: and
from the higheſt Term D let a Plane
95[Figure 95]
at pleaſure D F, be inclined to the
Circumference. I ſay that the De
ſcent along the Plane D F, and the
Fall along the Diameter B C, will
be paſſed by the ſame Moveable in
equal Times. For let F G be drawn
parallel to the Horizon A B, which
ſhall be perpendicular to the Diameter
D C, and let a Line conjoyn F and
C: and becauſe the Time of the Fall
along D C, is to the Time of the Fall along D G, as the Mean
Proportional between C D and D G, is to the ſaid D G; and the
perpendicular to the Horizon: and
from the higheſt Term D let a Plane
95[Figure 95]at pleaſure D F, be inclined to the
Circumference. I ſay that the De
ſcent along the Plane D F, and the
Fall along the Diameter B C, will
be paſſed by the ſame Moveable in
equal Times. For let F G be drawn
parallel to the Horizon A B, which
ſhall be perpendicular to the Diameter
D C, and let a Line conjoyn F and
C: and becauſe the Time of the Fall
along D C, is to the Time of the Fall along D G, as the Mean
Proportional between C D and D G, is to the ſaid D G; and the