THEOR. IX. PROP. IX.
If two Planes be inclined at pleaſure from a point
in a Line parallel to the Horizon, and be inter
ſected by a Line which may make Angles Al
ternately equal to the Angles contained be
tween the ſaid Planes and Horizontal Parallel,
the Motion along the parts cut off by the ſaid
Line, ſhall be performed in equal Times.
in a Line parallel to the Horizon, and be inter
ſected by a Line which may make Angles Al
ternately equal to the Angles contained be
tween the ſaid Planes and Horizontal Parallel,
the Motion along the parts cut off by the ſaid
Line, ſhall be performed in equal Times.
From off the point C of the Horizontal Line X, let any two Planes
be inclined at pleaſure C D and C E, and in any point of the
Line C D make the Angle C D F equal to the Angle X C E:
and let the Line D F cut the Plane C E in F, in ſuch a manner that
the Angles C D F and C F D may be equal to the Angles X C E, L C D
Alternately taken. I ſay, that
99[Figure 99]
the Times of the Deſcents along
C D and C F are equal. And
that (the Angle C D F being
ſuppoſed equal to the Angle
X C E) the Angle C F D is
equal to the Angle D C L, is
manifeſt. For the Common An
gle D C F being taken from the
three Angles of the Triangle
C D F equal to two Right An
gles, to which are equal all the Angles made with to the Line L X
at the point C, there remains in the Triangle two Angles C D F and
C F D, equal to the two Angles X C E and L C D: But it was ſup
poſed that C D F is equal to the Angle X C E: Therefore the remaining
Angle C F D is equal to the remaining angle D C L. Let the Plane
C E be ſuppoſed equal to the Plane C D, and from the points D and
E raiſe the Perpendiculars D A and E B, unto the Horizontal Paral
lel X L; and from C unto D F let fall the Perpendicular C G. And
becauſe the Angle C D G is equal to the Angle E C B; and becauſe
D G C and C B E are Right Angles; The Triangles C D G and
C B E ſhall be equiangled: And as D C is to C G, ſo let C E be
to E B: But D C is equal to C E: Therefore C G ſhall be equal to
E B. And inregard that of the Triangles D A C and C G F, the An
gles C and A are equal to the Angles F and G: Therefore as C D is to
D A, ſo ſhall F C be to C G; and Alternately, as D C is to C F, ſo
be inclined at pleaſure C D and C E, and in any point of the
Line C D make the Angle C D F equal to the Angle X C E:
and let the Line D F cut the Plane C E in F, in ſuch a manner that
the Angles C D F and C F D may be equal to the Angles X C E, L C D
Alternately taken. I ſay, that
99[Figure 99]the Times of the Deſcents along
C D and C F are equal. And
that (the Angle C D F being
ſuppoſed equal to the Angle
X C E) the Angle C F D is
equal to the Angle D C L, is
manifeſt. For the Common An
gle D C F being taken from the
three Angles of the Triangle
C D F equal to two Right An
gles, to which are equal all the Angles made with to the Line L X
at the point C, there remains in the Triangle two Angles C D F and
C F D, equal to the two Angles X C E and L C D: But it was ſup
poſed that C D F is equal to the Angle X C E: Therefore the remaining
Angle C F D is equal to the remaining angle D C L. Let the Plane
C E be ſuppoſed equal to the Plane C D, and from the points D and
E raiſe the Perpendiculars D A and E B, unto the Horizontal Paral
lel X L; and from C unto D F let fall the Perpendicular C G. And
becauſe the Angle C D G is equal to the Angle E C B; and becauſe
D G C and C B E are Right Angles; The Triangles C D G and
C B E ſhall be equiangled: And as D C is to C G, ſo let C E be
to E B: But D C is equal to C E: Therefore C G ſhall be equal to
E B. And inregard that of the Triangles D A C and C G F, the An
gles C and A are equal to the Angles F and G: Therefore as C D is to
D A, ſo ſhall F C be to C G; and Alternately, as D C is to C F, ſo
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