PROBL.XV. PROP. XXXVII.
A Perpendicular and Inclined Plane of the ſame
Elevation being given, to find a part in the In
clined Plane that is equal to the Perpendicu
lar, and paſſed in the ſame Time as the ſaid
Perpendicular.
Elevation being given, to find a part in the In
clined Plane that is equal to the Perpendicu
lar, and paſſed in the ſame Time as the ſaid
Perpendicular.
LET A B be the Perpendicular, and A C the Inclined Plane. It is
required in the Inclined to find a part equal to the Perpendicular
A B, that after Reſt in A may be paſſed in a Time equal to the
Time in which the Perpendicular is paſſed. Let A D be equal to A B,
and cut the Remainder B C in two equal parts in I; and as A C is to
142[Figure 142]
C I, ſo let C I be to another Line
A E; to which let D G be equal: It
is manifeſt that E G is equal to A D
and to A B. I ſay moreover, that
this ſame E G is the ſame that is
paſſed by the Moveable coming out
of Reſt in A in a Time equal to the
Time in which the Moveable fall eth along A B. For becauſe that as
A C is to C I, ſo is C I to A E, or I D to D G; Therefore by Converſion
of the proportion, as C A is to A I, ſo is D I to I G. And becauſe as the
whole C A is to the whole A I, ſo is the part taken away C I to the part
I G; therefore the Remaining part I A ſhall be to the Remainder A G,
as the whole C A is to the whole A I: Therefore A I is a Mean-propor
tional betwixt C A and A G; and C I a Mean-proportional betwixt
C A and A E: If therefore we ſuppoſe the Time along A B to be as A B;
A C ſhall be the Time along A C, and C I or I D the Time along A E:
And becauſe A I is a Mean-proportional betwixt C A and A G; and
C A is the Time along the whole A C: Therefore A I ſhall be the Time
along. A G; and the Remainder I C that along the Remainder G C: But
D I was the Time along A E: Therefore D I and I C are the Times
along both the Spaces A E and C G: Therefore the Remainder D A ſhall
be the Time along E G, to wit, equal to the Time along A B. Which was
to be done.
required in the Inclined to find a part equal to the Perpendicular
A B, that after Reſt in A may be paſſed in a Time equal to the
Time in which the Perpendicular is paſſed. Let A D be equal to A B,
and cut the Remainder B C in two equal parts in I; and as A C is to
142[Figure 142]C I, ſo let C I be to another Line
A E; to which let D G be equal: It
is manifeſt that E G is equal to A D
and to A B. I ſay moreover, that
this ſame E G is the ſame that is
paſſed by the Moveable coming out
of Reſt in A in a Time equal to the
Time in which the Moveable fall eth along A B. For becauſe that as
A C is to C I, ſo is C I to A E, or I D to D G; Therefore by Converſion
of the proportion, as C A is to A I, ſo is D I to I G. And becauſe as the
whole C A is to the whole A I, ſo is the part taken away C I to the part
I G; therefore the Remaining part I A ſhall be to the Remainder A G,
as the whole C A is to the whole A I: Therefore A I is a Mean-propor
tional betwixt C A and A G; and C I a Mean-proportional betwixt
C A and A E: If therefore we ſuppoſe the Time along A B to be as A B;
A C ſhall be the Time along A C, and C I or I D the Time along A E:
And becauſe A I is a Mean-proportional betwixt C A and A G; and
C A is the Time along the whole A C: Therefore A I ſhall be the Time
along. A G; and the Remainder I C that along the Remainder G C: But
D I was the Time along A E: Therefore D I and I C are the Times
along both the Spaces A E and C G: Therefore the Remainder D A ſhall
be the Time along E G, to wit, equal to the Time along A B. Which was
to be done.
COROLLARIE.
Hence it is manifeſt, that the Space required is an intermedial be
tween the upper and lower parts that are paſt in equal
Times.
tween the upper and lower parts that are paſt in equal
Times.
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