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.