Salusbury, Thomas, Mathematical collections and translations (Tome I), 1667

Table of figures

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