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

Table of figures

< >
[Figure 141]
Page: 894
[Figure 142]
Page: 895
[Figure 143]
Page: 896
[Figure 144]
Page: 900
[Figure 145]
Page: 901
[Figure 146]
Page: 903
[Figure 147]
Page: 909
[Figure 148]
Page: 911
[Figure 149]
Page: 912
[Figure 150]
Page: 914
[Figure 151]
Page: 916
[Figure 152]
Page: 917
[Figure 153]
Page: 919
[Figure 154]
Page: 922
[Figure 155]
Page: 923
[Figure 156]
Page: 924
[Figure 157]
Page: 926
[Figure 158]
Page: 927
[Figure 159]
Page: 927
[Figure 160]
Page: 928
[Figure 161]
Page: 929
[Figure 162]
Page: 930
[Figure 163]
Page: 932
[Figure 164]
Page: 938
[Figure 165]
Page: 941
[Figure 166]
Page: 942
[Figure 167]
Page: 944
[Figure 168]
Page: 945
[Figure 169]
Page: 948
[Figure 170]
Page: 949
< >
page |< < of 701 > >|
    <archimedes>
      <text>
        <body>
          <chap>
            <p type="main">
              <s>
                <pb xlink:href="040/01/901.jpg" pagenum="208"/>
              be ſuppoſed to paſſe by the Point E, which ſhall make in the Cone
                <lb/>
              a Circular Section, whoſe Diameter is G E H. </s>
              <s>And becauſe upon
                <lb/>
              the Diameter I K of the Circle I B K, B D is a Perpendicular, the
                <lb/>
              Square of B D ſhall be equal to the Rectangle made by the parts
                <lb/>
              I D and D K: And likewiſe in the upper Circle which is underſtood
                <lb/>
              to paſſe by the points G F H, the Square of the Line F E is equal
                <lb/>
              to the Rectangle of the parts G E H: Therefore the Square of B D
                <lb/>
              hath the ſame proportion to the Square of F E, that the Rectangle
                <lb/>
              I D K hath to the Rectangle G E H. </s>
              <s>And becauſe the Line E D is
                <lb/>
              Parallel to H K, E H ſhall be equal to D K, which alſo are Parallels:
                <lb/>
              And therefore the Rectangle I D K ſhall have the ſame proportion
                <lb/>
              to the Rectangle G E H, as I D hath to G E; that is, that D A hath
                <lb/>
              to A E: Therefore the Rectangle I D K to the Rectangle G E H,
                <lb/>
              that is, the Square B D to the Square F E, hath the ſame proportion
                <lb/>
              that the Axis D A hath to the part A E: Which was to be de­
                <lb/>
              monſtrated.</s>
            </p>
            <p type="main">
              <s>The other Propoſition, likewiſe neceſſary to the preſent Tract,
                <lb/>
              we will thus make out. </s>
              <s>Let us deſcribe the Parabola, of which let the
                <lb/>
              Axis C A be prolonged out unto D; and taking any point B, let the
                <lb/>
              Line B C be ſuppoſed to be continued out by the ſame Parallel un­
                <lb/>
                <figure id="id.040.01.901.1.jpg" xlink:href="040/01/901/1.jpg" number="145"/>
                <lb/>
              to the Baſe of the ſaid Parabola;
                <lb/>
              and let D A be ſuppoſed equal
                <lb/>
              to the part of the Axis C A. </s>
              <s>I ſay,
                <lb/>
              that the Right-Line drawn by
                <lb/>
              the points D and B, falleth not
                <lb/>
              within the Parabola, but without,
                <lb/>
              ſo as that it only toucheth the
                <lb/>
              ſame in the ſaid point B: For, if
                <lb/>
              it be poſſible for it to fall within,
                <lb/>
              it cutteth it above, or being pro­
                <lb/>
              longed, it cutteth it below. </s>
              <s>And
                <lb/>
              in that Line let any point G be
                <lb/>
              taken, by which paſſeth the Right
                <lb/>
              Line F G E. </s>
              <s>And becauſe the
                <lb/>
              Square F E is greater than the
                <lb/>
              Square G E, the ſaid Square F E
                <lb/>
              ſhall have greater proportion to
                <lb/>
              the Square B C, than the ſaid Square G E hath to the ſaid B C. </s>
              <s>And
                <lb/>
              becauſe, by the precedent, the Square F E is to the Square B C as
                <lb/>
              E A is to A C; therefore E A hath greater proportion to A C, than
                <lb/>
              the Square G E hath to the Square B C; that is, than the Square
                <lb/>
              E D hath to the Square D C: (becauſe in the Triangle D G E as
                <lb/>
              G E is to the Parallel B C, ſo is E
                <emph type="italics"/>
              D
                <emph.end type="italics"/>
              to
                <emph type="italics"/>
              D
                <emph.end type="italics"/>
              C:) But the Line E A to
                <lb/>
              A C, that is, to A
                <emph type="italics"/>
              D
                <emph.end type="italics"/>
              hath the ſame proportion that four Rectangles
                <lb/>
              E A
                <emph type="italics"/>
              D
                <emph.end type="italics"/>
              hath to four Squares of A
                <emph type="italics"/>
              D,
                <emph.end type="italics"/>
              that is, to the Square C
                <emph type="italics"/>
              D,
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
Impressum