Harriot, Thomas, Mss. 6785

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    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6785_f141" o="141" n="281"/>
          <div type="page_commentary" level="0" n="0">
            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> Harriot refers at the end to Euclid,
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI17.html"/>
              . </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI17.html"/>
                If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <p>
            <s xml:space="preserve"> Of three magnitudes in continuall proportion: the first being given & the
              <lb/>
            summe of the second & third; to find the second & </s>
          </p>
          <p>
            <s xml:space="preserve"> Let
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            be the first & the aggregate of the second & third
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            . Unto the
              <lb/>
            line
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            adde a line æquall which let be
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            : & unto them both add half of
              <lb/>
            the first magnitude which let be
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            . Then let
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            be perpendicular to
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                  <mi>E</mi>
                  <mo>&</mo>
                  <mo>æ</mo>
                  <mi>q</mi>
                  <mi>u</mi>
                  <mi>a</mi>
                  <mi>l</mi>
                  <mi>l</mi>
                </mstyle>
              </math>
              <lb/>
            & accomplishe the parallellogramme
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                  <mi>g</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . Betwixt the lines
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>f</mi>
                  <mi>g</mi>
                </mstyle>
              </math>
            or
              <lb/>
            which is all one
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            get a mean proportionall which let be
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            .
              <lb/>
            It is done by dividing the whole line
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            into two æquall partes in the poynt
              <math>
                <mstyle>
                  <mi>h</mi>
                </mstyle>
              </math>
              <lb/>
            & according to the distance
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
            or
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            describing the circle
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>i</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . & then producing
              <lb/>
            theline
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            to the periphery at
              <math>
                <mstyle>
                  <mi>i</mi>
                </mstyle>
              </math>
            .
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            is the meane proportionall befour [???] of
              <lb/>
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            which is æquall to half of the first magnitude subtrate from
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            & then
              <lb/>
              <math>
                <mstyle>
                  <mi>f</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            will remayne: & unto
              <math>
                <mstyle>
                  <mi>f</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            take a line æquall
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            . Then I say that
              <lb/>
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            is the second proportionall &
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            the third. Which is thus proved.
              <lb/>
            first accomplish the square
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>i</mi>
                  <mi>h</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            & form the poynt
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            draw the line
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>o</mi>
                </mstyle>
              </math>
            parallel to
              <lb/>
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
            . the figures
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>f</mi>
                  <mi>m</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>n</mi>
                  <mi>l</mi>
                  <mi>o</mi>
                </mstyle>
              </math>
            wilbe squares & their complements
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>n</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>i</mi>
                </mstyle>
              </math>
              <lb/>
            wilbe æquall. Let
              <math>
                <mstyle>
                  <mi>n</mi>
                  <mi>q</mi>
                </mstyle>
              </math>
            be æquall to
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>n</mi>
                </mstyle>
              </math>
            & the parallelogramm accomplished
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>q</mi>
                  <mi>p</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            . and from
              <lb/>
              <math>
                <mstyle>
                  <mi>h</mi>
                </mstyle>
              </math>
            let there be drawne the line
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>r</mi>
                </mstyle>
              </math>
            parallel to
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>p</mi>
                </mstyle>
              </math>
            or
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>q</mi>
                </mstyle>
              </math>
            . Befour I proceede farther I
              <lb/>
            say that the line
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            is æquall to
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            or
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            . which is manifest by this lemma. If
              <lb/>
            there be two magnitudes unæquall: the difference
              <emph style="st">betwixt</emph>
              <emph style="super">of</emph>
            there halfes is
              <emph style="st">the difference</emph>
              <lb/>
              <emph style="st">betwixt</emph>
            half the difference of the two wholes.
              <lb/>
            The whole lines are
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . their difference
              <lb/>
            is
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            . The half lines
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            are the halfes &
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            their difference, therefour
              <lb/>
            by the lemma
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            is is æquall to the half of
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            that is
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            . & therefour also
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>t</mi>
                </mstyle>
              </math>
            &
              <lb/>
              <math>
                <mstyle>
                  <mi>s</mi>
                  <mi>p</mi>
                </mstyle>
              </math>
            are squares & æquall, & either of them æquall to
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>m</mi>
                </mstyle>
              </math>
            .
              <lb/>
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
            divides the line
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            ito two æquall partes, therefour the parallelogramms
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>t</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>g</mi>
                </mstyle>
              </math>
              <lb/>
            are æquall: & let
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>z</mi>
                </mstyle>
              </math>
            be æquall to
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>t</mi>
                </mstyle>
              </math>
            . Therefour
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>p</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>l</mi>
                </mstyle>
              </math>
            are
              <lb/>
            æquall; but
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>p</mi>
                </mstyle>
              </math>
            is an oblonge made of
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>q</mi>
                </mstyle>
              </math>
            æquall
              <emph style="super">to</emph>
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            the first magnitude
              <lb/>
            & of
              <math>
                <mstyle>
                  <mi>q</mi>
                  <mi>p</mi>
                </mstyle>
              </math>
            æquall to
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            the last magnitude; which oblonge or parallelogram
              <lb/>
            when it is æquall to the square of the middest which is
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>l</mi>
                </mstyle>
              </math>
            the square
              <lb/>
            of
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>n</mi>
                </mstyle>
              </math>
            , æquall to
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            the middest: therefour (by the 17 of the 6th)
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
              <lb/>
            are three lines in continuall proportion which was required to be proved.
              <lb/>
            </s>
          </p>
        </div>
      </text>
    </echo>
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