Harriot, Thomas, Mss. 6784

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    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6784_f269" o="269" n="537"/>
          <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"> The problem pursued in this and many other folios in Add MS 6784 is that of 'inclination' or 'neusis', as set out in Pappus,
                <emph style="it">Mathematicae collectiones</emph>
                <ref id="pappus_1588"> (Pappus </ref>
              , Book 7. For a statement of the problem see Add MS 6784
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/XT0KZ8QC/&start=580&viewMode=image&pn=581"> f. </ref>
              . </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> De
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          On ]</head>
          <p>
            <s xml:space="preserve"> prop.
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Proposition ]</s>
          </p>
          <p>
            <s xml:space="preserve"> Sit semicirculus
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . et contingens
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            : oportet puncto
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            ducere rectam
              <lb/>
            lineam ad contingentem
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ita ut intersecta
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            sit æqualis datæ lineæ
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let there be a semicircle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and a tangent
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . It is required to draw a line from a point
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            to the tangent
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            so that the intercepted line
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to the given line
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p>
            <s xml:space="preserve"> Ducatur
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            perpendicularis ad
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . et producatur versus
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Fiat
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            Dimidium lineæ
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            . Agatur lineam
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            * cui fiat æqualis
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
            centro
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            intervallo
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            describatur periferia
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            quæ secabit perferiam
              <lb/>
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            in puncto
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Per
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            Ducatur lineam usque ad
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            est æqualis lineaæ
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Draw
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            perpendicular to
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and extend it towards
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Make
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            half of the line
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            . Construct the line
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            * and make
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            equal to it.
              <lb/>
            With centre
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            and radius
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            draw the circumference
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            which will cut the circumference
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            at the point
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Through
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            draw a line as far as
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to the line
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p>
            <s xml:space="preserve"> Aliter:
              <lb/>
            * a qua abscindatur
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            æqualis
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Inscribatur
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            æqualis
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            et producetur ad
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            est æqualis
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Another way:
              <lb/>
            * from which there must cut off
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            equal to
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Inscribe
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            equal to
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and extend it to
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to the given line. </s>
          </p>
          <p>
            <s xml:space="preserve"> Aliter: et optime.
              <lb/>
            Producatur
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            versus
              <math>
                <mstyle>
                  <mi>I</mi>
                </mstyle>
              </math>
            . et fiat
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            dimidium
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Agatur
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            et sit
              <math>
                <mstyle>
                  <mi>O</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            æqualis
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . centro
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            et intervallo
              <lb/>
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            describatur periferia
              <math>
                <mstyle>
                  <mi>K</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , quæ secabit periferiam
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            in puncto
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            . Per
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            Ducatur linea usque ad
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            est æqualis
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Anotoher way, and the best.
              <lb/>
            Extend
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            towards
              <math>
                <mstyle>
                  <mi>I</mi>
                </mstyle>
              </math>
            and make
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            half of
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Construct
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            and let
              <math>
                <mstyle>
                  <mi>O</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            be equal to
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . With centre
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            and radius
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            draw the circumference
              <math>
                <mstyle>
                  <mi>K</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , which will cut the circumference
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            at the point
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            . Through
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            draw a line as far as
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to the given line. </s>
          </p>
          <p>
            <s xml:space="preserve"> Nota quod Triangulum
              <lb/>
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            continet totum
              <lb/>
            effectionum huius
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Note that triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            contains all the construction of this equation. </s>
          </p>
        </div>
      </text>
    </echo>
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