Harriot, Thomas, Mss. 6785

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      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6785_f185" o="185" n="369"/>
          <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"> On this page Harriot gives a statement and diagram for Proposition 24 from
                <emph style="it">Supplementum geometriæ</emph>
                <ref id="Viete_1593c" target="http://www.e-rara.ch/zut/content/pageview/2684125"> (Viète 1593c, Prop </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Proposition XXIV.
                  <lb/>
                In dato circulo heptagonum æquilaterum & æquiangulum </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> To describe a regular heptagon in a given </s>
              </quote>
              <lb/>
              <s xml:space="preserve"> There are two references to equations found in connection with Proposition 19 of the
                <emph style="it">Supplementum</emph>
              ; these are to be found on Add MS 6785
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/KN1CRTZ2/&start=370&viewMode=image&pn=373"> f. </ref>
              . </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve"> Explicatio aeqationum quae habentur post 24 propositionem
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          An explanation of the equation to be found after Proposition 19 in the ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Sit triangulum æquicrurum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>N</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
              <lb/>
            cuius angulus ad verticem
              <math>
                <mstyle>
                  <mi>N</mi>
                </mstyle>
              </math>
              <lb/>
            sesquialter est utriusque angulorum
              <lb/>
            ad basim. oportet invenire
              <lb/>
            basis quantitatem
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            in
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>N</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            be an isosceles triangle with vertical angle
              <math>
                <mstyle>
                  <mi>N</mi>
                </mstyle>
              </math>
            , which is one and a half times either angle at the base.
              <lb/>
            There must be found
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , the length of the base, in numbers. </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> In 19
              <emph style="super">a</emph>
            propositione, secundum illatum ita est:
              <lb/>
            sit ergo pro basi
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , nota
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            . et pro cruro
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>N</mi>
                </mstyle>
              </math>
            cui æquatur
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            nota
              <math>
                <mstyle>
                  <mi>Z</mi>
                </mstyle>
              </math>
            . et forma æquationis ita erit.
              <lb/>
            Aliter per reductionem.
              <lb/>
            In eadem 19
              <emph style="it">a</emph>
            propositione demonstratur ista Analogia:
              <lb/>
            Notatur igitur loco
              <math>
                <mstyle>
                  <mn>3</mn>
                  <mi>I</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            vel
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mo>+</mo>
                  <mn>3</mn>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , litera
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            . Et pro
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>Z</mi>
                </mstyle>
              </math>
            et contra.
              <lb/>
            et analogia ita erit:
              <lb/>
            Ergo resoluta analogia æquatio ita
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            In Proposition 19, the second result is:
              <lb/>
            therfore let
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            be the base, denoted by
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            , and
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>N</mi>
                </mstyle>
              </math>
            the side, which is equal to
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , denoted by
              <math>
                <mstyle>
                  <mi>Z</mi>
                </mstyle>
              </math>
            .
              <lb/>
            And the form of the equation will be:
              <lb/>
            Otherwise, by reduction:
              <lb/>
            In the same Proposition 19, there is demonstrated this ratio:
              <lb/>
            Therefore there may be put the letter
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            in place of
              <math>
                <mstyle>
                  <mn>3</mn>
                  <mi>I</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            or
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mo>+</mo>
                  <mn>3</mn>
                  <mi>I</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            . And
              <math>
                <mstyle>
                  <mi>Z</mi>
                </mstyle>
              </math>
            for
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , and conversely.
              <lb/>
            And the ratio will be:
              <lb/>
            Therfore, having resolved the ratio, the equation will ]</s>
          </p>
        </div>
      </text>
    </echo>
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