Harriot, Thomas, Mss. 6785

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page |< < (303) of 882 > >|
    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6785_f303" o="303" n="605"/>
          <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 reference on this page is to a lemma proved by Commandino on page 198v of his edition of
                <emph style="it">Mathematicae collectiones</emph>
                <ref id="pappus_1588"> (Pappus </ref>
              . </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> pappus. pag. 198.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Pappus, page ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> vide digraphum in altera
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            see the diagram on the other ]
              <lb/>
            [
              <emph style="bf">Commentary: </emph>
            The other sheet is Add MS
              <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/KN1CRTZ2/&start=600&viewMode=image&pn=601"> f. </ref>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Sit triangulum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Ducantur perpendiculares
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            , quæ sese
              <lb/>
            intersecant in puncto
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            . et ab
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            ad
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            agatur recta usque ad
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            linea est perpendicularis lineæ
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Quoniam anguli
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            sunt recta: Quadrilaterum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            est
              <lb/>
            in circulo, qui desribatur. Deinde circa triangulum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            describatur
              <emph style="super">etiam</emph>
              <lb/>
            circulus et ducatur
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            usque ad
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            in perpheria,
              <lb/>
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , usque ad
              <math>
                <mstyle>
                  <mi>K</mi>
                </mstyle>
              </math>
            , et
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            ad
              <math>
                <mstyle>
                  <mi>I</mi>
                </mstyle>
              </math>
            . agatur etiam lineæ
              <lb/>
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>K</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let there be a triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Draw perpendiculars
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            which intersect each other at the point
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            . And from
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            to
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            construct a line as far as
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that the line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            is perpendicular to the line
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Because the angles
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            are right angles, the quadrilateral
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            is in a circle, which may be drawn. Then around the triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            there is also drawn a circle and
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            is drawn as far as
              <math>
                <mstyle>
                  <mi>H</mi>
                </mstyle>
              </math>
            on the circumference,
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            as far as
              <math>
                <mstyle>
                  <mi>K</mi>
                </mstyle>
              </math>
            , and
              <math>
                <mstyle>
                  <mi>C</mi>
                </mstyle>
              </math>
            to
              <math>
                <mstyle>
                  <mi>I</mi>
                </mstyle>
              </math>
            . ALso there are constructed the lines
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>K</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>K</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>I</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>I</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Demonstratio:
              <lb/>
            Quoniam
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            est quadrilaterum in circulo:
              <lb/>
            ergo:
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                  <mo>+</mo>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            Duobus rectis.
              <lb/>
            Sed:
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mo>-</mo>
                  <mi>C</mi>
                  <mi>K</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . quia insistunt super
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            et:
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mo>=</mo>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            . quia anguli ad verticem.
              <lb/>
            et:
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mo>+</mo>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            Duobus rectis.
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Demonstration
              <lb/>
            Because
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            is a quadrilateral in a circle, therefore
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                  <mo>+</mo>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            two right angles.
              <lb/>
            But
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mo>-</mo>
                  <mi>C</mi>
                  <mi>K</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            , because they stand on
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mo>=</mo>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            , because they are angles at the vertex.
              <lb/>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mo>+</mo>
                  <mi>C</mi>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            two right angles.
              <lb/>
            </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> * Atque inde:
              <math>
                <mstyle>
                  <mo>Δ</mo>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>F</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            . cum angulus æqualibus
              <lb/>
            Ergo etiam:
              <math>
                <mstyle>
                  <mo>Δ</mo>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            =
              <math>
                <mstyle>
                  <mo>Δ</mo>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>C</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            . cum angulis et lateribus.
              <lb/>
            Ergo:
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>G</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            recti.
              <lb/>
            Ergo:
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            perpendicularis est ad
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            . Quæsitum.
              <lb/>
            Ex eadem diagrammate nullis alijs modis conlusio infertur ut
              <lb/>
            inspicienti facile
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            * And thence triangle
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>F</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            , because of equal angles.
              <lb/>
            Therefore also triangle
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            = triangle
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>C</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            because of angles and sides.
              <lb/>
            Therefore
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>G</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            are right angles.
              <lb/>
            Therefore
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            is perpendicular to
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            . Sought.
              <lb/>
            From the same diagram, the conclusion is inferred in no other way, as is easily clear from ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Triangula igitur
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            sunt æquiangula.
              <lb/>
            […]
              <lb/>
            Sed
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            est rectus. ergo:
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>G</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            .
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Therefore the triangles
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            are equiangular.
              <lb/>
              <lb/>
            But
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            is a right angle, so therefore is
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>G</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . Sought. </s>
          </p>
        </div>
      </text>
    </echo>
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