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_f184" o="184" n="367"/>
          <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 investigates Propositions 20 and 21 from
                <emph style="it">Supplementum geometriæ</emph>
                <ref id="Viete_1593c" target="http://www.e-rara.ch/zut/content/pageview/2684123"> (Viète 1593c, Props 20, </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Proposition XX.
                  <lb/>
                Constituere triangulum aæquicrurum, ut differntia inter basin & alterum e cruribus fit ad basin, sicut quadratum cruris ad quadratum compositae ex crure & </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> To construct an isosceles triangle so that the difference between the base and either of the legs is to the base as the square of a leg is to the square of the sum of a leg and the base.</s>
              </quote>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Proposition XXI.
                  <lb/>
                Si fuerit triangulum aequicrurum, fit autem differentia inter basin & alterum e cruribus ad basin, sicut quadratum cruris ad quadratum compositæ ex crure & base: quae a termino basis ducetur ad crus linea recta ipsi cruri æquale, secabit bisariam angulum ad basin.</s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> If there is an isosceles triangle, and moreover the [ratio of the] difference between the base and either of the legs, to the base, is equal to the square of the leg to the square of the sum of a leg and the base, then a line drawn from the [end of the] base to the leg, equal [in length] to that leg, will bisect the angle at the </s>
              </quote>
              <lb/>
              <s xml:space="preserve"> There is a reference to Propostion 19 of the
                <emph style="it">Supplementum</emph>
              (see Add MS 6785
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/KN1CRTZ2/&start=370&viewMode=image&pn=371"> f. </ref>
              ). There are also references to Euclid's Propositions
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII20.html"/>
              ,
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII36.html"/>
              ,
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI12.html"/>
              , and
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI14.html"/>
              . </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII20.html"/>
                The angle at the centre of a circle is double the angle at the circumference, when they have the same part of the circumference for a base. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII36.html"/>
                If from a point without a circle two straight lines be drawn to it, one of which is a tangent to the circle, and the other cuts it; the rectangle under the whole cutting line and the external segment is equal to the square of the tangent. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI12.html"/>
                To find a fourth proportional to three given lines. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI14.html"/>
                Equal parallograms which have one angle each equal have the sides about the equal angles reciprocally proportional. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve"> prop. 20.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Proposition 20 from the ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Constituere triangulum aæquicrurum; ut differntia inter basin et alterum
              <lb/>
            e cruribus fit ad basin, sicut quadratum cruris ad quadratum compositæ
              <lb/>
            ex crure et
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            To construct an isosceles triangle so that the difference between the base and either of the legs is to the base, as the square of a leg is to the square of the sum of a leg and the base.</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> per 19,p fiat
              <lb/>
            Et ponatur in circumferentia,
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mo>=</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            vel
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Et ducantur recta
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
              <lb/>
            Triangulum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            est quod
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            By Proposition 19,
              <lb/>
            And in the circumference, put
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mo>=</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            or
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . and constructing the line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            , the triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>E</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is as required. </s>
          </p>
          <head xml:space="preserve"> prop.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Proposition ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Si fuerit triangulum aequicrurum: fit autem differentia inter basin et alterum
              <lb/>
            e cruribus ad basin; sicut quadratum cruris ad quadratum compositæ ex crure et base.
              <lb/>
            Quae a termino basis ducetur ad crus linea recta ipsi cruri æquale: secabit
              <lb/>
            bisariam angulum ad
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            If there is an isosceles triangle, and moreover the [ratio of the] difference between the base and either of the legs, to the base, is equal to the square of the leg to the square of the sum of a leg and the base, then a line drawn from the [end of the] base to the leg, equal [in length] to that leg, will bisect the angle at the ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve">
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            secat bisariam angulum
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Nam ex hypothesi.
              <lb/>
            […]
              <lb/>
            Et per 36,3 el
              <lb/>
            Ergo. per 14, 6 el
              <lb/>
            […]
              <lb/>
            Consequenter:
              <lb/>
            Et subducendo
              <lb/>
            Ergo: per 2,6: el:
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            sunt parallelæ
              <lb/>
            Et: Angulus
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            æqualis angulo
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Sed. per 20,3: Angulus
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            est duplus anguli
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            Hoc est:
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            Ergo angulus
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            sectis est bisariam a recta
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Quod erat
              <lb/>
            [
              <emph style="bf">Translation: </emph>
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            bisects the angle
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            For from the hypothesis
              <lb/>
              <lb/>
            And by Elements III.36
              <lb/>
            Therefore by Elements VI.14
              <lb/>
              <lb/>
            Consequently:
              <lb/>
            And subtracting
              <lb/>
            Therefore, by Elements VI.2,
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            are parallel.
              <lb/>
            And angle
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to angle
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            But by Elements III.20, angle
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is twice angle
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , that is
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Therefore angle
              <math>
                <mstyle>
                  <mi>E</mi>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is cut in two by the line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Which was to be ]</s>
          </p>
        </div>
      </text>
    </echo>
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