Harriot, Thomas, Mss. 6785

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          <pb file="add_6785_f187" o="187" n="373"/>
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            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> On this page Harriot investigates Proposition 19 from
                <emph style="it">Supplementum geometriæ</emph>
                <ref id="viete_1593c" target="http://www.e-rara.ch/zut/content/pageview/2684121"> (Viete 1593c, Prop </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Proposition XIX.
                  <lb/>
                Diametrum circuli ita continuare, ut fit continuatio ad semidiametrum adjunctum continuationi, sicut quadratum semidiametri ad quadratum continuatae diametri.</s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> To extend the diameter of a circle so that the extension is to the semidiameter together with the extension as the square of the semidiameter to the square of the extended diameter.</s>
              </quote>
              <lb/>
              <s xml:space="preserve"> There is a reference to Proposition 10 from the
                <emph style="it">Supplementum</emph>
              (see Add MS 6784
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/XT0KZ8QC/&start=700&viewMode=image&pn=707"> f. </ref>
              ), and there are also references to Euclid, Propositions
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII4.html"/>
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII12.html"/>
              . </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII4.html"/>
                If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the parts, together with twice the rectangle contained by the parts. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII12.html"/>
                If an equilateral triangle be inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
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          <head xml:space="preserve"> prop. 19.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Proposition 19 from the ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> per 16,p
              <lb/>
            […]
              <lb/>
            Ducantur per
              <lb/>
            Dividantur per 3:
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            If two triangles are each isosceles, equal to one another in their legs, By Proposition 16.
              <lb/>
              <lb/>
            Multiplying by
              <lb/>
            Dividing by 3, ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> per
              <lb/>
            12,13
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            by Proposition XIII.12 of the ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Ista æquatio [???] fit ex æquatione supra
              <lb/>
            Scilicet
              <lb/>
            Ducantibus per
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>L</mi>
                </mstyle>
              </math>
              <lb/>
            Itaque per istam, et primam
              <lb/>
            et 10
              <emph style="super">am</emph>
            æquationem supra:
              <lb/>
            Primum
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            This equation arises from the equation above, namely:
              <lb/>
            Having multiplied by
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>L</mi>
                </mstyle>
              </math>
              <lb/>
            Therfore by this, and the first, and the equation of the 10th above:
              <lb/>
            The first ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Ducantur per 27. Ergo:
              <lb/>
            Fiat reductio ad
              <lb/>
            analogiam et erunt:
              <lb/>
            per 4,2, el
              <lb/>
            […]
              <lb/>
            Ducantur per 9. et erunt:
              <lb/>
            per superiorem analogiam et
              <lb/>
            æquationis erunt:
              <lb/>
            resolutio Anaolgia: erunt:
              <lb/>
            Secundum
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Multiplying by 27, therefore:
              <lb/>
            Carry out the reduction to the ratio, and then:
              <lb/>
            by Proposition II.4 of the Elements
              <lb/>
              <lb/>
            Multiplying by 9, and then:
              <lb/>
            by the ratio above and the equation:
              <lb/>
            The second ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Fiat reductio ad analogiam: et erunt:
              <lb/>
            per 4,2, el
              <lb/>
            Ergo tandem
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Carry out the reduction of the ratio and then:
              <lb/>
            by Proposition II.4
              <lb/>
            Therefore finally the ]</s>
          </p>
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