Harriot, Thomas, Mss. 6785

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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 works on Proposition 15 from
                <emph style="it">Effectionum geometricarum canonica recensio</emph>
                <ref id="Viete_1593b" target="http://www.e-rara.ch/zut/content/pageview/2684105"> (Viète 1593b, Prop </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Propositio XV.
                  <lb/>
                Quadratum a media proportionali inter basin trianguli rectanguli & perpendiculum ejusdem, proportionale est inter quadratum basi, & quadratum hypotenusae multatum ipso basis quadrato. Vel etiam inter quadratum perpendiculi, & quadratum hypotenusae multatum ipso perpendiculi </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> The square of the mean proportional between the base of a right-angled triangle and its perpendicular, is the proportional between the square of the base and the square of the hypotenuse, reduced by the square of the base. Or also between the square of the perpendicular and the square of the hypotenuse, reduced by the square of the perpendicular.</s>
              </quote>
              <lb/>
              <s xml:space="preserve"> Viète demonstrated this proposition geometrically and showed that it can be represented by the quartic
                <math>
                  <mstyle>
                    <mrow>
                      <msup>
                        <mi>B</mi>
                        <mn>2</mn>
                      </msup>
                    </mrow>
                    <mrow>
                      <msup>
                        <mi>A</mi>
                        <mn>2</mn>
                      </msup>
                    </mrow>
                    <mo>-</mo>
                    <mrow>
                      <msup>
                        <mi>A</mi>
                        <mn>4</mn>
                      </msup>
                    </mrow>
                    <mo>=</mo>
                    <mrow>
                      <msup>
                        <mi>D</mi>
                        <mn>4</mn>
                      </msup>
                    </mrow>
                  </mstyle>
                </math>
              (in modern notation), where
                <math>
                  <mstyle>
                    <mi>A</mi>
                  </mstyle>
                </math>
              is the base or perpendicular,
                <math>
                  <mstyle>
                    <mi>B</mi>
                  </mstyle>
                </math>
              the hypotenuse, and
                <math>
                  <mstyle>
                    <mi>D</mi>
                  </mstyle>
                </math>
              the mean. As in the earlier pages in this set, Harriot works the other way round, beginning from the equation
                <math>
                  <mstyle>
                    <mi>b</mi>
                    <mi>b</mi>
                    <mi>a</mi>
                    <mi>a</mi>
                    <mo>-</mo>
                    <mi>a</mi>
                    <mi>a</mi>
                    <mi>a</mi>
                    <mi>a</mi>
                    <mo>=</mo>
                    <mi>d</mi>
                    <mi>d</mi>
                    <mi>d</mi>
                    <mi>d</mi>
                  </mstyle>
                </math>
              and then deriving the corresponding construction. </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> i.) Effectiones
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Geometrical ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> 3.)
              <math>
                <mstyle>
                  <mn>2</mn>
                  <mo>,</mo>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mo>-</mo>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mo>=</mo>
                  <mi>d</mi>
                  <mi>d</mi>
                  <mi>d</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
              <lb/>
            Et intelligatur.
              <math>
                <mstyle>
                  <mn>2</mn>
                  <mi>c</mi>
                  <mo>=</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            3.)
              <math>
                <mstyle>
                  <mn>2</mn>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>a</mi>
                  <mo>-</mo>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mi>a</mi>
                  <mo>=</mo>
                  <mi>d</mi>
                  <mi>d</mi>
                  <mi>d</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            ; and it may be understood that
              <math>
                <mstyle>
                  <mn>2</mn>
                  <mi>c</mi>
                  <mo>=</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Notatio pro effectione
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Notation for the geometric ]</s>
          </p>
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