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"> Harriot refers to Euclid,
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookV/propV16.html"/>
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookV/propV17.html"/>
              </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookV/propV16.html"/>
                If four magnitudes be proportional they will also be proportional alternately. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookV/propV17.html"/>
                If magnitudes be proportional componendo, they will also the proportional separando. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <p>
            <s xml:space="preserve"> 1.Of three magnitudes in continuall proportion; the summe of the first
              <lb/>
            & second being given & the summe of the second and third: to find the
              <lb/>
            </s>
          </p>
          <p>
            <s xml:space="preserve"> By the 17 of the 5th as the first & second hath to the second, so hath the second
              <lb/>
            & third to the third. Therefour (alterne) by the 16 of the 5th, as the summe
              <lb/>
            of the first and second hath unto the summe of the second & third,
              <lb/>
            so hath the second to the third. Then agayne (componendo) by the
              <lb/>
            same 17 prop. as the aggregate of the two summes hath to the
              <lb/>
            summe of the second & third, so hath the summe of the second & third
              <lb/>
            to the third. The third being known the rest are also known, &
              <lb/>
            therefour the proportion </s>
          </p>
          <p>
            <s xml:space="preserve"> 2. Of three proportionalls in continuall proportion: the difference
              <lb/>
            betwixt the first & last & the summe of the second & third
              <emph style="super">being given</emph>
            to find
              <lb/>
            the </s>
          </p>
          <p>
            <s xml:space="preserve"> It must be understood also which is the least terme &
              <emph style="super">let</emph>
            that for sake of
              <lb/>
            delivery be recconed the first. Then subtract that
              <emph style="st">first</emph>
            difference
              <lb/>
            out of the sayd summe of the second & third & then will remayne the
              <lb/>
            summe of the first & second as is thus proved. Suppose the
              <lb/>
            three magnitudes unkown be
              <lb/>
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            .
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            .
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . from
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
            towarde
              <math>
                <mstyle>
                  <mi>d</mi>
                </mstyle>
              </math>
            there
              <lb/>
            is a mgnitude æquall to
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            which
              <lb/>
            let be
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            . Then
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            wilbe the difference which is given. Which being
              <lb/>
            subtracted out of
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            the summe of second & third, leaveth
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            ; which
              <lb/>
            is æquall to
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            by the 2 common substrates, the summe of the first & second.
              <lb/>
            The proposition then is as that </s>
          </p>
          <p>
            <s xml:space="preserve"> This propostion wrought by algebra the
              <lb/>
            first being suppose
              <math>
                <mstyle>
                  <mn>1</mn>
                  <mi>a</mi>
                </mstyle>
              </math>
            is as </s>
          </p>
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