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/bookVI/propVI8.html"/>
              . </s>
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              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI8.html"/>
                If in a right-angle triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
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          <p>
            <s xml:space="preserve"> Of 3 magnitudes in continuall proportion:
              <lb/>
            There differences being known: to
              <emph style="st">find</emph>
              <emph style="super">lineate</emph>
            the </s>
          </p>
          <p>
            <s xml:space="preserve"> For sake of understanding, suppose it first don.
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            the first proportionall,
              <math>
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                  <mi>a</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            the
              <lb/>
            second &
              <math>
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                  <mi>d</mi>
                </mstyle>
              </math>
            the third. The difference betwixt the first & second
              <math>
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                  <mi>b</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            ; betwixt
              <lb/>
            the second & third
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . The difference of differences that is betwixt
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            </s>
            <lb/>
            <s xml:space="preserve"> The construction of the problem </s>
            <lb/>
            <s xml:space="preserve"> let the first difference be
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>e</mi>
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            : the second
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            , the difference of those differences
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
              <lb/>
            from the poynt
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            rayse a perpendicular
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            . In it let
              <math>
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                  <mi>i</mi>
                </mstyle>
              </math>
            be æquall to
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            . make
              <lb/>
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>h</mi>
                </mstyle>
              </math>
            æquall to
              <math>
                <mstyle>
                  <mi>k</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            . from
              <math>
                <mstyle>
                  <mi>h</mi>
                </mstyle>
              </math>
            to
              <math>
                <mstyle>
                  <mi>i</mi>
                </mstyle>
              </math>
            draw a right line. Make
              <math>
                <mstyle>
                  <mi>h</mi>
                  <mi>i</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
            , a rectangle.
              <lb/>
            & produce
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            till it meet with
              <math>
                <mstyle>
                  <mi>i</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
            . by the 8th of the 6th it is a meane
              <lb/>
            proportionall betwixt
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>h</mi>
                </mstyle>
              </math>
            &
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            . Therefour
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            is the first proportionall sought
              <lb/>
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>e</mi>
                </mstyle>
              </math>
            must therefour be the second or
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            . &
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            the third. as is manifest
              <lb/>
            by the corollaries in the paper </s>
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