Harriot, Thomas, Mss. 6784

List of thumbnails

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611
611 (306) 		612
612 (306v)
613
613 (307) 		614
614 (307v)
615
615 (308) 		616
616 (308v)
617
617 (309) 		618
618 (309v)
619
619 (310) 		620
620 (310v)
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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 16 from
                <emph style="it">Supplementum geometriæ</emph>
                <ref id="Viete_1593c" target="http://www.e-rara.ch/zut/content/pageview/2684118"> (Viète 1593c, Prop </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Proposition XVI.
                  <lb/>
                Si duo triangula fuerint aequicrura singula, & ipsa alterum alteri cruribus aequalia, angulus autem qui est ad basin secundi sit triplus anguli qui est ad basin primi: cubus ex base primi, minus triplo solido sub base primi & cruris communis quadrato, aequalis est solido sub base secundi & ejusdem cruris </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> If two triangles are each isosceles, the legs of one equal to the legs of the other, and moreover the angle at the base of the second is three times the angle at the base of the first, then the cube of the first base, minus three times the product of the base of the first and the square of the common side, is equal to the product of the second base and the square of the same side.</s>
              </quote>
              <lb/>
              <s xml:space="preserve"> The working contains a reference to Euclid's
                <emph style="it">Elements</emph>
              , Proposition
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII5.html"/>
              . </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII5.html"/>
                If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half that line. </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
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          <head xml:space="preserve"> prop. 16.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Proposition 16 from the ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Si duo triangula fuerint aequicrura singula,
              <lb/>
            et ipsa alterum alteri cruribus aequalia: angulus
              <lb/>
            autem qui est ad basin secundi sit triplus
              <lb/>
            anguli qui est ad basin primi. Cubus ex
              <lb/>
            base primi, minus triplo solido sub base primi
              <lb/>
            et cruris communis quadrato, aequalis
              <lb/>
            est solido sub base secundi et ejusdem
              <lb/>
            cruris
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            If two triangles are each isosceles, the legs of one equal to the legs of the other, and moreover the angle at the base of the second is three times the angle at the base of the first, then the cube of the first base, minus three times the product of the base of the first and the square of the common side, is equal to the product of the second base and the square of the same side.</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> per 5,2 el.
              <lb/>
            […]
              <lb/>
            Quia parallogramma æquialta
              <lb/>
            et sunt ut bases.
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            .
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            […]
              <lb/>
            vel per notas
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            simplices
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            Resoluatur analogia et erit:
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            by Elements II.5
              <lb/>
              <lb/>
            Because the parallelograms are of equal height and are as the bases
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
              <lb/>
            or in simple notation
              <lb/>
            The ratio is resolved, and hence the ]</s>
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