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<TEI.2><text><body><div1 n="10" type="book" org="uniform" sample="complete"><div2 n="Prop 2" type="type" org="uniform" sample="complete"><div3 id="elem.10.62" n="62" type="number" org="uniform" sample="complete">
      <head>PROPOSITION 62.</head>
      <p><hi rend="ital">The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial</hi>. </p>
      <p>Let <hi rend="ital">AB</hi> be a second bimedial straight line divided into its medials at <hi rend="ital">C</hi>, so that <hi rend="ital">AC</hi> is the greater segment; let <hi rend="ital">DE</hi> be any rational straight line, and to <hi rend="ital">DE</hi> let there be applied the parallelogram <hi rend="ital">DF</hi> equal to the square on <hi rend="ital">AB</hi> and producing <hi rend="ital">DG</hi> as its breadth; <figure /> I say that <hi rend="ital">DG</hi> is a third binomial straight line. </p>
      <p>Let the same construction be made as before shown. <pb n="138" /></p>
      <p>Then, since <hi rend="ital">AB</hi> is a second bimedial divided at <hi rend="ital">C</hi>, therefore <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi> are medial straight lines commensurable in square only and containing a medial rectangle, [<ref target="elem.10.38" targOrder="U">X. 38</ref>] so that the sum of the squares on <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi> is also medial. [<ref target="elem.10.15 elem.10.23.p.1" targOrder="U">X. 15 and 23 Por.</ref>] </p>
      <p>And it is equal to <hi rend="ital">DL</hi>; therefore <hi rend="ital">DL</hi> is also medial. </p>
      <p>And it is applied to the rational straight line <hi rend="ital">DE</hi>; therefore <hi rend="ital">MD</hi> is also rational and incommensurable in length with <hi rend="ital">DE</hi>. [<ref target="elem.10.22" targOrder="U">X. 22</ref>] </p>
      <p>For the same reason, <hi rend="ital">MG</hi> is also rational and incommensurable in length with <hi rend="ital">ML</hi>, that is, with <hi rend="ital">DE</hi>; therefore each of the straight lines <hi rend="ital">DM</hi>, <hi rend="ital">MG</hi> is rational and incommensurable in length with <hi rend="ital">DE</hi>. </p>
      <p>And, since <hi rend="ital">AC</hi> is incommensurable in length with <hi rend="ital">CB</hi>, and, as <hi rend="ital">AC</hi> is to <hi rend="ital">CB</hi>, so is the square on <hi rend="ital">AC</hi> to the rectangle <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi>, therefore the square on <hi rend="ital">AC</hi> is also incommensurable with the rectangle <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi>. [<ref target="elem.10.11" targOrder="U">X. 11</ref>] </p>
      <p>Hence the sum of the squares on <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi> is incommensurable with twice the rectangle <hi rend="ital">AC</hi>, <hi rend="ital">CB</hi>, [<ref target="elem.10.12 elem.10.13" targOrder="U">X. 12, 13</ref>] that is, <hi rend="ital">DL</hi> is incommensurable with <hi rend="ital">MF</hi>, so that <hi rend="ital">DM</hi> is also incommensurable with <hi rend="ital">MG</hi>. [<ref target="elem.6.1" targOrder="U">VI. 1</ref>, <ref target="elem.10.11" targOrder="U">X. 11</ref>] </p>
      <p>And they are rational; therefore <hi rend="ital">DG</hi> is binomial. [<ref target="elem.10.36" targOrder="U">X. 36</ref>] </p>
      <p>It is to be proved that it is also a third binomial straight line. </p>
      <p>In manner similar to the foregoing we may conclude that <hi rend="ital">DM</hi> is greater than <hi rend="ital">MG</hi>, and that <hi rend="ital">DK</hi> is commensurable with <hi rend="ital">KM</hi>. </p>
      <p>And the rectangle <hi rend="ital">DK</hi>, <hi rend="ital">KM</hi> is equal to the square on <hi rend="ital">MN</hi>; therefore the square on <hi rend="ital">DM</hi> is greater than the square on <hi rend="ital">MG</hi> by the square on a straight line commensurable with <hi rend="ital">DM</hi>. </p>
      <p>And neither of the straight lines <hi rend="ital">DM</hi>, <hi rend="ital">MG</hi> is commensurable in length with <hi rend="ital">DE</hi>. </p>
      <p>Therefore <hi rend="ital">DG</hi> is a third binomial straight line. [<ref target="elem.10.def.2.3" targOrder="U">X. Deff. II. 3</ref>] Q. E. D.<pb n="139" /></p>
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