Heron Alexandrinus, Mechanica, 1999

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    <archimedes>
      <text>
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          <chap n="1">
            <pb n="11">
              <s id="A18-1.11.01">[11] How to find two mean proportionals to two given lines, however, we want to explain now with the aid of an instrument, whereby we do not need solid figures; and we want to show the easiest method for this.</s>
              <s id="A18-1.11.02">Let the two given lines be the lines <ab> and <bg>; one is perpendicular to the other one and let the two be the lines for which we want to find the two mean proportionals.</s>
              <s id="A18-1.11.03">Let us now complete the rectangle <abgd> by drawing the two lines <dg> and <da>.Let us further connect <b> with <d> and <g> with <a> and put a ruler to point <b> which intersects the lines <de> and <az>, turn it until the line starting at point <h> towards the intersection of <ge> equals the line starting at point <h> towards the intersection of <az>.</s>
              <s id="A18-1.11.04">Let the position of the ruler be <ebz> and the two lines <eh> and <hz> be equal; then I say that the two lines <az> and <ge> are the two mean proportionals to <ab> and <bg>, <ab> being the first, <za> the second, <ge> the third and <bg> the fourth proportional.</s>
              <s id="A18-1.11.05">Proof: because the quadrangle <abgd> is a rectangle, the four lines <dh>, <ha>, <hb> and <hg> are equal; and because line <hd> equals line <ha>, moreover the line <hz> is drawn,</s>
              <s id="A18-1.11.06"> thus is<dz> x <za> + <ah>2 = <hz>2.Also<de> x <eg> + <gh>2 = <he>2.</s>
              <s id="A18-1.11.07">The lines <eh> and <hz> are, however, equal; accordingly is<dz> x <za> + <ah>2 = <de> x <eg> + <hg>2.</s>
              <s id="A18-1.11.08">It is, however,<ah>2 = <hg>2Therefore by subtraction:<dz> x <za> = <de> x <eg></s>
              <s id="A18-1.11.09">Then the line <ed> relates to <dz> as the line <za> to <ge>. The line <ed>, however, relates to <dz> as the line <ba> to <az> and as the line <eg> to <gb>.Then the line <za> relates to <ge> and the line <ge> and <gb> as the line <ab> to <az> (i.e. <ab>/<az> = <az>/<ge> = <ge>/<gb>).</s>
              <s id="A18-1.11.10">Thus we have constructed two mean proportionals to the two lines <ab> and <bg>, namely the lines <az> and <ge>. q.e.d.</s>
            </pb>
          </chap>
        </body>
      </text>
    </archimedes>
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