Gravesande, Willem Jacob 's, An essay on perspective

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        <div xml:id="echoid-div49" type="section" level="1" n="23">
          <pb o="12" file="0032" n="33" rhead="An ESSAY"/>
        </div>
        <div xml:id="echoid-div50" type="section" level="1" n="24">
          <head xml:id="echoid-head26" xml:space="preserve">
            <emph style="sc">Theorem</emph>
          VI.</head>
          <p style="it">
            <s xml:id="echoid-s419" xml:space="preserve">19. </s>
            <s xml:id="echoid-s420" xml:space="preserve">Let A C be a Line inclined to the Geometrical
              <lb/>
            Plane, and O D another Line drawn parallel to
              <lb/>
            A C, from the Eye to the perſpective Plane. </s>
            <s xml:id="echoid-s421" xml:space="preserve">Now
              <lb/>
              <note position="left" xlink:label="note-0032-01" xlink:href="note-0032-01a" xml:space="preserve">Fig. 6.</note>
            if B A be drawn in the Geometrical Plane, pa-
              <lb/>
            rallel to the baſe Line, and likewiſe D E, in the
              <lb/>
            perſpective Plane, parallel to the ſaid Line, ſo that
              <lb/>
            B A be to A C, as E d to D O. </s>
            <s xml:id="echoid-s422" xml:space="preserve">I ſay, the Ap-
              <lb/>
            pearance of the Line B C, paſſing through the Point
              <lb/>
            B, and the Extremity of the Line A C, being con-
              <lb/>
            tinued, will meet the Point E.</s>
            <s xml:id="echoid-s423" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s424" xml:space="preserve">Now to prove this; </s>
            <s xml:id="echoid-s425" xml:space="preserve">it is evident, that
              <note symbol="*" position="left" xlink:label="note-0032-02" xlink:href="note-0032-02a" xml:space="preserve">13.</note>
            need but demonſtrate, that O E is parallel to
              <lb/>
            B C: </s>
            <s xml:id="echoid-s426" xml:space="preserve">And this may be done in the following
              <lb/>
            Manner:</s>
            <s xml:id="echoid-s427" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s428" xml:space="preserve">A B is parallel to E D, and A C to O D;
              <lb/>
            </s>
            <s xml:id="echoid-s429" xml:space="preserve">whence the Angle (E D O) of the Triangle
              <lb/>
            O E D, is equal to the Angle (B A C) of the
              <lb/>
            Triangle A C B: </s>
            <s xml:id="echoid-s430" xml:space="preserve">And ſo theſe two Triangles
              <lb/>
            are ſimilar; </s>
            <s xml:id="echoid-s431" xml:space="preserve">becauſe they have alſo their Sides
              <lb/>
            Proportional. </s>
            <s xml:id="echoid-s432" xml:space="preserve">But ſince theſe two ſimilar Tri-
              <lb/>
            angles, have two of their Sides parallel, the
              <lb/>
            third B C is alſo parallel to O E; </s>
            <s xml:id="echoid-s433" xml:space="preserve">which was to be
              <lb/>
            demonſtrated.</s>
            <s xml:id="echoid-s434" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div53" type="section" level="1" n="25">
          <head xml:id="echoid-head27" xml:space="preserve">
            <emph style="sc">Corollary</emph>
          .</head>
          <p>
            <s xml:id="echoid-s435" xml:space="preserve">20. </s>
            <s xml:id="echoid-s436" xml:space="preserve">If A B be made equal to A C, and E D to D O,
              <lb/>
            the Appearance of B C will paſs thro’ the Point E,</s>
          </p>
        </div>
        <div xml:id="echoid-div54" type="section" level="1" n="26">
          <head xml:id="echoid-head28" xml:space="preserve">CHAP. III.</head>
          <p style="it">
            <s xml:id="echoid-s437" xml:space="preserve">The Practice of Perſpective upon the Per-
              <lb/>
            ſpective Plane, ſuppoſed to be perpendicu-
              <lb/>
            lar, or upright.</s>
            <s xml:id="echoid-s438" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s439" xml:space="preserve">IN order to give a diſtinct Idea of the Theory, I
              <lb/>
            have hitherto conſider’d the Geometrical Plane,
              <lb/>
            as it were the Ground upon which the </s>
          </p>
        </div>
      </text>
    </echo>
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