Gravesande, Willem Jacob 's, An essay on perspective

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        <div xml:id="echoid-div140" type="section" level="1" n="76">
          <p>
            <s xml:id="echoid-s1020" xml:space="preserve">
              <pb o="39" file="0077" n="87" rhead="on PERSPECTIVE."/>
            the Station Point, and A the Point wherein the
              <lb/>
            Perpendicular meets the Geometrical Plane.</s>
            <s xml:id="echoid-s1021" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div142" type="section" level="1" n="77">
          <head xml:id="echoid-head83" xml:space="preserve">
            <emph style="sc">Operation</emph>
          ,
            <lb/>
          Without Compaſſes.</head>
          <p>
            <s xml:id="echoid-s1022" xml:space="preserve">Having firſt found the Perſpective a of
              <note symbol="*" position="right" xlink:label="note-0077-01" xlink:href="note-0077-01a" xml:space="preserve">31.</note>
            Point A, draw the Line A S cutting the Baſe
              <lb/>
            Line in E, through which Point E draw the
              <lb/>
            Line Ea; </s>
            <s xml:id="echoid-s1023" xml:space="preserve">then from the Point B draw a Line
              <lb/>
            B a to the Point a, cutting the Horizontal Line
              <lb/>
            in F. </s>
            <s xml:id="echoid-s1024" xml:space="preserve">Again through F draw a Line to the
              <lb/>
            Point L, cutting E a in I; </s>
            <s xml:id="echoid-s1025" xml:space="preserve">and a I is the Repre-
              <lb/>
            ſentation ſought.</s>
            <s xml:id="echoid-s1026" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div144" type="section" level="1" n="78">
          <head xml:id="echoid-head84" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <p>
            <s xml:id="echoid-s1027" xml:space="preserve">To prove this, let G N be a Perpendicular to
              <lb/>
            the Baſe Line drawn from the Point G, wherein
              <lb/>
            the ſaid Baſe Line is cut by the Line B F; </s>
            <s xml:id="echoid-s1028" xml:space="preserve">alſo
              <lb/>
            let G D be equal to the Perpendicular whoſe Ap-
              <lb/>
            pearance is ſought, and a H parallel to the Baſe
              <lb/>
            Line.</s>
            <s xml:id="echoid-s1029" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1030" xml:space="preserve">It is plain that the Perſpective of E A is
              <lb/>
            E a: </s>
            <s xml:id="echoid-s1031" xml:space="preserve">But E A paſſes through the Station Point;
              <lb/>
            </s>
            <s xml:id="echoid-s1032" xml:space="preserve">and conſequently its Repreſentation is
              <note symbol="*" position="right" xlink:label="note-0077-02" xlink:href="note-0077-02a" xml:space="preserve">41.</note>
            dicular to the Baſe Line; </s>
            <s xml:id="echoid-s1033" xml:space="preserve">therefore we are
              <note symbol="*" position="right" xlink:label="note-0077-03" xlink:href="note-0077-03a" xml:space="preserve">56.</note>
            to prove, that a I is equal to a H.</s>
            <s xml:id="echoid-s1034" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1035" xml:space="preserve">Now the Triangles B G C and B F M are ſimi-
              <lb/>
            lar; </s>
            <s xml:id="echoid-s1036" xml:space="preserve">and ſo</s>
          </p>
          <p style="it">
            <s xml:id="echoid-s1037" xml:space="preserve">B C : </s>
            <s xml:id="echoid-s1038" xml:space="preserve">B M :</s>
            <s xml:id="echoid-s1039" xml:space="preserve">: B G: </s>
            <s xml:id="echoid-s1040" xml:space="preserve">B F.</s>
            <s xml:id="echoid-s1041" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1042" xml:space="preserve">But B M by Conſtruction is the double of B C;
              <lb/>
            </s>
            <s xml:id="echoid-s1043" xml:space="preserve">whence B F is alſo the double of B G, which,
              <lb/>
            conſequently, is equal to G F.</s>
            <s xml:id="echoid-s1044" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1045" xml:space="preserve">Becauſe the Triangles F G N and F B L are
              <lb/>
            ſimilar, therefore</s>
          </p>
          <p style="it">
            <s xml:id="echoid-s1046" xml:space="preserve">F G : </s>
            <s xml:id="echoid-s1047" xml:space="preserve">F B : </s>
            <s xml:id="echoid-s1048" xml:space="preserve">: </s>
            <s xml:id="echoid-s1049" xml:space="preserve">G N : </s>
            <s xml:id="echoid-s1050" xml:space="preserve">B L.</s>
            <s xml:id="echoid-s1051" xml:space="preserve"/>
          </p>
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