Gravesande, Willem Jacob 's, An essay on perspective

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        <div xml:id="echoid-div224" type="section" level="1" n="118">
          <pb o="64" file="0116" n="133" rhead="An ESSAY"/>
          <p>
            <s xml:id="echoid-s1530" xml:space="preserve">Whence</s>
          </p>
          <p style="it">
            <s xml:id="echoid-s1531" xml:space="preserve">G I: </s>
            <s xml:id="echoid-s1532" xml:space="preserve">G C:</s>
            <s xml:id="echoid-s1533" xml:space="preserve">: G F: </s>
            <s xml:id="echoid-s1534" xml:space="preserve">G O.</s>
            <s xml:id="echoid-s1535" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1536" xml:space="preserve">Again, becauſe the Triangles G I E and G C D
              <lb/>
            are ſimilar, we have</s>
          </p>
          <p style="it">
            <s xml:id="echoid-s1537" xml:space="preserve">G I: </s>
            <s xml:id="echoid-s1538" xml:space="preserve">G C:</s>
            <s xml:id="echoid-s1539" xml:space="preserve">: G E: </s>
            <s xml:id="echoid-s1540" xml:space="preserve">G D.</s>
            <s xml:id="echoid-s1541" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1542" xml:space="preserve">And conſequently</s>
          </p>
          <p style="it">
            <s xml:id="echoid-s1543" xml:space="preserve">G F: </s>
            <s xml:id="echoid-s1544" xml:space="preserve">G O:</s>
            <s xml:id="echoid-s1545" xml:space="preserve">: G E: </s>
            <s xml:id="echoid-s1546" xml:space="preserve">G D.</s>
            <s xml:id="echoid-s1547" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1548" xml:space="preserve">And ſo the Triangles G F E, and G O D are
              <lb/>
            ſimilar; </s>
            <s xml:id="echoid-s1549" xml:space="preserve">and the Line F E A is parallel to O D:
              <lb/>
            </s>
            <s xml:id="echoid-s1550" xml:space="preserve">Whence it follows , that the Perſpective
              <note symbol="*" position="left" xlink:label="note-0116-01" xlink:href="note-0116-01a" xml:space="preserve">13.</note>
            E A, is a Part of E a D. </s>
            <s xml:id="echoid-s1551" xml:space="preserve">We demonſtrate in
              <lb/>
            the ſame Manner, that B a is the Perſpective
              <lb/>
            of B A, and ſo the Perſpective of the Point A,
              <lb/>
            the common Section of E A and B A, is a, the
              <lb/>
            Interſection of the Appearances of the ſaid two
              <lb/>
            Lines.</s>
            <s xml:id="echoid-s1552" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div227" type="section" level="1" n="119">
          <head xml:id="echoid-head125" xml:space="preserve">
            <emph style="sc">Prob</emph>
          . IV.</head>
          <p style="it">
            <s xml:id="echoid-s1553" xml:space="preserve">80. </s>
            <s xml:id="echoid-s1554" xml:space="preserve">To find the Repreſentation of a Line, per-
              <lb/>
            pendicular to the Geometrical Plane, when the per-
              <lb/>
            ſpective Plane is above the Eye.</s>
            <s xml:id="echoid-s1555" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1556" xml:space="preserve">In the Baſe Line B E, aſſume the Line E D,
              <lb/>
              <note position="left" xlink:label="note-0116-02" xlink:href="note-0116-02a" xml:space="preserve">Fig. 43.</note>
            equal in Length to the propoſed Perpendicular;
              <lb/>
            </s>
            <s xml:id="echoid-s1557" xml:space="preserve">and draw C L, parallel to the Baſe Line, and
              <lb/>
            diſtant therefrom (for Example) {1/4} of the Height
              <lb/>
            of the Eye; </s>
            <s xml:id="echoid-s1558" xml:space="preserve">make F L equal to {3/4} of D E, and
              <lb/>
            draw the Lines E L and D F. </s>
            <s xml:id="echoid-s1559" xml:space="preserve">Note, if the
              <lb/>
            Diſtance from C L to B E, had been aſſumed
              <lb/>
            equal to a fifth Part of the Height of the Eye,
              <lb/>
            F L muſt have been aſſumed equal to {4/5} Parts of
              <lb/>
            E D. </s>
            <s xml:id="echoid-s1560" xml:space="preserve">Now let a be the Perſpective of the Foot
              <lb/>
            of the propoſed Perpendicular; </s>
            <s xml:id="echoid-s1561" xml:space="preserve">through which
              <lb/>
            draw a H parallel to the Baſe Line, and a I per-
              <lb/>
            pendicular to the ſaid Line; </s>
            <s xml:id="echoid-s1562" xml:space="preserve">then make a I equal
              <lb/>
            to G H, and the propoſed Perſpective will be
              <lb/>
            had. </s>
            <s xml:id="echoid-s1563" xml:space="preserve">The Demonſtration of this Operation is
              <lb/>
            manifeſt , in conſidering that D F and E
              <note symbol="*" position="left" xlink:label="note-0116-03" xlink:href="note-0116-03a" xml:space="preserve">56.</note>
            </s>
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