Gravesande, Willem Jacob 's, An essay on perspective

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            the Circular Baſe of the Cone will be deter-
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            min’d.</s>
            <s xml:id="echoid-s870" xml:space="preserve"/>
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        <div xml:id="echoid-div126" type="section" level="1" n="69">
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            <emph style="sc">Demonstration</emph>
          .</head>
          <p style="it">
            <s xml:id="echoid-s871" xml:space="preserve">To prove this, draw the Lines B C and L F, cut-
              <lb/>
            ting the Line A S in the Points N and M; </s>
            <s xml:id="echoid-s872" xml:space="preserve">and make
              <lb/>
            the Line G n equal to A N, and draw the Line
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            n D m. </s>
            <s xml:id="echoid-s873" xml:space="preserve">It is now manifeſt, that if the Cone be
              <lb/>
            continued out above its Vertex, (that is, if the oppo-
              <lb/>
            ſite Cone be form’d) it will cut the Horizontal Plane
              <lb/>
            in a Circle equal to B E C, whoſe Seat will be BEC:
              <lb/>
            </s>
            <s xml:id="echoid-s874" xml:space="preserve">So that the Point S, in reſpect of B E C, is in the
              <lb/>
            ſame Situation as the Eye hath, with reſpect to the
              <lb/>
            Circle form’d in the Horizontal Plane, by the Conti-
              <lb/>
            nuation of the Cone. </s>
            <s xml:id="echoid-s875" xml:space="preserve">Whence it follows, that B C
              <lb/>
            is the Seat of the viſible Portion of that Circle. </s>
            <s xml:id="echoid-s876" xml:space="preserve">For,
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            by Conſtruction, B and C are the Points of Contact
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            of the Tangents to the Circle B E C, which paſs
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            thro’ the Point S; </s>
            <s xml:id="echoid-s877" xml:space="preserve">becauſe the Angle ABS, which
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            is in a Semicircle, is a right one.</s>
            <s xml:id="echoid-s878" xml:space="preserve"/>
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            <s xml:id="echoid-s879" xml:space="preserve">Now, if a Plane be conceiv’d, as paſſing thro’ ſome
              <lb/>
            Points in the Horizontal Plane, whoſe Seats are
              <lb/>
            B and C, and which cuts the two oppoſite Cones
              <lb/>
            thro’ their Vertex; </s>
            <s xml:id="echoid-s880" xml:space="preserve">it is evident, that this Plane
              <lb/>
            continued, will cut the Geometrical Plane in a Line
              <lb/>
            parallel to B N C; </s>
            <s xml:id="echoid-s881" xml:space="preserve">and that this Line upon the
              <lb/>
            ſaid Plane, will determine the viſible Part of the
              <lb/>
            Cone’s Baſe. </s>
            <s xml:id="echoid-s882" xml:space="preserve">So, ſince G n was made equal to
              <lb/>
            A N, we have only to prove, that P m is equal to
              <lb/>
            A M: </s>
            <s xml:id="echoid-s883" xml:space="preserve">For, it follows from thence, that L M F is
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            the Common Section of the Geometrical Plane, and
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            the Plane which we have here imagin’d.</s>
            <s xml:id="echoid-s884" xml:space="preserve"/>
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          <p style="it">
            <s xml:id="echoid-s885" xml:space="preserve">The Triangles D Q P and G H D are ſimilar, whence
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            D G: </s>
            <s xml:id="echoid-s886" xml:space="preserve">D P:</s>
            <s xml:id="echoid-s887" xml:space="preserve">: G H: </s>
            <s xml:id="echoid-s888" xml:space="preserve">P Q.</s>
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