Apollonius <Pergaeus>; Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

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        <div xml:id="echoid-div46" type="section" level="1" n="46">
          <p>
            <s xml:id="echoid-s820" xml:space="preserve">
              <pb o="(25)" file="0037" n="37"/>
            which RUM by Lemma II. </s>
            <s xml:id="echoid-s821" xml:space="preserve">is equal to a rectangle under UO and a line
              <lb/>
            drawn through the points U and O to the further ſurface of the ſphere YN.
              <lb/>
            </s>
            <s xml:id="echoid-s822" xml:space="preserve">Therefore the point Q is in the ſurface of the ſphere YN; </s>
            <s xml:id="echoid-s823" xml:space="preserve">it is therefore
              <lb/>
            common to the ſpheres YN and OTS; </s>
            <s xml:id="echoid-s824" xml:space="preserve">and I ſay that theſe ſpheres touch in
              <lb/>
            the ſaid point Q. </s>
            <s xml:id="echoid-s825" xml:space="preserve">For from the point U let a line UZ be drawn in any
              <lb/>
            plane of the ſphere OTS, and being produced let it cut the three
              <lb/>
            ſpheres in the points Z, D, H, K, P, B. </s>
            <s xml:id="echoid-s826" xml:space="preserve">The rectangle ZUB in the
              <lb/>
            ſphere OTS is by Lemma I. </s>
            <s xml:id="echoid-s827" xml:space="preserve">and II. </s>
            <s xml:id="echoid-s828" xml:space="preserve">equal to the rectangle DUP terminated
              <lb/>
            by the ſpheres XM and YN. </s>
            <s xml:id="echoid-s829" xml:space="preserve">But DU is greater than ZU, becauſe the
              <lb/>
            ſpheres XM and OTS touch in the point O, and therefore any other line from
              <lb/>
            U but UO muſt meet the ſurface of OTS before it meets the ſurface XM. </s>
            <s xml:id="echoid-s830" xml:space="preserve">
              <lb/>
            Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs
              <lb/>
            than UB, and the point B will fall without the ſphere YN; </s>
            <s xml:id="echoid-s831" xml:space="preserve">and by the
              <lb/>
            fame reaſon, all other points in the ſurface of the ſphere OTS, except the
              <lb/>
            point Q.</s>
            <s xml:id="echoid-s832" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s833" xml:space="preserve">
              <emph style="sc">The</emph>
            Demonſtration is ſimilar and equally eaſy in all caſes, whether the
              <lb/>
            ſpheres touch exlernally or internally.</s>
            <s xml:id="echoid-s834" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div47" type="section" level="1" n="47">
          <head xml:id="echoid-head54" xml:space="preserve">LEMMA IV.</head>
          <p>
            <s xml:id="echoid-s835" xml:space="preserve">
              <emph style="sc">Let</emph>
            there be a plane AC, and a ſphere FGD through whoſe center O let
              <lb/>
            FODB be drawn perpendicular to the plane, and from F any right line
              <lb/>
            FGA cutting the ſphere in G and the plane in A; </s>
            <s xml:id="echoid-s836" xml:space="preserve">I ſay that the rectangle
              <lb/>
            AFG = the rectangle BFD. </s>
            <s xml:id="echoid-s837" xml:space="preserve">For let the given ſphere and plane be cut by
              <lb/>
            the plane of the triangle ABF, the ſection of the one will be the circle GDF,
              <lb/>
            and of the other the right line ABC. </s>
            <s xml:id="echoid-s838" xml:space="preserve">Since the line FB is perpendicular
              <lb/>
            to the plane AC, it will be alſo to the right line AC. </s>
            <s xml:id="echoid-s839" xml:space="preserve">Having then a circle
              <lb/>
            FDB, and a right line AC in the ſame plane; </s>
            <s xml:id="echoid-s840" xml:space="preserve">and a line FDB paſſing thro'
              <lb/>
            the center perpendicular to AC, join D and G, and in the quadrilateral
              <lb/>
            figure ABDG the angles at B and G being both right ones, it will be in a
              <lb/>
            circle, and the rectangle AFG = the rectangle BFD; </s>
            <s xml:id="echoid-s841" xml:space="preserve">and the ſame may be
              <lb/>
            proved in any other ſection of the ſphere.</s>
            <s xml:id="echoid-s842" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div48" type="section" level="1" n="48">
          <head xml:id="echoid-head55" xml:space="preserve">LEMMA V.</head>
          <p>
            <s xml:id="echoid-s843" xml:space="preserve">
              <emph style="sc">Let</emph>
            there be a plane ABD and a ſphere EGF, through whoſe center O
              <lb/>
            let FOEC be drawn perpendicular to the plane, and in any other plane let
              <lb/>
            FHI be drawn ſo that the rectangle IFH = the rectangle CFE: </s>
            <s xml:id="echoid-s844" xml:space="preserve">if </s>
          </p>
        </div>
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