Harriot, Thomas, Mss. 6789

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page |< < (17r) of 1074 > >|
    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="0033.jpg" o="17r" n="33"/>
          <div type="page_commentary" level="0" n="0">
            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> This folio contains Harriot's proof of the angle preserving properties of stereographic projection. For the diagram see Add MS 6789, f. 18
                <lb/>
              For further details, including a transcript and translation, see Jon Pepper, 'Harriot's calculation of the meridional parts as logarithimic tangents', Archive for History of Exact Sciences, 4(1968), 359–413 (pages 366–367, and </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> Quod rumbus in planispærio nostro facit angulum cum meridiano
            <lb/>
          æqualem angulo facto a rumbo cum meridiano in
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          That a rhumb in our planisphere makes an angle with the meridian equal to the angle made by the rhumb with the meridian on the sphere.</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Sit
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            semicirculus pro
              <lb/>
            meridiano quolibet in globo
              <lb/>
            terrestri.
              <lb/>
            Cuius centrum
              <math>
                <mstyle>
                  <mi>e</mi>
                </mstyle>
              </math>
            . poli
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
              <emph style="st">In illo meridiano sumatur quoduis </emph>
              <lb/>
              <emph style="st">punctum
                <math>
                  <mstyle>
                    <mi>B</mi>
                  </mstyle>
                </math>
              .</emph>
              <lb/>
            Sit,
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            , communis sectio
              <emph style="st">plani</emph>
              <lb/>
            æquatoris et plani meridiani,
              <lb/>
            quæ meridianus est in nostro
              <lb/>
            planisphærio correlatiuus meridiano
              <lb/>
            in globo terrestri.
              <lb/>
            In semicirculo pro meridiano
              <lb/>
            sumatur quodvis punctum
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Agatur recta
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            producta quæ
              <lb/>
            secabit,
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            , in puncto
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            ,
              <lb/>
            et punctum
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            in planisphærio erit
              <lb/>
            correllatium puncto
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            in superficie
              <lb/>
            sphæræ.
              <lb/>
            Agatur etiam recta,
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , et
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
              <lb/>
            tangens peripheriam in puncto,
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico primo quod anguli
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            sunt æquales.
              <lb/>
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>+</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            recto.
              <lb/>
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>+</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            recto.
              <lb/>
            Ergo:
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>+</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                  <mi>e</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>+</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Sed:
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            ergo:
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            be a semicircle on any meridian of the terrestrial sphere, with centre
              <math>
                <mstyle>
                  <mi>e</mi>
                </mstyle>
              </math>
            , and poles
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            . Let
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            be the intersection of the equator and the meridional plane, which meridian in our planisphere corresponds to the meridian in the terrestrial sphere.
              <lb/>
            In the semicircle take any point
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            on the meridian. Construct the extended line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            which will cut
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            in the point
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            , and the point
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            in the planisphere will correspond to the point
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            on the surface of the sphere. Also construct the lines
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            touching the circumference in the point
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say first that the angles
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            amd
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            are equal.
              <lb/>
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>+</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            a right angle.
              <lb/>
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>+</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mo>=</mo>
                </mstyle>
              </math>
            a right angle.
              <lb/>
            Therefore
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>+</mo>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                  <mi>e</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>+</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            But
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>e</mi>
                  <mo>=</mo>
                  <mi>e</mi>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Therefore
              <math>
                <mstyle>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Iam intelligitur
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            punctum esse centrum
              <lb/>
            circuli nautici qui vulgo compassus
              <lb/>
            dicitur; sive circuli horizontis visibilis.
              <lb/>
            cuius plano
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , est ad angulos rectos.
              <lb/>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            est communis sectio plani meridiani
              <lb/>
            et
              <emph style="super">illius</emph>
            circuli nautici, et est linea dicta
              <lb/>
            meridiana in plano
              <emph style="super">horizontis</emph>
              <emph style="st">circuli nautici</emph>
            .
              <lb/>
            Agatur
              <emph style="super">quælibet</emph>
            linea
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <emph style="super">in plano horizontis</emph>
            faciens
              <emph style="st">quemvis</emph>
            angulum
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
              <lb/>
              <emph style="st">cum meridiana
                <math>
                  <mstyle>
                    <mi>B</mi>
                    <mi>G</mi>
                  </mstyle>
                </math>
              </emph>
              <lb/>
            et in illa linea sumatur quodvis punctum
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            et Agatur
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            perpendicularis ad
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Conectantur puncta
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            : et
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . et constituatur
              <lb/>
            pyramis cuius basis
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , et vertex,
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Quoniam
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , perpendicularis est in plano
              <lb/>
            erecto ad planum meridionis circuli productum
              <lb/>
            erit etiam perpendicularis ad
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , et planum
              <lb/>
            trianguli
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            erit erectum ad planum
              <lb/>
            meridiani.
              <lb/>
            Linea
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , secat,
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            , in puncto
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
              <lb/>
            (non opus est ducere lineas
              <lb/>
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            , et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . ideo
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Now it is to be understood that the point
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            is the centre of a nautical circle, which is commonly called the compass; or circle of the visible horizon, which is at right angles to the plane of
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            is common to the plane of the meridian and that nautical circle and is said to be the meridional line in the horizontal plane.
              <lb/>
            Construct any line
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            in the horizontal plane making any angle
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            with the meridian
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            and in that lane take any point
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            , and construct
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            perpendicular to
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>H</mi>
                </mstyle>
              </math>
            . Join the points
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            to make a pyramid whose base is
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            and vertex
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            . Since
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is a perpendicular in a plane at right angles to the plane of the circle of the meridion, it is also perpendicular to
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , and the plane of triangle
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            will be at right angles to the plane of the meridian.
              <lb/>
            The line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            cuts
              <math>
                <mstyle>
                  <mi>e</mi>
                  <mi>X</mi>
                </mstyle>
              </math>
            in the point
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
            .
              <lb/>
            (there is no need to draw the lines
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            , therefore they are omitted). </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Et in plano
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            a puncto
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
              <lb/>
            in linea
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            erigatur,
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            , ad
              <lb/>
            angulos rectos, quæ secabit
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            in puncto
              <math>
                <mstyle>
                  <mi>d</mi>
                </mstyle>
              </math>
            . Et erit communis
              <lb/>
            sectio æquatoris et trianguli
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Connectantur puncta
              <math>
                <mstyle>
                  <mi>d</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod angulus
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            in plano
              <lb/>
            æquatoris est qualis angulo
              <lb/>
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            in plano horizontis
              <lb/>
            sive circuli nauticis.
              <lb/>
            producatur
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            et sit
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            perpen-
              <lb/>
            dicularis ad illam.
              <lb/>
            fiat
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>P</mi>
                  <mo>=</mo>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            et agantur
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , et
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Quoniam
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            est perpendicularis
              <lb/>
            ad planum meridiani, facit
              <lb/>
            rectos angulos cum
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            ,
              <lb/>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Cum etiam
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            , sunt æquales
              <lb/>
            triangula rectangula
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            et
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , sunt æqualia et
              <lb/>
            æquiangula, et angulus
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            æqualis est
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Sed angulus
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
              <emph style="super">in æquatore</emph>
            est æqualis
              <lb/>
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , ob parallelismum
              <lb/>
            triangulorm
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            nam
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            est parallela
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
              <lb/>
            et
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            , quia:
              <lb/>
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>A</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>P</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            ergo tertium latus
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
              <lb/>
            est parallelum
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
              <lb/>
            ergo plana triang-
              <lb/>
            ulorum
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            et CPD
              <math>
                <mstyle/>
              </math>
              <lb/>
            sunt parallela. Et
              <lb/>
            similia, et angulus
              <lb/>
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Ergo
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            quod demonstrandum
              <lb/>
            fuit.
              <lb/>
            Aliter
              <lb/>
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            est parall.
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>P</mi>
                  <mo>,</mo>
                  <mi>q</mi>
                  <mi>u</mi>
                  <mi>i</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            And in the plane
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , there is constructed
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            from point
              <math>
                <mstyle>
                  <mi>c</mi>
                </mstyle>
              </math>
            at right angles to the line
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , which will cut
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            in the point
              <math>
                <mstyle>
                  <mi>d</mi>
                </mstyle>
              </math>
            . And it will be common to the equatorial plane and the triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . Join the points
              <math>
                <mstyle>
                  <mi>d</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>b</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that the angle
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            in the equatorial plane is equal to angle
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            in the horizontal plane, or nautical circle.
              <lb/>
            Let
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            be extended, and let
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            be perpendicular to it. Make
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>P</mi>
                  <mo>=</mo>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and construct
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            , and
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Since
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            is perpendicular to the plane of the meridian, it makes right angles with
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            , and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Since also
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            are equal, the right-angled triangles
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>P</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            are equal and equiangluar, and the angle
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . But the angle
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            in the equatorial plane is equal to
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , because the triangles
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            are parallel, for
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            is parallel to
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , and so are
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            since
              <lb/>
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>A</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>A</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>P</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>B</mi>
                  <mi>b</mi>
                  <mo>=</mo>
                  <mi>H</mi>
                  <mi>b</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            Therefore the third side
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            is parallel to
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>P</mi>
                </mstyle>
              </math>
            , and so the planes of the triangles
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            are parallel. And similarly also angles
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>P</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Therefore
              <math>
                <mstyle>
                  <mi>c</mi>
                  <mi>b</mi>
                  <mi>d</mi>
                  <mo>=</mo>
                  <mi>C</mi>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , as was to be proved.
              <lb/>
            Another way,
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            is parallel to
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>P</mi>
                  <mo>,</mo>
                  <mi>s</mi>
                  <mi>i</mi>
                  <mi>n</mi>
                  <mi>c</mi>
                  <mi>e</mi>
                  <mo>:</mo>
                </mstyle>
              </math>
            </s>
          </p>
        </div>
      </text>
    </echo>
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