Harriot, Thomas, Mss. 6785

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    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6785_f301" o="301" n="601"/>
          <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"> The first reference on this page is to a lemma proved by Commandino on page 198v of his edition of
                <emph style="it">Mathematicae collectiones</emph>
                <ref id="pappus_1588"> (Pappus </ref>
              . Harriot's diagram imitates Commandino's except that he has changed some of the lettering.
                <lb/>
              Towards the end there are references to Euclid's
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV4.html"/>
                <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV5.html"/>
              . </s>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV4.html"/>
                In a given triangle to inscribe a circle. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve">
                  <ref target="http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV5.html"/>
                About a given triangle to circumscribe a circle. </s>
              </quote>
              <lb/>
              <s xml:space="preserve"> There is also a reference to
                <emph style="it">In duos Archimedis aequiponderantium libros</emph>
                <ref id="guidobaldi_1588"> (Guidobaldi </ref>
              . </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> Pappus. pag. </head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Quod hic construitur est in altera
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            What is here constructed is in another ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Sit triangulum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Ducantur
              <lb/>
            perpendicularis
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            , quæ sese
              <lb/>
            intersecant in puncto
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            . et ab
              <lb/>
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            ad
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            agatur recta usque ad
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Dico quod
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            est linea perpendicu-
              <lb/>
            laris lineæ
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            .
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let there be a triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . Draw the perpendiculars
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            , which intersect each other at the point
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            . And from
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            ad
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            construct a line as far as
              <math>
                <mstyle>
                  <mi>G</mi>
                </mstyle>
              </math>
            .
              <lb/>
            I say that
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            is a line perpendicular to the line
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            . * </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Demonstratio in
              <lb/>
            altera charta.
              <lb/>
            In huius demonstratione
              <lb/>
            Erravit
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            The demonstration is in another sheet.
              <lb/>
            In this demonstration Commndino has ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Demonstratio
              <lb/>
            * puncta
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            , sunt in
              <lb/>
            peripheria quoniam anguli ad
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
              <lb/>
            sunt recti. agatur igitur circulus,
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            puncta sunt etiam in perpheria
              <lb/>
            semicirculi
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , [???] causa [???]
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Demonstration
              <lb/>
            * The points
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>F</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>A</mi>
                </mstyle>
              </math>
            are on a circumference because the angles at
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            are right angles. Therefore construct the circle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            . The points
              <math>
                <mstyle>
                  <mi>D</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>E</mi>
                </mstyle>
              </math>
            are also on the circumference of the semicircle
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            , [???] because [???]. </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Anguli
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            sunt æqualis quia
              <lb/>
            sunt in peripheria [???] super eandem
              <lb/>
            basis
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . Tum
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            æqualis
              <lb/>
            quia insistunt super
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . æqualis igitur
              <lb/>
            anguli
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            .
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            sunt æqualis quia [???] ad verticem.
              <lb/>
            Ergo, tertius angulus
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            æquatur
              <lb/>
            tertio
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            recto. quod demonstrantur
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            The agnles
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            are equal because the are on a circumference [???] on the same base
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            . Then
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            are equal because they stand on
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . Therefore angles
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>A</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            are equal.
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>F</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            are euqal because [???] to the vertex.
              <lb/>
            Therefore, the third angle
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>B</mi>
                </mstyle>
              </math>
            is equal to the third
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>D</mi>
                  <mi>A</mi>
                </mstyle>
              </math>
            , a right angle. Which was to be proved. </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Nota.
              <lb/>
            Tres lineæ ductæ
              <lb/>
            per tres angulos et
              <lb/>
            Biscantes angulos: Euclid. el. 4,4.
              <lb/>
            Bisecantes latera: Archimed. de æqui:
              <lb/>
            perpenduclares ad latera Hic demonstratur.
              <lb/>
            Tres, perpendiculares in medijs laterum, El. 5,4. Sese
              <lb/>
            interse-
              <lb/>
            cant in
              <lb/>
            uno
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Note.
              <lb/>
            Three lines drawn through three angles.
              <lb/>
            Bisection of angules: Euclid, Elements, IV,4.
              <lb/>
            Bisection of sides: Archimedes, De æqui:
              <lb/>
            Perpendiculars to the sides: demonstrated here.
              <lb/>
            Three perpendiculars in the middle of the sides, Elements, IV.5. Intersecting each other in a single point.</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Alia propositio: seu conclusio
              <lb/>
            Superioribus constructis et demonstratis; fiat præterea triangulum
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            . Dico quod
              <lb/>
            eius anguli bisecantur per lineas
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , et
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Another proposition: or conclusion
              <lb/>
            By the above construction and demonstration; make also the triangle
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            . I say that its angles are bisected by the lines
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>B</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , and
              <math>
                <mstyle>
                  <mi>C</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Nam: Anguli
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            æquantur, quia in perpheria et insistunt super
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Angulus etiam
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            , æqualis
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            quia trainguli æquianguli;
              <lb/>
            duo enim recti et duo ad verticem; ergo tertius tertio æqualis.
              <lb/>
            Sed angulus
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            æqualis
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            quia in peripheria super
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . Ergo
              <lb/>
            anguli
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            et
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            sunt æqualis. Simili modo probatur bisectio
              <lb/>
            duorum
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            For the angles
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            are equal because they are on a circumference and stand on
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Also the angle
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal to
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            because in an equiangular triangle, for two are right angles and two are at the vertex. Therefore the third is equal to the third.
              <lb/>
            But the angle
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>C</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            is equal
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            because on the circumference standing on
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>F</mi>
                </mstyle>
              </math>
            . Therefore the angles
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>D</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>F</mi>
                  <mi>G</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            are equal. In a similar way there can be proved the bisection of the others. </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Etiam triangula:
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            sunt similia.
              <lb/>
            Tertio
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            est una linea reflexa ad
              <lb/>
            angulis æqualis intra triangluum
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Also the triangles
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>D</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>D</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>G</mi>
                  <mi>B</mi>
                  <mi>E</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            are similar.
              <lb/>
            Third,
              <math>
                <mstyle>
                  <mi>D</mi>
                  <mi>E</mi>
                  <mi>G</mi>
                </mstyle>
              </math>
            is a line of reflection at equal angles inside triangle
              <math>
                <mstyle>
                  <mi>A</mi>
                  <mi>B</mi>
                  <mi>C</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve">
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Most ]</s>
          </p>
        </div>
      </text>
    </echo>
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