DelMonte, Guidubaldo, Le mechaniche

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    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N13354">
            <p id="id.2.1.521.0.0" type="main">
              <s id="N13DFD">
                <pb pagenum="43" xlink:href="037/01/101.jpg"/>
                <emph type="italics"/>
              le à BF, ſarà BN minore di eſſa BM. </s>
              <s id="N13E08">ma che BM ſia minore di eſſa BA
                <lb/>
              è manifeſto, percioche BM, è minore di eſſa BF, laquale è eguale à BA. </s>
              <s id="N13E0C">che
                <lb/>
              biſognaua moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.523.0.0" type="margin">
              <s id="id.2.1.523.1.0">
                <margin.target id="note146"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              4.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.524.0.0" type="margin">
              <s id="id.2.1.524.1.0">
                <margin.target id="note147"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              8.
                <emph type="italics"/>
              del terzo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.525.0.0" type="margin">
              <s id="id.2.1.525.1.0">
                <margin.target id="note148"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              25.
                <emph type="italics"/>
              del
                <emph.end type="italics"/>
              5. </s>
            </p>
            <p id="id.2.1.526.0.0" type="margin">
              <s id="id.2.1.526.1.0">
                <margin.target id="note149"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              5.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.527.0.0" type="margin">
              <s id="id.2.1.527.1.0">
                <margin.target id="note150"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              26.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.528.0.0" type="main">
              <s id="id.2.1.528.1.0">
                <emph type="italics"/>
              Di più ſe tra BG BE ſi tiri à piacere vn'altra linea eguale à BG; & facciaſi l'ope
                <lb/>
              ratione, come di ſopra è stato detto, proueraſſi ſimilmente la linea BR eſſer mi­
                <lb/>
              nore di BN. </s>
              <s id="N13E79">& quanto più da vicino ſarà ad eſſa BE, ſarà anche ſempre minore.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.529.0.0" type="main">
              <s id="id.2.1.529.1.0">Che ſe i triangoli eguali BFH BGK foſſero di ſotto fra BC
                <lb/>
              BA collocati; & foſſero congiunte le linee HC KC, le­
                <lb/>
              quali tagliaſſero le linee BF BG allungate dalla parte di FG
                <lb/>
              ne' punti MN, ſarà
                <lb/>
              la BN maggiore del
                <lb/>
              la BM, & la BM di
                <lb/>
              eſſa BA. </s>
            </p>
            <p id="id.2.1.530.0.0" type="main">
              <s id="id.2.1.530.1.0">
                <emph type="italics"/>
              Imperoche allunghiſi CH CK
                <lb/>
              fin alla circonferenza in OP,
                <lb/>
              & congiunganſi BO BP;
                <lb/>
              con ſimile modo moſtreraſſi
                <lb/>
              la linea PK eſſere maggiore
                <lb/>
              ai OH, & l'angolo PKB eſ
                <lb/>
              ſere minore dell
                <expan abbr="ãgolo">angolo</expan>
              OHB.
                <lb/>
              </s>
              <s id="id.2.1.530.2.0">& percioche l'angolo BHF
                <lb/>
              è eguale dell' angolo BKG, ſa
                <lb/>
              rà tutto l'angolo PKG mi­
                <lb/>
              nore dell' angolo OHF. </s>
              <s id="id.2.1.530.3.0">Per
                <lb/>
              laqual coſa il reſtante GKN
                <lb/>
              ſarà maggiore del reſtante
                <lb/>
              FHM. </s>
              <s id="id.2.1.530.4.0">Se
                <expan abbr="dũque">dunque</expan>
              faraſſi l'an
                <lb/>
              golo GKQ eguale ad FHM
                <lb/>
              la linea KQ taglierà in modo
                <lb/>
              la GN, che GQ diuenterà
                <lb/>
              eguale ad FM. </s>
              <s id="id.2.1.530.5.0">Per laqual
                <lb/>
              coſa maggiore ſarà GN, che
                <lb/>
              FM; allequali ſe ſaranno ag
                <lb/>
              giunte le eguali BF BG, ſa­
                <lb/>
              rà BN maggiore di BM. </s>
              <s id="N13ED4">&
                <lb/>
              per eſſere BM maggiore di
                <lb/>
              FB, ſarà anco maggiore di
                <lb/>
              BA. </s>
              <s id="id.2.1.530.6.0">ſimilmente proueraſſi
                <lb/>
              che
                <expan abbr="quãto">quanto</expan>
              più da vicino ſarà
                <lb/>
              BG à BC, la linea BN ſem
                <lb/>
              pre ſarà maggiore.
                <emph.end type="italics"/>
              </s>
            </p>
            <figure id="id.037.01.101.1.jpg" xlink:href="037/01/101/1.jpg" number="95"/>
          </chap>
        </body>
      </text>
    </archimedes>
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