DelMonte, Guidubaldo, Le mechaniche

Table of figures

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          <chap id="N13354">
            <p id="id.2.1.535.0.0" type="main">
              <s id="id.2.1.535.5.0">
                <pb pagenum="44" xlink:href="037/01/103.jpg"/>
                <emph type="italics"/>
              eguale ad eſſo FHM, ſarà il triangolo GKQ eguale al triangolo FHM, &
                <lb/>
              illato GQ al lato FM eguale; ſarà dunque maggiore GN di eſſa FM; &
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              perciò BN maggiore ſarà di BM. </s>
              <s id="N13F80">& BM ſarà maggiore di BA; impe­
                <lb/>
              roche BM è maggiore di eſſa BF. </s>
              <s id="id.2.1.535.6.0">Che biſognaua moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.536.0.0" type="main">
              <s id="id.2.1.536.1.0">
                <emph type="italics"/>
              Et nel modo iſteſſo in tutto, quanto più da preſſo ſarà BG ad eſſa BE, ſempre la li­
                <lb/>
              nea BN ſi dimoſtrerà eſſer maggiore.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.537.0.0" type="main">
              <s id="id.2.1.537.1.0">Che ſe ſaranno poſti di
                <lb/>
              ſotto i triangoli BF
                <lb/>
              HB GK tra AB
                <lb/>
              BC, & ſiano tiratele
                <lb/>
              linee CHO GKP,
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              lequali taglino le li­
                <lb/>
              nee BF BG ne' pun
                <lb/>
              ti MN: ſarà la linea
                <lb/>
              BN minore di eſſa
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              BM, & BM di eſsa
                <lb/>
              BA. </s>
            </p>
            <figure id="id.037.01.103.1.jpg" xlink:href="037/01/103/1.jpg" number="97"/>
            <p id="id.2.1.539.0.0" type="main">
              <s id="id.2.1.539.1.0">
                <emph type="italics"/>
              Congiunganſi BO BP. </s>
              <s id="id.2.1.539.2.0">ſimilmen
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              te proueraſſi, che l'angolo P
                <lb/>
              KB è minore dell' angolo OH
                <lb/>
              B. </s>
              <s id="id.2.1.539.3.0">Hor percioche l'angolo F
                <lb/>
              HB è eguale all'angolo GKB;
                <lb/>
              ſarà l'angolo GKN maggio­
                <lb/>
              re dell'angolo FHM: per la
                <lb/>
              qual coſa la linea GN ſarà
                <lb/>
              maggiore di eſſa FM. </s>
              <s id="N13FD2">& per­
                <lb/>
              ciò la linea BN ſarà minore
                <lb/>
              della linea BM. </s>
              <s id="N13FD8">& concio­
                <lb/>
              ſia che maggiore ſia BF di
                <lb/>
              BM; ſarà BM minore di
                <lb/>
              BA. </s>
              <s id="id.2.1.539.4.0">& con ſimile modo
                <lb/>
              proueraßi, che quanto più B
                <lb/>
              G ſarà dapreſſo ad eſſa BC,
                <lb/>
              la linea BN ſempre ſarà
                <lb/>
              minore.
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
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