DelMonte, Guidubaldo, Le mechaniche

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              <pb pagenum="16" xlink:href="037/01/047.jpg"/>
              <s id="N1199F">
                <emph type="italics"/>
              Ma il lato ancora CM è eguale al lato CP, dunque il triangolo MCX è egua
                <lb/>
              le al triangolo PCN, & il lato MX eguale al lato NP. </s>
              <s id="id.2.1.203.6.0">Onde la linea PN
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              ſarà eguale ad LX. </s>
              <s id="id.2.1.203.7.0">Tiriſi oltre a ciò dal punto O la linea OT egualmente di­
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              ſtante da AC, laquale tagli NP in V. </s>
              <s id="N119AF">& ſia anco tirata dal punto P vna
                <lb/>
              linea a piombo di OT,
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              la quale per certo non
                <lb/>
              puote cadere tra OV,
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              perche eſſendo l'angolo
                <lb/>
              ONV retto, ſarà acu
                <emph.end type="italics"/>
                <arrow.to.target n="note51"/>
                <lb/>
                <emph type="italics"/>
              to lo OVN. </s>
              <s id="id.2.1.203.8.0">Per la
                <lb/>
              qualcoſa OVP ſarà
                <lb/>
              ottuſo. </s>
              <s id="id.2.1.203.9.0">Non caderà
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              dunque la linea tirata
                <lb/>
              dal punto P tra OV
                <lb/>
              à piombo di OT: pe­
                <lb/>
              roche due angoli d'uno
                <lb/>
                <expan abbr="triãgolo">triangolo</expan>
              ſarebbono l'u­
                <lb/>
              no retto, & l'altro ot­
                <lb/>
              tuſo, che è impoßibile.
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              </s>
              <s id="id.2.1.203.10.0">Caderà dunque nella li
                <lb/>
              nea OT nella parte di
                <lb/>
              VT, et ſia PT. </s>
              <s id="id.2.1.203.11.0">ſarà ſe
                <lb/>
              condo eſſi, PT la di
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.047.1.jpg" xlink:href="037/01/047/1.jpg" number="29"/>
                <lb/>
                <emph type="italics"/>
              ritta ſceſa della circonferenza OP. </s>
              <s id="id.2.1.203.12.0">Percioche dunque l'angolo ONV è retto,
                <emph.end type="italics"/>
                <arrow.to.target n="note52"/>
                <lb/>
                <emph type="italics"/>
              ſarà la linea OV maggiore della ON. </s>
              <s id="id.2.1.203.13.0">Onde la OT ſarà parimente maggiore
                <lb/>
              della ON. </s>
              <s id="id.2.1.203.14.0">& coſi diſtendendoſi la linea OP ſotto gli angoli retti ONP,
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              OTP, ſarà il quadrato di OP eguale alli quadrati ON NP inſieme preſi, ſi
                <emph.end type="italics"/>
                <arrow.to.target n="note53"/>
                <lb/>
                <emph type="italics"/>
              milmente eguale a i quadrati di OT TP inſieme. </s>
              <s id="id.2.1.203.15.0">per laqual coſa li quadrati inſie­
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              me di ON NP ſaranno eguali a i quadrati inſieme di OT TP. </s>
              <s id="id.2.1.203.16.0">Ma il quadrato
                <lb/>
              di OT è maggiore del quadrato di ON, per eſſere maggiore la linea OT della
                <lb/>
              ON. </s>
              <s id="id.2.1.203.17.0">Adunque il quadrato di NP ſara maggiore del quadrato TP & perciò la
                <lb/>
              linea TP ſarà minore della linea PN, & della linea LX. </s>
              <s id="id.2.1.203.18.0">Meno obliqua
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              dunque ſarà la ſceſa dell'arco LA, che dell'arco OP. </s>
              <s id="id.2.1.203.19.0">Dunque il peſo po­
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              sto in L, per i loro detti, ſarà piu graue, che in O, il che, per le coſe, che di
                <lb/>
              ſopra habbiamo detto, è manifeſtamente falſo. </s>
              <s id="id.2.1.203.20.0">concioſia, che il peſo poſto in O
                <lb/>
              ſia piu graue, che in L. </s>
              <s id="id.2.1.203.21.0">Non ſi puote dunque raccogliere dal piu diritto, &
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              piu torto mouimento in quel modo pigliato, eſſere il peſo tanto piu graue ſecon­
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              do il ſito, quanto nel medeſimo ſito è meno torta la ſceſa. </s>
              <s id="id.2.1.203.22.0">& quinci naſce tutto
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              quaſi il ſuo errore & inganno in coteſta coſa. </s>
              <s id="id.2.1.203.23.0">Imperoche quantunque per acciden­
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              te alle volte dalle coſe falſe ne ſegua il vero, tutta via per ſe ſteſſe principalmente
                <lb/>
              dalle falſe ne ſegue il falſo, ſi come dalle vere ſempre il vero ne ſegue. </s>
              <s id="id.2.1.203.24.0">Non è pero
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              da marauigliarſi, ſe mentre eſſi prendono coſe falſe, & ſtanno ſopra quelle, come ve
                <emph.end type="italics"/>
              </s>
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