DelMonte, Guidubaldo, Le mechaniche

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          <chap id="N13354">
            <pb xlink:href="037/01/100.jpg"/>
            <p id="id.2.1.521.0.0" type="main">
              <s id="id.2.1.521.1.0">
                <emph type="italics"/>
              Congiunganſi BD BH B
                <emph.end type="italics"/>
              K,
                <emph type="italics"/>
              & percioche due linee DA AB ſono eguali à due
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note146"/>
                <emph type="italics"/>
              HF FB, & l'angolo DAB retto è anco eguale al retto HFB; ſaranno i
                <lb/>
              reſtanti angoli eguali à i reſtanti angoli, & HB eguale ad eſſa DB. </s>
              <s id="id.2.1.521.2.0">Similmen­
                <lb/>
              te moſtreraſſi il triangolo BKG eſſere eguale al triangolo BHF. </s>
              <s id="id.2.1.521.3.0">Per laqual co
                <lb/>
              ſa co'l centro B, & con l'in
                <lb/>
              teruallo di vna di eſſe deſcri­
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              uaſi il cerchio DH
                <emph.end type="italics"/>
              K
                <emph type="italics"/>
              E, il
                <lb/>
              quale tagli le linee CH CK
                <lb/>
              ne' punti OP; & congiun­
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              ganſi OB PB. </s>
              <s id="id.2.1.521.4.0">Percioche
                <lb/>
              dunque il punto K è più vi­
                <lb/>
              cino ad E, che H; ſarà la
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note147"/>
                <emph type="italics"/>
              linea CK maggiore di CH,
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              & CP minore di CO: dun
                <lb/>
              que PK ſarà maggiore di
                <lb/>
              OH. </s>
              <s id="id.2.1.521.5.0">Ma perche il triangolo
                <lb/>
              BKP di due lati eguali ha i
                <lb/>
              ſuoi lati BK BP eguali à i
                <lb/>
              lati BH BO del triangolo
                <lb/>
              BHO di due lati eguali, ma
                <lb/>
              ben la baſe KP maggiore
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              della baſe HO, ſarà l'ango­
                <lb/>
              lo KBP maggiore dell' an
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note148"/>
                <emph type="italics"/>
              golo HBO. </s>
              <s id="N13DB3">dunque i restan
                <lb/>
              ti angoli alla baſe, cioè KPB
                <lb/>
              PKB preſi inſieme, i quali
                <lb/>
              tra loro ſono eguali, ſaranno
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              minori de i reſtanti angoli al­
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              la baſe poſti, cioè OHB
                <lb/>
              HOB, iquali etiandio tra lo
                <lb/>
              ro ſono eguali eſſendo che tut
                <lb/>
              ti gli angoli di ciaſcuno trian
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note149"/>
                <emph type="italics"/>
              golo ſiano eguali à due angoli
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              retti. </s>
              <s id="id.2.1.521.6.0">Per laqual coſa anche
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              le metà di queſti, cioè NKB
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              ſarà minore di MHB. </s>
              <s id="id.2.1.521.7.0">Et
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note150"/>
                <emph type="italics"/>
              concioſia, che l'angolo BKG
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.100.1.jpg" xlink:href="037/01/100/1.jpg" number="94"/>
                <lb/>
                <emph type="italics"/>
              ſia eguale all'angolo BHF, ſarà NKG maggiore di MHF. </s>
              <s id="id.2.1.521.8.0">Se dunque nel
                <lb/>
              punto K ſi faccia l'angolo GKQ eguale ad FHM ſi ſarà il triangolo GKQ
                <lb/>
              eguale al triangolo FHM; Imperoche due angoli in FH di vno ſono eguali à
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              due in GK d'vn'altro, & il lato FH è eguale al lato GK, ſarà GQ eguale
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              ad FM. </s>
              <s id="id.2.1.521.9.0">Adunque GN ſarà maggiore di FM. </s>
              <s id="N13DFD">& coſi per eſſere BG egua­
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
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