DelMonte, Guidubaldo, Le mechaniche
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              Ponganſi le coſe isteſſe
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              & da i punti DE
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              ſiano tirate le linee
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              DHE
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              K
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              à piombo
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              dell'orizonte, & ſia
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              vn'altro cerchio L
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              DM, il cui centro
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              ſia N, ilquale toc
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              chi FDG nel pun
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              to D, & ſia eguale
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              ad FDG. </s>
              <s id="id.2.1.130.2.0">Sarà
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              NC linea retta: &
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              perche l'angolo
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              K
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              EC è eguale all'an­
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              golo HDN, &
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              l'angolo CEG è pa
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              rimente eguale al­
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              l'angolo NDM,
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              peroche egli è contenuto da mezi diametri, & da circonferenze eguali: ſarà il re­
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              stante angolo & miſto
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              K
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              EG eguale al reſtante angolo & miſto HDM. </s>
              <s id="id.2.1.130.3.0">Et per­
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              cioche preſuppongono, che quanto è minore l'angolo contenuto dalla linea tirata à
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              piombo dell'orizonte, & dalla circonferenza, tanto in quel ſito eſſere anco più gra
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              ue il peſo. </s>
              <s id="id.2.1.130.4.0">Talche ſi come l'angolo contenuto da HD, & dalla circonferenza
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              DG, è minore dell'angolo
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              K
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              EG, cioè dell'angolo HDM, coſi ſecondo queſta
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              proportione il peſo poſto in D ſia più graue di quello che ſtà in E. </s>
              <s id="id.2.1.130.5.0">Ma la pro­
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              portione dell'angolo MHD all'angolo HDG è minore di qual ſi voglia altra
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              proportione, che ſi troui tra la maggiore, & minore quantità: Adunque la pro­
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              portione de i peſi DE ſarà la minima di tutte le proportioni, anzi non ſarà quaſi
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              ne anche proportione, eſſendo la minima di tutte le proportioni. </s>
              <s id="id.2.1.130.6.0">Che la propor­
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              tione di MDH verſo HDG ſia di tutte la minima, moſtrano con queſta ne­
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              ceſſaria ragione, peroche MHD ſupera HDG con angolo di linea curua, che
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              è MGD, ilquale angolo è il minimo di tutti gli angoli fatti di linee rette: ne po­
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              tendoſi dare angolo minore di MGD ſarà la proportione di MDH verſo HDG
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              la minima di tutte le proportioni. </s>
              <s id="id.2.1.130.7.0">Laqual ragione pare eſſere grandemente friuo­
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              la, peroche quantunque l'angolo MDG ſia di tutti gli angoli fatti di linee rette
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              il minore, non perciò ſegue totalmente egli eſſere di tutti gli angoli il minimo, im­
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              peroche ſia dal punto D tirata la linea DO à piombo di NC, ambedue que­
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              ste toccheranno le circonferenze LDMFDG nel punto D. </s>
              <s id="id.2.1.130.8.0">Ma percioche le
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              circonferenze ſono eguali, ſarà l'angolo MDO misto eguale all'angolo ODG mi­
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              ſto. </s>
              <s id="id.2.1.130.9.0">L'vno de gli angoli dunque, cioè ODG ſarà minore di MDG, cioè minore
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              del minimo. </s>
              <s id="id.2.1.130.10.0">Dapoi l'angolo ODH ſarà minore dell'angolo MDH. </s>
              <s id="id.2.1.130.11.0">Per laqual coſa
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              ODH haurà proportione minore all'angolo HDG, che MDH all'iſteſſo
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