DelMonte, Guidubaldo, In duos Archimedis aequeponderantium libros Paraphrasis : scholijs illustrata

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    <archimedes>
      <text>
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          <chap id="N10019">
            <p id="N176FE" type="main">
              <s id="N17700">
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              què AB ipſam DG in H. Quoniam enim parabole
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                <lb/>
              ſeſquitertia eſt trianguli ABC, itidemquè parabole DBE
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              trianguli DBE ſeſquitertia exiſtit, erit parabole ABC ad trian
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              gulum ABC, vt parabole DBE ad triangulum DBE. &
                <arrow.to.target n="marg377"/>
              per­
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              mutando parabole ABC ad parabolen DBE, vt triangulum
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              ABC ad triangulum DBE. Quoniam autem AC ordina­
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              tim eſt applicata, vnde AF ipſi FC eſt æqualis, ac per conſe­
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              〈que〉ns AF eſt ipſius AC dimidia. </s>
              <s id="N17721">erit triangulum ABF dimi­
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              dium trianguli ABC. cùm vtraquè ſub eadem ſint
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              eademquè ratione triangulum DBG trianguli DBE dimi­
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              dium exiſtit. </s>
              <s id="N1772C">quare vt triangulum ABF ad triangulum
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              DBG, ita eſt triangulum ABC ad DBE triangulum, ac pro­
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              pterea triangulum ABF ad DBG triangulum eſt, vt parabo­
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              le ABC ad parabolen DBE. Cùm autem ſit HG æquidiſtans
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              ipſi AF, ſiquidem ſunt ordinatim applicatæ, ob
                <expan abbr="triangulorũ">triangulorum</expan>
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              ſimilitudinem ABF HBG, ita erit FB ad BG, vt AF ad HG
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              vt autem FB ad BG, ita quadratum ex AF ad quadratum
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                <lb/>
              DG, erit igitur quadratum ex AF ad quadratum ex DG, vt AF
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              ad HG. quare lineę AF DG HG ſunt proportionales. </s>
              <s id="N17748">Pro­
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              ducatur FB, ducaturquè à puncto D ipſi AB æquidiſtans
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              DK, erit vtiquè triangulorum ABF DKG anguli ABF
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              DHG æquales, & angulus AFB angulo DGK eſt æqualis, erit
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              igitur, & reliquus BAF reliquo KDG æqualis, ac propterea
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              triangulum ABF eſt triangulo DKG ſimile. </s>
              <s id="N17754">quare triangu­
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              lum ABF ad triangulum DKG eam habet proportionem,
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              quàm AF ad DG duplicatam, hoc eſt quàm AF ad HG, quę
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              eſt ea, quàm habet FB ad BG. atqui triangulum ABF ad
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              DKG eam quo〈que〉 habet proportionem, quam FB ad GK
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              duplicatam. </s>
              <s id="N17760">tres igitur lineę FB GK GB ſunt proportiona­
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              les. </s>
              <s id="N17764">ex quibus ſequiturita eſſe FB ad GK, vt AF ad DG; &
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              GK ad GB, vt DG ad GH. ſed quoniam triangulum GDK
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              ad GDB (cùm ſint ſub eadem altitudine) ita eſt, vt KG
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              BG, ſi igitur fiat HG ad L, vt KG ad BG, erit triangulum
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              GDK ad triangulum GDB, vt HG ad L. Cùm autem ſit
                <expan abbr="triã">triam</expan>
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              gulum ABF ad DKG, vt AF ad HG, eſtquè
                <expan abbr="triangulũ">triangulum</expan>
              DKG
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              ad DBG, vt HG ad L, erit ex ęquali triangulum ABF ad
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              triangulum DBG, vt AF ad L. ac propterea parabole ABC </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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