Casati, Paolo, Fabrica, et uso del compasso di proportione, dove insegna à gli artefici il modo di fare in esso le necessarie divisioni, e con varij problemi ...

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              <pb o="28" file="0040" n="41" rhead="CAPO II."/>
            mità A, taglieranno il diametro AB ad angoli retti in O, M,
              <lb/>
            P &</s>
            <s xml:id="echoid-s536" xml:space="preserve">c. </s>
            <s xml:id="echoid-s537" xml:space="preserve">E così le linee per-
              <lb/>
            pendicolari alla AB ſa-
              <lb/>
              <figure xlink:label="fig-0040-01" xlink:href="fig-0040-01a" number="14">
                <image file="0040-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0040-01"/>
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            ranno parallele trà di lo-
              <lb/>
            ro, & </s>
            <s xml:id="echoid-s538" xml:space="preserve">ordinatamente
              <lb/>
            applicate così al diame-
              <lb/>
            tro del circolo, come all’
              <lb/>
            Aſſe maggiore dell’ El-
              <lb/>
            lipſi.</s>
            <s xml:id="echoid-s539" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s540" xml:space="preserve">Mettanſi dunque ciaſ-
              <lb/>
            cuna delle applicate nel
              <lb/>
            circolo ad vn numero
              <lb/>
            della linea Aritmetica,
              <lb/>
            che habbia vn’altro nu-
              <lb/>
            mero, à cui ella ſia come
              <lb/>
            5 à 3, come ſaria 50, 50;
              <lb/>
            </s>
            <s xml:id="echoid-s541" xml:space="preserve">e 30, 30: </s>
            <s xml:id="echoid-s542" xml:space="preserve">perche il ſecon-
              <lb/>
            do interuallo 30, 30, darà l’Applicata dell’Ellipſi: </s>
            <s xml:id="echoid-s543" xml:space="preserve">Così OR
              <lb/>
            ad OV; </s>
            <s xml:id="echoid-s544" xml:space="preserve">MF ad MN; </s>
            <s xml:id="echoid-s545" xml:space="preserve">PI à PQ, e così ſuſſeguentemente,
              <lb/>
            ſaranno come 5 à 3, e pigliaraſſi ad OV vguale OG, & </s>
            <s xml:id="echoid-s546" xml:space="preserve">à
              <lb/>
            MN vguale MK &</s>
            <s xml:id="echoid-s547" xml:space="preserve">c. </s>
            <s xml:id="echoid-s548" xml:space="preserve">perche la linea tirata per li punti
              <lb/>
            Q, N, V, A, G, K, &</s>
            <s xml:id="echoid-s549" xml:space="preserve">c. </s>
            <s xml:id="echoid-s550" xml:space="preserve">ſarà Elliptica.</s>
            <s xml:id="echoid-s551" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s552" xml:space="preserve">Ciò ſi demoſtra, perche nell’ Ellipſi i Quadrati delle Ap-
              <lb/>
            plicate hanno la proportione delli rettangoli fatti dalli ſeg-
              <lb/>
            menti del diametro, à cui ſono Applicate: </s>
            <s xml:id="echoid-s553" xml:space="preserve">e nel circolo i
              <lb/>
            Quadrati delle perpendicolari OR, MF ſono vguali alli ret-
              <lb/>
            tangoli AOB, AMB fatti dalli ſteſſi ſegmenti: </s>
            <s xml:id="echoid-s554" xml:space="preserve">dunque co-
              <lb/>
            me il Quadrato di OV al Quadrato di MN, così il Quadra-
              <lb/>
            to di OR al Quadrato di MF. </s>
            <s xml:id="echoid-s555" xml:space="preserve">Dunque per la 22. </s>
            <s xml:id="echoid-s556" xml:space="preserve">del 6. </s>
            <s xml:id="echoid-s557" xml:space="preserve">co-
              <lb/>
            me OV ad MN, così OR ad MF, e permutando come </s>
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