Apollonius <Pergaeus>; Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

Table of handwritten notes

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[Handwritten note 1]
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[Handwritten note 2]
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[Handwritten note 2]
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[Handwritten note 4]
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[Handwritten note 5]
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          <head xml:id="echoid-head96" xml:space="preserve">LEMMA IV.</head>
          <p>
            <s xml:id="echoid-s2248" xml:space="preserve">If EK and IY (Fig. </s>
            <s xml:id="echoid-s2249" xml:space="preserve">27.) </s>
            <s xml:id="echoid-s2250" xml:space="preserve">be any perpendiculars to the diameter AU of
              <lb/>
            a circle AYUV, terminating in the circumference, and if KY be drawn,
              <lb/>
            on which, from U, the perpendicular UF is demitted; </s>
            <s xml:id="echoid-s2251" xml:space="preserve">then will KF be a
              <lb/>
            mean proportional between AI and EU, alſo YF a mean proportional
              <lb/>
            between AE and IU.</s>
            <s xml:id="echoid-s2252" xml:space="preserve"/>
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          <p>
            <s xml:id="echoid-s2253" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s2254" xml:space="preserve">Draw UY, UK, KA and AY, the angles I and F
              <lb/>
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            being right by conſtruction, and the angles IDU, and FKV equal, being
              <lb/>
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            both equal to the angle UAY, the triangles IYU and FKU are ſimilar,
              <lb/>
            and conſequently IY is to UY as KF is to UK; </s>
            <s xml:id="echoid-s2255" xml:space="preserve">or (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2256" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s2257" xml:space="preserve">22.) </s>
            <s xml:id="echoid-s2258" xml:space="preserve">the
              <lb/>
            ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare
              <lb/>
            on UK: </s>
            <s xml:id="echoid-s2259" xml:space="preserve">now the ſquare on IY is (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2260" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s2261" xml:space="preserve">8. </s>
            <s xml:id="echoid-s2262" xml:space="preserve">17.) </s>
            <s xml:id="echoid-s2263" xml:space="preserve">equal to the rectangle
              <lb/>
            contained by AI and IU, the ſquare on UY to the rectangle contained by
              <lb/>
            AU and IU, and the ſquare on UK to the rectangle contained by AU and
              <lb/>
            EU; </s>
            <s xml:id="echoid-s2264" xml:space="preserve">wherefore the rectangle contained by AI and IU is to that contained
              <lb/>
            by AU and IU as the ſquare on KF is to the rectangle contained by AU
              <lb/>
            and UE, whence (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2265" xml:space="preserve">V. </s>
            <s xml:id="echoid-s2266" xml:space="preserve">15.) </s>
            <s xml:id="echoid-s2267" xml:space="preserve">AI is to AU as the ſquare on KF is to the
              <lb/>
            rectangle contained by AU and UE, or the rectangle contained by AI and
              <lb/>
            UE is to that contained by AU and UE as the ſquare on KF is to the
              <lb/>
            rectangle contained by AU and UE; </s>
            <s xml:id="echoid-s2268" xml:space="preserve">ſeeing then that the conſequents are
              <lb/>
            here the ſame, the antecedents muſt be equal, and therefore AI is to KF
              <lb/>
            as KF is to UE.</s>
            <s xml:id="echoid-s2269" xml:space="preserve"/>
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          <p>
            <s xml:id="echoid-s2270" xml:space="preserve">Again, the angle AKE is equal to AUK, which is equal to the angle
              <lb/>
            AYK, of which the angle UYF is the complement, becauſe AYU is a
              <lb/>
            right angle; </s>
            <s xml:id="echoid-s2271" xml:space="preserve">and therefore as the angles F and E are both right, the tri-
              <lb/>
            angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF,
              <lb/>
            wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU
              <lb/>
            is to the ſquare on YF: </s>
            <s xml:id="echoid-s2272" xml:space="preserve">but the ſquare on AK is equal to the rectangle
              <lb/>
            contained by AU and AE, and the ſquare on YU is equal to the rectangle
              <lb/>
            contained by AU and IU, conſequently the rectangle contained by AU
              <lb/>
            and AE is to the ſquare on AE as the rectangle contained by AU and IU
              <lb/>
            is to the ſquare on YF; </s>
            <s xml:id="echoid-s2273" xml:space="preserve">whence (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2274" xml:space="preserve">V. </s>
            <s xml:id="echoid-s2275" xml:space="preserve">15.) </s>
            <s xml:id="echoid-s2276" xml:space="preserve">AU is to AE as the rect-
              <lb/>
            angle contained by AU and IU is to the ſquare on YF, or the rectangle
              <lb/>
            contained by AU and IU is to that contained by AE and IU as the </s>
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