Harriot, Thomas, Mss. 6785

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                <emph style="bf">Commentary:</emph>
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              <s xml:space="preserve"> Some working for Lemma II, preceding Problem IX of
                <emph style="it">Apollonius Gallus</emph>
                <ref id="Viete_1600a" target="http://www.e-rara.ch/zut/content/pageview/2684204"> (Viete 1600a, Prob 9, Lemma </ref>
              . Harriot notes some cases missed by Viète. </s>
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                <s xml:space="preserve"> Lemma II.
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                Sint duo circuli, unius ABCD, alter EFGH; jungens autem eorum centra KL secet circulum primum in A, D; secundum vero in E, H; & in ea sumatur M punctum, a quo acta MGFCB recta secet circulm primum in B, C, secundum in F, G, & sint similia segmenta, & puncta quidem sectionum A, B sint remotior a ipsis C, & puncta F, E ipsis C, H. Ajo id quod fit sub MG, MB aequari id quod fit sub MH, </s>
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                <s xml:space="preserve"> Lemma II. Let there be two circles ABCD and EFGH; moreover the line KL joining their centres cuts the first circle in A and D, and the second in E and H; and in that line there is taken the point M, from which the straight line MGFCB cuts the first circle in B and C, and the second in F and G, and the segments are similar, and the points A, B are further away than C, D, and the points F, E than C, H. I say that the rectangle formed by MG, MB is equal to that formed by MH, </s>
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          <head xml:space="preserve" xml:lang="lat"> Apol: Gallus. Lemma 2. pag.
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          [
            <emph style="bf">Translation: </emph>
          Apollonius Gallus, Lemma II, page ]</head>
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