Salusbury, Thomas, Mathematical collections and translations (Tome I), 1667

Table of figures

< >
[Figure 31]
Page: 447
[Figure 32]
Page: 447
[Figure 33]
Page: 447
[Figure 34]
Page: 589
[Figure 35]
Page: 592
[Figure 36]
Page: 593
[Figure 37]
Page: 594
[Figure 38]
Page: 602
[Figure 39]
Page: 606
[Figure 40]
Page: 607
[Figure 41]
Page: 608
[Figure 42]
Page: 610
[Figure 43]
Page: 611
[Figure 44]
Page: 612
[Figure 45]
Page: 614
[Figure 46]
Page: 617
[Figure 47]
Page: 620
[Figure 48]
Page: 625
[Figure 49]
Page: 628
[Figure 50]
Page: 659
[Figure 51]
Page: 662
[Figure 52]
Page: 698
[Figure 53]
Page: 700
[Figure 54]
Page: 701
[Figure 55]
Page: 704
[Figure 56]
Page: 710
[Figure 57]
Page: 715
[Figure 58]
Page: 723
[Figure 59]
Page: 729
[Figure 60]
Page: 733
< >
page |< < of 701 > >|
    <archimedes>
      <text>
        <body>
          <chap>
            <p type="main">
              <s>
                <pb xlink:href="040/01/729.jpg" pagenum="37"/>
              ſect A E, the Angle A being acute: Let the interſection be in E,
                <lb/>
              from whence let fall a Perpendicular to A E, which produced, will
                <lb/>
              meet with A B infinitely prolonged in F. </s>
              <s>I ſay, firſt, that the
                <lb/>
              Right-lines F E, and F C are equal: ſo that drawing the Line
                <lb/>
              E C, we ſhall, in the
                <lb/>
                <figure id="id.040.01.729.1.jpg" xlink:href="040/01/729/1.jpg" number="59"/>
                <lb/>
              two Triangles D E C,
                <lb/>
              B E C, have the two
                <lb/>
              Sides of the one, D E,
                <lb/>
              and C E, equal to the
                <lb/>
              two Sides of the other
                <lb/>
              B E, and E C; the
                <lb/>
              two Sides, D E, and
                <lb/>
              E B, being Tangents
                <lb/>
              to the Circle D B,
                <lb/>
              and the Baſes D C,
                <lb/>
              and C B, are likewiſe
                <lb/>
              equal: wherefore the
                <lb/>
              two Angles D E C,
                <lb/>
              and B E C, ſhall be
                <lb/>
              equal. </s>
              <s>And becauſe the Angle B C E wanteth of being a Right­
                <lb/>
              Angle, as much as the Angle B E C; and the Angle C E F, to
                <lb/>
              make it a Right-Angle, wants the Angle C E D, thoſe Supple­
                <lb/>
              ments being equal, the Angles F C E, and F E C ſhall be equal,
                <lb/>
              and ſo conſequently the Sides F E, and F C; wherefore making
                <lb/>
              the point F a Center, and at the diſtance F E, deſcribing a Circle,
                <lb/>
              it ſhall paſs by the point C. </s>
              <s>Deſcribe it, and let it be C E G. </s>
              <s>I ſay,
                <lb/>
              that this is the Circle required, by any point of the Circumfe­
                <lb/>
              rence of which, any two Lines that ſhall interſect, departing from
                <lb/>
              the terms A and B, ſhall be in proportion to each other, as are the
                <lb/>
              two parts A C, and B C, which beſore did concur in the point C.
                <lb/>
              </s>
              <s>This is manifeſt in the two that concur or interſect in the point E,
                <lb/>
              that is A E, and B E; the Angle E of the Triangle A E B being
                <lb/>
              divided in the midſt by C E; ſo that as A C is to C B, ſo is A E
                <lb/>
              to B E. </s>
              <s>The ſame we prove in the two A G, and B G, determined
                <lb/>
              in the point G. </s>
              <s>Therefore being (by the Similitude of the Tri­
                <lb/>
              angles A F E, and E F B) that as A F is to E F, ſo is E F to F B;
                <lb/>
              that is, as A F is to F C, ſo is C F to F B: So by Diviſion; as A C
                <lb/>
              is to C F, (that is, to F G) ſo is C B to B F; and the whole A B
                <lb/>
              is to the whole B G, as the part C B to the part B F: and by Com­
                <lb/>
              poſition; as A G is to G B, ſo is C F to F B; that is, as E F to
                <lb/>
              F B, that is, as A E to E B, and A C to C B: Which was to be de­
                <lb/>
              monſtrated. </s>
              <s>Again, let any other Point be taken in the Circum­
                <lb/>
              ference, as H; in which the two Lines A H and B H concur. </s>
              <s>I ſay, in
                <lb/>
              like manner as before, that as A C is to C B, ſo is A H to B H.
                <lb/>
              </s>
              <s>Continue H B untill it interſect the Circumference in I, and draw </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
Impressum