Again, if the arcs EK, KF, &c., be bisected, and their extremities be joined; and so on: there will at length remain segments of the circle, which taken together are less than the excess of the circle EFGH above the space S. For, by the foregoing lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there will at length remain a magnitude less than the least of the proposed magnitudes. Let the segments EK, KF, FL, LG, GM, MH, HN, and NE be those which remain, and are together less than the excess of the circle EFGH above S; therefore 4. The rest of the circle, viz., the polygon EKFLGMHN is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN; therefore 5. As the square on BD is to the square on FH, so is the polygon AXBOCPDR to the polygon EKFLGMHN (XII. 1); but the square on BD is also to the square on FH as the circle ABCD is to the space S (hyp.); therefore 6. As the circle ABCD is to the space S, so is the polygon AXBOCPDR to the polygon EKFLGMHN (V. 11). But the circle ABCD is greater than the polygon contained in it; therefore 7. The space S is greater than the polygon EKFLGMHN (V. 14); but it is also less, as has been proved, which is impossible; therefore 8. The square on BD is not to the square on FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner it may be demonstrated that the square on FH is not to the square on BD, as the circle EFGH is to any space less than the circle ABCD. Neither is the square on BD to the square on FH, as the circle ABCD is to any space greater than the circle EFGH. For, if possible, let the square on BD be to the square on FH, as the circle ABCD is to T, a space greater than the circle EFGH; therefore, inversely, 1. The square on FH is to the square on BD, as the space T is to the circle ABCD; but the space 7 is to the circle ABCD, as the circle EFGH is to some space which is less than the circle ABCD (V. 14); because the space T is greater than the circle EFGH (hyp.); therefore 2 As the square on FH is to the square on BD, so is the circle EFGH to a space less than the circle ABCD, which has been proved to be impossible; therefore 3. The square on BD is not to the square on FH as the circle ABCD is to any space greater than the circle EFGH. But it has been proved, that the square on BD is not to the square on FH, as the circle ABCD is to any space less than the circle EFGH; therefore 4. The square on BD is to the square on FH, as the circle ABCD is to the circle EFGH. Wherefore, circles, &c. Q.E.D. PRINTED BY G. PHILIP AND SON, LIVERPOOL. |