Demonstration. Because the polygon ABCDE is similar to the polygon FGHKL; therefore 1. The angle BAE is equal to the angle GFL; and BA is to AE as GF is to FL; therefore the two triangles BAB, GFL are equiangular (VI. 6); and therefore 2. The angle A EB is equal to the angle FLG. But the angle AEB is equal to the angle AMB, because they stand upon the same arc (III. 21). For the same reason, the angle FLG is equal to the angle ÀNG; therefore also 3. The angle AMB is equal to the angle FNG; And the angle BAM is equal to the angle GFN (III. 31); therefore 4. The remaining angles in the triangles ABM and FGN are equal, and they are equiangular to one another; wherefore BM is to GN as BA is to GF (VI. 4); therefore 5. The duplicate ratio of BM to GN is the same with the duplicate ratio of BA to GF (V. Def. 10 and V. 22); but the ratio of the square on BM to the square on GN is the duplicate ratio of that which BM has to GN (VI, 20); and the ratio of the polygon ABCDE to the polygon FÄHKL is the duplicate of that which BA has to GF (VI, 20); therefore 6. As the polygon ABCDE is to the polygon FGHKL so is the square on BM to the square on GN. Wherefore, similar polygons, &c. Q.E.D. PROPOSITION 2.-Theorem. Circles are to one another as the squares on their diameters. Let ABCD and EFGH be two circles, and BD and FH their diameters. Then the square on BD is to the square on FH as the circle ABCD is to the circle EFGI. For, if not, the square on BD is to the square on FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it. First, let it be to a space S less than the circle EFGH; and in the circle EFGH inscribe the square EFGH (IV. 6). This square is greater than half of the circle EFGH; because, if through the points E, F, G, and H, tangents be drawn to the circle, the square EFGH is half of the square described about the circle (I. 41). But the circle is less than the square described about it; therefore 1. The square EFGH is greater than half the circle. Bisect the arcs EF, FG, GH, and HE at the points K, L, M, and N, and join EK, KF, FL, LG. GM, MH, HN, and NE; then 2. Each of the triangles EKF and FLG is greater than half of the segment which contains it. For if straight lines touching the circle be drawn through the points K, L, M, and N, and the parallelograms upon the straight lines EF, FG, GH, and HE be completed, each of the triangles EKF, FLG, GMH, and HNE is the half of the parallelogram which contains it (I. 41); but every segment is less than the parallelogram which contains it; therefore 3. Each of the triangles EKF, FLG, GMA, and HNE is greater than half the segment which contains it. 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 $. 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 ERFLGMHN (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 T 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. |