PROPOSITION II. THEOREM.-Circles (AC, EG) are to one another as the squares on their diameters. DEMONSTRATION. For if it be not so, the square on BD must be to the square on FH, as the circle AC is to some space either less than the circle EG, or greater than it. First let it be to a space S less than the circle EG; and in the circle EG describe the square EFGH (a). This square is greater than half of the circle EG; because, if, through the points E, F, G, H there be drawn tangents to the circle, the square EFGH is half of the square described about the circle (b): and the circle is less than the square described about it, therefore the square EFGH is greater than half of the circle. Divide the circumferences EF, FG, GH, HE each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: therefore each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines EF, FG, GH, HE be completed, each of the triangles EKF, FLG, GMH, HNE is the half of the parallelogram in which it is (6); but every segment is less than the parallelogram in which it is: wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it. Again, if the remaining circumferences be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length remain segments of the circle, which together are less than the excess of the circle EG above the space S; because, by the preceding 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 shall at length remain a magnitude less than the least of the proposed magnitudes. Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those that remain, and are together less than the excess of the circle EG above S: therefore the rest of the circle, viz. the polygon EKFLGMHN, is greater than the space S. Describe likewise in the circle AC the volygon AXBOCPDR similar to the polygon EKFLGMHN: as therefore the square on BD is to the square on FH, so is the polygon AXBOCPDR to the polygon EKFLGMHŃ (c): but the square on BD is also to the square on FH, as the circle AC is to the space S (d); therefore as the circle AC is to the space S, so is the polygon AXBOCPDR to the polygon EKFLGMÂN (e): but the circle AC is greater than the polygon contained in it; wherefore the space S is greater than the polygon EKFLGMHN (ƒ): but it is likewise less, as has been demonstrated; which is impossible: therefore the square on BD is not to the square on FH, as the circle AC is to any space less than the circle EG. In the same manner it may be demonstrated, that neither is the square on FH to the square on BD, as the circle EG is to any space less than the circle AC. Nor is the square on BD to the square on FH, as the circle AC is to any space greater than the circle EG. For if possible, let it be so to T, a space greater than the circle EG: therefore inversely, as the square on FH is to the square on BD, so is the space T to the circle AC: but as the space T is to the circle AC, so is the circle EG to some space, which must be less than the circle AC (f), because the space T is greater, by hypothesis, than the circle EG; therefore as the square on FH is to the square on BD, so is the circle EG to a space less than the circle AC, which has been demonstrated to be impossible: therefore the square on BD is not to the square on FH, as the circle AC is to any space greater than the circle EG: and it has been demonstrated that neither is the square on BD to the square on FH, as the circle AC to any space less than the circle EG: wherefore, as the square on BD is to the square on FH, so is the circle AC to the circle EG. PROPOSITION III. THEOREM. Every pyramid having a triangular base (ABC) may be divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid; and into two equal prisms which together are greater than half of the whole pyramid. B K D E FC (a) VI. 2. (b) I. 34. DEMONSTRATION. Divide AB, BC, CA, AD, DB, DC each into two equal parts in the points E, F, G, H, K, L, and join EH, EG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel to DB (a): for the same reason, HK is parallel to AB; therefore HEBK is a parallelogram, and HK equal to EB (6): but EB is equal to AE; therefore also AE is equal to HK: and AH is equal to HD; wherefore EA, AH are equal to KH, HD, each to each; and the angle EAH is equal to the angle KHD (c); therefore the base EH is equal to the base KD, and the triangle AEH equal and similar to the triangle HKD (d): for the same reason, the triangle AGH is equal and similar to the triangle HLD. Again, because the two straight lines EH, HG, which meet one another, are parallel to KD, DL, that meet one another and are not in the same plane with them, they contain equal angles (e); therefore the angle EHG is equal to the angle KDL: and because EH, HG are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL; therefore the base EG is equal to the base KL; and the triangle EHG equal and similar to the triangle KDL (d): for the same reason, the triangle AEG is also equal and similar to the triangle HKL: therefore the pyramid, of which the base is the triangle AEG, and of which the vertex is the point H, is equal and similar to the pyramid, the base of which is the triangle KHL, and vertex the point ́D (ƒ). · And because HK is parallel to AB, a side of the triangle ADB, the triangle ADB is equiangular to the triangle HDK, and their sides are proportionals (g); therefore the triangle ADB is similar to the triangle HDK: and for the same reason, the triangle DBC is similar to the triangle DKL; and the triangle ADC to the triangle HDL; and also the triangle ABC to the triangle AEG: but the triangle AEG (k) I. 41. is similar to the triangle HKL, as before was proved; therefore the triangle ABC is similar to the triangle HKL (h): and therefore the pyramid of which the base is the triangle ABC, and vertex the point D, is similar to the pyramid of which the base is the triangle HKL, and vertex the same point D (2): but the pyramid of which the base is the triangle HKL, and vertex the point D, is similar, as has been proved, to the pyramid the base of which is the triangle AEG, and vertex the point H; wherefore the pyramid, the base of which is the triangle ABC, and vertex the point D, is similar to the pyramid of which the base is the triangle AEG and vertex H: therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD. And because BF is equal to FC, the parallelogram EBFG is double of the triangle GFC (k): but when there are two prisms of the same altitude, of which one has a parallelogram for its base, and the other a triangle that is half of the parallelogram, these prisms are equal to one another (); therefore the prism having the parallelogram EBFG for its base, and the straight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it; for they are of the same altitude, because they are between the parallel planes ABC, HKL (m): and it is manifest that each of these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the bases, and the vertices the points H, D; because, if EF be joined, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K: but this pyramid is equal to the pyramid, the base of which is the triangle AEG, and vertex the point H (f); because they are contained by equal and similar planes: wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triangle AEG, and vertex the point H: and the prism of which the base is the parallelogram EBFG, and opposite side KH, is equal to the prism having the triangle GFC for its base, and HKL the triangle opposite to it; and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D: therefore the two prisms before mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore, the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids similar to one another, and to the whole pyramid; and into two equal prisms; and the two prisms are together greater than half of the whole pyramid. 1 PROPOSITION IV. THEOREM.-If there be two pyramids (ABCG, DEFH) of the same altitude upon triangular bases (ABC, DEF), and each of them be divided into two equal pyramids similar to the whole pyramid, and also into two equal prisms; and if each of these pyramids be divided in the same manner as the first two, and so on: as the base (ABC) of one of the first two pyramids is to the base (DEF) of the other, so shall all the prisms in one of them (ABCG) be to all the prisms in the other (DEFH), that are produced by the same number of divisions. DEMONSTRATION. Make the same construction as in the foregoing proposition: and because BX is equal to XC, and AL to LC, therefore XL is parallel to AB (a), and the triangle ABC similar to the triangle LXC: for the same reason, the triangle DEF is similar to RVF. And because BC is double of CX, and EF double of FV; therefore BC is to CX, as EF is to FV (b) : and upon BC, CX are described the similar and similarly situated rectilineal figures ABC, LXC: and upon EF, FV, in like manner are described the similar figures DEF, RVF: therefore, as the triangle ABC is to the tri angle LXC, so is the triangle DEF to the triangle RVF (c), and, by permutation, as the triangle ABC is to the triangle DEF, so is the triangle LXC to the triangle RVF. And because the planes ABC, OMN, as also the planes DEF, STY, are parallel (d), the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the hypothesis, are equal to one another, shall be cut each into two equal parts by the planes OMN, STY (e), because the straight lines GC, HF are cut into two equal parts in the points N, Y, by the same planes: therefore the prisms LXCOMN, RVFSTY are of the same altitude; and therefore, as |