Similar polygons inscribed in circles, are to one another as the squares on their diameters. Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles: as the polygon ABCDE is to the polygon FGHKL, so shall the square on BM be to the square on GN. And because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: therefore the two triangles BAE, GFL having one angle in one equal to one angle in the other, and the sides about the equal angles proportionals, are equiangular; and therefore the angle AEB is equal to the angle FLG: but AEB is equal to AMB, because they stand upon the same circumference: (III. 21.) and the angle FLG is, for the same reason, equal to the angle FNG: therefore also the angle AMB is equal to FNG: and the right angle BAM is equal to the right angle GFN; (ш. 31.) wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another: therefore as BM to GN, so is BA to GF; (vI. 4.) and therefore 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 FGHKL is the duplicate of that which BA has to GF: (vi. 20.) therefore 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. Circles are to one another as the squares on their diameters. Let ABCD, EFGH be two circles, and BD, FH their diameters. As the square on BD to the square on FH, so shall the circle ABCD be to the circle EFGH. For, if it be not so, the square on BD must be to the square on FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it. the First, if possible, let it be to a space S less than a circle EFGII; 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, H, there be drawn tangents to the circle, square EFGH is half of the square described about the circle: (1. 47.) 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, HM, 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: (1. 41.) 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 EFGH 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 EFGH above_S: therefore 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: as therefore 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 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; wherefore the space S is greater than the polygon EKFLGMHN: (v. 14.) 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 ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that neither is the square on FH to the square on BD, as the circle EFGH is to any space less than the circle ABCD. Nor 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 it be so to T, a space greater than the circle EFGH: K N B therefore, inversely, as the square on FH to the square on BD, so is the space to the circle ABCD; but as the space T is to the circle ABCD, so is the circle EFGH to some space, which must be less than the circle ABCD, (v. 14.) because the space T'is greater, by hypothesis, than the circle EFGH; therefore 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 demonstrated to be impossible; therefore the square on BD is not to the square on FH, as the circle ABCD is to any space greater than the circle EFGH: and it has been demonstrated, that neither is the square on BD to the square on FH, as the circle ABCD to any space less than the circle EFGH: wherefore, as the square on BD to the square on FH, so is the circle ABCD to the circle EFGH. Circles, therefore, are, &c. q. E.D. PROPOSITION III. THEOREM. Every pyramid having a triangular base, 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. PROPOSITION IV. THEOREM. If there be two pyramids of the same altitude, upon triangular bases, 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 of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other, that are produced by the same number of divisions. PROPOSITION V. THEOREM. Pyramids of the same altitude which have triangular bases, are to one another as their bases. PROPOSITION VI. THEOREM. Pyramids of the same altitude which have polygons for their bases, are to one another as their bases. Every prism having a triangular base may be divided into three pyramids that have triangular bases, and are equal to one another. : COR. 1. From this it is manifest, that every pyramid is the third part of a prism which has the same base, and is of an equal altitude with it: for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases. COR. 2. Prisms of equal altitudes are to one another as their bases; because the pyramids upon the same bases, and of the same altitude, are to one another as their bases. Similar pyramids, having triangular bases, are one to another in the triplicate ratio of that of their homologous side. COR. From this it is evident, that similar pyramids which have multangular bases, are likewise to one another in the triplicate ratio of their homologous sides: for they may be divided into similar pyramids having triangular bases, because the similar polygons which are their bases, may be divided into the same number of similar triangles homologous to the whole polygons: therefore, as one of the triangular pyramids in the first multangular pyramid is to one of the triangular pyramids in the other, so are all the triangular pyramids in the first to all the triangular pyramids in the other; that is, so is the first multangular pyramid to the other: but one triangular pyramid is to its similar triangular pyramid, in the triplicate ratio of their homologous sides; and therefore the first multangular pyramid has to the other, the triplicate ratio of that which one of the sides of the first has to the homologous side of the other. PROPOSITION IX. THEOREM. The bases and altitudes of equal pyramids having triangular bases are reciprocally proportional: and, conversely, triangular pyramids of which the bases and altitudes are reciprocally proportionals, are equal to one another: PROPOSITION X. THEOREM. Every cone is the third part of a cylinder which has the same base, and is of an equal altitude with it. PROPOSITION XI. THEOREM. Cones and cylinders of the same altitude are to one another as thoir bases. PROPOSITION XII. THEOREM. Similar cones and cylinders have to one another the triplicate ratio of that which the diameters of their bases have. PROPOSITION XIII. THEOREM. If a cylinder be cut by a plane parallel to its opposite planes, or bases; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. PROPOSITION XIV. THEOREM. Cones and cylinders upon equal bases are to another as their altitudes. PROPOSITION XV. THEOREM. The bases and altitudes of equal cones and cylinders, are reciprocally proportional; and, conversely, if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. PROPOSITION XVI. PROBLEM. In the greater of two circles that have the same center, to inscribe a polygon of an even number of equal sides, that shall not meet the less circle. LEMMA II. If two trapeziums ABCD, EFGH be inscribed in the circles, the centers of which are the points K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal to one another; but the side AB greater than EF, and DC greater than HG; the straight line KA from the center of the circle in which the greater sides are, is greater than the straight line LE drawn from the center to the circumference of the other circle. PROPOSITION XVII. PROBLEM. In the greater of two spheres which have the same center, to describe a solid polyhedron, the superficies of which shall not meet the less sphere. COR. And if in the less sphere there be inscribed a solid polyhedron, by drawing straight lines betwixt the points in which the straight lines from the center of the sphere drawn to all the angles of the solid polyhedron in the greater sphere meet the superficies of the less; in the same order in which are joined the points in which the same lines from the center meet the superficies of the greater sphere; the solid polyhedron in the sphere BCDE has to this other solid polyhedron the triplicate ratio of that which the diameter of the sphere BCDE has to the diameter of the other sphere. For if these two solids be divided into the same number of pyramids, and in the same order, the pyramids shall be similar to one another, each to each: because they have the solid angles at their common vertex, the center of the sphere, the same in each pyramid, and their other solid angles at the bases equal to one another, each to each, because they are contained by three plane angles, each equal to each; and the pyramids are contained by the same number of similar planes; and are therefore similar to one another, each to each: but similar pyramids have to one another the triplicate ratio of their homologous sides. Therefore the pyramid of which the base is the quadrilateral KBOS, and vertex 4, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous sides, that is, of that ratio which AB from the center of the greater sphere has to the straight line from the same center to the superficies of the less sphere, and in like manner, each pyramid in the greater sphere has to each of the same order in the less, the triplicate ratio of that which AB has to the semi-diameter of the less sphere. And as one antecedent is to its consequent, so are all the antecedents to all the consequents. Wherefore the whole solid polyhedron in the greater sphere has to the whole solid polyhedron in the other, the triplicate ratio of that which AB the semi-diameter of the first has to the semi-diameter of the other; that is, which the diameter BD of the greater has to the diameter of the other sphere. PROPOSITION XVIII. THEOREM. Spheres have to one another the triplicate ratio of that which their diameters have. |