Let the strait line F H be drawn of any length, making right angles with A F at F; and drawing the strait line A H, let the triangle A F H be made. Then upon the same centre A let any arch I K be drawn, cutting A D in L; and its chord, cutting A D in M; and dividing M L equally in N, let N O be drawn parallel to the strait line F H, and meeting with the strait line A H in O. Seeing now B D C is the spherical superficies of the portion cut off with a plane passing through B C, and cutting the axis at right angles; and seeing F H divides E D, the axis of the portion, into two equal parts in F; the centre of equiponderation of the superficies B D C will be in F (by art. 8); and for the same reason the centre of equiponderation of the superficies I L K, K being in the strait line A C, will be in N. And in like manner, if there were drawn, between the centre of the sphere A and the outermost spherical superficies of the sector, arches infinite in number, the centres of equiponderation of the spherical superficies, in which those arches are, would be found to be in that part of the axis, which is intercepted between the superficies itself and a plane passing along by the chord of the arch, and cutting the axis in the middle at right angles. Let it now be supposed that the moment of the outermost spherical superficies B D C is F H. Seeing therefore the superficies B D C is to the superficies I L K in proportion duplicate to that of the arch B D C to the arch I L K, that is, of B E to I M, that is, of F H to N O; let it be as F H to N O, so N O to another N P; and again, as N O to N P, so N P to another N Q; and let this be done in all the strait lines parallel to the base F H that can possibly be drawn between the base and the vertex of the triangle A F H. If then through all the points Q there be drawn the crooked line A Q H, the figure A F H Q A will be the complement of the first three-sided figure of two means; and the same will also be the moment of all the spherical superficies, of which the solid sector A B C D is compounded; and by consequent, the moment of the sector itself. Let now F H be understood to be the semidiameter of the base of a right cone, whose side is A H, and axis A F Wherefore, seeing the bases of the cones, which pass through F and N and the rest of the points of the axis, are in proportion duplicate to that of the strait lines F H and N O, &c., the moment of all the bases together, that is, of the whole cone, will be the figure itself A F H Q A; and therefore the centre of equiponderation of the cone A F H is the same with that of the solid sector. Wherefore, seeing A G is ¾ of the axis A F, the centre of equiponderation of the cone A F H is in G; and therefore the centre of the solid sector is in G also, and divides the part A F of the axis so that A G is triple to G F; that is, A G is to A F as 3 to 4; which was to be demonstrated.