But the straight line AZ may be demonstrated to be greater Book XII. AB, that is, to the squares of AZ, ZB ; but the square of BZ s less than the square of GU; therefore the square of AZ is greater than the square of AG, and the straight line AZ conlequently greater than the straight line AG. Cor. And if in the lesser sphere there be described a solid polyhedron by drawing straight lines betwixt the points in which the straight lines from the centre of the sphere drawn to all the angles of the folid polyhedron in the greater sphere meet the superficies of the leffer; in the same order in which tre joined the points in which the same lines from the centre meet the superficies of the greater sphere ; the folid 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 tolids be divided into the same number of pyramids, and in the fame order ; the pyramids shall be fimilar to one another, each to each : Because they have the solid angles at their common vertex, the centre of the sphere, the same in each py, ramid, and their other folid angle at the bases equal to one another, each to cach “, because they are contained by three a B. 11. plane angles equal each to each ; and the pyramids are contained py the same number of similar planes; and are therefore fimilar bb 11. def. o one another, each to each : But fimilar pyramids have to ne another the triplicate ratio of their homologous fides.c Cor. Therefore the pyramid of which the base is the quadrilateral 12. BOS, and vertex A, has to the pyramid in the other sphere f the fame order, the triplicate ratio of their homologous T 2 fides ; 11. Book XII. fides ; that is, of that ratio which AB from the centre of the greater sphere has to the straight line from the same centre to the superficies of the lefser sphere. And in like manner, each pyramid in the greater sphere has to each of the same order in the leffer, the triplicate ratio of that which AB has to the femidiameter of the lefser 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 semidiameter of the first has to the semidiameter of the other ; that is, which the diameter BD of chc greater has to the diameter of the other sphere. SPheres have to one another the triplicate ratio of that which their diameters have. Let ABC, DEF be two spheres of which the diameters are BC, EF. The sphere ABC has to the sphere DEF the triplicate ratio of that which BC has to EF. For, if it has not, the sphere ABC shall have to a sphere cither less or greater than DEF, the triplicate ratio of that which BC has to EF. First, let it have that ratio to a less, viz. to the sphere GHK; and let the sphere DEF have the fame a 15. 12. ceatre with GHK; and in the greater sphere DEF describe a folid polyhedron, the superficies of which does not meet the lefler sphere GHK; and in the sphere ABC describe another fimilar to that in the sphere DEF: Therefore the folid polybe. dron in the sphere ABC has to the folid polyhedron in the sphere DEF, the triplicate ratio of that which BC has to EF. But the sphere ABC has to the sphere GHK, the triplicate ra. tio fed of pyramids, the bases of which are the aforesaid qua- Book XII. drilateral figures, and the triangle TRX, and those formed in the like manner in the rest of the sphere, the common vertex of them all being the point A : And the superficies of this solid polyhedron does not meet the lefser sphere in which is the circle FGH; For, from the point A draw AZ perpendicular a 11. 11. to the plane of the quadrilaterał KBOS meeting it in Z, and join BZ, ZK :. And because AZ is perpendicular to the plane KBOS, it makes right angles with every straight line meeting it in that plane ; therefore AZ is perpendicular to BZ and ZK: And because AB is equal to AK, and that the squares of AZ, ZB, are equal to the square of AB ; and the squares of Az, ZK to the square of AK'; therefore the squares of AZ, ZB 1.47. 11 are equal to the squares of AZ, ZK : Take from these equals the square of AZ, the remaining square of BZ is equal to the remaining square of ZK ; and therefore the straight line BZ is equal to ZK : In the like manner it may be demonstrated, that the straight lines drawn from the point Z to the points O, S are equal to BZ or ZK: Therefore the circle described from the centre Z, and distance ZB shall pass through the points K, O, S, and KBOS shall be a quadrilateral figure in the circle: And because KB is greater than QV, and QV equal to SO, therefore KB is greater than SO : But KB is equal to each of the straight lines BO, KS; wherefore each of the circumferences cut off by KB, BO, KS is greater than that cut off by OS ; and these three circumferences, together with a fourth equal to one of them, are greater than the same three together with that cut off by OS ; that is, than the whole circumference of the circle; therefore the circumference fubtended by KB is greater than the fourth part of the whole circumference of the circle KBOS, and consequently the angle BZK at the centre is greater than a right angle : And because the angle BZK is obruse, the square of BK is greater than the squares of BZ, ZK ;c 14. 2. thar is , greater than twice the square of Bz. Join KV, and because in the triangles KBV, OBV, KB, BV are equal to OB, BV, and that they contain equal angles; the angle KVB is equal to the angle OVB: And OVB is a right angle ; thered 4. s. fore also KVB is a right angle; And becaule BD is less than twice DV, the rectangle contained by DB, BV is less than twice the rectangle DVB; that is e, the square of KB is lesse 6. 6. than twice the square of KV: But the square of KB is greater than twice the square of BZ ; therefore the square of KV is T greater Book XII. greater than the square of BZ: And because BA is equal to AK, and that the squares of BZ, ZA are equal together to the square of BA, and the fquares of KV, VA to the square of AK; therefore the squares of BZ, ZA are equal to the squares of KV, VA: and of these the square of KV is greater than the square of BZ; therefore the square of VA is less than the square of ZA, and the straight line AZ greater than VA: Much more then is AZ greater than AG ; because, in the preceding propofition, it was fhewn that KV falls without the circle FGH: And AZ is perpendicular to the plane KBOS, and is therefore the shortest of all the straight lines that can be drawn from A, the centre of the sphere to that plane. Therefore the plane KBOS does not meet the leffer sphere. And that the other planes between the quadrants BX, KX fall without the leffer sphere, is thus demonstrated : From the point A draw AI perpendicular to the plane of the quadrilateral SOPT, and join 10; and, as was demonstrated of the plane KBOS and the point Z, in the same way it may be shewn that the point I is the centre of a circle described about SOPT: and that OS is greater than PT; and PT was fhewn to be parallel to OS : Therefore, because the two trapeziums KBOS, SOPT inscribed in circles have their lides BK, OS parallel, as alfo OS, PT ; and their other sides BO, KS, OP, ST all equal to one another, and that BK is greater than OS, and OS 12. Lem. greater than PT, therefore the straight line ZB is greater than 10. Join AO which will be equal to AB; and because AIO, AZB are right angles, the squares of AI, 10 are equal to the square of AO or of AB; that is, to the squares of AZ, ZB ; and the square of ZB is greater than the square of 10, therefore the square of AZ is less than the square of Al; and the straight line AZ less than the straight line Al : And it was proved that AZ is greater than AG; much more then is al greater than AG: Therefore the plane SOPT falls wholly without the lesser sphere: In the same manner it may be demonstrated that the plane TPRY falls without the fame sphere, as also the triangle YRX, viz. by the Cor. of 2d Lemma. And after the same way it may be demonstrated that all the planes which contain the solid polyhedron, fall without the leffer sphere. Therefore in the greater of two spheres which have the fame centre, a folid polyhedron is described, the superfices of which does not meet the letter sphere. Which was to be done. But given a great deal to do, both to ancient and modern geome. Book I. ters :, It seems not to be properly placed among the Axioms, as, indeed, it is not felf-evident; but it may be demonstrated thus : 2. DEFINITION 1. The distance of a point from a straight line, is the perpendicular drawn to it from the point, DEF. One straight line is said to go nearer to, or further from, an. other straight line, when the distance of the points of the first from the other straight line become less or greater than they were ; and two straight lines are said to keep the same distance from one another, when the distance of the points of one of them from the other is always the same. A XI Ο Μ. A straight line cannot first come nearer to another straight line, and then go further from А it, before it cuts it; and, in like B manner, a straight line cannot D E go further from another straight line, and then come nearer to F G H it; nor can a straight line keep the same distance from another straight line, and then come nearer to it, or go further from it; for a straight line keeps always the same direction. For example, the straight line ABC cannot first come nearer to the straight line DE, as from the point A to the point A B See the bo gure above B, and then, from the point B D to the point C, go further from the same DE: And, in like man. G H per, the straight line FGH can. go further from DE, as from F to G, and then, from G to H, come nearer to the fame DE: And so in the last case, as in CE丑 not fig. 2. PRO P. 1. If two equal straight lines AC, BD, be each at right angles to the fame straight line AB; if the points C, D be joined by the straight line CD, the straight line EF drawn from any point E in AB unto CD, at right angles to AB, shall be equal to AC, Or BD If EF be not equal to AC, one of them must be greater than the other; let AC be the greater; then, because FE is |