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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 folid polyhedron does not meet the leffer sphere in which is the circle FGH: For, from the point A draw* AZ perpendicular a It, Ir. to the plane of the quadrilateral 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 of AZ, ZB, are equal to the square of AB; and the squares of AZ, ZK to the square of AKO; therefore the squares of AZ, ZB 47. I. are equal to the squares of AZ, ZK: Take from thefe equals the square of AZ, the remaining square of BZ is equal to the remaining square of ZK ; and therefore the straigh is equal to ZK: In the like manner it may be demonstrated, that the straight lines drawn from the point Z to the points , S are equal to BZ or ZK: Therefore the circle described from the centre Z, and distance ZB shall pats through the points K, 0, S, and KLO, fhall be a quadrilateral figure in the circle : And becaule KB is greater than QV, and V equal to SO, therefore KB is greater than 80: But KB is equal to each of the ftraight 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 subtended 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 greatter than a right angle : 'And because the angle BZK is obtule, the square of BK is greater than the squares of BZ, ZK ; c 12. 2. that is, greater than twice the square of BZ. Join KV, and becaute 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; there- d 4.1, tore also KVB is a right angle: And becaute BD is less than twice DV, the rectangle contained by DB, BV is less than twice the rectangle DVB; that is, the square of KB is less
c 8. 6. than twice the square of KV : But the square of KB is greater tban twice the Iquare of BZ; therefore the square of KV is T
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 squares of KV, VÀ 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 fquare of ZA, and the straight line AZ greater than VA: Much more then is AZ greater than AG; because, in the preceding proposition, it was shown that KV falls without the circle FGH: And AZ is perpendicular to the plane KBOS, and is therefore the shorteft 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 lefser fphere.
And that the other planes between the quadrants BX, KX fall without the leffer sphere, is thus demonftrated : From the point A draw Al 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 shown that the point I is the centre of a circle described about SOPT: and that OS is greater than PT; and PT was shown to be parallel to OS: Therefore, because the two trapeziums KBOS, SOPT inscribed in circles have their fides BK, OS parallel, as also OS, PT; and their other lides BO, KS, OP, ST all equal
to one another, and that BK is greater than Os, and OS a 2. 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 Al, 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 AI: And it was proved that AZ is greater than AG; much more then is AI greater than AG: Therefore the plane SOPT falls wholly without the lefser sphere : In the same manner it may be demonstrated that the plane TPRY falls without the fame sphere, as also the triangle TRX, viz. by the Cor. of ad Lemma. And after the same way it may be demonstrated that all the planes which contain the folid polyhedron, fall without the leffer sphere. Therefore in the greater of two spheres which have the same centre, a folid polyhedron is described, the superficies of which does not meet the lefser sphere. Which was to be done.
But the straight line AZ may be demonstrated to be greater Book XII. than AG otherwise, and in a shorter manner, without the help of Prop. 16. as follows. From the point G draw GU at right angles to AG and join AU. If then the circumference BE he bisected, and its half again bisected, and so on, there will at length be left a circumference less than the circumference which is fubtended by a straight line equal to GU inscribed in the circle BCDE: Let this be the circumference KB: Therefore the straight line KB is less than GU: And because the angle BZK is obtuse, as was proved in the preceding, therefore BK is greater than BZ: But GU is greater than BK; much more then is GU greater than BZ, and the square of GU than the square of BZ; and AU is equal to AB ; therefore the square of AU, that is, the squares of AG, GU are equal to the Iguare of AB, that is, to the squares of AZ, ZB; but the square of BZ is lets than the square of GU ; therefore the square of AZ is greater than the square of AG, and the straight line AZ confequently greater than the straight line AG.
Cor. And if in the lefser 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 solid polyhedron in the greater sphere meet the superficies of the lefler; in the same order in which are 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 solids be divided into the same number of pyramids, and in the same order; the pyramids shall be similar to one ano. ther, each to each: Because they have the folid angles at their common vertex, the centre of the sphere, the same in each pysamid, and their other solid angle at the bases equal to one another, each to each“, because they are contained by three a B. 11, plane angles equal each to each; and the pyramids are contained by the same number of similar planes ; and are therefore similar b b 11. def. to one another, each to each : But similar pyramids have to one another the triplicate ratio of their homologous fides. c Cor. 11. Therefore the pyramid of which the base is the quadrilateral KBOS, and vertex A, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous
Τ Η Ε Ε L Ε Μ Ε Ν Τ S
Book XII. Gdes ; that is, of that ratio which AB from the centre of the
greater sphere has to the straight line from the fame centre to the superficies of the leffer sphere. And in like manner, each pyramid in the greater sphere has to each of the same order in the lefer, the triplicate ratio of that which AB has to the semidiameter 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 firft has to the semidiameter of the other ; that is, which the diameter BD of the greater has to the diameter of the other sphere.
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 either 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 same centre with GHK ; and in the greater sphere DEF describe
* 17. 12.
a solid polyhedron, the superficies of which does not meet the leffer sphere GHK; and in the fphere ABC describe another similar to that in the sphere DEF: Therefore the folid polyhe
dron in the sphere ABC has to the solid polyhedron in the Cor. 17. fphere DEF, the triplicate ratio of that which BC has to EF. 12. But the sphere ABC has to the sphere GHK, the triplicate ra.
tio of that which BC has to EF; therefore, as the sphere ABC Book X!l. to the sphere GHK, so is the solid polyhedron in the sphere ABC to the solid polyhedron in the sphere DEF: But the sphere ABC is greater than the folid polyhedron in it ; therefore al.C 14. 5. so the sphere GHK is greater than the solid polyhedron in the sphere DEF : But it is also less, because it is contained within it, which is impossible: Therefore the sphere ABC has not to any sphere less than DEF, the triplicate ratio of that which BC has to EF. In the same manner, it may be demonstrated, that the sphere DEF has not to any sphere less than ABC, the triplicate ratio of that which EF has to BC. Nor can the sphere ABC have to any sphere greater than DEF, the triplicate ratio of that which BC has to EF: For, if it can, let it have that ratio to a greater sphere LMN : Therefore, by inverfion, the sphere LMN has to the sphere ABC, the triplicate ratio of that which the diameter EF has to the diameter BC. But, as the sphere LMN to ABC, so is the sphere DEP to some sphere, which must be less than the sphere ABC, because the fphere LMN is greater than the sphere DEF: Therefore the spberc DEF has to a sphere less than ABC the triplicate ratio of that which EF has to BC; which was thewn to be impoffible: Therefore the sphere ABC has not to any sphere greater than DEF the triplicate ratio of that which BC has to EF : And it was demonstrated, that neither has it that ratio to any sphere Jess than Der. Therefore the sphere ABC has to the sphere DEF, tbe triplicate ratio of that which BC has to EF. Q. E. D.