« ΠροηγούμενηΣυνέχεια »
Book XII. BCDE, meeting the superficies of the sphere in the point X; and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, will produce great circles on the superficies of the sphere, and let BXD, KXN be the semicircles thus made upon the diameters BD, KN: Therefore, because XA is at right angles to the plane of the circle BCDE, every plane which passes through XA is at c 18. 11. right angles to the plane of the circle BCDE; wherefore the semicircles BXD, KXN are at right angles to that plane: And because the semicircles BED, BXD, KXN, upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, are equal to one another: Therefore, as many sides of the polygon as are in BE, so many there are in BX, KX equal to the sides BK, KL, LM, ME: Let these polygons be described, and their sides be BO, OP, PR, RX; KS, ST, TY, YX, and join OS, PT, RY; and from the points O, S draw OV, SQ, perpendiculars to AB, AK: And because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, ÖV is drawn perpendicular to AB the common section of the planes, therefore a 4. def. 11. OV is perpendicular to the plane BCDE: For the same reason, SQ is perpendicular to the same plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ; and because in the equal semicircles BXD, KXN the circumferences BO, KS are equal, and OV, SQ, are perpendicular to their diameters, therefore OV is equal to SQ, and BV equal to KQ. But the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder QA: As therefore BV is to VA, so is KQ to QA; wherefore VQ is parallel to BK: And because OV, SQ are each of them at f 6. 11. right angles to the plane of the circle BCDE, OV is parallelf to SQ; and it has been proved, that it is also equal to it; thereg 33. 1. fore QV, SO are equal and parallel: And because QV is h 9. 11. parallel to SO, and also to KB; OS is parallel to BK; and
d 26. 1.
e 2. 6.
therefore BO, KS which join them are in the same plane in which these parallels are, and the quadrilateral figure KBOS is in one plane: And if PB, TK be joined, and perpendiculars be drawn from the points P, T to the straight lines AB, AK, it may be demonstrated, that TP is parallel to KB in the very same way that SO was shown to be parallel to the same KB; wherefore TP is parallel to SO, and the quadrilateral figure SOPT is in one plane: For the same reason, the quadrilateral TPRY is in one plane: And the figure YRX is also in one b 2. 11. plane. Therefore, if from the points O, S, P, T, R, Y, there be drawn straight lines to the point A, there shall be formed a
á 9. 11.
solid polyhedron between the circumferences BX, KX, com- Book XII. posed of pyramids, the bases of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX, and of which the common vertex is the point A: And if the same construction be made upon each of the sides KL, LM, ME, as has been done upon BK, and the like be done also in the other three quadrants, and in the other hemisphere; there will be formed a solid polyhedron described in the sphere, compo
sed of pyramids, the bases of which are the aforesaid quadrilateral figures; and the triangle YRX, 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 less sphere in which is the circle FGH: For, from the point A draw a AZ perpendicular a 11. 11. 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
b 47. 1.
Book XII. ZK: And because AB is equal to AK, and 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 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 demonstra ed, 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, will pass through the points K, O, S, and KBOS will 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 withi 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 cir
cumference of the circle; therefore the circumference sub- Book XII. tended 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 obtuse, the square of BK is greater than c 12. 2. the squares of BZ, ZK; that is, greater than twice the square of BŻ. Join KV, and because in the triangles KBV, OBV, KB, BV are equal to OB, BV, and they contain equal angles; the angle KVB is equal to the angle OVB: And d 4. 1. OVB is a right angle; therefore also KVB is a right angle: And because 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 than twice the square of KV: e 8. 6. But the square of KB is greater than twice the square of BZ; therefore the square of KV is 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, 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 proposition, it was shown 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 less sphere.
And that the other planes between the quadrants BX, KX fall without the less sphere is thus demonstrated: From the point A draw AI perpendicular to the plane of the quadrilateral SOPT, and join IO; 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 sides BK, OS parallel, as also OS, PT; and their other sides BO, KS, OP, ST, all equal to one another, and BK is greater than OS, and OS greater than PT, therefore the straight line ZB is greater than IO. Join AO which will be equal to AB; and a 2. Lemma because AIO, AZB, are right angles, the squares of AI, IO 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 IO, therefore the square of AZ is less than the square of
Book XII. AI: 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 Al greater than AG: Therefore the plane SOPT falls wholly without the less sphere: In the same manner, it may be demonstrated, that the plane TPRY falls without the same 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 less sphere. Therefore in the greater of two spheres, which have the same centre, a solid polyhedron is described, the superficies of which does not meet the less sphere. Which was to be done.
But the straight line AZ may be demonstrated to be greater 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 be bisected, and its half again bisected, and so on, there will at length be left a circumference less than the circumference which is subtended 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 square of AB, that is, to the squares of AZ, ZB; but the square of BZ is less than the square of GU; therefore the square of AZ is greater than the square of AG, and the straight line AZ consequently greater than the straight line AG.
COR. And if in the less 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 less; 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 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 will be similar 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 pyramid, and their other solid angle at the bases equal to one