Book XII.

See N.

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PROP. XVII. PRO В.

10 describe in the greater of two spheres which have the fame center, a solid polyhedron, the superficies of which shall not meet the leffer sphere.

Let there be two spheres about the same center A; it is re quired to describe in the greater a folid polyhedron the superficies of which shall not meet the lefsser sphere.

Let the spheres be cut by a plane passing thro' the center; the common sections of it with the spheres shall be circles; because the sphere is described by the revolution of a semicircle about the diameter remaining unmoveable, so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the

15. 3. sphere which is likewise the diameter of the circle, is greater a than any straight line in the circle or sphere. let then the circle made by • the section of the plane with the greater sphere be BCDE, and with the leffer sphere be FGH; and draw the two diameters BD, CE at right angles to one another. and in BCDE the greater of the two

b. 16. 12. circles describeba polygon of an even number of equal fides not meeting the lesser circle FGH; and let its fides, in BE the fourth part of the circle, be BK, KL, LM, ME; join KA and produce it to N; and from A draw AX at right angles to the plane of the circle BCDE meeting the superficies of the sphere in the point X; and let planes pafs thro' AX and each of the straight lines BD, KN, which, from what has been faid, shall produce great circles on the superficies of the sphere, and let BXD, KXN be the femicircles 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

C. 18. 11. thro' XA is at right angles to the plane of the circle BCDE; wherefore the femicircles BXD, KXN are at right angles to that plane. and because the femicircles BED, BXD, KXN, upon the equal diameters BD, KN are equal to one another, their fourth parts BE, BX, KX are equal to one another. therefore as many fides of the polygon as are in the fourth part BE, so many there are in BX, KX equal to the fides BK, KL, LM, ME. let these polygons be described, and their fides be BO, OP, PR, RX; KS, ST, TY,

YX, and join OS, PT, RY; and from the points O, S draw OV, Book XII. SQ perpendiculars to AB, AK. and because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, OV is drawn perpendicular to AB the common section of the planes, therefore OV is perpendicular d to the plane BCDE. for the fame d.4.Def.11. reason SQ is perpendicular to the fame 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 * 26. r. whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder QA. as therefore BV is to VA, fo is KQ to QA, wherefore VQ is parallel to BK. and because OV, SQ are e. 2. 6. each of them at right angles to the plane of the circle BCDE, OV is parallel to SQ; and it has been proved that it is also equal to it; f. 6. rr. therefore QV, SO are equal and parallel 8. and because QV is pag. 33. 1. rallel to SO, and also to KB; OS is parallel to BK; and therefore h. 9. 11.

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h. 9. II.

Book XII. BO, KS which join them are in the fame 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 fame way that SO was shewn to be parallel to the fame KB; wherefore h TP is parallel to SO, and the quadrilateral figure SOPT is in one plane. for the fame reason the quadrilateral TPRY is in one plane. and the figure YRX is also in one

1. 2.-11.

done also in the other three quadrants, and in the other hemisphere; there shall be formed a folid polyhedron described in the sphere,

plane i. therefore, if from the points O, S, P, T, R, Y there be drawn ftraight lines to the point A, there shall be formed a folid polyhedron between the circumferences BX, KX composed 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 fame construction be made upon each of the fides KL, LM, ME, as has been done upon BK, and the like be

composed of pyramids the bases of which are the aforesaid quadri- Book XII. lateral 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 lesser sphere in which is the circle FGH. for from the point A draw * AZ perpendicular to the plane of the quadrilateral k. 11. 11. 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, * 47. Ι. 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 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 center Z, and distance ZB shall pass thro' 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 SQ, 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 center is greater than a right angle. and because the angle BZK is obtuse, the square of BK is greater 1 than 1. 12. 2. the squares of BZ, ZK; that 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 m to the angle OVB. and OVB is a right angle; m. 4. 1. therefore also KVBis a right angle. and because BDis 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 n. 8. 6. square of KV. 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

Book XII. 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 AZ is greater than AG, because in the preceding Proposition it was shewn 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 center of the sphere to that plane. Therefore the plane KBOS does not meet the lesser sphere.

12.

And that the other planes between the quadrants BX, KX fall without the lesser 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 fame way it may be shewn that the point I is the center of a circle described about SOPT, and that OS is greater than PT; and PT was shewn to be parallel to OS. therefore because the two trapeziums KBOS, SOPT inscribed in circles have their fides BK, OS parallel, as alfo OS, PT; and their other fides BO, KS, OP,

ST all equal to one another, and that BK is greater than OS, and 0. 2. Lem. OS greater than PT, therefore the straight line ZB is greater than IO. Join AO which will be equal to AB; and 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 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 AI greater than AG. therefore the plane SOPT falls wholly without the lesser sphere. in the fame 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 leffer sphere. therefore in the greater of two spheres which have the fame center, a folid polyhedron is described the superficies of which does not meet the lesser 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