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But let KA be less than LE, and make LM equal to
terthe straighn ference Breater than
the circumference, AD is greater than MP; for the same
Cor. And if there be an isosceles triangle, the sides of
than the straight line drawn from the centre to the cir. i cumference of the circle described about the triangle.
See N. To describe in the greater of two spheres which 14. have the same centre, a solid polyhedron, the su
perficies of which shall not meet the lesser sphere.
Let there be two spheres about the same centre A; it is required to describe in the greater a solid polyhedron, the . superficies of which shall not meet the lesser sphere.
Let the spheres be cut by a plane passing through the centre; 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 sphere, which is likewise the diameter of the circle, is * 15. 3. 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 lesser sphere be FGH; and draw the two diameters BD, CE at right angles
to one another; and in BCDE, the greater of the two cir116. 12. cles, describe b a polygon of an even number of equal sides
not meeting the lesser circle FGH; and let its sides, in BE the fourth part of the circle, be BK, KL, LM, ME; joiu 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 pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere, and let BXD, KÄN be the semicir. cles thus made upon the diameters BD, KN: Therefore, be
cause XA is at right angles to the plane of the circle BCDE, • 18. 11. every plane which passes through XA is at rightc 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 diame
· ters 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 Book XII. 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, OV is drawn perpendicular to AB the common section of the planes, therefore OV is perpendiculara to the plane BCDE: For 4 Def. 11. 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
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PE. 90 29136
-11911198 51 40 BXD, KXN, the circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore d d 26. 1, 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 e to BK: And • 2. 6. because OV, SQ are each of them at right angles to the plane of the circle BCDE, OV is parallelf to SQ; and it f 6. 11. has been proved, that it is also equal to it; therefore QV, SO are equal and parallel 8: And because QV is parallel to * 33. 1. SO, and also to KB; OS is parallel b to BK; and there. h 9. 11.
Boox Xļl. fore BO, KS, which join them are in the same plane jina
which these parallels are, and the quadrilateral figuseu KBOS is in one plane : And if BP, 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 part * 9. 11. rallel to the same KB; wherefore a TP is parallel to SO,
and the quadrilateral figure SOPT is in one plane: For thei'. same reason the quadrilateral TPRY is in one plane; and
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THON SA 919 39
09. 11, the figure YRX is also in one planeb. Therefore, if froins,
the points O, S, P, T, R, Y, there be drawn straight lines
be formed a solid polyhedron described in the sphere, com- Buos XIL posed 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 thein 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 drawa AZ perpendicular to the plane of the quadrilateral* 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 mecting it in that plane; therefore AZ is perpendicular to BZ and ZK : And because AB is equal to Åk, and that the squares of AZ, ZB, are : equal to the square of AB; and the squares of AZ, ZK, to the square of AKb; therefore the squares of AZ, ZB, are 47. 1. equal to the squares of AZ, ZK: Take from these equals the square of ÅZ, the remaining square of BZ is equal to. the reigaiping 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 0, S, are equal to BZ or ZK: Therefore the circle described froin the centre Z, and distance ZB, shall pass througli.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 thian 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 subtended by KB is grcater 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 the squares 1%. af 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 augles; the angle KVB is equald to the angle OVB : And" 4. 2. UVB is a right angle; therefore also KVB is a right angle: . And because BD is less than twice DV; the rectangle con." tained by DB, BV, is less than twice the rectangle DVB; that is, the square of KB is less than twice the square of 8.6 KV; But the square of KB is greater than twice the square