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a 16. 3.
Book XII. C; therefore AC touches the circle EFGH: Then, if the circumference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumb Lemma. ference less than AD: Let this be LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN. Therefore
LD is equal to DN; and be-
Ir two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L; and if the sides. AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal to one another; but the side AB greater than EF, and DC greater than HG; the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle.
If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less. First let KA be equal to LE: Therefore, because in two equal circles AD, BC, in the one, are equal to EH, FG in the other, the cir28. 3. cumferences AD, BC, are equal to the circumferences EH, FG; but because the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG: Therefore the whole eircumference ABCD is greater than the whole EFGH: but it is also
equal to it, which is impossible: Therefore the straight Book XII. line KA is not equal to LE.
But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP. PM, which are recspectively parallela to and less than EF, FG, GH, HE: * 2. 6. Then because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore
the circumference AD is greater than MP; for the same
COR. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle described about the triangle,
PROP. XVII. PROB.
See N. To describe in the greater of two spheres which
have the same centre, a solid polyhedron, the superficies 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 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 cir16. 12. cles, describe 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; 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 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, 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, * 18. 11. every plane which passes through XA is at 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 Book XI. the points O, S, draw OV, SQ perpendiculars to AB, AR 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 adt gothervigo TOPS COTES X
BXD, KXN, the circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefored 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 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 6. 11. has been proved, that it is also equal to it; therefore QV, SO are equal and parallels: And because QV is parallel to s 33. 1. SO, and also to KB; OS is parallelh to BK; and there- h 9. 11.
BOOK XII. fore 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 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 pa9. 11. rallel to the same KB; wherefore a TP is parallel to SO, and the quadrilateral figure SOPP is in one plane: For the same reason the quadrilateral TPRY is in one plane: and 14 saulets on elosionsgru a of drop yang diwa/206 Best SA or faups a d
2.11. the figure YRX is also in one 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 solid 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 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 shall