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MP or KL, so is the cylinder EO to the cylinder ES: Book XII. Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the same ES: Whence the cylinder AX is equal to the cylinder EO: and the same reasoning holds in cones. Q. E. D.
PROP. XVI. PROB.
To describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the less circle.
Let ABCD, EFGH be two given circles having the same centre K; it is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the less circle.
Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the less circle, draw GA at right angles to BD, and produce it to C; therefore AC touches the circle EFGG: Then, if the circum- a 16. 3. ference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumference less b b Lemma. than AD: Let this be LD; and from the point L draw LM per
A pendicular to BD, and produce
H it to N; and join LD, DN. Therefore LD is equal to DN; Е,
B and because LN is parallel to
If 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 circumferences 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 circumference ABCD is greater than the whole EFGH; but it is also equal to it, wbich is impossible : Therefore the straight 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
a 28. 3.
a 2. 6.
MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM, which are respectively parallel a to, and less than EF, FG, GH, HE: 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 reason, the Book XII. circumference BC is greater than NO; and because the straight line AB is greater than EF, which is greater than MN, much more is A B greater than MN: Therefore the circumference AB is greater than MN; and, for the same reason, the circumference DC is greater than PO: Therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewise equal to it, which is impossible: Therefore KA is not less than LE: nor is it equal to it; the straight line KA must therefore be greater than LE. Q. E. D.
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.
To describe in the greater of two spheres which See N. have the same centre, a solid polyhedron, the superficies of which shall not meet the less 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 less sphere.
Let the spheres be cut by a plane passing through the centres; the common sections of this plane with the spheres will 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 greatera a 15. 3. 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 less sphere be FGH; and draw the two diameters BD, CE, at right angles to one another: and in BCDE, the greater of the two circles, describe b a polygon b 16. 12. of an even number of equal sides not meeting the less 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
Bouk 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, OV is drawn per
pendicular to A B the common section of the planes, therefore a 4.def. 1 1. OV is perpendicular a 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 d 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 because OV, SQ are each of them at f 6. 11. 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; thereg 33. 1. fore QV, SO are equal and parallel 8 : And because QV is h 9. 11. parallel to SO, and also to KB; OS is parallel to · BK; and
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; a 9. 11. wherefore a 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. planeb. 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
d 26. 1.
e 2. 6.
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 bas 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 ll. 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