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Book XII. Cirele, draw GA at right angles to BD, and produce it to C;
therefore AC touches a the circle EFGH : then, if the cira 26. 3. cumference BAD be bisected, and the half of it be again bisecb Lemma. ted, and so on, there must at length remain a circumference less b than AD; let this be LD; and
K GM fore LD is equal to DN: and be- B cause LN is parallel to AC, and that AC, touches the circle EFGH; therefore LN does not meet the circle EFGH: and much less shall the straight lines LD, DN meet the, circle EFGH: so that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be described in the circle à polygon of an even number of equal sides not meeting the lesser circle. Which was to be done.
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.
a 28. 3.
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, AB, BC in the one, are equal to EH, FG in the other, the circumferences AD, BC are equal a to the circumferences EH, FG ; but because the straight lines AB, DC are respeotively 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, which is
impossible; therefore the straight line KA is not equal to Book Xir. 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 respectively parallel to and less than EF, FG, GH, HE: then, because EÚ a 2. B. is greater than MP, AD is greater than MP; and the circles
ABCD, MNOP are equal; therefore the circumference AD ís greater than MP: for the same reason, the circumference BC is greater than NO; and because the straight line AB is greater than EF, which is greater than MN, much more is AB 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 de scribed about the triangle.
PROP. XVII. PROB.
See Note. 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.
à 15. 3.
> 16. 12.
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 inwhichitiswith 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 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 Iesser 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 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 froin 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 BXZ, 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 right © angles to the plane of the circle BCDE; wherefore the semicircies BXD, KXN are at right angles to that plane; and because the semicircles BED, BX D, 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 Book XIV. to-'AB, AK: and because the plane BOXD iş at right angies 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 the a 4. def. 11. 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, d 26. 1. 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 Oy, SQ are cach ofe 2. 6. them at right angles to the plane of the circle BCDE, OV is parallelf to SQ; and it has been proved that it is also equal f 6. 12 to it; therefore QV, SO are equal and parallel 5: and because g 39. 1. QV is parallel to SO, and also to KB, OS is parallel h to BK; h 3-11. and therefore BO, KS which join them are in the same plane
Book XII. in which these parallels are, and the quadrilateral figure KBO
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 b 2. 11. TPRYisin one plane; and the figure YRX is also in one plane be
Therefore, if from the points O, S, P, T, R, Y there be dratvn 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, L:M, 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 be formed a solid polyhedron described in the sphere, .compo