a. 16. 3. Book XII. GA at right angles to BD, and produce it to C; therefore AC touches a the circle EFGH. then if the circumference BAD be bifccted, and the half of it be again bifceted, and so on, there b. Lemma. must at length remain a circunference less b than AD. let this be LD; and from the point L draw A H L E K GM D that AC touches the circle EFGH; therefore LN does not meet the circle EFGH. and much less thall I N the straight lines LD, DN meet the circle EFGH. fo 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 a polygon of an even number of equal lides not meeting the leffer circle. Which was to be done. LE M M A II. IF two trapeziums ABCD, EFGH be inscribed in the circles the centers 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 llG. the straight line KA from the center of the circle in which the greater sides are, is greater than the straight line LE drawn from the center to the circumference of the other circle. to If it be possible, let KA be not greater than LE ; then KA muft be either equal to it, or less. First, let KA be equal to LE. there fore because in two equal circles, AD, BC in the one are equal to di 28. 3. EH, FG in the other, the circumferences AD, BC are equal the circumferences EH, FG; but because the straight lines AB, which is imposible. therefore the straight line KA is not equal to Book XII. LE. But let KA be less than LE, and make LM equal to KA, and from the center 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 o to, and b. 2. W 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 fame reason, the circumference BC is greater than NO; and be cause 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 there, fore be greater than LE. 0. E. D. Cor. And if there be an Iofceles triangle the sides of which are equal to AD, BC, but its base less than AB the greater of the two fides AB, DC; the straight line KA may, in the fame manner, be demonstrated to be greater than the straight line drawn rom the center to the circumference of the circle described abour he triangle. Book XII. PRO P. XVII. PROB. To describe in the greater of two spheres which have the same center, a solid polyhedron, the superficies of which shall not meet the lesser sphere. Let there be two spheres about the fame center A; it is required to describe in the greater a solid polyhedron the superficies of which shall not meet the leller sphere. Let the spheres be cut by a plane paffing thro' the center; the common sections of it with the spheres shall be circles; because the fphere 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 a. 15. 3. sphere which is likewise the diameter of the circle, is 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 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 6. 16. 12. circles describe b a 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 pass thro'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 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 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 sides of the polygon as are in the fourth part BE, fo 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 fection of the planes, therefore OV is perpendicular d to the plane BCDE. for the same day Def.nr. 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 femicircies BXD, KXN the circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore * OV is equal to SQ, an1 BV equ:l to KQ. but the • 26.1, 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 VO is parallel to o to BK. and tecause OV,SQ are e, 3. 6. each of them at right angles to the plane of the circle BCDE, OV is parallel f to SQ; and it has been proved that it is also equal to it ; f. 6. 17, therefore QV; SO are equal and parallel 8. and because QV is pa- 8, 13. 1, rallel to SO, and also to KB; OS is parallel - to BK; and therefore hi 9 t. Book XII. BO, KS which join them are in the same plane in which these paral lels are, and the quadrilateral figure KBOS is in one plane. and if PD, 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 K5; wherefore h TP is parallel to SO, and the quadilateral figure SOPT is in one plane. for the same reason the quadrilateral TPRY is in one plane. and the figure TRX is also in one plane i. therefore, if fruin 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 compofed of pyramids the balcs 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 farne construction be made upon each of the fides KL, LN, ME, as has been done upon BK, and the like be done also in the other three quadrants, and in the other hemisphere; there full be formed a solid polyhedron described in the sphere, |