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Fook XII. GA at right angles to BD, and produce it to C; therefore AC touches the circle EFGH. then if the circumference BAD be bia. 16. 3. fected, and the half of it be again bisected, and fo on, there must b. Lemma. at length remain a circumference lefs b 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 becaufe LN is parallel to AC, and that B

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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. fo that if straight lines equal to LD be applied ia
the circle ABCD from the point L around to N, there fhall be de-
fcribed in the circle a polygon of an even number of equal fides not
meeting the leffer circle. Which was to be done.

IF

LEMMA II.

F two trapeziums ABCD, EFGH be infcribed in the circles the centers of which are the points K, L; and if the fides AB, DC be parallel, as alfo EF, HG; and the other four fides AD, BC, EH, FG be all equal to one another; but the fide AB greater than EF, and DC greater than HG. the ftraight line KA from the center of the circle in which the greater fides are, is greater than the ftraight line LE drawn from the center to the cir cumference of the other circle.

If it be poffible, let KA be not greater than LE; then KA must be either equal to it, or lefs. First, let KA be equal to LE. therefore because in two equal circles, AD, BC in the one are equal to 28. 3. EH, FG in the other, the circumferences AD, BC are equal to the circumferences EH, FG; but because the straight lines AB, DC are refpectively 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 impoffible. therefore the straight line KA is not equal Book XH. to LE.

But let KA be lefs than LE, and make LM equal to KA, and from the center L, and distance LM defcribe 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 b to, and less b. 2. 6... than EF, FG, GH, HE. then, because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal,

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therefore the circumference AD is greater than MP; for the fame reafon, the circumference BC is greater than NO; and becanfe the ftraight 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 fame reafon, the circumference DC is greater than PO. therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewife equal to it, which is impoffible. therefore KA is not lefs 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 Ifofceles triangle the fides of which are equal to AD, BC, but its bafe lefs 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 ftraight line drawn from the center to the circumference of the circle defcribed about the triangle.

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Book XII,

See N.

PROP. XVII. PROB.

To defcribe in the greater of two spheres which have the fame center, a folid polyhedron, the superficies

of which fhall not meet the leffer sphere.

Let there be two spheres about the fame center A; it is required to delcribe in the greater a folid polyhedron the fuperficies of which fhall not meet the leffer fphere.

Let the spheres be cut by a plane paffing thro' the center; the common/fections of it with the fpheres fhall be circles; because the fphere is defcribed by the revolution of a femicircle about the diameter remaining unmovcable; fo that in whatever pofition the semicircle be conceived, the common fection of the plane in which it is with the fuperficies of the fphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the 3. 15. 3. fphere which is likewife the diameter of the circle, is greater than any fhaight line in the circle or sphere. let then the circle made by the fection 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 b. 16. 11. circles. defcribeba polygon of an even number of equal fides pot

2.

meeting the leffer 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 fuperficies of the fphere in the point X; and let planes pafs thro' AX and each of the straight lines BD, KN, which, from what has been faid, fhall produce great circles on the fuperficies of the ff here. and let BXD, KXN be the femicircles thus made upon the diameters BD, KN. therefore, becaufe XA is at right angles to the plane of the circle BCDE, every plane which pafc. 18. 11. fs thro' XA is at right angles to the plane of the circle BCDE;

wherefore the femicicles BXD, KXN are at right angles to that plane. and because the femicircles EED, EXD, KAN, upon the equal dien eters ID, KN are equal to one another, their fourth parts FF, FX, KX are equal to one another. therefore as many fides of the polygon as are in the fourth part BE, fo many there are in BX, FX equal to the fides BK, KL, LM, ME. let thefe polygons be deforibed, 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. S2 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 to the plane BCDE, for the famed 4.Def.11. reafon SQ is perpendicular to the fame plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ, and becaufe in the equal femicircles BXD, KXN the circumferences BO,

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KS are equal, and OV, SQ are perpendicular to their diameters, therefore OV is equal to SQ, and BV equal 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 VQ is parallel to BK. and becaufe OV, SQ are e. 2. 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 alfo equal to it; f. 6. 11. therefore QV, SO are equal and parallel . and becaufe QV is pa- g. 33. 1. rallel to SO, and alfo to KB; OS is parallel to BK; and therefore h. 9. 11.

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LO,

Book XII. BO, KS which join them are in the fame 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 ftraight lines AB, AK, it may be demonftrated that TP is parallel to KB in the very fame way that SO was fhewn to be parallel to the fame KB; wherefore h TP is parallel to SO, and the qua drilateral figure SOPT is in one plane. for the fame reafon the qua drilateral TPRY is in one plane. and the figure YRX is alfo in one

2. 11.

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2.1. plane. therefore, if from the points O, S, P, T, R, Y there be drawn ftraight lines to the point A, there fhall be formed a folid polyhedron between the circumferences BX, KX compofed of pyramids the bafes 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 fame conftruction be made upon each of the fides KL, LM, ME, as has been done upon BK, and the like be done alfo in the other three quadrants, and in the other hemifphere; there fhall be formed a folid polyhedron defcribed in the fphere,

compofed

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