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Fouk XII. GA at right angles to BD, and produce it to C; therefore AC mtouches the circle EFGH. then if the circumference BAD be bi

a. 16. 3. fected, and the half of it be again bisected, and so on, there must b. Lemma. at length remain a circumference less b than AD. let this be LD;

and from the point L draw LM
perpendicular to BD, and produce

L
it to N; and join LD, DN. there-
fore LD is equal to DN, and be-

E
cause LN is parallel to AC, and that B
AC touches the circle EFGH;
therefore LN does not meet the
circle EFGH. and much less shall

F
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 de-
scribed in the circle a polygon of an even number of equal sides not
meeting the lesser circle. Which was to be done.

K GMD

N

L E M M A II.

IF
F 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 fides 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 center of the circle in which the greater fides are, is greater than the straight line LE drawn from the center to the cir. cumference of the other circle.

If it be posible, let KA be not greater than LE; then KA must 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 23. 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 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,

which is impossible. therefore the straight line KA is not equal Cook 11. to 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 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 same reason, the circumference BC is greater than NO; and becerise 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 PG. 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 K A 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 fides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the center to the circumference of the circle described about the triangle.

S 3

PROP.

Book XII.

PROP. XVII. PROB.

See N.

T O describe in the greater of two spheres which have

the fame center, a solid polyhedron, the superficies of which shall not meet the lesser sphere.

Let there be two spheres about the same center A; it is required to delcribe in the greater a folid polyhedron the superficies of which Mall not meet the lefser sphere.

Let the spheres be cut by a plane passing thro' the center; the commonfections of it with the spheres shall be circles; because the sphere is described by the revolution of a femicircle about the dia. meter remainiig unmovcable; fo 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 2. 15. 3. spliere which is likewise the diameter of the circle, is greater than

any ít aight 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 íphere 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. 12. circies. describe 5 a polygon of an even number of equal fides pot

mecting the leser 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 thro' AX and cach 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 pasC.18. 11. fis thro'XA is at right © angles to the plane of the circle BCDE;

whercfore te femicircles BXD, KXN are at right angles to that sare, and becar se the îi micircles EED, EXD, KXN, upon the equal dion eteis ID, KN are equal io one anoiter, their fourth parts 1.7, 1X, KX are equal to one another. therefore as many fides of the polygon as are in the fourth part BE, so many there are in BX, EX et val to the sides BK, KL, LM, ME. let thefe polygons be duluivod, 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. S 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 section of the planes, therefore OV is perpendiculard to the plane BCDE, for the famed 4. Def.11. reason SQ is perpendicular to the fame plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ, and be. cause 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 * 28, 1. whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder Q4. as therefore BV is to VA, fo is KO to QA, wherefore VQ is parallel e to BK. and because OV, SQ are e. 2. 6. each of 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 to it; f. 6.15. therefore QV, SO are equal and parallel 5, and because QV is pa- &. 33. 1. rallel to SO, and also to KB; OS is parallel h to BK; and therefore h. 9.11.

Book XI!. BO, KS which join them are in the same plane in which these paral. mlels 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 fame way that SO was shęwn to be paral12.11. lel to the fame KB; wherefore h TP is parallel to $O, and the qua

drilateral figure SOPT is in one plane. for the same reason the qua. crilateral TPRY is in one plane. and the figure YRX is also in one

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1.2.1. plane i, 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 TRX, and of which the common vertex is the point A. and if the fame construction be made upon each of the lides 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 hall be formed a solid polyhedron described in the sphere,

composed

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